diff --git a/data/handouts/Images/angles-90-60-30.asy b/data/handouts/Images/angles-90-60-30.asy new file mode 100644 index 00000000..4fc357fa --- /dev/null +++ b/data/handouts/Images/angles-90-60-30.asy @@ -0,0 +1,26 @@ +import _common; + +real a = 55; +pair B = (0, 0); +pair A = (a, 0); +pair Bp = (2 * a, 0); +pair C = (a, a * sqrt(3)); + +AngleMark(A, B, C, LightBlue); +RightAngleMark(B, A, C); + +Draw(B, A); +Draw(B, C); +Draw(C, A); +Draw(A, Bp); +Draw(Bp, C); + +LabeledDot(B, "B", SW); +LabeledDot(C, "C", N); +LabeledDot(Bp, "B'", SE); +LabeledDot(A, "A", S); + +EdgeLabel(B, C, "2a", W, distanceOffset = 8); +EdgeLabel(C, Bp, "2a", E, distanceOffset = 8); +EdgeLabel(B, A, "a", S, distanceOffset = 8); +EdgeLabel(A, Bp, "a", S, distanceOffset = 8); diff --git a/data/handouts/Images/angles-90-60-30.pdf b/data/handouts/Images/angles-90-60-30.pdf new file mode 100644 index 00000000..422f3b7e Binary files /dev/null and b/data/handouts/Images/angles-90-60-30.pdf differ diff --git a/data/handouts/Images/angles-90-60-30.svg b/data/handouts/Images/angles-90-60-30.svg new file mode 100644 index 00000000..677799cb --- /dev/null +++ b/data/handouts/Images/angles-90-60-30.svg @@ -0,0 +1,298 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-Ssa-proof-isosceles.asy b/data/handouts/Images/angles-Ssa-proof-isosceles.asy new file mode 100644 index 00000000..0bc624ce --- /dev/null +++ b/data/handouts/Images/angles-Ssa-proof-isosceles.asy @@ -0,0 +1,25 @@ +import _common; + +// Isosceles BCA with |CB| = |CA| = base; angle ABC = alpha. Apex angle at C +// equals 180 − 2α, so from C (with B at direction 180°) point A sits at +// direction 2α (CCW from +x). +real base = 50, alpha = 40, farExtra = 35; +pair B = (0, 0); +pair C = (base, 0); +real swingR = base; +pair A = Polar(C, 2 * alpha, swingR); +pair Bfar = ExtendPast(B, A, farExtra); + +AngleMark(C, B, A, LightRed, "\beta", labelPen = Red); + +Circle(C, swingR, Red); +Draw(B, C, Blue); +Draw(C, A, Red); +Draw(B, Bfar); + +LabeledDot(B, "B", SW); +LabeledDot(C, "C", SE); +LabeledDot(A, "A", N, Red); + +EdgeLabel(B, C, "a", S, color = Blue); +EdgeLabel(C, A, "b", E, color = Red); diff --git a/data/handouts/Images/angles-Ssa-proof-isosceles.pdf b/data/handouts/Images/angles-Ssa-proof-isosceles.pdf new file mode 100644 index 00000000..9e037d77 Binary files /dev/null and b/data/handouts/Images/angles-Ssa-proof-isosceles.pdf differ diff --git a/data/handouts/Images/angles-Ssa-proof-isosceles.svg b/data/handouts/Images/angles-Ssa-proof-isosceles.svg new file mode 100644 index 00000000..50568305 --- /dev/null +++ b/data/handouts/Images/angles-Ssa-proof-isosceles.svg @@ -0,0 +1,225 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-Ssa-proof-one-solution.asy b/data/handouts/Images/angles-Ssa-proof-one-solution.asy new file mode 100644 index 00000000..06225fb9 --- /dev/null +++ b/data/handouts/Images/angles-Ssa-proof-one-solution.asy @@ -0,0 +1,27 @@ +import _common; + +// A on ray from B at angle alpha at the unique forward intersection with +// circle(C, swingR). Law of cosines on triangle BCA gives: +// |BA|² − 2·base·cos(α)·|BA| + (base² − swingR²) = 0. +// Since swingR > base, B is inside the circle and only the positive root +// corresponds to a forward intersection. +real base = 40, swingR = 58, alpha = 45; +pair B = (0, 0); +pair C = (base, 0); +real cA = cos(radians(alpha)); +real BA = base * cA + sqrt((base * cA)^2 - (base^2 - swingR^2)); +pair A = Polar(B, alpha, BA); + +AngleMark(C, B, A, LightRed, "\beta", labelPen = Red); + +Circle(C, swingR, Red); +Draw(B, C, Blue); +Draw(C, A, Red); +Draw(B, A); + +LabeledDot(B, "B", SW); +LabeledDot(C, "C", SE); +LabeledDot(A, "A", NE, Red); + +EdgeLabel(B, C, "a", S, color = Blue); +EdgeLabel(C, A, "b", E, color = Red); diff --git a/data/handouts/Images/angles-Ssa-proof-one-solution.pdf b/data/handouts/Images/angles-Ssa-proof-one-solution.pdf new file mode 100644 index 00000000..a501f339 Binary files /dev/null and b/data/handouts/Images/angles-Ssa-proof-one-solution.pdf differ diff --git a/data/handouts/Images/angles-Ssa-proof-one-solution.svg b/data/handouts/Images/angles-Ssa-proof-one-solution.svg new file mode 100644 index 00000000..47fd9a79 --- /dev/null +++ b/data/handouts/Images/angles-Ssa-proof-one-solution.svg @@ -0,0 +1,225 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-Ssa-proof-two-solutions.asy b/data/handouts/Images/angles-Ssa-proof-two-solutions.asy new file mode 100644 index 00000000..2127c11a --- /dev/null +++ b/data/handouts/Images/angles-Ssa-proof-two-solutions.asy @@ -0,0 +1,32 @@ +import _common; + +// Both quadratic roots are positive when swingR < base and α is small enough, +// giving the two-solution case. +real base = 110, swingR = 65, alpha = 35, farExtra = 50; +pair B = (0, 0); +pair C = (base, 0); +real cA = cos(radians(alpha)); +real disc = sqrt((base * cA)^2 - (base^2 - swingR^2)); +real BAfar = base * cA + disc; +real BAnear = base * cA - disc; +pair A = Polar(B, alpha, BAfar); +pair Ap = Polar(B, alpha, BAnear); +pair Bfar = ExtendPast(B, A, farExtra); + +AngleMark(C, B, A, LightRed, "\beta", labelPen = Red); + +Circle(C, swingR, Red); +Draw(B, C, Blue); +Draw(C, A, Red); +Draw(C, Ap, Red); +Draw(B, A); +Draw(A, Bfar); + +LabeledDot(B, "B", SW); +LabeledDot(C, "C", SE); +LabeledDot(A, "A", N); +LabeledDot(Ap, "A'", NW); + +EdgeLabel(B, C, "a", S, color = Blue); +EdgeLabel(C, A, "b", NE, color = Red); +EdgeLabel(C, Ap, "b", SW, color = Red); diff --git a/data/handouts/Images/angles-Ssa-proof-two-solutions.pdf b/data/handouts/Images/angles-Ssa-proof-two-solutions.pdf new file mode 100644 index 00000000..185a2600 Binary files /dev/null and b/data/handouts/Images/angles-Ssa-proof-two-solutions.pdf differ diff --git a/data/handouts/Images/angles-Ssa-proof-two-solutions.svg b/data/handouts/Images/angles-Ssa-proof-two-solutions.svg new file mode 100644 index 00000000..67763e51 --- /dev/null +++ b/data/handouts/Images/angles-Ssa-proof-two-solutions.svg @@ -0,0 +1,300 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-alternate.asy b/data/handouts/Images/angles-alternate.asy new file mode 100644 index 00000000..98befa81 --- /dev/null +++ b/data/handouts/Images/angles-alternate.asy @@ -0,0 +1,18 @@ +import _common; + +real sep = 50, theta = 45; +real upperLeftPad = 70, lowerRightPad = 80; +pair L = (0, 0); +pair U = (sep / tan(radians(theta)), sep); +pair Lleft = U + (-upperLeftPad, 0); +pair Rright = L + (lowerRightPad, 0); + +AngleMark(Lleft, U, L, LightRed); +AngleMark(Rright, L, U, LightRed); + +Draw(Lleft, U); +Draw(L, Rright); +Draw(L, U); + +ParallelMark(Lleft, U); +ParallelMark(L, Rright); diff --git a/data/handouts/Images/angles-alternate.pdf b/data/handouts/Images/angles-alternate.pdf new file mode 100644 index 00000000..d996fbea Binary files /dev/null and b/data/handouts/Images/angles-alternate.pdf differ diff --git a/data/handouts/Images/angles-alternate.svg b/data/handouts/Images/angles-alternate.svg new file mode 100644 index 00000000..9395c1f6 --- /dev/null +++ b/data/handouts/Images/angles-alternate.svg @@ -0,0 +1,135 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-asa-proof.asy b/data/handouts/Images/angles-asa-proof.asy new file mode 100644 index 00000000..09d0b4bb --- /dev/null +++ b/data/handouts/Images/angles-asa-proof.asy @@ -0,0 +1,23 @@ +import _common; + +real base = 120, height = 65; +pair B = (0, 0); +pair C = (base, 0); +pair A = (base / 2, height); +pair Ap = ReflectAcross(A, B, C); + +AngleMark(C, B, A, LightBlue, "\alpha"); +AngleMark(Ap, B, C, LightBlue, "\alpha"); +AngleMark(A, C, B, LightRed, "\beta"); +AngleMark(B, C, Ap, LightRed, "\beta"); + +Draw(B, C, Green); +Draw(B, A); +Draw(A, C); +Draw(B, Ap); +Draw(Ap, C); + +LabeledDot(B, "B", W, Blue); +LabeledDot(C, "C", E, Red); +LabeledDot(A, "A", N, Green); +LabeledDot(Ap, "A'", S, Green); diff --git a/data/handouts/Images/angles-asa-proof.pdf b/data/handouts/Images/angles-asa-proof.pdf new file mode 100644 index 00000000..78116532 Binary files /dev/null and b/data/handouts/Images/angles-asa-proof.pdf differ diff --git a/data/handouts/Images/angles-asa-proof.svg b/data/handouts/Images/angles-asa-proof.svg new file mode 100644 index 00000000..b69629a5 --- /dev/null +++ b/data/handouts/Images/angles-asa-proof.svg @@ -0,0 +1,290 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-complementary-in-triangle.asy b/data/handouts/Images/angles-complementary-in-triangle.asy new file mode 100644 index 00000000..b8313acc --- /dev/null +++ b/data/handouts/Images/angles-complementary-in-triangle.asy @@ -0,0 +1,19 @@ +import _common; + +real base = 72, alphaB = 85, gammaC = 55, extra = 65; +pair B = (0, 0); +pair C = (base, 0); +pair A = extension(B, Polar(B, alphaB, 1), C, Polar(C, 180 - gammaC, 1)); +pair Bottom_right = ExtendPast(B, C, extra); + +AngleMark(B, A, C, LightRed, "\alpha"); +AngleMark(C, B, A, LightGreen, "\beta"); +AngleMark(Bottom_right, C, A, LightBlue, "?"); + +Draw(A, B); +Draw(A, C); +Draw(B, Bottom_right); + +LabeledDot(A, "A", N); +LabeledDot(B, "B", SW); +LabeledDot(C, "C", SE); diff --git a/data/handouts/Images/angles-complementary-in-triangle.pdf b/data/handouts/Images/angles-complementary-in-triangle.pdf new file mode 100644 index 00000000..efedbc4d Binary files /dev/null and b/data/handouts/Images/angles-complementary-in-triangle.pdf differ diff --git a/data/handouts/Images/angles-complementary-in-triangle.svg b/data/handouts/Images/angles-complementary-in-triangle.svg new file mode 100644 index 00000000..5f73deaf --- /dev/null +++ b/data/handouts/Images/angles-complementary-in-triangle.svg @@ -0,0 +1,226 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-complementary-parallel-solution.asy b/data/handouts/Images/angles-complementary-parallel-solution.asy new file mode 100644 index 00000000..b617129b --- /dev/null +++ b/data/handouts/Images/angles-complementary-parallel-solution.asy @@ -0,0 +1,26 @@ +import _common; + +real sep = 55, theta = 70, apexExtra = 35; +real lowerRightPad = 75, upperRightPad = 65; +pair D = (0, 0); +pair A = (sep / tan(radians(theta)), sep); +pair Atop = ExtendPast(D, A, apexExtra); +pair Dright = (lowerRightPad, 0); +pair B = A + (upperRightPad, 0); + +AngleMark(Dright, D, A, LightRed, "\alpha", labelFraction=1.25); +AngleMark(D, A, B, LightBlue, "\beta", labelFraction=1.3); +AngleMark(B, A, Atop, LightRed, "\alpha", labelFraction=1.3); + +Draw(D, A); +Draw(A, Atop); +Draw(D, Dright); +Draw(A, B); + +ParallelMark(D, Dright); +ParallelMark(A, B); + +VertexDots(new pair[] {D, A, B}); +PointLabel(D, "D", SW); +PointLabel(A, "A", NW); +PointLabel(B, "B", E); diff --git a/data/handouts/Images/angles-complementary-parallel-solution.pdf b/data/handouts/Images/angles-complementary-parallel-solution.pdf new file mode 100644 index 00000000..91f242a9 Binary files /dev/null and b/data/handouts/Images/angles-complementary-parallel-solution.pdf differ diff --git a/data/handouts/Images/angles-complementary-parallel-solution.svg b/data/handouts/Images/angles-complementary-parallel-solution.svg new file mode 100644 index 00000000..3d06ef68 --- /dev/null +++ b/data/handouts/Images/angles-complementary-parallel-solution.svg @@ -0,0 +1,198 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-complementary-parallel-statement.asy b/data/handouts/Images/angles-complementary-parallel-statement.asy new file mode 100644 index 00000000..b467e3bf --- /dev/null +++ b/data/handouts/Images/angles-complementary-parallel-statement.asy @@ -0,0 +1,23 @@ +import _common; + +real sep = 40, theta = 70; +real lowerRightPad = 70, upperRightPad = 60; +pair B = (0, 0); +pair A = (sep / tan(radians(theta)), sep); +pair lowerRight = (lowerRightPad, 0); +pair upperRight = A + (upperRightPad, 0); + +AngleMark(lowerRight, B, A, LightRed, radius=Radius2); +AngleMark(B, A, upperRight, LightBlue, radius=Radius2); + +Draw(B, lowerRight); +Draw(A, upperRight); +Draw(B, A); + +ParallelMark(B, lowerRight); +ParallelMark(A, upperRight); + +VertexDots(new pair[] {B, A}); + +PointLabel(B, "B", SW); +PointLabel(A, "A", N); diff --git a/data/handouts/Images/angles-complementary-parallel-statement.pdf b/data/handouts/Images/angles-complementary-parallel-statement.pdf new file mode 100644 index 00000000..88ab9416 Binary files /dev/null and b/data/handouts/Images/angles-complementary-parallel-statement.pdf differ diff --git a/data/handouts/Images/angles-complementary-parallel-statement.svg b/data/handouts/Images/angles-complementary-parallel-statement.svg new file mode 100644 index 00000000..bc01782b --- /dev/null +++ b/data/handouts/Images/angles-complementary-parallel-statement.svg @@ -0,0 +1,183 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.asy b/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.asy new file mode 100644 index 00000000..18daa3ba --- /dev/null +++ b/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.asy @@ -0,0 +1,22 @@ +import _common; + +real sep = 40, theta = 40; +real upperLeftPad = 32, upperRightPad = 50, lowerRightPad = 75; +pair Plower = (0, 0); +pair Pupper = (sep / tan(radians(theta)), sep); +pair Tup = Polar(Pupper, theta, 32); +pair Hupper_left = Pupper + (-upperLeftPad, 0); +pair Hupper_right = Pupper + (upperRightPad, 0); +pair Hlower_left = Plower; +pair Hlower_right = Plower + (lowerRightPad, 0); + +AngleMark(Hupper_right, Pupper, Tup, LightRed, "\alpha"); +AngleMark(Hlower_right, Plower, Tup, LightRed, "\gamma"); +AngleMark(Hupper_left, Pupper, Plower, LightRed, "\beta"); + +Draw(Hupper_left, Hupper_right); +Draw(Hlower_left, Hlower_right); +Draw(Plower, Tup); + +ParallelMark(Hupper_left, Hupper_right, placement = 0.7); +ParallelMark(Hlower_left, Hlower_right); diff --git a/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.pdf b/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.pdf new file mode 100644 index 00000000..67395687 Binary files /dev/null and b/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.pdf differ diff --git a/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.svg b/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.svg new file mode 100644 index 00000000..079deefb --- /dev/null +++ b/data/handouts/Images/angles-corresponding-alternate-supplementary-connection.svg @@ -0,0 +1,193 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-corresponding.asy b/data/handouts/Images/angles-corresponding.asy new file mode 100644 index 00000000..27f97aa4 --- /dev/null +++ b/data/handouts/Images/angles-corresponding.asy @@ -0,0 +1,19 @@ +import _common; + +real sep = 30, theta = 40; +real lowerRightPad = 100, upperRightPad = 70; +pair L = (0, 0); +pair U = (sep / tan(radians(theta)), sep); +pair Ttop = Polar(U, theta, 50); +pair Lright = (lowerRightPad, 0); +pair Uright = U + (upperRightPad, 0); + +AngleMark(Lright, L, U, LightRed); +AngleMark(Uright, U, Ttop, LightRed); + +Draw(L, Lright); +Draw(U, Uright); +Draw(L, Ttop); + +ParallelMark(U, Uright); +ParallelMark(L, Lright); diff --git a/data/handouts/Images/angles-corresponding.pdf b/data/handouts/Images/angles-corresponding.pdf new file mode 100644 index 00000000..12efb54a Binary files /dev/null and b/data/handouts/Images/angles-corresponding.pdf differ diff --git a/data/handouts/Images/angles-corresponding.svg b/data/handouts/Images/angles-corresponding.svg new file mode 100644 index 00000000..a53ef0c4 --- /dev/null +++ b/data/handouts/Images/angles-corresponding.svg @@ -0,0 +1,135 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-equilateral.asy b/data/handouts/Images/angles-equilateral.asy new file mode 100644 index 00000000..a8d1bb0b --- /dev/null +++ b/data/handouts/Images/angles-equilateral.asy @@ -0,0 +1,18 @@ +import _common; + +real base = 75; +pair B = (0, 0); +pair C = (base, 0); +pair A = (base / 2, base * sqrt(3) / 2); + +AngleMark(C, B, A, LightBlue, "60^\circ", radius = Radius2, labelFraction=1.7); +AngleMark(A, C, B, LightBlue, "60^\circ", radius = Radius2, labelFraction=1.7); +AngleMark(B, A, C, LightBlue, "60^\circ", radius = Radius2); + +Draw(B, C); +Draw(B, A); +Draw(A, C); + +LabeledDot(B, "B", SW); +LabeledDot(A, "A", N); +LabeledDot(C, "C", SE); diff --git a/data/handouts/Images/angles-equilateral.pdf b/data/handouts/Images/angles-equilateral.pdf new file mode 100644 index 00000000..7a4cfdd2 Binary files /dev/null and b/data/handouts/Images/angles-equilateral.pdf differ diff --git a/data/handouts/Images/angles-equilateral.svg b/data/handouts/Images/angles-equilateral.svg new file mode 100644 index 00000000..92d39872 --- /dev/null +++ b/data/handouts/Images/angles-equilateral.svg @@ -0,0 +1,275 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-isosceles-right-triangle.asy b/data/handouts/Images/angles-isosceles-right-triangle.asy new file mode 100644 index 00000000..c51a93c1 --- /dev/null +++ b/data/handouts/Images/angles-isosceles-right-triangle.asy @@ -0,0 +1,19 @@ +import _common; + +real leg = 65; +pair A = (0, 0); +pair B = (leg, 0); +pair C = (0, leg); + +RightAngleMark(B, A, C); + +AngleMark(A, C, B, LightBlue, "\beta"); +AngleMark(C, B, A, LightBlue, "\beta"); + +Draw(A, B); +Draw(C, B); +Draw(C, A); + +LabeledDot(A, "A", SW); +LabeledDot(B, "B", SE); +LabeledDot(C, "C", NW); diff --git a/data/handouts/Images/angles-isosceles-right-triangle.pdf b/data/handouts/Images/angles-isosceles-right-triangle.pdf new file mode 100644 index 00000000..181fa7cd Binary files /dev/null and b/data/handouts/Images/angles-isosceles-right-triangle.pdf differ diff --git a/data/handouts/Images/angles-isosceles-right-triangle.svg b/data/handouts/Images/angles-isosceles-right-triangle.svg new file mode 100644 index 00000000..cf5a074b --- /dev/null +++ b/data/handouts/Images/angles-isosceles-right-triangle.svg @@ -0,0 +1,193 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-isosceles-triangle.asy b/data/handouts/Images/angles-isosceles-triangle.asy new file mode 100644 index 00000000..6917865b --- /dev/null +++ b/data/handouts/Images/angles-isosceles-triangle.asy @@ -0,0 +1,19 @@ +import _common; + +real base = 85, height = 80; +pair B = (0, 0); +pair C = (base, 0); +pair M = Midpoint(B, C); +pair A = (M.x, height); + +RightAngleMark(B, M, A, color = LightBlue); + +Draw(B, A); +Draw(A, C); +Draw(C, B); +Draw(A, M); + +LabeledDot(B, "B", SW); +LabeledDot(A, "A", N); +LabeledDot(C, "C", SE); +LabeledDot(M, "M", S); diff --git a/data/handouts/Images/angles-isosceles-triangle.pdf b/data/handouts/Images/angles-isosceles-triangle.pdf new file mode 100644 index 00000000..9db3faeb Binary files /dev/null and b/data/handouts/Images/angles-isosceles-triangle.pdf differ diff --git a/data/handouts/Images/angles-isosceles-triangle.svg b/data/handouts/Images/angles-isosceles-triangle.svg new file mode 100644 index 00000000..bf11ea4e --- /dev/null +++ b/data/handouts/Images/angles-isosceles-triangle.svg @@ -0,0 +1,199 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-polygon-fan.asy b/data/handouts/Images/angles-polygon-fan.asy new file mode 100644 index 00000000..125d73c8 --- /dev/null +++ b/data/handouts/Images/angles-polygon-fan.asy @@ -0,0 +1,43 @@ +import _common; + +real S = 1.8; +pair A1 = S * (0, 0); +pair A2 = S * (-1, 22); +pair A3 = S * (18, 39); +pair A4 = S * (48, 40); +pair A5 = S * (55, 20); +pair A6 = S * (45, 0); + +AngleMark(A3, A1, A2, LightBlue, radius=Radius2); +AngleMark(A1, A2, A3, LightBlue, radius=Radius2); +AngleMark(A2, A3, A1, LightBlue, radius=Radius2); + +AngleMark(A4, A1, A3, LightRed, radius=Radius2); +AngleMark(A1, A3, A4, LightRed, radius=Radius2); +AngleMark(A3, A4, A1, LightRed, radius=Radius2); + +AngleMark(A5, A1, A4, LightGreen, radius=Radius2); +AngleMark(A1, A4, A5, LightGreen, radius=Radius2); +AngleMark(A4, A5, A1, LightGreen, radius=Radius2); + +AngleMark(A6, A1, A5, LightPurple, radius=Radius2); +AngleMark(A1, A5, A6, LightPurple, radius=Radius2); +AngleMark(A5, A6, A1, LightPurple, radius=Radius2); + +Draw(A1, A2, Blue); +Draw(A2, A3, Blue); +Draw(A3, A4, Red); +Draw(A4, A5, Green); +Draw(A5, A6, Purple); +Draw(A6, A1, Purple); + +Draw(A1, A3); +Draw(A1, A4); +Draw(A1, A5); + +LabeledDot(A1, "A_1", SW); +LabeledDot(A2, "A_2", W); +LabeledDot(A3, "A_3", N); +LabeledDot(A4, "A_4", NE); +LabeledDot(A5, "A_5", E); +LabeledDot(A6, "A_6", SE); diff --git a/data/handouts/Images/angles-polygon-fan.pdf b/data/handouts/Images/angles-polygon-fan.pdf new file mode 100644 index 00000000..5c1cd9bb Binary files /dev/null and b/data/handouts/Images/angles-polygon-fan.pdf differ diff --git a/data/handouts/Images/angles-polygon-fan.svg b/data/handouts/Images/angles-polygon-fan.svg new file mode 100644 index 00000000..ba507549 --- /dev/null +++ b/data/handouts/Images/angles-polygon-fan.svg @@ -0,0 +1,452 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-polygon-interior.asy b/data/handouts/Images/angles-polygon-interior.asy new file mode 100644 index 00000000..fb1307a2 --- /dev/null +++ b/data/handouts/Images/angles-polygon-interior.asy @@ -0,0 +1,60 @@ +import _common; + +pen LightOrange = rgb(1, 0.75, 0.5); +pen Orange = rgb(1, 0.5, 0); + +real scale = 1.8; +pair A1 = scale * (0, 0); +pair A2 = scale * (-1, 22); +pair A3 = scale * (18, 39); +pair A4 = scale * (48, 40); +pair A5 = scale * (55, 20); +pair A6 = scale * (45, 0); + +pair O = (A1 + A2 + A3 + A4 + A5 + A6) / 6; + +AngleMark(A2, O, A1, LightBlue, radius=Radius2); +AngleMark(O, A1, A2, LightBlue, radius=Radius2); +AngleMark(A1, A2, O, LightBlue, radius=Radius2); + +AngleMark(A3, O, A2, LightRed, radius=Radius2); +AngleMark(O, A2, A3, LightRed, radius=Radius2); +AngleMark(A2, A3, O, LightRed, radius=Radius2); + +AngleMark(A4, O, A3, LightGreen, radius=Radius2); +AngleMark(O, A3, A4, LightGreen, radius=Radius2); +AngleMark(A3, A4, O, LightGreen, radius=Radius2); + +AngleMark(A5, O, A4, LightPurple, radius=Radius2); +AngleMark(O, A4, A5, LightPurple, radius=Radius2); +AngleMark(A4, A5, O, LightPurple, radius=Radius2); + +AngleMark(A6, O, A5, LightPink, radius=Radius2); +AngleMark(O, A5, A6, LightPink, radius=Radius2); +AngleMark(A5, A6, O, LightPink, radius=Radius2); + +AngleMark(A1, O, A6, LightOrange, radius=Radius2); +AngleMark(O, A6, A1, LightOrange, radius=Radius2); +AngleMark(A6, A1, O, LightOrange, radius=Radius2); + +Draw(A1, A2, Blue); +Draw(A2, A3, Red); +Draw(A3, A4, Green); +Draw(A4, A5, Purple); +Draw(A5, A6, Pink); +Draw(A6, A1, Orange); + +Draw(O, A1); +Draw(O, A2); +Draw(O, A3); +Draw(O, A4); +Draw(O, A5); +Draw(O, A6); + +LabeledDot(A1, "A_1", SW); +LabeledDot(A2, "A_2", W); +LabeledDot(A3, "A_3", N); +LabeledDot(A4, "A_4", NE); +LabeledDot(A5, "A_5", E); +LabeledDot(A6, "A_6", SE); +LabeledDot(O, "O", S, offset=(-1,2.5)); diff --git a/data/handouts/Images/angles-polygon-interior.pdf b/data/handouts/Images/angles-polygon-interior.pdf new file mode 100644 index 00000000..f49d39e8 Binary files /dev/null and b/data/handouts/Images/angles-polygon-interior.pdf differ diff --git a/data/handouts/Images/angles-polygon-interior.svg b/data/handouts/Images/angles-polygon-interior.svg new file mode 100644 index 00000000..b3e5dad8 --- /dev/null +++ b/data/handouts/Images/angles-polygon-interior.svg @@ -0,0 +1,554 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-polygon.asy b/data/handouts/Images/angles-polygon.asy new file mode 100644 index 00000000..432c1b49 --- /dev/null +++ b/data/handouts/Images/angles-polygon.asy @@ -0,0 +1,35 @@ +import _common; + +real S = 1.8; +pair A1 = S * (0, 0); +pair A2 = S * (-1, 22); +pair A3 = S * (18, 39); +pair A4 = S * (48, 40); +pair A5 = S * (55, 20); +pair A6 = S * (45, 0); + + +AngleMark(A5, A1, A2, LightPink, radius=Radius2); +AngleMark(A1, A2, A3, LightPink, radius=Radius2); +AngleMark(A2, A3, A4, LightPink, radius=Radius2); +AngleMark(A3, A4, A5, LightPink, radius=Radius2); +AngleMark(A4, A5, A1, LightPink, radius=Radius2); + +AngleMark(A6, A1, A5, LightBlue, radius=Radius2); +AngleMark(A1, A5, A6, LightBlue, radius=Radius2); +AngleMark(A5, A6, A1, LightBlue, radius=Radius2); + +Draw(A1, A6, Blue); +Draw(A6, A5, Blue); +Draw(A1, A2, Purple); +Draw(A2, A3, Purple); +Draw(A3, A4, Purple); +Draw(A4, A5, Purple); +Draw(A1, A5); + +LabeledDot(A1, "A_1", SW); +LabeledDot(A2, "A_2", W); +LabeledDot(A3, "A_3", N); +LabeledDot(A4, "A_4", NE); +LabeledDot(A5, "A_5", E); +LabeledDot(A6, "A_6", SE); diff --git a/data/handouts/Images/angles-polygon.pdf b/data/handouts/Images/angles-polygon.pdf new file mode 100644 index 00000000..3c8f4302 Binary files /dev/null and b/data/handouts/Images/angles-polygon.pdf differ diff --git a/data/handouts/Images/angles-polygon.svg b/data/handouts/Images/angles-polygon.svg new file mode 100644 index 00000000..2c7de3b6 --- /dev/null +++ b/data/handouts/Images/angles-polygon.svg @@ -0,0 +1,418 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-sas-proof.asy b/data/handouts/Images/angles-sas-proof.asy new file mode 100644 index 00000000..7cec0ebf --- /dev/null +++ b/data/handouts/Images/angles-sas-proof.asy @@ -0,0 +1,25 @@ +import _common; + +real base = 100, legC = 72, alpha = 65; +pair A = (0, 0); +pair B = (base, 0); +pair C = Polar(A, alpha, legC); +pair Cp = Polar(A, -alpha, legC); + +Circle(A, legC, Blue); + +AngleMark(B, A, C, LightRed); +AngleMark(Cp, A, B, Red); + +Draw(A, B, Green); +Draw(A, C, Blue); +Draw(A, Cp, Blue); +Draw(C, B, black + dashedPen); + +LabeledDot(A, "A", W, Blue); +LabeledDot(B, "B", E, Red); +LabeledDot(C, "C", N, Green); +LabeledDot(Cp, "C'", S, Green); + +EdgeLabel(A, C, "c", W, color = Blue, distanceOffset = 8); +EdgeLabel(A, Cp, "c", W, color = Blue, distanceOffset = 8); diff --git a/data/handouts/Images/angles-sas-proof.pdf b/data/handouts/Images/angles-sas-proof.pdf new file mode 100644 index 00000000..75337406 Binary files /dev/null and b/data/handouts/Images/angles-sas-proof.pdf differ diff --git a/data/handouts/Images/angles-sas-proof.svg b/data/handouts/Images/angles-sas-proof.svg new file mode 100644 index 00000000..21b760c5 --- /dev/null +++ b/data/handouts/Images/angles-sas-proof.svg @@ -0,0 +1,253 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-sas-warning.asy b/data/handouts/Images/angles-sas-warning.asy new file mode 100644 index 00000000..8d8e0cf9 --- /dev/null +++ b/data/handouts/Images/angles-sas-warning.asy @@ -0,0 +1,31 @@ +import _common; + +real base = 70, leg = 40, beta = 40, gap = 65; +pair B = (0, 0); +pair C = (base, 0); +pair A = Polar(B, beta, leg); + +// A' is the OTHER intersection of ray B'A' with the swing circle around C' of +// radius |C'A'| = |CA|, giving a non-congruent triangle. +pair shift = (base + gap, 0); +pair Bp = B + shift; +pair Cp = C + shift; +real BAp = 2 * base * cos(radians(beta)) - leg; +pair Ap = Polar(Bp, beta, BAp); + +AngleMark(C, B, A, LightRed); +AngleMark(Cp, Bp, Ap, LightRed); + +Draw(B, C, Green); +Draw(A, C, Red); +Draw(B, A); +Draw(Bp, Cp, Green); +Draw(Ap, Cp, Red); +Draw(Bp, Ap); + +LabeledDot(B, "B", SW); +LabeledDot(C, "C", SE); +LabeledDot(A, "A", N); +LabeledDot(Bp, "B'", SW); +LabeledDot(Cp, "C'", SE); +LabeledDot(Ap, "A'", N); diff --git a/data/handouts/Images/angles-sas-warning.pdf b/data/handouts/Images/angles-sas-warning.pdf new file mode 100644 index 00000000..38bf5d0d Binary files /dev/null and b/data/handouts/Images/angles-sas-warning.pdf differ diff --git a/data/handouts/Images/angles-sas-warning.svg b/data/handouts/Images/angles-sas-warning.svg new file mode 100644 index 00000000..f18cdd2c --- /dev/null +++ b/data/handouts/Images/angles-sas-warning.svg @@ -0,0 +1,311 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-sss-proof.asy b/data/handouts/Images/angles-sss-proof.asy new file mode 100644 index 00000000..f2c8fd10 --- /dev/null +++ b/data/handouts/Images/angles-sss-proof.asy @@ -0,0 +1,26 @@ +import _common; + +real base = 65, rB = 60, rC = 55; +pair B = (0, 0); +pair C = (base, 0); +// A and A' are the two intersections of circle(B, rB) and circle(C, rC); pick +// the upper one for A and reflect across BC for A'. +pair[] meets = intersectionpoints(circle(B, rB), circle(C, rC)); +pair A = meets[0].y > meets[1].y ? meets[0] : meets[1]; +pair Ap = ReflectAcross(A, B, C); + +Circle(B, rB, Blue); +Circle(C, rC, Red); + +Draw(B, C, Green); +Draw(A, B, Blue); +Draw(C, A, Red); + +LabeledDot(B, "B", W, Blue); +LabeledDot(C, "C", E, Red); +LabeledDot(A, "A", N, Green); +LabeledDot(Ap, "A'", S, Green); + +EdgeLabel(B, A, "c", W, color = Blue, distanceOffset = 8); +EdgeLabel(C, A, "b", E, color = Red, distanceOffset = 8); +EdgeLabel(B, C, "a", S, distanceOffset = 4, color = Green); diff --git a/data/handouts/Images/angles-sss-proof.pdf b/data/handouts/Images/angles-sss-proof.pdf new file mode 100644 index 00000000..55c39f5a Binary files /dev/null and b/data/handouts/Images/angles-sss-proof.pdf differ diff --git a/data/handouts/Images/angles-sss-proof.svg b/data/handouts/Images/angles-sss-proof.svg new file mode 100644 index 00000000..548b2566 --- /dev/null +++ b/data/handouts/Images/angles-sss-proof.svg @@ -0,0 +1,273 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-supplementary-derive-vertical.asy b/data/handouts/Images/angles-supplementary-derive-vertical.asy new file mode 100644 index 00000000..4631d347 --- /dev/null +++ b/data/handouts/Images/angles-supplementary-derive-vertical.asy @@ -0,0 +1,15 @@ +import _common; + +real ang1 = 30, ang2 = 135, len = 55; +pair X = (0, 0); +pair L1a = Polar(X, ang1 + 180, len); +pair L1b = Polar(X, ang1, len); +pair L2a = Polar(X, ang2 + 180, len); +pair L2b = Polar(X, ang2, len); + +AngleMark(L1b, X, L2b, LightRed, "\alpha"); +AngleMark(L2b, X, L1a, LightBlue, "\alpha'"); +AngleMark(L1a, X, L2a, LightRed, "\alpha"); + +Draw(L1a, L1b); +Draw(L2a, L2b); diff --git a/data/handouts/Images/angles-supplementary-derive-vertical.pdf b/data/handouts/Images/angles-supplementary-derive-vertical.pdf new file mode 100644 index 00000000..6a60ed47 Binary files /dev/null and b/data/handouts/Images/angles-supplementary-derive-vertical.pdf differ diff --git a/data/handouts/Images/angles-supplementary-derive-vertical.svg b/data/handouts/Images/angles-supplementary-derive-vertical.svg new file mode 100644 index 00000000..a2653d3b --- /dev/null +++ b/data/handouts/Images/angles-supplementary-derive-vertical.svg @@ -0,0 +1,131 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-supplementary.asy b/data/handouts/Images/angles-supplementary.asy new file mode 100644 index 00000000..743cce98 --- /dev/null +++ b/data/handouts/Images/angles-supplementary.asy @@ -0,0 +1,13 @@ +import _common; + +real base = 85, theta = 70, rayLen = 50; +pair Lleft = (0, 0); +pair Lright = (base, 0); +pair V = Midpoint(Lleft, Lright); +pair Rtop = Polar(V, theta, rayLen); + +AngleMark(Lright, V, Rtop, LightRed); +AngleMark(Rtop, V, Lleft, LightBlue); + +Draw(Lleft, Lright); +Draw(Rtop, V); diff --git a/data/handouts/Images/angles-supplementary.pdf b/data/handouts/Images/angles-supplementary.pdf new file mode 100644 index 00000000..1fba2a22 Binary files /dev/null and b/data/handouts/Images/angles-supplementary.pdf differ diff --git a/data/handouts/Images/angles-supplementary.svg b/data/handouts/Images/angles-supplementary.svg new file mode 100644 index 00000000..c207b0ae --- /dev/null +++ b/data/handouts/Images/angles-supplementary.svg @@ -0,0 +1,91 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-triangle.asy b/data/handouts/Images/angles-triangle.asy new file mode 100644 index 00000000..b87c727c --- /dev/null +++ b/data/handouts/Images/angles-triangle.asy @@ -0,0 +1,30 @@ +import _common; + +real base = 150, alphaB = 70, gammaC = 40; +pair B = (0, 0); +pair C = (base, 0); +pair A = extension(B, Polar(B, alphaB, 1), C, Polar(C, 180 - gammaC, 1)); + +real apexLeftPad = 40, apexRightPad = 60; +pair Hleft = (A.x - apexLeftPad, A.y); +pair Hright = (A.x + apexRightPad, A.y); + +AngleMark(C, B, A, LightRed, "\beta"); +AngleMark(A, C, B, LightPink, "\gamma"); +AngleMark(B, A, C, LightBlue, "\alpha"); +AngleMark(Hleft, A, B, LightRed, "\beta"); +AngleMark(C, A, Hright, LightPink, "\gamma", labelFraction=1.4); + +Draw(B, A); +Draw(A, C); +Draw(B, C); +Draw(Hleft, A); +Draw(A, Hright); + +ParallelMark(B, C); +ParallelMark(A, Hright, placement=0.7); + +LabeledDot(A, "A", N); +LabeledDot(B, "B", SW); +LabeledDot(C, "C", SE); +LabeledDot(O, "O", S); diff --git a/data/handouts/Images/angles-triangle.pdf b/data/handouts/Images/angles-triangle.pdf new file mode 100644 index 00000000..2f209ed4 Binary files /dev/null and b/data/handouts/Images/angles-triangle.pdf differ diff --git a/data/handouts/Images/angles-triangle.svg b/data/handouts/Images/angles-triangle.svg new file mode 100644 index 00000000..b6549c90 --- /dev/null +++ b/data/handouts/Images/angles-triangle.svg @@ -0,0 +1,313 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-vertical-derive-supplementary.asy b/data/handouts/Images/angles-vertical-derive-supplementary.asy new file mode 100644 index 00000000..be103f8f --- /dev/null +++ b/data/handouts/Images/angles-vertical-derive-supplementary.asy @@ -0,0 +1,16 @@ +import _common; + +real ang1 = 30, ang2 = 135, len = 55; +pair X = (0, 0); +pair L1a = Polar(X, ang1 + 180, len); +pair L1b = Polar(X, ang1, len); +pair L2a = Polar(X, ang2 + 180, len); +pair L2b = Polar(X, ang2, len); + +AngleMark(L1b, X, L2b, LightRed, "\alpha"); +AngleMark(L2b, X, L1a, LightBlue, "\beta"); +AngleMark(L1a, X, L2a, LightRed, "\gamma"); +AngleMark(L2a, X, L1b, LightBlue, "\delta"); + +Draw(L1a, L1b); +Draw(L2a, L2b); diff --git a/data/handouts/Images/angles-vertical-derive-supplementary.pdf b/data/handouts/Images/angles-vertical-derive-supplementary.pdf new file mode 100644 index 00000000..5a96bffc Binary files /dev/null and b/data/handouts/Images/angles-vertical-derive-supplementary.pdf differ diff --git a/data/handouts/Images/angles-vertical-derive-supplementary.svg b/data/handouts/Images/angles-vertical-derive-supplementary.svg new file mode 100644 index 00000000..9216968d --- /dev/null +++ b/data/handouts/Images/angles-vertical-derive-supplementary.svg @@ -0,0 +1,149 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-vertical.asy b/data/handouts/Images/angles-vertical.asy new file mode 100644 index 00000000..ca21b222 --- /dev/null +++ b/data/handouts/Images/angles-vertical.asy @@ -0,0 +1,14 @@ +import _common; + +real ang1 = 35, ang2 = 140, len = 50; +pair I = (0, 0); +pair A = Polar(I, ang1 + 180, len); +pair B = Polar(I, ang1, len); +pair D = Polar(I, ang2, len); +pair C = Polar(I, ang2 + 180, len); + +AngleMark(B, I, D, LightRed, radius=Radius2); +AngleMark(A, I, C, LightRed, radius=Radius2); + +Draw(A, B); +Draw(D, C); diff --git a/data/handouts/Images/angles-vertical.pdf b/data/handouts/Images/angles-vertical.pdf new file mode 100644 index 00000000..79165e4a Binary files /dev/null and b/data/handouts/Images/angles-vertical.pdf differ diff --git a/data/handouts/Images/angles-vertical.svg b/data/handouts/Images/angles-vertical.svg new file mode 100644 index 00000000..eab389ff --- /dev/null +++ b/data/handouts/Images/angles-vertical.svg @@ -0,0 +1,69 @@ + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/angle-basics-1.cs.tex b/data/handouts/angle-basics-1.cs.tex new file mode 100644 index 00000000..389dd319 --- /dev/null +++ b/data/handouts/angle-basics-1.cs.tex @@ -0,0 +1,304 @@ +\input _template + +\setlanguage{CS} + +\Title{Základy počítání úhlů} + +\MathcompsLink{zaklady-pocitani-uhlu-1} + +\Author{Patrik Bak} + +\sec Úvod + +Geometrie je nejvizuálnější část matematiky -- umíme si ji kreslit na papír, v~softwaru jako \Link[https://www.geogebra.org/]{GeoGebra} a~můžeme se těšit, když na vlastní oči vidíme, že věci v~ní fungují. Proč tomu tak je však může být záhadou. Abychom tato tajemství pochopili, potřebujeme kousek po kousku budovat základnu znalostí a~vizuální intuici. + +Jedna z~nejzákladnějších technik je rozumět světu úhlů a~umět jejich vlastnosti s~úspěchem používat v~různých typech úloh. Jen málo zajímavých příkladů úhly nepoužívá. Teorie kolem úhlů přitom není složitá, je však důležité ji budovat pomalu a~detailně. Začneme s~nejjednoduššími vlastnostmi, které ani nevyžadují práci s~kružnicemi. Už ty se však dají s~úspěchem využít v~mnoha úlohách. + +V~tomto materiálu se budeme věnovat úhlům, ale záměrně vynecháme úhly na kružnicích, kterým se patří věnovat samostatnou pozornost. Na chvíli zapomeňme, že kružnice existují. + +Na závěr úvodu jedna menší poznámka o~konvenci. V~tomto materiálu uvidíme mnoho obrázků trojúhelníků, ve kterých je bod~$A$ nakreslen nahoře. Ukazuje se, že tento způsob kreslení je mezinárodní standard, zejména na poli matematické olympiády a~podobných soutěží se to tak zvykne dělat. V~Česku a~na~Slovensku je však ve školách stále zvykem kreslit~$C$ nahoře. V~některých zemích lze vidět i~$B$ nahoře, např. na Ukrajině. Z~hlediska řešení úloh je vždy dobré nakreslit si nahoru ten bod, u~kterého obrázek vypadá co nejsymetričtěji, snadněji v~něm jdou vidět další symetrické věci. V~tomto materiálu budou úlohy zadávány tak, že ta symetrie půjde vidět více při~$A$ nahoře \SlightSmile. + +\sec Základy světa úhlů + +V~této sekci si odvodíme ty nejjednodušší vlastnosti související s~úhly. Pevně věřím, že klíčem ke~zvládnutí geometrie je porozumění věcem od základů a~do hloubky, proto se těmto jednoduchým vlastnostem budeme věnovat více. + +Základní vlastnosti úhlů, které běžně používáme skoro bez přemýšlení při řešení úloh: + +\Highlight{ + \begitems \style i + \i Vrcholové úhly jsou shodné: + \Image{angles-vertical.pdf} + \i Souhlasné úhly jsou shodné: + \Image{angles-corresponding.pdf} + \i Střídavé úhly jsou shodné: + \Image{angles-alternate.pdf} + \i Vedlejší úhly mají součet $180^\circ$: + \Image{angles-supplementary.pdf} + \enditems +} + +Tyto vlastnosti spolu velmi přirozeně souvisí, menší cvičení k~zamyšlení: + +\Exercise{}{ + Uvědomte si, že: + \begitems \style a + \i z~vlastnosti o~vedlejších úhlech vyplývá vlastnost vrcholových úhlů (a~naopak) + \i z~kterékoli dvojice vlastností o~vrcholových, souhlasných, střídavých úhlech vyplývá ta třetí vlastnost + \enditems +}{ + \textbf{(a)} Vedlejší úhel $\alpha'$ k~danému úhlu $\alpha$ má velikost $180^\circ - \alpha$. Úhel vedlejší k~$\alpha'$ má zase velikost $180^\circ - (180^\circ - \alpha) = \alpha$. Ten je však zároveň vrcholový úhel k~$\alpha$, tedy jsou vrcholové úhly shodné. + + \Image{angles-supplementary-derive-vertical.pdf} + + Naopak, předpokládejme vlastnost o~vrcholových úhlech a~označme po sobě jdoucí úhly v~průsečíku dvou přímek $\alpha$, $\beta$, $\gamma$, $\delta$. Ze shodnosti vrcholových úhlů $\alpha = \gamma$ a~$\beta = \delta$. Součet všech čtyř je plný úhel, tedy $\alpha + \beta + \gamma + \delta = 360^\circ$. Po dosazení dostáváme $2\alpha + 2\beta = 360^\circ$, čili $\alpha + \beta = 180^\circ$ -- to je právě vlastnost vedlejších úhlů. + + \Image{angles-vertical-derive-supplementary.pdf} + + \textbf{(b)} Všechny tři vlastnosti říkají, že jistá dvojice úhlů má stejnou velikost. + + Podívejme se na trojici úhlů $\alpha$, $\beta$, $\gamma$ z~obrázku a~na to, co o~jejich vztazích tvrdí jednotlivé vlastnosti. Pro dvojici $\alpha$, $\beta$ při horním průsečíku tvrdí vlastnost vrcholových úhlů $\alpha = \beta$. Pro dvojici $\alpha$, $\gamma$ tvrdí vlastnost souhlasných úhlů $\alpha = \gamma$. Konečně pro dvojici $\beta$, $\gamma$ tvrdí vlastnost střídavých úhlů $\beta = \gamma$. + + \Image{angles-corresponding-alternate-supplementary-connection.pdf} + + Tři vlastnosti tedy tvrdí tři rovnosti mezi týmiž třemi úhly, jen přes různé dvojice. Jakmile platí libovolné dvě, třetí plyne tranzitivitou: např. pokud víme (V) a~(S), tak $\alpha = \beta$ a~$\alpha = \gamma$, takže $\beta = \gamma$, čili (St). Ostatní dvojice analogicky. +} + +Následující tvrzení všichni známe, umíte ho však dokázat? + +\Theorem{}{ + Součet úhlů v~trojúhelníku je $180^\circ$. +}{ + Veďme přes vrchol~$A$ přímku rovnoběžnou s~$BC$. Protože $AB$ je příčka rovnoběžek, úhel $\beta$ při~$A$ a~úhel $\beta$ při~$B$ jsou střídavé, tedy stejné. Analogicky přes příčku $AC$ jsou úhly $\gamma$ při~$A$ a~při~$C$ střídavé. Úhly $\beta$, $\alpha$, $\gamma$ leží vedle sebe podél přímky přes~$A$, tedy $\alpha + \beta + \gamma = 180^\circ.$ + + \Image{angles-triangle.pdf} +} + +Toto tvrzení se rozhodně dá zobecnit. Víme například, že součet úhlů ve~čtyřúhelníku je $360^\circ$. Jak je to v~5-úhelníku? A~jak v~67-úhelníku? Odpovědí je následující tvrzení: + +\Theorem{}{ + Nechť $n \ge 3$ je přirozené číslo. Součet úhlů v~konvexním $n$-úhelníku je roven $(n-2) \cdot 180^\circ$. +}{ + \NamedProof{Důkaz 1 (matematická indukce).} Základ $n=3$: součet úhlů trojúhelníku je $180^\circ = (3-2)\cdot 180^\circ$. Indukční krok: nechť $A_1 A_2 \ldots A_{n+1}$ je konvexní $(n+1)$-úhelník. Úhlopříčkou $A_1 A_n$ jej rozdělíme na trojúhelník $A_1 A_n A_{n+1}$ a~konvexní $n$-úhelník $A_1 A_2 \ldots A_n$: úhly při vrcholech $A_2, \ldots, A_{n-1}$ patří celé $n$-úhelníku, úhel při $A_{n+1}$ patří celý trojúhelníku a~úhly při $A_1$, $A_n$ se rozdělí mezi oba útvary tak, aby jejich části daly původní vnitřní úhly $(n+1)$-úhelníku. Součet úhlů $(n+1)$-úhelníku je tedy + $$ + \underbrace{180^\circ}_{\text{trojúhelník}} + \underbrace{(n-2) \cdot 180^\circ}_{n\text{-úhelník}} = (n-1) \cdot 180^\circ = \bigl((n+1) - 2\bigr) \cdot 180^\circ. + $$ + + \Image{angles-polygon.pdf} + + \NamedProof{Důkaz 2 (vějířová triangulace).} Veďme z~vrcholu $A_1$ úhlopříčky do každého nesousedního vrcholu $A_3, A_4, \ldots, A_{n-1}$. Vznikne $n-2$ trojúhelníků pokrývajících vnitřek $n$-úhelníku bez překrytí. Při každém vrcholu $A_k$ se úhly sousedních trojúhelníků při $A_k$ přesně poskládají na vnitřní úhel $n$-úhelníku při $A_k$, takže součet úhlů všech trojúhelníků se rovná součtu vnitřních úhlů $n$-úhelníku, čili $(n-2)\cdot 180^\circ$. + + \Image{angles-polygon-fan.pdf} + + \Remark Důkaz~2 je \uv{rozbalený} důkaz~1: indukce odřezává trojúhelníky jeden po druhém, zde je vidíme všechny najednou. + + \NamedProof{Důkaz 3 (vnitřní bod).} Zvolme bod $O$ uvnitř $n$-úhelníku a~spojme ho s~každým vrcholem. Vznikne $n$ trojúhelníků s~celkovým součtem úhlů $n\cdot 180^\circ$. Úhly trojúhelníků při každém vrcholu $A_k$ spolu tvoří vnitřní úhel $n$-úhelníku při~$A_k$; úhly trojúhelníků při~$O$ spolu tvoří plný úhel $360^\circ$. Tedy součet vnitřních úhlů $n$-úhelníku je + $$ + n\cdot 180^\circ - 360^\circ = (n-2)\cdot 180^\circ. + $$ + + \Image{angles-polygon-interior.pdf} + + \Remark{nekonvexní mnohoúhelníky} Vzorec $(n-2)\cdot 180^\circ$ platí i~pro navzájem se neprotínající nekonvexní mnohoúhelníky, pokud všechny vnitřní úhly měříme směrem dovnitř -- tedy i~úhly větší než $180^\circ$ počítáme jako takové. + + Vějířová triangulace z~jednoho vrcholu nemusí fungovat ani z~konvexního vrcholu -- některé úhlopříčky mohou procházet mimo mnohoúhelník. Platí však, že každý jednoduchý $n$-úhelník se dá rozdělit na $n-2$ trojúhelníků (ne nutně vějířovitě), což důkaz zachrání. Důvodem je indukce přes trojúhelník tvořený nějakým vrcholem $V$ a~jeho sousedy $U$, $W$, který leží celý uvnitř našeho $n$-úhelníku. Odříznutím $V$ úhlopříčkou $UW$ dostaneme $(n-1)$-úhelník; opakováním tedy $n-2$ trojúhelníků. Dá se dokázat, že takový trojúhelník vždy existuje (dokonce dva), viz \Link[https://en.wikipedia.org/wiki/Two_ears_theorem]{věta o~dvou uších}. + + K~důkazu~3: funguje jen pokud existuje vnitřní bod $O$ viditelný ze všech vrcholů (tj. úsečky $OA_i$ leží celé uvnitř). Pro některé nekonvexní mnohoúhelníky takový $O$ nemusí existovat (zkuste takový nakreslit) -- v~takovém případě se důkaz~3 nedá přímo zachránit a~je třeba sáhnout po triangulaci z~důkazu~2. +} + +Teď uveďme ještě dvě jednoduchá, ale užitečná pomocná tvrzení: + +\Exercise{}{ + Dokažte, že součet úhlů na obrázku je roven $180^\circ$. + + \Image{angles-complementary-parallel-statement.pdf} +}{ + Označme $\alpha$ úhel svíraný příčkou $AD$ s~rovnoběžkami na straně $B$ (jako na obrázku). Souhlasné úhly $\alpha$ při~$A$ a~$\alpha$ při~$D$ jsou shodné. Při vrcholu $A$ jsou $\alpha$ a~$\beta$ vedlejší, tedy + $$ + \beta + \alpha = 180^\circ. + $$ + + \Image{angles-complementary-parallel-solution.pdf} +} + +\Exercise{}{ + Dokažte, že úhel s~otazníkem na obrázku je roven $\alpha+\beta$. + + \Image{angles-complementary-in-triangle.pdf} +}{ + Ze součtu úhlů v~trojúhelníku $|\angle ACB| = 180^\circ - \alpha - \beta$. Vnější úhel při $C$ je vedlejší k~$|\angle ACB|$: + $$ + 180^\circ - |\angle ACB| = \alpha + \beta. + $$ + + \Image{angles-complementary-in-triangle.pdf} +} + +Poslední dvě cvičení možná vypadala velmi triviálně. Realita však je, že v~praktických úlohách jsou velmi užitečná -- když máme obrovský netriviální obrázek a~máme provést za sebou velký počet úhlových operací, tak je opravdu velmi výhodné si i~jen kousek té práce zjednodušit -- v~případě rovnoběžek nemusíme uvažovat pomocný souhlasný úhel a~v~případě trojúhelníku nemusíme pracovat s~$180^\circ$. + +\sec Základy světa délek + +Zatím jsme fungovali pouze ve světě úhlů a~vůbec jsme neřešili délky. Geometrie však začne být zajímavá, když tyto světy začneme propojovat. Základem pro nás bude {\em shodnost trojúhelníků}. Připomeňme si kritéria. + +\Highlight{ + Věty o~shodnosti trojúhelníků: + \begitems \style i + \i Věta $sss$: dva trojúhelníky jsou shodné, pokud se shodují ve všech třech stranách. + \i Věta $sus$: dva trojúhelníky jsou shodné, pokud se shodují ve dvou stranách a~úhlu, který tyto strany svírají. + \i Věta $usu$: dva trojúhelníky jsou shodné, pokud se shodují v~jedné straně a~nějakých dvou úhlech. + \i Věta $Ssu$: dva trojúhelníky jsou shodné, pokud se shodují ve dvou stranách a~úhlu naproti delší z~těchto stran. + \enditems +} + +Zdůrazněme, že u~věty $sus$ je opravdu důležité, že shodný úhel je ten svíraný oběma stranami -- bez tohoto předpokladu shodnost nemusí fungovat, viz obrázek. Na něm máme dva trojúhelníky $ABC$ a~$A'B'C'$, pro které platí $|BC|=|B'C'|$, $|AC|=|A'C'|$, a~$\beta=\beta'$, avšak zjevně nejsou shodné. + +\Image{angles-sas-warning.pdf} + +Věta $Ssu$ by byla naše záchrana -- zde ji však aplikovat neumíme, protože strana, naproti které leží úhel $\beta$, tedy $|AC|$, není nejdelší, neboť je zjevně kratší než $|BC|$ \SmileWithTear. + +Než půjdeme dál, uvědomme si, co tyto věty znamenají. Z~mého pohledu je dobrý pohled na shodnost takový: chceme sestrojit trojúhelník, když máme dány nějaké tři jeho elementy -- budou všechny sestrojitelné trojúhelníky shodné? + +\Example{}{ + Přesvědčte se o~platnosti věty $sss$ o~shodnosti. +}{ + Představme si, že máme dány tři úsečky délek $a$, $b$, $c$, ze kterých umíme sestrojit trojúhelník. Začneme např. úsečkou~$BC$ délky~$a$, a~následně sestrojíme dvě kružnice: (a) kružnici se středem v~$B$ a~poloměrem $c$; (b) kružnici se středem v~$C$ a~poloměrem $b$. Ty se buď neprotnou (když např. $b+c < a$), nebo se protnou v~jediném bodě na $BC$ (když $b+c=a$), nebo se protnou ve dvou bodech $A$ a~$A'$. + + \Image{angles-sss-proof.pdf} + + Na konci si uvědomme, že trojúhelníky $ABC$ a~$A'BC$ jsou však zřejmě vzájemně zrcadlovými obrazy podle $BC$. Také pokud bychom začali jinou úsečkou, tak vytvoříme pouze posunutí/otočení této konfigurace. + + \Remark Uvědomme si, že jsme vlastně po cestě dokázali~trojúhelníkovou nerovnost: součet dvou stran trojúhelníku je větší než třetí -- náš trojúhelník je sestrojitelný a~nedegenerovaný~\fnote{nemá úhel rovný $180^\circ$} právě tehdy, když $b+c>a$. +} + +Podobně umíme zdůvodnit další kritéria: + +\Exercise{}{ + Přesvědčte se o~platnosti věty $sus$ o~shodnosti. +}{ + Nechť jsou dány strany $b = |AC|$, $c = |AB|$ a~úhel $|\angle BAC| = \alpha$ mezi nimi. Sestrojme vrchol~$A$ a~úsečku $AB$ délky~$c$ -- ta je dána až na shodné zobrazení. Bod $C$ musí ležet na polopřímce z~$A$ svírající s~$AB$ úhel~$\alpha$ a~zároveň ve vzdálenosti~$b$ od~$A$. Polopřímky svírající s~$AB$ úhel $\alpha$ jsou dvě (po jedné na každé straně $AB$): na každé leží právě jeden bod ve vzdálenosti $b$ od~$A$, tedy dostaneme body $C$ a~$C'$. Trojúhelníky $ABC$ a~$ABC'$ jsou zrcadlovými obrazy podle $AB$, tedy shodné. + + \Image{angles-sas-proof.pdf} +} + +\Exercise{}{ + Přesvědčte se o~platnosti věty $usu$ o~shodnosti. +}{ + V~prvé řadě si uvědomme, že nezáleží na tom, které dva úhly jsou shodné -- pokud se dva trojúhelníky shodují ve dvou úhlech, tak třetí je určen jednoznačně, neboť všechny tři mají součet úhlů $180^\circ$. + + Pro účely našeho důkazu uvažme tedy, že jsou shodné právě úhly přiléhající ke shodné straně. Nechť je to úhel $\beta$ při vrcholu $B$, $\gamma$ při vrcholu $C$ a~strana $a = |BC|$ mezi nimi. Sestrojme úsečku $BC$ délky~$a$ -- ta je dána až na shodné zobrazení. Bod $A$ musí ležet na polopřímce z~$B$ svírající s~$BC$ úhel $\beta$ a~zároveň na polopřímce z~$C$ svírající s~$CB$ úhel $\gamma$. Součet vnitřních úhlů při $B$ a~$C$ v~trojúhelníku je menší než $180^\circ$, čili $\beta + \gamma < 180^\circ$, a~proto nejsou obě polopřímky rovnoběžné -- protnou se v~jediném bodě~$A$. Trojúhelník $ABC$ je tedy určen jednoznačně (až na shodná zobrazení). + + \Image{angles-asa-proof.pdf} +} + +Méně známé tvrzení $Ssu$ je už trošičku těžší, proto k~němu budou i~návody \SlightSmile. + +\Problem{0}{shodnost $Ssu$}{ + Přesvědčte se o~platnosti věty $Ssu$ o~shodnosti. Navíc si uvědomte, kde by se konstrukce pokazila, kdyby úhel nebyl naproti větší ze stran -- co by se stalo, kdyby byly obě strany stejně dlouhé? +}{ + Začněte konstrukci konstrukcí kratší úsečky. Jak pokračuje konstrukce po jejím sestrojení? +}{ + Po sestrojení kratší úsečky sestrojíme polopřímku pod naším daným shodným úhlem. Zbývá poslední krok konstrukce. Rozmyslete, kdy dostaneme žádný, kdy jeden a~kdy dva vyhovující trojúhelníky. +}{ + Nechť jsou dány strany $a = |BC|$, $b = |AC|$ a~úhel $|\angle ABC| = \beta$ naproti delší straně $b$. Začneme sestrojením kratší úsečky $BC$ délky~$a$. Bod $A$ musí ležet na polopřímce z~$B$ svírající s~$BC$ úhel $\beta$ a~zároveň ve vzdálenosti $b$ od~$C$, čili na kružnici $k$ se středem~$C$ a~poloměrem $b$. + + Vzdálenost bodu~$B$ od středu kružnice je $|BC| = a$. Protože $a < b$, bod~$B$ leží \textit{uvnitř} kružnice $k$. Polopřímka z~bodu~$B$ tak začíná uvnitř kružnice a~pokračuje směrem ven, takže kružnici protne právě jednou. Bod $A$ je tedy určen jednoznačně, a~trojúhelník $ABC$ je rovněž jednoznačný (až na shodné zobrazení). + + \Image{angles-Ssa-proof-one-solution.pdf} + + Podívejme se, proč je podmínka $b > a$ podstatná. Pokud by bylo $a = b$, bod~$B$ by ležel přímo \textit{na} kružnici~$k$. Polopřímka z~$B$ by ji protínala v~bodě~$B$ samotném (což nedává trojúhelník) a~v~právě jednom dalším bodě -- ten je hledaným~$A$. Konstrukce tedy stále vede k~jednoznačnému (rovnoramennému) trojúhelníku, takže věta $Ssu$ formálně platí i~v~tomto hraničním případě. Pod $Ssu$ ji však samostatně nezařazujeme: pokud víme, že trojúhelník je rovnoramenný se $|BC|=|AC|$ a~známe jeden úhel $\beta$, tak ostatní dva úhly jsou už díky rovnoramennosti určeny (oba při základně mají velikost~$\beta$). Tatáž konfigurace je proto pokryta větami $sus$ či $sss$ a~$Ssu$ k~ní nepřidává žádnou novou informaci. + + \Image{angles-Ssa-proof-isosceles.pdf} + + Pokud by bylo $b < a$, bod $B$ by ležel \textit{vně} kružnice $k$. Polopřímka by ji mohla protnout ve dvou bodech -- vznikly by tak dva různé (neshodné) trojúhelníky -- nebo by ji minula úplně, což nevede k~žádnému trojúhelníku. + + \Image{angles-Ssa-proof-two-solutions.pdf} + + \Remark Důkaz je možné provést i~tak, že nejprve sestrojíme delší úsečku. To však vyžaduje znalost množiny bodů nad pevnou úsečkou majících pevný úhel -- k~tomu se dostaneme v~jiném materiálu. +} + +Na naší geometrické cestě se průběžně setkáme se všemi těmito tvrzeními. Prozatím si však ukažme nějaké konkrétní aplikace. Začneme tím nejzřejmějším tvrzením, které se ale také sluší a~patří dokázat: + +\Theorem{rovnoramenný trojúhelník}{ + Dokažte, že pokud pro trojúhelník~$ABC$ platí $|AB|=|AC|$, tak $|\angle ABC|=|\angle ACB|$. Dokažte také opačnou implikaci (tedy že z~rovnosti úhlů vyplývá rovnost délek). +}{ + \textit{Přímá implikace.} Předpokládejme $|AB| = |AC|$. Nechť $M$ je střed $BC$. Trojúhelníky $ABM$ a~$ACM$ jsou shodné podle $sss$ ($|AB|=|AC|$ z~předpokladu, $|BM|=|CM|$ ze~středu, společná strana $AM$), odkud $|\angle ABC| = |\angle ACB|$. + + \Image{angles-isosceles-triangle.pdf} + + \textit{Opačná implikace.} Předpokládejme $|\angle ABC| = |\angle ACB| = \beta$. Nechť $M$ je pata kolmice z~$A$ na $BC$. Podle věty $usu$ je $\triangle ABM \cong \triangle ACM$, neboť máme dva shodné úhly $|\angle ABM| = |\angle ACM|$ a~$|\angle AMB| = |\angle AMC| = 90^\circ$ a~společnou stranu~$AM$, tedy $|AB| = |AC|$. + + \Remark Jiný vtipný důkaz je založen na tom, že dokážeme $\triangle ABC \cong \triangle ACB$ (všimněte si různého pořadí vrcholů). V~případě, že známe stejné strany, použijeme větu $sus$ nebo dokonce $sss$ -- ta nám následně dá shodné úhly. V~případě, že známe stejné úhly, to zase bude věta $usu$, a~shodnost nám dá stejné strany. +} + +Vyzkoušejte si exaktně dokázat tyto jednoduché poznatky, to už bude jednodušší: + +\Exercise{}{ + Rovnostranný trojúhelník má tři shodné úhly rovny $60^\circ$ +}{ + V~$\triangle ABC$ s~$|AB| = |AC| = |BC|$ dává věta o~rovnoramenném trojúhelníku $|\angle ABC| = |\angle ACB|$ (z~$|AB|=|AC|$); analogicky z~$|AB|=|BC|$ je $|\angle BAC| = |\angle BCA|$. Všechny tři úhly jsou tedy shodné, a~ze součtu $180^\circ$ je každý $60^\circ$. + + \Image{angles-equilateral.pdf} +} + +\Exercise{}{ + Rovnoramenný pravoúhlý trojúhelník má tři úhly rovny $90^\circ$, $45^\circ$, $45^\circ$. +}{ + Nechť je pravý úhel při vrcholu~$A$, tedy $|\angle BAC| = 90^\circ$, a~odvěsny jsou shodné, $|AB| = |AC|$. Podle věty o~rovnoramenném trojúhelníku platí $|\angle ABC| = |\angle ACB|$. Označme tuto společnou velikost $\beta$. Ze součtu úhlů v~trojúhelníku + $$ + 90^\circ + \beta + \beta = 180^\circ, + $$ + odkud $2\beta = 90^\circ$, čili $\beta = 45^\circ$. + + \Image{angles-isosceles-right-triangle.pdf} +} + +Velmi šikovná věc použitelná v~nelehkých úlohách je následující úloha. Jeden možný důkaz je přes trigonometrii. + +\Problem{0}{}{ + Pravoúhlý trojúhelník má zbývající dva úhly rovny $60^\circ$ a~$30^\circ$ právě tehdy, když je jeho přepona dvakrát delší než kratší odvěsna. +}{ + Řekněme, že $BC$ je přepona a~je dvakrát delší než odvěsna $AB$. Trikem je uvážit bod $B'$ takový, že $A$ je střed úsečky $BB'$. +}{ + Nechť náš trojúhelník má pravý úhel při vrcholu~$A$. + + \textit{($\Rightarrow$)} Předpokládejme $|BC| = 2|AB|$. Nechť $B'$ je takový bod, že $A$ je středem úsečky $BB'$. Potom $|AB'| = |AB|$ a~$A$ leží mezi $B$, $B'$, takže $|\angle B'AC|$ je vedlejší k~$|\angle BAC| = 90^\circ$, čili také $90^\circ$. Platí $\triangle ABC \cong \triangle AB'C$ z~věty $sus$, neboť úhly při~$A$ jsou oba pravé, strana~$AC$ je společná, a~$|AB|=|AB'|$. + + \Image{angles-90-60-30.pdf} + + Protože $A$ je středem $BB'$, je $|BB'| = 2|AB|$, a~ze zadání $|BC| = 2|AB|$. Spolu + $$ + |BB'| = |BC| = |B'C|, + $$ + takže $\triangle BB'C$ je rovnostranný a~všechny jeho úhly jsou $60^\circ$. Speciálně $|\angle ABC| = |\angle B'BC| = 60^\circ$. Ze součtu úhlů v~$\triangle ABC$ dopočítáme $|\angle BCA| = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. + + \textit{($\Leftarrow$)} Není těžké rozmyslet si, že úvahy z~předchozího odstavce umíme snadno obrátit -- klíčovou shodnost $\triangle ABC \cong \triangle AB'C$ tentokrát dostaneme z~věty $usu$. + + \Remark Jiné řešení je uvážit, že podle Thaletovy věty je střed~$O$ kružnice opsané trojúhelníku $ABC$ zároveň středem přepony $BC$. K~tomuto řešení se podrobněji vrátíme, až se budeme bavit o~kružnicích. +} + +\sec Co si zapamatovat + +\secc Techniky + +\begitems \style N +\i Při rovnoběžkách hledáme shodné střídavé a~souhlasné úhly, případně úhly se součtem~$180^\circ$. +\i Je užitečné nahlížet na vnější úhel trojúhelníku jako na součet dvou úhlů. +\i Pro propojení světa délek a~úhlů se hodí shodnost dvou trojúhelníků. +\i Pomocné body (střed úsečky, zrcadlový obraz, prodloužení) často odhalí shodnost nebo speciální trojúhelník. +\enditems + +\secc Užitečná fakta + +\begitems \style N +\i Součet úhlů v~trojúhelníku je $180^\circ$, obecně v~konvexním $n$-úhelníku $(n-2) \cdot 180^\circ$. +\i Věty o~shodnosti trojúhelníků: $sss$, $sus$, $usu$, $Ssu$. +\i V~trojúhelníku jsou dvě strany shodné právě když jsou shodné jim přilehlé úhly. +\i Pravoúhlý trojúhelník má přeponu dvakrát delší než odvěsnu právě když má úhly $90^\circ, 60^\circ, 30^\circ$. +\enditems + +\DisplayExerciseSolutions + +\DisplayHints + +\DisplayProblemSolutions + +\bye diff --git a/data/handouts/angle-basics-1.en.tex b/data/handouts/angle-basics-1.en.tex new file mode 100644 index 00000000..323d7769 --- /dev/null +++ b/data/handouts/angle-basics-1.en.tex @@ -0,0 +1,304 @@ +\input _template + +\setlanguage{EN} + +\Title{Angle Chasing Basics} + +\MathcompsLink{angle-basics-1} + +\Author{Patrik Bak} + +\sec Introduction + +Geometry is the most visual part of mathematics -- we can draw it on paper, in software like \Link[https://www.geogebra.org/]{GeoGebra}, and we can enjoy seeing with our own eyes that things in it work. Why this is so, however, can be a mystery. To uncover these secrets, we need to build up a foundation of knowledge and visual intuition piece by piece. + +One of the most fundamental techniques is to understand the world of angles and to know how to use their properties successfully in various types of problems. Only a few interesting examples don't use angles. The theory around angles is not complicated, but it is important to build it up slowly and in detail. We will start with the simplest properties, which don't even require working with circles. Even these can already be used successfully in many problems. + +In this handout we will focus on angles, but we will deliberately leave out angles in circles, which deserve separate attention. For a moment, let's forget that circles exist. + +To close the introduction, a small note about convention. In this handout we will see many diagrams of triangles in which vertex~$A$ is drawn at the top. It turns out that this way of drawing is the international standard, especially in the world of the Mathematical Olympiad and similar competitions. In Czechia and Slovakia, however, schools still customarily draw~$C$ at the top. In some countries one can also see~$B$ at the top, e.g. in Ukraine. From the point of view of solving problems, it is always a good idea to put at the top the vertex that makes the diagram look as symmetric as possible -- it is then easier to spot further symmetric things in it. In this handout the problems will be posed so that the symmetry is more visible with~$A$ at the top \SlightSmile. + +\sec Basics of the World of Angles + +In this section we will derive the simplest properties related to angles. I firmly believe that the key to mastering geometry is understanding things from the ground up and in depth, so we will spend more time on these simple properties. + +Basic properties of angles, which we commonly use almost without thinking when solving problems: + +\Highlight{ + \begitems \style i + \i Vertical angles are equal: + \Image{angles-vertical.pdf} + \i Corresponding angles are equal: + \Image{angles-corresponding.pdf} + \i Alternate angles are equal: + \Image{angles-alternate.pdf} + \i Angles in a linear pair sum to $180^\circ$: + \Image{angles-supplementary.pdf} + \enditems +} + +These properties are very naturally related to one another; a small exercise to ponder: + +\Exercise{}{ + Convince yourself that: + \begitems \style a + \i the linear-pair property implies the vertical-angles property (and vice versa) + \i any two of the three properties about vertical, corresponding, and alternate angles imply the third + \enditems +}{ + \textbf{(a)} The angle $\alpha'$ in a linear pair with a given angle $\alpha$ has measure $180^\circ - \alpha$. The angle in a linear pair with $\alpha'$ in turn has measure $180^\circ - (180^\circ - \alpha) = \alpha$. But this last angle is also vertical to $\alpha$, so vertical angles are equal. + + \Image{angles-supplementary-derive-vertical.pdf} + + Conversely, assume the vertical-angles property and label the consecutive angles at the intersection of two lines $\alpha$, $\beta$, $\gamma$, $\delta$. By the equality of vertical angles, $\alpha = \gamma$ and $\beta = \delta$. The sum of all four is a full angle, i.e. $\alpha + \beta + \gamma + \delta = 360^\circ$. Substituting, we get $2\alpha + 2\beta = 360^\circ$, so $\alpha + \beta = 180^\circ$ -- which is exactly the linear-pair property. + + \Image{angles-vertical-derive-supplementary.pdf} + + \textbf{(b)} All three properties say that a certain pair of angles has the same measure. + + Look at the three angles $\alpha$, $\beta$, $\gamma$ in the figure and at what each property says about their relationships. For the pair $\alpha$, $\beta$ at the upper intersection, the vertical-angles property gives $\alpha = \beta$. For the pair $\alpha$, $\gamma$, the corresponding-angles property gives $\alpha = \gamma$. Finally, for the pair $\beta$, $\gamma$, the alternate-angles property gives $\beta = \gamma$. + + \Image{angles-corresponding-alternate-supplementary-connection.pdf} + + The three properties therefore assert three equalities among the same three angles, just via different pairs. As soon as any two of them hold, the third follows by transitivity: e.g. if (V) and (C) hold, then $\alpha = \beta$ and $\alpha = \gamma$, so $\beta = \gamma$, which is (A). The other pairs are analogous. +} + +The following statement is familiar to all of us, but can you prove it? + +\Theorem{}{ + The sum of the angles in a triangle is $180^\circ$. +}{ + Through vertex~$A$ draw a line parallel to $BC$. Since $AB$ is a transversal of the parallel lines, the angle $\beta$ at~$A$ and the angle $\beta$ at~$B$ are alternate, hence equal. Similarly, via the transversal $AC$, the angles $\gamma$ at~$A$ and at~$C$ are alternate. The angles $\beta$, $\alpha$, $\gamma$ lie next to each other along the line through~$A$, so $\alpha + \beta + \gamma = 180^\circ.$ + + \Image{angles-triangle.pdf} +} + +This statement can certainly be generalized. We know, for example, that the sum of the angles in a quadrilateral is $360^\circ$. What about a pentagon? And a 67-gon? The answer is the following statement: + +\Theorem{}{ + Let $n \ge 3$ be a natural number. The sum of the angles in a convex $n$-gon equals $(n-2) \cdot 180^\circ$. +}{ + \NamedProof{Proof 1 (mathematical induction).} Base case $n=3$: the sum of the angles of a triangle is $180^\circ = (3-2)\cdot 180^\circ$. Inductive step: let $A_1 A_2 \ldots A_{n+1}$ be a convex $(n+1)$-gon. The diagonal $A_1 A_n$ splits it into a triangle $A_1 A_n A_{n+1}$ and a convex $n$-gon $A_1 A_2 \ldots A_n$: the angles at vertices $A_2, \ldots, A_{n-1}$ belong entirely to the $n$-gon, the angle at $A_{n+1}$ belongs entirely to the triangle, and the angles at $A_1$, $A_n$ are split between the two figures so that their parts add up to the original interior angles of the $(n+1)$-gon. The angle sum of the $(n+1)$-gon is therefore + $$ + \underbrace{180^\circ}_{\text{triangle}} + \underbrace{(n-2) \cdot 180^\circ}_{n\text{-gon}} = (n-1) \cdot 180^\circ = \bigl((n+1) - 2\bigr) \cdot 180^\circ. + $$ + + \Image{angles-polygon.pdf} + + \NamedProof{Proof 2 (fan triangulation).} From vertex $A_1$ draw diagonals to every non-adjacent vertex $A_3, A_4, \ldots, A_{n-1}$. This produces $n-2$ triangles that cover the interior of the $n$-gon without overlap. At each vertex $A_k$, the angles of the adjacent triangles at $A_k$ assemble exactly into the interior angle of the $n$-gon at $A_k$, so the sum of the angles of all triangles equals the sum of the interior angles of the $n$-gon, i.e. $(n-2)\cdot 180^\circ$. + + \Image{angles-polygon-fan.pdf} + + \Remark Proof~2 is an \uv{unrolled} version of Proof~1: induction cuts off triangles one by one, and here we see them all at once. + + \NamedProof{Proof 3 (interior point).} Choose a point $O$ in the interior of the $n$-gon and connect it to every vertex. This produces $n$ triangles with a total angle sum of $n\cdot 180^\circ$. At each vertex $A_k$ of the $n$-gon, the triangle angles at $A_k$ together form the entire interior angle of the $n$-gon there; the triangle angles at~$O$ together form a full angle of $360^\circ$. So the sum of the interior angles of the $n$-gon is + $$ + n\cdot 180^\circ - 360^\circ = (n-2)\cdot 180^\circ. + $$ + + \Image{angles-polygon-interior.pdf} + + \Remark{non-convex polygons} The formula $(n-2)\cdot 180^\circ$ also holds for non-self-intersecting non-convex polygons, provided we measure all interior angles toward the inside -- so we count angles greater than $180^\circ$ as such. + + A fan triangulation from a single vertex may fail even from a convex vertex -- some diagonals can pass outside the polygon. It is still true, however, that every simple $n$-gon can be split into $n-2$ triangles (not necessarily as a fan), which rescues the proof. The reason is induction via a triangle formed by some vertex $V$ and its neighbors $U$, $W$ that lies entirely inside our $n$-gon. Cutting $V$ off along the diagonal $UW$ gives an $(n-1)$-gon; iterating yields $n-2$ triangles. One can prove that such a triangle always exists (in fact, two of them); see the \Link[https://en.wikipedia.org/wiki/Two_ears_theorem]{two ears theorem}. + + As for Proof~3: it works only if there is an interior point $O$ visible from every vertex (i.e., the segments $OA_i$ lie entirely inside). For some non-convex polygons no such $O$ exists (try drawing one) -- in that case Proof~3 cannot be salvaged directly and one must fall back on the triangulation from Proof~2. +} + +Now let's state two more simple but useful auxiliary claims: + +\Exercise{}{ + Prove that the sum of the angles in the figure equals $180^\circ$. + + \Image{angles-complementary-parallel-statement.pdf} +}{ + Let $\alpha$ denote the angle that the transversal $AD$ makes with the parallel lines on the side of $B$ (as in the figure). The corresponding angles $\alpha$ at~$A$ and $\alpha$ at~$D$ are equal. At vertex $A$ the angles $\alpha$ and $\beta$ form a linear pair, so + $$ + \beta + \alpha = 180^\circ. + $$ + + \Image{angles-complementary-parallel-solution.pdf} +} + +\Exercise{}{ + Prove that the angle marked with a question mark in the figure equals $\alpha+\beta$. + + \Image{angles-complementary-in-triangle.pdf} +}{ + From the angle sum in the triangle, $\angle ACB = 180^\circ - \alpha - \beta$. The exterior angle at $C$ is in a linear pair with $\angle ACB$: + $$ + 180^\circ - \angle ACB = \alpha + \beta. + $$ + + \Image{angles-complementary-in-triangle.pdf} +} + +The last two exercises may have looked very trivial. The reality, however, is that in practical problems they are very useful -- when we have a huge non-trivial diagram and have to chain together a large number of angle operations, it is genuinely advantageous to simplify even a small piece of that work. In the parallel-lines case we don't have to introduce an auxiliary corresponding angle, and in the triangle case we don't have to work with~$180^\circ$. + +\sec Basics of the World of Lengths + +So far we have only operated in the world of angles and have not dealt with lengths at all. Geometry, however, becomes interesting once we start connecting these worlds. Our foundation will be \textit{triangle congruence}. Let us recall the criteria. + +\Highlight{ + Triangle congruence theorems: + \begitems \style i + \i $SSS$: two triangles are congruent if they agree on all three sides. + \i $SAS$: two triangles are congruent if they agree on two sides and the angle between them. + \i $ASA$: two triangles are congruent if they agree on one side and any two angles. + \i $SsA$: two triangles are congruent if they agree on two sides and the angle opposite the longer of these sides. + \enditems +} + +Let us emphasize that in $SAS$ it really is important that the equal angle be the one between the two sides -- without this assumption, congruence may fail; see the figure. There we have two triangles $ABC$ and $A'B'C'$ for which $BC = B'C'$, $AC = A'C'$, and $\beta = \beta'$, yet they are clearly not congruent. + +\Image{angles-sas-warning.pdf} + +The $SsA$ theorem would be our rescue -- but here we cannot apply it, because the side opposite the angle $\beta$, namely $AC$, is not the longest, since it is clearly shorter than $BC$ \SmileWithTear. + +Before we go further, let us understand what these theorems mean. In my view, a good way to look at congruence is the following: we want to construct a triangle when given some three of its elements -- will all the constructible triangles be congruent? + +\Example{}{ + Convince yourself of the validity of the $SSS$ congruence theorem. +}{ + Imagine we are given three segments of lengths $a$, $b$, $c$ from which a triangle can be built. Start, say, with the segment~$BC$ of length~$a$, and then construct two circles: (a) the circle centered at~$B$ with radius $c$; (b) the circle centered at~$C$ with radius $b$. They either fail to meet (when e.g. $b+c < a$), or they meet at a single point on $BC$ (when $b+c=a$), or they meet at two points $A$ and $A'$. + + \Image{angles-sss-proof.pdf} + + Finally, observe that the triangles $ABC$ and $A'BC$ are clearly mirror images of each other across $BC$. Likewise, had we started from a different segment, we would only produce a translation/rotation of this configuration. + + \Remark Note that along the way we essentially proved the triangle inequality: the sum of two sides of a triangle is greater than the third -- our triangle is constructible and non-degenerate~\fnote{has no angle equal to $180^\circ$} if and only if $b+c>a$. +} + +We can justify the other criteria similarly: + +\Exercise{}{ + Convince yourself of the validity of the $SAS$ congruence theorem. +}{ + Suppose the sides $b = AC$, $c = AB$ and the angle $\angle BAC = \alpha$ between them are given. Construct vertex~$A$ and the segment $AB$ of length~$c$ -- this is determined up to congruence. Point $C$ must lie on a ray from~$A$ making angle~$\alpha$ with $AB$ and at distance~$b$ from~$A$. There are two rays from~$A$ making angle $\alpha$ with $AB$ (one on each side of $AB$): on each there is exactly one point at distance $b$ from~$A$, giving points $C$ and $C'$. The triangles $ABC$ and $ABC'$ are mirror images across $AB$, hence congruent. + + \Image{angles-sas-proof.pdf} +} + +\Exercise{}{ + Convince yourself of the validity of the $ASA$ congruence theorem. +}{ + First, note that it doesn't matter which two angles are equal -- if two triangles agree on two angles, then the third is determined uniquely, since all three angles sum to $180^\circ$. + + For the purposes of our proof, suppose the equal angles are exactly the ones adjacent to the equal side. Let them be the angle $\beta$ at vertex $B$, $\gamma$ at vertex $C$, and the side $a = BC$ between them. Construct the segment $BC$ of length~$a$ -- this is determined up to congruence. Point $A$ must lie on the ray from~$B$ making angle $\beta$ with $BC$ and on the ray from~$C$ making angle $\gamma$ with $CB$. The sum of the interior angles at $B$ and $C$ in a triangle is less than $180^\circ$, i.e. $\beta + \gamma < 180^\circ$, so the two rays are not parallel -- they meet at a single point~$A$. The triangle $ABC$ is therefore determined uniquely (up to congruences). + + \Image{angles-asa-proof.pdf} +} + +The less well-known $SsA$ theorem is a touch harder, so we will provide hints \SlightSmile. + +\Problem{0}{$SsA$ congruence}{ + Convince yourself of the validity of the $SsA$ congruence theorem. In addition, figure out where the construction would break down if the angle were not opposite the longer of the two sides -- what would happen if both sides were equal? +}{ + Begin the construction with the shorter segment. How does the construction proceed once it is built? +}{ + After constructing the shorter segment, draw a ray at our given equal angle. One last step of the construction remains. Think about when we get no triangle, when one, and when two satisfying triangles. +}{ + Suppose the sides $a = BC$, $b = AC$ and the angle $\angle ABC = \beta$ opposite the longer side $b$ are given. We begin by constructing the shorter segment $BC$ of length~$a$. Point $A$ must lie on the ray from~$B$ making angle $\beta$ with $BC$ and at distance $b$ from~$C$, i.e. on the circle $k$ centered at~$C$ with radius $b$. + + The distance from~$B$ to the center of the circle is $BC = a$. Since $a < b$, the point~$B$ lies \textit{inside} the circle $k$. The ray from~$B$ thus starts inside the circle and continues outward, so it crosses the circle exactly once. Point $A$ is therefore determined uniquely, and the triangle $ABC$ is also unique (up to congruence). + + \Image{angles-Ssa-proof-one-solution.pdf} + + Let's see why the condition $b > a$ is essential. If $a = b$, the point~$B$ would lie directly \textit{on} the circle~$k$. The ray from~$B$ would meet it at $B$ itself (which does not give a triangle) and at exactly one further point -- which is the desired~$A$. The construction therefore still leads to a unique (isosceles) triangle, so the $SsA$ theorem formally holds even in this borderline case. We do not, however, list it under $SsA$ as a separate case: if we know that the triangle is isosceles with $BC=AC$ and we know one angle $\beta$, the other two angles are already determined by the isosceles property (both base angles equal~$\beta$). The same configuration is therefore covered by $SAS$ or $SSS$, and $SsA$ adds no new information. + + \Image{angles-Ssa-proof-isosceles.pdf} + + If $b < a$, the point $B$ would lie \textit{outside} the circle $k$. The ray could meet it at two points -- giving two different (non-congruent) triangles -- or miss it entirely, which gives no triangle at all. + + \Image{angles-Ssa-proof-two-solutions.pdf} + + \Remark The proof can also be done by first constructing the longer segment. That, however, requires knowing the locus of points subtending a fixed angle from a fixed segment -- we will get to this in another handout. +} + +Throughout our geometric journey we will encounter all of these statements. For now, however, let us show some concrete applications. We start with the most obvious statement, which still deserves to be proved: + +\Theorem{isosceles triangle}{ + Prove that if a triangle~$ABC$ satisfies $AB=AC$, then $\angle ABC=\angle ACB$. Prove also the converse implication (i.e., that equality of angles implies equality of lengths). +}{ + \textit{Forward implication.} Assume $AB = AC$. Let $M$ be the midpoint of $BC$. The triangles $ABM$ and $ACM$ are congruent by $SSS$ ($AB=AC$ by assumption, $BM=CM$ by definition of the midpoint, common side $AM$), from which $\angle ABC = \angle ACB$. + + \Image{angles-isosceles-triangle.pdf} + + \textit{Converse implication.} Assume $\angle ABC = \angle ACB = \beta$. Let $M$ be the foot of the perpendicular from~$A$ to $BC$. By $ASA$, $\triangle ABM \cong \triangle ACM$, since we have two equal angles $\angle ABM = \angle ACM$ and $\angle AMB = \angle AMC = 90^\circ$ and the common side~$AM$, so $AB = AC$. + + \Remark Another cute proof is based on showing that $\triangle ABC \cong \triangle ACB$ (note the different order of vertices). In the case where we know the equal sides, we use $SAS$ or even $SSS$ -- which then gives equal angles. In the case where we know the equal angles, it will instead be $ASA$, and the congruence gives equal sides. +} + +Try proving these simple facts rigorously -- it will be easier this time: + +\Exercise{}{ + An equilateral triangle has three equal angles of $60^\circ$. +}{ + In $\triangle ABC$ with $AB = AC = BC$, the isosceles triangle theorem gives $\angle ABC = \angle ACB$ (from $AB=AC$); analogously, from $AB=BC$ we get $\angle BAC = \angle BCA$. All three angles are therefore equal, and from the angle sum of $180^\circ$ each is $60^\circ$. + + \Image{angles-equilateral.pdf} +} + +\Exercise{}{ + An isosceles right triangle has angles $90^\circ$, $45^\circ$, $45^\circ$. +}{ + Suppose the right angle is at vertex~$A$, i.e. $\angle BAC = 90^\circ$, and the legs are equal, $AB = AC$. By the isosceles triangle theorem, $\angle ABC = \angle ACB$. Denote this common measure $\beta$. From the angle sum in the triangle, + $$ + 90^\circ + \beta + \beta = 180^\circ, + $$ + so $2\beta = 90^\circ$, i.e. $\beta = 45^\circ$. + + \Image{angles-isosceles-right-triangle.pdf} +} + +A very handy fact useful in non-trivial problems is the following. One possible proof goes via trigonometry. + +\Problem{0}{}{ + A right triangle has its other two angles equal to $60^\circ$ and $30^\circ$ if and only if its hypotenuse is twice as long as the shorter leg. +}{ + Say $BC$ is the hypotenuse and is twice as long as the leg $AB$. The trick is to introduce a point $B'$ such that $A$ is the midpoint of segment $BB'$. +}{ + Let our triangle have its right angle at vertex~$A$. + + \textit{($\Rightarrow$)} Assume $BC = 2\cdot AB$. Let $B'$ be the point such that $A$ is the midpoint of segment $BB'$. Then $AB' = AB$ and $A$ lies between $B$ and $B'$, so $\angle B'AC$ is in a linear pair with $\angle BAC = 90^\circ$, hence also equals $90^\circ$. We have $\triangle ABC \cong \triangle AB'C$ by $SAS$, since the angles at~$A$ are both right, the side~$AC$ is common, and $AB=AB'$. + + \Image{angles-90-60-30.pdf} + + Since $A$ is the midpoint of $BB'$, we have $BB' = 2\cdot AB$, and from the assumption $BC = 2\cdot AB$. Together, + $$ + BB' = BC = B'C, + $$ + so $\triangle BB'C$ is equilateral and all its angles are $60^\circ$. In particular, $\angle ABC = \angle B'BC = 60^\circ$. From the angle sum in $\triangle ABC$ we compute $\angle BCA = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. + + \textit{($\Leftarrow$)} It is not hard to see that the reasoning of the previous paragraph can easily be reversed -- this time the key congruence $\triangle ABC \cong \triangle AB'C$ comes from $ASA$. + + \Remark Another solution is to note that by Thales' theorem, the circumcenter~$O$ of triangle $ABC$ is also the midpoint of the hypotenuse $BC$. We will return to this solution in more detail when we discuss circles. +} + +\sec What to Remember + +\secc Techniques + +\begitems \style N +\i With parallel lines, look for equal alternate and corresponding angles, or angles summing to~$180^\circ$. +\i It is useful to view the exterior angle of a triangle as the sum of the two non-adjacent interior angles. +\i To bridge the worlds of lengths and angles, congruence of two triangles is the right tool. +\i Auxiliary points (midpoint of a segment, mirror image, extension) often reveal a congruence or a special triangle. +\enditems + +\secc Useful Facts + +\begitems \style N +\i The angle sum in a triangle is $180^\circ$, in general in a convex $n$-gon $(n-2) \cdot 180^\circ$. +\i Triangle congruence theorems: $SSS$, $SAS$, $ASA$, $SsA$. +\i In a triangle, two sides are equal if and only if the angles opposite them are equal. +\i A right triangle has its hypotenuse twice as long as a leg if and only if its angles are $90^\circ, 60^\circ, 30^\circ$. +\enditems + +\DisplayExerciseSolutions + +\DisplayHints + +\DisplayProblemSolutions + +\bye diff --git a/data/handouts/angle-basics-1.sk.tex b/data/handouts/angle-basics-1.sk.tex new file mode 100644 index 00000000..2b919832 --- /dev/null +++ b/data/handouts/angle-basics-1.sk.tex @@ -0,0 +1,304 @@ +\input _template + +\setlanguage{SK} + +\Title{Základy počítania uhlov} + +\MathcompsLink{zaklady-pocitania-uhlov-1} + +\Author{Patrik Bak} + +\sec Úvod + +Geometria je najvizuálnejšia časť matematiky -- vieme si ju kresliť na papier, v~softwari ako \Link[https://www.geogebra.org/]{GeoGebra} a~môžeme sa tešiť, keď na vlastné oči vidíme, že veci v~nej fungujú. Prečo tomu tak je však vie byť záhadou. Aby sme tieto tajomstvá pochopili, potrebujeme kúsok po kúsky budovať základňu znalostí a~vizuálnu intuíciu. + +Jedna z~najzákladnejších techník je rozumieť svetu uhlov a~vedieť ich vlastnosti s~úspechom používať v~rôznych typoch úloh. Len málo zaujímavých príkladov uhly nepoužíva. Teória okolo uhlov pritom nie je zložitá, je však dôležité budovať ju pomaly a detailne. Začneme s~najjednoduchšími vlastnosťami, ktoré ani nevyžadujú prácu s kružnicami. Už tieto sa však dajú s~úspechom využiť vo veľa úlohách. + +V~tomto materiáli sa budeme venovať uhlom, ale zámerne vynecháme uhly na kružniciach, ktoré si zaslúžia separátnu pozornosť. Na chvíľu zabudnime, že kružnice existujú. + +Na záver úvodu jedna menšia poznámka o konvencii. V~tomto materiáli uvidíme veľa obrázkov trojuholníkov, v~ktorých je bod~$A$ nakreslený hore. Ukazuje sa, že tento spôsob kreslenia je medzinárodný štandard, najmä na poli matematickej olympiády a podobných súťaží sa to tak zvykne robiť. V~Česku a na~Slovensku je však v~školách stále zvykom kresliť~$C$ hore. V~niektorých krajinách možno vidieť aj~$B$ hore, napr. na Ukrajine. Z~hľadiska riešenia úloh je vždy dobré nakresliť si hore ten bod, pri ktorom obrázok vyzerá čo najsymetrickejšie, ľahšie v~ňom vidieť ďalšie symetrické veci. V~tomto materiáli budú úlohy zadávané tak, že tá symetria bude vidieť viac pri~$A$ hore \SlightSmile. + +\sec Základy sveta uhlov + +V~tejto sekcii si odvodíme tie najjednoduchšie vlastnosti súvisiace s uhlami. Pevne verím, že kľúčom k~zvládnutiu geometrie je porozumenie vecí od základov a do hĺbky, preto sa týmto jednoduchým vlastnostiam budeme venovať viac. + +Základné vlastnosti uhlov, ktoré bežne používame skoro bez rozmýšľania pri riešení úloh: + +\Highlight{ + \begitems \style i + \i Vrcholové uhly sú zhodné: + \Image{angles-vertical.pdf} + \i Súhlasné uhly sú zhodné: + \Image{angles-corresponding.pdf} + \i Striedavé uhly sú zhodné: + \Image{angles-alternate.pdf} + \i Vedľajšie uhly majú súčet $180^\circ$: + \Image{angles-supplementary.pdf} + \enditems +} + +Tieto vlastnosti spolu veľmi prirodzene súvisia, menšie cvičenie na rozmyslenie: + +\Exercise{}{ + Uvedomte si, že: + \begitems \style a + \i z~vlastnosti o~vedľajších uhloch vyplýva vlastnosť vrcholových uhlov (a~naopak) + \i z~ktorejkoľvek dvojice vlastností o~vrcholových, súhlasných, striedavých uhloch vyplýva tá tretia vlastnosť + \enditems +}{ + \textbf{(a)} Vedľajší uhol $\alpha'$ k~danému uhlu $\alpha$ má veľkosť $180^\circ - \alpha$. Uhol vedľajší k~$\alpha'$ má zasa veľkosť $180^\circ - (180^\circ - \alpha) = \alpha$. Ten je však zároveň vrcholový uhol k~$\alpha$, teda sú vrcholové uhly zhodné. + + \Image{angles-supplementary-derive-vertical.pdf} + + Naopak, predpokladajme vlastnosť o~vrcholových uhloch a~označme po sebe idúce uhly v~priesečníku dvoch priamok $\alpha$, $\beta$, $\gamma$, $\delta$. Zo zhodnosti vrcholových uhlov $\alpha = \gamma$ a~$\beta = \delta$. Súčet všetkých štyroch je plný uhol, teda $\alpha + \beta + \gamma + \delta = 360^\circ$. Po dosadení dostávame $2\alpha + 2\beta = 360^\circ$, čiže $\alpha + \beta = 180^\circ$ -- to je práve vlastnosť vedľajších uhlov. + + \Image{angles-vertical-derive-supplementary.pdf} + + \textbf{(b)} Všetky tri vlastnosti hovoria, že istá dvojica uhlov má rovnakú veľkosť. + + Pozrime sa na trojicu uhlov $\alpha$, $\beta$, $\gamma$ z~obrázka a~na to, čo o~ich vzťahoch tvrdia jednotlivé vlastnosti. Pre dvojicu $\alpha$, $\beta$ pri hornom priesečníku tvrdí vlastnosť vrcholových uhlov $\alpha = \beta$. Pre dvojicu $\alpha$, $\gamma$ tvrdí vlastnosť súhlasných uhlov $\alpha = \gamma$. Konečne pre dvojicu $\beta$, $\gamma$ tvrdí vlastnosť striedavých uhlov $\beta = \gamma$. + + \Image{angles-corresponding-alternate-supplementary-connection.pdf} + + Tri vlastnosti teda tvrdia tri rovnosti medzi tými istými troma uhlami, len cez rôzne dvojice. Akonáhle platia ľubovoľné dve, tretia plynie tranzitivitou: napr. ak vieme (V) a~(S), tak $\alpha = \beta$ a~$\alpha = \gamma$, takže $\beta = \gamma$, čiže (St). Ostatné dvojice analogicky. +} + +Nasledujúce tvrdenie všetci poznáme, viete ho však dokázať? + +\Theorem{}{ + Súčet uhlov v~trojuholníku je $180^\circ$. +}{ + Veďme cez vrchol~$A$ priamku rovnobežnú s~$BC$. Pretože $AB$ je priečka rovnobežiek, uhol $\beta$ pri~$A$ a~uhol $\beta$ pri~$B$ sú striedavé, teda rovnaké. Analogicky cez priečku $AC$ sú uhly $\gamma$ pri~$A$ a~pri~$C$ striedavé. Uhly $\beta$, $\alpha$, $\gamma$ ležia vedľa seba pozdĺž priamky cez~$A$, teda $\alpha + \beta + \gamma = 180^\circ.$ + + \Image{angles-triangle.pdf} +} + +Toto tvrdenie sa rozhodne dá zovšeobecniť. Vieme napríklad, že súčet uhlov v~štvoruholníku je $360^\circ$. Ako je to v 5-uholníku? A ako v~67-uholníku? Odpoveďou je nasledujúce tvrdenie: + +\Theorem{}{ + Nech $n \ge 3$ je prirodzené číslo. Súčet uhlov v~konvexnom $n$-uholníku je rovný $(n-2) \cdot 180^\circ$. +}{ + \NamedProof{Dôkaz 1 (matematická indukcia).} Základ $n=3$: súčet uhlov trojuholníka je $180^\circ = (3-2)\cdot 180^\circ$. Indukčný krok: nech $A_1 A_2 \ldots A_{n+1}$ je konvexný $(n+1)$-uholník. Uhlopriečkou $A_1 A_n$ ho rozdelíme na trojuholník $A_1 A_n A_{n+1}$ a~konvexný $n$-uholník $A_1 A_2 \ldots A_n$: uhly pri vrcholoch $A_2, \ldots, A_{n-1}$ patria celé $n$-uholníku, uhol pri $A_{n+1}$ patrí celý trojuholníku a~uhly pri $A_1$, $A_n$ sa rozdelia medzi oba útvary tak, aby ich časti dali pôvodné vnútorné uhly $(n+1)$-uholníka. Súčet uhlov $(n+1)$-uholníka je teda + $$ + \underbrace{180^\circ}_{\text{trojuholník}} + \underbrace{(n-2) \cdot 180^\circ}_{n\text{-uholník}} = (n-1) \cdot 180^\circ = \bigl((n+1) - 2\bigr) \cdot 180^\circ. + $$ + + \Image{angles-polygon.pdf} + + \NamedProof{Dôkaz 2 (vejárová triangulácia).} Veďme z~vrcholu $A_1$ uhlopriečky do každého nepriliehajúceho vrcholu $A_3, A_4, \ldots, A_{n-1}$. Vznikne $n-2$ trojuholníkov pokrývajúcich vnútro $n$-uholníka bez prekrytia. Pri každom vrchole $A_k$ sa uhly susedných trojuholníkov pri $A_k$ presne poskladajú na vnútorný uhol $n$-uholníka pri $A_k$, takže súčet uhlov všetkých trojuholníkov sa rovná súčtu vnútorných uhlov $n$-uholníka, čiže $(n-2)\cdot 180^\circ$. + + \Image{angles-polygon-fan.pdf} + + \Remark Dôkaz~2 je \uv{rozbalený} dôkaz~1: indukcia odrezáva trojuholníky jeden po druhom, tu ich vidíme všetky naraz. + + \NamedProof{Dôkaz 3 (vnútorný bod).} Zvoľme bod $O$ vo vnútri $n$-uholníka a~spojme ho s~každým vrcholom. Vznikne $n$ trojuholníkov s~celkovým súčtom uhlov $n\cdot 180^\circ$. Uhly trojuholníkov pri každom vrchole $A_k$ spolu tvoria vnútorný uhol $n$-uholníka pri~$A_k$; uhly trojuholníkov pri~$O$ spolu tvoria plný uhol $360^\circ$. Teda súčet vnútorných uhlov $n$-uholníka je + $$ + n\cdot 180^\circ - 360^\circ = (n-2)\cdot 180^\circ. + $$ + + \Image{angles-polygon-interior.pdf} + + \Remark{nekonvexné mnohouholníky} Vzorec $(n-2)\cdot 180^\circ$ platí aj pre navzájom sa nepretínajúce nekonvexné mnohouholníky, ak všetky vnútorné uhly meriame smerom dovnútra -- teda aj uhly väčšie než $180^\circ$ rátame ako také. + + Vejárová triangulácia z~jedného vrcholu nemusí fungovať ani z~konvexného vrcholu -- niektoré uhlopriečky môžu prechádzať mimo mnohouholníka. Platí však, že každý jednoduchý $n$-uholník sa dá rozdeliť na $n-2$ trojuholníkov (nie nutne vejárovo), čo dôkaz zachráni. Dôvod je indukcia cez trojuholník tvorený nejakým vrcholom $V$ a~jeho susedmi $U$, $W$, ktorý leží celý vo vnútri nášho $n$-uholníka. Odrezaním $V$ uhlopriečkou $UW$ dostaneme $(n-1)$-uholník; opakovaním teda $n-2$ trojuholníkov. Dá sa dokázať, že takýto trojuholník vždy existuje (dokonca dva), viď \Link[https://en.wikipedia.org/wiki/Two_ears_theorem]{veta o~dvoch ušiach}. + + K~dôkazu~3: funguje len ak existuje vnútorný bod $O$ viditeľný zo všetkých vrcholov (t.~j. úsečky $OA_i$ ležia celé vo vnútri). Pre niektoré nekonvexné mnohouholníky taký $O$ nemusí existovať (skúste taký nakresliť) -- v~takom prípade sa dôkaz~3 nedá priamo zachrániť a~treba siahnuť po triangulácii z~dôkazu~2. +} + +Teraz uveďme ešte dve jednoduché ale užitočné pomocné tvrdenia: + +\Exercise{}{ + Dokážte, že súčet uhlov na obrázku je rovný $180^\circ$. + + \Image{angles-complementary-parallel-statement.pdf} +}{ + Označme $\alpha$ uhol zvieraný priečkou $AD$ s~rovnobežkami na strane $B$ (ako na obrázku). Súhlasné uhly $\alpha$ pri~$A$ a~$\alpha$ pri~$D$ sú zhodné. Pri vrchole $A$ sú $\alpha$ a~$\beta$ vedľajšie, teda + $$ + \beta + \alpha = 180^\circ. + $$ + + \Image{angles-complementary-parallel-solution.pdf} +} + +\Exercise{}{ + Dokážte, že uhol s~otázníkom na obrázku je rovný $\alpha+\beta$. + + \Image{angles-complementary-in-triangle.pdf} +}{ + Zo súčtu uhlov v~trojuholníku $|\angle ACB| = 180^\circ - \alpha - \beta$. Vonkajší uhol pri $C$ je vedľajší k~$|\angle ACB|$: + $$ + 180^\circ - |\angle ACB| = \alpha + \beta. + $$ + + \Image{angles-complementary-in-triangle.pdf} +} + +Posledné dve cvičenia možno vyzerali veľmi triviálne. Realita však je, že v~praktických úlohách sú veľmi užitočné -- keď máme obrovský netriviálny obrázok a~máme previesť za sebou veľký počet uhlových operácií, tak je naozaj veľmi výhodné si čo i len kúsok tej práce zjednodušiť -- v~prípade rovnobežiek nemusíme uvažovať pomocný súhlasný uhol a~v~prípade trojuholníka nemusíme pracovať s~$180^\circ$. + +\sec Základy sveta dĺžok + +Zatiaľ sme fungovali iba vo svete uhlov a~vôbec sme neriešili dĺžky. Geometria však začne byť zaujímavá, keď tieto svety začneme prepájať. Základom pre nás bude {\em zhodnosť trojuholníkov}. Pripomeňme si kritériá. + +\Highlight{ + Vety o~zhodnosti trojuholníkov: + \begitems \style i + \i Veta $sss$: dva trojuholníky sú zhodné, ak sa zhodujú vo všetkých troch stranách. + \i Veta $sus$: dva trojuholníky sú zhodné, ak sa zhodujú v~dvoch stranách a~uhle, ktorý tieto strany zvierajú. + \i Veta $usu$: dva trojuholníky sú zhodné, ak sa zhodujú v~jednej strane a~nejakých dvoch uhloch. + \i Veta $Ssu$: dva trojuholníky sú zhodné, ak sa zhodujú v~dvoch stranách a~uhle oproti dlhšej z~týchto strán. + \enditems +} + +Zdôraznime, že pri vete $sus$ je naozaj dôležité, že zhodný uhol je ten zvieraný dvoma stranami -- bez tohto predpokladu zhodnosť nemusí fungovať, viď obrázok. Na ňom máme dva trojuholníky $ABC$ a $A'B'C'$, pre ktoré platí $|BC|=|B'C'|$, $|AC|=|A'C'|$, a $\beta=\beta'$, avšak zjavne nie sú zhodné. + +\Image{angles-sas-warning.pdf} + +Veta $Ssu$ by bola naša záchrana -- tu ju však aplikovať nevieme, lebo strana, oproti ktorej je uhol $\beta$, teda $|AC|$, nie je najdlhšia, keďže je zjavne kratšia než $|BC|$ \SmileWithTear. + +Než pôjdeme ďalej, uvedomme si, čo tieto vety znamenajú. Za mňa dobrý pohľad na zhodnosť je takýto: chceme zostrojiť trojuholník, keď máme dané nejaké tri jeho elementy -- budú všetky zostrojiteľné trojuholníky zhodné? + +\Example{}{ + Presvedčte sa o~platnosti vety $sss$ o~zhodnosti. +}{ + Predstavme si, že máme dané tri úsečky dĺžok $a$, $b$, $c$, z~ktorých vieme zostrojiť trojuholník. Začneme napr. úsečkou~$BC$ dĺžky~$a$, a~následne zostrojíme dve kružnice: (a) kružnicu so stredom v~$B$ a~polomerom $c$; (b) kružnicu so stredom v~$C$ a~polomerom $b$. Tie sa buď nepretnú (keď napr. $b+c < a$), alebo sa pretnú v~jedinom bode na $BC$ (keď $b+c=a$), alebo sa pretnú v~dvoch bodoch $A$ a~$A'$. + + \Image{angles-sss-proof.pdf} + + Na konci si uvedomme, že trojuholníky $ABC$ a~$A'BC$ sú však zrejme vzájomné zrkadlové obrazy podľa $BC$. Taktiež ak by sme začali inou úsečkou, tak vytvoríme iba posunutie/otočenie tejto konfigurácie. + + \Remark Uvedomme si, že sme vlastne po ceste dokázali~trojuholníkovú nerovnosť: súčet dvoch strán trojuholníka je väčší než tretia -- náš trojuholník je zostrojiteľný a~nedegenerovaný~\fnote{nemá uhol rovný $180^\circ$} práve vtedy, keď $b+c>a$. +} + +Podobne vieme zdôvodniť ďalšie kritériá: + +\Exercise{}{ + Presvedčte sa o~platnosti vety $sus$ o~zhodnosti. +}{ + Nech sú dané strany $b = |AC|$, $c = |AB|$ a~uhol $|\angle BAC| = \alpha$ medzi nimi. Zostrojme vrchol~$A$ a~úsečku $AB$ dĺžky~$c$ -- tá je daná až na zhodné zobrazenie. Bod $C$ musí ležať na polopriamke z~$A$ zvierajúcej s~$AB$ uhol~$\alpha$ a~zároveň vo vzdialenosti~$b$ od~$A$. Polopriamky zvierajúce s~$AB$ uhol $\alpha$ sú dve (po jednej na každej strane $AB$): na každej leží práve jeden bod vo vzdialenosti $b$ od~$A$, teda dostaneme body $C$ a~$C'$. Trojuholníky $ABC$ a~$ABC'$ sú zrkadlové obrazy podľa $AB$, teda zhodné. + + \Image{angles-sas-proof.pdf} +} + +\Exercise{}{ + Presvedčte sa o~platnosti vety $usu$ o~zhodnosti. +}{ + V~prvom rade si uvedomme, že nezáleží na tom, ktoré dva uhly sú zhodné -- ak sa dva trojuholníky zhodujú v~dvoch uhloch, tak tretí je určený jednoznačne, keďže všetky tri majú súčet uhlov $180^\circ$. + + Pre účely nášho dôkazu uvážme teda, že sú zhodné práve uhly prihliehajúce zhodnej strane. Nech je to uhol $\beta$ pri vrchole $B$, $\gamma$ pri vrchole $C$ a~strana $a = |BC|$ medzi nimi. Zostrojme úsečku $BC$ dĺžky~$a$ -- tá je daná až na zhodné zobrazenie. Bod $A$ musí ležať na polopriamke z~$B$ zvierajúcej s~$BC$ uhol $\beta$ a~zároveň na polopriamke z~$C$ zvierajúcej s~$CB$ uhol $\gamma$. Súčet vnútorných uhlov pri $B$ a~$C$ v~trojuholníku je menší než $180^\circ$, čiže $\beta + \gamma < 180^\circ$, a~preto nie sú obe polopriamky rovnobežné -- pretnú sa v~jedinom bode~$A$. Trojuholník $ABC$ je teda určený jednoznačne (až na zhodné zobrazenia). + + \Image{angles-asa-proof.pdf} +} + +Menej známe tvrdenie $Ssu$ je už máličko ťažšie, preto k~nemu budú aj návody \SlightSmile. + +\Problem{0}{zhodnosť $Ssu$}{ + Presvedčte sa o~platnosti vety $Ssu$ o~zhodnosti. Navyše, uvedomte si, kde by sa konštrukcia pokazila, keby uhol nebol voči väčšej zo strán -- čo by sa stalo, keby boli obe strany rovnako dlhé? +}{ + Začnite konštrukciu konštrukciou kratšej úsečky. Ako pokračuje konštrukcia po jej zostrojení? +}{ + Po zostrojení kratšej úsečky zostrojíme polpriamku pod našim daným zhodným uhlom. Zostáva posledný krok konštrukcie. Rozmyslite, kedy dostaneme žiaden, kedy jeden a~kedy dva vyhovujúce trojuholníky. +}{ + Nech sú dané strany $a = |BC|$, $b = |AC|$ a~uhol $|\angle ABC| = \beta$ oproti dlhšej strane $b$. Začneme zostrojením kratšej úsečky $BC$ dĺžky~$a$. Bod $A$ musí ležať na polopriamke z~$B$ zvierajúcej s~$BC$ uhol $\beta$ a~zároveň vo vzdialenosti $b$ od~$C$, čiže na kružnici $k$ so stredom~$C$ a~polomerom $b$. + + Vzdialenosť bodu~$B$ od stredu kružnice je $|BC| = a$. Pretože $a < b$, bod~$B$ leží \textit{vnútri} kružnice $k$. Polopriamka z~bodu~$B$ tak začína vnútri kružnice a~pokračuje smerom von, takže kružnicu pretne práve raz. Bod $A$ je teda určený jednoznačne, a~trojuholník $ABC$ je takisto jednoznačný (až na zhodné zobrazenie). + + \Image{angles-Ssa-proof-one-solution.pdf} + + Pozrime sa, prečo je podmienka $b > a$ podstatná. Ak by bolo $a = b$, bod~$B$ by ležal priamo \textit{na} kružnici~$k$. Polopriamka z~$B$ by ju pretínala v~bode~$B$ samotnom (čo nedáva trojuholník) a~v~práve jednom ďalšom bode -- ten je hľadaným~$A$. Konštrukcia teda stále vedie k~jednoznačnému (rovnoramennému) trojuholníku, takže veta $Ssu$ formálne platí aj v~tomto hraničnom prípade. Pod $Ssu$ ho však samostatne nezaraďujeme: ak vieme, že trojuholník je rovnoramenný so $|BC|=|AC|$ a~poznáme jeden uhol $\beta$, tak ostatné dva uhly sú už vďaka rovnoramennosti určené (oba pri základni majú veľkosť~$\beta$). Tá istá konfigurácia je preto pokrytá vetami $sus$ či $sss$ a~$Ssu$ k~nej nepridáva žiadnu novú informáciu. + + \Image{angles-Ssa-proof-isosceles.pdf} + + Ak by bolo $b < a$, bod $B$ by ležal \textit{vonku} kružnice $k$. Polopriamka by ju mohla pretnúť v~dvoch bodoch -- vznikli by tak dva rôzne (nezhodné) trojuholníky -- alebo by ju minula celkom, čo nevedie k~žiadnemu trojuholníku. + + \Image{angles-Ssa-proof-two-solutions.pdf} + + \Remark Dôkaz je možné spraviť aj tak, že najprv zostrojíme dlhšiu úsečku. To však vyžaduje znalosť množiny bodov nad pevnou úsečkou majúcich pevný uhol -- k~tomuto sa dostaneme v~inom materiáli. +} + +V našej geometrickej ceste sa priebežne stretneme so všetkými týmito tvrdeniami. Nateraz si však ukážme nejaké konkrétne aplikácie. Začneme tým najzrejmejším tvrdením, ktoré sa ale tiež sluší a patrí dokázať: + +\Theorem{rovnoramenný trojuholník}{ + Dokážte, že ak pre trojuholník~$ABC$ platí $|AB|=|AC|$, tak $|\angle ABC|=|\angle ACB|$. Dokážte tiež opačnú implikáciu (teda že z~rovnosti uhlov vyplýva rovnosť dĺžok). +}{ + \textit{Priama implikácia.} Predpokladajme $|AB| = |AC|$. Nech $M$ je stred $BC$. Trojuholníky $ABM$ a~$ACM$ sú zhodné podľa $sss$ ($|AB|=|AC|$ z~predpokladu, $|BM|=|CM|$ zo~stredu, spoločná strana $AM$), odkiaľ $|\angle ABC| = |\angle ACB|$. + + \Image{angles-isosceles-triangle.pdf} + + \textit{Opačná implikácia.} Predpokladajme $|\angle ABC| = |\angle ACB| = \beta$. Nech $M$ je päta kolmice z~$A$ na $BC$. Podľa vety $usu$ je $\triangle ABM \cong \triangle ACM$, keďže máme dva zhodné uhly $|\angle ABM| = |\angle ACM|$ a $|\angle AMB| = |\angle AMC| = 90^\circ$ a~spoločnú stranu~$AM$, teda $|AB| = |AC|$. + + \Remark Iný vtipný dôkaz je založený na tom, že dokážeme $\triangle ABC \cong \triangle ACB$ (všimnite si rôzne poradie vrcholov). V~prípade, že poznáme rovnaké strany, tak použijeme vetu $sus$ alebo dokonca $sss$ -- tá nám následne dá zhodné uhly. V~prípade, že poznáme rovnaké uhly, to zas bude veta $usu$, a~zhodnosť nám dá rovnaké strany. +} + +Vyskúšajte si exaktne dokázať tieto jednoduché poznatky, to už bude jednoduchšie: + +\Exercise{}{ + Rovnostranný trojuholník má tri zhodné uhly rovné $60^\circ$ +}{ + V~$\triangle ABC$ s~$|AB| = |AC| = |BC|$ dáva veta o~rovnoramennom trojuholníku $|\angle ABC| = |\angle ACB|$ (z~$|AB|=|AC|$); analogicky z~$|AB|=|BC|$ je $|\angle BAC| = |\angle BCA|$. Všetky tri uhly sú teda zhodné, a~zo súčtu $180^\circ$ je každý $60^\circ$. + + \Image{angles-equilateral.pdf} +} + +\Exercise{}{ + Rovnoramenný pravouhlý trojuholník má tri uhly rovné $90^\circ$, $45^\circ$, $45^\circ$. +}{ + Nech je pravý uhol pri vrchole~$A$, teda $|\angle BAC| = 90^\circ$, a~odvesny sú zhodné, $|AB| = |AC|$. Podľa vety o~rovnoramennom trojuholníku platí $|\angle ABC| = |\angle ACB|$. Označme túto spoločnú veľkosť $\beta$. Zo súčtu uhlov v~trojuholníku + $$ + 90^\circ + \beta + \beta = 180^\circ, + $$ + odkiaľ $2\beta = 90^\circ$, čiže $\beta = 45^\circ$. + + \Image{angles-isosceles-right-triangle.pdf} +} + +Veľmi šikovná vec použiteľná v~neľahkých úlohách je nasledovná úloha. Jeden možný dôkaz je cez trigonometriu. + +\Problem{0}{}{ + Pravouhlý trojuholník má zvyšné dva uhly rovné $60^\circ$ a~$30^\circ$ práve vtedy, keď je jeho prepona dvakrát dlhšia než kratšia odvesna. +}{ + Povedzme, že $BC$ je prepona a~je dvakrát dlhšia než odvesna $AB$. Trikom je uvážiť bod $B'$ taký, že $A$ je stred úsečky $BB'$. +}{ + Nech náš trojuholník má pravý uhol pri vrchole~$A$. + + \textit{($\Rightarrow$)} Predpokladajme $|BC| = 2|AB|$. Nech $B'$ je taký bod, že $A$ je stredom úsečky $BB'$. Potom $|AB'| = |AB|$ a~$A$ leží medzi $B$, $B'$, takže $|\angle B'AC|$ je vedľajší k~$|\angle BAC| = 90^\circ$, čiže tiež $90^\circ$. Platí $\triangle ABC \cong \triangle AB'C$ z~vety $sus$, keďže uhly pri~$A$ sú oba pravé, strana~$AC$ je spoločná, a $|AB|=|AB'|$. + + \Image{angles-90-60-30.pdf} + + Pretože $A$ je stredom $BB'$, je $|BB'| = 2|AB|$, a~zo zadania $|BC| = 2|AB|$. Spolu + $$ + |BB'| = |BC| = |B'C|, + $$ + takže $\triangle BB'C$ je rovnostranný a~všetky jeho uhly sú $60^\circ$. Špeciálne $|\angle ABC| = |\angle B'BC| = 60^\circ$. Zo súčtu uhlov v~$\triangle ABC$ dopočítame $|\angle BCA| = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. + + \textit{($\Leftarrow$)} Nie je ťažké rozmyslieť si, že úvahy z~predošlého odstavca vieme ľahko obrátiť -- kľúčovú zhodnosť $\triangle ABC \cong \triangle AB'C$ tentokrát dostaneme z~vety $usu$. + + \Remark Iné riešenie je uvážiť, že podľa Tálesovej vety je stred~$O$ kružnice opísanej trojuholníku $ABC$ zároveň stredom prepony $BC$. K~tomuto riešeniu sa podrobnejšie vrátime, keď sa budeme baviť o~kružniciach. +} + +\sec Čo si zapamätať + +\secc Techniky + +\begitems \style N +\i Pri rovnobežkách hľadáme zhodné striedavé a súhlasné uhly, prípadne uhly so súčtom~$180^\circ$. +\i Je užitočné nahliadať na vonkajší uhol trojuholníka ako na súčet dvoch uhlov. +\i Na prepojenie sveta dĺžok a~uhlov sa hodí zhodnosť dvoch trojuholníkov. +\i Pomocné body (stred úsečky, zrkadlový obraz, predĺženie) často odhalia zhodnosť alebo špeciálny trojuholník. +\enditems + +\secc Užitočné fakty + +\begitems \style N +\i Súčet uhlov v~trojuholníku je $180^\circ$, všeobecne v~konvexnom $n$-uholníku $(n-2) \cdot 180^\circ$. +\i Vety o~zhodnosti trojuholníkov: $sss$, $sus$, $usu$, $Ssu$. +\i V trojuholníku sú dve strany zhodné práve keď sú zhodné im priľahlé uhly. +\i Pravouhlý trojuholník má preponu dvakrát dlhšiu než odvesnu práve keď má uhly $90^\circ, 60^\circ, 30^\circ$. +\enditems + +\DisplayExerciseSolutions + +\DisplayHints + +\DisplayProblemSolutions + +\bye \ No newline at end of file diff --git a/web/src/content/handouts.json b/web/src/content/handouts.json index 532e1c05..44ef30ff 100644 --- a/web/src/content/handouts.json +++ b/web/src/content/handouts.json @@ -236,6 +236,28 @@ "en": "A guide to the technique of adding auxiliary points in olympiad geometry — midpoints, reflections of orthocentres, second circle intersections, and more — followed by 24 problems from national olympiads, MEMO, ISL, and IMO." }, "authors": ["Patrik Bak"] + }, + { + "status": "ready", + "source": "matikaCesku", + "id": "ZgFNhI74gQeJl7cEMggw7", + "languages": ["sk", "cs", "en"], + "slug": { + "sk": "zaklady-pocitania-uhlov-1", + "cs": "zaklady-pocitani-uhlu-1", + "en": "angle-basics-1" + }, + "title": { + "sk": "Základy počítania uhlov", + "cs": "Základy počítání úhlů", + "en": "Angle Chasing Basics" + }, + "description": { + "sk": "Základy sveta uhlov bez použitia kružníc. Pokrýva vrcholové, súhlasné, striedavé a vedľajšie uhly, súčet uhlov v mnohouholníku a vety o zhodnosti trojuholníkov.", + "cs": "Základy světa úhlů bez použití kružnic. Pokrývá vrcholové, souhlasné, střídavé a vedlejší úhly, součet úhlů v mnohoúhelníku a věty o shodnosti trojúhelníků.", + "en": "Foundations of the world of angles without circles. Covers vertical, corresponding, alternate, and supplementary angles, the angle sum in polygons, and the triangle congruence theorems." + }, + "authors": ["Patrik Bak"] } ] }, diff --git a/web/src/content/handouts/angle-basics-1.cs.json b/web/src/content/handouts/angle-basics-1.cs.json new file mode 100644 index 00000000..83788369 --- /dev/null +++ b/web/src/content/handouts/angle-basics-1.cs.json @@ -0,0 +1,4965 @@ +{ + "document": { + "title": "Základy počítání úhlů", + "sections": [ + { + "title": "Úvod", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Geometrie je nejvizuálnější část matematiky – umíme si ji kreslit na papír, v\u00A0softwaru jako " + }, + { + "type": "link", + "url": "https://www.geogebra.org/", + "content": [ + { + "type": "text", + "text": "GeoGebra" + } + ] + }, + { + "type": "text", + "text": " a\u00A0můžeme se těšit, když na vlastní oči vidíme, že věci v\u00A0ní fungují. Proč tomu tak je však může být záhadou. Abychom tato tajemství pochopili, potřebujeme kousek po kousku budovat základnu znalostí a\u00A0vizuální intuici." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Jedna z\u00A0nejzákladnějších technik je rozumět světu úhlů a\u00A0umět jejich vlastnosti s\u00A0úspěchem používat v\u00A0různých typech úloh. Jen málo zajímavých příkladů úhly nepoužívá. Teorie kolem úhlů přitom není složitá, je však důležité ji budovat pomalu a\u00A0detailně. Začneme s\u00A0nejjednoduššími vlastnostmi, které ani nevyžadují práci s\u00A0kružnicemi. Už ty se však dají s\u00A0úspěchem využít v\u00A0mnoha úlohách." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0tomto materiálu se budeme věnovat úhlům, ale záměrně vynecháme úhly na kružnicích, kterým se patří věnovat samostatnou pozornost. Na chvíli zapomeňme, že kružnice existují." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Na závěr úvodu jedna menší poznámka o\u00A0konvenci. V\u00A0tomto materiálu uvidíme mnoho obrázků trojúhelníků, ve kterých je bod\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " nakreslen nahoře. Ukazuje se, že tento způsob kreslení je mezinárodní standard, zejména na poli matematické olympiády a\u00A0podobných soutěží se to tak zvykne dělat. V\u00A0Česku a\u00A0na\u00A0Slovensku je však ve školách stále zvykem kreslit\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " nahoře. V\u00A0některých zemích lze vidět i\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " nahoře, např. na Ukrajině. Z\u00A0hlediska řešení úloh je vždy dobré nakreslit si nahoru ten bod, u\u00A0kterého obrázek vypadá co nejsymetričtěji, snadněji v\u00A0něm jdou vidět další symetrické věci. V\u00A0tomto materiálu budou úlohy zadávány tak, že ta symetrie půjde vidět více při\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " nahoře \uD83D\uDE42." + } + ], + "highligted": false + } + ] + } + }, + { + "title": "Základy světa úhlů", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0této sekci si odvodíme ty nejjednodušší vlastnosti související s\u00A0úhly. Pevně věřím, že klíčem ke\u00A0zvládnutí geometrie je porozumění věcem od základů a\u00A0do hloubky, proto se těmto jednoduchým vlastnostem budeme věnovat více." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Základní vlastnosti úhlů, které běžně používáme skoro bez přemýšlení při řešení úloh:" + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vrcholové úhly jsou shodné: " + }, + { + "type": "image", + "id": "angle-basics-1.cs-1.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Souhlasné úhly jsou shodné: " + }, + { + "type": "image", + "id": "angle-basics-1.cs-2.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Střídavé úhly jsou shodné: " + }, + { + "type": "image", + "id": "angle-basics-1.cs-3.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vedlejší úhly mají součet " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ": " + }, + { + "type": "image", + "id": "angle-basics-1.cs-4.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerRomanParens" + } + ], + "highligted": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Tyto vlastnosti spolu velmi přirozeně souvisí, menší cvičení k\u00A0zamyšlení:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Uvědomte si, že:" + } + ], + "highligted": false + }, + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "z\u00A0vlastnosti o\u00A0vedlejších úhlech vyplývá vlastnost vrcholových úhlů (a\u00A0naopak)" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "z\u00A0kterékoli dvojice vlastností o\u00A0vrcholových, souhlasných, střídavých úhlech vyplývá ta třetí vlastnost" + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerAlphaParens" + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "bold", + "content": [ + { + "type": "text", + "text": "(a)" + } + ] + }, + { + "type": "text", + "text": " Vedlejší úhel " + }, + { + "type": "math", + "text": "\\alpha'", + "isDisplay": false + }, + { + "type": "text", + "text": " k\u00A0danému úhlu " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " má velikost " + }, + { + "type": "math", + "text": "180^\\circ - \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ". Úhel vedlejší k\u00A0" + }, + { + "type": "math", + "text": "\\alpha'", + "isDisplay": false + }, + { + "type": "text", + "text": " má zase velikost " + }, + { + "type": "math", + "text": "180^\\circ - (180^\\circ - \\alpha) = \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ten je však zároveň vrcholový úhel k\u00A0" + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", tedy jsou vrcholové úhly shodné." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-5.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Naopak, předpokládejme vlastnost o\u00A0vrcholových úhlech a\u00A0označme po sobě jdoucí úhly v\u00A0průsečíku dvou přímek " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\delta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ze shodnosti vrcholových úhlů " + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "\\beta = \\delta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Součet všech čtyř je plný úhel, tedy " + }, + { + "type": "math", + "text": "\\alpha + \\beta + \\gamma + \\delta = 360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Po dosazení dostáváme " + }, + { + "type": "math", + "text": "2\\alpha + 2\\beta = 360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čili " + }, + { + "type": "math", + "text": "\\alpha + \\beta = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " – to je právě vlastnost vedlejších úhlů." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-6.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "bold", + "content": [ + { + "type": "text", + "text": "(b)" + } + ] + }, + { + "type": "text", + "text": " Všechny tři vlastnosti říkají, že jistá dvojice úhlů má stejnou velikost." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podívejme se na trojici úhlů " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " z\u00A0obrázku a\u00A0na to, co o\u00A0jejich vztazích tvrdí jednotlivé vlastnosti. Pro dvojici " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " při horním průsečíku tvrdí vlastnost vrcholových úhlů " + }, + { + "type": "math", + "text": "\\alpha = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Pro dvojici " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " tvrdí vlastnost souhlasných úhlů " + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Konečně pro dvojici " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " tvrdí vlastnost střídavých úhlů " + }, + { + "type": "math", + "text": "\\beta = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-7.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Tři vlastnosti tedy tvrdí tři rovnosti mezi týmiž třemi úhly, jen přes různé dvojice. Jakmile platí libovolné dvě, třetí plyne tranzitivitou: např. pokud víme (V) a\u00A0(S), tak " + }, + { + "type": "math", + "text": "\\alpha = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže " + }, + { + "type": "math", + "text": "\\beta = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", čili (St). Ostatní dvojice analogicky." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Následující tvrzení všichni známe, umíte ho však dokázat?" + } + ], + "highligted": false + }, + { + "type": "theorem", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Součet úhlů v\u00A0trojúhelníku je " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Veďme přes vrchol\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " přímku rovnoběžnou s\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Protože " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " je příčka rovnoběžek, úhel " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " při\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0úhel " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " při\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou střídavé, tedy stejné. Analogicky přes příčku " + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou úhly " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " při\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0při\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " střídavé. Úhly " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " leží vedle sebe podél přímky přes\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", tedy " + }, + { + "type": "math", + "text": "\\alpha + \\beta + \\gamma = 180^\\circ.", + "isDisplay": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-8.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Toto tvrzení se rozhodně dá zobecnit. Víme například, že součet úhlů ve\u00A0čtyřúhelníku je " + }, + { + "type": "math", + "text": "360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Jak je to v\u00A05-úhelníku? A\u00A0jak v\u00A067-úhelníku? Odpovědí je následující tvrzení:" + } + ], + "highligted": false + }, + { + "type": "theorem", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nechť " + }, + { + "type": "math", + "text": "n \\ge 3", + "isDisplay": false + }, + { + "type": "text", + "text": " je přirozené číslo. Součet úhlů v\u00A0konvexním " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku je roven " + }, + { + "type": "math", + "text": "(n-2) \\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Důkaz 1 (matematická indukce)." + } + ] + }, + { + "type": "text", + "text": " Základ " + }, + { + "type": "math", + "text": "n=3", + "isDisplay": false + }, + { + "type": "text", + "text": ": součet úhlů trojúhelníku je " + }, + { + "type": "math", + "text": "180^\\circ = (3-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Indukční krok: nechť " + }, + { + "type": "math", + "text": "A_1 A_2 \\ldots A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " je konvexní " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelník. Úhlopříčkou " + }, + { + "type": "math", + "text": "A_1 A_n", + "isDisplay": false + }, + { + "type": "text", + "text": " jej rozdělíme na trojúhelník " + }, + { + "type": "math", + "text": "A_1 A_n A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0konvexní " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelník " + }, + { + "type": "math", + "text": "A_1 A_2 \\ldots A_n", + "isDisplay": false + }, + { + "type": "text", + "text": ": úhly při vrcholech " + }, + { + "type": "math", + "text": "A_2, \\ldots, A_{n-1}", + "isDisplay": false + }, + { + "type": "text", + "text": " patří celé " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku, úhel při " + }, + { + "type": "math", + "text": "A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " patří celý trojúhelníku a\u00A0úhly při " + }, + { + "type": "math", + "text": "A_1", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "A_n", + "isDisplay": false + }, + { + "type": "text", + "text": " se rozdělí mezi oba útvary tak, aby jejich části daly původní vnitřní úhly " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku. Součet úhlů " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku je tedy" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\underbrace{180^\\circ}_{\\text{trojúhelník}} + \\underbrace{(n-2) \\cdot 180^\\circ}_{n\\text{-úhelník}} = (n-1) \\cdot 180^\\circ = \\bigl((n+1) - 2\\bigr) \\cdot 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-9.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Důkaz 2 (vějířová triangulace)." + } + ] + }, + { + "type": "text", + "text": " Veďme z\u00A0vrcholu " + }, + { + "type": "math", + "text": "A_1", + "isDisplay": false + }, + { + "type": "text", + "text": " úhlopříčky do každého nesousedního vrcholu " + }, + { + "type": "math", + "text": "A_3, A_4, \\ldots, A_{n-1}", + "isDisplay": false + }, + { + "type": "text", + "text": ". Vznikne " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " trojúhelníků pokrývajících vnitřek " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku bez překrytí. Při každém vrcholu " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " se úhly sousedních trojúhelníků při " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " přesně poskládají na vnitřní úhel " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku při " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže součet úhlů všech trojúhelníků se rovná součtu vnitřních úhlů " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku, čili " + }, + { + "type": "math", + "text": "(n-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-10.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Důkaz\u00A02 je " + }, + { + "type": "quote", + "content": [ + { + "type": "text", + "text": "rozbalený" + } + ] + }, + { + "type": "text", + "text": " důkaz\u00A01: indukce odřezává trojúhelníky jeden po druhém, zde je vidíme všechny najednou." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Důkaz 3 (vnitřní bod)." + } + ] + }, + { + "type": "text", + "text": " Zvolme bod " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " uvnitř " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku a\u00A0spojme ho s\u00A0každým vrcholem. Vznikne " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": " trojúhelníků s\u00A0celkovým součtem úhlů " + }, + { + "type": "math", + "text": "n\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Úhly trojúhelníků při každém vrcholu " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " spolu tvoří vnitřní úhel " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku při\u00A0" + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": "; úhly trojúhelníků při\u00A0" + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " spolu tvoří plný úhel " + }, + { + "type": "math", + "text": "360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Tedy součet vnitřních úhlů " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku je" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "n\\cdot 180^\\circ - 360^\\circ = (n-2)\\cdot 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-11.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka (nekonvexní mnohoúhelníky)." + } + ] + }, + { + "type": "text", + "text": " Vzorec " + }, + { + "type": "math", + "text": "(n-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " platí i\u00A0pro navzájem se neprotínající nekonvexní mnohoúhelníky, pokud všechny vnitřní úhly měříme směrem dovnitř – tedy i\u00A0úhly větší než " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " počítáme jako takové." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vějířová triangulace z\u00A0jednoho vrcholu nemusí fungovat ani z\u00A0konvexního vrcholu – některé úhlopříčky mohou procházet mimo mnohoúhelník. Platí však, že každý jednoduchý " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelník se dá rozdělit na " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " trojúhelníků (ne nutně vějířovitě), což důkaz zachrání. Důvodem je indukce přes trojúhelník tvořený nějakým vrcholem " + }, + { + "type": "math", + "text": "V", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0jeho sousedy " + }, + { + "type": "math", + "text": "U", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "W", + "isDisplay": false + }, + { + "type": "text", + "text": ", který leží celý uvnitř našeho " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku. Odříznutím " + }, + { + "type": "math", + "text": "V", + "isDisplay": false + }, + { + "type": "text", + "text": " úhlopříčkou " + }, + { + "type": "math", + "text": "UW", + "isDisplay": false + }, + { + "type": "text", + "text": " dostaneme " + }, + { + "type": "math", + "text": "(n-1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelník; opakováním tedy " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " trojúhelníků. Dá se dokázat, že takový trojúhelník vždy existuje (dokonce dva), viz " + }, + { + "type": "link", + "url": "https://en.wikipedia.org/wiki/Two_ears_theorem", + "content": [ + { + "type": "text", + "text": "věta o\u00A0dvou uších" + } + ] + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "K\u00A0důkazu\u00A03: funguje jen pokud existuje vnitřní bod " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " viditelný ze všech vrcholů (tj. úsečky " + }, + { + "type": "math", + "text": "OA_i", + "isDisplay": false + }, + { + "type": "text", + "text": " leží celé uvnitř). Pro některé nekonvexní mnohoúhelníky takový " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " nemusí existovat (zkuste takový nakreslit) – v\u00A0takovém případě se důkaz\u00A03 nedá přímo zachránit a\u00A0je třeba sáhnout po triangulaci z\u00A0důkazu\u00A02." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Teď uveďme ještě dvě jednoduchá, ale užitečná pomocná tvrzení:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dokažte, že součet úhlů na obrázku je roven " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-12.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Označme " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " úhel svíraný příčkou " + }, + { + "type": "math", + "text": "AD", + "isDisplay": false + }, + { + "type": "text", + "text": " s\u00A0rovnoběžkami na straně " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " (jako na obrázku). Souhlasné úhly " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " při\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " při\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou shodné. Při vrcholu " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " vedlejší, tedy" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\beta + \\alpha = 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-13.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dokažte, že úhel s\u00A0otazníkem na obrázku je roven " + }, + { + "type": "math", + "text": "\\alpha+\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-14.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Ze součtu úhlů v\u00A0trojúhelníku " + }, + { + "type": "math", + "text": "|\\angle ACB| = 180^\\circ - \\alpha - \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Vnější úhel při " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " je vedlejší k\u00A0" + }, + { + "type": "math", + "text": "|\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": ":" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "180^\\circ - |\\angle ACB| = \\alpha + \\beta.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-14.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Poslední dvě cvičení možná vypadala velmi triviálně. Realita však je, že v\u00A0praktických úlohách jsou velmi užitečná – když máme obrovský netriviální obrázek a\u00A0máme provést za sebou velký počet úhlových operací, tak je opravdu velmi výhodné si i\u00A0jen kousek té práce zjednodušit – v\u00A0případě rovnoběžek nemusíme uvažovat pomocný souhlasný úhel a\u00A0v\u00A0případě trojúhelníku nemusíme pracovat s\u00A0" + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + } + }, + { + "title": "Základy světa délek", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Zatím jsme fungovali pouze ve světě úhlů a\u00A0vůbec jsme neřešili délky. Geometrie však začne být zajímavá, když tyto světy začneme propojovat. Základem pro nás bude " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "shodnost trojúhelníků" + } + ] + }, + { + "type": "text", + "text": ". Připomeňme si kritéria." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Věty o\u00A0shodnosti trojúhelníků:" + } + ], + "highligted": false + }, + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Věta " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": ": dva trojúhelníky jsou shodné, pokud se shodují ve všech třech stranách." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Věta " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": ": dva trojúhelníky jsou shodné, pokud se shodují ve dvou stranách a\u00A0úhlu, který tyto strany svírají." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Věta " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": ": dva trojúhelníky jsou shodné, pokud se shodují v\u00A0jedné straně a\u00A0nějakých dvou úhlech." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Věta " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": ": dva trojúhelníky jsou shodné, pokud se shodují ve dvou stranách a\u00A0úhlu naproti delší z\u00A0těchto stran." + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerRomanParens" + } + ], + "highligted": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Zdůrazněme, že u\u00A0věty " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " je opravdu důležité, že shodný úhel je ten svíraný oběma stranami – bez tohoto předpokladu shodnost nemusí fungovat, viz obrázek. Na něm máme dva trojúhelníky " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A'B'C'", + "isDisplay": false + }, + { + "type": "text", + "text": ", pro které platí " + }, + { + "type": "math", + "text": "|BC|=|B'C'|", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "|AC|=|A'C'|", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0" + }, + { + "type": "math", + "text": "\\beta=\\beta'", + "isDisplay": false + }, + { + "type": "text", + "text": ", avšak zjevně nejsou shodné." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-15.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Věta " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " by byla naše záchrana – zde ji však aplikovat neumíme, protože strana, naproti které leží úhel " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", tedy " + }, + { + "type": "math", + "text": "|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ", není nejdelší, neboť je zjevně kratší než " + }, + { + "type": "math", + "text": "|BC|", + "isDisplay": false + }, + { + "type": "text", + "text": " \uD83E\uDD72." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Než půjdeme dál, uvědomme si, co tyto věty znamenají. Z\u00A0mého pohledu je dobrý pohled na shodnost takový: chceme sestrojit trojúhelník, když máme dány nějaké tři jeho elementy – budou všechny sestrojitelné trojúhelníky shodné?" + } + ], + "highligted": false + }, + { + "type": "example", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Přesvědčte se o\u00A0platnosti věty " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": " o\u00A0shodnosti." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Představme si, že máme dány tři úsečky délek " + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": ", ze kterých umíme sestrojit trojúhelník. Začneme např. úsečkou\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " délky\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0následně sestrojíme dvě kružnice: (a) kružnici se středem v\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0poloměrem " + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": "; (b) kružnici se středem v\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0poloměrem " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ty se buď neprotnou (když např. " + }, + { + "type": "math", + "text": "b+c < a", + "isDisplay": false + }, + { + "type": "text", + "text": "), nebo se protnou v\u00A0jediném bodě na " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " (když " + }, + { + "type": "math", + "text": "b+c=a", + "isDisplay": false + }, + { + "type": "text", + "text": "), nebo se protnou ve dvou bodech " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A'", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-16.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Na konci si uvědomme, že trojúhelníky " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A'BC", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou však zřejmě vzájemně zrcadlovými obrazy podle " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Také pokud bychom začali jinou úsečkou, tak vytvoříme pouze posunutí/otočení této konfigurace." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Uvědomme si, že jsme vlastně po cestě dokázali\u00A0trojúhelníkovou nerovnost: součet dvou stran trojúhelníku je větší než třetí – náš trojúhelník je sestrojitelný a\u00A0nedegenerovaný\u00A0" + }, + { + "type": "footnote", + "content": [ + { + "type": "text", + "text": "nemá úhel rovný " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + } + ] + }, + { + "type": "text", + "text": " právě tehdy, když " + }, + { + "type": "math", + "text": "b+c>a", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podobně umíme zdůvodnit další kritéria:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Přesvědčte se o\u00A0platnosti věty " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " o\u00A0shodnosti." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nechť jsou dány strany " + }, + { + "type": "math", + "text": "b = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "c = |AB|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0úhel " + }, + { + "type": "math", + "text": "|\\angle BAC| = \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " mezi nimi. Sestrojme vrchol\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0úsečku " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " délky\u00A0" + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": " – ta je dána až na shodné zobrazení. Bod " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " musí ležet na polopřímce z\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " svírající s\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " úhel\u00A0" + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0zároveň ve vzdálenosti\u00A0" + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " od\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". Polopřímky svírající s\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " úhel " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou dvě (po jedné na každé straně " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": "): na každé leží právě jeden bod ve vzdálenosti " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " od\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", tedy dostaneme body " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "C'", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trojúhelníky " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "ABC'", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou zrcadlovými obrazy podle " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ", tedy shodné." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-17.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Přesvědčte se o\u00A0platnosti věty " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": " o\u00A0shodnosti." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0prvé řadě si uvědomme, že nezáleží na tom, které dva úhly jsou shodné – pokud se dva trojúhelníky shodují ve dvou úhlech, tak třetí je určen jednoznačně, neboť všechny tři mají součet úhlů " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pro účely našeho důkazu uvažme tedy, že jsou shodné právě úhly přiléhající ke shodné straně. Nechť je to úhel " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " při vrcholu " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " při vrcholu " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0strana " + }, + { + "type": "math", + "text": "a = |BC|", + "isDisplay": false + }, + { + "type": "text", + "text": " mezi nimi. Sestrojme úsečku " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " délky\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": " – ta je dána až na shodné zobrazení. Bod " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " musí ležet na polopřímce z\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " svírající s\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " úhel " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0zároveň na polopřímce z\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " svírající s\u00A0" + }, + { + "type": "math", + "text": "CB", + "isDisplay": false + }, + { + "type": "text", + "text": " úhel " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Součet vnitřních úhlů při " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " v\u00A0trojúhelníku je menší než " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čili " + }, + { + "type": "math", + "text": "\\beta + \\gamma < 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0proto nejsou obě polopřímky rovnoběžné – protnou se v\u00A0jediném bodě\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trojúhelník " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " je tedy určen jednoznačně (až na shodná zobrazení)." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-18.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Méně známé tvrzení " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " je už trošičku těžší, proto k\u00A0němu budou i\u00A0návody \uD83D\uDE42." + } + ], + "highligted": false + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "shodnost " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Přesvědčte se o\u00A0platnosti věty " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " o\u00A0shodnosti. Navíc si uvědomte, kde by se konstrukce pokazila, kdyby úhel nebyl naproti větší ze stran – co by se stalo, kdyby byly obě strany stejně dlouhé?" + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Začněte konstrukci konstrukcí kratší úsečky. Jak pokračuje konstrukce po jejím sestrojení?" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Po sestrojení kratší úsečky sestrojíme polopřímku pod naším daným shodným úhlem. Zbývá poslední krok konstrukce. Rozmyslete, kdy dostaneme žádný, kdy jeden a\u00A0kdy dva vyhovující trojúhelníky." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nechť jsou dány strany " + }, + { + "type": "math", + "text": "a = |BC|", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "b = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0úhel " + }, + { + "type": "math", + "text": "|\\angle ABC| = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " naproti delší straně " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": ". Začneme sestrojením kratší úsečky " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " délky\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ". Bod " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " musí ležet na polopřímce z\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " svírající s\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " úhel " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0zároveň ve vzdálenosti " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " od\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": ", čili na kružnici " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": " se středem\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0poloměrem " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vzdálenost bodu\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " od středu kružnice je " + }, + { + "type": "math", + "text": "|BC| = a", + "isDisplay": false + }, + { + "type": "text", + "text": ". Protože " + }, + { + "type": "math", + "text": "a < b", + "isDisplay": false + }, + { + "type": "text", + "text": ", bod\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " leží " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "uvnitř" + } + ] + }, + { + "type": "text", + "text": " kružnice " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". Polopřímka z\u00A0bodu\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " tak začíná uvnitř kružnice a\u00A0pokračuje směrem ven, takže kružnici protne právě jednou. Bod " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " je tedy určen jednoznačně, a\u00A0trojúhelník " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovněž jednoznačný (až na shodné zobrazení)." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-19.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podívejme se, proč je podmínka " + }, + { + "type": "math", + "text": "b > a", + "isDisplay": false + }, + { + "type": "text", + "text": " podstatná. Pokud by bylo " + }, + { + "type": "math", + "text": "a = b", + "isDisplay": false + }, + { + "type": "text", + "text": ", bod\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " by ležel přímo " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "na" + } + ] + }, + { + "type": "text", + "text": " kružnici\u00A0" + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". Polopřímka z\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " by ji protínala v\u00A0bodě\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " samotném (což nedává trojúhelník) a\u00A0v\u00A0právě jednom dalším bodě – ten je hledaným\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". Konstrukce tedy stále vede k\u00A0jednoznačnému (rovnoramennému) trojúhelníku, takže věta " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " formálně platí i\u00A0v\u00A0tomto hraničním případě. Pod " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " ji však samostatně nezařazujeme: pokud víme, že trojúhelník je rovnoramenný se " + }, + { + "type": "math", + "text": "|BC|=|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0známe jeden úhel " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", tak ostatní dva úhly jsou už díky rovnoramennosti určeny (oba při základně mají velikost\u00A0" + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": "). Tatáž konfigurace je proto pokryta větami " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " či " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " k\u00A0ní nepřidává žádnou novou informaci." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-20.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pokud by bylo " + }, + { + "type": "math", + "text": "b < a", + "isDisplay": false + }, + { + "type": "text", + "text": ", bod " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " by ležel " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "vně" + } + ] + }, + { + "type": "text", + "text": " kružnice " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". Polopřímka by ji mohla protnout ve dvou bodech – vznikly by tak dva různé (neshodné) trojúhelníky – nebo by ji minula úplně, což nevede k\u00A0žádnému trojúhelníku." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-21.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Důkaz je možné provést i\u00A0tak, že nejprve sestrojíme delší úsečku. To však vyžaduje znalost množiny bodů nad pevnou úsečkou majících pevný úhel – k\u00A0tomu se dostaneme v\u00A0jiném materiálu." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Na naší geometrické cestě se průběžně setkáme se všemi těmito tvrzeními. Prozatím si však ukažme nějaké konkrétní aplikace. Začneme tím nejzřejmějším tvrzením, které se ale také sluší a\u00A0patří dokázat:" + } + ], + "highligted": false + }, + { + "type": "theorem", + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "rovnoramenný trojúhelník" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dokažte, že pokud pro trojúhelník\u00A0" + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " platí " + }, + { + "type": "math", + "text": "|AB|=|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ", tak " + }, + { + "type": "math", + "text": "|\\angle ABC|=|\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Dokažte také opačnou implikaci (tedy že z\u00A0rovnosti úhlů vyplývá rovnost délek)." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Přímá implikace." + } + ] + }, + { + "type": "text", + "text": " Předpokládejme " + }, + { + "type": "math", + "text": "|AB| = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nechť " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": " je střed " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trojúhelníky " + }, + { + "type": "math", + "text": "ABM", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "ACM", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou shodné podle " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "|AB|=|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": " z\u00A0předpokladu, " + }, + { + "type": "math", + "text": "|BM|=|CM|", + "isDisplay": false + }, + { + "type": "text", + "text": " ze\u00A0středu, společná strana " + }, + { + "type": "math", + "text": "AM", + "isDisplay": false + }, + { + "type": "text", + "text": "), odkud " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-22.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Opačná implikace." + } + ] + }, + { + "type": "text", + "text": " Předpokládejme " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle ACB| = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nechť " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": " je pata kolmice z\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " na " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Podle věty " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": " je " + }, + { + "type": "math", + "text": "\\triangle ABM \\cong \\triangle ACM", + "isDisplay": false + }, + { + "type": "text", + "text": ", neboť máme dva shodné úhly " + }, + { + "type": "math", + "text": "|\\angle ABM| = |\\angle ACM|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|\\angle AMB| = |\\angle AMC| = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0společnou stranu\u00A0" + }, + { + "type": "math", + "text": "AM", + "isDisplay": false + }, + { + "type": "text", + "text": ", tedy " + }, + { + "type": "math", + "text": "|AB| = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Jiný vtipný důkaz je založen na tom, že dokážeme " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": " (všimněte si různého pořadí vrcholů). V\u00A0případě, že známe stejné strany, použijeme větu " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " nebo dokonce " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": " – ta nám následně dá shodné úhly. V\u00A0případě, že známe stejné úhly, to zase bude věta " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0shodnost nám dá stejné strany." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vyzkoušejte si exaktně dokázat tyto jednoduché poznatky, to už bude jednodušší:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Rovnostranný trojúhelník má tři shodné úhly rovny " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0" + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " s\u00A0" + }, + { + "type": "math", + "text": "|AB| = |AC| = |BC|", + "isDisplay": false + }, + { + "type": "text", + "text": " dává věta o\u00A0rovnoramenném trojúhelníku " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": " (z\u00A0" + }, + { + "type": "math", + "text": "|AB|=|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": "); analogicky z\u00A0" + }, + { + "type": "math", + "text": "|AB|=|BC|", + "isDisplay": false + }, + { + "type": "text", + "text": " je " + }, + { + "type": "math", + "text": "|\\angle BAC| = |\\angle BCA|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Všechny tři úhly jsou tedy shodné, a\u00A0ze součtu " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " je každý " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-23.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Rovnoramenný pravoúhlý trojúhelník má tři úhly rovny " + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nechť je pravý úhel při vrcholu\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", tedy " + }, + { + "type": "math", + "text": "|\\angle BAC| = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0odvěsny jsou shodné, " + }, + { + "type": "math", + "text": "|AB| = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Podle věty o\u00A0rovnoramenném trojúhelníku platí " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Označme tuto společnou velikost " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ze součtu úhlů v\u00A0trojúhelníku" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "90^\\circ + \\beta + \\beta = 180^\\circ,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "odkud " + }, + { + "type": "math", + "text": "2\\beta = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čili " + }, + { + "type": "math", + "text": "\\beta = 45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-24.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Velmi šikovná věc použitelná v\u00A0nelehkých úlohách je následující úloha. Jeden možný důkaz je přes trigonometrii." + } + ], + "highligted": false + }, + { + "type": "problem", + "difficulty": 0, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pravoúhlý trojúhelník má zbývající dva úhly rovny " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " právě tehdy, když je jeho přepona dvakrát delší než kratší odvěsna." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Řekněme, že " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " je přepona a\u00A0je dvakrát delší než odvěsna " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trikem je uvážit bod " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": " takový, že " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " je střed úsečky " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nechť náš trojúhelník má pravý úhel při vrcholu\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "(" + }, + { + "type": "math", + "text": "\\Rightarrow", + "isDisplay": false + }, + { + "type": "text", + "text": ")" + } + ] + }, + { + "type": "text", + "text": " Předpokládejme " + }, + { + "type": "math", + "text": "|BC| = 2|AB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nechť " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": " je takový bod, že " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " je středem úsečky " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": ". Potom " + }, + { + "type": "math", + "text": "|AB'| = |AB|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " leží mezi " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže " + }, + { + "type": "math", + "text": "|\\angle B'AC|", + "isDisplay": false + }, + { + "type": "text", + "text": " je vedlejší k\u00A0" + }, + { + "type": "math", + "text": "|\\angle BAC| = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čili také " + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Platí " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle AB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " z\u00A0věty " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": ", neboť úhly při\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou oba pravé, strana\u00A0" + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": " je společná, a\u00A0" + }, + { + "type": "math", + "text": "|AB|=|AB'|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.cs-25.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Protože " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " je středem " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": ", je " + }, + { + "type": "math", + "text": "|BB'| = 2|AB|", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0ze zadání " + }, + { + "type": "math", + "text": "|BC| = 2|AB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Spolu" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|BB'| = |BC| = |B'C|,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "takže " + }, + { + "type": "math", + "text": "\\triangle BB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnostranný a\u00A0všechny jeho úhly jsou " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Speciálně " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle B'BC| = 60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ze součtu úhlů v\u00A0" + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " dopočítáme " + }, + { + "type": "math", + "text": "|\\angle BCA| = 180^\\circ - 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "(" + }, + { + "type": "math", + "text": "\\Leftarrow", + "isDisplay": false + }, + { + "type": "text", + "text": ")" + } + ] + }, + { + "type": "text", + "text": " Není těžké rozmyslet si, že úvahy z\u00A0předchozího odstavce umíme snadno obrátit – klíčovou shodnost " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle AB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " tentokrát dostaneme z\u00A0věty " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Jiné řešení je uvážit, že podle Thaletovy věty je střed\u00A0" + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " kružnice opsané trojúhelníku " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " zároveň středem přepony " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". K\u00A0tomuto řešení se podrobněji vrátíme, až se budeme bavit o\u00A0kružnicích." + } + ], + "highligted": false + } + ] + } + ] + } + }, + { + "title": "Co si zapamatovat", + "level": 1, + "text": { + "content": [] + } + }, + { + "title": "Techniky", + "level": 2, + "text": { + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Při rovnoběžkách hledáme shodné střídavé a\u00A0souhlasné úhly, případně úhly se součtem\u00A0" + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Je užitečné nahlížet na vnější úhel trojúhelníku jako na součet dvou úhlů." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pro propojení světa délek a\u00A0úhlů se hodí shodnost dvou trojúhelníků." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pomocné body (střed úsečky, zrcadlový obraz, prodloužení) často odhalí shodnost nebo speciální trojúhelník." + } + ], + "highligted": false + } + ] + ], + "styleType": "NumberParens" + } + ] + } + }, + { + "title": "Užitečná fakta", + "level": 2, + "text": { + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Součet úhlů v\u00A0trojúhelníku je " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", obecně v\u00A0konvexním " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-úhelníku " + }, + { + "type": "math", + "text": "(n-2) \\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Věty o\u00A0shodnosti trojúhelníků: " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0trojúhelníku jsou dvě strany shodné právě když jsou shodné jim přilehlé úhly." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pravoúhlý trojúhelník má přeponu dvakrát delší než odvěsnu právě když má úhly " + }, + { + "type": "math", + "text": "90^\\circ, 60^\\circ, 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "styleType": "NumberParens" + } + ] + } + } + ] + }, + "images": [ + { + "contentId": "angle-basics-1.cs-1.svg", + "originalId": "angles-vertical.pdf", + "width": "82.61pt", + "height": "65.04pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-2.svg", + "originalId": "angles-corresponding.pdf", + "width": "106.36pt", + "height": "66.59pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-3.svg", + "originalId": "angles-alternate.pdf", + "width": "100.63pt", + "height": "58.17pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-4.svg", + "originalId": "angles-supplementary.pdf", + "width": "85.68pt", + "height": "47.81pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-5.svg", + "originalId": "angles-supplementary-derive-vertical.pdf", + "width": "95.91pt", + "height": "78.49pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-6.svg", + "originalId": "angles-vertical-derive-supplementary.pdf", + "width": "95.91pt", + "height": "78.49pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-7.svg", + "originalId": "angles-corresponding-alternate-supplementary-connection.pdf", + "width": "98.31pt", + "height": "65.02pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-8.svg", + "originalId": "angles-triangle.pdf", + "width": "182.63pt", + "height": "128.67pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-9.svg", + "originalId": "angles-polygon.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-10.svg", + "originalId": "angles-polygon-fan.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-11.svg", + "originalId": "angles-polygon-interior.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-12.svg", + "originalId": "angles-complementary-parallel-statement.pdf", + "width": "91.52pt", + "height": "72.46pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-13.svg", + "originalId": "angles-complementary-parallel-solution.pdf", + "width": "85.7pt", + "height": "92.24pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-14.svg", + "originalId": "angles-complementary-in-triangle.pdf", + "width": "153.73pt", + "height": "123.67pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-15.svg", + "originalId": "angles-sas-warning.pdf", + "width": "240.45pt", + "height": "68.01pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-16.svg", + "originalId": "angles-sss-proof.pdf", + "width": "180.33pt", + "height": "129.76pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-17.svg", + "originalId": "angles-sas-proof.pdf", + "width": "190.6pt", + "height": "165.53pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-18.svg", + "originalId": "angles-asa-proof.pdf", + "width": "156.76pt", + "height": "165.02pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-19.svg", + "originalId": "angles-Ssa-proof-one-solution.pdf", + "width": "116.57pt", + "height": "129.18pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-20.svg", + "originalId": "angles-Ssa-proof-isosceles.pdf", + "width": "116.86pt", + "height": "122.28pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-21.svg", + "originalId": "angles-Ssa-proof-two-solutions.pdf", + "width": "191.58pt", + "height": "154.75pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-22.svg", + "originalId": "angles-isosceles-triangle.pdf", + "width": "117.87pt", + "height": "114.32pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-23.svg", + "originalId": "angles-equilateral.pdf", + "width": "107.91pt", + "height": "97.32pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-24.svg", + "originalId": "angles-isosceles-right-triangle.pdf", + "width": "97.95pt", + "height": "95.36pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.cs-25.svg", + "originalId": "angles-90-60-30.pdf", + "width": "146.09pt", + "height": "131.24pt", + "scale": 1.0 + } + ] +} diff --git a/web/src/content/handouts/angle-basics-1.en.json b/web/src/content/handouts/angle-basics-1.en.json new file mode 100644 index 00000000..fd94dbbc --- /dev/null +++ b/web/src/content/handouts/angle-basics-1.en.json @@ -0,0 +1,4971 @@ +{ + "document": { + "title": "Angle Chasing Basics", + "sections": [ + { + "title": "Introduction", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Geometry is the most visual part of mathematics – we can draw it on paper, in software like " + }, + { + "type": "link", + "url": "https://www.geogebra.org/", + "content": [ + { + "type": "text", + "text": "GeoGebra" + } + ] + }, + { + "type": "text", + "text": ", and we can enjoy seeing with our own eyes that things in it work. Why this is so, however, can be a mystery. To uncover these secrets, we need to build up a foundation of knowledge and visual intuition piece by piece." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "One of the most fundamental techniques is to understand the world of angles and to know how to use their properties successfully in various types of problems. Only a few interesting examples don't use angles. The theory around angles is not complicated, but it is important to build it up slowly and in detail. We will start with the simplest properties, which don't even require working with circles. Even these can already be used successfully in many problems." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In this handout we will focus on angles, but we will deliberately leave out angles in circles, which deserve separate attention. For a moment, let's forget that circles exist." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "To close the introduction, a small note about convention. In this handout we will see many diagrams of triangles in which vertex\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " is drawn at the top. It turns out that this way of drawing is the international standard, especially in the world of the Mathematical Olympiad and similar competitions. In Czechia and Slovakia, however, schools still customarily draw\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " at the top. In some countries one can also see\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " at the top, e.g. in Ukraine. From the point of view of solving problems, it is always a good idea to put at the top the vertex that makes the diagram look as symmetric as possible – it is then easier to spot further symmetric things in it. In this handout the problems will be posed so that the symmetry is more visible with\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " at the top \uD83D\uDE42." + } + ], + "highligted": false + } + ] + } + }, + { + "title": "Basics of the World of Angles", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In this section we will derive the simplest properties related to angles. I firmly believe that the key to mastering geometry is understanding things from the ground up and in depth, so we will spend more time on these simple properties." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Basic properties of angles, which we commonly use almost without thinking when solving problems:" + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vertical angles are equal: " + }, + { + "type": "image", + "id": "angle-basics-1.en-1.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Corresponding angles are equal: " + }, + { + "type": "image", + "id": "angle-basics-1.en-2.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Alternate angles are equal: " + }, + { + "type": "image", + "id": "angle-basics-1.en-3.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Angles in a linear pair sum to " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ": " + }, + { + "type": "image", + "id": "angle-basics-1.en-4.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerRomanParens" + } + ], + "highligted": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "These properties are very naturally related to one another; a small exercise to ponder:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Convince yourself that:" + } + ], + "highligted": false + }, + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "the linear-pair property implies the vertical-angles property (and vice versa)" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "any two of the three properties about vertical, corresponding, and alternate angles imply the third" + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerAlphaParens" + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "bold", + "content": [ + { + "type": "text", + "text": "(a)" + } + ] + }, + { + "type": "text", + "text": " The angle " + }, + { + "type": "math", + "text": "\\alpha'", + "isDisplay": false + }, + { + "type": "text", + "text": " in a linear pair with a given angle " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " has measure " + }, + { + "type": "math", + "text": "180^\\circ - \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ". The angle in a linear pair with " + }, + { + "type": "math", + "text": "\\alpha'", + "isDisplay": false + }, + { + "type": "text", + "text": " in turn has measure " + }, + { + "type": "math", + "text": "180^\\circ - (180^\\circ - \\alpha) = \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ". But this last angle is also vertical to " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", so vertical angles are equal." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-5.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Conversely, assume the vertical-angles property and label the consecutive angles at the intersection of two lines " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\delta", + "isDisplay": false + }, + { + "type": "text", + "text": ". By the equality of vertical angles, " + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "\\beta = \\delta", + "isDisplay": false + }, + { + "type": "text", + "text": ". The sum of all four is a full angle, i.e. " + }, + { + "type": "math", + "text": "\\alpha + \\beta + \\gamma + \\delta = 360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Substituting, we get " + }, + { + "type": "math", + "text": "2\\alpha + 2\\beta = 360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", so " + }, + { + "type": "math", + "text": "\\alpha + \\beta = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " – which is exactly the linear-pair property." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-6.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "bold", + "content": [ + { + "type": "text", + "text": "(b)" + } + ] + }, + { + "type": "text", + "text": " All three properties say that a certain pair of angles has the same measure." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Look at the three angles " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " in the figure and at what each property says about their relationships. For the pair " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " at the upper intersection, the vertical-angles property gives " + }, + { + "type": "math", + "text": "\\alpha = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". For the pair " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", the corresponding-angles property gives " + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Finally, for the pair " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", the alternate-angles property gives " + }, + { + "type": "math", + "text": "\\beta = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-7.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The three properties therefore assert three equalities among the same three angles, just via different pairs. As soon as any two of them hold, the third follows by transitivity: e.g. if (V) and (C) hold, then " + }, + { + "type": "math", + "text": "\\alpha = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", so " + }, + { + "type": "math", + "text": "\\beta = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", which is (A). The other pairs are analogous." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The following statement is familiar to all of us, but can you prove it?" + } + ], + "highligted": false + }, + { + "type": "theorem", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The sum of the angles in a triangle is " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Through vertex\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " draw a line parallel to " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Since " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " is a transversal of the parallel lines, the angle " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " at\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " and the angle " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " at\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " are alternate, hence equal. Similarly, via the transversal " + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": ", the angles " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " at\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " and at\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " are alternate. The angles " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " lie next to each other along the line through\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", so " + }, + { + "type": "math", + "text": "\\alpha + \\beta + \\gamma = 180^\\circ.", + "isDisplay": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-8.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "This statement can certainly be generalized. We know, for example, that the sum of the angles in a quadrilateral is " + }, + { + "type": "math", + "text": "360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". What about a pentagon? And a 67-gon? The answer is the following statement:" + } + ], + "highligted": false + }, + { + "type": "theorem", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Let " + }, + { + "type": "math", + "text": "n \\ge 3", + "isDisplay": false + }, + { + "type": "text", + "text": " be a natural number. The sum of the angles in a convex " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon equals " + }, + { + "type": "math", + "text": "(n-2) \\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Proof 1 (mathematical induction)." + } + ] + }, + { + "type": "text", + "text": " Base case " + }, + { + "type": "math", + "text": "n=3", + "isDisplay": false + }, + { + "type": "text", + "text": ": the sum of the angles of a triangle is " + }, + { + "type": "math", + "text": "180^\\circ = (3-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Inductive step: let " + }, + { + "type": "math", + "text": "A_1 A_2 \\ldots A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " be a convex " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon. The diagonal " + }, + { + "type": "math", + "text": "A_1 A_n", + "isDisplay": false + }, + { + "type": "text", + "text": " splits it into a triangle " + }, + { + "type": "math", + "text": "A_1 A_n A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " and a convex " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon " + }, + { + "type": "math", + "text": "A_1 A_2 \\ldots A_n", + "isDisplay": false + }, + { + "type": "text", + "text": ": the angles at vertices " + }, + { + "type": "math", + "text": "A_2, \\ldots, A_{n-1}", + "isDisplay": false + }, + { + "type": "text", + "text": " belong entirely to the " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon, the angle at " + }, + { + "type": "math", + "text": "A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " belongs entirely to the triangle, and the angles at " + }, + { + "type": "math", + "text": "A_1", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "A_n", + "isDisplay": false + }, + { + "type": "text", + "text": " are split between the two figures so that their parts add up to the original interior angles of the " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon. The angle sum of the " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon is therefore" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\underbrace{180^\\circ}_{\\text{triangle}} + \\underbrace{(n-2) \\cdot 180^\\circ}_{n\\text{-gon}} = (n-1) \\cdot 180^\\circ = \\bigl((n+1) - 2\\bigr) \\cdot 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-9.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Proof 2 (fan triangulation)." + } + ] + }, + { + "type": "text", + "text": " From vertex " + }, + { + "type": "math", + "text": "A_1", + "isDisplay": false + }, + { + "type": "text", + "text": " draw diagonals to every non-adjacent vertex " + }, + { + "type": "math", + "text": "A_3, A_4, \\ldots, A_{n-1}", + "isDisplay": false + }, + { + "type": "text", + "text": ". This produces " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " triangles that cover the interior of the " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon without overlap. At each vertex " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": ", the angles of the adjacent triangles at " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " assemble exactly into the interior angle of the " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon at " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": ", so the sum of the angles of all triangles equals the sum of the interior angles of the " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon, i.e. " + }, + { + "type": "math", + "text": "(n-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-10.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Proof\u00A02 is an " + }, + { + "type": "quote", + "content": [ + { + "type": "text", + "text": "unrolled" + } + ] + }, + { + "type": "text", + "text": " version of Proof\u00A01: induction cuts off triangles one by one, and here we see them all at once." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Proof 3 (interior point)." + } + ] + }, + { + "type": "text", + "text": " Choose a point " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " in the interior of the " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon and connect it to every vertex. This produces " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": " triangles with a total angle sum of " + }, + { + "type": "math", + "text": "n\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". At each vertex " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " of the " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon, the triangle angles at " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " together form the entire interior angle of the " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon there; the triangle angles at\u00A0" + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " together form a full angle of " + }, + { + "type": "math", + "text": "360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". So the sum of the interior angles of the " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon is" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "n\\cdot 180^\\circ - 360^\\circ = (n-2)\\cdot 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-11.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka (non-convex polygons)." + } + ] + }, + { + "type": "text", + "text": " The formula " + }, + { + "type": "math", + "text": "(n-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " also holds for non-self-intersecting non-convex polygons, provided we measure all interior angles toward the inside – so we count angles greater than " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " as such." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "A fan triangulation from a single vertex may fail even from a convex vertex – some diagonals can pass outside the polygon. It is still true, however, that every simple " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon can be split into " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " triangles (not necessarily as a fan), which rescues the proof. The reason is induction via a triangle formed by some vertex " + }, + { + "type": "math", + "text": "V", + "isDisplay": false + }, + { + "type": "text", + "text": " and its neighbors " + }, + { + "type": "math", + "text": "U", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "W", + "isDisplay": false + }, + { + "type": "text", + "text": " that lies entirely inside our " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon. Cutting " + }, + { + "type": "math", + "text": "V", + "isDisplay": false + }, + { + "type": "text", + "text": " off along the diagonal " + }, + { + "type": "math", + "text": "UW", + "isDisplay": false + }, + { + "type": "text", + "text": " gives an " + }, + { + "type": "math", + "text": "(n-1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon; iterating yields " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " triangles. One can prove that such a triangle always exists (in fact, two of them); see the " + }, + { + "type": "link", + "url": "https://en.wikipedia.org/wiki/Two_ears_theorem", + "content": [ + { + "type": "text", + "text": "two ears theorem" + } + ] + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "As for Proof\u00A03: it works only if there is an interior point " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " visible from every vertex (i.e., the segments " + }, + { + "type": "math", + "text": "OA_i", + "isDisplay": false + }, + { + "type": "text", + "text": " lie entirely inside). For some non-convex polygons no such " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " exists (try drawing one) – in that case Proof\u00A03 cannot be salvaged directly and one must fall back on the triangulation from Proof\u00A02." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Now let's state two more simple but useful auxiliary claims:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Prove that the sum of the angles in the figure equals " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-12.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Let " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " denote the angle that the transversal " + }, + { + "type": "math", + "text": "AD", + "isDisplay": false + }, + { + "type": "text", + "text": " makes with the parallel lines on the side of " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " (as in the figure). The corresponding angles " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " at\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " at\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " are equal. At vertex " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " the angles " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " form a linear pair, so" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\beta + \\alpha = 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-13.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Prove that the angle marked with a question mark in the figure equals " + }, + { + "type": "math", + "text": "\\alpha+\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-14.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "From the angle sum in the triangle, " + }, + { + "type": "math", + "text": "\\angle ACB = 180^\\circ - \\alpha - \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". The exterior angle at " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " is in a linear pair with " + }, + { + "type": "math", + "text": "\\angle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": ":" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "180^\\circ - \\angle ACB = \\alpha + \\beta.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-14.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The last two exercises may have looked very trivial. The reality, however, is that in practical problems they are very useful – when we have a huge non-trivial diagram and have to chain together a large number of angle operations, it is genuinely advantageous to simplify even a small piece of that work. In the parallel-lines case we don't have to introduce an auxiliary corresponding angle, and in the triangle case we don't have to work with\u00A0" + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + } + }, + { + "title": "Basics of the World of Lengths", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "So far we have only operated in the world of angles and have not dealt with lengths at all. Geometry, however, becomes interesting once we start connecting these worlds. Our foundation will be " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "triangle congruence" + } + ] + }, + { + "type": "text", + "text": ". Let us recall the criteria." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Triangle congruence theorems:" + } + ], + "highligted": false + }, + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "math", + "text": "SSS", + "isDisplay": false + }, + { + "type": "text", + "text": ": two triangles are congruent if they agree on all three sides." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": ": two triangles are congruent if they agree on two sides and the angle between them." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "math", + "text": "ASA", + "isDisplay": false + }, + { + "type": "text", + "text": ": two triangles are congruent if they agree on one side and any two angles." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": ": two triangles are congruent if they agree on two sides and the angle opposite the longer of these sides." + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerRomanParens" + } + ], + "highligted": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Let us emphasize that in " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": " it really is important that the equal angle be the one between the two sides – without this assumption, congruence may fail; see the figure. There we have two triangles " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "A'B'C'", + "isDisplay": false + }, + { + "type": "text", + "text": " for which " + }, + { + "type": "math", + "text": "BC = B'C'", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "AC = A'C'", + "isDisplay": false + }, + { + "type": "text", + "text": ", and " + }, + { + "type": "math", + "text": "\\beta = \\beta'", + "isDisplay": false + }, + { + "type": "text", + "text": ", yet they are clearly not congruent." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-15.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The " + }, + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": " theorem would be our rescue – but here we cannot apply it, because the side opposite the angle " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", namely " + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": ", is not the longest, since it is clearly shorter than " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " \uD83E\uDD72." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Before we go further, let us understand what these theorems mean. In my view, a good way to look at congruence is the following: we want to construct a triangle when given some three of its elements – will all the constructible triangles be congruent?" + } + ], + "highligted": false + }, + { + "type": "example", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Convince yourself of the validity of the " + }, + { + "type": "math", + "text": "SSS", + "isDisplay": false + }, + { + "type": "text", + "text": " congruence theorem." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Imagine we are given three segments of lengths " + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": " from which a triangle can be built. Start, say, with the segment\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " of length\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ", and then construct two circles: (a) the circle centered at\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " with radius " + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": "; (b) the circle centered at\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " with radius " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": ". They either fail to meet (when e.g. " + }, + { + "type": "math", + "text": "b+c < a", + "isDisplay": false + }, + { + "type": "text", + "text": "), or they meet at a single point on " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " (when " + }, + { + "type": "math", + "text": "b+c=a", + "isDisplay": false + }, + { + "type": "text", + "text": "), or they meet at two points " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "A'", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-16.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Finally, observe that the triangles " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "A'BC", + "isDisplay": false + }, + { + "type": "text", + "text": " are clearly mirror images of each other across " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Likewise, had we started from a different segment, we would only produce a translation/rotation of this configuration." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Note that along the way we essentially proved the triangle inequality: the sum of two sides of a triangle is greater than the third – our triangle is constructible and non-degenerate\u00A0" + }, + { + "type": "footnote", + "content": [ + { + "type": "text", + "text": "has no angle equal to " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + } + ] + }, + { + "type": "text", + "text": " if and only if " + }, + { + "type": "math", + "text": "b+c>a", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "We can justify the other criteria similarly:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Convince yourself of the validity of the " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": " congruence theorem." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Suppose the sides " + }, + { + "type": "math", + "text": "b = AC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "c = AB", + "isDisplay": false + }, + { + "type": "text", + "text": " and the angle " + }, + { + "type": "math", + "text": "\\angle BAC = \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " between them are given. Construct vertex\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " and the segment " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " of length\u00A0" + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": " – this is determined up to congruence. Point " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " must lie on a ray from\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " making angle\u00A0" + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " with " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " and at distance\u00A0" + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " from\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". There are two rays from\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " making angle " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " with " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " (one on each side of " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": "): on each there is exactly one point at distance " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " from\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", giving points " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "C'", + "isDisplay": false + }, + { + "type": "text", + "text": ". The triangles " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "ABC'", + "isDisplay": false + }, + { + "type": "text", + "text": " are mirror images across " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ", hence congruent." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-17.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Convince yourself of the validity of the " + }, + { + "type": "math", + "text": "ASA", + "isDisplay": false + }, + { + "type": "text", + "text": " congruence theorem." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "First, note that it doesn't matter which two angles are equal – if two triangles agree on two angles, then the third is determined uniquely, since all three angles sum to " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "For the purposes of our proof, suppose the equal angles are exactly the ones adjacent to the equal side. Let them be the angle " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " at vertex " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " at vertex " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": ", and the side " + }, + { + "type": "math", + "text": "a = BC", + "isDisplay": false + }, + { + "type": "text", + "text": " between them. Construct the segment " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " of length\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": " – this is determined up to congruence. Point " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " must lie on the ray from\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " making angle " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " with " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " and on the ray from\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " making angle " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " with " + }, + { + "type": "math", + "text": "CB", + "isDisplay": false + }, + { + "type": "text", + "text": ". The sum of the interior angles at " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " in a triangle is less than " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", i.e. " + }, + { + "type": "math", + "text": "\\beta + \\gamma < 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", so the two rays are not parallel – they meet at a single point\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". The triangle " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " is therefore determined uniquely (up to congruences)." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-18.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The less well-known " + }, + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": " theorem is a touch harder, so we will provide hints \uD83D\uDE42." + } + ], + "highligted": false + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": " congruence" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Convince yourself of the validity of the " + }, + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": " congruence theorem. In addition, figure out where the construction would break down if the angle were not opposite the longer of the two sides – what would happen if both sides were equal?" + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Begin the construction with the shorter segment. How does the construction proceed once it is built?" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "After constructing the shorter segment, draw a ray at our given equal angle. One last step of the construction remains. Think about when we get no triangle, when one, and when two satisfying triangles." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Suppose the sides " + }, + { + "type": "math", + "text": "a = BC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "b = AC", + "isDisplay": false + }, + { + "type": "text", + "text": " and the angle " + }, + { + "type": "math", + "text": "\\angle ABC = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " opposite the longer side " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " are given. We begin by constructing the shorter segment " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " of length\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ". Point " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " must lie on the ray from\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " making angle " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " with " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " and at distance " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " from\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": ", i.e. on the circle " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": " centered at\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " with radius " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The distance from\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " to the center of the circle is " + }, + { + "type": "math", + "text": "BC = a", + "isDisplay": false + }, + { + "type": "text", + "text": ". Since " + }, + { + "type": "math", + "text": "a < b", + "isDisplay": false + }, + { + "type": "text", + "text": ", the point\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " lies " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "inside" + } + ] + }, + { + "type": "text", + "text": " the circle " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". The ray from\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " thus starts inside the circle and continues outward, so it crosses the circle exactly once. Point " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " is therefore determined uniquely, and the triangle " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " is also unique (up to congruence)." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-19.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Let's see why the condition " + }, + { + "type": "math", + "text": "b > a", + "isDisplay": false + }, + { + "type": "text", + "text": " is essential. If " + }, + { + "type": "math", + "text": "a = b", + "isDisplay": false + }, + { + "type": "text", + "text": ", the point\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " would lie directly " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "on" + } + ] + }, + { + "type": "text", + "text": " the circle\u00A0" + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". The ray from\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " would meet it at " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " itself (which does not give a triangle) and at exactly one further point – which is the desired\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". The construction therefore still leads to a unique (isosceles) triangle, so the " + }, + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": " theorem formally holds even in this borderline case. We do not, however, list it under " + }, + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": " as a separate case: if we know that the triangle is isosceles with " + }, + { + "type": "math", + "text": "BC=AC", + "isDisplay": false + }, + { + "type": "text", + "text": " and we know one angle " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", the other two angles are already determined by the isosceles property (both base angles equal\u00A0" + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": "). The same configuration is therefore covered by " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": " or " + }, + { + "type": "math", + "text": "SSS", + "isDisplay": false + }, + { + "type": "text", + "text": ", and " + }, + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": " adds no new information." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-20.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "If " + }, + { + "type": "math", + "text": "b < a", + "isDisplay": false + }, + { + "type": "text", + "text": ", the point " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " would lie " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "outside" + } + ] + }, + { + "type": "text", + "text": " the circle " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". The ray could meet it at two points – giving two different (non-congruent) triangles – or miss it entirely, which gives no triangle at all." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-21.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " The proof can also be done by first constructing the longer segment. That, however, requires knowing the locus of points subtending a fixed angle from a fixed segment – we will get to this in another handout." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Throughout our geometric journey we will encounter all of these statements. For now, however, let us show some concrete applications. We start with the most obvious statement, which still deserves to be proved:" + } + ], + "highligted": false + }, + { + "type": "theorem", + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "isosceles triangle" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Prove that if a triangle\u00A0" + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " satisfies " + }, + { + "type": "math", + "text": "AB=AC", + "isDisplay": false + }, + { + "type": "text", + "text": ", then " + }, + { + "type": "math", + "text": "\\angle ABC=\\angle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Prove also the converse implication (i.e., that equality of angles implies equality of lengths)." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Forward implication." + } + ] + }, + { + "type": "text", + "text": " Assume " + }, + { + "type": "math", + "text": "AB = AC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Let " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": " be the midpoint of " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". The triangles " + }, + { + "type": "math", + "text": "ABM", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "ACM", + "isDisplay": false + }, + { + "type": "text", + "text": " are congruent by " + }, + { + "type": "math", + "text": "SSS", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "AB=AC", + "isDisplay": false + }, + { + "type": "text", + "text": " by assumption, " + }, + { + "type": "math", + "text": "BM=CM", + "isDisplay": false + }, + { + "type": "text", + "text": " by definition of the midpoint, common side " + }, + { + "type": "math", + "text": "AM", + "isDisplay": false + }, + { + "type": "text", + "text": "), from which " + }, + { + "type": "math", + "text": "\\angle ABC = \\angle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-22.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Converse implication." + } + ] + }, + { + "type": "text", + "text": " Assume " + }, + { + "type": "math", + "text": "\\angle ABC = \\angle ACB = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Let " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": " be the foot of the perpendicular from\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " to " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". By " + }, + { + "type": "math", + "text": "ASA", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\triangle ABM \\cong \\triangle ACM", + "isDisplay": false + }, + { + "type": "text", + "text": ", since we have two equal angles " + }, + { + "type": "math", + "text": "\\angle ABM = \\angle ACM", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "\\angle AMB = \\angle AMC = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " and the common side\u00A0" + }, + { + "type": "math", + "text": "AM", + "isDisplay": false + }, + { + "type": "text", + "text": ", so " + }, + { + "type": "math", + "text": "AB = AC", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Another cute proof is based on showing that " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": " (note the different order of vertices). In the case where we know the equal sides, we use " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": " or even " + }, + { + "type": "math", + "text": "SSS", + "isDisplay": false + }, + { + "type": "text", + "text": " – which then gives equal angles. In the case where we know the equal angles, it will instead be " + }, + { + "type": "math", + "text": "ASA", + "isDisplay": false + }, + { + "type": "text", + "text": ", and the congruence gives equal sides." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Try proving these simple facts rigorously – it will be easier this time:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "An equilateral triangle has three equal angles of " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In " + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " with " + }, + { + "type": "math", + "text": "AB = AC = BC", + "isDisplay": false + }, + { + "type": "text", + "text": ", the isosceles triangle theorem gives " + }, + { + "type": "math", + "text": "\\angle ABC = \\angle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": " (from " + }, + { + "type": "math", + "text": "AB=AC", + "isDisplay": false + }, + { + "type": "text", + "text": "); analogously, from " + }, + { + "type": "math", + "text": "AB=BC", + "isDisplay": false + }, + { + "type": "text", + "text": " we get " + }, + { + "type": "math", + "text": "\\angle BAC = \\angle BCA", + "isDisplay": false + }, + { + "type": "text", + "text": ". All three angles are therefore equal, and from the angle sum of " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " each is " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-23.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "An isosceles right triangle has angles " + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Suppose the right angle is at vertex\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", i.e. " + }, + { + "type": "math", + "text": "\\angle BAC = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", and the legs are equal, " + }, + { + "type": "math", + "text": "AB = AC", + "isDisplay": false + }, + { + "type": "text", + "text": ". By the isosceles triangle theorem, " + }, + { + "type": "math", + "text": "\\angle ABC = \\angle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Denote this common measure " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". From the angle sum in the triangle," + } + ], + "highligted": false + }, + { + "type": "math", + "text": "90^\\circ + \\beta + \\beta = 180^\\circ,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "so " + }, + { + "type": "math", + "text": "2\\beta = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", i.e. " + }, + { + "type": "math", + "text": "\\beta = 45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-24.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "A very handy fact useful in non-trivial problems is the following. One possible proof goes via trigonometry." + } + ], + "highligted": false + }, + { + "type": "problem", + "difficulty": 0, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "A right triangle has its other two angles equal to " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " if and only if its hypotenuse is twice as long as the shorter leg." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Say " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " is the hypotenuse and is twice as long as the leg " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". The trick is to introduce a point " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": " such that " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " is the midpoint of segment " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Let our triangle have its right angle at vertex\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "(" + }, + { + "type": "math", + "text": "\\Rightarrow", + "isDisplay": false + }, + { + "type": "text", + "text": ")" + } + ] + }, + { + "type": "text", + "text": " Assume " + }, + { + "type": "math", + "text": "BC = 2\\cdot AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Let " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": " be the point such that " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " is the midpoint of segment " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": ". Then " + }, + { + "type": "math", + "text": "AB' = AB", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " lies between " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": ", so " + }, + { + "type": "math", + "text": "\\angle B'AC", + "isDisplay": false + }, + { + "type": "text", + "text": " is in a linear pair with " + }, + { + "type": "math", + "text": "\\angle BAC = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", hence also equals " + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". We have " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle AB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " by " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": ", since the angles at\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " are both right, the side\u00A0" + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": " is common, and " + }, + { + "type": "math", + "text": "AB=AB'", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.en-25.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Since " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " is the midpoint of " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": ", we have " + }, + { + "type": "math", + "text": "BB' = 2\\cdot AB", + "isDisplay": false + }, + { + "type": "text", + "text": ", and from the assumption " + }, + { + "type": "math", + "text": "BC = 2\\cdot AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Together," + } + ], + "highligted": false + }, + { + "type": "math", + "text": "BB' = BC = B'C,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "so " + }, + { + "type": "math", + "text": "\\triangle BB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " is equilateral and all its angles are " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". In particular, " + }, + { + "type": "math", + "text": "\\angle ABC = \\angle B'BC = 60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". From the angle sum in " + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " we compute " + }, + { + "type": "math", + "text": "\\angle BCA = 180^\\circ - 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "(" + }, + { + "type": "math", + "text": "\\Leftarrow", + "isDisplay": false + }, + { + "type": "text", + "text": ")" + } + ] + }, + { + "type": "text", + "text": " It is not hard to see that the reasoning of the previous paragraph can easily be reversed – this time the key congruence " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle AB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " comes from " + }, + { + "type": "math", + "text": "ASA", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Another solution is to note that by Thales' theorem, the circumcenter\u00A0" + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " of triangle " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " is also the midpoint of the hypotenuse " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". We will return to this solution in more detail when we discuss circles." + } + ], + "highligted": false + } + ] + } + ] + } + }, + { + "title": "What to Remember", + "level": 1, + "text": { + "content": [] + } + }, + { + "title": "Techniques", + "level": 2, + "text": { + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "With parallel lines, look for equal alternate and corresponding angles, or angles summing to\u00A0" + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "It is useful to view the exterior angle of a triangle as the sum of the two non-adjacent interior angles." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "To bridge the worlds of lengths and angles, congruence of two triangles is the right tool." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Auxiliary points (midpoint of a segment, mirror image, extension) often reveal a congruence or a special triangle." + } + ], + "highligted": false + } + ] + ], + "styleType": "NumberParens" + } + ] + } + }, + { + "title": "Useful Facts", + "level": 2, + "text": { + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The angle sum in a triangle is " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", in general in a convex " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-gon " + }, + { + "type": "math", + "text": "(n-2) \\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Triangle congruence theorems: " + }, + { + "type": "math", + "text": "SSS", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "ASA", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "SsA", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In a triangle, two sides are equal if and only if the angles opposite them are equal." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "A right triangle has its hypotenuse twice as long as a leg if and only if its angles are " + }, + { + "type": "math", + "text": "90^\\circ, 60^\\circ, 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "styleType": "NumberParens" + } + ] + } + } + ] + }, + "images": [ + { + "contentId": "angle-basics-1.en-1.svg", + "originalId": "angles-vertical.pdf", + "width": "82.61pt", + "height": "65.04pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-2.svg", + "originalId": "angles-corresponding.pdf", + "width": "106.36pt", + "height": "66.59pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-3.svg", + "originalId": "angles-alternate.pdf", + "width": "100.63pt", + "height": "58.17pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-4.svg", + "originalId": "angles-supplementary.pdf", + "width": "85.68pt", + "height": "47.81pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-5.svg", + "originalId": "angles-supplementary-derive-vertical.pdf", + "width": "95.91pt", + "height": "78.49pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-6.svg", + "originalId": "angles-vertical-derive-supplementary.pdf", + "width": "95.91pt", + "height": "78.49pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-7.svg", + "originalId": "angles-corresponding-alternate-supplementary-connection.pdf", + "width": "98.31pt", + "height": "65.02pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-8.svg", + "originalId": "angles-triangle.pdf", + "width": "182.63pt", + "height": "128.67pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-9.svg", + "originalId": "angles-polygon.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-10.svg", + "originalId": "angles-polygon-fan.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-11.svg", + "originalId": "angles-polygon-interior.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-12.svg", + "originalId": "angles-complementary-parallel-statement.pdf", + "width": "91.52pt", + "height": "72.46pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-13.svg", + "originalId": "angles-complementary-parallel-solution.pdf", + "width": "85.7pt", + "height": "92.24pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-14.svg", + "originalId": "angles-complementary-in-triangle.pdf", + "width": "153.73pt", + "height": "123.67pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-15.svg", + "originalId": "angles-sas-warning.pdf", + "width": "240.45pt", + "height": "68.01pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-16.svg", + "originalId": "angles-sss-proof.pdf", + "width": "180.33pt", + "height": "129.76pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-17.svg", + "originalId": "angles-sas-proof.pdf", + "width": "190.6pt", + "height": "165.53pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-18.svg", + "originalId": "angles-asa-proof.pdf", + "width": "156.76pt", + "height": "165.02pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-19.svg", + "originalId": "angles-Ssa-proof-one-solution.pdf", + "width": "116.57pt", + "height": "129.18pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-20.svg", + "originalId": "angles-Ssa-proof-isosceles.pdf", + "width": "116.86pt", + "height": "122.28pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-21.svg", + "originalId": "angles-Ssa-proof-two-solutions.pdf", + "width": "191.58pt", + "height": "154.75pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-22.svg", + "originalId": "angles-isosceles-triangle.pdf", + "width": "117.87pt", + "height": "114.32pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-23.svg", + "originalId": "angles-equilateral.pdf", + "width": "107.91pt", + "height": "97.32pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-24.svg", + "originalId": "angles-isosceles-right-triangle.pdf", + "width": "97.95pt", + "height": "95.36pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.en-25.svg", + "originalId": "angles-90-60-30.pdf", + "width": "146.09pt", + "height": "131.24pt", + "scale": 1.0 + } + ] +} diff --git a/web/src/content/handouts/angle-basics-1.sk.json b/web/src/content/handouts/angle-basics-1.sk.json new file mode 100644 index 00000000..212a214c --- /dev/null +++ b/web/src/content/handouts/angle-basics-1.sk.json @@ -0,0 +1,4965 @@ +{ + "document": { + "title": "Základy počítania uhlov", + "sections": [ + { + "title": "Úvod", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Geometria je najvizuálnejšia časť matematiky – vieme si ju kresliť na papier, v\u00A0softwari ako " + }, + { + "type": "link", + "url": "https://www.geogebra.org/", + "content": [ + { + "type": "text", + "text": "GeoGebra" + } + ] + }, + { + "type": "text", + "text": " a\u00A0môžeme sa tešiť, keď na vlastné oči vidíme, že veci v\u00A0nej fungujú. Prečo tomu tak je však vie byť záhadou. Aby sme tieto tajomstvá pochopili, potrebujeme kúsok po kúsky budovať základňu znalostí a\u00A0vizuálnu intuíciu." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Jedna z\u00A0najzákladnejších techník je rozumieť svetu uhlov a\u00A0vedieť ich vlastnosti s\u00A0úspechom používať v\u00A0rôznych typoch úloh. Len málo zaujímavých príkladov uhly nepoužíva. Teória okolo uhlov pritom nie je zložitá, je však dôležité budovať ju pomaly a detailne. Začneme s\u00A0najjednoduchšími vlastnosťami, ktoré ani nevyžadujú prácu s kružnicami. Už tieto sa však dajú s\u00A0úspechom využiť vo veľa úlohách." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0tomto materiáli sa budeme venovať uhlom, ale zámerne vynecháme uhly na kružniciach, ktoré si zaslúžia separátnu pozornosť. Na chvíľu zabudnime, že kružnice existujú." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Na záver úvodu jedna menšia poznámka o konvencii. V\u00A0tomto materiáli uvidíme veľa obrázkov trojuholníkov, v\u00A0ktorých je bod\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " nakreslený hore. Ukazuje sa, že tento spôsob kreslenia je medzinárodný štandard, najmä na poli matematickej olympiády a podobných súťaží sa to tak zvykne robiť. V\u00A0Česku a na\u00A0Slovensku je však v\u00A0školách stále zvykom kresliť\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " hore. V\u00A0niektorých krajinách možno vidieť aj\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " hore, napr. na Ukrajine. Z\u00A0hľadiska riešenia úloh je vždy dobré nakresliť si hore ten bod, pri ktorom obrázok vyzerá čo najsymetrickejšie, ľahšie v\u00A0ňom vidieť ďalšie symetrické veci. V\u00A0tomto materiáli budú úlohy zadávané tak, že tá symetria bude vidieť viac pri\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " hore \uD83D\uDE42." + } + ], + "highligted": false + } + ] + } + }, + { + "title": "Základy sveta uhlov", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0tejto sekcii si odvodíme tie najjednoduchšie vlastnosti súvisiace s uhlami. Pevne verím, že kľúčom k\u00A0zvládnutiu geometrie je porozumenie vecí od základov a do hĺbky, preto sa týmto jednoduchým vlastnostiam budeme venovať viac." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Základné vlastnosti uhlov, ktoré bežne používame skoro bez rozmýšľania pri riešení úloh:" + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vrcholové uhly sú zhodné: " + }, + { + "type": "image", + "id": "angle-basics-1.sk-1.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Súhlasné uhly sú zhodné: " + }, + { + "type": "image", + "id": "angle-basics-1.sk-2.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Striedavé uhly sú zhodné: " + }, + { + "type": "image", + "id": "angle-basics-1.sk-3.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vedľajšie uhly majú súčet " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ": " + }, + { + "type": "image", + "id": "angle-basics-1.sk-4.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerRomanParens" + } + ], + "highligted": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Tieto vlastnosti spolu veľmi prirodzene súvisia, menšie cvičenie na rozmyslenie:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Uvedomte si, že:" + } + ], + "highligted": false + }, + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "z\u00A0vlastnosti o\u00A0vedľajších uhloch vyplýva vlastnosť vrcholových uhlov (a\u00A0naopak)" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "z\u00A0ktorejkoľvek dvojice vlastností o\u00A0vrcholových, súhlasných, striedavých uhloch vyplýva tá tretia vlastnosť" + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerAlphaParens" + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "bold", + "content": [ + { + "type": "text", + "text": "(a)" + } + ] + }, + { + "type": "text", + "text": " Vedľajší uhol " + }, + { + "type": "math", + "text": "\\alpha'", + "isDisplay": false + }, + { + "type": "text", + "text": " k\u00A0danému uhlu " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " má veľkosť " + }, + { + "type": "math", + "text": "180^\\circ - \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ". Uhol vedľajší k\u00A0" + }, + { + "type": "math", + "text": "\\alpha'", + "isDisplay": false + }, + { + "type": "text", + "text": " má zasa veľkosť " + }, + { + "type": "math", + "text": "180^\\circ - (180^\\circ - \\alpha) = \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ten je však zároveň vrcholový uhol k\u00A0" + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", teda sú vrcholové uhly zhodné." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-5.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Naopak, predpokladajme vlastnosť o\u00A0vrcholových uhloch a\u00A0označme po sebe idúce uhly v\u00A0priesečníku dvoch priamok " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\delta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Zo zhodnosti vrcholových uhlov " + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "\\beta = \\delta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Súčet všetkých štyroch je plný uhol, teda " + }, + { + "type": "math", + "text": "\\alpha + \\beta + \\gamma + \\delta = 360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Po dosadení dostávame " + }, + { + "type": "math", + "text": "2\\alpha + 2\\beta = 360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čiže " + }, + { + "type": "math", + "text": "\\alpha + \\beta = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " – to je práve vlastnosť vedľajších uhlov." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-6.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "bold", + "content": [ + { + "type": "text", + "text": "(b)" + } + ] + }, + { + "type": "text", + "text": " Všetky tri vlastnosti hovoria, že istá dvojica uhlov má rovnakú veľkosť." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pozrime sa na trojicu uhlov " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " z\u00A0obrázka a\u00A0na to, čo o\u00A0ich vzťahoch tvrdia jednotlivé vlastnosti. Pre dvojicu " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " pri hornom priesečníku tvrdí vlastnosť vrcholových uhlov " + }, + { + "type": "math", + "text": "\\alpha = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Pre dvojicu " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " tvrdí vlastnosť súhlasných uhlov " + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Konečne pre dvojicu " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " tvrdí vlastnosť striedavých uhlov " + }, + { + "type": "math", + "text": "\\beta = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-7.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Tri vlastnosti teda tvrdia tri rovnosti medzi tými istými troma uhlami, len cez rôzne dvojice. Akonáhle platia ľubovoľné dve, tretia plynie tranzitivitou: napr. ak vieme (V) a\u00A0(S), tak " + }, + { + "type": "math", + "text": "\\alpha = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "\\alpha = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže " + }, + { + "type": "math", + "text": "\\beta = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", čiže (St). Ostatné dvojice analogicky." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nasledujúce tvrdenie všetci poznáme, viete ho však dokázať?" + } + ], + "highligted": false + }, + { + "type": "theorem", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Súčet uhlov v\u00A0trojuholníku je " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Veďme cez vrchol\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " priamku rovnobežnú s\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Pretože " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " je priečka rovnobežiek, uhol " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " pri\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0uhol " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " pri\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " sú striedavé, teda rovnaké. Analogicky cez priečku " + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": " sú uhly " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " pri\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0pri\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " striedavé. Uhly " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " ležia vedľa seba pozdĺž priamky cez\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", teda " + }, + { + "type": "math", + "text": "\\alpha + \\beta + \\gamma = 180^\\circ.", + "isDisplay": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-8.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Toto tvrdenie sa rozhodne dá zovšeobecniť. Vieme napríklad, že súčet uhlov v\u00A0štvoruholníku je " + }, + { + "type": "math", + "text": "360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ako je to v 5-uholníku? A ako v\u00A067-uholníku? Odpoveďou je nasledujúce tvrdenie:" + } + ], + "highligted": false + }, + { + "type": "theorem", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nech " + }, + { + "type": "math", + "text": "n \\ge 3", + "isDisplay": false + }, + { + "type": "text", + "text": " je prirodzené číslo. Súčet uhlov v\u00A0konvexnom " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníku je rovný " + }, + { + "type": "math", + "text": "(n-2) \\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Dôkaz 1 (matematická indukcia)." + } + ] + }, + { + "type": "text", + "text": " Základ " + }, + { + "type": "math", + "text": "n=3", + "isDisplay": false + }, + { + "type": "text", + "text": ": súčet uhlov trojuholníka je " + }, + { + "type": "math", + "text": "180^\\circ = (3-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Indukčný krok: nech " + }, + { + "type": "math", + "text": "A_1 A_2 \\ldots A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " je konvexný " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholník. Uhlopriečkou " + }, + { + "type": "math", + "text": "A_1 A_n", + "isDisplay": false + }, + { + "type": "text", + "text": " ho rozdelíme na trojuholník " + }, + { + "type": "math", + "text": "A_1 A_n A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0konvexný " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholník " + }, + { + "type": "math", + "text": "A_1 A_2 \\ldots A_n", + "isDisplay": false + }, + { + "type": "text", + "text": ": uhly pri vrcholoch " + }, + { + "type": "math", + "text": "A_2, \\ldots, A_{n-1}", + "isDisplay": false + }, + { + "type": "text", + "text": " patria celé " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníku, uhol pri " + }, + { + "type": "math", + "text": "A_{n+1}", + "isDisplay": false + }, + { + "type": "text", + "text": " patrí celý trojuholníku a\u00A0uhly pri " + }, + { + "type": "math", + "text": "A_1", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "A_n", + "isDisplay": false + }, + { + "type": "text", + "text": " sa rozdelia medzi oba útvary tak, aby ich časti dali pôvodné vnútorné uhly " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka. Súčet uhlov " + }, + { + "type": "math", + "text": "(n+1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka je teda" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\underbrace{180^\\circ}_{\\text{trojuholník}} + \\underbrace{(n-2) \\cdot 180^\\circ}_{n\\text{-uholník}} = (n-1) \\cdot 180^\\circ = \\bigl((n+1) - 2\\bigr) \\cdot 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-9.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Dôkaz 2 (vejárová triangulácia)." + } + ] + }, + { + "type": "text", + "text": " Veďme z\u00A0vrcholu " + }, + { + "type": "math", + "text": "A_1", + "isDisplay": false + }, + { + "type": "text", + "text": " uhlopriečky do každého nepriliehajúceho vrcholu " + }, + { + "type": "math", + "text": "A_3, A_4, \\ldots, A_{n-1}", + "isDisplay": false + }, + { + "type": "text", + "text": ". Vznikne " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " trojuholníkov pokrývajúcich vnútro " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka bez prekrytia. Pri každom vrchole " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " sa uhly susedných trojuholníkov pri " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " presne poskladajú na vnútorný uhol " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka pri " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže súčet uhlov všetkých trojuholníkov sa rovná súčtu vnútorných uhlov " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka, čiže " + }, + { + "type": "math", + "text": "(n-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-10.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Dôkaz\u00A02 je " + }, + { + "type": "quote", + "content": [ + { + "type": "text", + "text": "rozbalený" + } + ] + }, + { + "type": "text", + "text": " dôkaz\u00A01: indukcia odrezáva trojuholníky jeden po druhom, tu ich vidíme všetky naraz." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Dôkaz 3 (vnútorný bod)." + } + ] + }, + { + "type": "text", + "text": " Zvoľme bod " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " vo vnútri " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka a\u00A0spojme ho s\u00A0každým vrcholom. Vznikne " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": " trojuholníkov s\u00A0celkovým súčtom uhlov " + }, + { + "type": "math", + "text": "n\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Uhly trojuholníkov pri každom vrchole " + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": " spolu tvoria vnútorný uhol " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka pri\u00A0" + }, + { + "type": "math", + "text": "A_k", + "isDisplay": false + }, + { + "type": "text", + "text": "; uhly trojuholníkov pri\u00A0" + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " spolu tvoria plný uhol " + }, + { + "type": "math", + "text": "360^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Teda súčet vnútorných uhlov " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka je" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "n\\cdot 180^\\circ - 360^\\circ = (n-2)\\cdot 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-11.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka (nekonvexné mnohouholníky)." + } + ] + }, + { + "type": "text", + "text": " Vzorec " + }, + { + "type": "math", + "text": "(n-2)\\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " platí aj pre navzájom sa nepretínajúce nekonvexné mnohouholníky, ak všetky vnútorné uhly meriame smerom dovnútra – teda aj uhly väčšie než " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " rátame ako také." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vejárová triangulácia z\u00A0jedného vrcholu nemusí fungovať ani z\u00A0konvexného vrcholu – niektoré uhlopriečky môžu prechádzať mimo mnohouholníka. Platí však, že každý jednoduchý " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholník sa dá rozdeliť na " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " trojuholníkov (nie nutne vejárovo), čo dôkaz zachráni. Dôvod je indukcia cez trojuholník tvorený nejakým vrcholom " + }, + { + "type": "math", + "text": "V", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0jeho susedmi " + }, + { + "type": "math", + "text": "U", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "W", + "isDisplay": false + }, + { + "type": "text", + "text": ", ktorý leží celý vo vnútri nášho " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníka. Odrezaním " + }, + { + "type": "math", + "text": "V", + "isDisplay": false + }, + { + "type": "text", + "text": " uhlopriečkou " + }, + { + "type": "math", + "text": "UW", + "isDisplay": false + }, + { + "type": "text", + "text": " dostaneme " + }, + { + "type": "math", + "text": "(n-1)", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholník; opakovaním teda " + }, + { + "type": "math", + "text": "n-2", + "isDisplay": false + }, + { + "type": "text", + "text": " trojuholníkov. Dá sa dokázať, že takýto trojuholník vždy existuje (dokonca dva), viď " + }, + { + "type": "link", + "url": "https://en.wikipedia.org/wiki/Two_ears_theorem", + "content": [ + { + "type": "text", + "text": "veta o\u00A0dvoch ušiach" + } + ] + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "K\u00A0dôkazu\u00A03: funguje len ak existuje vnútorný bod " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " viditeľný zo všetkých vrcholov (t.\u00A0j. úsečky " + }, + { + "type": "math", + "text": "OA_i", + "isDisplay": false + }, + { + "type": "text", + "text": " ležia celé vo vnútri). Pre niektoré nekonvexné mnohouholníky taký " + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " nemusí existovať (skúste taký nakresliť) – v\u00A0takom prípade sa dôkaz\u00A03 nedá priamo zachrániť a\u00A0treba siahnuť po triangulácii z\u00A0dôkazu\u00A02." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Teraz uveďme ešte dve jednoduché ale užitočné pomocné tvrdenia:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dokážte, že súčet uhlov na obrázku je rovný " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-12.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Označme " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " uhol zvieraný priečkou " + }, + { + "type": "math", + "text": "AD", + "isDisplay": false + }, + { + "type": "text", + "text": " s\u00A0rovnobežkami na strane " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " (ako na obrázku). Súhlasné uhly " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " pri\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " pri\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " sú zhodné. Pri vrchole " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " sú " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " vedľajšie, teda" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\beta + \\alpha = 180^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-13.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dokážte, že uhol s\u00A0otázníkom na obrázku je rovný " + }, + { + "type": "math", + "text": "\\alpha+\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-14.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Zo súčtu uhlov v\u00A0trojuholníku " + }, + { + "type": "math", + "text": "|\\angle ACB| = 180^\\circ - \\alpha - \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Vonkajší uhol pri " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " je vedľajší k\u00A0" + }, + { + "type": "math", + "text": "|\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": ":" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "180^\\circ - |\\angle ACB| = \\alpha + \\beta.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-14.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Posledné dve cvičenia možno vyzerali veľmi triviálne. Realita však je, že v\u00A0praktických úlohách sú veľmi užitočné – keď máme obrovský netriviálny obrázok a\u00A0máme previesť za sebou veľký počet uhlových operácií, tak je naozaj veľmi výhodné si čo i len kúsok tej práce zjednodušiť – v\u00A0prípade rovnobežiek nemusíme uvažovať pomocný súhlasný uhol a\u00A0v\u00A0prípade trojuholníka nemusíme pracovať s\u00A0" + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + } + }, + { + "title": "Základy sveta dĺžok", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Zatiaľ sme fungovali iba vo svete uhlov a\u00A0vôbec sme neriešili dĺžky. Geometria však začne byť zaujímavá, keď tieto svety začneme prepájať. Základom pre nás bude " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "zhodnosť trojuholníkov" + } + ] + }, + { + "type": "text", + "text": ". Pripomeňme si kritériá." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vety o\u00A0zhodnosti trojuholníkov:" + } + ], + "highligted": false + }, + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Veta " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": ": dva trojuholníky sú zhodné, ak sa zhodujú vo všetkých troch stranách." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Veta " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": ": dva trojuholníky sú zhodné, ak sa zhodujú v\u00A0dvoch stranách a\u00A0uhle, ktorý tieto strany zvierajú." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Veta " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": ": dva trojuholníky sú zhodné, ak sa zhodujú v\u00A0jednej strane a\u00A0nejakých dvoch uhloch." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Veta " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": ": dva trojuholníky sú zhodné, ak sa zhodujú v\u00A0dvoch stranách a\u00A0uhle oproti dlhšej z\u00A0týchto strán." + } + ], + "highligted": false + } + ] + ], + "styleType": "LowerRomanParens" + } + ], + "highligted": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Zdôraznime, že pri vete " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " je naozaj dôležité, že zhodný uhol je ten zvieraný dvoma stranami – bez tohto predpokladu zhodnosť nemusí fungovať, viď obrázok. Na ňom máme dva trojuholníky " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " a " + }, + { + "type": "math", + "text": "A'B'C'", + "isDisplay": false + }, + { + "type": "text", + "text": ", pre ktoré platí " + }, + { + "type": "math", + "text": "|BC|=|B'C'|", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "|AC|=|A'C'|", + "isDisplay": false + }, + { + "type": "text", + "text": ", a " + }, + { + "type": "math", + "text": "\\beta=\\beta'", + "isDisplay": false + }, + { + "type": "text", + "text": ", avšak zjavne nie sú zhodné." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-15.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Veta " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " by bola naša záchrana – tu ju však aplikovať nevieme, lebo strana, oproti ktorej je uhol " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", teda " + }, + { + "type": "math", + "text": "|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ", nie je najdlhšia, keďže je zjavne kratšia než " + }, + { + "type": "math", + "text": "|BC|", + "isDisplay": false + }, + { + "type": "text", + "text": " \uD83E\uDD72." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Než pôjdeme ďalej, uvedomme si, čo tieto vety znamenajú. Za mňa dobrý pohľad na zhodnosť je takýto: chceme zostrojiť trojuholník, keď máme dané nejaké tri jeho elementy – budú všetky zostrojiteľné trojuholníky zhodné?" + } + ], + "highligted": false + }, + { + "type": "example", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Presvedčte sa o\u00A0platnosti vety " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": " o\u00A0zhodnosti." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Predstavme si, že máme dané tri úsečky dĺžok " + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": ", z\u00A0ktorých vieme zostrojiť trojuholník. Začneme napr. úsečkou\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " dĺžky\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0následne zostrojíme dve kružnice: (a) kružnicu so stredom v\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0polomerom " + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": "; (b) kružnicu so stredom v\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0polomerom " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": ". Tie sa buď nepretnú (keď napr. " + }, + { + "type": "math", + "text": "b+c < a", + "isDisplay": false + }, + { + "type": "text", + "text": "), alebo sa pretnú v\u00A0jedinom bode na " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " (keď " + }, + { + "type": "math", + "text": "b+c=a", + "isDisplay": false + }, + { + "type": "text", + "text": "), alebo sa pretnú v\u00A0dvoch bodoch " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A'", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-16.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Na konci si uvedomme, že trojuholníky " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A'BC", + "isDisplay": false + }, + { + "type": "text", + "text": " sú však zrejme vzájomné zrkadlové obrazy podľa " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Taktiež ak by sme začali inou úsečkou, tak vytvoríme iba posunutie/otočenie tejto konfigurácie." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Uvedomme si, že sme vlastne po ceste dokázali\u00A0trojuholníkovú nerovnosť: súčet dvoch strán trojuholníka je väčší než tretia – náš trojuholník je zostrojiteľný a\u00A0nedegenerovaný\u00A0" + }, + { + "type": "footnote", + "content": [ + { + "type": "text", + "text": "nemá uhol rovný " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + } + ] + }, + { + "type": "text", + "text": " práve vtedy, keď " + }, + { + "type": "math", + "text": "b+c>a", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podobne vieme zdôvodniť ďalšie kritériá:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Presvedčte sa o\u00A0platnosti vety " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " o\u00A0zhodnosti." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nech sú dané strany " + }, + { + "type": "math", + "text": "b = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "c = |AB|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0uhol " + }, + { + "type": "math", + "text": "|\\angle BAC| = \\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " medzi nimi. Zostrojme vrchol\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0úsečku " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " dĺžky\u00A0" + }, + { + "type": "math", + "text": "c", + "isDisplay": false + }, + { + "type": "text", + "text": " – tá je daná až na zhodné zobrazenie. Bod " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " musí ležať na polopriamke z\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " zvierajúcej s\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " uhol\u00A0" + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0zároveň vo vzdialenosti\u00A0" + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " od\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". Polopriamky zvierajúce s\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " uhol " + }, + { + "type": "math", + "text": "\\alpha", + "isDisplay": false + }, + { + "type": "text", + "text": " sú dve (po jednej na každej strane " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": "): na každej leží práve jeden bod vo vzdialenosti " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " od\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", teda dostaneme body " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "C'", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trojuholníky " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "ABC'", + "isDisplay": false + }, + { + "type": "text", + "text": " sú zrkadlové obrazy podľa " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ", teda zhodné." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-17.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Presvedčte sa o\u00A0platnosti vety " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": " o\u00A0zhodnosti." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0prvom rade si uvedomme, že nezáleží na tom, ktoré dva uhly sú zhodné – ak sa dva trojuholníky zhodujú v\u00A0dvoch uhloch, tak tretí je určený jednoznačne, keďže všetky tri majú súčet uhlov " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pre účely nášho dôkazu uvážme teda, že sú zhodné práve uhly prihliehajúce zhodnej strane. Nech je to uhol " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " pri vrchole " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " pri vrchole " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0strana " + }, + { + "type": "math", + "text": "a = |BC|", + "isDisplay": false + }, + { + "type": "text", + "text": " medzi nimi. Zostrojme úsečku " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " dĺžky\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": " – tá je daná až na zhodné zobrazenie. Bod " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " musí ležať na polopriamke z\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " zvierajúcej s\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " uhol " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0zároveň na polopriamke z\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " zvierajúcej s\u00A0" + }, + { + "type": "math", + "text": "CB", + "isDisplay": false + }, + { + "type": "text", + "text": " uhol " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Súčet vnútorných uhlov pri " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " v\u00A0trojuholníku je menší než " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čiže " + }, + { + "type": "math", + "text": "\\beta + \\gamma < 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0preto nie sú obe polopriamky rovnobežné – pretnú sa v\u00A0jedinom bode\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trojuholník " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " je teda určený jednoznačne (až na zhodné zobrazenia)." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-18.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Menej známe tvrdenie " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " je už máličko ťažšie, preto k\u00A0nemu budú aj návody \uD83D\uDE42." + } + ], + "highligted": false + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "zhodnosť " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Presvedčte sa o\u00A0platnosti vety " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " o\u00A0zhodnosti. Navyše, uvedomte si, kde by sa konštrukcia pokazila, keby uhol nebol voči väčšej zo strán – čo by sa stalo, keby boli obe strany rovnako dlhé?" + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Začnite konštrukciu konštrukciou kratšej úsečky. Ako pokračuje konštrukcia po jej zostrojení?" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Po zostrojení kratšej úsečky zostrojíme polpriamku pod našim daným zhodným uhlom. Zostáva posledný krok konštrukcie. Rozmyslite, kedy dostaneme žiaden, kedy jeden a\u00A0kedy dva vyhovujúce trojuholníky." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nech sú dané strany " + }, + { + "type": "math", + "text": "a = |BC|", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "b = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0uhol " + }, + { + "type": "math", + "text": "|\\angle ABC| = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " oproti dlhšej strane " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": ". Začneme zostrojením kratšej úsečky " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " dĺžky\u00A0" + }, + { + "type": "math", + "text": "a", + "isDisplay": false + }, + { + "type": "text", + "text": ". Bod " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " musí ležať na polopriamke z\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " zvierajúcej s\u00A0" + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " uhol " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0zároveň vo vzdialenosti " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": " od\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": ", čiže na kružnici " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": " so stredom\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0polomerom " + }, + { + "type": "math", + "text": "b", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vzdialenosť bodu\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " od stredu kružnice je " + }, + { + "type": "math", + "text": "|BC| = a", + "isDisplay": false + }, + { + "type": "text", + "text": ". Pretože " + }, + { + "type": "math", + "text": "a < b", + "isDisplay": false + }, + { + "type": "text", + "text": ", bod\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " leží " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "vnútri" + } + ] + }, + { + "type": "text", + "text": " kružnice " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". Polopriamka z\u00A0bodu\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " tak začína vnútri kružnice a\u00A0pokračuje smerom von, takže kružnicu pretne práve raz. Bod " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " je teda určený jednoznačne, a\u00A0trojuholník " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " je takisto jednoznačný (až na zhodné zobrazenie)." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-19.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pozrime sa, prečo je podmienka " + }, + { + "type": "math", + "text": "b > a", + "isDisplay": false + }, + { + "type": "text", + "text": " podstatná. Ak by bolo " + }, + { + "type": "math", + "text": "a = b", + "isDisplay": false + }, + { + "type": "text", + "text": ", bod\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " by ležal priamo " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "na" + } + ] + }, + { + "type": "text", + "text": " kružnici\u00A0" + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". Polopriamka z\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " by ju pretínala v\u00A0bode\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " samotnom (čo nedáva trojuholník) a\u00A0v\u00A0práve jednom ďalšom bode – ten je hľadaným\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ". Konštrukcia teda stále vedie k\u00A0jednoznačnému (rovnoramennému) trojuholníku, takže veta " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " formálne platí aj v\u00A0tomto hraničnom prípade. Pod " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " ho však samostatne nezaraďujeme: ak vieme, že trojuholník je rovnoramenný so " + }, + { + "type": "math", + "text": "|BC|=|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0poznáme jeden uhol " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ", tak ostatné dva uhly sú už vďaka rovnoramennosti určené (oba pri základni majú veľkosť\u00A0" + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": "). Tá istá konfigurácia je preto pokrytá vetami " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " či " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": " k\u00A0nej nepridáva žiadnu novú informáciu." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-20.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Ak by bolo " + }, + { + "type": "math", + "text": "b < a", + "isDisplay": false + }, + { + "type": "text", + "text": ", bod " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": " by ležal " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "vonku" + } + ] + }, + { + "type": "text", + "text": " kružnice " + }, + { + "type": "math", + "text": "k", + "isDisplay": false + }, + { + "type": "text", + "text": ". Polopriamka by ju mohla pretnúť v\u00A0dvoch bodoch – vznikli by tak dva rôzne (nezhodné) trojuholníky – alebo by ju minula celkom, čo nevedie k\u00A0žiadnemu trojuholníku." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-21.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Dôkaz je možné spraviť aj tak, že najprv zostrojíme dlhšiu úsečku. To však vyžaduje znalosť množiny bodov nad pevnou úsečkou majúcich pevný uhol – k\u00A0tomuto sa dostaneme v\u00A0inom materiáli." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V našej geometrickej ceste sa priebežne stretneme so všetkými týmito tvrdeniami. Nateraz si však ukážme nejaké konkrétne aplikácie. Začneme tým najzrejmejším tvrdením, ktoré sa ale tiež sluší a patrí dokázať:" + } + ], + "highligted": false + }, + { + "type": "theorem", + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "rovnoramenný trojuholník" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dokážte, že ak pre trojuholník\u00A0" + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " platí " + }, + { + "type": "math", + "text": "|AB|=|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ", tak " + }, + { + "type": "math", + "text": "|\\angle ABC|=|\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Dokážte tiež opačnú implikáciu (teda že z\u00A0rovnosti uhlov vyplýva rovnosť dĺžok)." + } + ], + "highligted": false + } + ], + "proof": [ + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Priama implikácia." + } + ] + }, + { + "type": "text", + "text": " Predpokladajme " + }, + { + "type": "math", + "text": "|AB| = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nech " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": " je stred " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trojuholníky " + }, + { + "type": "math", + "text": "ABM", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "ACM", + "isDisplay": false + }, + { + "type": "text", + "text": " sú zhodné podľa " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "|AB|=|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": " z\u00A0predpokladu, " + }, + { + "type": "math", + "text": "|BM|=|CM|", + "isDisplay": false + }, + { + "type": "text", + "text": " zo\u00A0stredu, spoločná strana " + }, + { + "type": "math", + "text": "AM", + "isDisplay": false + }, + { + "type": "text", + "text": "), odkiaľ " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-22.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Opačná implikácia." + } + ] + }, + { + "type": "text", + "text": " Predpokladajme " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle ACB| = \\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nech " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": " je päta kolmice z\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " na " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Podľa vety " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": " je " + }, + { + "type": "math", + "text": "\\triangle ABM \\cong \\triangle ACM", + "isDisplay": false + }, + { + "type": "text", + "text": ", keďže máme dva zhodné uhly " + }, + { + "type": "math", + "text": "|\\angle ABM| = |\\angle ACM|", + "isDisplay": false + }, + { + "type": "text", + "text": " a " + }, + { + "type": "math", + "text": "|\\angle AMB| = |\\angle AMC| = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0spoločnú stranu\u00A0" + }, + { + "type": "math", + "text": "AM", + "isDisplay": false + }, + { + "type": "text", + "text": ", teda " + }, + { + "type": "math", + "text": "|AB| = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Iný vtipný dôkaz je založený na tom, že dokážeme " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": " (všimnite si rôzne poradie vrcholov). V\u00A0prípade, že poznáme rovnaké strany, tak použijeme vetu " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " alebo dokonca " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": " – tá nám následne dá zhodné uhly. V\u00A0prípade, že poznáme rovnaké uhly, to zas bude veta " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0zhodnosť nám dá rovnaké strany." + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vyskúšajte si exaktne dokázať tieto jednoduché poznatky, to už bude jednoduchšie:" + } + ], + "highligted": false + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Rovnostranný trojuholník má tri zhodné uhly rovné " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0" + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " s\u00A0" + }, + { + "type": "math", + "text": "|AB| = |AC| = |BC|", + "isDisplay": false + }, + { + "type": "text", + "text": " dáva veta o\u00A0rovnoramennom trojuholníku " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": " (z\u00A0" + }, + { + "type": "math", + "text": "|AB|=|AC|", + "isDisplay": false + }, + { + "type": "text", + "text": "); analogicky z\u00A0" + }, + { + "type": "math", + "text": "|AB|=|BC|", + "isDisplay": false + }, + { + "type": "text", + "text": " je " + }, + { + "type": "math", + "text": "|\\angle BAC| = |\\angle BCA|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Všetky tri uhly sú teda zhodné, a\u00A0zo súčtu " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " je každý " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-23.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "exercise", + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Rovnoramenný pravouhlý trojuholník má tri uhly rovné " + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nech je pravý uhol pri vrchole\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": ", teda " + }, + { + "type": "math", + "text": "|\\angle BAC| = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0odvesny sú zhodné, " + }, + { + "type": "math", + "text": "|AB| = |AC|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Podľa vety o\u00A0rovnoramennom trojuholníku platí " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Označme túto spoločnú veľkosť " + }, + { + "type": "math", + "text": "\\beta", + "isDisplay": false + }, + { + "type": "text", + "text": ". Zo súčtu uhlov v\u00A0trojuholníku" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "90^\\circ + \\beta + \\beta = 180^\\circ,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "odkiaľ " + }, + { + "type": "math", + "text": "2\\beta = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čiže " + }, + { + "type": "math", + "text": "\\beta = 45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-24.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Veľmi šikovná vec použiteľná v\u00A0neľahkých úlohách je nasledovná úloha. Jeden možný dôkaz je cez trigonometriu." + } + ], + "highligted": false + }, + { + "type": "problem", + "difficulty": 0, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pravouhlý trojuholník má zvyšné dva uhly rovné " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " práve vtedy, keď je jeho prepona dvakrát dlhšia než kratšia odvesna." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Povedzme, že " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " je prepona a\u00A0je dvakrát dlhšia než odvesna " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trikom je uvážiť bod " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": " taký, že " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " je stred úsečky " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nech náš trojuholník má pravý uhol pri vrchole\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "(" + }, + { + "type": "math", + "text": "\\Rightarrow", + "isDisplay": false + }, + { + "type": "text", + "text": ")" + } + ] + }, + { + "type": "text", + "text": " Predpokladajme " + }, + { + "type": "math", + "text": "|BC| = 2|AB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nech " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": " je taký bod, že " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " je stredom úsečky " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": ". Potom " + }, + { + "type": "math", + "text": "|AB'| = |AB|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " leží medzi " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "B'", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže " + }, + { + "type": "math", + "text": "|\\angle B'AC|", + "isDisplay": false + }, + { + "type": "text", + "text": " je vedľajší k\u00A0" + }, + { + "type": "math", + "text": "|\\angle BAC| = 90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čiže tiež " + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Platí " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle AB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " z\u00A0vety " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": ", keďže uhly pri\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " sú oba pravé, strana\u00A0" + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": " je spoločná, a " + }, + { + "type": "math", + "text": "|AB|=|AB'|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1.sk-25.svg", + "scale": 1.0, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pretože " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " je stredom " + }, + { + "type": "math", + "text": "BB'", + "isDisplay": false + }, + { + "type": "text", + "text": ", je " + }, + { + "type": "math", + "text": "|BB'| = 2|AB|", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0zo zadania " + }, + { + "type": "math", + "text": "|BC| = 2|AB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Spolu" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|BB'| = |BC| = |B'C|,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "takže " + }, + { + "type": "math", + "text": "\\triangle BB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnostranný a\u00A0všetky jeho uhly sú " + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Špeciálne " + }, + { + "type": "math", + "text": "|\\angle ABC| = |\\angle B'BC| = 60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Zo súčtu uhlov v\u00A0" + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " dopočítame " + }, + { + "type": "math", + "text": "|\\angle BCA| = 180^\\circ - 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "(" + }, + { + "type": "math", + "text": "\\Leftarrow", + "isDisplay": false + }, + { + "type": "text", + "text": ")" + } + ] + }, + { + "type": "text", + "text": " Nie je ťažké rozmyslieť si, že úvahy z\u00A0predošlého odstavca vieme ľahko obrátiť – kľúčovú zhodnosť " + }, + { + "type": "math", + "text": "\\triangle ABC \\cong \\triangle AB'C", + "isDisplay": false + }, + { + "type": "text", + "text": " tentokrát dostaneme z\u00A0vety " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "Poznámka." + } + ] + }, + { + "type": "text", + "text": " Iné riešenie je uvážiť, že podľa Tálesovej vety je stred\u00A0" + }, + { + "type": "math", + "text": "O", + "isDisplay": false + }, + { + "type": "text", + "text": " kružnice opísanej trojuholníku " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " zároveň stredom prepony " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ". K\u00A0tomuto riešeniu sa podrobnejšie vrátime, keď sa budeme baviť o\u00A0kružniciach." + } + ], + "highligted": false + } + ] + } + ] + } + }, + { + "title": "Čo si zapamätať", + "level": 1, + "text": { + "content": [] + } + }, + { + "title": "Techniky", + "level": 2, + "text": { + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pri rovnobežkách hľadáme zhodné striedavé a súhlasné uhly, prípadne uhly so súčtom\u00A0" + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Je užitočné nahliadať na vonkajší uhol trojuholníka ako na súčet dvoch uhlov." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Na prepojenie sveta dĺžok a\u00A0uhlov sa hodí zhodnosť dvoch trojuholníkov." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pomocné body (stred úsečky, zrkadlový obraz, predĺženie) často odhalia zhodnosť alebo špeciálny trojuholník." + } + ], + "highligted": false + } + ] + ], + "styleType": "NumberParens" + } + ] + } + }, + { + "title": "Užitočné fakty", + "level": 2, + "text": { + "content": [ + { + "type": "list", + "items": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Súčet uhlov v\u00A0trojuholníku je " + }, + { + "type": "math", + "text": "180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", všeobecne v\u00A0konvexnom " + }, + { + "type": "math", + "text": "n", + "isDisplay": false + }, + { + "type": "text", + "text": "-uholníku " + }, + { + "type": "math", + "text": "(n-2) \\cdot 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Vety o\u00A0zhodnosti trojuholníkov: " + }, + { + "type": "math", + "text": "sss", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "usu", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "Ssu", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V trojuholníku sú dve strany zhodné práve keď sú zhodné im priľahlé uhly." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pravouhlý trojuholník má preponu dvakrát dlhšiu než odvesnu práve keď má uhly " + }, + { + "type": "math", + "text": "90^\\circ, 60^\\circ, 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "styleType": "NumberParens" + } + ] + } + } + ] + }, + "images": [ + { + "contentId": "angle-basics-1.sk-1.svg", + "originalId": "angles-vertical.pdf", + "width": "82.61pt", + "height": "65.04pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-2.svg", + "originalId": "angles-corresponding.pdf", + "width": "106.36pt", + "height": "66.59pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-3.svg", + "originalId": "angles-alternate.pdf", + "width": "100.63pt", + "height": "58.17pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-4.svg", + "originalId": "angles-supplementary.pdf", + "width": "85.68pt", + "height": "47.81pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-5.svg", + "originalId": "angles-supplementary-derive-vertical.pdf", + "width": "95.91pt", + "height": "78.49pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-6.svg", + "originalId": "angles-vertical-derive-supplementary.pdf", + "width": "95.91pt", + "height": "78.49pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-7.svg", + "originalId": "angles-corresponding-alternate-supplementary-connection.pdf", + "width": "98.31pt", + "height": "65.02pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-8.svg", + "originalId": "angles-triangle.pdf", + "width": "182.63pt", + "height": "128.67pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-9.svg", + "originalId": "angles-polygon.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-10.svg", + "originalId": "angles-polygon-fan.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-11.svg", + "originalId": "angles-polygon-interior.pdf", + "width": "146.67pt", + "height": "106.43pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-12.svg", + "originalId": "angles-complementary-parallel-statement.pdf", + "width": "91.52pt", + "height": "72.46pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-13.svg", + "originalId": "angles-complementary-parallel-solution.pdf", + "width": "85.7pt", + "height": "92.24pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-14.svg", + "originalId": "angles-complementary-in-triangle.pdf", + "width": "153.73pt", + "height": "123.67pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-15.svg", + "originalId": "angles-sas-warning.pdf", + "width": "240.45pt", + "height": "68.01pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-16.svg", + "originalId": "angles-sss-proof.pdf", + "width": "180.33pt", + "height": "129.76pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-17.svg", + "originalId": "angles-sas-proof.pdf", + "width": "190.6pt", + "height": "165.53pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-18.svg", + "originalId": "angles-asa-proof.pdf", + "width": "156.76pt", + "height": "165.02pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-19.svg", + "originalId": "angles-Ssa-proof-one-solution.pdf", + "width": "116.57pt", + "height": "129.18pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-20.svg", + "originalId": "angles-Ssa-proof-isosceles.pdf", + "width": "116.86pt", + "height": "122.28pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-21.svg", + "originalId": "angles-Ssa-proof-two-solutions.pdf", + "width": "191.58pt", + "height": "154.75pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-22.svg", + "originalId": "angles-isosceles-triangle.pdf", + "width": "117.87pt", + "height": "114.32pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-23.svg", + "originalId": "angles-equilateral.pdf", + "width": "107.91pt", + "height": "97.32pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-24.svg", + "originalId": "angles-isosceles-right-triangle.pdf", + "width": "97.95pt", + "height": "95.36pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1.sk-25.svg", + "originalId": "angles-90-60-30.pdf", + "width": "146.09pt", + "height": "131.24pt", + "scale": 1.0 + } + ] +}