diff --git a/data/handouts/Images/angles-equilateral.asy b/data/handouts/Images/angles-equilateral.asy index a8d1bb0..493c2d3 100644 --- a/data/handouts/Images/angles-equilateral.asy +++ b/data/handouts/Images/angles-equilateral.asy @@ -5,10 +5,6 @@ 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); diff --git a/data/handouts/Images/angles-equilateral.pdf b/data/handouts/Images/angles-equilateral.pdf index b8c41c3..78f8348 100644 Binary files a/data/handouts/Images/angles-equilateral.pdf 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 index e656d0d..8782767 100644 --- a/data/handouts/Images/angles-equilateral.svg +++ b/data/handouts/Images/angles-equilateral.svg @@ -4,70 +4,34 @@ height="97.32pt" viewBox="0 0 107.91 97.32" version="1.1" - id="svg29" + id="svg14" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" xmlns:svg="http://www.w3.org/2000/svg"> + id="defs3"> + id="g3"> - - - - - - - - - + id="path2" /> + id="glyph-0-2"> + id="path3" /> - - - + id="path4" /> + id="path5" /> + id="path6" /> + id="path7" /> + id="path8" /> + id="path9" /> + id="path10" /> + id="path11" /> - - - - - - - - - - - - - - - - - - - - - - - - - - - + id="path12" /> + id="g12"> + id="use12" /> + id="g13"> + id="use13" /> + id="g14"> + id="use14" /> diff --git a/data/handouts/Images/angles-isosceles-chain.asy b/data/handouts/Images/angles-isosceles-chain.asy new file mode 100644 index 0000000..fe94ea8 --- /dev/null +++ b/data/handouts/Images/angles-isosceles-chain.asy @@ -0,0 +1,33 @@ +import _common; + +real s = 110; +pair A = (0, 0); +pair B = (s, 0); +real apexHeight = s/2 * tan(72 * pi / 180); +pair C = (s/2, apexHeight); +pair D = C + s * unit(B - C); +pair E = C + s * unit(A - C); + +real r = Radius2; +pen p = Font2; + +AngleMark(A, C, B, LightPink, "\gamma", radius = Radius3, labelFraction = 1.4, labelPen = p); +AngleMark(D, B, E, LightPink, "\gamma", radius = Radius3, labelFraction = 1.5, labelPen = p); +AngleMark(C, D, E, LightGreen, "2\gamma", radius = r, labelFraction = 1.6, labelPen = p); +AngleMark(D, E, C, LightGreen, "2\gamma", radius = r, labelFraction = 1.6, labelPen = p); +AngleMark(B, E, D, LightPink, "\gamma", radius = r, labelFraction = 1.9, labelPen = p); + +Draw(A, B, Blue); +Draw(B, D, Red); +Draw(D, C, Blue); +Draw(C, E, Blue); +Draw(E, A, Red + dashedPen); +Draw(B, E, Blue); +Draw(D, E, Red); +Draw(A, D, Blue + dashedPen); + +LabeledDot(A, "A", SW); +LabeledDot(B, "B", SE); +LabeledDot(C, "C", N); +LabeledDot(D, "D", (1, 0)); +LabeledDot(E, "E", W); diff --git a/data/handouts/Images/angles-isosceles-chain.pdf b/data/handouts/Images/angles-isosceles-chain.pdf new file mode 100644 index 0000000..127346f Binary files /dev/null and b/data/handouts/Images/angles-isosceles-chain.pdf differ diff --git a/data/handouts/Images/angles-isosceles-chain.svg b/data/handouts/Images/angles-isosceles-chain.svg new file mode 100644 index 0000000..18c4dff --- /dev/null +++ b/data/handouts/Images/angles-isosceles-chain.svg @@ -0,0 +1,384 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-pentagon.asy b/data/handouts/Images/angles-pentagon.asy new file mode 100644 index 0000000..ce76ef9 --- /dev/null +++ b/data/handouts/Images/angles-pentagon.asy @@ -0,0 +1,41 @@ +import _common; + +real alpha = 100; +real lenBC = 50; +real lenDE = 50; +real lenAB = lenBC + lenDE; + +pair A = (0, 0); +pair B = A + lenAB * dir(-90 - alpha/2); +pair E = A + lenAB * dir(-90 + alpha/2); +pair C = B + lenBC * (rotate(-alpha) * unit(A - B)); +pair D = E + lenDE * (rotate(alpha) * unit(A - E)); +pair P = A + (lenBC / lenAB) * (B - A); +pair Q = A + (lenDE / lenAB) * (E - A); + +fill(P -- B -- C -- cycle, LightYellow + opacity(0.5)); +fill(Q -- A -- P -- cycle, LightYellow + opacity(0.5)); +fill(D -- E -- Q -- cycle, LightYellow + opacity(0.5)); + +AngleMark(C, B, A, LightBlue, radius = Radius2); +AngleMark(B, A, E, LightBlue, radius = Radius2); +AngleMark(A, E, D, LightBlue, radius = Radius2); + +Draw(A, P, Green); +Draw(P, B, Red); +Draw(B, C, Green); +Draw(C, D); +Draw(D, E, Red); +Draw(E, Q, Green); +Draw(Q, A, Red); +Draw(C, P); +Draw(P, Q); +Draw(Q, D); + +LabeledDot(A, "A", N); +LabeledDot(B, "B", W); +LabeledDot(C, "C", SW); +LabeledDot(D, "D", SE); +LabeledDot(E, "E", (1, 0)); +LabeledDot(P, "P", NW); +LabeledDot(Q, "Q", NE); diff --git a/data/handouts/Images/angles-pentagon.pdf b/data/handouts/Images/angles-pentagon.pdf new file mode 100644 index 0000000..3db3dab Binary files /dev/null and b/data/handouts/Images/angles-pentagon.pdf differ diff --git a/data/handouts/Images/angles-pentagon.svg b/data/handouts/Images/angles-pentagon.svg new file mode 100644 index 0000000..83c7c3d --- /dev/null +++ b/data/handouts/Images/angles-pentagon.svg @@ -0,0 +1,394 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-quad-30-45.asy b/data/handouts/Images/angles-quad-30-45.asy new file mode 100644 index 0000000..3ff9cd8 --- /dev/null +++ b/data/handouts/Images/angles-quad-30-45.asy @@ -0,0 +1,44 @@ +import _common; + +real s = 240; +real angleA = 30; +real angleB = 45; +real factorC = 2; + +pair A = (0, 0); +pair B = (s, 0); + +pair T = extension(A, A + dir(angleA), B, B + dir(180 - angleB)); +pair C = A + factorC * (T - A); +pair Z = Foot(C, B, T); +pair M = extension(C, Z, A, B); +pair R = Polar(A, angleA, length(M - A)); +pair D = extension(M, R, B, B + dir(180 - angleB)); + +real r = Radius3; +pen p = Font2; + +AngleMark(B, A, C, LightRed, "30^\circ", radius = r, labelFraction = 1.5, labelPen = p); +AngleMark(D, B, A, LightBlue, "45^\circ", radius = r, labelFraction = 1.5, labelPen = p); +AngleMark(D, T, R, LightPurple, "75^\circ", radius = r, labelFraction = 1.5, labelPen = p); +AngleMark(A, R, M, LightPurple, "75^\circ", radius = r, labelFraction = 1.5, labelPen = p); +AngleMark(R, D, T, LightPink, "30^\circ", radius = Radius4, labelFraction = 1.5, labelPen = p); +RightAngleMark(C, Z, B, color = LightGreen, radius = Radius2); + +Draw(A, B); +Draw(B, C); +Draw(C, D); +Draw(D, A); +Draw(A, C); +Draw(B, D); +Draw(C, M); +Draw(M, D); + +LabeledDot(A, "A", SW); +LabeledDot(B, "B", SE); +LabeledDot(C, "C", NE); +LabeledDot(D, "D", NW); +LabeledDot(T, "T", N); +LabeledDot(R, "R", W, offset=(-2,3)); +LabeledDot(M, "M", S); +LabeledDot(Z, "Z", N); diff --git a/data/handouts/Images/angles-quad-30-45.pdf b/data/handouts/Images/angles-quad-30-45.pdf new file mode 100644 index 0000000..9253ffd Binary files /dev/null and b/data/handouts/Images/angles-quad-30-45.pdf differ diff --git a/data/handouts/Images/angles-quad-30-45.svg b/data/handouts/Images/angles-quad-30-45.svg new file mode 100644 index 0000000..e2f795d --- /dev/null +++ b/data/handouts/Images/angles-quad-30-45.svg @@ -0,0 +1,591 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-square-equilateral-shared.asy b/data/handouts/Images/angles-square-equilateral-shared.asy new file mode 100644 index 0000000..1c32743 --- /dev/null +++ b/data/handouts/Images/angles-square-equilateral-shared.asy @@ -0,0 +1,37 @@ +import _common; + +real s = 160; +pair A = (0, 0); +pair B = (s, 0); +pair C = (s, s); +pair D = (0, s); +pair X = EquilateralTriangle(A, B); +pair Y = EquilateralTriangle(C, D); + +void BaseFills() +{ + fill(A -- B -- X -- cycle, LightBlue + opacity(0.3)); + fill(C -- D -- Y -- cycle, LightGreen + opacity(0.3)); +} + +void BaseEdges() +{ + Draw(A, B); + Draw(B, C); + Draw(C, D); + Draw(D, A); + Draw(A, X); + Draw(B, X); + Draw(C, Y); + Draw(D, Y); +} + +void BaseDots() +{ + LabeledDot(A, "A", SW); + LabeledDot(B, "B", SE); + LabeledDot(C, "C", NE); + LabeledDot(D, "D", NW); + LabeledDot(X, "X", N); + LabeledDot(Y, "Y", S); +} diff --git a/data/handouts/Images/angles-square-equilateral-solution.asy b/data/handouts/Images/angles-square-equilateral-solution.asy new file mode 100644 index 0000000..b59fdfc --- /dev/null +++ b/data/handouts/Images/angles-square-equilateral-solution.asy @@ -0,0 +1,17 @@ +include "angles-square-equilateral-shared.asy"; + +BaseFills(); + +real r = Radius4; +pen p = Font2; + +AngleMark(X, A, D, LightGreen, "30^\circ", radius = r, labelFraction = 1.5, labelPen = p); +AngleMark(B, A, X, LightPurple, "60^\circ", radius = r, labelFraction = 1.5, labelPen = p); +AngleMark(A, D, Y, LightGreen, "30^\circ", radius = r, labelFraction = 1.5, labelPen = p); +AngleMark(Y, D, X, LightPink, "45^\circ", radius = r, labelFraction = 1.4, labelPen = p); +AngleMark(D, X, A, LightBlue, "75^\circ", radius = r, labelFraction = 1.5, labelPen = p); + +BaseEdges(); +Draw(D, X); + +BaseDots(); \ No newline at end of file diff --git a/data/handouts/Images/angles-square-equilateral-solution.pdf b/data/handouts/Images/angles-square-equilateral-solution.pdf new file mode 100644 index 0000000..8ab1a30 Binary files /dev/null and b/data/handouts/Images/angles-square-equilateral-solution.pdf differ diff --git a/data/handouts/Images/angles-square-equilateral-solution.svg b/data/handouts/Images/angles-square-equilateral-solution.svg new file mode 100644 index 0000000..15467db --- /dev/null +++ b/data/handouts/Images/angles-square-equilateral-solution.svg @@ -0,0 +1,548 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-square-equilateral-statement.asy b/data/handouts/Images/angles-square-equilateral-statement.asy new file mode 100644 index 0000000..6fc7d68 --- /dev/null +++ b/data/handouts/Images/angles-square-equilateral-statement.asy @@ -0,0 +1,16 @@ +include "angles-square-equilateral-shared.asy"; + +BaseFills(); + +AngleMark(Y, A, X, LightPink, radius = Radius4); +AngleMark(X, B, Y, LightPink, radius = Radius4); +AngleMark(X, C, Y, LightPink, radius = Radius4); +AngleMark(Y, D, X, LightPink, radius = Radius4); + +BaseEdges(); +Draw(A, Y); +Draw(B, Y); +Draw(C, X); +Draw(D, X); + +BaseDots(); diff --git a/data/handouts/Images/angles-square-equilateral-statement.pdf b/data/handouts/Images/angles-square-equilateral-statement.pdf new file mode 100644 index 0000000..78b2036 Binary files /dev/null and b/data/handouts/Images/angles-square-equilateral-statement.pdf differ diff --git a/data/handouts/Images/angles-square-equilateral-statement.svg b/data/handouts/Images/angles-square-equilateral-statement.svg new file mode 100644 index 0000000..ba4083e --- /dev/null +++ b/data/handouts/Images/angles-square-equilateral-statement.svg @@ -0,0 +1,383 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/Images/angles-trapezoid-bisectors.asy b/data/handouts/Images/angles-trapezoid-bisectors.asy new file mode 100644 index 0000000..adbf2cb --- /dev/null +++ b/data/handouts/Images/angles-trapezoid-bisectors.asy @@ -0,0 +1,27 @@ +import _common; + +pair A = (0, 0); +pair B = (210, 0); +pair D = (54, 72); +pair C = (114, 72); +pair P = (90, 0); + +AngleMark(A, D, P, LightBlue, radius = Radius3); +AngleMark(P, D, C, LightBlue, radius = Radius2); +AngleMark(D, C, P, LightGreen, radius = Radius2); +AngleMark(P, C, B, LightGreen, radius = Radius3); +AngleMark(D, P, A, LightBlue, radius = Radius3); +AngleMark(B, P, C, LightGreen, radius = Radius3); + +Draw(A, B); +Draw(B, C); +Draw(C, D); +Draw(D, A); +Draw(D, P, black + dashedPen); +Draw(C, P, black + dashedPen); + +LabeledDot(A, "A", SW); +LabeledDot(B, "B", SE); +LabeledDot(C, "C", NE); +LabeledDot(D, "D", NW); +LabeledDot(P, "P", S); diff --git a/data/handouts/Images/angles-trapezoid-bisectors.pdf b/data/handouts/Images/angles-trapezoid-bisectors.pdf new file mode 100644 index 0000000..d3891c9 Binary files /dev/null and b/data/handouts/Images/angles-trapezoid-bisectors.pdf differ diff --git a/data/handouts/Images/angles-trapezoid-bisectors.svg b/data/handouts/Images/angles-trapezoid-bisectors.svg new file mode 100644 index 0000000..d9f7ec7 --- /dev/null +++ b/data/handouts/Images/angles-trapezoid-bisectors.svg @@ -0,0 +1,286 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/data/handouts/angle-basics-1.cs.tex b/data/handouts/angle-basics-1.cs.tex index 13a50b3..d3a08c8 100644 --- a/data/handouts/angle-basics-1.cs.tex +++ b/data/handouts/angle-basics-1.cs.tex @@ -232,7 +232,7 @@ Vyzkoušejte si exaktně dokázat tyto jednoduché poznatky, to už bude jednodušší: \Exercise{}{ - Rovnostranný trojúhelník má tři shodné úhly rovny $60^\circ$ + 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 $60^\circ$. @@ -291,10 +291,153 @@ \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 V~trojúhelníku jsou dvě strany shodné právě když jsou shodné jim protilehlé úhly. \i Pravoúhlý trojúhelník má přeponu dvakrát delší než odvěsnu právě když má úhly $90^\circ, 60^\circ, 30^\circ$. \enditems +\sec Úlohy + +V~této sekci najdete různé úlohy, na které vám stačí poznatky z~tohoto materiálu, žádné pokročilé věci jako \textit{obvodové a~středové úhly} potřeba nejsou (na tyto úlohy se podíváme později). + + +\Problem{0}{MO okresní kolo Z8 2025}{ + V~trojúhelníku $ABC$ leží bod $D$ na straně $BC$, bod $E$ na straně $AC$, přičemž $|AB| = |BE| = |EC| = |CD|$ a~$|BD| = |DE|$. Určete velikosti úhlů $ACB$ a~$BAD$. +}{ + Označte $\gamma = |\angle ACB|$ a~snažte se všechny úhly v~obrázku vyjádřit pomocí~$\gamma$. Máme spoustu rovnoramenností. +}{ + Jedna možná cesta je postupně pomocí $\gamma$ vyjádřit úhly $CBE$, $BED$, $CDE$, $CED$. Neumíme teď někde napsat rovnici, ze které se dá vypočítat $\gamma$? +}{ + Podívejte se na součet úhlů v~trojúhelníku~$BEC$. Z~něj by mělo jít spočítat $\gamma$. +}{ + K~dopočítání~$BAD$ potřebujeme více kroků. První je spočítat co nejvíce úhlů a~něčeho si všimnout. +}{ + Když spočítáme úhly v~$ABC$, dostaneme, že $ABC$ je rovnoramenný. +}{ + Teď už umíme použít symetrii, jedna možnost je dokázat shodnost trojúhelníků $EAD$ a~$DBE$. +}{ + Označme $\gamma = |\angle ACB|$. + + V~rovnoramenném $\triangle BEC$ ($|BE|=|EC|$) je $|\angle EBC| = |\angle ECB| = \gamma$. Protože $D$ leží na $BC$, je $|\angle DBE| = |\angle EBC| = \gamma$, a~z~rovnoramennosti $\triangle BDE$ ($|BD|=|DE|$) také $|\angle DEB| = \gamma$. Věta o~vnějším úhlu v~$\triangle BDE$ při~$D$ dává + $$ + |\angle EDC| = |\angle DBE| + |\angle DEB| = 2\gamma, + $$ + a~v~rovnoramenném $\triangle CDE$ ($|CD|=|CE|$) tak $|\angle CED| = |\angle CDE| = 2\gamma$. Ze součtu úhlů v~něm $\gamma + 4\gamma = 180^\circ$, čili $\gamma = 36^\circ$. + + \Image{angles-isosceles-chain.pdf}{0.8} + + Nyní si vezměme, že $|\angle AEB|$ je vedlejší úhel k~$|\angle BEC|=3\gamma=108^\circ$, takže je roven $72^\circ$. Z~rovnoramennosti $\triangle BAE$ tak také $|\angle BAE|=72^\circ$. Z~$\triangle ABC$, ve kterém už máme při $C$ a~$A$ po řadě $36^\circ$ a~$72^\circ$, dopočítáme $|\angle CBA|=72^\circ$, takže trojúhelník $ABC$ je rovnoramenný. + + Dále z~$|CA|=|CB|$ a $|CE|=|CD|$ vlastně máme $|EA|=|DB|$. Vezměme trojúhelníky $EAD$ a $DBE$; ty jsou rovnoramenné se shodným úhlem při vrcholu (rovným $180^\circ-2\gamma$) a~shodnými rameny, takže podle $sus$ jsou shodné. To dává $|\angle DAE|=|\angle DBE|=36^\circ$. + + Na závěr tedy máme + $$ + |\angle BAD| = |\angle BAE| - |\angle DAE| = 2\gamma - \gamma = \gamma = 36^\circ. + $$ +} + +\Problem{0}{DuoGeo 2025}{ + Do čtverce $ABCD$ byly nakresleny rovnostranné trojúhelníky $ABX$ a~$CDY$. Určete součet vyznačených úhlů. + + \Image{angles-square-equilateral-statement.pdf}{0.7} +}{ + Zkuste nejprve spočítat co nejvíce úhlů na obrázku. +}{ + Klíčem k~dopočítání finálního úhlu je najít vhodný rovnoramenný trojúhelník přes stejné délky. +}{ + Podle symetrie jsou všechny čtyři vyznačené úhly shodné. Stačí tedy najít velikost jednoho z~nich -- zaměříme se na $|\angle YDX|$. + + Nejprve $|\angle XAD| = |\angle BAD| - |\angle BAX| = 90^\circ - 60^\circ = 30^\circ$; analogicky $|\angle ADY| = |\angle ADC| - |\angle YDC| = 90^\circ - 60^\circ = 30^\circ$. + + Dále $|DA| = |AB| = |AX|$ (první je strana čtverce, druhé strana rovnostranného trojúhelníka $ABX$), takže $\triangle AXD$ je rovnoramenný se základnou $DX$. Z~rovnosti $|\angle XAD| = 30^\circ$ a~ze součtu úhlů v~trojúhelníku + $$ + |\angle ADX| = \tfrac{1}{2}\bigl(180^\circ - 30^\circ\bigr) = 75^\circ. + $$ + Hledaný úhel je rozdílem dvou již vypočítaných: + $$ + |\angle YDX| = |\angle ADX| - |\angle ADY| = 75^\circ - 30^\circ = 45^\circ. + $$ + Součet všech čtyř vyznačených úhlů je tedy $4 \cdot 45^\circ = 180^\circ$. + + \Image{angles-square-equilateral-solution.pdf}{0.8} +} + +\Problem{0}{MO školní kolo C 2024}{ + V~lichoběžníku $ABCD$, kde $AB \parallel CD$, se osy vnitřních úhlů při vrcholech $C$ a~$D$ protínají na úsečce~$AB$. Dokažte, že $|AD| + |BC| = |AB|$. +}{ + Nechť~$P$ je náš společný průsečík. Máme rovnoběžnost, máme osu úhlu, to dává dost stejných úhlů. +}{ + Cílem je najít rovnoramenné trojúhelníky. +}{ + Označme $P$ společný průsečík os; podle zadání leží na úsečce~$AB$. Ukážeme, že + $$ + |AD| = |AP| \quad \hbox{a} \quad |BC| = |BP|, + $$ + odkud přímo $|AD| + |BC| = |AP| + |BP| = |AB|$. + + Protože $DP$ je osa vnitřního úhlu při~$D$, máme $|\angle ADP| = |\angle CDP|$. Z~rovnoběžnosti $AB \parallel CD$ jsou $|\angle CDP|$ a~$|\angle APD|$ střídavé úhly při příčce~$DP$, takže + $$ + |\angle APD| = |\angle CDP| = |\angle ADP|. + $$ + V~trojúhelníku $ADP$ jsou tedy úhly při vrcholech $D$ a~$P$ shodné, takže $|AD| = |AP|$. Analogickou úvahou při vrcholu~$C$ dostaneme $|BC| = |BP|$. + + \Image{angles-trapezoid-bisectors.pdf}{0.8} +} + +\Problem{0}{MO školní kolo A 2023}{ + V~konvexním pětiúhelníku $ABCDE$ platí $|\angle CBA| = |\angle BAE| = |\angle AED|$. Na stranách $AB$ a~$AE$ existují po řadě body $P$ a~$Q$ tak, že $|AP| = |BC| = |QE|$ a~$|AQ| = |BP| = |DE|$. Dokažte, že $CD \parallel PQ$. +}{ + Máme spoustu stejných úhlů a~stran, zkuste najít shodné trojúhelníky. +}{ + Klíčové jsou shodné trojúhelníky $PBC$, $QAP$ a~$DEQ$. Z~nich dostaneme užitečné shodné vlastnosti. Pokračujte v~hledání shodností. +}{ + Finální shodnost k~dokázání je $CPQ$ a $DQP$. Proč to stačí? +}{ + Protože $|BC| = |AP| = |EQ|$, $|BP| = |AQ| = |ED|$ a~$|\angle CBP| = |\angle PAQ| = |\angle QED|$, jsou podle věty $sus$ trojúhelníky $PBC$, $QAP$ a~$DEQ$ navzájem shodné. + + \Image{angles-pentagon.pdf}{0.8} + + Odtud plyne $|CP| = |PQ| = |QD|$ a~také + $$ + \gather + |\angle CPQ| = 180^\circ - |\angle BPC| - |\angle APQ| = \\ = 180^\circ - |\angle PQA| - |\angle EQD| = |\angle PQD|. + \endgather + $$ + Podle věty $sus$ jsou tedy shodné i~rovnoramenné trojúhelníky $CPQ$ a~$DQP$. Z~toho vyplývá shodnost jejich výšek z~vrcholů $C$ a~$D$ na společnou stranu $PQ$; tyto výšky jsou navíc rovnoběžné (obě kolmé na $PQ$), takže $CD \parallel PQ$. +} + +\Problem{0}{DuoGeo 2025}{ + Je dán čtyřúhelník $ABCD$ s~průsečíkem úhlopříček $T$. Předpokládejme, že velikosti úhlů $BAC$ a~$DBA$ jsou po řadě $30^\circ$ a~$45^\circ$. Na úsečce $BT$ leží bod $Z$ takový, že $CZ \perp BT$. Předpokládejme, že přímka $CZ$ protne úsečku $AB$ v~bodě $M$. Nechť $R$ je průsečík úseček $AT$ a~$MD$. Předpokládejme, že $|AM| = |AR|$ a~$|MR| + |TD| = 14$. Určete velikost úsečky $|BZ|$. +}{ + Dopočítávejte úhly, dokud nenajdeme rovnoramenný trojúhelník. +}{ + Trpělivým počítáním úhlů se dá dojít k~tomu, že $DTR$ je rovnoramenný. To by mělo vnést světlo do podmínky $|MR|+|TD|=14$. +}{ + Podmínka se po dokázání rovnoramenností přeloží jako $|MD|=14$. To pomůže později -- aktuálně už víc úhlů nespočítáme a~musíme ještě něco najít ze světa délek. Klíčem je najít pěkný pravoúhlý trojúhelník. +}{ + Dá se spočítat, že úhly trojúhelníku $MDZ$ jsou $90-60-30$. Známe jeho přeponu $MD$. Naše skvěle dokázané pomocné tvrzení nám teď dává další délku. Pak je to krok od řešení. +}{ + V~trojúhelníku $ATB$ známe úhly při vrcholech $A$ a~$B$, a~sice $30^\circ$ a~$45^\circ$. Tím pádem je vnější úhel při~$T$ roven součtu těchto úhlů, konkrétně + $$ + |\angle DTR|=|\angle TAB|=|\angle TBA|=30^\circ+45^\circ=75^\circ. + $$ + V~rovnoramenném trojúhelníku $AMR$ s~$|AM| = |AR|$ je úhel při vrcholu $A$ roven $30^\circ$, takže oba úhly při základně $MR$ mají velikost $\tfrac{1}{2}(180^\circ - 30^\circ) = 75^\circ$. Speciálně $|\angle ARM| = 75^\circ$, a~tedy i~jeho vrcholový úhel $|\angle DRT|$ má velikost $75^\circ$. + + \Image{angles-quad-30-45.pdf}{0.8} + + Trojúhelník $DTR$ má tedy dva úhly velikosti $75^\circ$, čili je rovnoramenný se~základnou $TR$ a~$|DT| = |DR|$. Odtud + $$ + |MD| = |MR| + |RD| = |MR| + |TD| = 14. + $$ + Navíc $|\angle TDR| = 180^\circ - 2 \cdot 75^\circ = 30^\circ$. + + Podívejme se nyní na trojúhelník $MDZ$. Protože $T$, $Z$ leží na úsečce $BD$ a~$R$ leží na úsečce $MD$, je $|\angle MDZ| = |\angle RDT| = 30^\circ$. Trojúhelník $MDZ$ je tedy pravoúhlý (při~$Z$) s~přeponou $MD$ a~úhlem $30^\circ$ při~$D$, takže podle dřívějšího tvrzení o~$30^\circ$-$60^\circ$-$90^\circ$ trojúhelníku + $$ + |MZ| = \tfrac{1}{2} |MD| = 7. + $$ + + Konečně trojúhelník $MZB$ má pravý úhel při~$Z$ a~úhel $45^\circ$ při~$B$, takže i~$|\angle BMZ| = 45^\circ$. Je tedy rovnoramenný pravoúhlý a~$|BZ| = |MZ| = 7$. +} + \DisplayExerciseSolutions \DisplayHints diff --git a/data/handouts/angle-basics-1.en.tex b/data/handouts/angle-basics-1.en.tex index 6acea6e..c4fc208 100644 --- a/data/handouts/angle-basics-1.en.tex +++ b/data/handouts/angle-basics-1.en.tex @@ -295,6 +295,149 @@ \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 +\sec Problems + +In this section you will find various problems for which the knowledge from this handout is sufficient -- no advanced topics like \textit{inscribed and central angles} are needed (we will look at those problems later). + + +\Problem{0}{MO district round Z8 2025}{ + In a triangle $ABC$, point $D$ lies on side $BC$ and point $E$ on side $AC$, with $AB = BE = EC = CD$ and $BD = DE$. Determine the measures of the angles $\angle ACB$ and $\angle BAD$. +}{ + Set $\gamma = \angle ACB$ and try to express every angle in the figure in terms of~$\gamma$. We have plenty of isosceles triangles. +}{ + One possible path is to express the angles $\angle CBE$, $\angle BED$, $\angle CDE$, $\angle CED$ in terms of $\gamma$ in turn. Can we now write down an equation somewhere from which $\gamma$ can be computed? +}{ + Look at the angle sum in triangle~$BEC$. It should let us compute $\gamma$. +}{ + Computing~$\angle BAD$ takes more steps. The first is to compute as many angles as possible and notice something. +}{ + Once we compute the angles in $ABC$, we see that $ABC$ is isosceles. +}{ + Now we can exploit symmetry; one option is to prove the congruence of the triangles $EAD$ and $DBE$. +}{ + Set $\gamma = \angle ACB$. + + In the isosceles $\triangle BEC$ ($BE=EC$) we have $\angle EBC = \angle ECB = \gamma$. Since $D$ lies on $BC$, $\angle DBE = \angle EBC = \gamma$, and by the isosceles property of $\triangle BDE$ ($BD=DE$) also $\angle DEB = \gamma$. The exterior-angle theorem in $\triangle BDE$ at~$D$ gives + $$ + \angle EDC = \angle DBE + \angle DEB = 2\gamma, + $$ + and in the isosceles $\triangle CDE$ ($CD=CE$) we have $\angle CED = \angle CDE = 2\gamma$. From its angle sum $\gamma + 4\gamma = 180^\circ$, so $\gamma = 36^\circ$. + + \Image{angles-isosceles-chain.pdf}{0.8} + + Now note that $\angle AEB$ is in a linear pair with $\angle BEC = 3\gamma = 108^\circ$, so it equals $72^\circ$. By the isosceles property of $\triangle BAE$ we also get $\angle BAE = 72^\circ$. From $\triangle ABC$, where the angles at $C$ and $A$ are $36^\circ$ and $72^\circ$ respectively, we compute $\angle CBA = 72^\circ$, so triangle $ABC$ is isosceles. + + Now observe that from $CA = CB$ and $CE = CD$ we get $EA = DB$. Consider the triangles $EAD$ and $DBE$: they are isosceles with equal apex angle (equal to $180^\circ - 2\gamma$) and equal legs, so by $SAS$ they are congruent. This gives $\angle DAE = \angle DBE = 36^\circ$. + + Finally, + $$ + \angle BAD = \angle BAE - \angle DAE = 2\gamma - \gamma = \gamma = 36^\circ. + $$ +} + +\Problem{0}{DuoGeo 2025}{ + Inside the square $ABCD$, equilateral triangles $ABX$ and $CDY$ have been drawn. Determine the sum of the marked angles. + + \Image{angles-square-equilateral-statement.pdf}{0.7} +}{ + Try first to compute as many angles as possible in the figure. +}{ + The key to computing the final angle is to find a suitable isosceles triangle using equal lengths. +}{ + By symmetry, all four marked angles are equal. So it suffices to find the measure of one of them -- we focus on $\angle YDX$. + + First, $\angle XAD = \angle BAD - \angle BAX = 90^\circ - 60^\circ = 30^\circ$; analogously $\angle ADY = \angle ADC - \angle YDC = 90^\circ - 60^\circ = 30^\circ$. + + Next, $DA = AB = AX$ (the first is a side of the square, the second a side of the equilateral triangle $ABX$), so $\triangle AXD$ is isosceles with base $DX$. From $\angle XAD = 30^\circ$ and the angle sum of the triangle, + $$ + \angle ADX = \tfrac{1}{2}\bigl(180^\circ - 30^\circ\bigr) = 75^\circ. + $$ + The desired angle is the difference of the two already computed: + $$ + \angle YDX = \angle ADX - \angle ADY = 75^\circ - 30^\circ = 45^\circ. + $$ + The sum of all four marked angles is therefore $4 \cdot 45^\circ = 180^\circ$. + + \Image{angles-square-equilateral-solution.pdf}{0.8} +} + +\Problem{0}{MO school round C 2024}{ + In a trapezoid $ABCD$ with $AB \parallel CD$, the interior angle bisectors at vertices $C$ and $D$ meet on segment~$AB$. Prove that $AD + BC = AB$. +}{ + Let~$P$ be the common intersection. We have parallel lines and angle bisectors, which gives plenty of equal angles. +}{ + The goal is to find isosceles triangles. +}{ + Let $P$ be the common intersection of the bisectors; by hypothesis it lies on segment~$AB$. We will show that + $$ + AD = AP \quad \hbox{and} \quad BC = BP, + $$ + from which $AD + BC = AP + BP = AB$ directly. + + Since $DP$ is the bisector of the interior angle at~$D$, we have $\angle ADP = \angle CDP$. By $AB \parallel CD$, $\angle CDP$ and $\angle APD$ are alternate angles cut by the transversal~$DP$, so + $$ + \angle APD = \angle CDP = \angle ADP. + $$ + In triangle $ADP$ the angles at vertices $D$ and $P$ are equal, so $AD = AP$. An analogous argument at vertex~$C$ yields $BC = BP$. + + \Image{angles-trapezoid-bisectors.pdf}{0.8} +} + +\Problem{0}{MO school round A 2023}{ + In a convex pentagon $ABCDE$ we have $\angle CBA = \angle BAE = \angle AED$. On the sides $AB$ and $AE$ there exist points $P$ and $Q$ respectively such that $AP = BC = QE$ and $AQ = BP = DE$. Prove that $CD \parallel PQ$. +}{ + We have plenty of equal angles and sides; try to find congruent triangles. +}{ + The key congruences are among triangles $PBC$, $QAP$, and $DEQ$. They yield useful equal pieces. Keep looking for more congruences. +}{ + The final congruence to prove is between $CPQ$ and $DQP$. Why does that suffice? +}{ + Since $BC = AP = EQ$, $BP = AQ = ED$, and $\angle CBP = \angle PAQ = \angle QED$, the triangles $PBC$, $QAP$, and $DEQ$ are mutually congruent by $SAS$. + + \Image{angles-pentagon.pdf}{0.8} + + From this $CP = PQ = QD$, and also + $$ + \gather + \angle CPQ = 180^\circ - \angle BPC - \angle APQ = \\ = 180^\circ - \angle PQA - \angle EQD = \angle PQD. + \endgather + $$ + Hence, by $SAS$, the isosceles triangles $CPQ$ and $DQP$ are also congruent. From this, the altitudes from vertices $C$ and $D$ to the common side $PQ$ are equal; these altitudes are also parallel (both perpendicular to $PQ$), so $CD \parallel PQ$. +} + +\Problem{0}{DuoGeo 2025}{ + A quadrilateral $ABCD$ is given, with $T$ the intersection of its diagonals. Assume that the angles $\angle BAC$ and $\angle DBA$ measure $30^\circ$ and $45^\circ$ respectively. On segment $BT$ there is a point $Z$ such that $CZ \perp BT$. Assume that line $CZ$ meets segment $AB$ at point $M$. Let $R$ be the intersection of segments $AT$ and $MD$. Suppose that $AM = AR$ and $MR + TD = 14$. Determine the length of segment $BZ$. +}{ + Keep computing angles until we find an isosceles triangle. +}{ + Patient angle chasing leads to $DTR$ being isosceles. That should shed light on the condition $MR + TD = 14$. +}{ + Once the isosceles relations are proved, the condition translates to $MD = 14$. That will help later -- for now we cannot compute more angles and must find something from the world of lengths. The key is to find a nice right triangle. +}{ + One can show that the angles of triangle $MDZ$ are $90$-$60$-$30$. We know its hypotenuse $MD$. Our nicely proved auxiliary statement now gives another length. From there it is one step to the solution. +}{ + In triangle $ATB$ we know the angles at vertices $A$ and $B$, namely $30^\circ$ and $45^\circ$. Hence the exterior angle at~$T$ equals the sum of these angles; concretely + $$ + \angle DTR = \angle TAB = \angle TBA = 30^\circ + 45^\circ = 75^\circ. + $$ + In the isosceles triangle $AMR$ with $AM = AR$, the angle at vertex $A$ equals $30^\circ$, so both angles at the base $MR$ have measure $\tfrac{1}{2}(180^\circ - 30^\circ) = 75^\circ$. In particular $\angle ARM = 75^\circ$, and so its vertical angle $\angle DRT$ also has measure $75^\circ$. + + \Image{angles-quad-30-45.pdf}{0.8} + + Triangle $DTR$ therefore has two angles of measure $75^\circ$, so it is isosceles with base $TR$ and $DT = DR$. Hence + $$ + MD = MR + RD = MR + TD = 14. + $$ + Moreover, $\angle TDR = 180^\circ - 2 \cdot 75^\circ = 30^\circ$. + + Now consider triangle $MDZ$. Since $T$, $Z$ lie on segment $BD$ and $R$ lies on segment $MD$, we have $\angle MDZ = \angle RDT = 30^\circ$. Triangle $MDZ$ is therefore right-angled (at~$Z$) with hypotenuse $MD$ and an angle of $30^\circ$ at~$D$, so by the earlier statement about the $30^\circ$-$60^\circ$-$90^\circ$ triangle + $$ + MZ = \tfrac{1}{2} MD = 7. + $$ + + Finally, triangle $MZB$ has a right angle at~$Z$ and an angle of $45^\circ$ at~$B$, so $\angle BMZ = 45^\circ$. It is therefore an isosceles right triangle and $BZ = MZ = 7$. +} + \DisplayExerciseSolutions \DisplayHints diff --git a/data/handouts/angle-basics-1.sk.tex b/data/handouts/angle-basics-1.sk.tex index 41d012b..ae442b4 100644 --- a/data/handouts/angle-basics-1.sk.tex +++ b/data/handouts/angle-basics-1.sk.tex @@ -2,6 +2,8 @@ \setlanguage{SK} +\ReviewModetrue + \Title{Základy počítania uhlov} \MathcompsLink{zaklady-pocitania-uhlov-1} @@ -184,7 +186,7 @@ }{ 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 polpriamke z~$B$ zvierajúcej s~$BC$ uhol $\beta$ a~zároveň na polpriamke 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 polpriamky rovnobežné -- pretnú sa v~jedinom bode~$A$. Trojuholník $ABC$ je teda určený jednoznačne (až na zhodné zobrazenia). + Pre účely nášho dôkazu uvážme teda, že sú zhodné práve uhly priliehajúce 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 polpriamke z~$B$ zvierajúcej s~$BC$ uhol $\beta$ a~zároveň na polpriamke 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 polpriamky rovnobežné -- pretnú sa v~jedinom bode~$A$. Trojuholník $ABC$ je teda určený jednoznačne (až na zhodné zobrazenia). \Image{angles-asa-proof.pdf} } @@ -232,7 +234,7 @@ Vyskúšajte si exaktne dokázať tieto jednoduché poznatky, to už bude jednoduchšie: \Exercise{}{ - Rovnostranný trojuholník má tri zhodné uhly rovné $60^\circ$ + 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 $60^\circ$. @@ -291,10 +293,153 @@ \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 V trojuholníku sú dve strany zhodné práve keď sú zhodné im protiľahlé uhly. \i Pravouhlý trojuholník má preponu dvakrát dlhšiu než odvesnu práve keď má uhly $90^\circ, 60^\circ, 30^\circ$. \enditems +\sec Úlohy + +V~tejto sekcii nájdete rôzne úlohy, na ktoré vám stačia poznatky z~tohto materiálu, žiadne pokročilé veci ako \textit{obvodové a~stredové uhly} potrebné nie sú (na tieto úlohy sa pozrieme neskôr). + + +\Problem{0}{MO okresné kolo Z8 2025}{ + V~trojuholníku $ABC$ leží bod $D$ na strane $BC$, bod $E$ na strane $AC$, pričom $|AB| = |BE| = |EC| = |CD|$ a~$|BD| = |DE|$. Určte veľkosti uhlov $ACB$ a~$BAD$. +}{ + Označte $\gamma = |\angle ACB|$ a~snažte sa všetky uhly v~obrázku vyjadriť cez~$\gamma$. Máme veľa rovnoramenností. +}{ + Jedna možná cesta je postupne pomocou $\gamma$ vyjadriť uhly $CBE$, $BED$, $CDE$, $CED$. Nevieme teraz niekde napísať rovnicu, z~ktorej sa dá vypočítať $\gamma$? +}{ + Pozrite sa na súčet uhlov v~trojuholníku~$BEC$. Z~neho by malo ísť spočítať $\gamma$. +}{ + K~dopočítaniu~$BAD$ potrebujeme viac krokov. Prvý je spočítať čo najviac uhlov a~niečo si všimnúť. +}{ + Keď spočítame uhly v~$ABC$, tak dostaneme, že $ABC$ je rovnoramenný. +}{ + Teraz už vieme použiť symetriu, jedna možnosť je dokázať zhodnosť trojuholníkov $EAD$ a~$DBE$. +}{ + Označme $\gamma = |\angle ACB|$. + + V~rovnoramennom $\triangle BEC$ ($|BE|=|EC|$) je $|\angle EBC| = |\angle ECB| = \gamma$. Keďže $D$ leží na $BC$, je $|\angle DBE| = |\angle EBC| = \gamma$, a~z~rovnoramennosti $\triangle BDE$ ($|BD|=|DE|$) aj $|\angle DEB| = \gamma$. Veta o~vonkajšom uhle v~$\triangle BDE$ pri~$D$ dáva + $$ + |\angle EDC| = |\angle DBE| + |\angle DEB| = 2\gamma, + $$ + a~v~rovnoramennom $\triangle CDE$ ($|CD|=|CE|$) tak $|\angle CED| = |\angle CDE| = 2\gamma$. Zo súčtu uhlov v~ňom $\gamma + 4\gamma = 180^\circ$, čiže $\gamma = 36^\circ$. + + \Image{angles-isosceles-chain.pdf}{0.8} + + Teraz si vezmime, že $|\angle AEB|$ je vedľajší uhol k~$|\angle BEC|=3\gamma=108^\circ$, takže je rovný $72^\circ$. Z~rovnoramennosti $\triangle BAE$ tak aj $|\angle BAE|=72^\circ$. Z~$\triangle ABC$, v~ktorom už máme pri $C$ a~$A$ rovné postupne $36^\circ$ a~$72^\circ$ dopočítame $|\angle CBA|=72^\circ$, takže trojuholník $ABC$ je rovnoramenný. + + Teraz si vezmime, že z~$|CA|=|CB|$ a $|CE|=|CD|$ vlastne máme $|EA|=|DB|$. Vezmime si trojuholníky $EAD$ a $DBE$, tie sú rovnoramenné so zhodným uhlom pri vrchole (rovným $180^\circ-2\gamma$) a~zhodnými ramenami, takže podľa $sus$ sú zhodné. To dáva $|\angle DAE|=|\angle DBE|=36^\circ$. + + Na záver teda máme + $$ + |\angle BAD| = |\angle BAE| - |\angle DAE| = 2\gamma - \gamma = \gamma = 36^\circ. + $$ +} + +\Problem{0}{DuoGeo 2025}{ + Do štvorca $ABCD$ boli nakreslené rovnostranné trojuholníky $ABX$ a~$CDY$. Určte súčet vyznačených uhlov. + + \Image{angles-square-equilateral-statement.pdf}{0.7} +}{ + Skúste najprv spočítať čo najviac uhlov na obrázku. +}{ + Kľúčom k~dopočítaniu finálneho uhla je nájsť vhodný rovnoramenný trojuholník cez rovnaké dĺžky. +}{ + Podľa symetrie sú všetky štyri vyznačené uhly zhodné. Stačí teda nájsť veľkosť jedného z~nich -- zameriame sa na $|\angle YDX|$. + + Najprv $|\angle XAD| = |\angle BAD| - |\angle BAX| = 90^\circ - 60^\circ = 30^\circ$; analogicky $|\angle ADY| = |\angle ADC| - |\angle YDC| = 90^\circ - 60^\circ = 30^\circ$. + + Ďalej $|DA| = |AB| = |AX|$ (prvé je strana štvorca, druhé strana rovnostranného trojuholníka $ABX$), takže $\triangle AXD$ je rovnoramenný so základňou $DX$. Z~rovnosti $|\angle XAD| = 30^\circ$ a~zo súčtu uhlov v~trojuholníku + $$ + |\angle ADX| = \tfrac{1}{2}\bigl(180^\circ - 30^\circ\bigr) = 75^\circ. + $$ + Hľadaný uhol je rozdielom dvoch už vypočítaných: + $$ + |\angle YDX| = |\angle ADX| - |\angle ADY| = 75^\circ - 30^\circ = 45^\circ. + $$ + Súčet všetkých štyroch vyznačených uhlov je teda $4 \cdot 45^\circ = 180^\circ$. + + \Image{angles-square-equilateral-solution.pdf}{0.8} +} + +\Problem{0}{MO školské kolo C 2024}{ + V~lichobežníku $ABCD$, kde $AB \parallel CD$, sa osi vnútorných uhlov pri vrcholoch $C$ a~$D$ pretínajú na úsečke~$AB$. Dokážte, že $|AD| + |BC| = |AB|$. +}{ + Nech~$P$ je náš spoločný priesečník. Máme rovnobežnosť, máme os uhla, to dáva dosť rovnakých uhlov. +}{ + Cieľ je nájsť rovnoramenné trojuholníky. +}{ + Označme $P$ spoločný priesečník osí; podľa zadania leží na úsečke~$AB$. Ukážeme, že + $$ + |AD| = |AP| \quad \hbox{a} \quad |BC| = |BP|, + $$ + odkiaľ priamo $|AD| + |BC| = |AP| + |BP| = |AB|$. + + Pretože $DP$ je os vnútorného uhla pri~$D$, máme $|\angle ADP| = |\angle CDP|$. Z~rovnobežnosti $AB \parallel CD$ sú $|\angle CDP|$ a~$|\angle APD|$ striedavé uhly pri priečke~$DP$, takže + $$ + |\angle APD| = |\angle CDP| = |\angle ADP|. + $$ + V~trojuholníku $ADP$ sú teda uhly pri vrcholoch $D$ a~$P$ zhodné, takže $|AD| = |AP|$. Analogickou úvahou pri vrchole~$C$ dostaneme $|BC| = |BP|$. + + \Image{angles-trapezoid-bisectors.pdf}{0.8} +} + +\Problem{0}{MO školské kolo A 2023}{ + V~konvexnom päťuholníku $ABCDE$ platí $|\angle CBA| = |\angle BAE| = |\angle AED|$. Na stranách $AB$ a~$AE$ existujú po rade body $P$ a~$Q$ tak, že $|AP| = |BC| = |QE|$ a~$|AQ| = |BP| = |DE|$. Dokážte, že $CD \parallel PQ$. +}{ + Máme veľa rovnakých uhlov a~strán, skúste nájsť zhodné trojuholníky. +}{ + Kľúčové sú zhodné trojuholníky $PBC$, $QAP$ a~$DEQ$. Z~nich dostaneme užitočné zhodné vecičky. Pokračujte v~hľadaní zhodnosti. +}{ + Finálna zhodnosť na dokázanie je $CPQ$ a $DQP$. Prečo to stačí? +}{ + Keďže $|BC| = |AP| = |EQ|$, $|BP| = |AQ| = |ED|$ a~$|\angle CBP| = |\angle PAQ| = |\angle QED|$, sú podľa vety $sus$ trojuholníky $PBC$, $QAP$ a~$DEQ$ navzájom zhodné. + + \Image{angles-pentagon.pdf}{0.8} + + Odtiaľ plynie $|CP| = |PQ| = |QD|$ a~tiež + $$ + \gather + |\angle CPQ| = 180^\circ - |\angle BPC| - |\angle APQ| = \\ = 180^\circ - |\angle PQA| - |\angle EQD| = |\angle PQD|. + \endgather + $$ + Podľa vety $sus$ sú teda zhodné aj rovnoramenné trojuholníky $CPQ$ a~$DQP$. Z~toho vyplýva zhodnosť ich výšok z~vrcholov $C$ a~$D$ na spoločnú stranu $PQ$; tieto výšky sú zároveň rovnobežné (obe kolmé na $PQ$), takže $CD \parallel PQ$. +} + +\Problem{0}{DuoGeo 2025}{ + Je daný štvoruholník $ABCD$ s~priesečníkom uhlopriečok $T$. Predpokladajme, že veľkosti uhlov $BAC$ a~$DBA$ sú postupne $30^\circ$ a~$45^\circ$. Na úsečke $BT$ leží bod $Z$ taký, že $CZ \perp BT$. Predpokladajme, že priamka $CZ$ pretne úsečku $AB$ v~bode $M$. Nech $R$ je priesečník úsečiek $AT$ a~$MD$. Predpokladajme, že $|AM| = |AR|$ a~$|MR| + |TD| = 14$. Určte veľkosť úsečky $|BZ|$. +}{ + Dopočítavajte uhly, až kým nenájdeme rovnoramenný trojuholník. +}{ + Trpezlivým počítaním uhlov sa dá dôjsť k~tomu, že $DTR$ je rovnoramenný. To by malo vniesť svetlo do podmienky $|MR|+|TD|=14$. +}{ + Podmienka sa po dokázaní rovnoramenností preloží ako $|MD|=14$. To pomôže neskôr -- aktuálne už viac uhlov nespočítame a~musíme ešte niečo nájsť zo sveta dĺžok. Kľúčom je nájsť pekný pravouhlý trojuholník. +}{ + Dá sa spočítať, že uhly trojuholníka $MDZ$ sú $90-60-30$. Poznáme jeho preponu $MD$. Naše skvelé dokázané pomocné tvrdenie nám teraz dáva ďalšiu dĺžku. Potom je to krok od riešenia. +}{ + V~trojuholníku $ATB$ poznáme uhly pri vrcholoch $A$ a~$B$, a~síce $30^\circ$ a~$45^\circ$. Tým pádom je vonkajší uhol pri~$T$ rovný súčtu týchto uhlov, konkrétne + $$ + |\angle DTR|=|\angle TAB|=|\angle TBA|=30^\circ+45^\circ=75^\circ. + $$ + V~rovnoramennom trojuholníku $AMR$ s~$|AM| = |AR|$ je uhol pri vrchole $A$ rovný $30^\circ$, takže oba uhly pri základni $MR$ majú veľkosť $\tfrac{1}{2}(180^\circ - 30^\circ) = 75^\circ$. Špeciálne $|\angle ARM| = 75^\circ$, a~teda aj jeho vrcholový uhol $|\angle DRT|$ má veľkosť $75^\circ$. + + \Image{angles-quad-30-45.pdf}{0.8} + + Trojuholník $DTR$ má teda dva uhly veľkosti $75^\circ$, čiže je rovnoramenný so~základňou $TR$ a~$|DT| = |DR|$. Odtiaľ + $$ + |MD| = |MR| + |RD| = |MR| + |TD| = 14. + $$ + Navyše $|\angle TDR| = 180^\circ - 2 \cdot 75^\circ = 30^\circ$. + + Pozrime sa teraz na trojuholník $MDZ$. Pretože $T$, $Z$ ležia na úsečke $BD$ a~$R$ leží na úsečke $MD$, je $|\angle MDZ| = |\angle RDT| = 30^\circ$. Trojuholník $MDZ$ je teda pravouhlý (pri~$Z$) s~preponou $MD$ a~uhlom $30^\circ$ pri~$D$, takže podľa skoršieho tvrdenia o~$30^\circ$-$60^\circ$-$90^\circ$ trojuholníku + $$ + |MZ| = \tfrac{1}{2} |MD| = 7. + $$ + + Konečne trojuholník $MZB$ má pravý uhol pri~$Z$ a~uhol $45^\circ$ pri~$B$, takže aj $|\angle BMZ| = 45^\circ$. Je teda rovnoramenný pravouhlý a~$|BZ| = |MZ| = 7$. +} + \DisplayExerciseSolutions \DisplayHints diff --git a/web/src/content/handouts/angle-basics-1.cs.json b/web/src/content/handouts/angle-basics-1.cs.json index 6883070..ade2811 100644 --- a/web/src/content/handouts/angle-basics-1.cs.json +++ b/web/src/content/handouts/angle-basics-1.cs.json @@ -4113,6 +4113,10 @@ "type": "math", "text": "60^\\circ", "isDisplay": false + }, + { + "type": "text", + "text": "." } ], "highligted": false @@ -5047,7 +5051,7 @@ "content": [ { "type": "text", - "text": "V\u00A0trojúhelníku jsou dvě strany shodné právě když jsou shodné jim přilehlé úhly." + "text": "V\u00A0trojúhelníku jsou dvě strany shodné právě když jsou shodné jim protilehlé úhly." } ], "highligted": false @@ -5079,184 +5083,2692 @@ } ] } - } - ] - }, - "images": [ - { - "contentId": "angle-basics-1/angles-vertical.svg", - "originalId": "angles-vertical.pdf", - "width": "82.61pt", - "height": "65.04pt", - "scale": 1.0 - }, - { - "contentId": "angle-basics-1/angles-corresponding.svg", - 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"height": "78.4pt", - "scale": 1.0 - }, - { - "contentId": "angle-basics-1/angles-90-60-30.svg", - "originalId": "angles-90-60-30.pdf", - "width": "146.09pt", - "height": "131.24pt", - "scale": 1.0 + }, + { + "title": "Úlohy", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0této sekci najdete různé úlohy, na které vám stačí poznatky z\u00A0tohoto materiálu, žádné pokročilé věci jako " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "obvodové a\u00A0středové úhly" + } + ] + }, + { + "type": "text", + "text": " potřeba nejsou (na tyto úlohy se podíváme později)." + } + ], + "highligted": false + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "MO okresní kolo Z8 2025" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0trojúhelníku " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " leží bod " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " na straně " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ", bod " + }, + { + "type": "math", + "text": "E", + "isDisplay": false + }, + { + "type": "text", + "text": " na straně " + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": ", přičemž " + }, + { + "type": "math", + "text": "|AB| = |BE| = |EC| = |CD|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|BD| = |DE|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Určete velikosti úhlů " + }, + { + "type": "math", + "text": "ACB", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "BAD", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Označte " + }, + { + "type": "math", + "text": "\\gamma = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0snažte se všechny úhly v\u00A0obrázku vyjádřit pomocí\u00A0" + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Máme spoustu rovnoramenností." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Jedna možná cesta je postupně pomocí " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " vyjádřit úhly " + }, + { + "type": "math", + "text": "CBE", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "BED", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "CDE", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "CED", + "isDisplay": false + }, + { + "type": "text", + "text": ". Neumíme teď někde napsat rovnici, ze které se dá vypočítat " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": "?" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podívejte se na součet úhlů v\u00A0trojúhelníku\u00A0" + }, + { + "type": "math", + "text": "BEC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0něj by mělo jít spočítat " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "K\u00A0dopočítání\u00A0" + }, + { + "type": "math", + "text": "BAD", + "isDisplay": false + }, + { + "type": "text", + "text": " potřebujeme více kroků. První je spočítat co nejvíce úhlů a\u00A0něčeho si všimnout." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Když spočítáme úhly v\u00A0" + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": ", dostaneme, že " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnoramenný." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Teď už umíme použít symetrii, jedna možnost je dokázat shodnost trojúhelníků " + }, + { + "type": "math", + "text": "EAD", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DBE", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Označme " + }, + { + "type": "math", + "text": "\\gamma = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0rovnoramenném " + }, + { + "type": "math", + "text": "\\triangle BEC", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "|BE|=|EC|", + "isDisplay": false + }, + { + "type": "text", + "text": ") je " + }, + { + "type": "math", + "text": "|\\angle EBC| = |\\angle ECB| = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Protože " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " leží na " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ", je " + }, + { + "type": "math", + "text": "|\\angle DBE| = |\\angle EBC| = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0z\u00A0rovnoramennosti " + }, + { + "type": "math", + "text": "\\triangle BDE", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "|BD|=|DE|", + "isDisplay": false + }, + { + "type": "text", + "text": ") také " + }, + { + "type": "math", + "text": "|\\angle DEB| = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Věta o\u00A0vnějším úhlu v\u00A0" + }, + { + "type": "math", + "text": "\\triangle BDE", + "isDisplay": false + }, + { + "type": "text", + "text": " při\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " dává" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle EDC| = |\\angle DBE| + |\\angle DEB| = 2\\gamma,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "a\u00A0v\u00A0rovnoramenném " + }, + { + "type": "math", + "text": "\\triangle CDE", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "|CD|=|CE|", + "isDisplay": false + }, + { + "type": "text", + "text": ") tak " + }, + { + "type": "math", + "text": "|\\angle CED| = |\\angle CDE| = 2\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ze součtu úhlů v\u00A0něm " + }, + { + "type": "math", + "text": "\\gamma + 4\\gamma = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čili " + }, + { + "type": "math", + "text": "\\gamma = 36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-isosceles-chain.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nyní si vezměme, že " + }, + { + "type": "math", + "text": "|\\angle AEB|", + "isDisplay": false + }, + { + "type": "text", + "text": " je vedlejší úhel k\u00A0" + }, + { + "type": "math", + "text": "|\\angle BEC|=3\\gamma=108^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže je roven " + }, + { + "type": "math", + "text": "72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0rovnoramennosti " + }, + { + "type": "math", + "text": "\\triangle BAE", + "isDisplay": false + }, + { + "type": "text", + "text": " tak také " + }, + { + "type": "math", + "text": "|\\angle BAE|=72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0" + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": ", ve kterém už máme při " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " po řadě " + }, + { + "type": "math", + "text": "36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", dopočítáme " + }, + { + "type": "math", + "text": "|\\angle CBA|=72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže trojúhelník " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnoramenný." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dále z\u00A0" + }, + { + "type": "math", + "text": "|CA|=|CB|", + "isDisplay": false + }, + { + "type": "text", + "text": " a " + }, + { + "type": "math", + "text": "|CE|=|CD|", + "isDisplay": false + }, + { + "type": "text", + "text": " vlastně máme " + }, + { + "type": "math", + "text": "|EA|=|DB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Vezměme trojúhelníky " + }, + { + "type": "math", + "text": "EAD", + "isDisplay": false + }, + { + "type": "text", + "text": " a " + }, + { + "type": "math", + "text": "DBE", + "isDisplay": false + }, + { + "type": "text", + "text": "; ty jsou rovnoramenné se shodným úhlem při vrcholu (rovným " + }, + { + "type": "math", + "text": "180^\\circ-2\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ") a\u00A0shodnými rameny, takže podle " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou shodné. To dává " + }, + { + "type": "math", + "text": "|\\angle DAE|=|\\angle DBE|=36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Na závěr tedy máme" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle BAD| = |\\angle BAE| - |\\angle DAE| = 2\\gamma - \\gamma = \\gamma = 36^\\circ.", + "isDisplay": true + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "DuoGeo 2025" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Do čtverce " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": " byly nakresleny rovnostranné trojúhelníky " + }, + { + "type": "math", + "text": "ABX", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "CDY", + "isDisplay": false + }, + { + "type": "text", + "text": ". Určete součet vyznačených úhlů." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-square-equilateral-statement.svg", + "scale": 0.7, + "isInline": false + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Zkuste nejprve spočítat co nejvíce úhlů na obrázku." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Klíčem k\u00A0dopočítání finálního úhlu je najít vhodný rovnoramenný trojúhelník přes stejné délky." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podle symetrie jsou všechny čtyři vyznačené úhly shodné. Stačí tedy najít velikost jednoho z\u00A0nich – zaměříme se na " + }, + { + "type": "math", + "text": "|\\angle YDX|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nejprve " + }, + { + "type": "math", + "text": "|\\angle XAD| = |\\angle BAD| - |\\angle BAX| = 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "; analogicky " + }, + { + "type": "math", + "text": "|\\angle ADY| = |\\angle ADC| - |\\angle YDC| = 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dále " + }, + { + "type": "math", + "text": "|DA| = |AB| = |AX|", + "isDisplay": false + }, + { + "type": "text", + "text": " (první je strana čtverce, druhé strana rovnostranného trojúhelníka " + }, + { + "type": "math", + "text": "ABX", + "isDisplay": false + }, + { + "type": "text", + "text": "), takže " + }, + { + "type": "math", + "text": "\\triangle AXD", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnoramenný se základnou " + }, + { + "type": "math", + "text": "DX", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0rovnosti " + }, + { + "type": "math", + "text": "|\\angle XAD| = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0ze součtu úhlů v\u00A0trojúhelníku" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle ADX| = \\tfrac{1}{2}\\bigl(180^\\circ - 30^\\circ\\bigr) = 75^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Hledaný úhel je rozdílem dvou již vypočítaných:" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle YDX| = |\\angle ADX| - |\\angle ADY| = 75^\\circ - 30^\\circ = 45^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Součet všech čtyř vyznačených úhlů je tedy " + }, + { + "type": "math", + "text": "4 \\cdot 45^\\circ = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-square-equilateral-solution.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "MO školní kolo C 2024" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0lichoběžníku " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": ", kde " + }, + { + "type": "math", + "text": "AB \\parallel CD", + "isDisplay": false + }, + { + "type": "text", + "text": ", se osy vnitřních úhlů při vrcholech " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " protínají na úsečce\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Dokažte, že " + }, + { + "type": "math", + "text": "|AD| + |BC| = |AB|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nechť\u00A0" + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " je náš společný průsečík. Máme rovnoběžnost, máme osu úhlu, to dává dost stejných úhlů." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Cílem je najít rovnoramenné trojúhelníky." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Označme " + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " společný průsečík os; podle zadání leží na úsečce\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ukážeme, že" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|AD| = |AP| \\quad \\hbox{a} \\quad |BC| = |BP|,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "odkud přímo " + }, + { + "type": "math", + "text": "|AD| + |BC| = |AP| + |BP| = |AB|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Protože " + }, + { + "type": "math", + "text": "DP", + "isDisplay": false + }, + { + "type": "text", + "text": " je osa vnitřního úhlu při\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": ", máme " + }, + { + "type": "math", + "text": "|\\angle ADP| = |\\angle CDP|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0rovnoběžnosti " + }, + { + "type": "math", + "text": "AB \\parallel CD", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou " + }, + { + "type": "math", + "text": "|\\angle CDP|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|\\angle APD|", + "isDisplay": false + }, + { + "type": "text", + "text": " střídavé úhly při příčce\u00A0" + }, + { + "type": "math", + "text": "DP", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle APD| = |\\angle CDP| = |\\angle ADP|.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0trojúhelníku " + }, + { + "type": "math", + "text": "ADP", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou tedy úhly při vrcholech " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " shodné, takže " + }, + { + "type": "math", + "text": "|AD| = |AP|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Analogickou úvahou při vrcholu\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " dostaneme " + }, + { + "type": "math", + "text": "|BC| = |BP|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-trapezoid-bisectors.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "MO školní kolo A 2023" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0konvexním pětiúhelníku " + }, + { + "type": "math", + "text": "ABCDE", + "isDisplay": false + }, + { + "type": "text", + "text": " platí " + }, + { + "type": "math", + "text": "|\\angle CBA| = |\\angle BAE| = |\\angle AED|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Na stranách " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "AE", + "isDisplay": false + }, + { + "type": "text", + "text": " existují po řadě body " + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "Q", + "isDisplay": false + }, + { + "type": "text", + "text": " tak, že " + }, + { + "type": "math", + "text": "|AP| = |BC| = |QE|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|AQ| = |BP| = |DE|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Dokažte, že " + }, + { + "type": "math", + "text": "CD \\parallel PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Máme spoustu stejných úhlů a\u00A0stran, zkuste najít shodné trojúhelníky." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Klíčové jsou shodné trojúhelníky " + }, + { + "type": "math", + "text": "PBC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "QAP", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DEQ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0nich dostaneme užitečné shodné vlastnosti. Pokračujte v\u00A0hledání shodností." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Finální shodnost k\u00A0dokázání je " + }, + { + "type": "math", + "text": "CPQ", + "isDisplay": false + }, + { + "type": "text", + "text": " a " + }, + { + "type": "math", + "text": "DQP", + "isDisplay": false + }, + { + "type": "text", + "text": ". Proč to stačí?" + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Protože " + }, + { + "type": "math", + "text": "|BC| = |AP| = |EQ|", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "|BP| = |AQ| = |ED|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|\\angle CBP| = |\\angle PAQ| = |\\angle QED|", + "isDisplay": false + }, + { + "type": "text", + "text": ", jsou podle věty " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " trojúhelníky " + }, + { + "type": "math", + "text": "PBC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "QAP", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DEQ", + "isDisplay": false + }, + { + "type": "text", + "text": " navzájem shodné." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-pentagon.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Odtud plyne " + }, + { + "type": "math", + "text": "|CP| = |PQ| = |QD|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0také" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\begin{gather*}\n|\\angle CPQ| = 180^\\circ - |\\angle BPC| - |\\angle APQ| = \\\\ = 180^\\circ - |\\angle PQA| - |\\angle EQD| = |\\angle PQD|.\n\\end{gather*}", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podle věty " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou tedy shodné i\u00A0rovnoramenné trojúhelníky " + }, + { + "type": "math", + "text": "CPQ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DQP", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0toho vyplývá shodnost jejich výšek z\u00A0vrcholů " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " na společnou stranu " + }, + { + "type": "math", + "text": "PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "; tyto výšky jsou navíc rovnoběžné (obě kolmé na " + }, + { + "type": "math", + "text": "PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "), takže " + }, + { + "type": "math", + "text": "CD \\parallel PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "DuoGeo 2025" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Je dán čtyřúhelník " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": " s\u00A0průsečíkem úhlopříček " + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": ". Předpokládejme, že velikosti úhlů " + }, + { + "type": "math", + "text": "BAC", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DBA", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou po řadě " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Na úsečce " + }, + { + "type": "math", + "text": "BT", + "isDisplay": false + }, + { + "type": "text", + "text": " leží bod " + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " takový, že " + }, + { + "type": "math", + "text": "CZ \\perp BT", + "isDisplay": false + }, + { + "type": "text", + "text": ". Předpokládejme, že přímka " + }, + { + "type": "math", + "text": "CZ", + "isDisplay": false + }, + { + "type": "text", + "text": " protne úsečku " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " v\u00A0bodě " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nechť " + }, + { + "type": "math", + "text": "R", + "isDisplay": false + }, + { + "type": "text", + "text": " je průsečík úseček " + }, + { + "type": "math", + "text": "AT", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ". Předpokládejme, že " + }, + { + "type": "math", + "text": "|AM| = |AR|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|MR| + |TD| = 14", + "isDisplay": false + }, + { + "type": "text", + "text": ". Určete velikost úsečky " + }, + { + "type": "math", + "text": "|BZ|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dopočítávejte úhly, dokud nenajdeme rovnoramenný trojúhelník." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Trpělivým počítáním úhlů se dá dojít k\u00A0tomu, že " + }, + { + "type": "math", + "text": "DTR", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnoramenný. To by mělo vnést světlo do podmínky " + }, + { + "type": "math", + "text": "|MR|+|TD|=14", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podmínka se po dokázání rovnoramenností přeloží jako " + }, + { + "type": "math", + "text": "|MD|=14", + "isDisplay": false + }, + { + "type": "text", + "text": ". To pomůže později – aktuálně už víc úhlů nespočítáme a\u00A0musíme ještě něco najít ze světa délek. Klíčem je najít pěkný pravoúhlý trojúhelník." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dá se spočítat, že úhly trojúhelníku " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": " jsou " + }, + { + "type": "math", + "text": "90-60-30", + "isDisplay": false + }, + { + "type": "text", + "text": ". Známe jeho přeponu " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ". Naše skvěle dokázané pomocné tvrzení nám teď dává další délku. Pak je to krok od řešení." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0trojúhelníku " + }, + { + "type": "math", + "text": "ATB", + "isDisplay": false + }, + { + "type": "text", + "text": " známe úhly při vrcholech " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0sice " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Tím pádem je vnější úhel při\u00A0" + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": " roven součtu těchto úhlů, konkrétně" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle DTR|=|\\angle TAB|=|\\angle TBA|=30^\\circ+45^\\circ=75^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0rovnoramenném trojúhelníku " + }, + { + "type": "math", + "text": "AMR", + "isDisplay": false + }, + { + "type": "text", + "text": " s\u00A0" + }, + { + "type": "math", + "text": "|AM| = |AR|", + "isDisplay": false + }, + { + "type": "text", + "text": " je úhel při vrcholu " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " roven " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže oba úhly při základně " + }, + { + "type": "math", + "text": "MR", + "isDisplay": false + }, + { + "type": "text", + "text": " mají velikost " + }, + { + "type": "math", + "text": "\\tfrac{1}{2}(180^\\circ - 30^\\circ) = 75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Speciálně " + }, + { + "type": "math", + "text": "|\\angle ARM| = 75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0tedy i\u00A0jeho vrcholový úhel " + }, + { + "type": "math", + "text": "|\\angle DRT|", + "isDisplay": false + }, + { + "type": "text", + "text": " má velikost " + }, + { + "type": "math", + "text": "75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-quad-30-45.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Trojúhelník " + }, + { + "type": "math", + "text": "DTR", + "isDisplay": false + }, + { + "type": "text", + "text": " má tedy dva úhly velikosti " + }, + { + "type": "math", + "text": "75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čili je rovnoramenný se\u00A0základnou " + }, + { + "type": "math", + "text": "TR", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|DT| = |DR|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Odtud" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|MD| = |MR| + |RD| = |MR| + |TD| = 14.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Navíc " + }, + { + "type": "math", + "text": "|\\angle TDR| = 180^\\circ - 2 \\cdot 75^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podívejme se nyní na trojúhelník " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Protože " + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " leží na úsečce " + }, + { + "type": "math", + "text": "BD", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "R", + "isDisplay": false + }, + { + "type": "text", + "text": " leží na úsečce " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ", je " + }, + { + "type": "math", + "text": "|\\angle MDZ| = |\\angle RDT| = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trojúhelník " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": " je tedy pravoúhlý (při\u00A0" + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": ") s\u00A0přeponou " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0úhlem " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " při\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže podle dřívějšího tvrzení o\u00A0" + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "-" + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "-" + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " trojúhelníku" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|MZ| = \\tfrac{1}{2} |MD| = 7.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Konečně trojúhelník " + }, + { + "type": "math", + "text": "MZB", + "isDisplay": false + }, + { + "type": "text", + "text": " má pravý úhel při\u00A0" + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0úhel " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " při\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže i\u00A0" + }, + { + "type": "math", + "text": "|\\angle BMZ| = 45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Je tedy rovnoramenný pravoúhlý a\u00A0" + }, + { + "type": "math", + "text": "|BZ| = |MZ| = 7", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + } + ] + } + } + ] + }, + "images": [ + { + "contentId": "angle-basics-1/angles-vertical.svg", + "originalId": "angles-vertical.pdf", + "width": "82.61pt", + "height": "65.04pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-corresponding.svg", + "originalId": "angles-corresponding.pdf", + "width": "106.36pt", + "height": "66.59pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-alternate.svg", + "originalId": "angles-alternate.pdf", + "width": "100.63pt", + "height": "58.17pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-supplementary.svg", + "originalId": "angles-supplementary.pdf", + "width": "85.68pt", + "height": "47.81pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-supplementary-derive-vertical.svg", + "originalId": 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b/web/src/content/handouts/angle-basics-1.en.json index 97322d5..f10cd39 100644 --- a/web/src/content/handouts/angle-basics-1.en.json +++ b/web/src/content/handouts/angle-basics-1.en.json @@ -5085,6 +5085,2490 @@ } ] } + }, + { + "title": "Problems", + "level": 1, + "text": { + "content": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In this section you will find various problems for which the knowledge from this handout is sufficient – no advanced topics like " + }, + { + "type": "italic", + "content": [ + { + "type": "text", + "text": "inscribed and central angles" + } + ] + }, + { + "type": "text", + "text": " are needed (we will look at those problems later)." + } + ], + "highligted": false + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "MO district round Z8 2025" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In a triangle " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": ", point " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " lies on side " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": " and point " + }, + { + "type": "math", + "text": "E", + "isDisplay": false + }, + { + "type": "text", + "text": " on side " + }, + { + "type": "math", + "text": "AC", + "isDisplay": false + }, + { + "type": "text", + "text": ", with " + }, + { + "type": "math", + "text": "AB = BE = EC = CD", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "BD = DE", + "isDisplay": false + }, + { + "type": "text", + "text": ". Determine the measures of the angles " + }, + { + "type": "math", + "text": "\\angle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "\\angle BAD", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Set " + }, + { + "type": "math", + "text": "\\gamma = \\angle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": " and try to express every angle in the figure in terms of\u00A0" + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". We have plenty of isosceles triangles." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "One possible path is to express the angles " + }, + { + "type": "math", + "text": "\\angle CBE", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\angle BED", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\angle CDE", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\angle CED", + "isDisplay": false + }, + { + "type": "text", + "text": " in terms of " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " in turn. Can we now write down an equation somewhere from which " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " can be computed?" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Look at the angle sum in triangle\u00A0" + }, + { + "type": "math", + "text": "BEC", + "isDisplay": false + }, + { + "type": "text", + "text": ". It should let us compute " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Computing\u00A0" + }, + { + "type": "math", + "text": "\\angle BAD", + "isDisplay": false + }, + { + "type": "text", + "text": " takes more steps. The first is to compute as many angles as possible and notice something." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Once we compute the angles in " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": ", we see that " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " is isosceles." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Now we can exploit symmetry; one option is to prove the congruence of the triangles " + }, + { + "type": "math", + "text": "EAD", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "DBE", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Set " + }, + { + "type": "math", + "text": "\\gamma = \\angle ACB", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In the isosceles " + }, + { + "type": "math", + "text": "\\triangle BEC", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "BE=EC", + "isDisplay": false + }, + { + "type": "text", + "text": ") we have " + }, + { + "type": "math", + "text": "\\angle EBC = \\angle ECB = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Since " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " lies on " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\angle DBE = \\angle EBC = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", and by the isosceles property of " + }, + { + "type": "math", + "text": "\\triangle BDE", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "BD=DE", + "isDisplay": false + }, + { + "type": "text", + "text": ") also " + }, + { + "type": "math", + "text": "\\angle DEB = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". The exterior-angle theorem in " + }, + { + "type": "math", + "text": "\\triangle BDE", + "isDisplay": false + }, + { + "type": "text", + "text": " at\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " gives" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\angle EDC = \\angle DBE + \\angle DEB = 2\\gamma,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "and in the isosceles " + }, + { + "type": "math", + "text": "\\triangle CDE", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "CD=CE", + "isDisplay": false + }, + { + "type": "text", + "text": ") we have " + }, + { + "type": "math", + "text": "\\angle CED = \\angle CDE = 2\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". From its angle sum " + }, + { + "type": "math", + "text": "\\gamma + 4\\gamma = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", so " + }, + { + "type": "math", + "text": "\\gamma = 36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-isosceles-chain.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Now note that " + }, + { + "type": "math", + "text": "\\angle AEB", + "isDisplay": false + }, + { + "type": "text", + "text": " is in a linear pair with " + }, + { + "type": "math", + "text": "\\angle BEC = 3\\gamma = 108^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", so it equals " + }, + { + "type": "math", + "text": "72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". By the isosceles property of " + }, + { + "type": "math", + "text": "\\triangle BAE", + "isDisplay": false + }, + { + "type": "text", + "text": " we also get " + }, + { + "type": "math", + "text": "\\angle BAE = 72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". From " + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": ", where the angles at " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " are " + }, + { + "type": "math", + "text": "36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " respectively, we compute " + }, + { + "type": "math", + "text": "\\angle CBA = 72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", so triangle " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " is isosceles." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Now observe that from " + }, + { + "type": "math", + "text": "CA = CB", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "CE = CD", + "isDisplay": false + }, + { + "type": "text", + "text": " we get " + }, + { + "type": "math", + "text": "EA = DB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Consider the triangles " + }, + { + "type": "math", + "text": "EAD", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "DBE", + "isDisplay": false + }, + { + "type": "text", + "text": ": they are isosceles with equal apex angle (equal to " + }, + { + "type": "math", + "text": "180^\\circ - 2\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ") and equal legs, so by " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": " they are congruent. This gives " + }, + { + "type": "math", + "text": "\\angle DAE = \\angle DBE = 36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Finally," + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\angle BAD = \\angle BAE - \\angle DAE = 2\\gamma - \\gamma = \\gamma = 36^\\circ.", + "isDisplay": true + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "DuoGeo 2025" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Inside the square " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": ", equilateral triangles " + }, + { + "type": "math", + "text": "ABX", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "CDY", + "isDisplay": false + }, + { + "type": "text", + "text": " have been drawn. Determine the sum of the marked angles." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-square-equilateral-statement.svg", + "scale": 0.7, + "isInline": false + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Try first to compute as many angles as possible in the figure." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The key to computing the final angle is to find a suitable isosceles triangle using equal lengths." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "By symmetry, all four marked angles are equal. So it suffices to find the measure of one of them – we focus on " + }, + { + "type": "math", + "text": "\\angle YDX", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "First, " + }, + { + "type": "math", + "text": "\\angle XAD = \\angle BAD - \\angle BAX = 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "; analogously " + }, + { + "type": "math", + "text": "\\angle ADY = \\angle ADC - \\angle YDC = 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Next, " + }, + { + "type": "math", + "text": "DA = AB = AX", + "isDisplay": false + }, + { + "type": "text", + "text": " (the first is a side of the square, the second a side of the equilateral triangle " + }, + { + "type": "math", + "text": "ABX", + "isDisplay": false + }, + { + "type": "text", + "text": "), so " + }, + { + "type": "math", + "text": "\\triangle AXD", + "isDisplay": false + }, + { + "type": "text", + "text": " is isosceles with base " + }, + { + "type": "math", + "text": "DX", + "isDisplay": false + }, + { + "type": "text", + "text": ". From " + }, + { + "type": "math", + "text": "\\angle XAD = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " and the angle sum of the triangle," + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\angle ADX = \\tfrac{1}{2}\\bigl(180^\\circ - 30^\\circ\\bigr) = 75^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The desired angle is the difference of the two already computed:" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\angle YDX = \\angle ADX - \\angle ADY = 75^\\circ - 30^\\circ = 45^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The sum of all four marked angles is therefore " + }, + { + "type": "math", + "text": "4 \\cdot 45^\\circ = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-square-equilateral-solution.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "MO school round C 2024" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In a trapezoid " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": " with " + }, + { + "type": "math", + "text": "AB \\parallel CD", + "isDisplay": false + }, + { + "type": "text", + "text": ", the interior angle bisectors at vertices " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " meet on segment\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Prove that " + }, + { + "type": "math", + "text": "AD + BC = AB", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Let\u00A0" + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " be the common intersection. We have parallel lines and angle bisectors, which gives plenty of equal angles." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The goal is to find isosceles triangles." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Let " + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " be the common intersection of the bisectors; by hypothesis it lies on segment\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". We will show that" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "AD = AP \\quad \\hbox{and} \\quad BC = BP,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "from which " + }, + { + "type": "math", + "text": "AD + BC = AP + BP = AB", + "isDisplay": false + }, + { + "type": "text", + "text": " directly." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Since " + }, + { + "type": "math", + "text": "DP", + "isDisplay": false + }, + { + "type": "text", + "text": " is the bisector of the interior angle at\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": ", we have " + }, + { + "type": "math", + "text": "\\angle ADP = \\angle CDP", + "isDisplay": false + }, + { + "type": "text", + "text": ". By " + }, + { + "type": "math", + "text": "AB \\parallel CD", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "\\angle CDP", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "\\angle APD", + "isDisplay": false + }, + { + "type": "text", + "text": " are alternate angles cut by the transversal\u00A0" + }, + { + "type": "math", + "text": "DP", + "isDisplay": false + }, + { + "type": "text", + "text": ", so" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\angle APD = \\angle CDP = \\angle ADP.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In triangle " + }, + { + "type": "math", + "text": "ADP", + "isDisplay": false + }, + { + "type": "text", + "text": " the angles at vertices " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " are equal, so " + }, + { + "type": "math", + "text": "AD = AP", + "isDisplay": false + }, + { + "type": "text", + "text": ". An analogous argument at vertex\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " yields " + }, + { + "type": "math", + "text": "BC = BP", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-trapezoid-bisectors.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "MO school round A 2023" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In a convex pentagon " + }, + { + "type": "math", + "text": "ABCDE", + "isDisplay": false + }, + { + "type": "text", + "text": " we have " + }, + { + "type": "math", + "text": "\\angle CBA = \\angle BAE = \\angle AED", + "isDisplay": false + }, + { + "type": "text", + "text": ". On the sides " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "AE", + "isDisplay": false + }, + { + "type": "text", + "text": " there exist points " + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "Q", + "isDisplay": false + }, + { + "type": "text", + "text": " respectively such that " + }, + { + "type": "math", + "text": "AP = BC = QE", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "AQ = BP = DE", + "isDisplay": false + }, + { + "type": "text", + "text": ". Prove that " + }, + { + "type": "math", + "text": "CD \\parallel PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "We have plenty of equal angles and sides; try to find congruent triangles." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The key congruences are among triangles " + }, + { + "type": "math", + "text": "PBC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "QAP", + "isDisplay": false + }, + { + "type": "text", + "text": ", and " + }, + { + "type": "math", + "text": "DEQ", + "isDisplay": false + }, + { + "type": "text", + "text": ". They yield useful equal pieces. Keep looking for more congruences." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "The final congruence to prove is between " + }, + { + "type": "math", + "text": "CPQ", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "DQP", + "isDisplay": false + }, + { + "type": "text", + "text": ". Why does that suffice?" + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Since " + }, + { + "type": "math", + "text": "BC = AP = EQ", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "BP = AQ = ED", + "isDisplay": false + }, + { + "type": "text", + "text": ", and " + }, + { + "type": "math", + "text": "\\angle CBP = \\angle PAQ = \\angle QED", + "isDisplay": false + }, + { + "type": "text", + "text": ", the triangles " + }, + { + "type": "math", + "text": "PBC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "QAP", + "isDisplay": false + }, + { + "type": "text", + "text": ", and " + }, + { + "type": "math", + "text": "DEQ", + "isDisplay": false + }, + { + "type": "text", + "text": " are mutually congruent by " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-pentagon.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "From this " + }, + { + "type": "math", + "text": "CP = PQ = QD", + "isDisplay": false + }, + { + "type": "text", + "text": ", and also" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\begin{gather*}\n\\angle CPQ = 180^\\circ - \\angle BPC - \\angle APQ = \\\\ = 180^\\circ - \\angle PQA - \\angle EQD = \\angle PQD.\n\\end{gather*}", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Hence, by " + }, + { + "type": "math", + "text": "SAS", + "isDisplay": false + }, + { + "type": "text", + "text": ", the isosceles triangles " + }, + { + "type": "math", + "text": "CPQ", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "DQP", + "isDisplay": false + }, + { + "type": "text", + "text": " are also congruent. From this, the altitudes from vertices " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " to the common side " + }, + { + "type": "math", + "text": "PQ", + "isDisplay": false + }, + { + "type": "text", + "text": " are equal; these altitudes are also parallel (both perpendicular to " + }, + { + "type": "math", + "text": "PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "), so " + }, + { + "type": "math", + "text": "CD \\parallel PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "DuoGeo 2025" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "A quadrilateral " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": " is given, with " + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": " the intersection of its diagonals. Assume that the angles " + }, + { + "type": "math", + "text": "\\angle BAC", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "\\angle DBA", + "isDisplay": false + }, + { + "type": "text", + "text": " measure " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " respectively. On segment " + }, + { + "type": "math", + "text": "BT", + "isDisplay": false + }, + { + "type": "text", + "text": " there is a point " + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " such that " + }, + { + "type": "math", + "text": "CZ \\perp BT", + "isDisplay": false + }, + { + "type": "text", + "text": ". Assume that line " + }, + { + "type": "math", + "text": "CZ", + "isDisplay": false + }, + { + "type": "text", + "text": " meets segment " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " at point " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": ". Let " + }, + { + "type": "math", + "text": "R", + "isDisplay": false + }, + { + "type": "text", + "text": " be the intersection of segments " + }, + { + "type": "math", + "text": "AT", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ". Suppose that " + }, + { + "type": "math", + "text": "AM = AR", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "MR + TD = 14", + "isDisplay": false + }, + { + "type": "text", + "text": ". Determine the length of segment " + }, + { + "type": "math", + "text": "BZ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Keep computing angles until we find an isosceles triangle." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Patient angle chasing leads to " + }, + { + "type": "math", + "text": "DTR", + "isDisplay": false + }, + { + "type": "text", + "text": " being isosceles. That should shed light on the condition " + }, + { + "type": "math", + "text": "MR + TD = 14", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Once the isosceles relations are proved, the condition translates to " + }, + { + "type": "math", + "text": "MD = 14", + "isDisplay": false + }, + { + "type": "text", + "text": ". That will help later – for now we cannot compute more angles and must find something from the world of lengths. The key is to find a nice right triangle." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "One can show that the angles of triangle " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": " are " + }, + { + "type": "math", + "text": "90", + "isDisplay": false + }, + { + "type": "text", + "text": "-" + }, + { + "type": "math", + "text": "60", + "isDisplay": false + }, + { + "type": "text", + "text": "-" + }, + { + "type": "math", + "text": "30", + "isDisplay": false + }, + { + "type": "text", + "text": ". We know its hypotenuse " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ". Our nicely proved auxiliary statement now gives another length. From there it is one step to the solution." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In triangle " + }, + { + "type": "math", + "text": "ATB", + "isDisplay": false + }, + { + "type": "text", + "text": " we know the angles at vertices " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", namely " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Hence the exterior angle at\u00A0" + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": " equals the sum of these angles; concretely" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\angle DTR = \\angle TAB = \\angle TBA = 30^\\circ + 45^\\circ = 75^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "In the isosceles triangle " + }, + { + "type": "math", + "text": "AMR", + "isDisplay": false + }, + { + "type": "text", + "text": " with " + }, + { + "type": "math", + "text": "AM = AR", + "isDisplay": false + }, + { + "type": "text", + "text": ", the angle at vertex " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " equals " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", so both angles at the base " + }, + { + "type": "math", + "text": "MR", + "isDisplay": false + }, + { + "type": "text", + "text": " have measure " + }, + { + "type": "math", + "text": "\\tfrac{1}{2}(180^\\circ - 30^\\circ) = 75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". In particular " + }, + { + "type": "math", + "text": "\\angle ARM = 75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", and so its vertical angle " + }, + { + "type": "math", + "text": "\\angle DRT", + "isDisplay": false + }, + { + "type": "text", + "text": " also has measure " + }, + { + "type": "math", + "text": "75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-quad-30-45.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Triangle " + }, + { + "type": "math", + "text": "DTR", + "isDisplay": false + }, + { + "type": "text", + "text": " therefore has two angles of measure " + }, + { + "type": "math", + "text": "75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", so it is isosceles with base " + }, + { + "type": "math", + "text": "TR", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "DT = DR", + "isDisplay": false + }, + { + "type": "text", + "text": ". Hence" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "MD = MR + RD = MR + TD = 14.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Moreover, " + }, + { + "type": "math", + "text": "\\angle TDR = 180^\\circ - 2 \\cdot 75^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Now consider triangle " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Since " + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " lie on segment " + }, + { + "type": "math", + "text": "BD", + "isDisplay": false + }, + { + "type": "text", + "text": " and " + }, + { + "type": "math", + "text": "R", + "isDisplay": false + }, + { + "type": "text", + "text": " lies on segment " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ", we have " + }, + { + "type": "math", + "text": "\\angle MDZ = \\angle RDT = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Triangle " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": " is therefore right-angled (at\u00A0" + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": ") with hypotenuse " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": " and an angle of " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " at\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": ", so by the earlier statement about the " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "-" + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "-" + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " triangle" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "MZ = \\tfrac{1}{2} MD = 7.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Finally, triangle " + }, + { + "type": "math", + "text": "MZB", + "isDisplay": false + }, + { + "type": "text", + "text": " has a right angle at\u00A0" + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " and an angle of " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " at\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", so " + }, + { + "type": "math", + "text": "\\angle BMZ = 45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". It is therefore an isosceles right triangle and " + }, + { + "type": "math", + "text": "BZ = MZ = 7", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + } + ] + } } ] }, @@ -5263,6 +7747,48 @@ "width": "146.09pt", "height": "131.24pt", "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-isosceles-chain.svg", + "originalId": "angles-isosceles-chain.pdf", + "width": "142.28pt", + "height": "201.25pt", + "scale": 0.8 + }, + { + "contentId": "angle-basics-1/angles-square-equilateral-statement.svg", + "originalId": "angles-square-equilateral-statement.pdf", + "width": "193.47pt", + "height": "190pt", + "scale": 0.7 + }, + { + "contentId": "angle-basics-1/angles-square-equilateral-solution.svg", + "originalId": "angles-square-equilateral-solution.pdf", + "width": "193.47pt", + "height": "190pt", + "scale": 0.8 + }, + { + "contentId": "angle-basics-1/angles-trapezoid-bisectors.svg", + "originalId": "angles-trapezoid-bisectors.pdf", + "width": "241.91pt", + "height": "104.34pt", + "scale": 0.8 + }, + { + "contentId": "angle-basics-1/angles-pentagon.svg", + "originalId": "angles-pentagon.pdf", + "width": "189.99pt", + "height": "139.79pt", + "scale": 0.8 + }, + { + "contentId": "angle-basics-1/angles-quad-30-45.svg", + "originalId": "angles-quad-30-45.pdf", + "width": "335.58pt", + "height": "207.65pt", + "scale": 0.8 } ] } diff --git a/web/src/content/handouts/angle-basics-1.sk.json b/web/src/content/handouts/angle-basics-1.sk.json index 6ee46fe..4136ea4 100644 --- a/web/src/content/handouts/angle-basics-1.sk.json +++ b/web/src/content/handouts/angle-basics-1.sk.json @@ -2663,7 +2663,7 @@ "content": [ { "type": "text", - "text": "Pre účely nášho dôkazu uvážme teda, že sú zhodné práve uhly prihliehajúce zhodnej strane. Nech je to uhol " + "text": "Pre účely nášho dôkazu uvážme teda, že sú zhodné práve uhly priliehajúce zhodnej strane. Nech je to uhol " }, { "type": "math", @@ -4113,6 +4113,10 @@ "type": "math", "text": "60^\\circ", "isDisplay": false + }, + { + "type": "text", + "text": "." } ], "highligted": false @@ -5047,7 +5051,7 @@ "content": [ { "type": "text", - "text": "V trojuholníku sú dve strany zhodné práve keď sú zhodné im priľahlé uhly." + "text": "V trojuholníku sú dve strany zhodné práve keď sú zhodné im protiľahlé uhly." } ], "highligted": false @@ -5079,184 +5083,2692 @@ } ] } - } - ] - }, - "images": [ - { - "contentId": "angle-basics-1/angles-vertical.svg", - "originalId": "angles-vertical.pdf", - "width": "82.61pt", - "height": "65.04pt", - "scale": 1.0 - }, - { - "contentId": "angle-basics-1/angles-corresponding.svg", - "originalId": "angles-corresponding.pdf", - "width": "106.36pt", - "height": "66.59pt", - "scale": 1.0 - }, - { - "contentId": "angle-basics-1/angles-alternate.svg", - "originalId": "angles-alternate.pdf", - "width": "100.63pt", - "height": "58.17pt", - "scale": 1.0 - }, - { - "contentId": "angle-basics-1/angles-supplementary.svg", - 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Určte veľkosti uhlov " + }, + { + "type": "math", + "text": "ACB", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "BAD", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Označte " + }, + { + "type": "math", + "text": "\\gamma = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0snažte sa všetky uhly v\u00A0obrázku vyjadriť cez\u00A0" + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Máme veľa rovnoramenností." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Jedna možná cesta je postupne pomocou " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": " vyjadriť uhly " + }, + { + "type": "math", + "text": "CBE", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "BED", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "CDE", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "CED", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nevieme teraz niekde napísať rovnicu, z\u00A0ktorej sa dá vypočítať " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": "?" + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pozrite sa na súčet uhlov v\u00A0trojuholníku\u00A0" + }, + { + "type": "math", + "text": "BEC", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0neho by malo ísť spočítať " + }, + { + "type": "math", + "text": "\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "K\u00A0dopočítaniu\u00A0" + }, + { + "type": "math", + "text": "BAD", + "isDisplay": false + }, + { + "type": "text", + "text": " potrebujeme viac krokov. Prvý je spočítať čo najviac uhlov a\u00A0niečo si všimnúť." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Keď spočítame uhly v\u00A0" + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": ", tak dostaneme, že " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnoramenný." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Teraz už vieme použiť symetriu, jedna možnosť je dokázať zhodnosť trojuholníkov " + }, + { + "type": "math", + "text": "EAD", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DBE", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Označme " + }, + { + "type": "math", + "text": "\\gamma = |\\angle ACB|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0rovnoramennom " + }, + { + "type": "math", + "text": "\\triangle BEC", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "|BE|=|EC|", + "isDisplay": false + }, + { + "type": "text", + "text": ") je " + }, + { + "type": "math", + "text": "|\\angle EBC| = |\\angle ECB| = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Keďže " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " leží na " + }, + { + "type": "math", + "text": "BC", + "isDisplay": false + }, + { + "type": "text", + "text": ", je " + }, + { + "type": "math", + "text": "|\\angle DBE| = |\\angle EBC| = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0z\u00A0rovnoramennosti " + }, + { + "type": "math", + "text": "\\triangle BDE", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "|BD|=|DE|", + "isDisplay": false + }, + { + "type": "text", + "text": ") aj " + }, + { + "type": "math", + "text": "|\\angle DEB| = \\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Veta o\u00A0vonkajšom uhle v\u00A0" + }, + { + "type": "math", + "text": "\\triangle BDE", + "isDisplay": false + }, + { + "type": "text", + "text": " pri\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " dáva" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle EDC| = |\\angle DBE| + |\\angle DEB| = 2\\gamma,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "a\u00A0v\u00A0rovnoramennom " + }, + { + "type": "math", + "text": "\\triangle CDE", + "isDisplay": false + }, + { + "type": "text", + "text": " (" + }, + { + "type": "math", + "text": "|CD|=|CE|", + "isDisplay": false + }, + { + "type": "text", + "text": ") tak " + }, + { + "type": "math", + "text": "|\\angle CED| = |\\angle CDE| = 2\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ". Zo súčtu uhlov v\u00A0ňom " + }, + { + "type": "math", + "text": "\\gamma + 4\\gamma = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čiže " + }, + { + "type": "math", + "text": "\\gamma = 36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-isosceles-chain.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Teraz si vezmime, že " + }, + { + "type": "math", + "text": "|\\angle AEB|", + "isDisplay": false + }, + { + "type": "text", + "text": " je vedľajší uhol k\u00A0" + }, + { + "type": "math", + "text": "|\\angle BEC|=3\\gamma=108^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže je rovný " + }, + { + "type": "math", + "text": "72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0rovnoramennosti " + }, + { + "type": "math", + "text": "\\triangle BAE", + "isDisplay": false + }, + { + "type": "text", + "text": " tak aj " + }, + { + "type": "math", + "text": "|\\angle BAE|=72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0" + }, + { + "type": "math", + "text": "\\triangle ABC", + "isDisplay": false + }, + { + "type": "text", + "text": ", v\u00A0ktorom už máme pri " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " rovné postupne " + }, + { + "type": "math", + "text": "36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " dopočítame " + }, + { + "type": "math", + "text": "|\\angle CBA|=72^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže trojuholník " + }, + { + "type": "math", + "text": "ABC", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnoramenný." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Teraz si vezmime, že z\u00A0" + }, + { + "type": "math", + "text": "|CA|=|CB|", + "isDisplay": false + }, + { + "type": "text", + "text": " a " + }, + { + "type": "math", + "text": "|CE|=|CD|", + "isDisplay": false + }, + { + "type": "text", + "text": " vlastne máme " + }, + { + "type": "math", + "text": "|EA|=|DB|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Vezmime si trojuholníky " + }, + { + "type": "math", + "text": "EAD", + "isDisplay": false + }, + { + "type": "text", + "text": " a " + }, + { + "type": "math", + "text": "DBE", + "isDisplay": false + }, + { + "type": "text", + "text": ", tie sú rovnoramenné so zhodným uhlom pri vrchole (rovným " + }, + { + "type": "math", + "text": "180^\\circ-2\\gamma", + "isDisplay": false + }, + { + "type": "text", + "text": ") a\u00A0zhodnými ramenami, takže podľa " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " sú zhodné. To dáva " + }, + { + "type": "math", + "text": "|\\angle DAE|=|\\angle DBE|=36^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Na záver teda máme" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle BAD| = |\\angle BAE| - |\\angle DAE| = 2\\gamma - \\gamma = \\gamma = 36^\\circ.", + "isDisplay": true + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "DuoGeo 2025" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Do štvorca " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": " boli nakreslené rovnostranné trojuholníky " + }, + { + "type": "math", + "text": "ABX", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "CDY", + "isDisplay": false + }, + { + "type": "text", + "text": ". Určte súčet vyznačených uhlov." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-square-equilateral-statement.svg", + "scale": 0.7, + "isInline": false + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Skúste najprv spočítať čo najviac uhlov na obrázku." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Kľúčom k\u00A0dopočítaniu finálneho uhla je nájsť vhodný rovnoramenný trojuholník cez rovnaké dĺžky." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podľa symetrie sú všetky štyri vyznačené uhly zhodné. Stačí teda nájsť veľkosť jedného z\u00A0nich – zameriame sa na " + }, + { + "type": "math", + "text": "|\\angle YDX|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Najprv " + }, + { + "type": "math", + "text": "|\\angle XAD| = |\\angle BAD| - |\\angle BAX| = 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "; analogicky " + }, + { + "type": "math", + "text": "|\\angle ADY| = |\\angle ADC| - |\\angle YDC| = 90^\\circ - 60^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Ďalej " + }, + { + "type": "math", + "text": "|DA| = |AB| = |AX|", + "isDisplay": false + }, + { + "type": "text", + "text": " (prvé je strana štvorca, druhé strana rovnostranného trojuholníka " + }, + { + "type": "math", + "text": "ABX", + "isDisplay": false + }, + { + "type": "text", + "text": "), takže " + }, + { + "type": "math", + "text": "\\triangle AXD", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnoramenný so základňou " + }, + { + "type": "math", + "text": "DX", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0rovnosti " + }, + { + "type": "math", + "text": "|\\angle XAD| = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0zo súčtu uhlov v\u00A0trojuholníku" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle ADX| = \\tfrac{1}{2}\\bigl(180^\\circ - 30^\\circ\\bigr) = 75^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Hľadaný uhol je rozdielom dvoch už vypočítaných:" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle YDX| = |\\angle ADX| - |\\angle ADY| = 75^\\circ - 30^\\circ = 45^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Súčet všetkých štyroch vyznačených uhlov je teda " + }, + { + "type": "math", + "text": "4 \\cdot 45^\\circ = 180^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-square-equilateral-solution.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "MO školské kolo C 2024" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0lichobežníku " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": ", kde " + }, + { + "type": "math", + "text": "AB \\parallel CD", + "isDisplay": false + }, + { + "type": "text", + "text": ", sa osi vnútorných uhlov pri vrcholoch " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " pretínajú na úsečke\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Dokážte, že " + }, + { + "type": "math", + "text": "|AD| + |BC| = |AB|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Nech\u00A0" + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " je náš spoločný priesečník. Máme rovnobežnosť, máme os uhla, to dáva dosť rovnakých uhlov." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Cieľ je nájsť rovnoramenné trojuholníky." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Označme " + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " spoločný priesečník osí; podľa zadania leží na úsečke\u00A0" + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": ". Ukážeme, že" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|AD| = |AP| \\quad \\hbox{a} \\quad |BC| = |BP|,", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "odkiaľ priamo " + }, + { + "type": "math", + "text": "|AD| + |BC| = |AP| + |BP| = |AB|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pretože " + }, + { + "type": "math", + "text": "DP", + "isDisplay": false + }, + { + "type": "text", + "text": " je os vnútorného uhla pri\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": ", máme " + }, + { + "type": "math", + "text": "|\\angle ADP| = |\\angle CDP|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0rovnobežnosti " + }, + { + "type": "math", + "text": "AB \\parallel CD", + "isDisplay": false + }, + { + "type": "text", + "text": " sú " + }, + { + "type": "math", + "text": "|\\angle CDP|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|\\angle APD|", + "isDisplay": false + }, + { + "type": "text", + "text": " striedavé uhly pri priečke\u00A0" + }, + { + "type": "math", + "text": "DP", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle APD| = |\\angle CDP| = |\\angle ADP|.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0trojuholníku " + }, + { + "type": "math", + "text": "ADP", + "isDisplay": false + }, + { + "type": "text", + "text": " sú teda uhly pri vrcholoch " + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " zhodné, takže " + }, + { + "type": "math", + "text": "|AD| = |AP|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Analogickou úvahou pri vrchole\u00A0" + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " dostaneme " + }, + { + "type": "math", + "text": "|BC| = |BP|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-trapezoid-bisectors.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "MO školské kolo A 2023" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0konvexnom päťuholníku " + }, + { + "type": "math", + "text": "ABCDE", + "isDisplay": false + }, + { + "type": "text", + "text": " platí " + }, + { + "type": "math", + "text": "|\\angle CBA| = |\\angle BAE| = |\\angle AED|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Na stranách " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "AE", + "isDisplay": false + }, + { + "type": "text", + "text": " existujú po rade body " + }, + { + "type": "math", + "text": "P", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "Q", + "isDisplay": false + }, + { + "type": "text", + "text": " tak, že " + }, + { + "type": "math", + "text": "|AP| = |BC| = |QE|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|AQ| = |BP| = |DE|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Dokážte, že " + }, + { + "type": "math", + "text": "CD \\parallel PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Máme veľa rovnakých uhlov a\u00A0strán, skúste nájsť zhodné trojuholníky." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Kľúčové sú zhodné trojuholníky " + }, + { + "type": "math", + "text": "PBC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "QAP", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DEQ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0nich dostaneme užitočné zhodné vecičky. Pokračujte v\u00A0hľadaní zhodnosti." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Finálna zhodnosť na dokázanie je " + }, + { + "type": "math", + "text": "CPQ", + "isDisplay": false + }, + { + "type": "text", + "text": " a " + }, + { + "type": "math", + "text": "DQP", + "isDisplay": false + }, + { + "type": "text", + "text": ". Prečo to stačí?" + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Keďže " + }, + { + "type": "math", + "text": "|BC| = |AP| = |EQ|", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "|BP| = |AQ| = |ED|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|\\angle CBP| = |\\angle PAQ| = |\\angle QED|", + "isDisplay": false + }, + { + "type": "text", + "text": ", sú podľa vety " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " trojuholníky " + }, + { + "type": "math", + "text": "PBC", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "QAP", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DEQ", + "isDisplay": false + }, + { + "type": "text", + "text": " navzájom zhodné." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-pentagon.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Odtiaľ plynie " + }, + { + "type": "math", + "text": "|CP| = |PQ| = |QD|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0tiež" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "\\begin{gather*}\n|\\angle CPQ| = 180^\\circ - |\\angle BPC| - |\\angle APQ| = \\\\ = 180^\\circ - |\\angle PQA| - |\\angle EQD| = |\\angle PQD|.\n\\end{gather*}", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podľa vety " + }, + { + "type": "math", + "text": "sus", + "isDisplay": false + }, + { + "type": "text", + "text": " sú teda zhodné aj rovnoramenné trojuholníky " + }, + { + "type": "math", + "text": "CPQ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DQP", + "isDisplay": false + }, + { + "type": "text", + "text": ". Z\u00A0toho vyplýva zhodnosť ich výšok z\u00A0vrcholov " + }, + { + "type": "math", + "text": "C", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": " na spoločnú stranu " + }, + { + "type": "math", + "text": "PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "; tieto výšky sú zároveň rovnobežné (obe kolmé na " + }, + { + "type": "math", + "text": "PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "), takže " + }, + { + "type": "math", + "text": "CD \\parallel PQ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + }, + { + "type": "problem", + "difficulty": 0, + "title": { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "DuoGeo 2025" + } + ], + "highligted": false + }, + "body": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Je daný štvoruholník " + }, + { + "type": "math", + "text": "ABCD", + "isDisplay": false + }, + { + "type": "text", + "text": " s\u00A0priesečníkom uhlopriečok " + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": ". Predpokladajme, že veľkosti uhlov " + }, + { + "type": "math", + "text": "BAC", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "DBA", + "isDisplay": false + }, + { + "type": "text", + "text": " sú postupne " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Na úsečke " + }, + { + "type": "math", + "text": "BT", + "isDisplay": false + }, + { + "type": "text", + "text": " leží bod " + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " taký, že " + }, + { + "type": "math", + "text": "CZ \\perp BT", + "isDisplay": false + }, + { + "type": "text", + "text": ". Predpokladajme, že priamka " + }, + { + "type": "math", + "text": "CZ", + "isDisplay": false + }, + { + "type": "text", + "text": " pretne úsečku " + }, + { + "type": "math", + "text": "AB", + "isDisplay": false + }, + { + "type": "text", + "text": " v\u00A0bode " + }, + { + "type": "math", + "text": "M", + "isDisplay": false + }, + { + "type": "text", + "text": ". Nech " + }, + { + "type": "math", + "text": "R", + "isDisplay": false + }, + { + "type": "text", + "text": " je priesečník úsečiek " + }, + { + "type": "math", + "text": "AT", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ". Predpokladajme, že " + }, + { + "type": "math", + "text": "|AM| = |AR|", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|MR| + |TD| = 14", + "isDisplay": false + }, + { + "type": "text", + "text": ". Určte veľkosť úsečky " + }, + { + "type": "math", + "text": "|BZ|", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + "hints": [ + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dopočítavajte uhly, až kým nenájdeme rovnoramenný trojuholník." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Trpezlivým počítaním uhlov sa dá dôjsť k\u00A0tomu, že " + }, + { + "type": "math", + "text": "DTR", + "isDisplay": false + }, + { + "type": "text", + "text": " je rovnoramenný. To by malo vniesť svetlo do podmienky " + }, + { + "type": "math", + "text": "|MR|+|TD|=14", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Podmienka sa po dokázaní rovnoramenností preloží ako " + }, + { + "type": "math", + "text": "|MD|=14", + "isDisplay": false + }, + { + "type": "text", + "text": ". To pomôže neskôr – aktuálne už viac uhlov nespočítame a\u00A0musíme ešte niečo nájsť zo sveta dĺžok. Kľúčom je nájsť pekný pravouhlý trojuholník." + } + ], + "highligted": false + } + ], + [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Dá sa spočítať, že uhly trojuholníka " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": " sú " + }, + { + "type": "math", + "text": "90-60-30", + "isDisplay": false + }, + { + "type": "text", + "text": ". Poznáme jeho preponu " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ". Naše skvelé dokázané pomocné tvrdenie nám teraz dáva ďalšiu dĺžku. Potom je to krok od riešenia." + } + ], + "highligted": false + } + ] + ], + "solution": [ + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0trojuholníku " + }, + { + "type": "math", + "text": "ATB", + "isDisplay": false + }, + { + "type": "text", + "text": " poznáme uhly pri vrcholoch " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0síce " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Tým pádom je vonkajší uhol pri\u00A0" + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": " rovný súčtu týchto uhlov, konkrétne" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|\\angle DTR|=|\\angle TAB|=|\\angle TBA|=30^\\circ+45^\\circ=75^\\circ.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "V\u00A0rovnoramennom trojuholníku " + }, + { + "type": "math", + "text": "AMR", + "isDisplay": false + }, + { + "type": "text", + "text": " s\u00A0" + }, + { + "type": "math", + "text": "|AM| = |AR|", + "isDisplay": false + }, + { + "type": "text", + "text": " je uhol pri vrchole " + }, + { + "type": "math", + "text": "A", + "isDisplay": false + }, + { + "type": "text", + "text": " rovný " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže oba uhly pri základni " + }, + { + "type": "math", + "text": "MR", + "isDisplay": false + }, + { + "type": "text", + "text": " majú veľkosť " + }, + { + "type": "math", + "text": "\\tfrac{1}{2}(180^\\circ - 30^\\circ) = 75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Špeciálne " + }, + { + "type": "math", + "text": "|\\angle ARM| = 75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", a\u00A0teda aj jeho vrcholový uhol " + }, + { + "type": "math", + "text": "|\\angle DRT|", + "isDisplay": false + }, + { + "type": "text", + "text": " má veľkosť " + }, + { + "type": "math", + "text": "75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "image", + "id": "angle-basics-1/angles-quad-30-45.svg", + "scale": 0.8, + "isInline": false + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Trojuholník " + }, + { + "type": "math", + "text": "DTR", + "isDisplay": false + }, + { + "type": "text", + "text": " má teda dva uhly veľkosti " + }, + { + "type": "math", + "text": "75^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ", čiže je rovnoramenný so\u00A0základňou " + }, + { + "type": "math", + "text": "TR", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "|DT| = |DR|", + "isDisplay": false + }, + { + "type": "text", + "text": ". Odtiaľ" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|MD| = |MR| + |RD| = |MR| + |TD| = 14.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Navyše " + }, + { + "type": "math", + "text": "|\\angle TDR| = 180^\\circ - 2 \\cdot 75^\\circ = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Pozrime sa teraz na trojuholník " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Pretože " + }, + { + "type": "math", + "text": "T", + "isDisplay": false + }, + { + "type": "text", + "text": ", " + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " ležia na úsečke " + }, + { + "type": "math", + "text": "BD", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0" + }, + { + "type": "math", + "text": "R", + "isDisplay": false + }, + { + "type": "text", + "text": " leží na úsečke " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": ", je " + }, + { + "type": "math", + "text": "|\\angle MDZ| = |\\angle RDT| = 30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Trojuholník " + }, + { + "type": "math", + "text": "MDZ", + "isDisplay": false + }, + { + "type": "text", + "text": " je teda pravouhlý (pri\u00A0" + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": ") s\u00A0preponou " + }, + { + "type": "math", + "text": "MD", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0uhlom " + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " pri\u00A0" + }, + { + "type": "math", + "text": "D", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže podľa skoršieho tvrdenia o\u00A0" + }, + { + "type": "math", + "text": "30^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "-" + }, + { + "type": "math", + "text": "60^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": "-" + }, + { + "type": "math", + "text": "90^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " trojuholníku" + } + ], + "highligted": false + }, + { + "type": "math", + "text": "|MZ| = \\tfrac{1}{2} |MD| = 7.", + "isDisplay": true + }, + { + "type": "paragraph", + "content": [ + { + "type": "text", + "text": "Konečne trojuholník " + }, + { + "type": "math", + "text": "MZB", + "isDisplay": false + }, + { + "type": "text", + "text": " má pravý uhol pri\u00A0" + }, + { + "type": "math", + "text": "Z", + "isDisplay": false + }, + { + "type": "text", + "text": " a\u00A0uhol " + }, + { + "type": "math", + "text": "45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": " pri\u00A0" + }, + { + "type": "math", + "text": "B", + "isDisplay": false + }, + { + "type": "text", + "text": ", takže aj " + }, + { + "type": "math", + "text": "|\\angle BMZ| = 45^\\circ", + "isDisplay": false + }, + { + "type": "text", + "text": ". Je teda rovnoramenný pravouhlý a\u00A0" + }, + { + "type": "math", + "text": "|BZ| = |MZ| = 7", + "isDisplay": false + }, + { + "type": "text", + "text": "." + } + ], + "highligted": false + } + ] + } + ] + } + } + ] + }, + "images": [ + { + "contentId": "angle-basics-1/angles-vertical.svg", + "originalId": "angles-vertical.pdf", + "width": "82.61pt", + "height": "65.04pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-corresponding.svg", + "originalId": "angles-corresponding.pdf", + "width": "106.36pt", + "height": "66.59pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-alternate.svg", + "originalId": "angles-alternate.pdf", + "width": "100.63pt", + "height": "58.17pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-supplementary.svg", + "originalId": "angles-supplementary.pdf", + "width": "85.68pt", + "height": "47.81pt", + "scale": 1.0 + }, + { + "contentId": "angle-basics-1/angles-supplementary-derive-vertical.svg", + "originalId": 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