Difference between revisions of "2021 CIME I Problems/Problem 14"

Latest revision as of 13:10, 13 June 2025

Problem

Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$ . The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects lines $BH$ and $CH$ at $X$ and $Y$ , and $BY\parallel CX$ . Let line $AO$ intersect $\overline{BC}$ at $D$ . Suppose that $AO=25, BC=49$ , and $AD=a-b\sqrt{c}$ for positive integers $a, b, c,$ where $c$ is not divisible by the square of any prime. Find $a+b+c$ .

Solution 1 by TheUltimate123

Let $H$ be the orthocenter of $\triangle ABC$ , and let $E$ , $F$ be the feet of the altitudes from $B, C$ . Also let $A'$ be the antipode of $A$ on the circumcircle and let $S=\overline{AH}\cap\overline{EF}$ , as shown below: $[asy] size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=heavycyan; pen qua=paleblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,H,X,Y,EE,F,Ap,SS,D; O=(0,0); A=dir(147.55); B=dir(195); C=dir(345); H=A+B+C; X=extension(B,H,A,A+rotate(90)*A); Y=extension(C,H,A,X); EE=foot(B,C,A); F=foot(C,A,B); Ap=-A; SS=extension(A,H,EE,F); D=extension(A,O,B,C); draw(A--Ap,qua+Dotted); draw(B--Ap--C,qua); draw(B--X,tri); draw(C--Y,tri); draw(EE--F,sec+Dotted); draw(X--Y,sec); draw(B--Y,sec); draw(C--X,sec); filldraw(circumcircle(B,C,X),sfil,sec); filldraw(A--B--C--cycle,fil,pri); filldraw(circle(O,1),fil,pri); dot("$A$",A,A); dot("$B$",B,dir(210)); dot("$C$",C,dir(-30)); dot("$H$",H,dir(300)); dot("$X$",X,N); dot("$Y$",Y,W); dot("$E$",EE,dir(30)); dot("$F$",F,dir(120)); dot("$S$",SS,N); dot("$A'$",Ap,Ap); dot("$D$",D,S);[/asy]$ Disregarding the condition $\overline{BY}\parallel\overline{CX}$ , we contend:

$\textbf{Claim}:$ In general, $BCXY$ is cyclic.

$\textbf{Proof}.$ Recall that $\overline{AA}\parallel\overline{EF}$ , so the claim follows from Reims' theorem on $BCEF, BCXY. \blacksquare$

With $\overline{BY}\parallel\overline{CX}$ , it follows that $BCXY$ is an isosceles trapezoid. In particular, $HB=HY$ and $HC=HX$ . Since $\overline{SF}\parallel\overline{AY}$ , we have $\frac{HS}{HA}=\frac{HF}{HY}=\frac{HF}{HB}=\cos A.$ But note that $\triangle AEF\cup H\sim\triangle ABC\cup A'$ , so $\frac{AD}{2R}=1-\frac{A'D}{2R}=1-\frac{HS}{HA}=1-\cos A,$ i.e.\ $AD=2R(1-\cos A)$ . We are given $R=25$ , and by the law of sines, $\sin A=\frac{49}{50}$ , so $\cos A=\frac{3\sqrt{11}}{50}$ , and $AD=50-3\sqrt{11}$ , so $50+3+11=\boxed{064}$ .

Solution 2

Let $E$ and $F$ be the feet of the altitudes from $B$ to $\overline{AC}$ and from $C$ to $\overline{AB}$ , respectively. It is clear that $\overline{EF} \parallel \overline{XY}$ , as $\angle XAE = \angle AEF = \angle B$ and $\angle YAF = \angle AFE = \angle C$ . We conjecture that quadrilateral $BCXY$ is cyclic because it looks cyclic. Indeed, note that for this to be true we need $\angle XYC = \angle XBC$ , but $\overline{EF} \parallel \overline{XY}$ reduces this condition to $\angle EFC = \angle EBC$ , which is obvious as $BCEF$ is cyclic with $\angle BEC = \angle BFC = 90^{\circ}$ .

$[asy] import olympiad; size(7cm); defaultpen(linewidth(0.7)); usepackage("mathptmx"); pair A = (-15, 20), B = (-24, -7), C = (24, -7), O = (0, 0), H = A + B + C, Ap = -A, X = extension(B, H, A, A + rotate(90, O) * A), Y = extension(C, H, A, A + rotate(90, O) * A), D = extension(A, O, B, C), E = extension(A, C, B, X), F = extension(A, B, C, Y); draw(A--B--C--cycle, blue); draw(B--X^^C--Y, fuchsia); draw(B--Y--X--C^^E--F, lightblue); draw(H--A--D, lightblue+dotted); draw(circle(O, 25), blue); draw(circumcircle(B, C, X), lightblue); dot("$A$", A, dir(127)); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$H$", H, dir(285)); dot("$X$", X, dir(80)); dot("$Y$", Y, dir(175)); dot("$D$", D, dir(270)); dot("$E$", E, dir(15)); dot("$F$", F, dir(210)); dot("$O$", O, dir(210)); dot(extension(A, H, E, F)); draw(anglemark(C, Y, X, 105), pink); draw(anglemark(C, F, E, 105), pink); draw(anglemark(C, B, E, 105), pink); [/asy]$

Thus, note that $BCXY$ cyclic along with $\overline{BY} \parallel \overline{CX}$ are enough to imply that it is an isosceles trapezoid and $XY = BC = 49$ . Then, going back to $\overline{EF} \parallel \overline{XY}$ , it is evident that $\triangle HEF$ and $\triangle HXY$ are directly similar and the ratio in which they are similar is given by $\frac{EF}{XY} =\frac{EF}{BC} =\frac{a \cos A}{a} = \cos A.$ Consider the geometric transformation consisting of a homothety centered at $A$ with ratio $\cos A$ followed by a reflection about the bisector of $\angle A$ , under which $\triangle ABC$ maps to $\triangle AEF$ . Then, point $D$ (which lies on both $\overline{BC}$ and $\overline{AO}$ ) maps to the point $D^{\prime}$ on $\overline{EF}$ such that $\overline{AO}$ and $\overline{AD^{\prime}}$ are isogonal wrt. $\angle A$ . But we know that the circumcenter and orthocenter of $\triangle ABC$ are isogonal conjugates, and this is enough to imply $D^{\prime}$ lies on $\overline{AH}$ . Now $\frac{HD^{\prime}}{HA} = \cos A \implies\frac{AD^{\prime}}{AH} = 1 - \cos A \implies AD^{\prime} = 2R \cos A(1 - \cos A) \implies AD = 2R(1 - \cos A)$ where the final step comes from undoing the homothety and reflection, thereby dividing all lengths by the ratio $\cos A$ . We are given $a = 49$ and $R = 25$ , thus $\sin A = \tfrac{a}{2R} = \tfrac{49}{50}$ . It follows that, as $\triangle ABC$ is an acute triangle, $\cos A = \sqrt{1 - \left(\tfrac{49}{50}\right)^{2}} = \tfrac{3\sqrt{11}}{50}$ . Finally, $50\left(1 - \tfrac{3\sqrt{11}}{50}\right) = 50 - 3\sqrt{11}$ and the answer is $50 + 3 + 11 = \boxed{64}$ .

~StressedPineapple

2021 CIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2021 CIME I Problems/Problem 14"

Latest revision as of 13:10, 13 June 2025

Contents

Problem

Solution 1 by TheUltimate123

Solution 2

See also

@@ Line 1: / Line 1: @@
+==Problem==
 Let <math>ABC</math> be an acute triangle with orthocenter <math>H</math> and circumcenter <math>O</math>. The tangent to the circumcircle of <math>\triangle ABC</math> at <math>A</math> intersects lines <math>BH</math> and <math>CH</math> at <math>X</math> and <math>Y</math>, and <math>BY\parallel CX</math>. Let line <math>AO</math> intersect <math>\overline{BC}</math> at <math>D</math>. Suppose that <math>AO=25, BC=49</math>, and <math>AD=a-b\sqrt{c}</math> for positive integers <math>a, b, c,</math> where <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>.
-==Solution by TheUltimate123==
+==Solution 1 by TheUltimate123==
-Let \(H\) be the orthocenter of \(\triangle ABC\), and let \(E\), \(F\) be the feet of the altitudes from \(B\), \(C\). Also let \(A'\) be the antipode of \(A\) on the circumcircle and let \(S=\overline{AH}\cap\overline{EF}\).
+Let <math>H</math> be the orthocenter of <math>\triangle ABC</math>, and let <math>E</math>, <math>F</math> be the feet of the altitudes from <math>B, C</math>. Also let <math>A'</math> be the antipode of <math>A</math> on the circumcircle and let <math>S=\overline{AH}\cap\overline{EF}</math>, as shown below:
 <asy>
 size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=heavycyan; pen qua=paleblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible;
@@ Line 11: / Line 12: @@
 dot("\(A\)",A,A); dot("\(B\)",B,dir(210)); dot("\(C\)",C,dir(-30)); dot("\(H\)",H,dir(300)); dot("\(X\)",X,N); dot("\(Y\)",Y,W); dot("\(E\)",EE,dir(30)); dot("\(F\)",F,dir(120)); dot("\(S\)",SS,N); dot("\(A'\)",Ap,Ap); dot("\(D\)",D,S);</asy>
-Disregarding the condition \(\overline{BY}\parallel\overline{CX}\), we contend:
+Disregarding the condition <math>\overline{BY}\parallel\overline{CX}</math>, we contend:
-<math>\textbf{Claim}:</math> In general, \(BCXY\) is cyclic.
+<math>\textbf{Claim}:</math> In general, <math>BCXY</math> is cyclic.
-Proof. Recall that \(\overline{AA}\parallel\overline{EF}\), so the claim follows from Reim's theorem on \(BCEF\), \(BCXY\). \(\blacksquare\)
+<math>\textbf{Proof}.</math> Recall that <math>\overline{AA}\parallel\overline{EF}</math>, so the claim follows from Reims' theorem on <math>BCEF, BCXY. \blacksquare</math>
-With \(\overline{BY}\parallel\overline{CX}\), it follows that \(BCXY\) is an isosceles trapezoid. In particular, \(HB=HY\) and \(HC=HX\). Since \(\overline{SF}\parallel\overline{AY}\), we have<cmath>\frac{HS}{HA}=\frac{HF}{HY}=\frac{HF}{HB}=\cos A.</cmath>But note that \(\triangle AEF\cup H\sim\triangle ABC\cup A'\), so<cmath>\frac{AD}{2R}=1-\frac{A'D}{2R}=1-\frac{HS}{HA}=1-\cos A,</cmath>i.e.\ \(AD=2R(1-\cos A)\). We are given \(R=25\), and by the law of sines, \(\sin A=\frac{49}{50}\), so \(\cos A=\frac{3\sqrt{11}}{50}\), and \(AD=50-3\sqrt{11}\), so <math>50+3+11=\boxed{064}</math>.
+With <math>\overline{BY}\parallel\overline{CX}</math>, it follows that <math>BCXY</math> is an isosceles trapezoid. In particular, <math>HB=HY</math> and <math>HC=HX</math>. Since <math>\overline{SF}\parallel\overline{AY}</math>, we have <cmath>\frac{HS}{HA}=\frac{HF}{HY}=\frac{HF}{HB}=\cos A.</cmath> But note that <math>\triangle AEF\cup H\sim\triangle ABC\cup A'</math>, so <cmath>\frac{AD}{2R}=1-\frac{A'D}{2R}=1-\frac{HS}{HA}=1-\cos A,</cmath> i.e.\ <math>AD=2R(1-\cos A)</math>. We are given <math>R=25</math>, and by the law of sines, <math>\sin A=\frac{49}{50}</math>, so <math>\cos A=\frac{3\sqrt{11}}{50}</math>, and <math>AD=50-3\sqrt{11}</math>, so <math>50+3+11=\boxed{064}</math>.
+==Solution 2==
+Let <math>E</math> and <math>F</math> be the feet of the altitudes from <math>B</math> to <math>\overline{AC}</math> and from <math>C</math> to <math>\overline{AB}</math>, respectively. It is clear that <math>\overline{EF} \parallel \overline{XY}</math>, as <math>\angle XAE = \angle AEF = \angle B</math> and <math>\angle YAF = \angle AFE = \angle C</math>. We conjecture that quadrilateral <math>BCXY</math> is cyclic because it looks cyclic. Indeed, note that for this to be true we need <math>\angle XYC = \angle XBC</math>, but <math>\overline{EF} \parallel \overline{XY}</math> reduces this condition to <math>\angle EFC = \angle EBC</math>, which is obvious as <math>BCEF</math> is cyclic with <math>\angle BEC = \angle BFC = 90^{\circ}</math>.
+<asy>
+import olympiad;
+size(7cm); defaultpen(linewidth(0.7));
+usepackage("mathptmx");
+pair A = (-15, 20), B = (-24, -7), C = (24, -7), O = (0, 0), H = A + B + C, Ap = -A, X = extension(B, H, A, A + rotate(90, O) * A), Y = extension(C, H, A, A + rotate(90, O) * A), D = extension(A, O, B, C), E = extension(A, C, B, X), F = extension(A, B, C, Y);
+draw(A--B--C--cycle, blue);
+draw(B--X^^C--Y, fuchsia);
+draw(B--Y--X--C^^E--F, lightblue);
+draw(H--A--D, lightblue+dotted);
+draw(circle(O, 25), blue);
+draw(circumcircle(B, C, X), lightblue);
+dot("$A$", A, dir(127));
+dot("$B$", B, dir(225));
+dot("$C$", C, dir(315));
+dot("$H$", H, dir(285));
+dot("$X$", X, dir(80));
+dot("$Y$", Y, dir(175));
+dot("$D$", D, dir(270));
+dot("$E$", E, dir(15));
+dot("$F$", F, dir(210));
+dot("$O$", O, dir(210));
+dot(extension(A, H, E, F));
+draw(anglemark(C, Y, X, 105), pink);
+draw(anglemark(C, F, E, 105), pink);
+draw(anglemark(C, B, E, 105), pink);
+</asy>
+Thus, note that <math>BCXY</math> cyclic along with <math>\overline{BY} \parallel \overline{CX}</math> are enough to imply that it is an isosceles trapezoid and <math>XY = BC = 49</math>. Then, going back to <math>\overline{EF} \parallel \overline{XY}</math>, it is evident that <math>\triangle HEF</math> and <math>\triangle HXY</math> are directly similar and the ratio in which they are similar is given by <cmath>\frac{EF}{XY} =\frac{EF}{BC} =\frac{a \cos A}{a} = \cos A.</cmath> Consider the geometric transformation consisting of a homothety centered at <math>A</math> with ratio <math>\cos A</math> followed by a reflection about the bisector of <math>\angle A</math>, under which <math>\triangle ABC</math> maps to <math>\triangle AEF</math>. Then, point <math>D</math> (which lies on both <math>\overline{BC}</math> and <math>\overline{AO}</math>) maps to the point <math>D^{\prime}</math> on <math>\overline{EF}</math> such that <math>\overline{AO}</math> and <math>\overline{AD^{\prime}}</math> are isogonal wrt. <math>\angle A</math>. But we know that the circumcenter and orthocenter of <math>\triangle ABC</math> are isogonal conjugates, and this is enough to imply <math>D^{\prime}</math> lies on <math>\overline{AH}</math>. Now <cmath>\frac{HD^{\prime}}{HA} = \cos A \implies\frac{AD^{\prime}}{AH} = 1 - \cos A \implies AD^{\prime} = 2R \cos A(1 - \cos A) \implies AD = 2R(1 - \cos A)</cmath> where the final step comes from undoing the homothety and reflection, thereby dividing all lengths by the ratio <math>\cos A</math>. We are given <math>a = 49</math> and <math>R = 25</math>, thus <math>\sin A = \tfrac{a}{2R} = \tfrac{49}{50}</math>. It follows that, as <math>\triangle ABC</math> is an acute triangle, <math>\cos A = \sqrt{1 - \left(\tfrac{49}{50}\right)^{2}} = \tfrac{3\sqrt{11}}{50}</math>. Finally, <cmath>50\left(1 - \tfrac{3\sqrt{11}}{50}\right) = 50 - 3\sqrt{11}</cmath> and the answer is <math>50 + 3 + 11 = \boxed{64}</math>.
+~StressedPineapple
 ==See also==