2021 CIME I Problems/Problem 14

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)
Preceded by Problem 13	Followed by Problem 15
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2021 CIME I Problems/Problem 14

Contents

Problem

Solution 1 by TheUltimate123

Solution 2

See also