# 2021 CIME I Problems/Problem 14

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 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 $50+3+11=\boxed{064}$.