Problem 65: IMO Shortlist 2015, G3

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Joined: Mar 10, 2015

by henderson, May 7, 2017, 5:21 PM

$\color{red}\bf{Problem \ 65}$ Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$ , and let $H$ be the foot of the altitude from $C$ . A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$ . Let $P$ be the intersection point of the lines $BD$ and $CH$ . Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$ . Prove that the lines $CQ$ and $AD$ meet on $\omega$ .
$\color{red}\bf{Solution}$ $[asy] unitsize(3cm); pointfontpen=fontsize(10); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pair B=dir(0), A=-B, C=dir(110), H=foot(C,A,B), N=WP(D(C--H),0.85), D=2*N-A, P=extension(B,D,C,H), Q=tangent(P,(B+D)/2,length(B-D)/2,2), K=extension(A,D,C,Q), M=(B+D)/2; D(CR((0,0),1,0,180),heavygreen); D(arc(M,length(K-M),degrees(B-M),degrees(D-M),CCW),red); D(A--B--C--cycle,linewidth(1)+pathpen); DPA(D--B--Q--cycle^^Q--P--D^^H--C); D(C--K--A,dashed+pathpen); /*Angle marks and pathticks*/ markscalefactor=0.01; DPA(rightanglemark(B,C,A)^^rightanglemark(B,H,C)^^rightanglemark(B,Q,D)); /* Dot and label points */ D("A",A,W); D("B",B,E); D("C",C,dir(C)); D("D",D,SSE); D("H",H); D("K",K,dir(K)); D("P",P,dir(P)); D("Q",Q,dir(Q)); [/asy]$
Let $K=AD\cap CQ$ . We claim that $\triangle ABC\sim\triangle DBQ$ . Indeed, $\angle BCA=\angle BQD=90^{\circ}$ and since $\triangle PDQ\sim\triangle PQB$ and $CH$ bisects $\overline{AD}$ , $\frac{DQ^2}{BQ^2}=\frac{PQ^2}{PB^2}=\frac{PD}{PB}=\frac{AH}{BH}=\tan^2B=\frac{AC^2}{BC^2},$ and our claim holds. Thus $B$ is the center of the spiral similarity $\overline{AD}\to\overline{CQ}$ , so $\angle KDB=\angle KQB$ , and $K$ lies on $\omega$ .

This post has been edited 1 time. Last edited by henderson, Jul 3, 2017, 7:41 PM