Let $ABC$ be a triangle with $AH$ altitude. The point $K$ is chosen on the segment $AH$ as follows such that $AH =3KH$. Let $O$ be the center of the circle circumscribed around by triangle $ABC, M$ and $N$ be the midpoints of $AC$ and AB respectively. Lines $KO$ and $MN$ intersect at the point $Z$, a perpendicular to $OK$ passing through point $Z$ intersects lines $AC$ and $AB$ at points $X$ and $Y$ respectively. Prove that $\angle XKY =\angle CKB$.
Problem
Source: Ukraine TST 2018 p10
Tags: geometry, circumcircle, midpoints, altitude, equal angles
KST2003
30.04.2021 23:27
Let $\overline{KZ}$ meet $\overline{BC}$ at $D$, and let $E$ be the orthocenter of $\triangle ABC$. We will show that $\triangle KZX\sim \triangle KHB$, from which the conclusion easily follows. First notice that $ANOM$ and $OMZX$ are concyclic. Therefore by spiral similarity, we have \[\triangle OZX\sim\triangle ONA\sim\triangle EHB.\]Now it suffices to show that $\frac{EH}{KH}=\frac{OZ}{KZ}$. Let $G$ be the centroid of $\triangle ABC$, and let $G'$ be the point on segment $DZ$ such that $\frac{HG'}{G'Z}=2$. From the length condition, it is easy to see that $KG\parallel BC$. Moreover, \[\frac{DK}{KZ}=2=\frac{HG'}{G'Z},\]so $KG'\parallel BC$ as well, and hence $K,G,G'$ are collinear. Finally, since $\frac{EG}{GO}=2=\frac{HG'}{G'Z}$, we see that $EO\parallel HZ$ as desired. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pair A = dir(120); pair B = dir(222); pair C = dir(318); pair M = midpoint(A--C); pair N = midpoint(A--B); pair H = foot(A,B,C); pair K = (1/3)*A+(2/3)*H; pair O = circumcenter(A,B,C); pair Z = extension(M,N,O,K); pair D = extension(B,C,O,K); pair G = extension(B,M,C,N); pair G1 = (1/3)*H+(2/3)*Z; pair E = orthocenter(A,B,C); pair X = 2*foot(circumcenter(O,M,Z),A,C)-M; pair P = midpoint(B--C); draw(A--B--C--cycle); draw(A--H); draw(N--Z); draw(Z--D); draw(O--E); draw(Z--H); draw(D--B); draw(K--G1, dashed); draw(A--P, dotted); draw(circumcircle(A,B,C)); draw(circumcircle(A,M,N)); draw(circumcircle(O,M,Z)); draw(K--Z--X--cycle, fuchsia+1); draw(K--H--B--cycle, fuchsia+1); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,S); dot("$E$",E,W); dot("$D$",D,W); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); dot("$K$",K,dir(135)); dot("$O$",O,dir(135)); dot("$G$",G,(0,1)); dot("$G'$",G1,(0,1)); dot("$Z$",Z,dir(45)); dot("$X$",X,(1,0)); dot("$P$",P,dir(270)); [/asy][/asy]