Let $ABC$ be a triangle and let $H$ be the orthogonal projection of $A$ on the line $BC$. Let $K$ be a point on the segment $AH$ such that $AH = 3 KH$. Let $O$ be the circumcenter of triangle $ABC$ and let $M$ and $N$ be the midpoints of sides $AC$ and $AB$ respectively. The lines $KO$ and $MN$ meet at a point $Z$ and the perpendicular at $Z$ to $OK$ meets lines $AB, AC$ at $X$ and $Y$ respectively. Show that $\angle XKY = \angle CKB$. Italy
Problem
Source: RMM 2018 SL G1
Tags: geometry, circumcircle
aqwxderf
21.02.2019 20:23
Let $G$ be the orthocenter $\bigtriangleup ABC$. Let $AH$ cut $MN$ at $D$. Let the foot from $O$ to $MN$ be $E$. Some simple angle chasing gives: $\bigtriangleup BHC$ $\sim$ $\bigtriangleup XOY$. By proving $\frac{KZ}{OZ}$ $=$ $\frac{KH}{GH}$ we prove that $K$ is corespondent in triangles $\bigtriangleup BHC$ and $\bigtriangleup XOY$ hence implying the result. $\frac{KZ}{OZ}$ $=$ $\frac{KD}{OE}$ $=$ $\frac{KH}{GH}$. Done!
MathInfinite
21.02.2019 20:31
Angle chasing gives $KO \cap BC \equiv (XBO) \cap (YCO)$. Denote the intersection point by $T$. Thus, $\Delta ABC \sim \Delta XTY$. Now consider the spiral similarity mapping $\mathcal{H} : \Delta ABC \mapsto \Delta XTY$. Note that $K$ is invariant in this mapping. Thus the claim holds true.
yayups
27.03.2020 06:39
Relabel the foot from $A$ to $BC$ to $D$, and let $H$ be the orthocenter of $ABC$. Let $O'$ denote the projection of $O$ onto $AD$, and let $U=MN\cap AD$. The main observation is that $\angle OMY=\angle OZY=\pi/2$, so $OMZY$ is cyclic. Therefore, using directed angles mod $\pi$, \[\angle OYZ=\angle OMZ=\angle OMN=\pi/2-\angle BCA=\angle HBC.\]Similarly $\angle OXZ=\angle HCB$, so $\triangle OXY\sim\triangle HBC$. Now, it just suffices to show that $KO/KZ=KH/KD$, as that would imply the figures $KOZXY$ and $KHDCB$ are similar, which would finish. Indeed, $KO/KZ=KO'/KU$, so it suffices to show $KO'/KU=KH/KD$, or that $KO'/KH=KU/KD$. However, $KU/KD=1/2$, so it suffices to show $KO'/KH=1/2$. This follows since $K$ is the projection of the centroid $G$ onto $AD$ (by the length condition in the problem), so $KO'/KH=GO/GH=1/2$, so we're done.