AoPS - Problem

Source: Centroamerican and Caribbean Math Olympiad 2024 P3

Tags: geometry, collinearity, circumcircle

jgamer2768

18.10.2024 03:12

Let $ABC$ be a triangle, $H$ its orthocenter, and $\Gamma$ its circumcircle. Let $J$ be the point diametrically opposite to $A$ on $\Gamma$. The points $D$, $E$ and $F$ are the feet of the altitudes from $A$, $B$ and $C$, respectively. The line $AD$ intersects $\Gamma$ again at $P$. The circumcircle of $EFP$ intersects $\Gamma$ again at $Q$. Let $K$ be the second point of intersection of $JH$ with $\Gamma$. Prove that $K$, $D$ and $Q$ are collinear.

Let $T=AK\cap BC=EF\cap BC$ since $K$ is the $A$-queue point in $\triangle ABC$ and $S$ be the $A$-HM point. Note that $T$ is radical radical center of $(AEF)$, $(PEQ)$ and $(ABC)$, therefore $T$, $P$ and $Q$ are colinear. So, lets prove that $\angle DAT=\angle QAT$. As $ACQK$ is cyclic, we get that $\angle QCA=\angle TKQ$ But, we know that $\angle TDH=\angle HAT=90°$ so $TKHD$ is cyclic and then $\angle TKD=\angle THD=\angle SHA=\angle SKA$ Hence, since it´s known that $HK$ bissects $\angle DKS$, and $\angle HKT=90°$, we get that $\angle TKD=\angle SKA=\angle TKQ$. $\blacksquare$