Let $ABC$ be an acute triangle. Let $P$ be a point on the circle $(ABC)$, and $Q$ be a point on the segment $AC$ such that $AP\perp BC$ and $BQ\perp AC$. Lot $O$ be the circumcenter of triangle $APQ$. Find the angle $OBC$.
Problem
Source: VII Caucasus Olympiad, Senior, Day2 P6
Tags: geometry, circumcircle
Jalil_Huseynov
13.03.2022 22:42
Let $H$ be orthocenter of $\triangle ABC$ and $D$ be foot of $A \text{-altitude}$ in $\triangle ABC$. Define $E=BQ\cap (APQ)$, where $E\ne Q$. $\angle EQA=90 \implies O\in EA$. $\angle APE=\angle AQE=90\implies BD||EP$ and since $DH=DP$ we get $BH=BE$. So $OB$ is $E\text{-midline}$ in $\triangle AEH \implies OB||AH\perp BC \implies OB\perp BC \implies \angle OBC=90.$
VicKmath7
13.03.2022 22:46
Sniped, I was just writing that. I got motivated for this construction (of $E$) by taking the reflection of $H$ wrt $B$ and doing PoP.
Mahdi_Mashayekhi
02.04.2022 13:57
Let $H$ be orthocenter of $ABC$, $S$ be foot of $A$ on $BC$ and $QB$ meet $APQ$ at $K$. we have $\angle APK = \angle 90 = \angle ASB \implies BS || KP$. It's well known that $P$ is reflection of $H$ across $BC$ so $HB = BK$ so $B$ lies on perpendicular bisector of $KP$ so $OB \perp KP || BC \implies OB \perp BC \implies \angle OBC = \angle 90$.