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\title{Iberoamerican 2020/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 51}
\maketitle

\section*{Problem}
Let $ABC$ be an acute, scalene triangle.
Let $H$ be the orthocenter
and $O$ be the circumcenter of triangle $ABC$,
and let $P$ be a point interior to the segment $HO$.
The circle with center $P$ and radius $PA$
intersects the lines $AB$ and $AC$ again at $R$ and $S$, respectively.
Denote by $Q$ the symmetric point of $P$
with respect to the perpendicular bisector of $BC$.
Prove that points $P$, $Q$, $R$ and $S$ lie on the same circle.



\section*{External Link}
\url{https://aops.com/community/p18969935}

\newpage

\section*{Solution}
Let $P'$, $R'$, $S'$ be the midpoints of $AP$, $AR$, $AS$.
Apply \textbf{IMO Shortlist 2016 G5} to the circle centered
at $P'$ through $R'$ and $S'$ to deduce
that the circumcenter of $\triangle P'S'R'$ lies halfway between the $A$-altitude
and the perpendicular bisector of $\ol{BC}$.

Thus the circumcenter of $\triangle PSR$
lies on the perpendicular bisector of $\ol{BC}$.
So it passes through $Q$.

\end{document}
