\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{SJMO 2020/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 28}
\maketitle

\section*{Problem}
Let $O$ and $\Omega$ denote the circumcenter and circumcircle,
respectively, of scalene triangle $\triangle ABC$.
Furthermore, let $M$ be the midpoint of side $BC$.
The tangent to $\Omega$ at $A$ intersects $BC$ and $OM$
at points $X$ and $Y$, respectively.
If the circumcircle of triangle $\triangle OXY$
intersects $\Omega$ at two distinct points $P$ and $Q$,
prove that line $PQ$ bisects $\ol{AM}$.

\section*{Video}
\href{https://www.youtube.com/watch?v=bI_9MjmoRgE&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/bI\_9MjmoRgE}}



\newpage

\section*{Solution}
In other words, we want the midpoint of $\ol{AM}$
to lie on the radical axis of the two circles.

However, we have
\begin{align*}
  \opname{Pow}(M, \Omega) &= -\frac14 BC^2 \\
  \opname{Pow}(A, \Omega) &= 0 \\
  \opname{Pow}(M, OXY) &= MO \cdot MY \\
  \opname{Pow}(A, OXY) &= -AX \cdot AY.
\end{align*}
Since the function
$\opname{Pow}(\bullet, \Omega) -
\opname{Pow}(\bullet, OXY)$
is linear, it thus suffices to show that
\[ AX \cdot AY = \frac14 BC^2 + MO \cdot MY \]
where the lengths are signed.

Let $YA = h$, $AO = OB = R$, so $YO = \sqrt{R^2+h^2}$.
Let $YM = kh$, $YX = k\sqrt{r^2+h^2}$, and $XM = kh$.
Note that $\frac14BC^2 = BM^2 = R^2 - OM^2$,
and overall we find our desired relation reads
\[ \left( k\sqrt{R^2+h^2}-h \right) \cdot h
  = \left( R^2 - \left( kh - \sqrt{R^2+h^2} \right)^2 \right)
  + \left( k \cdot h - \sqrt{R^2 + h^2} \right) \cdot k \cdot h  \]
which is easily verified.

\end{document}
