\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{Shortlist 2008 G6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 8}
\maketitle
\section*{Problem}
Let $ABCD$ be a given convex quadrilateral.
Prove that there exists a point $P$
inside the quadrilateral such that
\begin{align*}
\angle PAB + \angle PDC
&= \angle PBC + \angle PAD
= \angle PCD + \angle PBA \\
&= \angle PDA + \angle PCB = 90^{\circ}
\end{align*}
if and only if the diagonals $AC$ and $BD$ are perpendicular.
\section*{Video}
\href{https://www.youtube.com/watch?v=okxnxCnAAas&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/okxnxCnAAas}}
\newpage
\section*{Solution}
The problem is killed by quoting the following theorem:
\begin{theorem}
If $X$ is a point such that
$\angle AXD + \angle BXC = 180\dg$,
then the isogonal conjugate of $X$ exists.
\end{theorem}
\begin{proof}
[Proof, for completeness only]
It's enough for the projections of $X$
to the sides to be cyclic,
by considering the six-point circle.
Let them be $X_1$, $X_2$, $X_3$, $X_4$.
\begin{align*}
\dang XX_4X_1 &= \dang XAX_1 \\
&\vdotswithin=
\end{align*}
and this means $\dang X_1X_2X_3 = \dang X_1X_4X_3$.
\end{proof}
Now in one direction,
if the diagonals are perpendicular and meet at $Q$,
then its isogonal conjugate exists,
and is seen to have the desired property.
Conversely, given such a point $P$,
it has an isogonal conjugate $Q$ which
satisfies $\angle AQB = \angle BQC = \angle CQD = \angle DQA = 90\dg$,
which implies that $Q$ is the perpendicular
intersection of the diagonals.
\end{document}