\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}}

\section*{External Link}
\url{https://aops.com/community/p1555924}

\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}
