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

\begin{document}
\title{CMIMC 2017 G9}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 48}
\maketitle

\section*{Problem}
Let $\triangle ABC$ be an acute triangle with circumcenter $O$,
and let $Q\neq A$ denote the point on $\odot (ABC)$ for which $AQ\perp BC$.
The circumcircle of $\triangle BOC$ intersects
lines $AC$ and $AB$ for the second time at $D$ and $E$ respectively.
Suppose that $AQ$, $BC$, and $DE$ are concurrent.
If $OD=3$ and $OE=7$, compute $AQ$.

\section*{Video}
\href{https://www.youtube.com/watch?v=-beq_Y_Npsw&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/-beq\_Y\_Npsw}}



\newpage

\section*{Solution}
Let $L$ be the concurrency point,
so by radical axis $AEQD$ is cyclic.
\begin{claim*}
  Line $DLE$ is the Simson line from $Q$ to $ABC$.
\end{claim*}
\begin{proof}
  Note
  \[ \dang BED = \dang BCD = \dang BCA = \dang BQA = \dang BQL \]
  so $BEQL$ is cyclic, hence $\dang QEB = \dang QLB = 90\dg$,
  as desired.
\end{proof}


\begin{center}
\begin{asy}
import graph; size(10cm);
pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((13.34740,4.06778)--(9.,-9.)--(24.,-9.)--cycle, linewidth(0.6) + zzttqq);

draw((13.34740,4.06778)--(9.,-9.), linewidth(0.6) + zzttqq);
draw((9.,-9.)--(24.,-9.), linewidth(0.6) + zzttqq);
draw((24.,-9.)--(13.34740,4.06778), linewidth(0.6) + zzttqq);
draw(circle((16.5,-4.23806), 8.88403), linewidth(0.6) + qqwuqq);
draw((13.34740,4.06778)--(13.34740,-12.54391), linewidth(0.6));
draw((21.48280,-5.91210)--(8.37170,-10.88859), linewidth(0.6));
draw(circle((13.34740,-4.23806), 8.30585), linewidth(0.6) + blue);
draw((9.,-9.)--(8.37170,-10.88859), linewidth(0.6));
draw(circle((16.5,-12.54391), 8.29513), linewidth(0.6));
draw((9.,-9.)--(16.5,-20.81242), linewidth(0.6) + qqwuqq);
draw((24.,-9.)--(16.5,-20.81242), linewidth(0.6) + qqwuqq);
dot("$A$", (13.34740,4.06778), dir((9.614, 24.036)));
dot("$B$", (9.,-9.), dir((9.308, 23.273)));
dot("$C$", (24.,-9.), dir((9.136, 23.273)));
dot("$O$", (16.5,-4.23806), dir((9.222, 18.178)));
dot("$Q$", (13.34740,-12.54391), dir((9.614, 19.532)));
dot("$E$", (8.37170,-10.88859), dir((9.645, 24.653)));
dot("$D$", (21.48280,-5.91210), dir((8.481, 24.543)));
dot("$L$", (13.34740,-9.), dir((9.614, 18.466)));
dot("$T$", (16.5,-20.81242), dir((9.222, 19.555)));
dot("$H$", (13.34740,-5.45608), dir((9.614, 19.802)));
\end{asy}
\end{center}

\begin{claim*}
  $OEQD$ is a parallelogram.
\end{claim*}
\begin{proof}
  Since $\dang OED = \dang OCD = 90\dg - A$, we find
  $\ol{OE} \perp \ol{AC}$.
  But $\ol{QD} \perp \ol{AC}$ too.
\end{proof}

The parallelogram law now gives
\[ DE^2 + OQ^2 = 2(OD^2 + OE^2). \]
Also since $O$ is the orthocenter of $\triangle ADE$ it is known that
\[ AO^2 = AQ^2 - DE^2. \]
Putting these two together and noting $AO = OQ$ gives
$AQ^2 = 2(OD^2 + OE^2) = 116$, hence $AQ = 2\sqrt{29}$.

\end{document}
