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

\begin{document}
\title{Shortlist 2004 G3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 7}
\maketitle

\section*{Problem}
Let $O$ be the circumcenter of an acute-angled triangle $ABC$
with $\angle B < \angle C$ and let $D = \ol{AO} \cap \ol{BC}$.
Let $E$ and $F$ denote the circumcenters
of triangles $ABD$ and $ACD$.
Extend the sides $BA$ and $CA$ beyond $A$,
and choose on the respective extensions points $G$ and $H$
such that $AG=AC$ and $AH=AB$.
Prove that the quadrilateral $EFGH$ is a rectangle
if and only if $\angle ACB - \angle ABC = 60\dg$.

\section*{Video}
\href{https://www.youtube.com/watch?v=3W4kkTPJup0&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/3W4kkTPJup0}}

\section*{External Link}
\url{https://aops.com/community/p205737}

\newpage

\section*{Solution}
We start with a few observations
which are always true regardless of the condition.
\begin{itemize}
  \ii Quadrilateral $HGCB$ is always an isosceles trapezoid,
  and in particular $BC = GH$.

  \ii By angle chasing $\ol{AO} \perp \ol{GH}$ always holds.
  (One clean way to see this
  is to note that $\ol{GH}$ and $\ol{BC}$
  are antiparallel through $\angle A$.)
  This implies $\ol{EF} \parallel \ol{GH}$.

  \ii By Salmon theorem,
  we always have
  \[ \triangle AEF \overset{+}{\sim} \triangle ABC. \]
\end{itemize}

\begin{center}
\begin{asy}
size(6cm);
pair B = dir(200);
pair C = dir(-20);
pair A = dir(30);
pair O = origin;
pair D = extension(A, O, B, C);
pair E = circumcenter(A, B, D);
pair F = circumcenter(A, C, D);

filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
draw(A--D, blue);
filldraw(A--E--F--cycle, opacity(0.1)+lightred, red);
pair G = A+(A-B)*abs(C-A)/abs(B-A);
pair H = A+(A-C)*abs(B-A)/abs(C-A);

filldraw(A--C--G--cycle, opacity(0.1)+lightcyan, blue);
filldraw(A--B--H--cycle, opacity(0.1)+lightcyan, blue);
filldraw(A--G--H--cycle, opacity(0.1)+lightblue, blue);

dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$A$", A, dir(330));
dot("$O$", O, dir(270));
dot("$D$", D, dir(270));
dot("$E$", E, dir(E));
dot("$F$", F, dir(295));
dot("$G$", G, dir(G));
dot("$H$", H, dir(H));

/* TSQ Source:

!size(10cm);
B = dir 200
C = dir -20
A = dir 30 R330
O = origin R270
D = extension A O B C R270
E = circumcenter A B D
F = circumcenter A C D R295

A--B--C--cycle 0.1 lightblue / blue
A--D blue
A--E--F--cycle 0.1 lightred / red
G = A+(A-B)*abs(C-A)/abs(B-A)
H = A+(A-C)*abs(B-A)/abs(C-A)

A--C--G--cycle 0.1 lightcyan / blue
A--B--H--cycle 0.1 lightcyan / blue
A--G--H--cycle 0.1 lightblue / blue

*/
\end{asy}
\end{center}

We begin now with:
\begin{claim*}
  We have $\triangle AEF \cong \triangle ABC$
  if and only if $\angle C - \angle B = 60\dg$.
\end{claim*}
\begin{proof}
  The congruence just means $FA = AC$.
  Since $FA = FC$ always,
  triangle $AFC$ is equilateral if and only if
  $\angle AFC = 60\dg \iff \angle ADC = 30\dg$.
  As $\angle ADC = (90\dg - \angle C) + \angle B$, and
  the result follows.
\end{proof}
\begin{claim*}
  We have $EFGH$ is a parallelogram
  if and only if $\triangle AEF \cong \triangle ABC$.
\end{claim*}
\begin{proof}
  Since we already know $\ol{EF} \parallel \ol{GH}$,
  the parallelogram condition is equivalent to $EF = GH$,
  but as $GH = BC$ we get the earlier congruence.
\end{proof}

It remains only to show that if $EFGH$ is a parallelogram
then it is also a rectangle.
In the situation of the claims,
note that $EG = FH$ by symmetry
through the oppositely congruent triangles
$\triangle AEF$ and $\triangle AHG$ as needed.

\end{document}
