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

\begin{document}
\title{BAMO 2003/5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 31}
\maketitle

\section*{Problem}
Let $ABCD$ be a square,
and let $E$ be an internal point on side $AD$.
Let $F$ be the foot of the perpendicular from $B$ to $CE$.
Suppose $G$ is a point such that $BG = FG$,
and the line through $G$ parallel to $BC$
passes through the midpoint of $EF$.
Prove that $AC < 2 FG$.

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

\section*{External Link}
\url{https://aops.com/community/p13027339}

\newpage

\section*{Solution}
Since the opposite sides are parallel,
it's clear that $GMCP$ is a parallelogram.
Let $H$ be the antipode of $F$.
Then
\[ FG = BG = MP = \half EH \]
where $H$ is the antipode of $F$.
\begin{center}
\begin{asy}
/* start ggb to asy preamble */
import graph; size(11.53966cm);
pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215,0.); pen ffdxqq = rgb(1.,0.84313,0.); pen qqffff = rgb(0.,1.,1.);
pair A = (9.,-3.), B = (9.,-8.), C = (14.,-8.), D = (14.,-3.), F = (13.38819,-6.36148), M = (12.76061,-4.68074), G = (10.26061,-4.68074), P = (11.5,-8.), H = (9.61180,-9.63851);
/* end preamble */
draw(A--B--C--D--cycle, linewidth(0.6) + zzttqq);
draw((13.18964,-6.43562)--(13.26377,-6.63417)--(13.46233,-6.56003)--F--cycle, linewidth(0.6) + qqwuqq);
draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--D, linewidth(0.6) + zzttqq);
draw(D--A, linewidth(0.6) + zzttqq);
draw(C--(12.13304,-3.), linewidth(0.6));
draw(B--F, linewidth(0.6));
draw(G--M, linewidth(0.6));
draw(F--G, linewidth(0.6) + qqwuqq);
draw(B--G, linewidth(0.6) + qqwuqq);
draw(A--C, linewidth(0.6));
draw(circle(P, 2.5), linewidth(0.6) + red);
draw(G--P, linewidth(0.6));
draw(P--M, linewidth(0.6) + ffdxqq);
draw(circle(P, 3.53553), linewidth(0.6) + qqffff);
draw((12.13304,-3.)--H, linewidth(0.6) + ffdxqq);
dot("$A$", A, dir((-18.104, 11.609)));
dot("$B$", B, dir((-28.095, -25.910)));
dot("$C$", C, dir((14.419, -14.920)));
dot("$D$", D, dir((4.428, 9.611)));
dot("$E$", (12.13304,-3.), dir(45));
dot("$F$", F, dir((3.665, 8.061)));
dot("$M$", M, dir((4.478, 7.837)));
dot("$G$", G, dir((-20.276, 13.831)));
dot("$N$", (11.19409,-7.18074), dir(45));
dot("$P$", P, dir((-7.338, -31.905)));
dot("$H$", H, dir((-22.335, -31.907)));
\end{asy}
\end{center}
So now,
\begin{align*}
  EH^2 &= EC^2 + CH^2 = BF^2 + CE^2
  \overset{?}{>} BC^2 + CD^2
\end{align*}
which is true because $\triangle BFC \sim \triangle CED$
and $BC < EC$.

\end{document}
