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

\begin{document}
\title{HMNT 2023 Gen10}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 136}
\maketitle

\section*{Problem}
Let $ABCD$ be a convex trapezoid such that $\angle ABC = \angle BCD = 90^\circ$,
$AB = 3$, $BC = 6$, and $CD = 12$.
Among all points $X$ inside the trapezoid satisfying $\angle XBC = \angle XDA$,
compute the minimum possible value of $CX$.

\section*{Video}
\href{https://www.youtube.com/watch?v=xR7rzNMpQlw&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/xR7rzNMpQlw}}

\section*{External Link}
\url{https://www.hmmt.org/www/archive/271}

\newpage

\section*{Solution}
Let $E = \ol{BC} \cap \ol{AD}$.
\begin{claim*}
  The point $X$ moves along the circumcircle of $\triangle EBD$.
\end{claim*}
\begin{proof}
  Because $\angle XBE = 180\dg - \angle CBX = 180\dg - \angle ADX = 180\dg - \angle EDX$.
\end{proof}

\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair C = (0.,0.);
pair B = (0.,6.);
pair A = (3.,6.);
pair D = (12.,0.);
pair E = (0.,8.);
pair X = (1.44631,2.30434);
pair K = (4.,0.);

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(A--B--C--D--cycle, linewidth(0.6) + zzttqq);
draw(arc(D,0.9,146.30993,167.68306)--D--cycle, linewidth(0.6) + qqwuqq);
draw(arc(B,0.9,-90.,-68.62686)--B--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(E--A, linewidth(0.6));
draw(B--(2.34812,0.), linewidth(0.6));
draw(D--(0.,2.62013), linewidth(0.6));
draw(circle((8.,7.), 8.06225), heavycyan);

dot("$C$", C, dir(225));
dot("$B$", B, dir(180));
dot("$A$", A, dir(45));
dot("$D$", D, dir(315));
dot("$E$", E, dir(180));
dot("$X$", X, dir(45));
dot("$K$", K, dir(270));
\end{asy}
\end{center}

It remains to do the calculation, which is surprisingly quick.
Work in a coordinate system with
\begin{align*}
  C &= (0,0) \\
  B &= (0,6) \\
  D &= (12,0) \\
  A &= (3,6) \\
  E &= (0,8).
\end{align*}
By power of a point, the second intersection of $(EBD)$ with $CD$ has
\[ CK = \frac{CB \cdot CE}{CD} = \frac{6 \cdot 8}{12} = 4. \]
In other words, $K = (4,0)$.
So the circumcenter of $\triangle EBD$ is located at \[ O = (8,7). \]
The radius of the circle is then $\sqrt{65}$, while $CO = \sqrt{113}$.
It follows the answer is $\sqrt{113} - \sqrt{65}$.

\end{document}
