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

\begin{document}
\title{Shortlist 2006 G2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 5}
\maketitle

\section*{Problem}
Let $ABCD$ be a trapezoid with parallel sides $AB > CD$.
Points $K$ and $L$ lie on the line segments $AB$ and $CD$,
respectively, so that $AK/KB=DL/LC$.
Suppose points $P$ and $Q$ on segment $KL$ obey
\[ \angle APB = \angle BCD \quad\text{and}\quad
  \angle CQD = \angle ABC.  \]
Prove that $P$, $Q$, $B$, $C$ are concyclic.

\section*{Video}
\href{https://www.youtube.com/watch?v=--mpFb3GbBA&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/--mpFb3GbBA}}

\section*{External Link}
\url{https://aops.com/community/p875014}

\newpage

\section*{Solution}
We begin with two simple observations:

\begin{claim*}
  Lines $BC$, $AD$, $KL$ are concurrent at a point $E$.
\end{claim*}
\begin{proof}
  Follows from $\ol{AB} \parallel \ol{CD}$ and $AK/KB=DL/LC$.
\end{proof}

\begin{claim*}
  Line $BC$ is tangent to $(ABP)$ and $(CQD)$.
\end{claim*}
\begin{proof}
  Follows from $\angle APB = \angle BCD = 180\dg - \angle EBA$
  and $\angle CQD = \angle ABC = \angle ECD$ respectively.
\end{proof}

\begin{center}
\begin{asy}
size(10cm);

pair C = dir(40);
pair D = dir(140);
pair Q = dir(290);
pair R = dir(95);
pair V = C+dir(90)*C;
pair E = extension(Q, R, C, V);

transform t = shift(E)*scale(2.6)*shift(-E);

pair B = t*C;
pair A = t*D;
pair P = t*R;

filldraw(D--C--Q--cycle, opacity(0.1)+lightred, red);
filldraw(D--C--R--cycle, opacity(0.1)+lightred, red);
filldraw(unitcircle, opacity(0.1)+yellow, red);

filldraw(circumcircle(B, P, A), opacity(0.1)+lightcyan, deepcyan);

draw(E--2*B-C, deepgreen);
draw(E--2*A-D, lightgreen);
pair S = t*Q;

filldraw(B--P--A--cycle, opacity(0.1)+blue, deepcyan);
filldraw(B--S--A--cycle, opacity(0.1)+blue, deepcyan);

pair L = extension(D, C, E, Q);
pair K = extension(B, A, E, Q);
draw(E--S, deepgreen);

clip(box((-8,-3),(3,2)));

dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$Q$", Q, dir(330));
dot("$R$", R, dir(70));
dot("$E$", E, dir(E));
dot("$B$", B, dir(0));
dot("$A$", A, dir(180));
dot("$P$", P, dir(45));
dot("$L$", L, dir(70));
dot("$K$", K, dir(225));

/* -----------------------------------------------------------------+
|                 TSQX: by CJ Quines and Evan Chen                  |
| https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py |
+-------------------------------------------------------------------+
!size(10cm);
C = dir 40
D = dir 140
Q 330 = dir 290
R 70 = dir 95
!pair V = C+dir(90)*C;
E = extension Q R C V
!transform t = shift(E)*scale(2.6)*shift(-E);
B 0 = t*C
A 180 = t*D
P 45 = t*R
D--C--Q--cycle / 0.1 lightred / red
D--C--R--cycle / 0.1 lightred / red
unitcircle / 0.1 yellow / red
circumcircle B P A / 0.1 lightcyan / deepcyan
E--(2*B-C) / deepgreen
E--(2*A-D) / lightgreen
S := t*Q
B--P--A--cycle / 0.1 blue / deepcyan
B--S--A--cycle / 0.1 blue / deepcyan
L 70 = extension D C E Q
K 225 = extension B A E Q
E--S / deepgreen
!clip(box((-8,-3),(3,2)));
*/
\end{asy}
\end{center}

Let $R$ denote the second intersection of $\ol{EKQ}$ with $(CQD)$.
Thus there is a homothety at $E$ which maps
$\triangle DRC$ to $\triangle APB$, say.

Since $ER \cdot EP = EB^2$ it follows now that
\[ EQ \cdot EP = EB \cdot EC \]
and we're done.

\end{document}
