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

\begin{document}
\title{JMO 2024/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 141}
\maketitle

\section*{Problem}
Let $ABCD$ be a cyclic quadrilateral with $AB=7$ and $CD=8$.
Points $P$ and $Q$ are selected on line segment $AB$ so that $AP=BQ=3$.
Points $R$ and $S$ are selected on line segment $CD$ so that $CR=DS=2$.
Prove that $PQRS$ is a cyclic quadrilateral.

\section*{Video}
\href{https://www.youtube.com/watch?v=N3WG9AY9HYY&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/N3WG9AY9HYY}}

\section*{External Link}
\url{https://aops.com/community/p30216434}

\newpage

\section*{Solution}
Here are two possible approaches.

\paragraph{The one-liner.}
The four points $P$, $Q$, $R$, $S$ have equal power $-12$ with respect to $(ABCD)$.
So in fact they're on a circle concentric with $(ABCD)$.

\paragraph{The external power solution.}
We distinguish between two cases.

\subparagraph{Case where $AB$ and $CD$ are not parallel.}
We let lines $AB$ and $CD$ meet at $T$.
Without loss of generality,
$A$ lies between $B$ and $T$ and $D$ lies between $C$ and $T$.
Let $x=TA$ and $y=TD$, as shown below.
\begin{center}
\begin{asy}
size(8cm);
pair A = dir(170);
pair B = dir(80);
pair C = dir(323.912853);
pair D = dir(216.087147);
pair T = extension(A, B, C, D);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(B--T--C--B, blue);
draw(A--D, blue);

pair P = 4*A/7+3*B/7;
pair Q = 3*A/7+4*B/7;
pair R = 6*C/8+2*D/8;
pair S = 2*C/8+6*D/8;

filldraw(circumcircle(P, Q, R), opacity(0.1)+yellow, dashed+red);
label("$x$", midpoint(T--A), dir(A+B), deepgreen);
label("$3$", midpoint(A--P), dir(A+B), deepgreen);
label("$1$", midpoint(P--Q), -dir(A+B), deepgreen);
label("$3$", midpoint(Q--B), dir(A+B), deepgreen);
label("$y$", midpoint(T--D), dir(C+D), deepgreen);
label("$2$", midpoint(D--S), dir(C+D), deepgreen);
label("$4$", midpoint(S--R), -dir(C+D), deepgreen);
label("$2$", midpoint(R--C), dir(C+D), deepgreen);

dot("$A$", A, dir(120));
dot("$B$", B, dir(B));
dot("$C$", C, dir(300));
dot("$D$", D, dir(240));
dot("$T$", T, dir(T));
dot("$P$", P, dir(160));
dot("$Q$", Q, dir(80));
dot("$R$", R, dir(270));
dot("$S$", S, dir(270));

/* -----------------------------------------------------------------+
|                 TSQX: by CJ Quines and Evan Chen                  |
| https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py |
+-------------------------------------------------------------------+
A 120 = dir 170
B = dir 80
C 300 = dir 323.912853
D 240 = dir 216.087147
T = extension A B C D
unitcircle / 0.1 lightcyan / blue
B--T--C--B / blue
A--D / blue
P 160 = 4*A/7+3*B/7
Q 80 = 3*A/7+4*B/7
R 270 = 6*C/8+2*D/8
S 270 = 2*C/8+6*D/8
circumcircle P Q R / 0.1 yellow / dashed red
!label("$x$", midpoint(T--A), dir(A+B), deepgreen);
!label("$3$", midpoint(A--P), dir(A+B), deepgreen);
!label("$1$", midpoint(P--Q), -dir(A+B), deepgreen);
!label("$3$", midpoint(Q--B), dir(A+B), deepgreen);
!label("$y$", midpoint(T--D), dir(C+D), deepgreen);
!label("$2$", midpoint(D--S), dir(C+D), deepgreen);
!label("$4$", midpoint(S--R), -dir(C+D), deepgreen);
!label("$2$", midpoint(R--C), dir(C+D), deepgreen);
*/
\end{asy}
\end{center}
By power of a point,
\begin{align*}
  \text{$ABCD$ cyclic} \iff x(x+7) &= y(y+8) \\
  \text{$PQRS$ cyclic} \iff (x+3)(x+4) &= (y+2)(y+6).
\end{align*}
However, the latter equation is just the former with $12$ added to both sides.
(That is, $(x+3)(x+4) = x(x+7)+12$ while $(y+2)(y+6)=y(y+8)+12$.)
So the conclusion is immediate.

\subparagraph{Case where $AB$ and $CD$ are parallel.}
In that case $ABCD$ is an isosceles trapezoid.
Then the entire picture is symmetric around the common perpendicular bisector
of the lines $AB$ and $CD$.
Now $PQRS$ is also an isosceles trapezoid, so it's cyclic too.
\begin{center}
\begin{asy}
size(4cm);
pair A = dir(135);
pair B = dir(45);
pair C = dir(323.912853);
pair D = dir(216.087147);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(A--B--C--D--cycle, blue);

pair P = 4*A/7+3*B/7;
pair Q = 3*A/7+4*B/7;
pair R = 6*C/8+2*D/8;
pair S = 2*C/8+6*D/8;

filldraw(circumcircle(P, Q, R), opacity(0.1)+yellow, dashed+red);

dot("$A$", A, dir(120));
dot("$B$", B, dir(B));
dot("$C$", C, dir(300));
dot("$D$", D, dir(240));
dot("$P$", P, dir(90));
dot("$Q$", Q, dir(80));
dot("$R$", R, dir(270));
dot("$S$", S, dir(270));
\end{asy}
\end{center}

\end{document}
