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

\begin{document}
\title{AMC 12A 2021/24}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 52}
\maketitle

\section*{Problem}
Semicircle $\Gamma$ has diameter $\ol{AB}$ of length $14$.
Circle $\Omega$ lies tangent to $\ol{AB}$ at a point $P$
and intersects $\Gamma$ at points $Q$ and $R$.
If $QR=3\sqrt3$ and $\angle QPR=60^\circ$,
find the area of $\triangle PQR$.

\section*{Video}
\href{https://www.youtube.com/watch?v=O-fh01O_Zao&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/O-fh01O\_Zao}}

\section*{External Link}
\url{https://aops.com/community/p23613274}

\newpage

\section*{Solution}
Let $S$ be the center of $\Omega$,
$O$ the center of $\Gamma$,
and $T = \ol{QR} \cap \ol{APOB}$.
Let $M = \ol{OS} \cap \ol{QR}$ be the midpoint of $\ol{QR}$.

\begin{center}
\begin{asy}
pair A = dir(180);
pair B = dir(0);
pair O = origin;
pair S = (-0.42,0.50);
pair P = (S.x,0);
pair Q = IP(CP(S, P), CP(O, A));
pair R = OP(CP(S, P), CP(O, A));
draw(arc(O, 1, 0, 180), blue);
draw(A--B, blue);
filldraw(CP(S, P), opacity(0.1)+yellow, blue);
pair T = extension(Q, R, A, O);
pair M = midpoint(R--Q);
draw(R--P--Q--cycle, red);
draw(R--S--Q, deepgreen);
draw(O--M, deepgreen);
draw(R--T, red);
draw(T--A, blue);
draw(P--S, deepgreen);

dot("$A$", A, dir(270));
dot("$B$", B, dir(270));
dot("$O$", O, dir(270));
dot("$S$", S, dir(225));
dot("$P$", P, dir(270));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
dot("$T$", T, dir(T));
dot("$M$", M, dir(90));

/* TSQ Source:

A = dir 180 R270
B = dir 0 R270
O = origin R270
S = (-0.42,0.50) R225
P = (S.x,0) R270
Q = IP CP S P CP O A
R = OP CP S P CP O A
arc O 1 0 180 blue
A--B blue
CP S P 0.1 yellow / blue
T = extension Q R A O
M = midpoint R--Q R90
R--P--Q--cycle red
R--S--Q deepgreen
O--M deepgreen
R--T red
T--A blue
P--S deepgreen

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

Since $QR = 3\sqrt3$ and $\angle QSR = 2 \angle QPR = 120\dg$,
it follows that $SP = SQ = SR = 3$.
Also, $SM = SQ \sin \angle QSM = 3\sin60\dg = \frac32$.
Therefore,
\begin{align*}
  OS &= OM - SM = \sqrt{OR^2 - \left( \half QR \right)^2} - SM \\
  &= \sqrt{7^2 - \frac{27}{4}} - \frac 32 = 5.
\end{align*}
So $\triangle OSP$ is a $3-4-5$ triangle.

Now, $\triangle OTM \sim \triangle OSP$, so one can compute
\begin{align*}
  TO &= \frac54 \cdot OM = \frac54 \cdot \frac{13}{2} = \frac{65}{8} \\
  TP &= TO - PO = \frac{65}{8} - 4 = \frac{33}{8}.
\end{align*}
So the height from $P$ to $\ol{TM}$ equals
\[ OM \cdot \frac{TP}{TO} = \frac{13}{2} \cdot \frac{33/8}{65/8}
  = \frac{33}{10}. \]
Finally, $[PQR] = \half QR \cdot \frac{33}{10} = \frac{99\sqrt3}{20}$.

\end{document}
