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

\begin{document}
\title{bissue 2022/15}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 120}
\maketitle

\section*{Problem}
Quadrilateral $ABCD$ is such that $2\angle A = \angle B$ and $2 \angle D = \angle C$.
If $AB=3$, $BC=4$, and $CD=5$, find the area of quadrilateral $ABCD$.

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

\section*{External Link}
\url{https://aops.com/community/p24130067}

\newpage

\section*{Solution}
\emph{This problem and solution were contributed by Benny Wang}.

Let $E$ be the intersection of the angle bisector of $\angle ABC$ with $\ol{AD}$;
let $F$ be the intersection of the angle bisector of $\angle BCD$ with $\ol{AD}$.
\begin{claim*}
We have $E = F$.
\end{claim*}

\begin{proof}
Let the perpendiculars from $E$ to $\ol{AB}$ and $\ol{BC}$ be $E_1$ and $E_2$.
Note that due to $\triangle ABE$ isosceles with $AE = BE$ (using the angle condition), it
follows that $E_1 B = \frac{AB}{2} = 1.5$. Then, since we defined $\ol{BE}$ as the
bisector of $\angle ABC$, we have that $E_2 B = E_1 B = 1.5$.

Similarly, if we define $F_1$ and $F_2$ as the perpendiculars from $F$ to $\ol{CD}$
and $\ol{BC}$, we have that $F_2 C = F_1 C = \frac{CD}{2} = 2.5$. Note that
$BE_2 + F_2 C = 4 = BC$, implying that $E_2 = F_2$, implying that $E = F$, as desired.
\end{proof}

\begin{center}
\begin{asy}
unitsize(1cm);
pair A = (0.03,-1.15);
pair B = (1.28, 1.57);
pair C = (5.18, 2.45);
pair D = (8.01, -1.67);
pair X = (3.68, -1.38);
pair G = (2.93, 1.95);
pair F1 = (0.66, 0.21);
pair F2 = (6.6, 0.39);
draw(A--B--C--D--cycle);
draw(B--X);
draw(C--X);

/*
markscalefactor=0.06;
draw(anglemark(X,A,B));
draw(anglemark(A,B,X));
draw(anglemark(X,B,C));

markscalefactor=0.055;
draw(anglemark(B,C,X));
draw(anglemark(X,C,D));
draw(anglemark(C,D,X));

markscalefactor=0.07;
draw(anglemark(B,C,X));
draw(anglemark(X,C,D));
draw(anglemark(C,D,X));
*/

dot("$A$", A, 1.5*dir(180), linewidth(4));
dot("$B$", B, 1.5*dir(135), linewidth(4));
dot("$C$", C, 1.5*dir(60), linewidth(4));
dot("$D$", D, 1.5*dir(0), linewidth(4));
dot("$X$", X, 1.5*dir(270), linewidth(4));
dot("$G$", G, 1.5*dir(110), linewidth(4));
dot("$F_1$", F1, 1.5*dir(150), linewidth(4));
dot("$F_2$", F2, 1.5*dir(30), linewidth(4));

draw(X--F1, dotted);
draw(X--G, dotted);
draw(X--F2, dotted);
\end{asy}
\end{center}

Now we compute a bunch.
Let $G$ be the point $E_2=F_2$, and let $X$ be the point $E=F$.

Consider $\triangle BXC$.
Then $G$ is defined as the foot of the altitude from $X$ to
$\ol{BC}$; we know $BG = 1.5$ and $GC = 2.5$.
We also know that $\angle BXC = 60^{\circ}$ by some straightforward angle chasing.
This is enough information to solve for $\ol{BXC}$; in particular, we desire
the length of $XG$; then, since $X$ is equidistant from $\ol{AB}$, $\ol{BC}$
and $\ol{CD}$, we could compute the area of $ABCD$ to
be $\frac{(AB + BC + CD) \times XG}{2}$.

To find $XG$ is a straightforward geometry problem now.
There are several ways we could go about this. Here is one:
let $M$ be the midpoint of $\ol{BC}$, and let $O$ be the circumcenter
of $\triangle BXC$.
\begin{center}
\begin{asy}
unitsize(1.2cm);
pair C = (0,0);
pair M = (2,0);
pair G = (2.5,0);
pair B = (4,0);
pair X = (2.5, 4.65);
pair O = (2, 1.15);
draw(B--X--C--cycle);

dot("$B$", B, 1.5*dir(320), linewidth(4));
dot("$C$", C, 1.5*dir(220), linewidth(4));
dot("$X$", X, 1.5*dir(90), linewidth(4));
dot("$M$", M, 1.5*dir(270), linewidth(4));
dot("$G$", G, 1.5*dir(270), linewidth(4));
dot("$O$", O, 2*dir(135), linewidth(4));

draw(M--O, dashed);
draw(O--C, dashed);
draw(O--X, dashed);
draw(X--G, dashed);

label("$2$", C--M, 1.5*dir(90));
label("$0.5$", M--G, 1.5*dir(90));
label("$1.5$", G--B, 1.5*dir(90));
\end{asy}
\end{center}
Note that $\angle BOC = 2 \angle BXC = 120^{\circ}$, so $\triangle BMO$
and $\triangle CMO$ are 30-60-90 triangles. Therefore, $MO = \frac{2}{\sqrt{3}}$
and $BO = OC = OX = \frac{4}{\sqrt{3}}$. Now, note that
\begin{align*}
GX &= \sqrt{XO^2 - GM^2} + MO \\
&= \sqrt{\left( \frac{4}{\sqrt{3}} \right)^2 - 0.5^2} + \frac{2}{\sqrt{3}} \\
&= \sqrt{\frac{61}{12}} + \frac{4}{\sqrt{12}} \\
&= \frac{4 + \sqrt{61}}{\sqrt{12}}.
\end{align*}
Therefore,
\[ [ABCD] = \frac{XG \cdot (AB + BC + CD)}{2} = \frac{4 + \sqrt{61}}{2 \sqrt{12}}
\cdot 12 = \boxed{4 \sqrt{3} + \sqrt{183}}.\]

\end{document}
