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

\begin{document}
\title{H3195318}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 133}
\maketitle

\section*{Problem}
Given two positive real numbers $x$, $y$ such that $x+y < \frac\pi2$
and $\sin(2y)+\sin(2x+2y)=2\sin(2x)$, determine, with proof, the value of
\[ \frac{\cos(x+y)\sin(y)}{\sin(x)}. \]

\section*{Video}
\href{https://www.youtube.com/watch?v=5A5Yu65JV_U&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/5A5Yu65JV\_U}}

\section*{External Link}
\url{https://aops.com/community/p29162889}

\newpage

\section*{Solution}
First manipulate both sides of the given equation to get
\begin{align*}
  \sin(2x+2y) + \sin(2y)
  &= \sin\left( (x+2y)+x \right) + \sin\left( (x+2y)-x \right) \\
  &= 2\sin(x+2y)\cos x \\
  2\sin(2x) &= 4 \sin x \cos x.
\end{align*}
So the problem condition is equivalent to
\begin{align*}
  \sin(x+2y) &= 2\sin(x) \\
  \sin\left( (x+y)+y \right) &= 2\sin \left( (x+y)-y \right) \\
  \sin(x+y)\cos(y) + \cos(x+y)\sin(y) &= 2\sin(x+y)\cos(y) - 2\cos(x+y)\sin(y) \\
  3 \cos(x+y)\sin(y) &= \sin(x+y)\cos(y).
\end{align*}
On the other hand, we could also write
\[ \sin(x+y)\cos(y) + \cos(x+y)\sin(y) = 2\sin(x). \]
Put together the two most recent equations to
eliminate the undesired $\sin(x+y)\cos(y)$ term gives
\[ 3\cos(x+y)\sin(y) = 2\sin(x)-\cos(x+y)\sin(y)
  \implies \frac{\cos(x+y)\sin(y)}{\sin(x)} = \boxed{\frac12}. \]

\end{document}
