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

\begin{document}
\title{Twitch 125.5}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 125}
\maketitle

\section*{Problem}
For which positive integers $n$ is it possible to
partition a square into $n$ acute triangles?

\section*{Video}
\href{https://www.youtube.com/watch?v=3D2kETEbaI3rM&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/3D2kETEbaI3rM}}

\section*{External Link}
\url{https://www.ics.uci.edu/~eppstein/junkyard/acute-square/}

\newpage

\section*{Solution}
The answer is $n \ge 8$ only.
The proof is divided into three parts.

\paragraph{Proof that $n \le 7$ fails.}
I didn't write this up because it's too annoying.
The idea is to classify a vertex as \emph{bursting}
if it lies strictly inside the square and has no $180\dg$ angle.
This gives at least five triangles already.

\paragraph{Construction for $n \in \{8,9,10\}$.}
See below.
\begin{center}
\begin{asy}
  size(12cm);
  picture p8;
  picture p9;
  picture p10;

  // n=8 construction
  pair A = (0.45,0.2);
  pair B = (0.55,0.2);
  draw(p8, (0,1)--A--(0.5,1)--B--(1,1));
  draw(p8, (0,0)--A--(0.5,0)--B--(1,0));
  draw(p8, A--B);

  // n=9 construction
  pair O = (0.75,0.55);
  pair M = (0.5,0.7);
  pair N = (0.6,1);
  pair L = (1,0.65);
  pair P = (0.75,0.75);
  draw(p9, (0,0)--O--(1,0));
  draw(p9, O--(0,1));
  draw(p9, M--N--P--(1,1));
  draw(p9, P--O--L--P--M);

  // n=10
  pair O = (0.5,0.8);
  draw(p10, (0,0)--O--(1,0));
  draw(p10, (0,1)--O--(1,1));
  pair M = 0.8*O+0.2*(0,1);
  pair N = 0.8*O+0.2*(1,1);
  pair A = (0.42,1);
  pair B = (0.58,1);
  pair G = (0.5,0.9);
  draw(p10, M--A--G--M);
  draw(p10, N--B--G--N);
  draw(p10, O--G);

  draw(p8, unitsquare, blue+1.5);
  draw(p9, unitsquare, blue+1.5);
  draw(p10, unitsquare, blue+1.5);
  label(p8, "$n=8$", (0.5,0), dir(-90), blue);
  label(p9, "$n=9$", (0.5,0), dir(-90), blue);
  label(p10, "$n=10$", (0.5,0), dir(-90), blue);

  add(p8);
  add(shift(1.3,0)*p9);
  add(shift(2.6,0)*p10);
\end{asy}
\end{center}


\paragraph{Inductive finish.}
Given a valid construction for $n$,
one can take any triangle and add its medial triangle
to get a valid construction for $n+3$.
This completes the proof.

\end{document}
