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

\begin{document}
\title{JMO 2021/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 65}
\maketitle

\section*{Problem}
An equilateral triangle $\Delta$ of side length $L > 0$ is given.
Suppose that $n$ equilateral triangles with side length $1$
and with non-overlapping interiors are drawn inside $\Delta$,
such that each unit equilateral triangle has sides parallel to $\Delta$,
but with opposite orientation.
Prove that \[ n \le \frac{2}{3} L^2. \]

\section*{Video}
\href{https://www.youtube.com/watch?v=9WNgDETHOlI&t=655}{\texttt{https://youtu.be/9WNgDETHOlI}}

\section*{External Link}
\url{https://aops.com/community/p21499596}

\newpage

\section*{Solution}
We present the approach of Andrew Gu.
For each triangle, we draw a green regular hexagon
of side length $1/2$ as shown below.

\begin{center}
\begin{asy}
  size(2cm);
  filldraw(dir(0)--dir(180)--(dir(240)+dir(300))--cycle,
    opacity(0.2)+lightred, black);
  filldraw(dir(0)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle,
  opacity(0.1)+green, deepgreen+1.5);
\end{asy}
\end{center}

\begin{claim*}
  All the hexagons are disjoint and lie inside $\Delta$.
\end{claim*}
\begin{proof}
  Annoying casework.
\end{proof}

Since each hexagon has area $\frac{3\sqrt{3}}{8}$ and
lies inside $\Delta$, we conclude
\[ \frac{3\sqrt3}{8} \cdot n \le \frac{\sqrt3}{4} L^2
  \implies n \le \frac 23 L^2. \]

\begin{remark*}
  The constant $\frac23$ is sharp and cannot be improved.
  The following tessellation shows how to achieve the $\frac23$ density.
  In the figure on the left,
  one of the green hexagons is drawn in for illustration.
  The version on the right has all the hexagons.
  \begin{center}
  \begin{asy}
    size(11cm);
    picture pic;
    filldraw(pic, dir(0)--dir(180)--(dir(240)+dir(300))--cycle,
      opacity(0.2)+lightred, black);
    pair v = dir(180)+dir(240);
    pair w = dir(0)+dir(-60);

    for (int i=0; i<=7; ++i) {
      for (int j=0; j<=7; ++j) {
        if (abs(i-j) <= 4) {
          add(shift(i*v+j*w)*pic);
        }
      }
    }

    filldraw(pic, (dir(0)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle),
      opacity(0.1)+lightgreen, deepgreen+1.5);

    for (int i=0; i<=7; ++i) {
      for (int j=0; j<=7; ++j) {
        if (abs(i-j) <= 4) {
          add(shift(15,0)*shift(i*v+j*w)*pic);
        }
      }
    }
    filldraw(shift(0,-4*3**0.5)*(dir(0)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle),
      opacity(0.1)+lightgreen, deepgreen+1.5);
  \end{asy}
  \end{center}
\end{remark*}

\end{document}
