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

\begin{document}
\title{JMO 2021/4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 65}
\maketitle

\section*{Problem}
Carina has three pins, labeled $A$, $B$, and $C$, respectively,
located at the origin of the coordinate plane.
In a \emph{move}, Carina may move a pin to
an adjacent lattice point at distance $1$ away.
What is the least number of moves that Carina can make
in order for triangle $ABC$ to have area $2021$?

\section*{Video}
\href{https://www.youtube.com/watch?v=9WNgDETHOlI&t=317}{\texttt{https://youtu.be/9WNgDETHOlI}}

\section*{External Link}
\url{https://aops.com/community/p21498566}

\newpage

\section*{Solution}
The answer is $128$.

Define the \textbf{bounding box} of triangle $ABC$
to be the smallest axis-parallel rectangle which
contains all three of the vertices $A$, $B$, $C$.

\begin{center}
\begin{asy}
  pair A = (0,0);
  pair X = (7,0);
  pair Z = (0,3);
  pair Y = X+Z-A;
  pair B = (7,1);
  pair C = (2,3);
  filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
  filldraw(A--X--Y--Z--cycle, opacity(0.1)+lightred, red);
  dot("$A$", A, dir(225), blue);
  dot("$B$", B, dir(0), blue);
  dot("$C$", C, dir(90), blue);
  dot("$X$", X, dir(315), red);
  dot("$Y$", Y, dir(45), red);
  dot("$Z$", Z, dir(135), red);
\end{asy}
\end{center}

\begin{lemma*}
  The area of a triangle $ABC$ is at
  most half the area of the bounding box.
\end{lemma*}
\begin{proof}
  This can be proven by explicit calculation in coordinates.
  Nonetheless, we outline a geometric approach.
  By considering the smallest/largest $x$ coordinate
  and the smallest/largest $y$ coordinate,
  one can check that some vertex of the triangle
  must coincide with a corner of the bounding box
  (there are four ``extreme'' coordinates
  across the $3\cdot2=6$ coordinates of our three points).

  So, suppose the bounding box is $AXYZ$.
  Imagine fixing $C$ and varying $B$ along
  the perimeter entire rectangle.
  The area is a linear function of $B$,
  so the maximal area should be achieved when $B$
  coincides with one of the vertices $\{A,X,Y,Z\}$.
  But obviously the area of $\triangle ABC$ is
  \begin{itemize}
    \ii exactly $0$ if $B=A$,
    \ii at most half the bounding box if $B\in\{X,Z\}$
    by one-half-base-height,
    \ii at most half the bounding box if $B=Y$,
    since $\triangle ABC$ is contained inside
    either $\triangle AYZ$ or $\triangle AXZ$. \qedhere
  \end{itemize}
\end{proof}

We now proceed to the main part of the proof.
\begin{claim*}
  If $n$ moves are made, the bounding box has area at most $(n/2)^2$.
  (In other words, a bounding box of area $A$ requires
  at least $\left\lceil 2\sqrt{A} \right\rceil$ moves.)
\end{claim*}
\begin{proof}
  The sum of the width and height of the bounding box
  increases by at most $1$ each move,
  hence the width and height have sum at most $n$.
  So, by AM-GM, their product is at most $(n/2)^2$.
\end{proof}

This immediately implies $n \ge 128$,
since the bounding box needs to have area at least $4042 > 63.5^2$.

On the other hand, if we start all the pins at
the point $(3,18)$ then we can reach the
following three points in $128$ moves:
\begin{align*}
  A &= (0,0) \\
  B &= (64,18) \\
  C &= (3,64)
\end{align*}
and indeed triangle $ABC$ has area exactly $2021$.

\end{document}
