\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 of the 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 $2021$.
\begin{remark*}
In fact, it can be shown that to obtain an area of $n/2$,
the bounding--box bound of $\left\lceil 2\sqrt n \right\rceil$ moves is best possible,
i.e.\ there will in fact exist a triangle with area $n/2$.
However, since this was supposed to be a JMO4 problem,
the committee made a choice to choose $n = 4042$
so that contestants only needed to give a single concrete triangle
rather than a general construction for all integers $n$.
\end{remark*}
\end{document}