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 $2021$?

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 positive integers $n$.
\end{remark*}