Carina has three pins, labeled $A, B$, and $C$, respectively, located at the origin of the coordinate plane. In a 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 $A B C$ to have area 2021 ?
The answer is 128 .
Define the bounding box of triangle $A B C$ to be the smallest axis-parallel rectangle which contains all three of the vertices $A, B, C$.
Lemma
The area of a triangle $A B C$ is at most half the area of the bounding box.

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 $3 \cdot 2=6$ coordinates of our three points).

So, suppose the bounding box is $A X Y Z$. 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 $\triangle A B C$ is
- exactly 0 if $B=A$,
- at most half the bounding box if $B \in\{X, Z\}$ by one-half-base-height,
- at most half the bounding box if $B=Y$, since $\triangle A B C$ is contained inside either $\triangle A Y Z$ or $\triangle A X Z$.
We now proceed to the main part of the proof.
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 $\lceil 2 \sqrt{A}\rceil$ moves.)

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 $(n / 2)^{2}$.

This immediately implies $n \geq 128$, 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:
$$
\begin{array}{l}
A=(0,0) \\
B=(64,18) \\
C=(3,64)
\end{array}
$$
and indeed triangle $A B C$ has area exactly 2021.
Remark. In fact, it can be shown that to obtain an area of $n / 2$, the bounding-box bound of $\lceil 2 \sqrt{n}\rceil$ 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 $n$.