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 \(ABC\) to be the smallest axis-parallel rectangle which contains all three of the vertices \(A, B, C\). ![](https://cdn.mathpix.com/cropped/2024_11_19_160003dbb585e37c9fb6g-06.jpg?height=401&width=804&top_left_y=1050&top_left_x=629) ## Lemma The area of a triangle \(ABC\) is at most half the area of the bounding box. 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 - 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 \(\triangle ABC\) is contained inside either \(\triangle AYZ\) or \(\triangle AXZ\). 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.) 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: \[ \begin{aligned} & A=(0,0) \\ & B=(64,18) \\ & C=(3,64) \end{aligned} \] and indeed triangle \(ABC\) 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 \(n\).