Berkeley Math Circle: Monthly Contest 2

Solution:
First, we claim $P$ must be convex to maximize its area. If not, let $A$ and $B$ be consecutive vertices on the perimeter of its convex hull that aren't consecutive vertices of $P$. Reflecting the path between $A$ and $B$ over line $AB$ must increase the area of $P$ as the new shape strictly contains $P$.

Thus we assume $P$ is convex. Let $\ell$ be the line containing the side with length not equal to $1$. Let $P'$ be the reflection of $P$ over $\ell$. By convexity, $P$ and $P'$ do not overlap, so the union of $P$ and $P'$ is a polygon, specifically an equilateral $4044$-gon with sides of length $1$.

The area of an equilateral polygon is maximized when it is regular, so this union has maximum area that of a regular $4044$-gon, which is $1011 \cot \dfrac{\pi}{4044}$. Thus, the answer is half of this, i.e.
$$\frac{1011}{2} \cdot \cot \frac{\pi}{4044}$$