HMMT February

Lemma: The area of any triangle inscribed in an $a$ by $b$ rectangle is at most $\frac{ab}{2}$. (Any triangle's area can be increased by moving one of its sides to a side of the rectangle). Given this, because any triangle in $T$ is inscribed in a $25 \times 25$ square, we know that the largest possible area of a triangle is $\frac{25^{2}}{2}$, and any triangle which does not use the full range of $x$ or $y$-values will have area no more than $\frac{25 \cdot 24}{2}=300$. There are $4 \cdot 25=100$ triangles of maximal area: pick a side of the square and pick one of the 26 vertices on the other side of our region; each triangle with three vertices at the corners of the square is double-counted once. To get areas between $\frac{25 \cdot 24}{2}$ and $\frac{25 \cdot 25}{2}$, we need to pick a vertex of the square $\left((0,0)\right.$ without loss of generality), as well as $(25, y)$ and $(x, 25)$. By Shoelace, this has area $\frac{25^{2}-xy}{2}$, and since $x$ and $y$ must both be integers, there are $d(n)$ ways to get an area of $\frac{25^{2}-n}{2}$ in this configuration, where $d(n)$ denotes the number of divisors of $n$. Since we can pick any of the four vertices to be our corner, there are then $4 d(n)$ triangles of area $\frac{25^{2}-n}{2}$ for $1 \leq n \leq 25$. So, we compute the answer to be $$\begin{aligned} |P| & =100+4(d(1)+\ldots+d(24)) \\ & =4 \sum_{k \leq 24}\left\lfloor\frac{24}{k}\right\rfloor \\ & =100+4(24+12+8+6+4+4+3+3+2 \cdot 4+1 \cdot 12) \\ & =436 \end{aligned}$$