2025 AMC 10A Problem 3

Split into the case where two sides equal $2025$ and the case where $2025$ is the unique longest side

In the second case the two equal sides $s$ satisfy $2s \gt 2025$ and $s \lt 2025$

Split into two cases. Say two sides both equal $2025.$ Then the third side can be any integer from $1$ to $2025,$ which is $2025$ triangles. Now suppose $2025$ is the unique longest side. The two equal legs $s$ must satisfy $2s \gt 2025$ by the triangle inequality, and $s \le 2024.$ So $s$ runs from $1013$ to $2024,$ giving $1012$ triangles. Adding up, $2025 + 1012 = 3037.$ Thus, D is the correct answer.