2025 AMC 10A Problem 3

Positive area means the triangle inequality must be strict. The longest side is $2025$, so every side is at most $2025$ and at least one side equals $2025$. An isosceles triangle has a pair of equal sides, and the two cases are decided by whether that pair is the longest side or not.

The equal pair is the longest side. The triangle is $(2025, 2025, b)$ with $1 \le b \le 2025$. The inequality $2025 + b > 2025$ is automatic and $2025 + 2025 > b$ holds for every such $b$, so all $2025$ values work. This case includes the equilateral triangle $b = 2025$.

The equal pair is shorter. The triangle is $(a, a, 2025)$ with $a < 2025$, and now the inequality bites: the two short sides must beat the long one, so
$$2a > 2025 \implies a > 1012.5 \implies a \ge 1013.$$
Hence $a$ runs from $1013$ to $2024$, giving $2024 - 1013 + 1 = 1012$ triangles.

The cases are disjoint, since the repeated side equals $2025$ in the first and is smaller in the second, so the total is
$$2025 + 1012 = 3037.$$

The answer is $\boxed{\textbf{(D)}\ 3037}$.

Fixing the longest side first is what makes the count finite and the casework short: once $2025$ is pinned down, an isosceles triangle is determined by a single other integer, and the only question is which values of it are legal. The triangle inequality is slack in one case and tight in the other, which is exactly why the two families have different sizes. The same argument with an odd longest side $n$ gives $n + \frac{n-1}{2}$ triangles in all.

• Counting only the triangles whose two equal sides are the longest, giving 2025, choice (A).
• Ignoring the triangle inequality in the second family and allowing every $a$ from $1$ to $2024$, which inflates the count towards $4050$, choice (E).
• Counting the equilateral triangle in both families and answering $3038$.