2025 AMC 10A Problem 3

1. The longest side is $2025$, so every side is a whole number from $1$ to $2025$. Isosceles means two sides are equal. Either the equal pair is the long side $2025$, or the equal pair is shorter. Count each case.

1. "Positive area" means the triangle is not flat, so the two shorter sides together must be longer than the longest side (equal is not allowed).

Case 1: two sides are $2025$. The triangle is $2025, 2025, b$ with $b$ from $1$ to $2025$.

1. Is it always a real triangle? The two short sides are $b$ and $2025$, and $b + 2025$ is bigger than $2025$ for any $b \ge 1$. So every $b$ works. That is $2025$ triangles (including the equilateral one, $b = 2025$).

Case 2: the two equal sides are shorter. The triangle is $a, a, 2025$ with $a$ smaller than $2025$.

1. Now the rule really matters. Try small numbers first: $a = 1000$ gives $1000 + 1000 = 2000$, too short to reach across $2025$. So $a$ must be big enough that $a + a > 2025$.

1. $2a > 2025$ means $a > 1012.5$, so $a$ is at least $1013$. Check the edge: $1013 + 1013 = 2026 > 2025$ works, while $1012 + 1012 = 2024$ fails.

1. So $a$ can be $1013, 1014, \ldots, 2024$. That is $2024 - 1013 + 1 = 1012$ triangles.

1. No triangle is counted twice: in Case 1 the repeated side is $2025$, in Case 2 it is less than $2025$.

1. Total: $2025 + 1012 = 3037$.

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

Fixing the longest side makes the list finite, and then each isosceles triangle is decided by one other number. The triangle rule (the two shorter sides must add to more than the longest) is easy in one case and does real work in the other, which is why the two cases have different sizes. When counting triangles, always ask which case each triangle falls in and apply that rule to it.

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 $\boxed{\textbf{(D)}\ 3037}$.