HMMT February

Geometry Difficulty 4.8 AIME Find the answer

A right triangle has side lengths $a, b$ , and $\sqrt{2016}$ in some order, where $a$ and $b$ are positive integers. Determine the smallest possible perimeter of the triangle.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

There are no integer solutions to $a^{2}+b^{2}=2016$ due to the presence of the prime 7 on the right-hand side (by Fermat's Christmas Theorem). Assuming $a<b$ , the minimal solution $(a, b)=(3,45)$ which gives the answer above.

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Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.