HMMT February

Geometry Difficulty 5.0 AIME Prove it United States

Problem:
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.

Solution

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: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.