Maths Olympiad Prep

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, 2017

Number theory Difficulty 5.0 AIME Prove it United States

Problem:
Let PP and AA denote the perimeter and area respectively of a right triangle with relatively prime integer side-lengths. Find the largest possible integral value of P2A\frac{P^{2}}{A}

Solution

Solution:
Assume WLOG that the side lengths of the triangle are pairwise coprime. Then they can be written as m2n2m^{2}-n^{2}, 2mn2 m n, m2+n2m^{2}+n^{2} for some coprime integers mm and nn where m>nm>n and mnm n is even. Then we obtain
P2A=4m(m+n)n(mn) \frac{P^{2}}{A}=\frac{4 m(m+n)}{n(m-n)}
But nn, mnm-n, mm, m+nm+n are all pairwise coprime so for this to be an integer we need n(mn)4n(m-n) \mid 4 and by checking each case we find that (m,n)=(5,4)(m, n)=(5,4) yields the maximum ratio of 4545.

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