Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Find the answer

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}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Assume WLOG that the side lengths of the triangle are pairwise coprime. Then they can be written as m2n2,2mn,m2+n2m^{2}-n^{2}, 2mn, m^{2}+n^{2} for some coprime integers mm and nn where m>nm>n and mnmn is even. Then we obtain P2A=4m(m+n)n(mn)\frac{P^{2}}{A}=\frac{4m(m+n)}{n(m-n)}. But n,mn,m,m+nn, m-n, m, m+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 45.

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