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Irish Mathematical Olympiad

Geometry Difficulty 5.2 AIME, harder Prove it Ireland

Find, with proof, all triples of integers $(a, b, c)$ such that $a$ , $b$ and $c$ are the lengths of the sides of a right angled triangle whose area is $a + b + c$ .

Solution

Assume for the moment that $a \leq b \leq c$ . So
$a^2 + b^2 = c^2. \quad (1)$
The condition on the area entails $ab = 2(a + b + c)$ which implies that
$4c^2 = (ab)^2 - 4ab(a + b) + 4(a^2 + b^2 + 2ab). \quad (2)$
Equations (1) and (2) imply that
$ab - 4(a + b) + 8 = 0$
or
$ab - 4(a + b) + 16 = 8.$
Thus
$(a - 4)(b - 4) = 8. \quad (3)$
Now, there are only two ways to factorise $8$ in positive integers. Thus, either $(a, b) = (5, 12)$ or $(a, b) = (6, 8)$ . Thus, there are (up to permutations) two triples that satisfy the given conditions namely, $(5, 12, 13)$ and $(6, 8, 10)$ . $\square$

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