An Equable Question

Let the legs of the unknown triangle be $x$ and $y$ . Then its hypotenuse is $z = \sqrt{x^2 + y^2}$ (by Pythagorean’s Theorem), its area is $A = \frac{1}{2}xy$ , and its perimeter is $P = x + y + \sqrt{x^2 + y^2}$ . Setting these equal to each other and solving for $y$ gives $y = \frac{4(x - 2)}{x - 4}$ . The only positive integer solutions ( $x$ , $y$ ) for this equation are (5, 12), (6, 8), (8, 6), and (12, 5), which correspond to two right triangles – one with legs 5 and 12 (given) and the other with legs of 6 and 8 (missing). The hypotenuse of the missing triangle is $z = \sqrt{6^2 + 8^2} = 10$ , its area is $A = \frac{1}{2} \cdot 6 \cdot 8 = 24$ , and its perimeter is $P = 6 + 8 + 10 = \boxed{24}$ .