2006 iTest Problems/Problem U7

Problem

Triangle $ABC$ has integer side lengths, including $BC = 696$ , and a right angle, $\angle ABC$ . Let $r$ and $s$ denote the inradius and semiperimeter of $ABC$ respectively. Find the perimeter of the triangle ABC which minimizes $\frac{s}{r}$ .

Solution

First, label the other leg $x$ and the hypotenuse $y$ . To minimize $\frac{s}{r}$ , $r$ must be maximize and $s$ must be minimized. Through logic, it becomes clear that the triangle must be as close to equilateral as possible to maximize $r$ and minimize $s$ (Think about stretching one vertice of an equilateral triangle. The perimeter increases “faster” than the inradius).

From the Pythagorean theorem, $y^2-x^2=696^2$ , applying difference of squares yields $(y-x)(y+x)=696^2$ . Since the question states $x$ and $y$ must be integers, we can find possible values of $x$ and $y$ by finding the prime factorization of $696^2$ , which is $2^6 \cdot 3 \cdot 29$ . The two values of $x$ and $y$ that are closest to each other are the values that satisfy $y-x=2^2 \cdot 3 \cdot 29$ , and $y+x=2^4 \cdot 3 \cdot 29$ . Solving the system yields $x = 522$ and $y = 870$ . Thus, the perimeter is $676+522+870=\boxed{2068}$

~Someonenumber011

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2006 iTest Problems/Problem U7

Problem

Solution