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 $2^6 \cdot 3^2 \cdot 29^2$. 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 $696+697+985=\boxed{2378}$

~Someonenumber011

Solution 2 (calculus)

As before, label the other leg $x$ and the hypotenuse $y$. Let the opposite angle to $x$ be $\theta$, and let $t:=\tan\frac{\theta}{2}$; let the area be $A$ and the semiperimeter $s$. Then we have $x = 696*\frac{2t}{1-t^2}, y = 696*\frac{1+t^2}{1-t^2}, A = 696^2*\frac{t}{1-t^2}, s = 696*\frac{1+t}{1-t^2} = 696*\frac{1}{1-t}$. This means that $\frac{s}{r} = \frac{s^2}{sr} = \frac{(\frac{1}{1-t})^2}{\frac{t}{1-t^2}} = \frac{1+t}{t(1-t)}$. By calculus, we know that this function is minimized at $t = \sqrt{2}-1$, which corresponds to $\theta = 45^\circ$ and $x = 696$; by geometry, we know that this function, expressed in terms of $\theta$, is symmetric around this point.

Then we proceed as before, searching for Diophantine solutions of $y^2 - x^2 = 696^2$ with $x$ closest to $696$, and we find that $x = 697, y = 985$ is the closest. (We can do so by noting that we would want $y - x \approx 696*(\sqrt{2}-1) = 288.29$.) Then the perimeter is $696+697+985=\boxed{2378}$ as before, and we are done.

~duck_master

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