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 $\frac{s}{r}$ .

Solutions

Solution 1 (credit to NikoIsLife)

Let $AB = x$ and $AC = y$ . By the Pythagorean Theorem, $y^2 - x^2 = 696^2$ , and applying difference of squares yields $(y-x)(y+x)=696^2$ . Because $x$ and $y$ have the same parity (due to being integers), both $y+x$ and $y-x$ are even.

Let $y+x = 2b$ abd $y-x = 2a$ ; then $ab = 348^2$ . Additionally, $r = \frac{x+696-y}{2} = 348 - a$ $s = \frac{x+696+y}{2} = 348+b.$ Therefore, $\begin{align*} \frac{s}{r} &= \frac{348+b}{348-a} \\ &= \frac{348 + b}{348 - \frac{348^2}{b}} \\ &= \frac{b^2 + 348b}{348b - 348^2} \\ &= \frac{b}{348} + 2 + \frac{2 \cdot 348^2}{348b - 348^2} \\ &= \frac{b}{348} + 2 + \frac{696}{b-348}. \end{align*}$ Note that because $y > 696$ , we must have $b > 696$ . We can do some optimization by using the derivative -- if we let $f(b) = \frac{b}{348} + 2 + \frac{696}{b-348}$ , then $f'(b) = \frac{1}{348} + \frac{-696}{(b-348)^2},$ which equals $0$ if $b = 348 + 348\sqrt{2} \approx 840.146$ . Since $f'(b) < 0$ if $696 < b < 348+348\sqrt{2}$ and $f'(b) > 0$ if $b > 348 + 348\sqrt{2}$ , we can confirm that $b = 348+348\sqrt{2}$ results in the absolute minimum of $\frac{s}{r}$ . However, the case where $b = 348+348\sqrt{2}$ does not happen if $x, y$ are integers, and since $b$ is a factor of $348^2$ , we need to test the largest factor of $348^2$ less than $348 + 348\sqrt{2}$ and the smallest factor of $348^2$ greater than $348 + 348\sqrt{2}$ .

The largest factor of $348^2$ less than $348 + 348\sqrt{2}$ is $696$ (which does not work), and the smallest factor of $348^2$ greater than $348+348\sqrt{2}$ is $841$ . Therefore, $b = 841$ , which means that $a = 144$ , $y = 985$ , and $x = 697$ . Our wanted perimeter is $696+697+985=\boxed{2378}$ .

Solution 2

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

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2006 iTest (Problems, Answer Key)
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2006 iTest Problems/Problem U7

Contents

Problem

Solutions

Solution 1 (credit to NikoIsLife)

Solution 2

See Also