# Radius of in-circle

The questions from geometry are most fascinating to me. As a parent I love to learn geometry and shapes from my son's textbook. I came across this statement about the in-circle:

Let $m, n$ that are integers and $m > n$. Let $a = 2mn$, $b = m^2 - n^2$ and $c = m^2 + n^2$ be the sides of the Pythagorean triangle. Then the radius of the in-circle $r$ will be $nm - n^2$.

How can this happen?

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A useful fact about triangles is $A = rs,$ where $r$ is the inradius and $s$ is the "semiperimeter" $$s := \frac{a+b+c}{2}.$$

Why? Draw in the segments between the incenter and the vertices (these are the angle bisectors). This cuts the triangle into three smaller triangles. Each has height $r$ and the bases are $a,b,$ and $c$, respectively. So the areas of these triangles are $\frac{ra}{2}, \frac{rb}{2},$ and $\frac{rc}{2}$; the sum of these areas is $rs$.

Now that you have this; what's another way of writing the area of a right triangle? We'd have $A = \frac{1}{2}(2mn)(m^2-n^2) = mn(m^2-n^2) = mn(m-n)(m+n)$. For this triangle, $s = \frac{1}{2}(2mn + (m^2-n^2) + (m^2+n^2)) = mn + m^2 = m(n+m)$.

Equating our two expressinos for the area,

$mn(m-n)(m+n) = rm(n+m)$, and cancelling gives $r = n(m-n) = mn - n^2$.

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Perhaps firstly convince yourself that the right triangle formed with those sides ($a, b$ and $c$ above) is geometrically allowed. Use Pythagoras to convinced yourself this triangle can be constructed.

Then (I haven't put this on paper this is merely conjecture) construct angle bisectors for each of the three vertices, these bisectors will slice through the inscribed circle. Does this help?

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Hint :

$$r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}} ~\text{where}~ s=\frac{a+b+c}{2}$$

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-1 I don't think a hint is sufficient for someone who might not have much mathematical knowledge. Nor do you go anywhere near explaining why we make this leap. –  Pureferret Mar 20 '12 at 11:33