**The Inradius of a Right Triangle With Integral Sides**
Bill Richardson

September 1999
Let *a = x*^{2} - y^{2}, b = 2xy, c = x^{2} + y^{2} with *0 < y < x, (x,y) = 1 *and *x* and *y *being of opposite parity. Then *(a, b, c)* is a primative Pythagorean triple. Let triangle ABC, in the figure below, be a right triangle with sides *a, b* and hypotenuse *c*. Let the circle with center I be the inscribed circle for this triangle. We will prove that the inradius, *r*, is an integer.

**Proof.** Let *r* be the inradius. Since the tangents to a circle from a point outside the circle are equal, we have the sides of triangle ABC configured as in the above figure. Thus, *c = (a - r) + (b - r) = a + b - 2r* and *r = (a + b - c)/2*. But *a = x*^{2} - y^{2}, b = 2xy, c = x^{2} + y^{2} which gives *r = (x*^{2} - y^{2} + 2xy - x^{2} - y^{2})/2 = (2xy - 2y^{2})/2 = y(x - y).
Thus, *r* is an integer given by *r = y(x - y)*.

An alternate proof uses the fact that

AreaTriangleABC = AreaTriangleAIB + AreaTriangleBIC + AreaTriangleAIC.

So, (½)*ab* = (½)*rc* + (½)*ra* + (½)*rb* and

*r = ab/(a + b + c) = (x*^{2} - y^{2})2xy/(x^{2} - y^{2} + 2xy + x^{2} + y^{2}) = [2xy(x - y)(x + y)]/[2x(x + y)] = y(x - y).

Thus, *r* is an integer given by *r = y(x - y)*.
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