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.
An alternate proof uses the fact that 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). |