Inscribed Circle

圖片參考：http://imgcld.yimg.com/8/n/HA00108663/o/20130116100132.jpg


Let s1 be half of the perimeter of the triangle with sides 2, 3 + r, 5 - r
thus s1 = [2 + (3 + r) + (5 - r)] / 2 = 5
area = sqrt[(5)(3)(2 - r)(r)] = sqrt(30r - 15r^2)

Let s2 be half of the perimeter of the triangle with sides 3, 2 + r, 5 - r
thus s1 = [3 + (2 + r) + (5 - r)] / 2 = 5
area = sqrt[(5)(2)(3 - r)(r)] = sqrt(30r - 10r^2)

since two triangle have same height,
the ratio of the area = the ratio of the base = 2 : 3
sqrt(30r - 15r^2) : sqrt(30r - 10r^2) = 2 : 3
3sqrt(30r - 15r^2) = 2sqrt(30r - 10r^2)
9(30r - 15r^2) = 4(30r - 10r^2)
270r - 135r^2 = 120r - 40r^2
150r = 95r^2
r = 30/19 or r = 0(rejected)