IMO 1964 - Problem 3

Problem: The circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Three tangents to the incircle are drawn, each of which is parallel to one side of the triangle $ABC$. These tangents form three smaller triangles (internal to $\triangle{ABC}$) with the sides of $\triangle{ABC}$. In each of these triangles an incircle is inscribed. Determine the sum of areas of all four incircles.
Solution: Let $r$ be the radius of the incircle of $\triangle{ABC},r_a,r_b,r_c$ the radius of the smaller circles corresponding to $A,B,C$, and $h_a,h_b,h_c$ the altitudes from $A,B,C$ respectively. The coefficient of similarity between the smaller triangle at $A$ and the triangle $ABC$ is $1-\frac{2r}{h_a}$, from which we easily obtain $r_a=\frac{(h_a-2r)r}{h_a}=\frac{(s-a)r}{s}$. Similarly, $r_b=\frac{(s-b)r}{s}$ and $r_c=\frac{(s-c)r}{s}$. Now a straightforward computation gives that the sum of areas of the four circles is given by $\sum =\frac{(b+c-a)(c+a-b)(a+b-c)(a^2+b^2+c^2)\pi}{(a+b+c)^3}$.