A circle is inscribed in a triangle ABC with sides a,b,c. Tangents to the circle parallel to the sides of the triangle are constructed. Each of these tangents cuts off a triangle from △ABC. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of a,b,c).
Let s be the semiperimeter of ABC, r be the inradius, and k be the area.
So I will start with just triangle ABC and the one tangent parallel to BC. Call that tangent segment B'C'.
The altitude of ABC passing through A is also the line containing an altitude of AB'C' ABC and AB'C' are similar so their scale factor is the same as the ratio of the altitudes.
The altitude of ABC can be expressed as 2k/a, then the altitude of AB'C' is then 2k/a-2r. The ratio of these altitudes is then (2k/a-2r)/(2k/a).
With help from the identity k=r*s the ratio simplifies to 1 - a/s.
This logic applies to all three smaller triangles when all three tangents are drawn in. So the ratios for the other two triangles are 1 - b/s and 1 - c/s.
The area of all four circles together can be expresses as
pi*r^2*[1+(1 - a/s)^2+(1 - b/s)^2+(1 - c/s)^2]
This looks like a mess but the part inside the square brackets can be simplified. After some algebra, and help from s=(a+b+c)/2, we get a simplified expression for the total area of all four circles is
pi*r^2*[a^2+b^2+c^2]/s^2
I could put in the explicit formulas for r and s, but this form seems elegant as each variable occurs exactly once and each is squared.
Solution
