# 1964 IMO Problems/Problem 3

## Problem

A circle is inscribed in a triangle with sides . Tangents to the circle parallel to the sides of the triangle are contructed. Each of these tangents cuts off a triangle from . In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of ).

## Solution

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Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let the points of contact of MH,FG,DE with the in circle be X,Y,Z respectively. Then perimeter of BHM = BH+HX+XM+MB=BH+HP+MR+BM=BP+BQ=2(s-b) and similar results follow!

Each of the triangles BHM,CGF,ADE are similar to ABC.

[BHM]/[BCA] = (perimeter of BHM/perimeter of BCA)^2 ={(2s-2b)/(a+b+c)}^2 ={(c+a-b)/(c+a+b)}^2

Thus, [BHM]+[CGF]+[ADE]=[ABC]/(a+b+c)^2 { ( b+c-a)^2 + (c+a-b)^2 + (a+b-c)^2}

Denote the area of and the perimeter of .

Then .

So .

We know, is the radius of the incircle of : .

Area of the incircle of Area of the incircle of :.

Sum of the area of the 4 incircles: