1981 IMO Problems/Problem 5

Problem

Three congruent circles have a common point $\displaystyle O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle and the point $\displaystyle O$ are collinear.

Solution

Let the triangle have vertices $\displaystyle A,B,C$ , and sides $\displaystyle a,b,c$ , respectively, and let the centers of the circles inscribed in the angles $\displaystyle A,B,C$ be denoted $\displaystyle O_A, O_B, O_C$ , respectively.

The triangles $\displaystyle O_A O_B O_C$ and $\displaystyle ABC$ are homothetic, as their corresponding sides are parallel. Furthermore, since $\displaystyle O_A$ lies on the bisector of angle $\displaystyle A$ and similar relations hold for the triangles' other corresponding points, the center of homothety is the incenter of both the triangles. Since $\displaystyle O$ is clearly the circumcenter of $\displaystyle O_A O_B O_C$ , $\displaystyle O$ is collinear with the incenter and circumcenter of $\displaystyle ABC$ , as desired.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1981 IMO Problems/Problem 5

Problem

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