# 1978 IMO Problems/Problem 4

## Problem

In a triangle $ABC$ we have $AB = AC.$ A circle which is internally tangent with the circumscribed circle of the triangle is also tangent to the sides $AB, AC$ in the points $P,$ respectively $Q.$ Prove that the midpoint of $PQ$ is the center of the inscribed circle of the triangle $ABC.$

## Solution

Denote $a = BC, b = AB = AC$ the triangle sides, $r, R$ the triangle inradius and circumradius and $s$, $\triangle$ the triangle semiperimeter and area. Let $S, T$ be the tangency points of the incircle (I) with the sides $AB, AC$. Let $K, L$ be the midpoints of of $BC, PQ$ and $M$ the midpoint of the arc $BC$ of the circumcircle (O) opposite to the vertex $A$. The isosceles triangles $\triangle AST \sim \triangle APQ \sim \triangle ABC$ are all centrally similar with the homothety center $a$. The homothety coefficient for the triangles $\triangle AST \sim \triangle ABC$ is $h_{13} = \frac{AS}{AB} = \frac{s - a}{b}$. The homothety coefficient for the triangles $\triangle AST \sim \triangle APQ$ is $h_{12} = \frac{AS}{AP}$. The point K is the tangency point of the incircle (I) with the side BC, while the point M is the tangency point of the given circle with the triangle circumcircle (O). It follows that the triangles $\triangle ASK \sim \triangle APM$ are also centrally similar with the homothety center A and the same homothety coefficient as the triangles $\triangle AST \sim \triangle APQ$: $h_{12} = \frac{AS}{AP} = \frac{AK}{AM} = \frac{h}{2R}$, where h = AK is the A-altitude of the triangle $\triangle ABC$. The homothety coefficient of the triangles $\triangle APQ \sim \triangle ABC$ is then $h_{23} = \frac{h_{12}}{h_{13}} = \frac{s - a}{b} \cdot \frac{2R}{h}$ Denote $h' = AL$ the $A$-altitude of the triangle $\triangle APQ$. Then

$\frac{h'}{h} = h_{23} = \frac{s - a}{b} \cdot \frac{2R}{h}$

$KL = h - h' = h - 2R\ \frac{s - a}{b}$

Substituting $h = \frac{2 \triangle}{a}$, $2R = \frac{ab^2}{2 \triangle}$ and $s - a = \frac{\triangle^2}{s(s - b)^2}$, we get

$KL = \frac{2 \triangle}{a} - \frac{ab^2}{2 \triangle} \cdot \frac{\triangle^2}{bs(s - b)^2} =$

(substituting 2s = 2b + a and $s - b = \frac a 2$ for an isosceles triangle)

$= \frac{\triangle}{s}\left(\frac{2s}{a} - \frac{ab}{2(s - b)^2}\right) = \frac{\triangle}{s} \left(\frac{2b}{a} + 1 - \frac{2b}{a}\right) = \frac{\triangle}{s} = r$

which means that the point $L \equiv I$ is identical with the incenter of the triangle $\triangle ABC$.

The above solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]