# 2008 IMO Problems/Problem 6

## Problem

Let $ABCD$ be a convex quadrilateral with $BA \ne BC$. Denote the incircles of triangles $ABC$ and $ADC$ by $\omega _1$ and $\omega _2$ respectively. Suppose that there exists a circle $\omega$ tangent to ray $BA$ beyond $A$ and to the ray $BC$ beyond $C$, which is also tangent to the lines $AD$ and $CD$. Prove that the common external tangents to $\omega _1$ and $\omega _2$ intersect on $\omega$

## Solution

Here are some hints:

Let B be the top vertex of triangle ABC, O and K are the centers of the incircles of triangles ABC and ADC with radii R and r, respectively. S is the center of the circumcircle tangential to the extensions of AB, AC and DC. And let

E be the foot of the projection of O to AB. U be the foot of the projection of O to AC. V be the foot of the projection of K to AC. M be the foot of the projection of S to DC. L be the foot of the projection of K to DC. L be the intercept of DC and AB.

We have

/_EOB = /_LSC /_AOU = /_SOC /_ASO = /_KSC /_ASK = /_OSC /_LSB = /_KCO UV = BC – AB AU = VC OA. AK. cos(/_OAK) = OC. KC. cos(/_OCK) OK**2 = (R + r)**2 + UV**2 SK**2 = (R' + r)**2 + ML**2

Use sin (90-x) = cos x and cos(90-x) = sinx and characteristic of triangle a**2 = b**2 + c**2 - 2.b.c.cosine(angle) to solve.

Vo Duc Dien