# 2004 Indonesia MO Problems/Problem 4

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## Problem

There exists 4 circles, $a,b,c,d$, such that $a$ is tangent to both $b$ and $d$, $b$ is tangent to both $a$ and $c$, $c$ is both tangent to $b$ and $d$, and $d$ is both tangent to $a$ and $c$. Show that all these tangent points are located on a circle.

## Solution

Let $A, B, C, D$ be the centers of circle $a, b, c,$ and $d,$ respectively. Also, let the tangent point of circles $a$ and $b$ be $W$, let the tangent point of circles $b$ and $c$ be $X$, let the tangent point of circles $c$ and $d$ be $Y$, and let the tangent point of circles $d$ and $a$ be $Z$. Finally, let $\angle A, \angle B, \angle C, \angle D$ be $w,x,y$ and $z,$ respectively.

Note that $\triangle AWZ, \triangle BWX, \triangle CXY, \triangle DYZ$ are isosceles triangles, so $\angle AWZ = 90 - \tfrac{w}{2}$ degrees, $\angle BWX = 90 - \tfrac{x}{2}$ degrees, $\angle CYX = 90 - \tfrac{y}{2}$ degrees, and $\angle DYZ = 90 - \tfrac{z}{2}$ degrees. The sum of the angles in a line is $180^\circ,$ so $\angle ZWX = \tfrac{w+x}{2}$ degrees and $\angle ZYX = \tfrac{y+z}{2}$ degrees.

That means $\angle ZWX + \angle ZYX = \tfrac{w+x+y+z}{2}$ degrees. Since $ABCD$ is a quadrilateral, $w+x+y+z=360^\circ,$ so $\angle ZWX + \angle ZYX = \tfrac{360}{2} = 180^\circ.$ Thus, all the tangent points of the four circles can be located in another circle.