Mongolian Mathematical Olympiad

We need following fact.

Lemma: Given a point $D$ on side $BC$ of the triangle $ABC$. Let ${\omega }_1(O_1,r_1),{\omega }_2(O_2,r_2)$ be incircles of triangles $ACD$, $ADB$ and ${\rho }_1(O_3,r_3),{\rho }_2(O_4,r_4)$ be circles inscribed in these triangles externally, which tangent at an internal point of $BC$. Then lines $O_1{O}_2$, $O_3{O}_4$, $BC$ pass through a point.



Proof of lemma: Denote $BC\cap O_3{O}_4=X$ and let's prove that $X\in O_1{O}_2$. From the well known property that a line passing centers of circles inscribed in an angle is bisector of the angle, it follows $A=(O_1{O}_3)\cap (O_2{O}_4)$, $D=(O_1{O}_4)\cap (O_2{O}_3)$. Since the line $O_3{O}_2$ intersects sides of the triangle $A{O}_1{O}_4$,

$$\text{by Menelaus theorem we get }1=\frac{O_3{O}_1}{O_{3}A}\cdot \frac{D{O}_4}{D{O}_1}\cdot \frac{O_{2}A}{O_2{O}_4}=\frac{O_{3}A-O_{1}A}{O_{3}A}\cdot \frac{D{O}_4}{D{O}_1}\cdot \frac{O_{2}A}{O_{4}A-O_{2}A}$$

$$=\left(1-\frac{O_{1}A}{O_{3}A}\right)\cdot \frac{D{O}_4}{D{O}_1}\cdot \left(\frac{1}{\frac{O_{4}A}{O_{2}A}-1}\right)=\left(1-\frac{r_1}{r_3}\right)\left(\frac{r_4}{r_1}\right)\left(\frac{1}{\frac{r_4}{r_2}-1}\right)$$

$$=\frac{(r_3-r_1)}{r_3}\cdot \frac{r_4}{r_1}\cdot \frac{r_2}{r_4-r_2}=\frac{(r_3-r_1)r_4{r}_2}{r_3{r}_1(r_4-r_2)}(\ast )$$

To prove $X\in (O_1{O}_2)$ is equivalent to proving by Menelaus theorem:

$$\frac{X{O}_3}{X{O}_4}\cdot \frac{O_{1}A}{O_1{O}_3}\cdot \frac{O_2{O}_4}{O_{2}A}=1.$$

Furthermore,

$$\frac{X{O}_3}{X{O}_4}=\frac{r_3}{r_4},\frac{O_{1}A}{O_1{O}_3}=\frac{O_{1}A}{O_{3}A-A{O}_1}=\frac{1}{\frac{O_{3}A}{O_{1}A}-1}=\frac{1}{\frac{r_3}{r_1}-1}=\frac{r_1}{r_3-r_1},$$

$$\frac{O_2{O}_4}{O_{2}A}=\frac{A{O}_4-A{O}_2}{O_{2}A}=\frac{A{O}_4}{A{O}_2}-1=\frac{r_4}{r_2}-1=\frac{r_4-r_2}{r_2}$$

and by $(\ast )$,

$$\frac{X{O}_3}{X{O}_4}\cdot \frac{O_{1}A}{O_1{O}_3}\cdot \frac{O_2{O}_4}{O_{2}A}=\frac{r_3}{r_4}\cdot \frac{r_1}{r_3-r_1}\cdot \frac{r_4-r_2}{r_2}=1.$$

This proves that $X\in O_1{O}_2$ and points $O_1{O}_2$, $O_3{O}_4$, $BC$ pass through a point.

Now solve the problem. Let $l$ be common tangent to circles ${\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4$ which is different from $BC$. Denote $l\cap BC=X$. Centers of circles ${\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4$ lie on the bisector of angle formed by $l$, $BC$ and the point $X$ lies on this bisector too.

Using the lemma for triangles $ABE$, $ADF$, $AEC$ we conclude that centers of circles ${\rho }_1,{\rho }_2,{\rho }_3,{\rho }_4$ and the point $X$ lie on a line. Reflecting $BC$ about this line we get a line which is common tangent to circles ${\rho }_1,{\rho }_2,{\rho }_3,{\rho }_4$. Thus we have the desired result.