Open Contests

Figure 6

Solution:

Let the radii of $c_1,c_2$ and $c$ be $r_1,r_2$ and $r$, respectively. Homothety of ratio $\frac{r}{r_1}$ with center $A_1$ takes circle $c_1$ to circle $c$ and points $E_1,F_1$ to points $B,D$, respectively. Thus it takes line $E_1{F}_1$ to line $BD$. Analogously, homothety of ratio $\frac{r}{r_2}$ with center $A_2$ takes line $E_2{F}_2$ to line $BD$. Consequently, lines $E_1{F}_1$ and $E_2{F}_2$ are parallel to line $BD$ (Fig. 6).

Furthermore, note that $|B{E}_1|\cdot |B{A}_1|=|BP|\cdot |BQ|$ and $|B{E}_2|\cdot |B{A}_2|=|BP|\cdot |BQ|$, implying $|B{E}_1|\cdot |B{A}_1|=|B{E}_2|\cdot |B{A}_2|$. Thus triangles $B{E}_1{E}_2$ and $B{A}_2{A}_1$ are similar and $$\angle B{E}_1{E}_2=\angle B{A}_2{A}_1=\angle BD{A}_1=\angle E_1{F}_1{A}_1=\frac{1}{2}\angle E_1{O}_1{A}_1=90^{\circ }-\angle O_1{E}_1{A}_1,(3)$$ whence $$\angle E_2{E}_1{O}_1=180^{\circ }-\angle B{E}_1{E}_2-O_1{E}_1{A}_1=90^{\circ }.(4)$$ Analogously, $\angle E_1{E}_2{O}_2=90^{\circ }$. Hence the quadrilateral $E_1{E}_2{O}_2{O}_1$ is a right-angled trapezoid (or rectangle in the case $r_1=r_2$) and the midpoint of the line segment $O_1{O}_2$ lies on the perpendicular bisector of the line segment $E_1{E}_2$, thus being equidistant from $E_1$ and $E_2$. Analogously, the midpoint of the line segment $O_1{O}_2$ is also equidistant from $F_1$ and $F_2$.

As line $O_1{O}_2$ is perpendicular to $BD$, line $O_1{O}_2$ is also perpendicular to $E_1{F}_1$. Thus the line segment $O_1{O}_2$ entirely lies on the perpendicular bisector of $E_1{F}_1$. This means that the midpoint of line segment $O_1{O}_2$ is equidistant from $E_1$ and $F_1$.

Altogether, we have shown that these four points lie on a circle with its center at the midpoint of the line segment $O_1{O}_2$.