Baltic Way 1991 · Problem 19

Solution:

First, note that the three straight lines $A_{1}A_{2}$, $B_{1}B_{2}$ and $C_{1}C_{2}$ intersect in a single point $O$. Indeed, each of the lines is the locus of points from which the tangents to two of the circles are of equal length (it is easy to check that this locus has the form of a straight line and obviously it contains the two intersection points of the circles).

Now, we have $|O A_{1}| \cdot |O A_{2}| = |O B_{1}| \cdot |O B_{2}|$ (as both of these products are equal to $|O T|^{2}$ where $O T$ is a tangent line to the circle containing $A_{1}, A_{2}, B_{1}, B_{2}$, and $T$ is the corresponding point of tangency). Hence

$$\frac{|O A_{1}|}{|O B_{2}|} = \frac{|O B_{1}|}{|O A_{2}|}$$

which implies that the triangles $O A_{1} B_{2}$ and $O B_{1} A_{2}$ are similar and

$$\frac{|A_{1} B_{2}|}{|A_{2} B_{1}|} = \frac{|O A_{1}|}{|O B_{1}|}.$$

Similarly we get

$$\frac{|B_{1} C_{2}|}{|B_{2} C_{1}|} = \frac{|O B_{1}|}{|O C_{1}|}$$

and

$$\frac{|C_{1} A_{2}|}{|C_{2} A_{1}|} = \frac{|O C_{1}|}{|O A_{1}|}.$$

Multiplying these three equalities gives the desired result.