, 2003

Solution:
Let $l$ be a line through $A$ different from $A B$ and join $B$ to $A, X, Y$ and $Z$ as in the above diagram. No matter how $l$ is chosen, the angles $A X B, A Y B$ and $A Z B$ always subtend the chord $A B$. For this reason the angles in the triangles $B X Y$ and $B X Z$ are the same for all such $l$. Thus the ratio $X Y: Y Z$ remains constant by similar triangles.

Note that this is true no matter how $X, Y$ and $Z$ lie in relation to $A$. Suppose $X, Y$ and $Z$ all lie on the same side of $A$ (as in the diagram) and that $\measuredangle A X B=\alpha, \measuredangle A Y B=\beta$ and $\measuredangle A Z B=\gamma$. Then $\measuredangle B X Y=180^\circ-\alpha, \measuredangle B Y X=\beta, \measuredangle B Y Z=180^\circ-\beta$ and $\measuredangle B Z Y=\gamma$. Now suppose $l$ is chosen so that $X$ is now on the opposite side of $A$ from $Y$ and $Z$. Now since $X$ is on the other side of the chord $A B, \measuredangle A X B=180^\circ-\alpha$, but it is still the case that $\measuredangle B X Y=180^\circ-\alpha$ and all other angles in the two pertinent triangles remain unchanged. If $l$ is chosen so that $X$ is identical with $A$, then $l$ is tangent to the first circle and it is still the case that $\measuredangle B X Y=180^\circ-\alpha$. All other cases can be checked in a similar manner.

Solution:
Let $m$ be the perpendicular bisector of $A B$ and let $O_{1}, O_{2}, O_{3}$ be the centres of the three circles. Since $A B$ is a chord common to all three circles, $O_{1}, O_{2}, O_{3}$ all lie on $m$. Let $l$ be a line through $A$ different from $A B$ and suppose that $X, Y, Z$ all lie on the same side of $A B$, as in the above diagram. Let perpendiculars from $O_{1}, O_{2}, O_{3}$ meet $l$ at $P, Q, R$, respectively. Since a line through the centre of a circle bisects any chord,
$$A X=2 A P, \quad A Y=2 A Q \quad \text{ and } \quad A Z=2 A R$$
Now
$$X Y=A Y-A X=2(A Q-A P)=2 P Q \quad \text{ and, similarly, } \quad Y Z=2 Q R.$$
Therefore $X Y: Y Z=P Q: Q R$. But $O_{1} P\parallel O_{2} Q\parallel O_{3} R$, so $P Q: Q R=O_{1} O_{2}: O_{2} O_{3}$. Since the centres of the circles are fixed, the ratio $X Y: Y Z=O_{1} O_{2}: O_{2} O_{3}$ does not depend on the choice of $l$.
If $X, Y, Z$ do not all lie on the same side of $A B$, we can obtain the same result with a similar proof. For instance, if $X$ and $Y$ are opposite sides of $A B$, then we will have $X Y=A Y+A X$, but since in this case $P Q=A Q+A P$, it is still the case that $X Y=2 P Q$ and result still follows, etc.