Consider an inversion of pole $A$. It turns circles into lines, so it turns the three circles into three lines concurrent at $B'$ (in general, let $T'$ be the image of $T$ under the inversion). For the lines $BA,BZ',BY',BX'$ we apply the cross-ratio and we get $\frac {AX'}{AZ'}:\frac {Y'X'}{Y'Z'}=$constant, so $\frac {AX'}{AZ'}:\frac {AX'-AY'}{AY'-AZ'}=$ constant. Now replace $AX',AY',AZ'$ by $\frac 1{AX},\frac 1{AY},\frac 1{AZ}$ respectively (remember the inversion). A brief computation gives what we want, namely $\frac {XY}{YZ}=$constant.

Grobber, your idea is nice, but the problem is much simpler. In fact, consider two points X and X' on the first circle. The line AX meets the other two circles at the points Y and Z (apart from A), and the line AX' meets the other two circles at the points Y' and Z' (apart from A again). Then we have to prove that XY / YZ = X'Y' / Y'Z'. Indeed, we have < AXB = < AX'B (cyclic) and thus < BXY = 180° - < AXB = 180° - < AX'B = < BX'Y'; also we have < AYB = < AY'B (cyclic) and therefore < BYX = < AYB = < AY'B = < BY'X'. Hence, from < BXY = < BX'Y' and < BYX = < BY'X', we conclude that the triangles BXY and BX'Y' are similar, so that XY / X'Y' = BY / BY'. Similarly, we find YZ / Y'Z' = BY / BY'. Therefore, XY / X'Y' = YZ / Y'Z', or, equivalently, XY / YZ = X'Y' / Y'Z'. Darij

Here is another proof Let O1,O2,O3 be three centers of the three circles. O1I1,O2I2,O3I3 are perpendicular to AX (I1,I2,I3 lie on AX ) We can easily prove that I1I2=XY/2 I2I3=YZ/2 I1I3=XZ/2 Therefore XY/YZ=I1I2/I2I3=O1O2/O2O3 XY/XZ=I1I2/I1I3=O1O2/O1O3

Iris Aliaj wrote: Three fixed circles pass through the points $A$ and $B$. Let $X$ be a variable point on the first circle different from $A$ and $B$. The line $AX$ intersects the other two circles at $Y$ and $Z$ (with $Y$ between $X$ and $Z$). Show that the ratio $\frac{XY}{YZ}$ is independent of the position of $X$. Just to note that this problem coincides with Baltic Way 2004 Problem 20: Three circular arcs $w_{1}$, $w_{2}$, $w_{3}$ with common endpoints A and B are on the same side of the line AB; hereby, $w_{2}$ lies between $w_{1}$ and $w_{3}$. Two rays emanating from B intersect these arcs at $M_{1}$, $M_{2}$, $M_{3}$ and $K_{1}$, $K_{2}$, $K_{3}$, respectively. Prove that $\frac{M_{1}M_{2}}{M_{2}M_{3}}=\frac{K_{1}K_{2}}{K_{2}K_{3}}$. See http://www.mathlinks.ro/Forum/viewtopic.php?t=20174 for a secondary discussion of this problem. Darij