Harvard-MIT Mathematics Tournament

Solution:

Answer: $\frac{115}{143}$. Denote the lengths $AB, BC, CD$, and $DA$ by $a, b, c$, and $d$ respectively. Because $ABCD$ is cyclic, $\triangle ABX \sim \triangle DCX$ and $\triangle ADX \sim \triangle BCX$. It follows that $\frac{AX}{DX} = \frac{BX}{CX} = \frac{a}{c}$ and $\frac{AX}{BX} = \frac{DX}{CX} = \frac{d}{b}$. Therefore we may write $AX = adk$, $BX = abk$, $CX = bck$, and $DX = cdk$ for some $k$.

Now, $\angle XDC = \angle BAX = \angle YXB$ and $\angle DCX = \angle XBY$, so $\triangle BXY \sim \triangle CDX$. Thus, $XY = DX \cdot \frac{BX}{CD} = cdk \cdot \frac{abk}{c} = abd k^2$. Analogously, $XZ = acd k^2$. Note that $XY / XZ = CB / CD$. Since $\angle YXZ = \pi - \angle ZAY = \angle BCD$, we have that $\triangle XYZ \sim \triangle CBD$. Thus, $YZ / BD = XY / CB = ad k^2$.

Finally, Ptolemy's theorem applied to $ABCD$ gives
$$(ad + bc)k \cdot (ab + cd)k = ac + bd$$
It follows that the answer is
$$\frac{ad(ac + bd)}{(ab + cd)(ad + bc)} = \frac{20 \cdot 23}{22 \cdot 26} = \frac{115}{143}$$