HMMT February

Solution:

Denote $E$ as the intersection point of $AD$ and $BC$. Let $x = EA$ and $y = EB$. Because $ABCD$ is a cyclic quadrilateral, $\triangle EAB$ is similar to $\triangle ECD$. Therefore,
$$\frac{y + 8}{x} = \frac{25}{10} \quad \text{and} \quad \frac{x + 12}{y} = \frac{25}{10}.$$
We get $x = \frac{128}{21}$ and $y = \frac{152}{21}$.

Note that $\omega$ is the $E$-excircle of $\triangle EAB$, so we may finish by standard calculations.

Indeed, first we compute the semiperimeter
$$s = \frac{EA + AB + BE}{2} = \frac{x + y + 10}{2} = \frac{35}{3}.$$
Now the radius of $\omega$ is (by Heron's formula for area)
$$r_E = \frac{[EAB]}{s - AB} = \sqrt{\frac{s(s - x)(s - y)}{s - 10}} = \sqrt{\frac{1209}{7}} = \frac{\sqrt{8463}}{7}.$$