Harvard-MIT Mathematics Tournament

Solution:
Let $t$ denote the radius of the excircle of triangle $ABC$ opposite $A$.

Lemma: Let $ABC$ be a triangle, and let $r$ and $r_A$ be the inradius and exradius opposite $A$. Then
$$\frac{r}{r_A} = \tan \frac{B}{2} \tan \frac{C}{2}$$
Proof. Let $I$ and $J$ denote the incenter and the excenter with respect to $A$. Let $D$ and $E$ be the foot of the perpendicular from $I$ and $J$ to $BC$, respectively. Then
$$\begin{aligned} r = ID & = BI \sin \frac{B}{2} \\ r_A = JE & = BJ \sin \frac{180^\circ - B}{2} = BJ \cos \frac{B}{2} \\ BI & = BJ \tan \angle AJB = BY \tan \frac{C}{2} . \end{aligned}$$
The last equation followed from
$$\angle AJB = 180^\circ - \angle ABJ - \angle JAB = \frac{180^\circ - B}{2} - \frac{A}{2} = \frac{C}{2} .$$
Hence
$$\frac{r}{r_A} = \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}} \cdot \frac{BI}{BJ} = \tan \frac{B}{2} \cdot \tan \frac{C}{2}$$
Noting $\tan \frac{\angle APB}{2} \tan \frac{\angle APQ}{2} = \tan \frac{\angle AQP}{2} \tan \frac{\angle AQC}{2} = 1$ and applying the lemma to $\triangle ABC$, $\triangle ABP$, $\triangle APQ$, and $\triangle AQC$ give
$$\begin{aligned} \frac{3/2}{t} & = \tan \frac{\angle ABC}{2} \cdot \tan \frac{\angle ACB}{2} \\ & = \left(\tan \frac{\angle ABC}{2} \cdot \tan \frac{\angle APB}{2}\right) \cdot \left(\tan \frac{\angle APQ}{2} \cdot \tan \frac{\angle AQP}{2}\right) \cdot \left(\tan \frac{\angle AQC}{2} \cdot \tan \frac{\angle ACB}{2}\right) \\ & = \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{1}{5} \end{aligned}$$
Therefore, $t = 135$.