Harvard-MIT Mathematics Tournament

Solution:
The distance between the point of tangency of the two circles and the nearest vertex of the triangle is seen to be both $r\left(\csc \frac{\pi}{8}-1\right)$ and $t\left(\csc \frac{\pi}{8}+1\right)$, so
$$\begin{aligned} r t & =\frac{r^{2}\left(\csc \frac{\pi}{8}-1\right)}{\csc \frac{\pi}{8}+1}=\frac{\left(1+\sin \frac{\pi}{8}\right)^{2}\left(1-\sin \frac{\pi}{8}\right)}{1+\sin \frac{\pi}{8}}=1-\sin ^{2} \frac{\pi}{8} \\ & =\frac{1}{2}+\frac{1-2 \sin ^{2} \frac{\pi}{8}}{2}=\frac{1}{2}+\frac{\cos \frac{\pi}{4}}{2}=\frac{1}{2}+\frac{\sqrt{2}}{4}=\frac{2+\sqrt{2}}{4} . \end{aligned}$$