13. Given a right triangle, one of whose acute angles is equal to $\alpha$. Find the ratio of the radii of the circumscribed and inscribed circles and determine for which $\alpha$ this ratio will be the smallest?
13. $2 R$ is equal to the hypotenuse of the triangle, but

\[
2 R=r \operatorname{ctg} \frac{\alpha}{2}+r \operatorname{ctg}\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)=\frac{2 r}{\sqrt{2} \cos \left(\alpha-\frac{\pi}{4}\right)-1}, \text{ hence }
\]

\[
\frac{R}{r}=\frac{1}{\sqrt{2} \cos \left(\alpha-\frac{\pi}{4}\right)-1}. \text{ This ratio will have the smallest value when } \cos \left(x-\frac{\pi}{4}\right)=1, \text{ i.e., } \alpha=\frac{\pi}{4}.
\]

Then \(\frac{R}{r}=\sqrt{2}+1\).