3.59. The acute angle of a right triangle is $\alpha$. Find the ratio of the radius of the inscribed circle to the radius of the circumscribed circle. For what value of $\alpha$ is this ratio the greatest?
3.59. By the condition, $\angle A C B=90^{\circ}$, $\angle C B A=\alpha$ (Fig. 3.61). Let $R$ and $r$ be the radii of the circumscribed and inscribed circles, respectively; then $A C=2 R \sin \alpha ; B C=2 R \cos \alpha$. Therefore, $S_{\triangle A B C}=\frac{1}{2} \cdot 4 R^{2} \sin \alpha \cos \alpha=2 R^{2} \sin \alpha \cos \alpha$.

Using formula (2.4), we find

$$
r=\frac{S}{p}=\frac{2 R^{2} \sin \alpha \cos \alpha \cdot 2}{2 R(1+\sin \alpha+\cos \alpha)}=\frac{2 R \sin \alpha \cos \alpha}{1+\sin \alpha+\cos \alpha}
$$

and, therefore,

$$
\begin{gathered}
\frac{r}{R}=\frac{4 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cos \alpha}{2 \cos ^{2} \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=\frac{2 \sin \frac{\alpha}{2}\left(\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}}= \\
=2 \sin \frac{\alpha}{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)=\sin \alpha-2 \sin ^{2} \frac{\alpha}{2}= \\
=\sin \alpha+\cos \alpha-1=\sqrt{2} \sin \left(\frac{\pi}{4}+\alpha\right)-1
\end{gathered}
$$

This ratio is the greatest when $\sin \left(\frac{\pi}{4}+\alpha\right)=1$, from which $\frac{\pi}{4}+\alpha=\frac{\pi}{2}$, i.e., $\alpha=\frac{\pi}{4}$.