12.132. 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?
Solution.

In $\triangle ABC$ (Fig. 12.5) $\angle ABC=90^{\circ}, \angle ABC=\alpha, O$ is the center of the inscribed circle, $D$ is the point of tangency with the hypotenuse $AB$.

Let the radius of the inscribed circle $OD=r$.

Then from $\triangle BDO\left(\angle BDO=90^{\circ}\right): BD=OD \operatorname{ctg} \angle OBD=r \operatorname{ctg} \frac{\alpha}{2}$.

From $\triangle ADO\left(\angle ADO=90^{\circ}\right): AD=OD \operatorname{ctg} \angle OAD-r \operatorname{ctg}\left(45^{\circ}-\frac{\alpha}{2}\right)$

The radius of the circumscribed circle will be equal to

$$
\begin{aligned}
& R=\frac{1}{2} AB=\frac{1}{2}(AD+BD)=\frac{1}{2} r\left(\operatorname{ctg}\left(45^{\circ}-\frac{\alpha}{2}\right)+\operatorname{ctg} \frac{\alpha}{2}\right)= \\
& =\frac{r}{2} \cdot \frac{\sin 45^{\circ}}{\sin \left(45^{\circ}-\frac{\alpha}{2}\right) \cdot \sin \frac{\alpha}{2}}=\frac{r}{2 \sqrt{2} \sin \left(45^{\circ}-\frac{\alpha}{2}\right) \sin \frac{\alpha}{2}}
\end{aligned}
$$

Then $\frac{r}{R}=2 \sqrt{2} \sin \left(45^{\circ}-\frac{\alpha}{2}\right) \sin \frac{\alpha}{2}$.

$2 \sqrt{2} \sin \left(45^{\circ}-\frac{\alpha}{2}\right) \sin \frac{\alpha}{2}=\sqrt{2}\left(\cos \left(45^{\circ}-\alpha\right)-\cos 45^{\circ}\right)=\sqrt{2} \cos \left(45^{\circ}-\alpha\right)-1$.

The obtained ratio takes the maximum value when $\cos \left(45^{\circ}-\alpha\right)=1$, that is, since $0<\alpha<90^{\circ}$, at $\alpha=45^{\circ}$.

Answer: $2 \sqrt{2} \sin \left(45^{\circ}-\frac{\alpha}{2}\right) \sin \frac{\alpha}{2} ; \alpha=45^{\circ}$.