17. Determine the angles of an isosceles triangle for which the ratio $\frac{r}{R}$ is the greatest (smallest).
17. Since $S=p r$, then $r=\frac{S}{p}=\frac{a \sin \frac{\alpha}{2} \cos \alpha}{\cos \frac{\alpha}{2}}$ (where $\alpha$ is the angle at the base). $\quad R=\frac{a}{2 \sin \alpha}$, but $b=2 a \cos \alpha$, then $\frac{r}{R}=4 \sin ^{2} \frac{x}{2}\left(1-2 \sin ^{2} \frac{\alpha}{2}\right)$.

Since $0<\alpha<\pi$, $2 \sin ^{2} \frac{\alpha}{2}+\left(1-2 \sin ^{2} \frac{\alpha}{2}\right)=1$ - a constant value. Therefore, the product $2 \sin ^{2} \frac{\alpha}{2}\left(1-2 \sin ^{2} \frac{\alpha}{2}\right)$, and thus $\frac{r}{R}$, will be the greatest when $2 \sin ^{2} \frac{\alpha}{2}=1-2 \sin ^{2} \frac{\alpha}{2}$.

From which $\alpha=60^{\circ}$. The minimum value of $\frac{r}{R}$ does not exist.