11.2 What is the maximum value that the ratio of the radius of the inscribed circle to the radius of the circumscribed circle of a right triangle can reach?
Answer $\sqrt{2}-1$. Solution. Let $c$ be the hypotenuse and $\alpha$ be the acute angle. Then $R=c / 2$ and $r=(c \cos \alpha+c \sin \alpha-c) / 2=(c \sqrt{2} \sin (\alpha+\pi / 4)-c) / 2 \leq c(\sqrt{2}-1) / 2$. Therefore, the maximum value of the ratio $r / R$ is $\sqrt{2}-1$ and is achieved when $\alpha=\pi / 4$, i.e., for an isosceles right triangle.