2.2. Among all right-angled triangles with a given circumradius $R$, find the sides of the one with the largest inradius.
2.2. $r=\frac{a+b-c}{2}, c=2 R=$ const. Therefore $r$ is maximal when the sum $a+b$ reaches its maximum value.

$$
a+b=\sqrt{(a+b)^{2}} \leq \sqrt{2\left(a^{2}+b^{2}\right)}=c \sqrt{2}
$$

Equality is achieved for $a=b=R \sqrt{2} . \quad c=2 R$.