Let $O A=x$. Then $A B=A D=2 x$. By the Pythagorean theorem, $O A^2+A B^2=O B^2$, or $x^2+4 x^2=R^2$, from which $x=\frac{R}{\sqrt{5}}$. Let $p$ be the semiperimeter of the isosceles triangle $O B C$, $S$ be the area, and $r$ be the radius of the inscribed circle. Then

$$
\begin{gathered}
S=\frac{1}{2} B C \cdot A B=\frac{1}{2} \cdot 2 x \cdot 2 x^2=x^2=\frac{2 R^{2}}{5} \\
p=O B+\frac{1}{2} B C=R+x=R+\frac{R}{\sqrt{5}}=\frac{R(1+\sqrt{5})}{\sqrt{5}}
\end{gathered}
$$

Therefore,

$$
r=\frac{S}{p}=\frac{\frac{2 R^{2}}{5}}{\frac{R(1+\sqrt{5})}{\sqrt{5}}}=\frac{2 R}{5+\sqrt{5}}=\frac{R(5-\sqrt{5})}{10}
$$

## Answer

$\frac{R(5-\sqrt{5})}{10}$.
In a semicircle of radius $R$ with center at point $O$, a square $A B C D$ is inscribed such that points $A$ and $D$ lie on the diameter, and points $B$ and $C$ lie on the circumference. Find the radius of the circle inscribed in triangle $O B C$.

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