In a semicircle of radius $R$, two circles of radius $r$ are inscribed, touching each other externally, and touching the semicircle's diameter and arc internally, as shown in the figure. Determine the radius $r$.

![](https://cdn.mathpix.com/cropped/2024_05_02_fa5672c2f6c68623d518g-1.jpg?height=231&width=455&top_left_y=275&top_left_x=824)
I. solution. Draw the common tangents of the circles with radii $r$ and $R$, and extend the given diameter of the circle with radius $R$ in both directions (Figure 1). Reflect the resulting triangle $BCD$ over the hypotenuse $BD$. The resulting quadrilateral $ABCD$ is clearly a square with side lengths of $2R$.

![](https://cdn.mathpix.com/cropped/2024_05_02_97fe0acb1b41eac05ee3g-1.jpg?height=436&width=458&top_left_y=278&top_left_x=823)

Figure 1

The area of the square: $T_{ABCD}=4R^2$.

The area of triangle $SBC$:

$$
T_{SBC \triangle}=\frac{T_{ABCD}}{4}=R^2
$$

Using the Pythagorean theorem for the diagonal of the square:

$$
AC=\sqrt{(2R)^2+(2R)^2}=\sqrt{8R^2}=2\sqrt{2}R
$$

from which

$$
SB=SC=\frac{AC}{2}=\sqrt{2}R
$$

$T_{SBC}=r \cdot s$, where $s$ is the semiperimeter:

$$
s=\frac{SB+SC+BC}{2}=\frac{\sqrt{2}R+\sqrt{2}R+2R}{2}=\sqrt{2}R+R=R(\sqrt{2}+1)
$$

Thus,

$$
r=\frac{T_{SBC \Delta}}{s}=\frac{R^2}{R(\sqrt{2}+1)}=\frac{R}{(\sqrt{2}+1)} \cdot \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}=\frac{(\sqrt{2}-1)R}{2-1}=(\sqrt{2}-1)R
$$

Therefore, the length of the radius $r$ is $(\sqrt{2}-1)R$.

II. solution. Let $e$ be the common tangent of the circles with center $O$ and radius $R$, and the circle with center $Q$ and radius $r$ (Figure 2). The radius drawn at the point of tangency is perpendicular to the tangent, so the point $Q$ lies on the segment $OD$:

$$
OD=OQ+QD=OQ+r=R
$$

$\angle BOC=90^\circ$, and the segments $OB$ and $OC$ are tangents to the circle with radius $r$, so $\angle OFQ=\angle OEQ=90^\circ$, and $EQ=FQ=r$, hence the quadrilateral $OFQE$ is a square. Using the Pythagorean theorem for the diagonal of the square:

$$
OQ=\sqrt{r^2+r^2}=\sqrt{2}r
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_97fe0acb1b41eac05ee3g-1.jpg?height=453&width=461&top_left_y=1926&top_left_x=821)

Figure 2

Thus,

$$
R=OQ+r=\sqrt{2}r+r=r \cdot(\sqrt{2}+1)
$$

from which

$$
r=\frac{R}{(\sqrt{2}+1)}=\frac{R}{(\sqrt{2}+1)} \cdot \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}=(\sqrt{2}-1)R
$$