10.259. On the segment $A C$ of length 12 cm, a point $B$ is constructed such that $A B=4 \text{ cm}$. On the segments $A B$ and $A C$ as diameters, semicircles are constructed in the same half-plane with boundary $A C$. Calculate the radius of the circle that is tangent to the constructed semicircles and $A C$.
## Solution.

Points $O_{1}$ and $O_{2}$ are the centers of semicircles with diameters $A B$ and $A C$ and radii $R_{1}=2$ cm and $R_{2}=6$ cm, respectively (Fig. 10.64). $O_{3}$ is the center of the circle with the unknown radius $x$ cm, $x>0$. Then $O_{1} O_{3}=(x+2)$ cm,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0754.jpg?height=276&width=507&top_left_y=191&top_left_x=126)

Fig. 10.64

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0754.jpg?height=432&width=618&top_left_y=63&top_left_x=642)

Fig. 10.65

$O_{1} O_{2}=R_{2}-R_{1}=4 \text{ cm}, O_{2} O_{3}=(6-x) \text{ cm}$. $D$ is the point of tangency of the circle with center $O_{3}$ and $A C$. Then $O_{3} D \perp A C, O_{3} D=x$ cm. In $\Delta O_{1} O_{2} O_{3}$, the semiperimeter $p=\frac{O_{1} O_{3}+O_{2} O_{3}+O_{1} O_{2}}{2}=6$ cm. Then its area $S=\sqrt{p\left(p-O_{1} O_{3}\right)\left(p-O_{1} O_{2}\right)\left(p-O_{2} O_{3}\right)}=\sqrt{6(6-x-2)(6-4)(6-6+x)}=$ $=\sqrt{12 x(4-x)}$.

On the other hand, $S=\frac{1}{2} O_{1} O_{2} \cdot O_{3} D=2 x$. Then $\sqrt{12 x(4-x)}=2 x$; $3 x(4-x)=x^{2} ; x=3$.

Answer: 3 cm.