10.391 On the segment $A C$ there is a point $B$, and $A B=14 \text{~cm}, B C=28 \text{~cm}$. On the segments $A B, B C$ and $A C$ as diameters, semicircles are constructed in the same half-plane relative to the boundary $\boldsymbol{A} \boldsymbol{B}$. Find the radius of the circle that is tangent to all three semicircles.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-238.jpg?height=266&width=469&top_left_y=1094&top_left_x=132)

Let $O_{1}, O_{2}, O_{3}$ be the centers of semicircles with radii $r_{1}, r_{2}, r_{3}$ and diameters $A B, B C$, and $A C$ respectively; $P$ be the center of the circle with radius $x$ that touches all three semicircles. Then

$$
\begin{aligned}
& P O_{1}=x+r_{1}, P O_{2}=x+r_{2}, P O_{3}=r_{3}-x \\
& O_{1} O_{3}=r_{3}-r_{1}, O_{2} O_{3}=r_{3}-r_{2} ; r_{1}=7
\end{aligned}
$$

$r_{2}=14, r_{3}=21$ (by the problem statement). Applying the cosine rule to $\Delta O_{1} \mathrm{PO}_{3}$ and $\Delta O_{3} \mathrm{PO}_{2}$,

we have: $\left\{\begin{array}{l}(7+x)^{2}=14^{2}+(21-x)^{2}-28 \cdot(21-x) \cdot \cos \angle O_{1} O_{3} P, \\ \left(14+x^{2}\right)=7^{2}+(21-x)^{2}-14 \cdot(21-x) \cdot \cos \left(180^{\circ}-\angle O_{1} O_{3} P\right),\end{array} \Rightarrow\right.$

$\Rightarrow(7+x)^{2}+2(14+x)^{2}=14^{2}+2 \cdot 7^{2}+3(21-x)^{2} \Rightarrow x=6$.

Answer: $6 \mathrm{~cm}$.