A semicircle with a radius of $3 \mathrm{~cm}$ has a diameter on which we placed a semicircle with a radius of $1 \mathrm{~cm}$ and another semicircle with a radius of $2 \mathrm{~cm}$ inside the larger semicircle, as shown in the figure. Calculate the radius of the circle that touches all three semicircular arcs.

![](https://cdn.mathpix.com/cropped/2024_05_02_77a100a78f70a9d1d0c1g-1.jpg?height=206&width=377&top_left_y=268&top_left_x=863)
Let $r$ denote the unknown radius, and the centers of the circles with radii $1 \mathrm{~cm}, 2 \mathrm{~cm}, 3 \mathrm{~cm}, r$ be denoted by $O_{1}, O_{2}, O_{3}, O_{r}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_18c835c588d133c20fecg-1.jpg?height=290&width=455&top_left_y=224&top_left_x=824)

The distance between the centers of tangent circles is the sum or the difference of their radii:

$$
O_{1} O_{r}=1+r ; \quad O_{2} O_{r}=2+r ; \quad O_{3} O_{r}=3-r
$$

Furthermore,

$$
O_{1} O_{3}=2 ; \quad O_{1} O_{2}=3(\mathrm{~cm})
$$

Let's apply the cosine rule to the triangles $O_{1} O_{2} O_{r}$ and $O_{1} O_{3} O_{r}$ $\left(\alpha=O_{2} O_{1} O_{r} \varangle\right)$:

$$
\begin{aligned}
& (2+r)^{2}=3^{2}+(1+r)^{2}-6(1+r) \cos \alpha \\
& (3-r)^{2}=2^{2}+(1+r)^{2}-4(1+r) \cos \alpha
\end{aligned}
$$

From the first equation, $\cos \alpha=\frac{3-r}{3(1+r)}$, and from the second equation, $\cos \alpha=\frac{2 r-1}{1+r}$. Equating these, we get $r=\frac{6}{7}$, so the radius of the sought circle is $\frac{6}{7}$.

Andrea Rácz (Dunaföldvár, Magyar L. Gymn., III. grade)

Remark. The problem can also be solved using the Pythagorean theorem or coordinate geometry methods.