1. Circle $k_{1}$ with radius 1 has an external tangency $\mathrm{s}$ with circle $k_{2}$ with radius 2. Each of the circles $k_{1}, k_{2}$ has an internal tangency with circle $k_{3}$ with radius 3. Calculate the radius of circle $k$, which has external tangency $\mathrm{s}$ with circles $k_{1}, k_{2}$ and internal tangency $\mathrm{s}$ with circle $k_{3}$.
1. Since the sum of the diameters of circles $k_{1}$ and $k_{2}$ equals the diameter of circle $k_{3}$, their centers $S_{1}, S_{2}$, and $S_{3}$ lie on a straight line. There are two congruent circles that satisfy the conditions of the problem, and they are symmetric with respect to the line $S_{1} S_{2}$. Let $k$ be one of them (Fig. 1), $S$ its center, and $r$ the corresponding radius.

![](https://cdn.mathpix.com/cropped/2024_04_17_25a061dd58330ed66a17g-2.jpg?height=614&width=622&top_left_y=618&top_left_x=383)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_04_17_25a061dd58330ed66a17g-2.jpg?height=385&width=574&top_left_y=841&top_left_x=1118)

Fig. 2

For the side lengths of triangle $S_{1} S_{2} S$, we have: $\left|S_{1} S\right|=1+r$, $\left|S_{2} S\right|=2+r$, $\left|S_{1} S_{2}\right|=3$, and $\left|S_{3} S\right|=3-r$. For point $S_{3}$, it also holds that $\left|S_{3} S_{1}\right|=2$ and $\left|S_{3} S_{2}\right|=1$. If we denote $P$ as the orthogonal projection of point $S$ onto the line $S_{1} S_{2}$ (Fig. 2) and $x=\left|S_{1} P\right|$, $y=|S P|$, we can write according to the Pythagorean theorem:

$$
\begin{aligned}
& (1+r)^{2}=x^{2}+y^{2} \\
& (2+r)^{2}=(3-x)^{2}+y^{2} \\
& (3-r)^{2}=(2-x)^{2}+y^{2}
\end{aligned}
$$

Subtracting the first equation from the second, we get $3+2 r=9-6 x$ or $2 r=6-6 x$. Subtracting the first equation from the third, we get $8-8 r=4-4 x$ or $2 r=1+x$. Comparing both results, we get the equation $6-6 x=1+x$, from which $x=\frac{5}{7}$, and $r=3-3 x=\frac{6}{7}$.

Note. With knowledge of the cosine rule, we can do without the auxiliary point $P$: it is sufficient to write the cosine rule for triangles $S_{1} S_{3} S$ and $S_{1} S_{2} S$. This gives us two equations:

$$
\begin{aligned}
& (3-r)^{2}=4+(1+r)^{2}-2 \cdot 2(1+r) \cos \omega, \\
& (2+r)^{2}=9+(1+r)^{2}-2 \cdot 3(1+r) \cos \omega,
\end{aligned}
$$

where $\omega=\left|\nless S_{2} S_{1} S\right|$. After rearranging and expressing $(1+r) \cos \omega$ from both equations, we get the equation $2 r-1=1-\frac{1}{3} r$, from which $r=\frac{6}{7}$.

For a complete solution, award 6 points. For identifying that the centers of circles $k_{1}, k_{2}$, and $k_{3}$ lie on a straight line, award 1 point. For setting up quadratic equations for the desired radius $r$, award 3 points. For calculating the radius $r$, award 2 points.