4. Circles $k, l, m$ touch each other externally in pairs and all three have a common tangent. The radii of circles $k, l$ are $3 \mathrm{~cm}$ and $12 \mathrm{~cm}$. Calculate the radius of circle $m$. Find all solutions.
SOLUTION. Let $R, S, T$ be the centers and $A, B, C$ the points of tangency of the circles $k, l, m$ on the common tangent, and let $r=3, s=12$ and $t$ be their radii (we will calculate lengths and areas without units for easier substitution). In the trapezoid (which is a rectangle in the case of $r=t$) $A R T C$ (Fig. 1), $|R T|=r+t$. If we denote by $U$ the intersection of the line $A R$ and the line through $T$ parallel to $A C$, then $|R U|=|r-t|$. From the right triangle $R U T$ it follows that $|U T|=|A C|=\sqrt{(r+t)^{2}-(r-t)^{2}}=2 \sqrt{r t}=2 \sqrt{3 t}$. Similarly, from the trapezoids $C T S B$ and $A R S B$ we get the relations $|B C|=2 \sqrt{s t}=4 \sqrt{3 t}$ and $|A B|=2 \sqrt{r s}=$ $=2 \sqrt{3 \cdot 12}=12$.

![](https://cdn.mathpix.com/cropped/2024_04_17_f82aeacbc4420b407496g-2.jpg?height=499&width=648&top_left_y=236&top_left_x=310)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_04_17_f82aeacbc4420b407496g-2.jpg?height=503&width=805&top_left_y=231&top_left_x=1025)

Fig. 2

First, consider the case where point $C$ lies between points $A$ and $B$. Then $2 \sqrt{3 t}+4 \sqrt{3 t}=$ $=12$, from which $t=\frac{4}{3}$. If point $A$ lies between points $C$ and $B$, we similarly get the equation $2 \sqrt{3 t}+12=4 \sqrt{3 t}$, from which $t=12$. The equation $12+4 \sqrt{3 t}=2 \sqrt{3 t}$, which we get for the position of point $B$ between points $A$ and $C$, clearly has no solution. That such a case is not possible is also evident from Fig. 2, because any circle that touches circle $k$ at a point $X$ different from $A$ and also contains point $C$ on the opposite ray to the ray $B A$, must contain the chord of circle $l$ (marked in the figure) in its interior, so it cannot touch it.

The radius of circle $m$ is therefore $\frac{4}{3} \mathrm{~cm}$ or $12 \mathrm{~cm}$.

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