Three circles, with radii measuring 1, 2, and $3 \mathrm{~cm}$, are pairwise externally tangent, as shown in the given figure. Determine the radius of the circle that is externally tangent to the three circles.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-035.jpg?height=374&width=374&top_left_y=2303&top_left_x=1503)
By connecting the centers of the three circles, we obtain the triangle $\triangle ABC$ with sides $AB=3, AC=4$, and $BC=5$. Since $3^{2}+4^{2}=5^{2}$, this triangle is a right triangle with hypotenuse $BC$.
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-121.jpg?height=336&width=1038&top_left_y=2076&top_left_x=590)

Now we construct the rectangle $ADCB$ with a congruent copy of the triangle $\triangle ABC$ and a common hypotenuse $BC$, as shown in the figure. Since $DC=AB=3$ and the circle centered at $C$ also has a radius of 3, we see that point $D$ lies on the circle. Finally, we connect point $D$ to each of the vertices of the triangle $\triangle ABC$ and extend these segments until they intersect the circles centered at $A, B$, and $C$, obtaining the points $P_{1}, P_{2}$, and $P_{3}$, as shown in the figure.
![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-122.jpg?height=534&width=1222&top_left_y=253&top_left_x=456)

Observe that

- $DP_{2}=DB+BP_{2}=CA+BP_{2}=4+2=6$,
- $DP_{1}=DA+AP_{1}=5+1=6$, and
- $DP_{3}=DC+CP_{3}=3+3=6$.

Thus, $DP_{1}=DP_{2}=DP_{3}=6$, and therefore, the circle centered at $D$ with radius 6 passes through $P_{1}, P_{2}$, and $P_{3}$. Furthermore, since the points $\{D, A, P_{1}\}, \{D, B, P_{2}\}$, and $\{D, C, P_{3}\}$ are collinear, it follows that the circle centered at $D$ with radius 6 is externally tangent to the given circles centered at $A, B$, and $C$.