The figure shows the logo of a company, formed by two concentric circles and four other circles of the same radius, each tangent to two of the others and to the two concentric circles. The radius of the smaller circle measures $1 \mathrm{~cm}$. What is, in centimeters, the radius of the larger circle?

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-008.jpg?height=260&width=257&top_left_y=1275&top_left_x=1576)
Let $r$ be the radius of the four equal circles.

Connecting the centers $A$ and $B$ of two of these circles to the center $O$ of the concentric circles, we obtain the triangle $\triangle O A B$, as shown in the figure. Remembering that the line connecting the centers of two tangent circles passes through the point of tangency, we see that $O A=O B=1+r$ and $A B=2 r$, all in $\mathrm{cm}$. The triangle $\triangle O A B$ is a right triangle at $O$ and the Pythagorean Theorem gives

$$
(2 r)^{2}=(1+r)^{2}+(1+r)^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-051.jpg?height=477&width=477&top_left_y=253&top_left_x=1406)

Thus, $4 r^{2}=2(1+r)^{2}$, or $2 r^{2}=(1+r)^{2}$, from which we get $r^{2}-2 r-1=0$. Therefore, $r=\frac{1}{2}[2 \pm \sqrt{8}]=1 \pm \sqrt{2}$. Since the radius $r$ is positive, we obtain $r=1+\sqrt{2}$. It follows that the radius of the larger circle measures $1+2 r=3+2 \sqrt{2} \mathrm{~cm}$.