Three circles each with a radius of 1 are placed such that each circle touches the other two circles, but none of the circles overlap. What is the exact value of the radius of the smallest circle that will enclose all three circles?
Suppose that the three small circles have centres $A, B$ and $C$.

For the larger circle to be as small as possible, these three small circles must just touch the larger circle (which has centre $O$ ) at $P, Q$ and $S$, respectively.

Note that $A P=B Q=C S=1$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-13.jpg?height=398&width=393&top_left_y=1666&top_left_x=931)

Let $r$ be the radius of the larger circle. Then $O P=O Q=O S=r$.

Since the larger circle is tangent to each of the smaller circles at $P, Q$ and $S$, this means that the circles share a common tangent at each of these points.

Consider the point $P$. Since $A P$ is perpendicular to the tangent to the smaller circle at $P$ and $O P$ is perpendicular to the tangent to the larger circle at $P$ and this tangent is common, then $A P$ and $O P$ must be parallel. Since $A P$ and $O P$ both pass through $P$, then $A P$ lies along $O P$. Since $A P=1$ and $O P=r$, then $O A=r-1$.

In a similar way, we can find that $O B=O C=r-1$.

Now since the circles with centres $A$ and $B$ just touch, then the distance $A B$ equals the sum of the radii of the circles, or $A B=2$.

Similarly, $A C=B C=2$.

Consider $\triangle A O B$. We have $O A=O B=r-1$ and $A B=2$.

By symmetry, $\angle A O B=\angle B O C=\angle C O A=\frac{1}{3}\left(360^{\circ}\right)=120^{\circ}$.

Let $M$ be the midpoint of $A B$. Thus, $A M=M B=1$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-14.jpg?height=192&width=417&top_left_y=419&top_left_x=911)

Since $\triangle A O B$ is isosceles, then $O M$ is perpendicular to $A B$ and bisects $\angle A O B$.

This means that $\triangle A O M$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Therefore, $\frac{A O}{A M}=\frac{2}{\sqrt{3}}$ and so $\frac{r-1}{1}=\frac{2}{\sqrt{3}}$ or $r=1+\frac{2}{\sqrt{3}}$.

Thus, the radius of the smallest circle enclosing the other three circles is $1+\frac{2}{\sqrt{3}}$

ANSWER: $1+\frac{2}{\sqrt{3}}$