6. Six unit disks $C_{1}, C_{2}, C_{3}, C_{4}, C_{5}, C_{6}$ are in the plane such that they don't intersect each other and $C_{i}$ is tangent to $C_{i+1}$ for $1 \leq i \leq 6$ (where $C_{7}=C_{1}$ ). Let $C$ be the smallest circle that contains all six disks. Let $r$ be the smallest possible radius of $C$, and $R$ the largest possible radius. Find $R-r$.
Answer: $\sqrt{3}-1$
The minimal configuration occurs when the six circles are placed with their centers at the vertices of a regular hexagon of side length 2 . This gives a radius of 3 .
The maximal configuration occurs when four of the circles are placed at the vertices of a square of side length 2. Letting these circles be $C_{1}, C_{3}, C_{4}, C_{6}$ in order, we place the last two so that $C_{2}$ is tangent to $C_{1}$ and $C_{3}$ and $C_{5}$ is tangent to $C_{4}$ and $C_{6}$. (Imagine pulling apart the last two circles on the plane; this is the configuration you end up with.) The resulting radius is $2+\sqrt{3}$, so the answer is $\sqrt{3}-1$.
Now we present the proofs for these configurations being optimal. First, we rephrase the problem: given an equilateral hexagon of side length 2 , let $r$ be the minimum radius of a circle completely containing the vertices of the hexagon. Find the difference between the minimum and maximum values in $r$. (Technically this $r$ is off by one from the actual problem, but since we want $R-r$ in the actual problem, this difference doesn't matter.)
Proof of minimality. We claim the minimal configuration stated above cannot be covered by a circle with radius $r60^{\circ}$ since $O_{1} O_{2}$ is the largest side of the triangle $O_{1} O O_{2}$, and similar for other angles $\angle O_{2} O O_{3}, \angle O_{3} O O_{4}, \ldots$, but we cannot have six angles greater than $60^{\circ}$ into $360^{\circ}$, contradiction. Therefore $r \geq 2$.
Proof of maximality. Let $A B C D E F$ be the hexagon, and choose the covering circle to be centered at $O$, the midpoint of $A D$, and radius $\sqrt{3}+1$. We claim the other vertices are inside this covering circle. First, we will show the claim for $B$. Let $M$ be the midpoint of $A C$. Since $A B C$ is isosceles and $A M \geq 1$, we must have $B M \leq \sqrt{4-1}=\sqrt{3}$. Furthermore, $M O$ is a midline of $A C D$, so $M O=\frac{C D}{2}=1$. Thus by the triangle inequality, $O B \leq M B+O M=\sqrt{3}+1$, proving the claim. A similar argument proves the claim for $C, E, F$. Finally, an analogous argument to above shows if we define $P$ as the midpoint of $B E$, then $A P \leq \sqrt{3}+1$ and $D P \leq \sqrt{3}+1$, so by triangle inequality $A D \leq 2(\sqrt{3}+1)$. Hence $O A=O D \leq \sqrt{3}+1$, proving the claim for $A$ and $D$. Thus the covering circle contains all six vertices of $A B C D E F$.