Consider two consecutive circles $\omega_{i}$ and $\omega_{i+1}$. Let $Q_{i}, Q_{i}^{\prime}$ be two points on $\omega_{i}$ and $Q_{i+1}, Q_{i+1}^{\prime}$ on $\omega_{i+1}$ such that $Q_{i}, P_{i}$ and $Q_{i+1}$ are colinear and also $Q_{i}^{\prime}, P_{i}$ and $Q_{i+1}^{\prime}$. Then $Q_{i} Q_{i}^{\prime}=2 \angle Q_{i} P_{i} Q_{i}^{\prime}=2 \angle Q_{i+1} P_{i} Q_{i+1}^{\prime}=\angle Q_{i+1} Q_{i+1}^{\prime}$. Refer to the center of $\omega_{i}$ as $O_{i}$. The previous result shows that the lines $O_{i} Q_{i}$ and $O_{i+1} Q_{i+1}$ meet at the same angle as the lines $O_{i} Q_{i}^{\prime}$ and $O_{i+1} Q_{i+1}^{\prime}$, call this angle $\psi_{i} . \psi_{i}$ is a function solely of the circles $\omega_{i}$ and $\omega_{i+1}$ and the distance between them (we have just showed that any two points $Q_{i}$ and $Q_{i}^{\prime}$ on $\omega_{i}$ give the same value of $\psi_{i}$, so $\psi_{i}$ can't depend on this.) Now, the geometry of $\omega_{i}$ and $\omega_{i+1}$ is the same for every $i$, so $\psi_{i}$ is simply a constant $\psi$ which depends only on $r$. We know $6 \psi=0 \bmod 2 \pi$ because $Q_{7}=Q_{1}$. We now compute $\psi$. It suffices to do the computaiton for some specific choice of $Q_{i}$. Take $Q_{i}$ to be the intersection of $O_{i} O_{i+1}$ and $\omega_{i}$ which is further from $O_{i+1}$. We are to compute the angle between $O_{i} Q_{i}$ and $O_{i+1} Q_{i+1}$ which is the same as $\angle O_{i} O_{i+1} Q_{i+1}$. Note the triangle $\triangle O_{i} P_{i} O_{i+1}$ is isosceles, call the base angle $\xi$. We have $\angle O_{i} O_{i+1} Q_{i+1}=\angle O_{i} O_{i+1} P_{i}+$ $\angle P_{i} O_{i+1} Q_{i+1}=\xi+\left(\pi-2 \angle O_{i+1} P_{i} Q_{i+1}\right)=\xi+\left(\pi-2\left(\pi-\angle Q_{i} O_{i+1} P_{i}-\angle P_{i} Q_{i} O_{i+1}\right)\right)=$ $\xi-\pi+2\left(\xi+(1 / 2) \angle P_{i} O_{i} O_{i+1}\right)=\xi-\pi+2(\xi+(1 / 2) \xi)=4 \xi-\pi$. So we get $6(4 \xi-\pi)=0 \bmod 2 \pi$. Noting that $\xi$ must be acute, $\xi=\pi / 12, \pi / 6, \pi / 4, \pi / 3$ or $5 \pi / 12$. $r$ is uniquely determined as $(1 / 2) \sec \xi$ so there are 5 possible values of $r$.
All subscripts in this problem are to be considered modulo 6 , that means for example that $\omega_{7}$ is the same as $\omega_{1}$. Let $\omega_{1}, \ldots \omega_{6}$ be circles of radius $r$, whose centers lie on a regular hexagon of side length 1 . Let $P_{i}$ be the intersection of $\omega_{i}$ and $\omega_{i+1}$ that lies further from the center of the hexagon, for $i=1, \ldots 6$. Let $Q_{i}, i=1 \ldots 6$, lie on $\omega_{i}$ such that $Q_{i}, P_{i}, Q_{i+1}$ are colinear. Find the number of possible values of $r$.