Harvard-MIT Math Tournament

Solution:

Consider two consecutive circles $\omega_{i}$ and $\omega_{i+1}$. Let $Q_{i}, Q_{i}'$ be two points on $\omega_{i}$ and $Q_{i+1}, Q_{i+1}'$ on $\omega_{i+1}$ such that $Q_{i}, P_{i}$ and $Q_{i+1}$ are colinear and also $Q_{i}', P_{i}$ and $Q_{i+1}'$. Then $Q_{i} Q_{i}' = 2 \angle Q_{i} P_{i} Q_{i}' = 2 \angle Q_{i+1} P_{i} Q_{i+1}' = \angle Q_{i+1} Q_{i+1}'$. 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 $O_{i} Q_{i}'$ and $O_{i+1} Q_{i+1}'$, 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 $Q_{i}'$ 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 computation 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 $r$.