, 2010

Assume that the length of the oval is one unit. Let $x(t)$ be the difference of distances that the slowest and the fastest skaters have skated by time $t$. Similarly, let $y(t)$ be the difference between the middle skater and the slowest skater. The path $(x(t), y(t))$ is a straight ray $R$ in $\mathbb{R}^2$, starting from the origin, with slope depending on $L$. By assumption, $0 < y(t) < x(t)$.

One skater passes another one when either $x(t) \in \mathbb{Z}$, $y(t) \in \mathbb{Z}$ or $x(t) - y(t) \in \mathbb{Z}$.

The race ends when both $x(t), y(t) \in \mathbb{Z}$.

Let $(a, b) \in \mathbb{Z}^2$ be the endpoint of the ray $R$. We need to find the number of such points satisfying:
(a) $0 < b < a$
(b) The ray $R$ intersects $\mathbb{Z}^2$ at endpoints only.
(c) The ray $R$ crosses $357$ times the lines $x \in \mathbb{Z}$, $y \in \mathbb{Z}$, $y - x \in \mathbb{Z}$.

The second condition says that $a$ and $b$ are relatively prime. The ray $R$ crosses $a-1$ of the lines $x \in \mathbb{Z}$, $b-1$ of the lines $y \in \mathbb{Z}$ and $a-b-1$ of the lines $x-y \in \mathbb{Z}$.

Thus, we need $(a-1) + (b-1) + (a-b-1) = 117$, or equivalently, $2a-3=117$.

That is $a=60$.

Now $b$ must be a positive integer less than and relatively prime to $60$. The number of such $b$ can be found using the Euler's $\phi$ function:
$$\phi(60) = \phi(2^2 \cdot 3 \cdot 5) = (2-1) \cdot 2 \cdot (3-1) \cdot (5-1) = 16.$$
Thus the answer is $16$. $\square$

First, let us name our skaters. From fastest to slowest, call them: A, B and C. (Abel, Bernoulli and Cayley?)

Now, it is helpful to consider the race from the viewpoint of C. Relative to C, both A and B complete a whole number of laps, since they both start and finish at C.

Let $n$ be the number of laps completed by A relative to C, and let $m$ be the number of laps completed by B relative to C. Note that: $n > m \in \mathbb{Z}^+$

Consider the number of minutes required to complete the race. Relative to C, A is moving with a speed of $3.14 - 1 = 2.14$ laps per minute and completes the race in $\frac{n}{2.14}$ minutes. Also relative to C, B is moving with a speed of $(L-1)$ laps per minute and completes the race in $\frac{m}{L-1}$ minutes. Since A and B finish the race together (when they both meet C):
$$\frac{n}{2.14} = \frac{m}{L-1} \quad \Rightarrow \quad L = 2.14 \left(\frac{m}{n}\right) + 1.$$
Hence, there is a one-to-one relation between values of $L$ and values of the positive proper fraction $\frac{m}{n}$. The fraction should be reduced, that is the pair $(m, n)$ should be relatively prime, or else, with $n = \text{gcd}(m, n)$, the race ends after $n/k$ laps for A and $m/k$ laps for B when they first meet C together.

It is also helpful to consider the race from the viewpoint of B. In this frame of reference, A completes only $n - m$ laps. Hence A passes B only $(n - m) - 1$ times, since the racers do not "pass" at the end of the race (nor at the beginning). Similarly A passes C only $n - 1$ times and B passes C only $m - 1$ times. The total number of passings is:
$$117 = (n-1) + (m-1) + (n-m-1) = 2n-3 \Rightarrow n=60$$
Hence the number of values of $L$ equals the number of $m$ for which the fraction $m/60$ is positive, proper and reduced. That is the number of positive integer values smaller than and relatively prime to $60$. One could simply count: $\{1,7,11,13,17,...\}$, but Euler's $\phi$ function gives this number:
$$\phi(60) = \phi(2^2 \cdot 3 \cdot 5) = (2-1) \cdot 2 \cdot (3-1) \cdot (5-1) = 16.$$
Therefore, there are $16$ values for $L$ which give the desired number of passings.

Note that the actual values for the speeds of A and C do not affect the result. They could be any values, rational or irrational, just so long as they are different, and there will be $16$ possible values for the speed of B between them. ☐