Jul 19, 2024
Maximum gap is 3.75km, and minimum breakaway group size is 10 (EC).
Explanation:
In my opinion, the original problem and the extra credit can be tackled together all at once by thinking about the setup in a generic way, as each of the problems is simply a special case. Consider the setup where there are \(N\) riders divided into two groups, where there are \(a \ge 1\) riders in the leading breakaway group and hence \((N - a)\) riders in the trailing peloton group. Let \(v(k)\) represent the average speed of all the riders in a group comprised of \(k\) riders. Furthermore, let \(d\) be the distance between the breakaway group and the finish line, and let \(g\) be the distance between the breakaway group and the peloton group.
The first part is to apply the classic [distance = speed \(\times\) time] concept in our setup using the speed function on the two rider groups. Here, we need \(v(N - a) > v(a)\), otherwise the peloton group can never catch up to the leading breakway group. Assuming this is the case, then the catchup time for the peloton group, denoted as \(t\) is given by:
\begin{align} t &= \frac{\text{distance}}{\text{speed}} \\ \\ t &= \frac{g}{v(N - a) - v(a)} \end{align}
By the time the peloton group catches up to the breakaway group in \(t\) units of time, the breakway group would've traveled:
\begin{align} \text{distance} &= \text{speed} \times \text{time} \\ \\ d_a &= v(a) \times t \\ \end{align}
Of course, for the breakaway group to win, \(d > d_a\), and vice versa for a peloton victory. Optimization questions in the Fiddler can be solved by making use of this fact.
The second part is to come up with an expression for \(v(k)\). Here, we make use of the constraints given in the Fiddler, namely, that the average power of all riders must be the same, and that the power is directly proportional to speed. Since in a group, each rider only uses full power as the lead and half-power elsewhere and each rider takes the lead in equal proportions of time, we have the following, letting the average power be 1:
\begin{align} &\left(\frac{v(k)}{k}\right) + \left(1 - \frac{1}{k}\right)\left(\frac{v(k)}{2}\right) = 1 \\ \\ &v(k) = \boxed{\frac{4k}{2k + 2}} \\ \end{align}
Finally, we investigate the relationship among \(d\), \(g\), \(v(a)\) and \(v(N - a)\) using the above derivations.
What happens when \(d = d_a\)? This scenario represents the fact that just as the peloton group catches up to the breakaway group, the latter crosses the finish line. Any inequalities on either side would result in one group finishing ahead of the other. When we have this equality, we have:
\begin{align} d &= v(a)t \\ \\ d &= \frac{g \times v(a)}{v(N - a) - v(a)} \\ \\ &\frac{v(a)}{v(N - a)} = \frac{d}{d + g} \\ \\ \end{align}
Now, we are ready to solve the two problems presented in the Fidder.
For the first problem, we have a lone rider in the lead and a peloton group of four riders. That is, \(a = 1\) and \(N = 5\). Furthermore, we are given that \(d + g = 10\) and we need to maximize g (i.e. minimize d). Therefore, using the above equation, we have:
\begin{align} \frac{v(a)}{v(N - a)} &= \frac{d}{d + g} \\ \\ \frac{v(1)}{v(4)} &= \frac{d}{10} \\ \\ \frac{1}{8/5} &= \frac{d}{10} \\ \\ d &= \frac{50}{8} \\ \\ d &= 6.25 \end{align}
Hence, \(g = 10 - d = 10 - 6.25 = \boxed{3.75}\) km behind as the maximum possible gap distance.
For the extra credit, we have \(N = 176\), \(d = 10\), and \(g = 1\). We are asked to minimize \(a\). Again, we use the equation:
\begin{align} \frac{v(a)}{v(N - a)} &= \frac{d}{d + g} \\ \\ \frac{v(a)}{v(176 - a)} &= \frac{10}{11} \\ \\ \frac{4a}{2a + 2} &= \frac{10 \times 4(176 - a)}{11 \times (2(176 - a) + 2)} \\ \\ \end{align}
After some messy simplifying, we get:
\begin{align} a^2 - 197a + 1760 &= 0 \\ \end{align}
This is a quadratic equation in terms of \(a\), of which we have two positive roots. Taking the lesser of the two roots, we have \(a \approx 9.38\). Since the size of the group must be an integer, and the average speed increases with the number of riders, the minimum value of \(a\) must be \(\lceil{9.38}\rceil = \boxed{10}\).
For extra extra credit, what happens when the total number of riders goes to infinity? What must be the minimum breakaway group? Well, let's start with \(v(N - a)\), which we know to be:
\begin{align} v(N - a) &= \frac{4(N - a)}{2(N - a) + 2} \\ \end{align}
As \(N\) goes to infinity, we have \(v(N - a) \to 2\). Applying this fact to our equation earlier, we get:
\begin{align} \frac{v(a)}{v(N - a)} &= \frac{d}{d + g} \\ \\ \frac{v(a)}{2} &= \frac{d}{d + g} \\ \\ \frac{4a}{2a + 2} &= \frac{2d}{d + g} \\ \\ 4a(d + g) &= 2d(2a + 2) \\ \\ 4da + 4ga &= 4da + 4d \\ \\ 4ga &= 4d \\ \\ a &= \frac{d}{g} \\ \end{align}
So, as \(N \to \infty\), \(a \to \lceil{\frac{d}{g}}\rceil\).