Can You Catch the Breakaway?

This Week's Fiddler

Because the power must be averaged to be the same, the pursuing group of four should alternate being leader at equal amounts (2.5 km each). If we say that the leader in this situation must use $p$ power, then the others must use $\frac{p}{2}$ power, making for an average of $\frac{p+3(\frac{p}{2})}{4}=\frac{5}{8}p$ power for each rider over the 10 km.

Because speed is directly proportional to power in this scenario, and because the lone rider maintains the same average speed as the group, the lone rider is only able to travel at $\frac{5}{8}$ the speed as the group. So if the lone rider is to beat the group, they must be at most $\frac{5}{8}$ the distance away from the finish line as the group is, which is $\frac{5}{8}\cdot10=6.25$ km, or $10-6.25=3.75$ km ahead of the group.

Answer: 3.75 km

Extra Credit

Let's generalize the process in the first part. Say $k$ riders are part of the breakaway group ($0\lt k\lt176$), meaning $176-k$ riders part of the peloton. We wish to find the minimum $k$ such that the breakaway group reaches the finish line before the peloton.

If $p$ is the average power needed to lead a group, then the breakaway group has a speed proportional to $v_b=\frac{(k+1)}{2k}p$, and the peloton has a speed proportional to $v_p=\frac{(176-k+1)}{2(176-k)}p$. These two groups need to cover different distances (say $d_b$ for the breakaway group and $d_p$ for the peloton group) within the same amount of time, so we can try to find $v_b\cdot d_b=v_p\cdot d_p$.

So:

$$\frac{(k+1)}{2k}\cdot 10=\frac{(176-k+1)}{2(176-k)}\cdot 11 \\ 10(k+1)(352-2k)=11(177-k)(2k) \\ k^2-197k+1760=0 \\ k=\frac{197\pm\sqrt{31769}}{2} \\ k\approx9.38\textrm{ or }184.62$$

Because $k\lt176$, this means $k$ must be about 9.38, meaning we need at least 10 riders in the breakaway group to win the race.

Answer: 10 riders