Tour de Fiddler fun

This time around, a lone rider in the Tour de Fiddler is being pursued by a group of four riders. The four riders have an advantage—they take equal turns being in the lead position, while the other three riders draft behind. At any given speed, being in the lead position (as well as riding solo) requires twice as much power as drafting.

Assume that every rider must maintain the exact same average power over time, whether they are the lone rider or in the pack of four. To be clear, their power can change over time, but the time-averaged value must be the same for every rider. Also, when leading a pursuing group or riding solo, one’s speed is directly proportional to one’s power. When drafting, one's speed matches that of the leader (again, at half the power output).

The pursuers just passed under a banner indicating there are 10 kilometers left in the stage. How far back of the lone rider can they afford to be, such that they still catch them at the finish line?

Let's first start with some preliminaries. Assume that all riders have average power output $\bar{P}$, such that, since speed is directly proportional to ones power, then the average speed of a single rider who has no one to draft off of is $v_1 = C\bar{P},$ for some constant of proportionality $C \gt 0.$

Let's now analyze what would happen if $k$ riders work together, taking equal turns being in the lead position. Since the average power must be $\bar{P},$ but any of the riders in this group spends only $\frac{1}{k}$ amount of time leading and the rest drafting at half-power, the amount of power spent while leading a pack of $k$ riders, say $P_k$, can be found in the equation $$\bar{P} = \frac{1}{k} P_k + \frac{k-1}{k} \frac{1}{2} P_k = \frac{k+1}{2k} P_k,$$ that is, $$P_k = \frac{2k}{k+1} \bar{P}.$$ So that means that the average velocity of a group of $k$ riders working together must be $$v_k = C P_k = C \frac{2k}{k+1} \hat{P} = \frac{2k}{k+1} v_1.$$

Ok, so with that in mind, the pursuers will finish in $T_4 = \frac{10}{v_4} = \frac{25}{4v_1},$ while if the leader has a lead of $x$ kilometers, then he will finish in $T_1 = \frac{10-x}{v_1}.$ Since we want $$T_4 = \frac{25}{4v_1} \leq \frac{10-x}{v_1} = T_1$$ in order for the pursuers to catch the leader, then the leader can be no more than $x \leq \frac{15}{4} = 3.75$ kilometers ahead of the pursuers if they want to catch him.

In today’s stage of the Tour de Fiddler, there are 176 total riders. Some riders are grouped together in a single breakaway, while the remainder are grouped together in the peloton. The breakaway group is 10 kilometers from the finish, while the peloton is one kilometer behind (i.e., 11 kilometers from the finish). What is the smallest number of riders that the breakaway needs to reach the finish line before the pursuing peloton?

Let's say there are $B$ riders in the breakaway group and $P$ riders in the peleton, where here $B+P = 176.$ The average speed of the peleton will be $$v_P = C\frac{2P}{P+1} v_1,$$ while the average speed of the breakaway is $$v_B = C\frac{2B}{B+1} v_1,$$ so the finishing times are $$T_P = \frac{11}{v_P} = \frac{11(P+1)}{2Pv_1}$$ and $$T_B = \frac{10}{v_B} = \frac{10(B+1)}{2Bv_1},$$ respectively. Since we want the breakaway group to win the stage, we must have $T_B \leq T_P,$ or equivalently $$\frac{10B+10}{B} \leq \frac{11P+11}{P},$$ which is again equivalent to $$BP+11B-10P \geq 0.$$ Since we have restriction that $B + P = 176,$ we are now looking for the minimum integer $B$ such that $$B(176-B) + 11B -10(176-B) = -B^2 + 197B - 1760 \geq 0.$$ Since the quadratic inequality holds for all (real) values of $$B \in \left[ \frac{197 -\sqrt{31769}}{2} , \frac{197 + \sqrt{31769}}{2} \right],$$ we see that the breakaway must have at least $$B_\min = \left\lceil \frac{197 - \sqrt{31769}}{2} \right\rceil = \lceil 9.3806979381.... \rceil = 10$$ riders in order to beat the peleton.