Our puzzlemaster Kurt Meyer discloses the solution to a challenging puzzle.
For the first time, our Numeracy puzzle from Fall 2018 resulted in zero reader subimissions of answers. Below you will find the puzzle and the solution.
The Puzzle
Blossom and Brian are in a foot race. At the starting line, both set out at the same time, each at a constant positive rate. Ever respectful of authority, Brian lets the head of school take the lead. The layout of the race is a straight track, out-and-back. Blossom wishes to gauge her speed so that, after making the U-turn at the far end of the race and facing the start-finish line, she maximizes the amount of time she is looking at Brian.
How fast should she run?
The Solution
Brian lets Blossom "take the lead" but he does not give her a head start; he simply allows her to run faster. The question asks us to describe how Blossom should "gauge her speed" in relation to Brian's so as to maximize their face-to-face time after she makes the turn. This is best done by representing Blossom's speed as some numerical factor times Brian's speed.
If we let Brian's speed be r then Blossom's speed can be represented as kr (k times r) with the understanding that k is greater than 1.0.
But we can further "abstract" the situation by representing Brian's speed as 1. [If you say it's not 1, I can respond, "well yes it is, if you change the units appropriately."] So in some units, Brian's speed is 1, and this makes Blossom's speed k.
The situation does not specify the distance to the turn-around point, so we can figure it makes no difference. So we'll again "simplify" the representation of the race by saying this distance is 1 (in some units).
Using the "theorem," RATE * TIME = DISTANCE, we are now ready to proceed ...
Blossom's time to the turn-around point is (distance divided by rate): 1/k
Brian's position at the time that Blossom turns around is (rate times time): 1 * (1/k)
The distance between the spouses at the moment Blossom makes the turn: 1 - 1/k
The TIME that elapses while the two are now approaching each other is (distance divided by rate): (1 - 1/k) / (1+k)
[Note that this "rate" is the sum of the rates of the two runners, as they are closing the distance between them, running in opposite directions, at their combined rate.]
So the puzzle asks us to maximize this time: t = (1 - 1/k) / (1+k) (remember k >1)
This can be done with calculus—or with a graph of the function by which one can easily see the "high point" of the graph (see image).
The vertical axis represents values of t and the horizontal axis represents values of k.
The graph shows that the maximum time apparently occurs just below 2.5*. This means that Blossom should run just shy of 2.5 times Brian's speed to maximize the time they gaze into each other's eyes.
*the "true" value is exactly 1 + sqrt(2), which is approximately 2.414