# Making the fastest track

This week’s Riddler Classic is a problem about minimum-time optimization.

While passing the time at home one evening, you decide to set up a marble race course. No Teflon is spared, resulting in a track that is effectively frictionless. The start and end of the track are 1 meter apart, and both positions are 10 centimeters off the floor. It’s up to you to design a speedy track. But the track must always be at floor level or higher — please don’t dig a tunnel through your floorboards. What’s the fastest track you can design, and how long will it take the marble to complete the course?

My solution:
[Show Solution]

## 5 thoughts on “Making the fastest track”

1. Hi Laurent,
Nice solution as always. I had a similar approach but one thing that bothered me is the possibility that a (partially) non cycloid curve that hits the ground beyond the ‘optimum’ point P in this solution could be net faster. The argument being that though this new curve gets to vertically above P (say Q) with a lower speed at that point, it could get to this point faster, allowing for a net faster time to floor than through P. I went ahead and showed that cannot be the case. Also an animation…
https://www.starvind.com/math/on-the-fast-track/

2. Guy Moore says:

I agree with your derivation but I think you made it harder than it needed to be in the interesting case that h < l/pi.
You consider portions of a brachistochrone which descend out to a shorter distance than (pi h / 2) and then have a hard corner and go flat. But if you think of that brachistochrone-portion and straight curve as a curve from (0,0) to (pi h/2,h) then it must take longer to get there than the half-brachistochrone.
So there was no need to do the second optimization, the best curve MUST be a half-brachistochrone, a straightaway, and a half-brachistochrone.

1. I guess you mean cycloid when you say brachistochrone. Technically, brachistochrone is just the shortest time path.
I’m not sure it’s obvious that the half cycloid to floor is faster than greater than half cycloid to point above floor and then meeting floor at a later point with a different curve. It turns out that half cycloid is the best but needs proof. (my comment above).

1. Guy Moore says:

Suppose you take a half-cycloid to the point (x,y) which is above the floor, and then some small additional path to (x’,y’). Then either I am continuing the half-cycloid, which goes higher and is therefore worse than going straight across, or the total path to (x’,y’) is not the cycloidal path to that point, and is therefore not optimal. Since (x’,y’) is still above the floor, the cycloidal path to (x’,y’) is better than the suggested curve.
So it’s clear that you don’t need to consider such constructions — it does not require calculation.

(Yes I should have said cycloid rather than brachistochrone, my bad)

1. Yes, fair enough argument without calculation, but still non obvious and needed to complete the answer.