Longest distance between 2 points on a cylinder

Question

Given a right circular cylinder with radius $R$ and height $H=kR$, where $k$ is a positive real number, what is the longest minimal distance between $2$ points when you can only travel along the surface of the cylinder?

I see $4$ cases to consider:

1. Both points on the same base of the cylinder
2. One point on a base and one on the side
3. One point on one base and one on the other
4. Both points on the side

After quite a bit of tinkering it seems to me that the worst case scenario is having one point on the perimeter of a base and the other point on the perimeter of the other base, diametrically opposite the first point. This gives the following answer:

a. $\; k \le \frac{\pi^2}{4}-1$

$Max = (2 + k)R$

b. $\; k \gt \frac{\pi^2}{4}-1$

$Max= \sqrt {\pi^2 + k^2}R$

Can anyone confirm or deny?

Answer 1

I agree with your results:
1) With a simple logical argument on the height of the cylinder, you can establish that the 2 points have to be on the opposite bases (otherwise they would be part of smaller cylinder)
2) Assume that they are also diametrically opposite (this can be proved after the calculation of the minimal distance of such points - see 9 below)
3) The shortest path has to be a combination of a "straight line on the side" and a straight line on the face
4) You can then compute the corresponding length as (using Pythagoras for the length on the side of the unfolded cylinder, and the length of a chord for the length on the base): $L_k(x)=2R (\sqrt{X^2+\frac{k^2}{4}}+cos(X))$ for $X \in [0,\frac{\pi}{2}]$
5) The function has a local minimum in $X=0$, and is increasing then decreasing over the interval (using derivation)
6) Therefore, the overall minimum is either for $X=0$ or $X=\frac{\pi}{2}$
7) You can find the value of k for which $L_k(0)=L_k(\frac{\pi}{2})$ : this gives $k=\frac{\pi^2}{4}-1$
8) This gives the values you found:
For $k \le \frac{\pi^2}{4}-1$ then $L_k(0)=R(k+2)$
For $k \ge \frac{\pi^2}{4}-1$ then $L_k(\frac{\pi}{2})=R \sqrt{\pi^2+k^2}$
NOTE: I think it is more correct to write less or equal in the first case (this means that for $k = \frac{\pi^2}{4}-1$ there are 2 different minimal paths)
9) You can then finally prove that the 2 points have to be diametrically opposite, by supposing the contrary and getting a contradiction.