# Ant on a Cylinder

The following question was asked in a mock test for CMI (Chennai Mathematical Institute):-

There is a glass cylinder with radius $$R$$. An ant is $$d$$ cm from the mouth of the cylinder. It is on the outer surface of the cylinder. There is a honey drop on the inner surface of cylinder that is also $$d$$ cm from the mouth of the cylinder and diametrically opposite to the ant. What is the shortest distance the ant has to travel to get to the honey drop? My attempt includes "unwrapping" the cylinder to get a rectangle with length equal to $$2πR$$ and breadth equal to $$H$$ which is the height of the cylinder. Now, the problem is reduced to finding the shortest path between two points on a rectangle given that the ant has to reach the edge first. I dropped a perpendicular from the ant's location to the edge nearer to it, and then drew a straight line from there straight to the honey. Using the Pythagorean Theorem, we get that the distance the ant travels is $$d+\sqrt{d^2+π^2R^2}$$. Many of the students on the mock test got the same answer, but, a few friends of mine got a different answer. Problem is, I see no error in their reasoning/method, and the distance they get is smaller than mine, too. So, I would like to know if I've made a mistake.

• PS:- I haven't given the solution they got so as to not bias anyone towards thinking my answer is right or wrong.
– user1059904
Nov 2, 2022 at 11:41
• What if the ant went to another point on the edge? You chose the point nearest the starting position. If you chose the point nearest the honey's position, then you have a different path with the same length. How about some point in between? Nov 2, 2022 at 11:47
• @JaapScherphuis Yes, that is similar to what they did, but as stated before, I'm not entirely sure who's right and who's wrong.
– user1059904
Nov 2, 2022 at 12:12
• See for example this question for a more general version of this problem, and how to solve it. Nov 2, 2022 at 12:28

Unwrapping, and using reflection, the ant should move to the edge point halfway between its original position and the position of the honey drop, then move towards the honey drop on the inside. The minimum distance is

$$d_{Min} = 2 \sqrt{ d^2 + (\dfrac{\pi}{2} R)^2 }$$

To come at this geometrically in the spirit of generalizing (while simultaneously simplifying!), we can not only unwrap the cylinder into a rectangle of width $$2\pi R$$, or an infinite strip of period $$2\pi R$$, but we can also "unfold" along the edge(s) of the cylinder, modeling the inside and outside surfaces in the same plane diagram.

In the "unwrapped" universe, the ant has infinitely many avatars, each a known distance from the edge, and with neighboring avatars separated by $$2\pi R$$. The honey similarly has infinitely many avatars at known distance from the edge, known angular (horizontal) offset from the ant, and with neighboring avatars separated by $$2\pi R$$. The shortest path from the ant to the honey is a line segment, represented geometrically by the shortest segment joining any ant avatar to any honey avatar. (Prospects are shown in blue.) This interpretation makes clear why proceeding straight to the edge, then straight to the honey, is generally not the shortest path. For the question given, where the ant and honey are "opposite" on the cylinder, the two blue paths have the same length.

The diagram shows the situation where both the ant and the honey are closer to one particular edge of the cylinder than to the other. Generalizing this is left as a straightforward exercise.

For posterity: The "two-sided cylinder" is a model of a flat torus. This model isn't mathematically literal (there is no distinction between the inside and outside of a cylinder solely in terms of points on the cylinder), but does informally capture the geometry of a flat torus.

• Thanks for the interesting general interpretation.
– user1059904
Nov 3, 2022 at 1:13

According to the Law of reflection the shortest path will be along the legs of isosceles triangle with height $$d$$ and base $$\pi R$$ which gives the shortest path of $$\sqrt{4d^2+\pi^2R^2}$$

You can also use calculus. Let $$\sqrt{x^2+d^2}$$ be the distance ants needs to travel to reach the top. Then the total distance $$f(x)=\sqrt{x^2+d^2}+\sqrt{(\pi R-x)^2+d^2}=\sqrt{x^2+d^2}+\sqrt{\pi^2R^2-2\pi R x+x^2+d^2}$$.
$$f'(x)=\frac{x}{\sqrt{x^2+d^2}}+\frac{x -\pi R}{\sqrt{\pi^2R^2-2\pi R x+x^2+d^2}}$$ $$f'(x)=0$$ gives $$(\pi R-x)\sqrt{x^2+d^2}=x\sqrt{\pi^2R^2-2\pi R x+x^2+d^2}$$ $$(\pi R-x)^2(x^2+d^2)=x^2(\pi^2R^2-2\pi R x+x^2+d^2)$$ $$(\pi^2R^2-2\pi Rx+x^2)(x^2+d^2)=x^2(\pi^2R^2-2\pi R x+x^2+d^2)$$ $$\pi^2R^2d^2-2\pi Rd^2x=0$$ Thus, $$x=\frac{\pi R}{2}$$ gives the minimum total distance.

• Why is the arclength distance on the cylinder the same as the Euclidean distance on the "unwrapped" rectangle? Nov 2, 2022 at 23:05
• @Argyll: Draw the line on the unwrapped rectangle then make a cylinder. During this process, the size of the segment does not change. Nov 3, 2022 at 2:31
• Why? How do you the line isn't stretched in the process? Nov 3, 2022 at 3:44