Conical Conundrum
Geometry; Difficulty: medium
Geodesic on a cone
The Riddler Classic from this week (Mar 4, 2022) is about the shortest path between points on the surface of a cone.
Two ants named Geo and Desik are racing along the surface of a cone. The circular base of the cone has a radius of 2 meters and a slant height of 4 meters. Geo and Desik both start the race on the base, a distance of 1 meter away from its center.
The race’s finish is halfway up the cone, 90 degrees around the cone’s central axis from the start, as shown in the following diagram:
Geo and Desik both want your help in strategizing for the race. What is the length of the shortest path from the start to the finish?
*spoiler alert* solution and answer follow in the sections below.
Solution
The ants’ names are a play on Geodesic: the shortest path between two points on a (curved) surface. Calculating this path in three dimensions on a cone is complicated. Fortunately for us, we can simplify the problem and solve it in the more familiar and comfortable two dimensions.
First, realize that the ant (either one), has to first travel on the base to the actual edge of the cone. For any eventual shortest path, this trip to the edge obviously has to be a straight line. Once at the edge, the ant has to travel along a curve to its destination on the cone. But any cone can be unfolded to a circular sector i.e. a two dimensional surface. The shortest path is now just a straight line on this surface! So the only decision for the ant to make is the direction of straight line travel to the edge.
See the figure below: a view from the top of the cone with the ant starting from point A and heading to point B on the edge with BOA making an angle \(\theta \) radians .
The distance AB covered by the ant in this first leg of the race can be calculated by applying the Law of Cosines on \( \triangle \) OAB.
\[
AB = \sqrt{2^2 + 1^2 – 4 cos\theta} = \sqrt{5 -4 cos\theta}
\]
The length of the arc BC is \(2\theta \)
Now let’s unfold the cone, making the cut along OE. The result is shown in the figure below. The total arc length of the unfolded sector of the circle is just the circumference of the base \( = 4\pi \). The sector has a radius equal to the slant height, i.e. 4m. Hence the central angle of this sector is \( \displaystyle{\frac{4\pi}{4} = \pi } \) radians i.e. it is a semicircle.
The sector OCE in the figure above corresponds to the sector OCE below, with an angle of \(45^{\large\circ} \) : one fourth of the unfolded cone sector.
The ant at the end of leg 1 is at edge point B. Its destination is point D. The shortest path is now again just the straight segment BD. We know length of arc CE = \( \pi \) and that of arc BC = \(2\theta \).
Therefore arc BE = \(\pi – 2\theta \). This implies \( \angle ABC = \displaystyle{\frac{\pi}{4} – \frac{\theta}{2}} \)
Now applying the Cosine rule to \( \triangle \) BOD:
\[
BD = \sqrt{4^2 + 2^2 – 16 cos(\frac{\pi}{4} – \frac{\theta}{2})} = \sqrt{20 -16 cos(\frac{\pi}{4} – \frac{\theta}{2})} \\
\]
The total distance the ant covers in both legs is:
\[
AB + BD = \sqrt{5 -4 cos\theta} + \sqrt{20 -16 cos(\frac{\pi}{4} – \frac{\theta}{2})}
\]
Plotting and solving in Desmos, the min distance is 3.718m and the value of \(\theta\) that minimizes this distance is \( \displaystyle{\frac{\pi}{6}} \). The individual segments are AB = 1.239m and BD = 2.479m
For the ant at point A (in the first figure) facing C, needs to head at an angle of \(53.79^{\large\circ} \) towards the edge of the cone on the side nearer to the race end point!