# Cone Trip

This week’s Riddler Classic asks us to find the shortest trip on the base and sides 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:”

We will redraw the cone in two separate pieces: a top view of the cone and base and an unwrapped view of the sides of the cone:

Given the dimensions of this cone it unwraps into a perfect semi-circle with radius 4. Each point on the perimeter of the semi-circle maps to a point on the perimeter on the circular base, with a central angle (theta/2) one-half that of that of the cicular base (theta).

We see that the trip around the cone has 2 pieces: first a traverse along the base to the “edge point” at the perimeter of the base, and then a trip up the side of the cone to the “finish” point. The total distance is the sum of these distances and we will calculate the first distance using the circular base drawing and the second using the semicircular drawing of the unwrapped sides:

Using the law of cosines we can see that the individual distances are:

$d1 = [5 - 4*cos(\theta)]^{1/2}$
$d2' = [20 - 16*cos(\pi/4 -\theta/2)]^{1/2}$

$d= d1+d2' = [5 - 4*cos(\theta)]^{1/2} + 2*[5 - 4*cos(\pi/4-\theta/2)]^{1/2}$

Taking the derivative with respect to theta, setting to zero, and re-organizing we see:

$\cfrac{sin(\theta)}{[5-4*cos(\theta)]^{1/2}} = \cfrac{sin(\pi/4-\theta/2)}{[5-4*cos(\pi/4-\theta/2)]^{1/2}}$

$\theta = \pi/4-\theta/2$

$\theta = \pi/6 = 30^{\circ}$

So we see that the minimum trip distance is:

$dmin = 3*[5-2*\sqrt{3}]^{1/2} \approx 3.717941 m$

An interesting note is that exactly 1/3 of the distance is traveled on the base, and 2/3 on the side of the cone.