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Problem 2 (Minimum Spanning Path on Tetrahedron)

The length of each edge of a tetrahedron (four triangular faces six edges) is one unit long. An ant wants to make a path on the surface of the tetrahedron such that each of the four vertices can be reached from any other vertex. What is the shortest path the ant can make a path ?

Solution
Let ABC and DBC represent two adjacent faces of the tetrahedron unfolded onto the plane, and let O be the midpoint of the line segment BC as shown in the diagram below.

We will consider the network of lines symmetric about the point O, consisting of PA, PB, QC, QD, and PQ as shown in the above drawing. We will choose these lines in such a way that the roads meet the points P and Q at angles 120 degrees.
We start by considering the triangle AOB, where we have assumed all sides of the tetrahedron have length 1, and hence i.e. AB = 1 and so BO = 1/2. We also know that the angle ABC is 60 degrees, and so from the triangle POB, we know that if the angle POB is alpha, then we have that the angle PBO = 60 - alpha. And since the angle ABC = 60 degrees, we also have that the angle PBA is alpha, and so the triangles POB and PBS are similar.
Now, letting PO = x, we have PB = 2x and PA = 4x. We now use the law of cosines applied to the triangle POB to get the relationship

x² + (2x)² + 2 (x) (2x) cos (120) = (1/2)²

which gives x = 1/(2 sqrt (7)). Hence, the total length of the network in the triangle AOB is

x + 2x + 4x = 7x

and so the length of the entire network on the surface of the tetrahedron connecting the four vertices is

4x = sqrt (7) = 2.645751311...

Note: We really haven't proven the path we found is the shortest network, but it is the shortest we have found. As far as we know, this is an unsolved problem. If you can find a shorter path than the one we've found, we would be glad to see it.

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Last modified on Tuesday, January 12, 1999