Show that for any point P on the surface of a regular tetrahedron we can find another point Q such that there are at least three different paths of minimal length from P to Q.
Solution
Let the tetrahedron be ABCD. Open the sides out flat. We get the large triangle AA'A" (the label on the top vertex should be A). For convenience we show two other copies of the face ABC: A"BC" and A'B'C. P is a point on ABC. Without loss of generality it is closer to A than to B or C. P' and P" are its images in A'B'C and A"BC". Note that we obtain P' by rotating P through 180o about C, and we obtain P" by rotating P through 180o about B. Let the perpendicular to PP' at C and the perpendicular to PP" at B meet at Q. Note that angle ABP ≤ 30o, so angle QBC ≤ 60o and similarly angle QCB ≤ 60o, so Q lies in the face BCD. Also Q lies on the perpendicular bisector of PP", so QP = QP". Similarly QP = QP'. So we have three equal paths from P to Q, corresponding to PQ, P'Q and P"Q. Since they are straight lines, they are also all of minimal length.
[In principle, it is possible that there is another image of P with a shorter distance. But it is easy to see that A"BCA' forms half a hexagon. The images of ABC are the triangles on the outside of the edges of the complete hexagon and it is clear that the other three images of P are all further from Q. Actually, that is not quite true. If we take P to be the midpoint of AB, then one of the other images becomes equidistant. In other words, if we take P to be the midpoint of an edge of the tetrahedron, then Q is the midpoint of the opposite edge and there are four minimal paths - go from P to the midpoint of any of the other four edges and then to Q.]
© John Scholes
jscholes@kalva.demon.co.uk
5 Oct 2002