A point is chosen on each edge of a tetrahedron so that the product of the distances from the point to each end of the edge is the same for each of the 6 points. Show that the 6 points lie on a sphere.
Solution
Solution by Ignacio Larrosa Cañestro.
Let the center of the circumsphere be O and its radius R. Let an edge be AB and a point on it P. Then it is well-known that PA·PB = R2 - OP2 (see below). So the six points given must all be the same distance from O. Hence they lie on a sphere center O.
[If chords AB, CD of a circle center O intersect at P, then triangles APC, BPD are similar, so PA/PC = PD/PB and hence PA·PB = PC·PD. If we take CD to be the chord perpendicular to OP, then angle OPD = 90o, so PC = PD and PC2 = OC2 - OP2. Hence PA·PB = OA2 - OP2.]
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002