53rd Putnam 1992 Problem A6 Four points are chosen independently and at random on the surface of a sphere (using the uniform distribution). What is the probability that the center of the sphere lies inside the resulting tetrahedron? Solution
: 1/8. Having placed 3 points ABC, the 4th D will enclose the center in the tetrahedron iff it lies in the spherical triangle A'B'C', where P' is directly opposite to P (so that the center lies on PP'). The probability of this is the area of ABC divided by the area of the sphere. So taking the area of the sphere as 1, we want to find the expected area of ABC. But the 8 triangles ABC, A'BC, AB'C, ABC', A'B'C, AB'C', A'BC', A'B'C' are all equally likely and between them partition the surface of the sphere. So the expected area of ABC, and hence the required probability, is just 1/8. 53rd Putnam 1992 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001