Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is independently chosen relative to a uniform distribution on the sphere.)

Guest Jun 5, 2018

#1**+1 **

Method 1: Logic

One the sphere, there exists a point A and its antipodal point, A'. This means that A' is the point with the farthest distance from A, the 3D version of the diameter, which means when you connect A and A', the line contains the center of the sphere. After one chooses the three points, W, X, Y, the region to place Z is the spherical traingle W'X'Y' opposite WXY. It is only then, the tetrahedron would contain the center of the sphere. This means that the probability the tetrahedron would contain the center of the sphere is the expected area of the triangle W'X'Y', which has the same expected area of WXY and 7 other triangles: WXY, W'XY, W'X'Y, W'X'Y', WX'Y, WX'Y', WXY', W'XY'. Since there are 8 such triangle, and the total area is set 1, the expected area is 1/8. The probabilty is 1/8.

Method 2: Mathematical Proof

We can proof the logic using the notion of barycentric coordidates: https://nicoguaro.github.io/posts/putnam_prob/

I think the website does a great job explaining the process, better than I can at least. If you have any questions though, about the website or my logic, ask away!

I hope this helped,

Gavin

GYanggg Jun 5, 2018