Probability With Spheres and Tetrahedrons
Probability
Level
3
Four points are chosen at random on the surface of the sphere. What is the probability that the center of the sphere lies on the inside of the tetrahedron whose vertices are at the four points?
1/8
5/9
3/4
1/1000000
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1 solution
Imagine choosing 3 random, distinct segments that pass through the center of the circle. Note that the segments divide the sphere into 8 sectors. Now imagine choosing a random point P 1 on the sphere (this cannot be any of the points where a line crosses the sphere). This point will land in one of the 8 sectors. On each segment, randomly choose one of the endpoints. There are 8 equally likely possibilities for where these points could be. Only one will make the 3 points directly opposite of P 1 , and this is the only way for the four points to form a tetrahedron which contains the center of the sphere. Therefore, no matter how large or small the sector P 1 lands on is, there will only be a 8 1 probability that the remaining 3 points will form a tetrahedron containing the center. Since every possible tetrahedron has an equally likely chance of being created as any other tetrahedron following this algorithm, this is the solution.
This problem was question A-6 on the 53rd Putnam Competition.