Solution
Tetrahedron Around the Centre
Show the problem again
Four points are chosen independently and uniformly at random on the surface of a sphere. What is the probability that the tetrahedron with these four points as vertices contains the centre of the sphere? Give a reduced fraction.
Worked solution
The answer is 1/8. Choose the points as follows: first pick three random lines through the centre and a fourth random point P₄; then for each line pick one of its two endpoints, ±P₁, ±P₂, ±P₃, by independent fair coin flips. This produces four independent uniform points. Fix the lines and P₄. The centre lies inside the tetrahedron (ε₁P₁, ε₂P₂, ε₃P₃, P₄) exactly when −P₄ lies in the cone spanned by ε₁P₁, ε₂P₂, ε₃P₃. The three lines cut space into eight such cones, one for each sign pattern, and (almost surely) −P₄ lies in exactly one of them. So exactly one of the eight equally likely coin outcomes produces a tetrahedron containing the centre, and the probability is 1/8. The same argument in the plane gives 1/4 for a triangle inscribed in a circle.
Source: Putnam 1992, Problem A6. Statement written for AxiomIQ.