Four points in a sphere

Question:

Consider a unit sphere. 4 points are randomly chosen on sphere, what is the probability that the center (of the sphere) lies within the tetrahedron (polygon) formed by those 4 points?

.

.

C

A

P

T

A

I

N

I

N

T

E

R

V

I

E

W

.

.

Solution:

Let A, B and C be random points on the sphere with Aa, Bb and Cc being diameters.

The spherical (minor) triangle ABC is common to the hemispheres abc, bca and cab (where the notation abc represents the hemisphere cut off by the great circle through a and b and containing the point c, etc), therefore the probability that a further random point, D, lies on this triangle is:

1/2 x 1/2 x 1/2 = 1/8

(For center to lie in the tetrahedron D should lie in the triangle i.e the opposite hemisphere of ABC)

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