A tetrahedron with its four vertices lying on a sphere contains the centre of the sphere iff a line from any vertex through the centre of the sphere passes through the spherical triangle formed by the other three vertices.
Consider the set of dispositions of the first three random points. Define on it an equivalence relation: two dispositions are equivalent iff one can be converted to the other by replacing some subset of its vertices by their antipodes. The equivalence classes have eight members. Each member of an equivalence class is equally probable.
Any two pairs of antipodal points specify a great circle. The three pairs of antipodal points which characterise an equivalence class lie on the intersections of three such great circles, which partition the sphere into eight (spherical) triangles. Wherever the fourth point is, the line from it through the centre of the sphere must continue through exactly one of the eight triangles. So, in any equivalence class, exactly one of the eight tetrahedra formed by the fourth point contains the centre of the sphere, and the probability of any set of four random points defining such a tetrahedron is ⅛.
Note: it is possible that three or more of the points lie on a single great circle. As this has probability 0, and the problem is to calculate a probability, we have ignored this possibility in the paragraphs above.
Generalisation of solution show
The solution above gives the logic for four points on the 2-sphere. The same logic applies to a sphere of any dimension.
For 2 random points on the 0-sphere, the probability is ½.
For 3 random points on the circle, the probability is ¼.
For 4 random points on the 2-sphere, the probability is ⅛.
For 5 random points on the 3-sphere, the probability is 1/16.
For n+2 random points on the n-sphere, the probability is 2–n–1.
Diagrams show
The first diagram shows the disposition described in the "Solution" section, for three points A B C and their antipodes A' B' C', with an arc connecting each point to the four other points not antipodal to it.
The projection (in the sense "map projection") is unspecified and arbitrary, but topologically correct. Any disposition of the points must looks like this, after some permutation of the labels applied to them.
The four arcs connecting each pair of antipodal pairs of points are shown in the same colour, and form a great circle. Because of the arbitrary projection, these great circular are shown as irregular shapes. Topologically, they form an octahedron.
The second diagram show four antipodal pairs of points { { A, A'}, { B, B'}, { C, C'}, { D, D'} } on a sphere, as for the first diagram but with points D, D' added.
There are sixteen ways of choosing one point from each antipodal pair. A set of four such points forms a tetrahedron containing the centre of the sphere iff, for each two of the four points, the great circle passing through them passes between the other two. Two of the sixteen options satisfy this: { A, B', C', D'} and {A', B, C, D } .
Comment show
It's easy to solve the 0-sphere and 1-sphere (circle) versions of
this problem without using symmetry.
I believe it's possible to solve the 2-sphere version using
trigonometry and calculus, maybe with assistance from
Mathematica. I wouldn't fancy doing it myself.
The solution above gives the logic for four points on the 2-sphere. The same logic applies to a sphere of any dimension.
For 2 random points on the 0-sphere, the probability is ½.For 3 random points on the circle, the probability is ¼.
For 4 random points on the 2-sphere, the probability is ⅛.
For 5 random points on the 3-sphere, the probability is 1/16.
For n+2 random points on the n-sphere, the probability is 2–n–1.
The first diagram shows the disposition described in the "Solution" section, for three points A B C and their antipodes A' B' C', with an arc connecting each point to the four other points not antipodal to it.
The projection (in the sense "map projection") is unspecified and arbitrary, but topologically correct. Any disposition of the points must looks like this, after some permutation of the labels applied to them.
The four arcs connecting each pair of antipodal pairs of points are shown in the same colour, and form a great circle. Because of the arbitrary projection, these great circular are shown as irregular shapes. Topologically, they form an octahedron.
The second diagram show four antipodal pairs of points
There are sixteen ways of choosing one point from each antipodal pair. A set of four such points forms a tetrahedron containing the centre of the sphere iff, for each two of the four points, the great circle passing through them passes between the other two. Two of the sixteen options satisfy this:
Comment show
It's easy to solve the 0-sphere and 1-sphere (circle) versions of
this problem without using symmetry.
I believe it's possible to solve the 2-sphere version using
trigonometry and calculus, maybe with assistance from
Mathematica. I wouldn't fancy doing it myself.
It's easy to solve the 0-sphere and 1-sphere (circle) versions of this problem without using symmetry.
I believe it's possible to solve the 2-sphere version using trigonometry and calculus, maybe with assistance from Mathematica. I wouldn't fancy doing it myself.