# tetrahedrons, points, and spheres

this problem is a tricky one that i can't really sum up as a title, but this is the basic jist: a few days ago, my friend gave me this geometry problem:

Given four random points on the surface of a sphere, what is the probability that the resulting tetrahedron made from connecting the 4 points contains the center?

the only issue is, i have no idea where to start. i thought of translating this from 3D to 2D, by making it:

given three random points on the edge of a circle, what is the probability that the triangle made from the four connecting points contains the center?

This reminds me of the "right triangle inscribed in a circle" theorem, stating that if the hypotenuse of a triangle is the diameter of a circle, then the intersection of the other two lines to form the right angle lies somewhere on the circle's circumference. It then gave me an idea: if you put two lines through a circle, pick 2 of the three points and put them on the endpoints of the lines, then only one of the four segmented sections will fulfill the problem if point 3 is put there. therefore, if you calculate the average percentage of the circle the arc comprises, we should be good to go, i think. this comes out to be a little less than $\frac{1}{4}$.

(this is actually part of paul's problem:here

but at this point, i hit a block. what comes next? how do i pull this off of the flat plane and into 3D? (this is usually the part where you tell me the answer or what i'm doing wrong) thanks in advance.

• it'd help more if i could add a picture. i can't :( – Alexander Day Dec 18 '17 at 15:12
• Lots of references if you search tetrahedron sphere center probability , Here's one: quora.com/… – Ethan Bolker Dec 18 '17 at 15:18