If you choose four random points in a sphere, and consider the tetrahedron with these points as its vertices, what is the probability that the center of the sphere is inside that tetrahedron?
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Answer:
I was working on points of concurrency and ‘centers’ of triangles with my geometry class when I came across the Riddler and I want to know what you think about this line of reasoning. If the triangle formed is an acute triangle, the center of the circle (circumcenter) will fall inside the triangle. If the triangle is obtuse, the center will fall outside. (A right triangle is a special case and will not impact the probability) The size of the largest angle dictates acute/obtuse and the range of values for the largest angle of a triangle is[60,180). Out of this range of 120 degrees, 30 will yield acute triangles; 30/120=1/4.
Answer:
This tantalizing, wonderful question was Problem A6 at the 53rd Putnam competition in 1992. It is hard, as A6’s are, but there’s an elegant and short solution requiring almost no calculation.
For a point X on a sphere, we will use X’ to denote the point antipodal to X. This is the point that lies the farthest from X, or more precisely it is the point where the sphere meets the line XO (O is the center of the sphere).
The short version of the solution is this:
- Once three points A, B, C are chosen, the region in which you need to pick D in order for ABCD to contain O is the spherical triangle A’B’C’ opposite ABC.
- Therefore, the probability of success is simply the expected area of the (spherical) triangle A’B’C’, normalized so that the surface of the sphere has area 1. This is clearly the same as the expected area of ABC, and in fact it is also the expected area of A’BC, A’BC’ and so on since all of these triangles are spanned by three uniformly chosen points on the sphere.
- Now, there are 8 such triangles, and their total area is 1, so the expected area of each one is 1/81/8 .
The answer, therefore, is simply 1/81/8 .
Step-by-step explanation:
may this ex helps you