tetrahedron ABCD is described on a ball. Center of the ball is point S. Ball's radius equals 1 and SA ≥ SB ≥ SC. Prove that SA > √5
Can someone show me the full solution to that problem, PLEASE?? I think it's more serious than it looks. Please it's important for me. And time is not on my side
Here are some thoughts to get you started. I don't have time right now for a full and proper solution, but this is the direction I think you'll need to go to get one. And since time is not on your side, I might not get to the solution before you need it...so here we are.
The points A, B, and C form one of the four triangles of the tetrahedron. (Kinda cool that adding one point creates three triangles...) Imagine that triangle with the sphere touching one face. The diameter of the sphere is 2 (note that ?5 is about 2.24). Now, in order for all three points to be able to connect to point D (on the opposite side of the sphere) with a straight line, there is some lower limit on how close the points can be to S.
To simplify the problem, I would let (SA = SB = SC) = minimum possible distance from S. That also makes ABC an equilateral triangle and ABCD a regular tetrahedron. (I'm not 100% sure that's true, but that's what my intuition says.) It also makes the triangles SAD, SAB, SAC, SBC, etc. not only similar but identical.
Now, consider the triangle SAD (don't let the name fool you, it's happy to be alive). Since SA = SD, you can calculate how long SA must be in order for the line AD to pass outside of the sphere. Just drop a perpendicular from S to the point where AD touches the wall of the sphere. This will give you right triangles with the length of one side and all of the angles known, which is enough information to find the lengths of all the other side.
Armed with this information, you should be able to solve the problem.
(And check to make sure my unproven intuitions are valid--I typed this rather quickly.)
El que pega primero pega dos veces.