Before tackling this one, take a look at
this one.
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A, B, C and D are nonadjacent vertices of a cube. There is a point P in space such that PA=3, PB=5, PC=7, and PD=6. Find the distance from P to the other four vertices and find the length of the edge of the cube.
There are two answers, one with P outside the cube and one with P inside the cube.
The maximum length of the face diagonal is 8. Spheres centered at A and B with radii 3 and 5, respectively, will not intersect if the face diagonal is larger than 8. Thus a side cannot be longer than (8/sqrt(2)), about 5.656854249.
Also, logically, the intersection of two spheres is a circle. The intersection of a circle and a sphere is (generally) two points, though it could be a circle or a single point in rare cases. The intersection of two points with a sphere is, well, you get the point. Assuming they intersect, either one or both points.
There's more math I don't feel like doing just this moment.

Posted by Eric
on 20040716 13:05:01 