There are three points on the surface of a sphere centered at origin. One has an x coordinate of 0, another has a y coordinate of 0, and the last has a z coordinate of 0.
What is the biggest possible equilateral triangle that can be made using these three points as the corners? How many equilateral triangles of this size are possible?
What if instead of a sphere, it is a regular octahedron centered at origin, with each of its vertices on an x, y, or z axis?
(In reply to
re: part 1 spoiler by Mindrod)
My solution to part 1 does in fact use all three points at the same time as the three corners of the triangle. For a unit sphere, one such set of coordinates is:
(0,sqrt(2)/2,sqrt(2)/2), (sqrt(2)/2,0,sqrt(2)/2), (sqrt(2)/2,sqrt(2)/2,0)
The distance separating each pair is sqrt(2/4 + 2/4 +2) = sqrt(3). Each has a different coordinate that is zero and all lie on the unit sphere. For nonunit spheres, multiply each dimension by a factor of r.
This particular set is based on the original spherical triangle formed by (1,0,0),(0,1,0),(0,0,1), with sides pinwheeled out clockwise (extended along a 45degree arc on the surface of the circle of intersection of the sphere with the appropriate coordinate=0 plane).

Posted by Charlie
on 20051206 09:27:16 