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 Tristan)
I take that to mean you're not convinced there are only 4 triangles of maximum size. I now see there are 8. Each original octanttype equilateral triangle on the sphere can be pinwheeled out clockwise or counterclockwise, and these correspond to the same directions as the opposite side's pinwheels. The adjacent original triangles do not duplicate their neighbors' as I had thought. So there are 8.
One set, as mentioned in my reply to MindRod was
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
The other sets of points retain the zeros, but the nonzeros in each column, that is for each coordinate, can be positive and negative in either order, so (all 8repeating the above one):
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
(0,sqrt(2)/2,sqrt(2)/2),
(sqrt(2)/2,0,sqrt(2)/2),
(sqrt(2)/2,sqrt(2)/2,0)
Of course, again, multiply all coordinates by r for nonunit circles.

Posted by Charlie
on 20051206 10:14:33 