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 Solid triangles (2) (Posted on 2005-12-05)
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?

 See The Solution Submitted by Tristan No Rating

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 part 2--the octahedron--spoiler. | Comment 2 of 7 |

If we follow the pinwheeling strategy from part 1, but along the edges of the octahedron, we can just compare the length of the edges when midway to the opposite vertex of the octahedron (for each of the vertices of the triangle).

One such midway triangle had vertices at (0,1/2,1/2), (1/2,0,-1/2) and (-1/2,-1/2,0).  The distance between any two of these is sqrt(3/2).  This is smaller than each edge of the original octahedron, so the eight faces of the original octahedron are the largest such triangles, with a side length equal to sqrt(2) -- assuming the octahedron is inscribed in a sphere of unit radius.

 Posted by Charlie on 2005-12-05 09:53:27

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