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Solid triangles (2) (Posted on 2005-12-05) Difficulty: 2 of 5
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    
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part 1 spoiler | Comment 1 of 7

For convenience, let's say the sphere has radius 1; it can always be scaled up by a ratio r for the actual radius.

The loci of x=0, y=0 and z=0 on the sphere are three mutually perpendicular circles, forming 8 equilateral right spherical triangles (octants of the sphere).  When the vertices of a spherical equilateral triangle are connected by chords of the sphere, a plane equilateral triangle is formed, and the larger the original spherical triangle is (in terms of side lengths, not of area), the larger the underlying plane triangle will be.

So start off with one of the eight octant triangles, and then pinwheel out the vertices along their respective x=0, y=0, z=0 loci, say always clockwise out as seen by the center of the octant triangle. The vertices will get farther apart until they contract together to form the opposite octant.  At one point the plane of the plane triangle will pass through the center of the sphere; at this point the plane triangle is a maximum, as the spherical triangle is just a circle in three equal segments (all its angles are 180 degrees).

The equilateral triangle, inscribed in this unit circle, has sides equal to sqrt(3). Scaled up that's sqrt(3) * r.

We could have started with any of the 8 octants, but each maximum triangle is shared with the opposite octant, so there are only 4.  One might think that going counterclockwise out would produce a different triangle from going clockwise out, and that would be true, but it would also be the same triangle as produced by going clockwise out from an adjacent octant.  It might be good to look at this from the perspective of the octahedron mentioned in part 2: there are only four planes going through the center that are parallel to two faces.

For part 2--the octahedron--the question is whether the pinwheeling out from the original vertices results in an enlarging or a shrinking of the triangles that start out as the faces of the octahedron.

  Posted by Charlie on 2005-12-05 09:37:06
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