When I was in high school pre-calc, must have been 11th grade, I was writing D&D campaign material into the wee hours many nights. One of my more manic projects was to number and assign a location to every magic item above a certain arbitrary power level in my game world. For example, one of the penultimate treasures for mages in AD&D 1E - Staves of the Magi. By some method factoring in both the xp value of the item and dice randomization, I determined there should be only 5, or maybe it was 7 or 12, whatever. (It was a big world.)

A corollary problem (with an easy solution) would be how to determine the radian (or degree) distance between points uniformly distributed on the circumference of the circle. Simple division: [degrees or radians in a circle] / [number of points] = distance between points.

The application for the even-spacing problem on a sphere was more for world-tour sorts of quests, where the players must solve a series of puzzles and endure story arcs to assemble a legendary rod of some arbitrary number of parts (7 is classical), or close ancient demon invasion portals or whatever.

But my world was a sphere. How do you calculate coordinates for N points evenly distributed on the surface of a sphere? For 2 points, they're at opposite poles. For three points, they define a plane bisecting the sphere (like the equator). Four points define a classical Platonic solid, the tetrahedron, and with a little brain power you can figure out coordinates for the vertices of a tetrahedron inscribed in a sphere.

Skipping the 5-point case for a moment, 6 vertices define another Platonic solid, the octahedron. Coordinates are easy, because it's just 4 points equidistant on a plane bisecting the sphere, and the remaining 2 points each at a pole; and the 8-point case defines a cube. It's an interesting feature of solid geometry that 8 vertices define a 6-sided solid and 6 vertices define an 8-sided solid; and even more interesting that the symmetry repeats with the 12-vertex/20-sided solid case reflecting the 20-vertex/12-sided solid case. The coordinate relationships of Platonic solids are well-studied.

I got that far from basic geometry and examination of handy little Platoic solids AKA dice. But back to the 5-point case: Initial intuition suggests a regression from the 6-point case, with 3 points defining a triangle on the equator and the remaining 2 points at the poles again. But if you try to envision a regular solid meeting those criteria, the projection of planar surfaces implies two more vertices and you end up with a cube.

So I turned to someone I figured should know: my 11th grade pre-calc teacher, Mrs. Burden. Sadly, she brushed me off with a "why don't you try to develop a solution yourself" sort of answer. Little did I know at that time, far more mathematical minds than mine had struggled to adequately define the problem and frame a general solution. In fact, to date there remains no general solution. I'm not sure whether Mrs. Burden should be lauded or castigated for keeping that fact to herself.