This is a problem we've been considering in our undergraduate math club, and I thought it would be nice to get further thoughts on the subject.
I will start with a two dimensional case and then extend the problem to 3 dimensions (which is the one we have been considering).
2 Dimensions
We want the minimum number of colors that can color the edge of a circle such that, when any two radii are drawn at a right angle to each other, they touch different colors on the edge of the circle.
One color obviously doesn't satisfy the condition, and we can see that two colors is a lower bound for this problem and four colors is trivially an upper bound. The solution is fairly trivial as well; the circle can be colored with two colors as shown below:
Note that the color at a boundary must be selected to meet the constraint, which isn't really shown in the picture.
3 Dimensions
Now consider the 3 dimensional case. We want the same constraint to hold, but now instead of there being just two points that touch right angled radii, we have an infinite number of points along a great circle.
The lower and upper bounds are not quite as obvious but are not much of a stretch.
- Lower Bound:
Pick a color for a single point on the sphere, and consider points on the great circle indicated by our constraint. Every point on the great circle must be different from the selected point, and we've already shown that a circle can be covered by two colors. Therefore the sphere requires at least 3 colors to cover it. - Upper Bound:
Divide the sphere into two hemispheres, with the upper hemisphere taking the boundary (so that the lower hemisphere is "open"). Being careful with the boundaries, you can color each quadrant of the hemisphere with just four colors, but it requires that the point in the center of each hemisphere be a fifth color.
A flattened illustration:
(Each color can only take one of the boundaries, but I didn't show that in the illustration)
We now know that the sphere can be colored with 3-5 colors, but we want the minimum, and have yet to show that 3 and 4 colors are not possible. We discussed the "inversion of a sphere" problem and considered that there might be some "tricky" coloring that could be done with just three or four colors. (Fully aware that the problems are not analagous).
Best Answer
This problem is solved here, with a minimum of four colours. See also here. (Christian Blatter is mentioned as having posed this problem; he's quite active on this site.) Note that there's also an interesting connection to the Kochen–Specker theorem.