[Math Puzzle]Day 6

Easy.

Choose two points and draw a line across these two points on the sphere. This cuts the sphere into two halves - call them A and B where A and B excludes the boundary. We will have the following cases:

1. No other points fall on the line. Then we have three points leftover, distributed in A or B. Therefore, A:B is 3:0, 2:1, 1:2, or 0:3, where, for example, 3:0 means A has three points and B has none, etc. Then we choose A or B that has at least two points, in addition to the two points on the boundary by construction, we get a solution.

2. There are three points on the line. Then A:B is 2:0, 1:1, or 0:2. Choose A or B that has at least one point.

3. The case where there are more than three points on the boundary is completely trivial.

5 dots
pick any 2, and draw a line connecting them.
trace this line across the entire sphere. this defines your hemispheres.

the other 3 will be on the line = 5 in one hemisphere
OR 2 on the line, 1 on a side = 5 in one hemisphere
OR 1 on the line, 2 on a side (2-0, 1-1, 0-2 <---whichever side doesn't matter) = 4 or 5 dots in one hemisphere
OR 0 on the line, 3 on a side (1-2, 2-1, 3-0, 0-3 <---whichever side doesn't matter) = 4 or 5 dots in one hemisphere

**sorry my explanation is kinda crappy -_-. im tired as hell for some reason and am going to bed now....

pick 2 points and draw a great circle. If a third point lies on the great circle, then you're done. So assume a third point does not lie on the great circle. Then either the other three will lie on one side (in which case you are done) or two will lie on one side (in which case you are also done).

Here's mine......

The opposite of a hemisphere (boundary included) containing 4 points is a hemisphere (boundary excluded) containing only 1 point.

Given any 2 points, the only case where there does not exist a hemisphere boundary excluded that contains both points is if they are polar opposites. And there can only be 1 polar opposite thus there always exist a hemisphere boundary included that contains 4 points.

Any two points on a sphere can be covered by a single hemisphere, because the maximum distance on the surface they can be from each other is opposite each other on the sphere. Any hemisphere with one point on the border will border the other.

Because those two points at maximum distance will form a line through the center, we can rotate the hemisphere along that axis and still contain the first two points. If you throw in a third point, we can rotate the hemisphere to contain it.

Throw in a fourth point, and we can still rotate the hemisphere along the same axis to contain both it and the third, as the maximum distance it will be away from the third point will be directly across it on the sphere, which means the hemisphere would border all four points.

Whether or not the fifth point is contained is irrelevant.

Take any two points, draw the circle through it with the same centre as the centre of the sphere. The circle divides the sphere into two hemispheres. There are 3 leftover points, by the pigeon hole principle two must lie on the same hemisphere. Including the original 2, there are 4 on that hemisphere.