Return to Answer

Finally, let's consider the case between my suggested constraint (no more than three in a plane) and that of loopy's answer (all five in a plane): exactly four line in a plane.

In this case the claim is:

true.

Case 1:

If the four points that share a plane lie on a great circle, then we trivially take the hemisphere, bounded by that circle, which does not have the fifth.

Case 2:

If the four points do not lie on a great circle, we can visualize their plane as a plane of latitude entirely north of an equator. This means none of the pairs are antipodes (since none are south of the equator) and no three of them share a great circle (since a circle can only intersect a plane (the plane with the four points) in two places if not entirely contained within it). Thus, we have 6 (4 choose 2) pairs, each with a distinct great circle. By the same reasoning as in the "Even more detail" paragraph above, at most two of these great circles can contain the fifth point, leaving four pairs of non-antipodal points whose great circle contains no other points. Pick any of those pairs and apply the procedure ("Now") at the end of the first half of this post.

In order to eliminate degenerate configurations, I suggest adding the constraint that no more than three points lie in the same plane. (This also implies all five points are distinct.) This would better model the case of actually drawing points on a real ball, since there will always be physical imperfections.

With that additional constraint, the claim is

true.

To see this, first:

pick two points that are not antipodes (exactly opposite each other) and whose great circle (which is uniquely determined by two points) does not contain a third point.

This is always possible (or else the problem immediately solved) because:

there can be at most one antipodal pair, because if there were two, then a great circle would contain all four, meaning they also lie on the same plane. If there is one antipodal pair, then if the other three all lie on the same great circle, then the original problem is immediately solved: that great circle plus either of its hemispheres contains four points. If there are no antipodes, then there can be at most two great circles containing three points (imagine five points arranged like an X), occupying 6 out of the 10 pairs and leaving 4 pairs that each determine a great circle that is not shared with any other point.

Now,

having selected two points whose great circle contains no other points, the remaining three points are either divided into two on one half and one on the other, or all three are on the same side. In the former case we select the hemisphere with two, which when combined with the two on the great circle yields four. In the latter case (all three on the same side), rotate the great circle by an infinitesimal amount around the axis that passes through one of the points it contains and the sphere's center (so it remains a great circle), so the other point previously on the circle is now on the opposite side from the three. Then, again, we have exactly four to a side.

The reasoning in my answer is somewhat similar to that in the answer by AxiomaticSystem (although developed independently), especially the last step. I think the approach based on great circles offers some additional clarity because the sphere is evenly divided into hemispheres at all stages, but this may just be a case of inventor's bias.

In order to eliminate degenerate configurations, I suggest adding the constraint that no more than three points lie in the same plane. (This also implies all five points are distinct.) This would better model the case of actually drawing points on a real ball, since there will always be physical imperfections.

With that additional constraint, the claim is

true.

To see this, first:

pick two points that are not antipodes (exactly opposite each other) and whose great circle (which is uniquely determined by two points) does not contain a third point.

This is always possible (or else the problem immediately solved) because:

there can be at most one antipodal pair, because if there were two, then a great circle would contain all four, meaning they also lie on the same plane. If there is one antipodal pair, then if the other three all lie on the same great circle, then the original problem is immediately solved: that great circle plus either of its hemispheres contains four points. If there are no antipodes, then there can be at most two great circles containing three points (imagine five points arranged like an X), occupying 6 out of the 10 pairs and leaving 4 pairs that each determine a great circle that is not shared with any other point.

Even more detail on the previous point (skip this at first):

If there are no antipodes, then there can be at most two great circles containing three points. Proof: Let A, B, C, D, and E be the points. Suppose we have one great circle containing A, B, and C. This circle cannot contain D or E (otherwise we would have four points on the same plane). Now suppose we have another great circle containing three points. It cannot have more than one point from among A, B, and C, since otherwise it would be the same great circle as the first. Thus, it has D, E, and (let's say) C. Now suppose there is a third great circle containing three points. Again, it cannot have more than one point from {A,B,C}, and it also cannot have more than one point from among {C,D,E}. But then it only has two points! So a third great circle containing three points cannot exist.

Now,

having selected two points whose great circle contains no other points, the remaining three points are either divided into two on one half and one on the other, or all three are on the same side. In the former case we select the hemisphere with two, which when combined with the two on the great circle yields four. In the latter case (all three on the same side), rotate the great circle by an infinitesimal amount around the axis that passes through one of the points it contains and the sphere's center (so it remains a great circle), so the other point previously on the circle is now on the opposite side from the three. Then, again, we have exactly four to a side.

The reasoning in my answer is somewhat similar to that in the answer by AxiomaticSystem (although developed independently), especially the last step. I think the approach based on great circles offers some additional clarity because the sphere is evenly divided into hemispheres at all stages, but this may just be a case of inventor's bias.

In order to eliminate degenerate configurations, I suggest adding the constraint that no more than three points lie in the same plane. (This also implies all five points are distinct.) This would better model the case of actually drawing points on a real ball, since there will always be physical imperfections.

With that additional constraint, the claim is

true.

To see this, first:

pick two points that are not antipodes (exactly opposite each other) and whose great circle (which is uniquely determined by two points) does not contain a third point.

This is always possible (or else the problem immediately solved) because:

there can be at most one antipodal pair, because if there were two, then a great circle would contain all four, meaning they also lie on the same plane. If there is one antipodal pair, then if the other three all lie on the same great circle, then the original problem is immediately solved: that great circle plus either of its hemispheres contains four points. If there are no antipodes, then there can be at most two great circles containing three points (imagine five points arranged like an X), occupying 6 out of the 10 pairs and leaving 4 pairs that each determine a great circle that is not shared with any other point.

Now,

having selected two points whose great circle contains no other points, the remaining three points are either divided into two on one half and one on the other, or all three are on the same side. In the former case we select the hemisphere with two, which when combined with the two on the great circle yields four. In the latter case (all three on the same side), twist the circle by an infinitesimal amount around one of the points it contains so the other point previously on the circle is now on the opposite side from the three. Then, again, we have exactly four to a side.

The reasoning in my answer is somewhat similar to that in the answer by AxiomaticSystem (although developed independently), especially the last step. I think the approach based on great circles offers some additional clarity because the sphere is evenly divided into hemispheres at all stages, but this may just be a case of inventor's bias.

In order to eliminate degenerate configurations, I suggest adding the constraint that no more than three points lie in the same plane. (This also implies all five points are distinct.) This would better model the case of actually drawing points on a real ball, since there will always be physical imperfections.

With that additional constraint, the claim is

true.

To see this, first:

pick two points that are not antipodes (exactly opposite each other) and whose great circle (which is uniquely determined by two points) does not contain a third point.

This is always possible (or else the problem immediately solved) because:

there can be at most one antipodal pair, because if there were two, then a great circle would contain all four, meaning they also lie on the same plane. If there is one antipodal pair, then if the other three all lie on the same great circle, then the original problem is immediately solved: that great circle plus either of its hemispheres contains four points. If there are no antipodes, then there can be at most two great circles containing three points (imagine five points arranged like an X), occupying 6 out of the 10 pairs and leaving 4 pairs that each determine a great circle that is not shared with any other point.

Now,

having selected two points whose great circle contains no other points, the remaining three points are either divided into two on one half and one on the other, or all three are on the same side. In the former case we select the hemisphere with two, which when combined with the two on the great circle yields four. In the latter case (all three on the same side), rotate the great circle by an infinitesimal amount around the axis that passes through one of the points it contains and the sphere's center (so it remains a great circle), so the other point previously on the circle is now on the opposite side from the three. Then, again, we have exactly four to a side.

The reasoning in my answer is somewhat similar to that in the answer by AxiomaticSystem (although developed independently), especially the last step. I think the approach based on great circles offers some additional clarity because the sphere is evenly divided into hemispheres at all stages, but this may just be a case of inventor's bias.