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Suppose that you take a pen and mark five points on a ball.

I claim that no matter where you draw those points, I can always slice the ball into two equal halves (two equal and closed hemispheres) such that one of the halves contain exactly four of those points. Is my claim true?

Source: this Mathematics Educators Stack Exchange question by user @MikePierce

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P.S: Please also check out the following answer given by Scott McPeak, where he discusses two, very interesting, variants of the above question: Scott's answer

Yet another variant of this question where we claim that we can always divide the sphere into two close hemispheres, such that one hemisphere contains at least 4 points, can be seen here :

At least 4 points are in a hemisphere

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The claim is

false.

Counter example: Choose an equator and draw the points on it at equal distance such that they form a regular pentagon.

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  • $\begingroup$ it is a closed hemisphere ...all the points have to belong to one hemisphere or the other. $\endgroup$ – Hemant Agarwal Mar 23 at 12:27
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    $\begingroup$ Doesn't matter. If the hemispheres align with the equator we have chosen both contain all five points. Othewise one contains three points, the other two or three $\endgroup$ – loopy walt Mar 23 at 12:30
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    $\begingroup$ They say "at least", you say "exactly". Big difference. $\endgroup$ – loopy walt Mar 23 at 12:41
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    $\begingroup$ @HemantAgarwal loopywalt is correct. The linked answer is "at least" four points so either four or five. The answer here shows an example where we can only do exactly five and not exactly four. $\endgroup$ – hexomino Mar 23 at 13:01
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    $\begingroup$ Any cut not along the pentagon's circle cuts it along a diameter: any four points on a pentagon define a quadrilateral containing the center and thus can't be isolated by a diameter. $\endgroup$ – AxiomaticSystem Mar 23 at 13:33
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False: the question does not state that the points are distinct. 5 coincident points are either all in a hemisphere or none is, hence it cannot be exactly 4.

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Kudos to loopy for getting the answer. Some reasoning to show that it's essentially best possible:
Take a sphere and cut it generally (i.e. through no points). If four points lie on a side, we are done.
If five lie on one side, then choose a polar axis through the cut such that there is a unique point with least longitude: there are finitely many cases where this is not possible - one for each side of the convex hull of the five points - so this is possible to do. Once an axis is found, simply rotate the cut plane around it until it passes the nearest point and the four others are isolated, as desired.
Assume now that three lie on one side and two on the other. Make a new cut plane through the two isolated points. If this plane divides the three such that two lie on one side, we are done. If three all lie on one side and none on the other, then we can twist the cut plane slightly such that one of the two moves to the other half, and we are done. The only remaining case is if the three all lie on the plane defined by the other two, in which case division of the sphere is only possible if a half of the circle containing all five points contains exactly four points.

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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.


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.

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  • $\begingroup$ "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." Question 1) If two points don't form an antipodal pair, then isn't there only one unique circle that can pass through two points on the surface of a sphere? Or is it really possible to twist the "circle" ? $\endgroup$ – Hemant Agarwal Mar 29 at 7:02
  • $\begingroup$ Question 2) Secondly, let's assume that more than one circle can be formed on a non antipodal pair. Then, my understanding is that only one of them will be a great circle. Therefore, if we twist this circle, then it is no longer dividing the sphere into 2 equal halves. $\endgroup$ – Hemant Agarwal Mar 29 at 7:05
  • $\begingroup$ @HemantAgarwal I just made an edit that I should should clarify. The great circle remains a great circle after the rotation (I removed the word "twist"). Also, there is only one great circle containing two non-antipodes, but infinitely many non-great circles containing them. The Euclidean 2D analogous statement is there is only one line containing two distinct points but infinitely many circles that do. $\endgroup$ – Scott McPeak Mar 29 at 7:37
  • $\begingroup$ Would you consider answering the following variant also, here or in your post above : Will my claim still be true if I now make the condition that exactly 4 points, drawn by you, have to lie in the same plane ? $\endgroup$ – Hemant Agarwal Mar 29 at 9:00
  • $\begingroup$ @HemantAgarwal Adding a constraint like that sounds a bit contrived to me. (Isn't it also too easy then? I just draw those 4 points on a great circle of my choosing.) I think a more interesting direction would be to explore increasing the number of points (e.g., can we do 5 out of 6?) and/or changing the shape of the object (maybe a cube?). One of these would be substantial enough to warrant its own question, but I (personally) would only post a question if I found (or suspected) an elegant way to solve it. $\endgroup$ – Scott McPeak Mar 29 at 9:04
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The claim is

True.

If you draw 4 points, you can grab any 3 of them and grab the hemisphere that contains that 3 points. So, if you draw 1 point, then you can grab 3 points so that the hemisphere that contains that 3 points can contain the last 1 point.

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    $\begingroup$ I think I can defend the first part ("for any three points, you can choose a hemisphere"): pick two of the three, cut along the great circle that has them both, and then the third point will be on one side or the other of that cut, so you pick the hemisphere the third one is on. But how do you defend the second part? $\endgroup$ – Daniel Wagner Mar 23 at 23:59
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The claim is

false.

Choose a diameter, and draw a point at each end.

If neither point is on the dividing equator, they must lie in opposite hemispheres.

If you have two such pairs, each hemisphere must have at least 2 points.

Since there are only 5 points total, each hemisphere can have at most 3 points.

Edit:

If we allow that points can lie on the dividing equator, the above will not suffice as a counterexample. See the comments below.

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  • $\begingroup$ That's a good argument, but you haven't quite finished it. What if you put the cut through one or both pairs of points? Maybe then it is possible to get 4 in one hemisphere. Eventually you want to use the argument to find a particular counterexample, such that the argument proves that it really is a counterexample to the claim. $\endgroup$ – Jaap Scherphuis Mar 24 at 8:39
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    $\begingroup$ @JaapScherphuis If a point is on the equator, do you consider it to be in both hemispheres? Or neither hemisphere? Or half a point in each hemisphere? $\endgroup$ – Azeezah M Mar 24 at 8:53
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    $\begingroup$ The question specifies that the hemispheres are closed, so they are included in both hemispheres. Actually only one of these matter, the one you want to have 4 of the 5 points. $\endgroup$ – Jaap Scherphuis Mar 24 at 8:59
  • $\begingroup$ @JaapScherphuis In that case, it's possible for a hemisphere to have 4 points. If exactly one pair lies on the equator, the hemisphere will have 2 points from this pair, 1 point from the other pair, and 1 from the 5th unpaired point. So my answer of 2 pairs of diametrically opposite points is not a counterexample. $\endgroup$ – Azeezah M Mar 24 at 9:41

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