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I draw an even number of points on a piece of paper. Is it possible to cut the paper into two pieces with a single straight cut, such that:

  1. Each piece gets the same number of points
  2. The cut does not go through any points
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    $\begingroup$ Weird that there were four answers to this question, nearly an hour after it was posted yet within sixty seconds of each other. $\endgroup$
    – Sneftel
    Sep 16, 2023 at 12:10
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    $\begingroup$ There is a more interesting version in which you have two kinds of points, say red and blue ones, and have to split both colours in half simultaneously. This is a discrete version of the Ham Sandwich Theorem. I think you then need to impose the constraint that no three points lie in a straight line, or else have rules about allowing your cut to go through one or more points. $\endgroup$ Sep 16, 2023 at 15:02
  • $\begingroup$ Here too, some of the drawn points may have to be unique to have such a cut... $\endgroup$ Sep 16, 2023 at 23:29
  • $\begingroup$ I found a similar question on StackOverflow: stackoverflow.com/questions/3106324/… $\endgroup$ Sep 17, 2023 at 11:18

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The answer is

yes, obviously.

Because

If you rotate the paper, there are only a finite number of angles where two points overlap on the X axis. There are plenty of ways to rotate the paper in a way that no two points share the same X coordinate.

Chose such a rotation and cut the paper vertically between the two middle points.

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Assuming no two points occupy the same place

it is always possible. There are only finitely many pairs of points, and thus finitely many exceptional slopes where some lines with that slope pass through more than one point.

So pick one of the (continuum-many) non-exceptional slopes and draw a line $l$ with the chosen slope not meeting any point. Beside each point write its signed distance to $l$; by non-exceptionality of the slope and our prior asssumption the distances will be distinct and will sort the points. A line $l'$ with the desired property is then any line parallel to $l$ where the two points with median distance are on opposite sides of $l'$.

This assumption is necessary:

Suppose there are just two points but they occupy the same spot. Then obviously no line satisfies the given constraints.

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  • $\begingroup$ You beat me for 11 seconds! Same idea. $\endgroup$
    – Florian F
    Sep 16, 2023 at 12:11
  • $\begingroup$ don't you want l' to be perpendicular to l? $\endgroup$ Sep 16, 2023 at 14:17
  • $\begingroup$ @DmitryKamenetsky Please rrad my answer carefully. $\endgroup$ Sep 16, 2023 at 14:20
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Only if

Just kidding! You can always do this. (Assuming you don't draw two points in the same location, which strikes me as unchivalrous behavior.)

First, sort the points by increasing x coordinate, with ties broken by increasing y coordinate, and draw a vertical line just to the right of point n/2. If this works, cut along that line, obviously. It won't work if x(n/2) happens to be equal to x(n/2-1) and/or x(n/2+1).

Otherwise, first move the line directly onto point n/2. Then pick a location between point n/2 and point n/2+1, then infinitesimally rotate the line counterclockwise about that location. Cut along that line. The result will have points 1 through n/2 on the left, and points n/2+1 through n on the right.

Note that

This is the same approach as Florian's and Parcly's, but involves a uniquely determined slope. (Which is great for proofs, but I wouldn't do it this way in real life.)

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Yes, it's possible.

For simplicity, we use the possibility that your even number was a 2 (because your puzzle simply asks if it's possible to do so after you have chosen an even number).

Follow these steps:

1. Choose 2 as your even number.
2. Draw two dots on a piece of paper.
3. Cut the paper in two such that there is a dot on both pieces.

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