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Original Q: Plot 24 points so that there is only one 23-sided polygon where 23 of those points are its vertices, and the other point is not on any edge of the polygon. A polygon, here, is defined as a simple closed curve that is made up of line segments. Therefore, any set of collinear points will always form only a single side (in a convex polygon). "Simple closed curve" means "closed curve that does not intersect itself".

Formal statement of Q

We say that a finite set $P$ of points in the Euclidean plane contains a 23-gon $Q$, if the following conditions are satisfied

  • $Q$ has 23 pairwise distinct corners
  • $Q$ has 23 pairwise non-crossing sides
  • At every corner of $Q$, the two incident sides enclose an angle $\ne180^{\circ}$
  • Every corner of $Q$ is contained in $P$

(Note that the interior of any side of $Q$ may contain an arbitrary number of points from $P$; there is no restriction on such points.)

Task: Plot a point set $P$ with $24$ points that contains exactly one 23-gon.

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  • $\begingroup$ I prefer i-corsika-itrigons, myself. $\endgroup$ – corsiKa Dec 4 '15 at 18:35
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    $\begingroup$ Wow, this went from being in understandable English to being accessible only to those who have studied a lot of math. Terminology like "pairwise distinct corners" and "incident sides" is not accessible to your average non-math-geek. I realize the point was probably to define the problem more precisely, but I don't know that doing so at the expense of accessibility is the best choice... $\endgroup$ – GentlePurpleRain Dec 4 '15 at 19:30
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    $\begingroup$ @GentlePurpleRain: It would appear that Gamow is trying to show off, at everybody else's expense (probably part of the point). $\endgroup$ – Lightness Races in Orbit Dec 4 '15 at 21:41
  • $\begingroup$ @GentlePurpleRain For clarity, I've retained both versions of the Q. $\endgroup$ – ghosts_in_the_code Dec 5 '15 at 7:56
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Inspired by Luke's idea:

enter image description here

The blue point is collinear with the two big black points above it, and also collinear with the black points on the right.

Let $P$ be a polygon made with some of these points, let $s$ be a side of this polygon, and let $\ell$ be the line containing $s$. Notice that there are no points in the interior of the shape formed by all 24 points. Thus, all the points of $P$ must lie either on $\ell$ or on one side of $\ell$ -- otherwise there would be a side that crosses $s$.

If $P$ is a icosikaitetragon and if the blue point is included, then $P$ must include three points in a line (either the blue point and the two above, or the blue point and the two on the right). WLOG, assume it is the blue point and the two above it. If the middle point is not an angle of $180^\circ$, then there is a side $s$ coming out of it that has points of $P$ on both sides, a contradiction.

Hence we must exclude the blue point. From there, the natural polygon is the only one, since any other option would again have a side $s$ with points of $P$ on both sides.

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  • $\begingroup$ Your answer looks right to me. I'll wait a few hours before accepting it, and posting my own solution as well. $\endgroup$ – ghosts_in_the_code Dec 4 '15 at 16:36
  • $\begingroup$ Just curious, wouldn't you be able to create a shape by omitting the bottom right point and going from there? the top left point would also work. Technically assuming your line was thin enough any of the points along the arc would as well. $\endgroup$ – Kingrames Dec 8 '15 at 15:04
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Edit: This is wrong. As Gamow points out, you can "Skip the southern-most black point. Start from northern-most black, go to next black, to central red, to blue, to black, and then follow the parabola of red points."

I think I have it! Pardon the terrible art:

enter image description here

Assume the black and blue points are exactly in a line. We cannot use the blue point. If we did, we'd still need to use at least three of the black points, which generates two adjacent sides (black-black-blue) forming a 180-degree angle no matter what.

Now all that we have to show is that there is only one 23-gon possible with the remaining points (that is, the one that's shaped like a horseshoe with the concavity on the left.) We run into a similar problem with the central red point as with the blue point: if we try to use it in some other way by removing the concavity, we'll end up with three black points forming two adjacent sides with a 180-degree angle.

I'm not a mathematician, so please tell me about any flaws you spot in my logic or any ways to make this clearer.

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  • $\begingroup$ Why not: Skip the southern-most black point. Start from northern-most black, go to next black, to central red, to blue, to black, and then follow the parabola of red points. (And a symmetric solution by skipping the northern-most black point.) $\endgroup$ – Gamow Dec 4 '15 at 14:40
  • $\begingroup$ Dang, you're right. $\endgroup$ – Luke Dec 4 '15 at 14:41
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@TylerSeacrest is right as well. Here is my solution which I thought of, even before posting the Q.


Take a regular $23$-gon. Take $4$ consecutive points, $A,B,C,D$. Extend $AB$ and $DC$, and name their intersection $X$. Now $A,B,C,D,E,\dots ,X$ is the required set of points.

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