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How many ways can regular (convex) polygons meet at a point (vertex), so there are no gaps or overlaps?

Here's an example with a square, hexagon, and dodecagon.

example

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  • $\begingroup$ Are infinite-sided regular convex polygons (circles) allowed? :) $\endgroup$
    – DanDan0101
    Aug 29, 2023 at 22:00
  • $\begingroup$ Part of the problem is proving the upper-bound on polygon sides. $\endgroup$
    – qwr
    Aug 29, 2023 at 22:03
  • $\begingroup$ Is a reflection considered distinct? $\endgroup$
    – RobPratt
    Aug 29, 2023 at 22:16
  • $\begingroup$ @RobPratt I wouldn't count them as distinct, but you can choose to count them if you wish $\endgroup$
    – qwr
    Aug 29, 2023 at 22:23
  • $\begingroup$ @qwr Please make the rules one way or the other. This ensure that answers can be judged according to the same criteria. $\endgroup$
    – bobble
    Aug 29, 2023 at 22:50

1 Answer 1

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Let nonnegative integer decision variable $x_k$ represent the number of $k$-gons. A necessary condition is that the internal angles sum to $360$ degrees: $$\sum_{k \ge 3} 180 \frac{k-2}{k} x_k = 360. \tag0\label0$$ I found

17 solutions to \eqref{0}:

\begin{align} &(x_{6}=3 )\tag{1}\label{1}\\ &(x_{5}=2 ,x_{10}=1 )\tag{2}\label{2}\\ &(x_{4}=1 ,x_{8}=2 )\tag{3}\label{3}\\ &(x_{4}=1 ,x_{6}=1 ,x_{12}=1 )\tag{4}\label{4}\\ &(x_{4}=1 ,x_{5}=1 ,x_{20}=1 )\tag{5}\label{5}\\ &(x_{4}=4 )\tag{6}\label{6}\\ &(x_{3}=1 ,x_{12}=2 )\tag{7}\label{7}\\ &(x_{3}=1 ,x_{10}=1 ,x_{15}=1 )\tag{8}\label{8}\\ &(x_{3}=1 ,x_{9}=1 ,x_{18}=1 )\tag{9}\label{9}\\ &(x_{3}=1 ,x_{8}=1 ,x_{24}=1 )\tag{10}\label{10}\\ &(x_{3}=1 ,x_{7}=1 ,x_{42}=1 )\tag{11}\label{11}\\ &(x_{3}=1 ,x_{4}=2 ,x_{6}=1 )\tag{12}\label{12}\\ &(x_{3}=2 ,x_{6}=2 )\tag{13}\label{13}\\ &(x_{3}=2 ,x_{4}=1 ,x_{12}=1 )\tag{14}\label{14}\\ &(x_{3}=3 ,x_{4}=2 )\tag{15}\label{15}\\ &(x_{3}=4 ,x_{6}=1 )\tag{16}\label{16}\\ &(x_{3}=6 )\tag{17}\label{17}\\ \end{align}

(Solution \eqref{4} corresponds to the example given in the question.)

Up to rotation and reflection, solutions \eqref{12} through \eqref{15} yield two arrangements, and the rest yield only one. So in total there are

$17+4=21$ arrangements.

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    $\begingroup$ The answer is correct, but there are nicer ways to show by hand these are the only solutions by casework on the number of polygons. $\endgroup$
    – qwr
    Aug 30, 2023 at 0:11
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    $\begingroup$ Yes: math.stackexchange.com/questions/4574012/… $\endgroup$
    – RobPratt
    Aug 30, 2023 at 0:15
  • $\begingroup$ Because of how the question is worded, there are quite a few other solutions in addition to these. For example, you can have a vertex with a square and two equilateral triangles, or a vertex with four triangles. Here's a picture illustrating both of these cases. :-) $\endgroup$
    – Bass
    Sep 1, 2023 at 2:41
  • $\begingroup$ That's true. I should've specified on the Cartesian plane. $\endgroup$
    – qwr
    Sep 18, 2023 at 2:13

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