7
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The 3x3 grid is given as below;

enter image description here

How many pentagons could be drawn by connecting the dots by lines in the grid?

Rules:

  • Pentagons could be convex or concave.
  • The lines cannot intersect each other.
  • If reflecting or rotating a pentagon forms the same pentagon you have already counted before, it should not be added.

Example:

enter image description here

Note: The other question referred as the duplicate question's answers are not right. and the OP is missing. That's why I believe this question should be solved here and accepted.

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  • $\begingroup$ The mathematician in me is begging to do this question. I shall oblige. :) $\endgroup$ – Quintec Jun 1 '17 at 18:04
  • $\begingroup$ This is a subset of an already answered puzzle. $\endgroup$ – Ian MacDonald Jun 1 '17 at 18:19
  • $\begingroup$ @IanMacDonald yes it is the same question, though it is more general, I missed it sorry, though there is no right answer in that question for the pentagon as far as I see. Someone might want to give the right answer here? $\endgroup$ – Oray Jun 1 '17 at 18:22
  • $\begingroup$ @IanMacDonald and it seems that --moti-- guy who asked the previous one is not active at all to accept an answer. If you want me to remove this that's fine though. $\endgroup$ – Oray Jun 1 '17 at 18:24
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    $\begingroup$ While it certainly is a duplicate, I think the posted question is much clearer and more appealing as a "puzzle" than the older post. I would vote against closing it. (And I very much like the nice answer post :c) ) $\endgroup$ – BmyGuest Jun 1 '17 at 20:07
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I get 23 pentagons. There are 126 ways of choosing 5 out of the 9 dots. After reducing for symmetry, only 23 are left, which you can check using Burnside's lemma. I drew these in the picture below. For any of these you can put a convex hull around the points. If that is a pentagon, then that is the unique one you can make from those points. If all the points are on the hull but it is a quadrilateral or triangle, then no pentagons are possible. If it has an interior point, then you have a choice of which side of the convex hull to replace by two sides connecting to the interior point.

enter image description here

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  • $\begingroup$ unfortunately, one of them is missing. $\endgroup$ – Oray Jun 1 '17 at 19:41
  • $\begingroup$ @Oray: You're right. I dismissed the case with a three points in a straight line on the convex hull but which also had an interior point. I overlooked that the interior point can be used to break up the straight line. Fixed it now. $\endgroup$ – Jaap Scherphuis Jun 1 '17 at 19:57
  • $\begingroup$ Follow-up question: how did you get 23 using Burnside's lemma? $\endgroup$ – Rand al'Thor Aug 23 '17 at 11:42

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