In how many ways can different numbers of right angles appear in convex heptagons?

Heptagons do not have to have right angles at all, of course. If a heptagon has exactly one right angle, only one arrangement is possible: a right angle, and six oblique angles (“oblique” means non-right, so it includes both acute and obtuse angles).

With two right angles, there are three possibilities. In the first one shown, the right angles are consecutive. In the second, one oblique angle appears between the right angles. In the third, two oblique angles appear between the right angles. Increasing two oblique angles to three is simply a repeat of the third three-right angle heptagon, so this set stops with exactly three members.

With three right angles, I have found four possibilities: (1) all three right angles appear consecutively; (2) two right angles are consecutive, and one of them has one oblique angle between it and the third right angle; (3) two right angles are consecutive, and each of them has two oblique angles between it and the third right angle; and (4) None of the three right angles is consecutive. In this heptagon, the number of oblique angles which appear between the three different right-angle pairs are one, one, and two.

I have found no others, and, after searching to exhaustion, I do not think any other arrangement of right and oblique angles in a convex heptagon is possible. However, this is a conjecture, not a proof, and may, in fact, be incorrect. If you can provide proof that this listing of possibilities is complete, or a counterexample to show that it is not, please leave a comment with details.

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I suppose that it’s possible to cook up some group theoretic/combinatorial proof here, but I feel that is overcomplicating things (unless the point is practicing using group theory/combinatorics). The simplest way for me seems to just list all possibilities. Looking over your answer quickly it seems to me that you have found all of possibilities. If you want me to take a closer look at the problem and elaborate feel free to ask.

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I feel that it is basically obvious that there can be no more than three right angles in convex n-gons with n greater than four. (trigons and tetragons can be counted manually as having 2 and 4 possibilities, respectively) For 0 and 1 right angle, there is only one possibility each, and for two/three right angles, the number of possibilities would be the number of combinations of two/three non-negative integers adding up to two/three less than n, respectively, which would be… ceil(n/2-1) and sum{i=0toceil(n/3-1)}ceil((n-3i)/2-1)? It’s very late as I’m typing this and I have no experience in combinatorics.

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After coming back to this, I’m pretty sure that the number of n-gons with three right angles is ceil(ceil((n/3-1)/2)/2+(2n-1)ceil(n/3)/4-3ceil(n/3)^2/4).

Interestingly, the difference between consecutive terms in this expression has a very regular pattern: 0 0 1 0 1 1 1 1 2 1 2 2 2 2 3 2 3 3 3 3 4 3 4 4 4 4 5 4 5 5 etc

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Coming back to this one last time, the number of ways right angles could appear in an n-gon is (6n^2+36n+83-27(-1)^n+16cos(2npi/3))/72 or, equivalently, 1+round(n/2)+round(n^2/12-7/72).

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