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I have noticed certain pattern in figural reasoning while preparing for the civil service entrance examination in my country. That is, given any polygon made by sticking together squares of the same size side-by-side, its number of internal angles of $90^o$ minus its number of internal angles of $270^o$ seems to be always $4$ (assuming that there are no 'hole' in the polygon, or put it in another way, it is simply connected).

For example, the first polygon in the following picture has $5$ internal angles of $90^o$ and $1$ internal angle of $270^o$.

enter image description here

I think this phenomenon may be proved by some elementary methods (induction, for example), but I wonder whether there is a more elegant explanation? In particular, I wonder whether it is related to Green's formula or de Rham cohomology (since I remember both require simple-connectedness).

P.S. I am just an amateur, so please do not laugh at me if I overthink :)

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    $\begingroup$ Why do you think this should be related to Green's formula or de Rham cohomology? It's totally not obvious to me if and how this should be the case, given that you're asking about angles, so you should substantiate why you think there should be a connection to these things in particular. $\endgroup$
    – Ben Steffan
    Commented Sep 13 at 10:46
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    $\begingroup$ If there is a proof of this fact using de Rham cohomology, it deserves a prominent place in Mathematics made difficult :-) $\endgroup$
    – Jean Abou Samra
    Commented Sep 13 at 11:06
  • $\begingroup$ @BenSteffan I apologize if I think too complicated. I think this should be related to things like de Rham cohomology because it seems to require the polygon to be simply connected in order to make the property holds. I did consider things simpler like the result concerning the sum of the exterior angles, as mentioned by Tzimmo, but I mistakenly assumed that it only held for convex polygons. Even if it holds for certain types of non-convex polygons (say, those that is not crossed, as remarked by Tzimmo), it seemingly still cannot explain the requirement of simple-connectedness. $\endgroup$
    – zyy
    Commented Sep 13 at 13:03

1 Answer 1

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You are thinking too complicated, I suppose. Elementary geometry should suffice for this question.

Your observation is just the fact that the sum of the exterior angles of a simple polygon is $360°$. Note that the exterior angles are signed, so an interior angle greater than $180˚$ yields a negative exterior angle. If the interior angle is $90°$, the exterior one is $90°$, too. If the interior angle is $270°$, the exterior one is $-90°$. See also Wikipedia.

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    $\begingroup$ I am not sure this is what OP is asking here. You are talking about the total , while OP is asking about the count. More work is necessary to match these two. $\endgroup$
    – Prem
    Commented Sep 13 at 10:54
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    $\begingroup$ @Prem As the only interior angles occurring are $90°$ and $270°$, all exterior angles are $\pm 90°$. The sum of all these angles is $90°$ times the difference of numbers of $90°$- and $270°$-angles. $\endgroup$
    – Tzimmo
    Commented Sep 13 at 10:58
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    $\begingroup$ No, it works as long as the polygon is not crossed. Just account for the signs when considering exterior angles and you're fine. $\endgroup$
    – Tzimmo
    Commented Sep 13 at 11:00
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    $\begingroup$ @Tzimmo You should include in your answer the content of your remark : for a non convex polygon, in the concavities, you have to consider negative angles in your totalization. This isn't at all evident ! $\endgroup$
    – Jean Marie
    Commented Sep 13 at 12:59
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    $\begingroup$ That is not a polygon. A polygon is bounded by a single chain of line segments. The square with middle removed needs one for the inside and another one for the outside. A non-crossed polygon is always simply connected. $\endgroup$
    – Tzimmo
    Commented Sep 13 at 13:00

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