# How many reflex angles can a polygon have? [closed]

I think the first of the polygons that can have a reflex angle is the pentagon. For a hexagon, a maximum of 2 reflex angles is possible. I tried to draw many concave polygons to find out a relation, but I got stuck. Is there a formula for this?

• What do you mean? A quadrilateral can have three obtuse angles. A hexagon can have all internal angles obtuse. Commented Nov 17, 2017 at 4:17
• I see. From your figure,you meant reflex angles. Commented Nov 17, 2017 at 4:18
• I am sorry ,I was too tired ,that I typed obtuse instead of reflexive angle .! Commented Nov 17, 2017 at 4:20
• The maximum number of reflex angles a simply connected $n$-gon can have is $n-3$. A quadrilateral is the smallest $n$-gon that can have a reflex internal angle. Commented Nov 17, 2017 at 4:20
• I think because sum of angles in a polygon is $180(n-2)$, the maximum number of reflex angles is $n-3$ Commented Nov 17, 2017 at 4:32

The maximum number of reflex angles a simply connected $n-$gon can have is $n-3$. This comes from the fact that the sum of the internal angles of a simply connected $n-$gon is $(n-2)180^\circ$ so there can be at most $n-3$ angles greater than $180^\circ$. To prove we can achieve this, take an equilateral triangle and bend one edge inward to make $n-3$ angles, all slightly greater than $180^\circ$. Below is a heptagon with four reflex angles. I think it is clear how to get as many reflex angles you want up to $n-3$

• @Khosrotash: Since you say (on your profile) you are a PhD candidate, I think you can do all that by yourself. Did you even try? Commented Nov 17, 2017 at 14:18
• I don’t know what you mean by “in mathematics language “. I showed the total of all the angles is too high at $n-2$ and a construction to achieve $n-3$ Commented Nov 17, 2017 at 15:53
• @Khosrotash: I'd say that I find it very unbelievable that a PhD student cannot write down a rigorous translation of Ross' answer. There are a hundred different ways: regular polygons, concave curve with vertical lines, iterative procedures, ... Commented Nov 17, 2017 at 18:23
• @Kevin: It's necessary for the theorem to be even true, since a 5-point star doesn't have the same sum of (directed) internal angles measured along the (directed) polygonal path. In fact, it is non-trivial to prove this theorem rigorously! In any case, Ross is referring to the interior, while you could instead say "simple (non-self-intersecting) polygon". Commented Nov 17, 2017 at 18:26
• @Kevin - in this case simply-connected just means that the boundary is not allowed to intersect itself, or be in more than one piece (for example, if one triangle is inside another, the region between them does not count as a hexagon). Occasionally people do allows such beasts to be called polygons, so Ross Millikan is just being careful in his wording to indicate that his analysis does not include them. Commented Nov 17, 2017 at 18:26

In non-Euclidean space, a digon can have both of its interior angles be reflex angles. On a sphere, it could look like a pac-man. This can be extended to any polygon which means you can have all sorts of polygons that are entirely reflex angles in large quantity.

In Euclidean space with $n_s$ polygon sides, you're limited to $n_s-3$ reflex angles. Why? First, consider that a reflex angle is between 180 and 360 degrees. Then consider the simplest possible polygon: a triangle. In Euclidean space, it is impossible for any of a triangle's interior angles to be larger than $180^\circ$, given the sum interior angle of a polygon is: $(n_s – 2)180$

If we take the lower limit of the reflex angle as $180^\circ$ in order to fit the most reflex angles possible into our polygon, we can then express this as an inequality. Given a number of reflex angles $n_r$,

$n_r 180 \lt (n_s-2) 180 \to n_r \lt n_s-2$

As we can only have integer numbers of angles, that means the maximum solution that doesn't violate the inequality is an increment:

$n_r \lt n_s-2$

∴ $n_r + 1 = n_s-2 \to n_r = n_s-3$

• On a sphere you can surround yourself with a polygon that has no reflex angles, and then declare yourself to be on the outside. (In hyperbolic space your best bet is to use Ross's construction on a polygon which is small enough to look near-Euclidean...) Commented Nov 17, 2017 at 15:26