# Talk:Regular polygon

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## Hexagon Formula

Can someone fix the hexagon formula? It's showing up as markup rather than a nice formula image and I don't know enough about the markup to know what's wrong. 66.162.50.98 (talk) 21:48, 12 November 2008 (UTC)

## Angles

How do you find the interior and exterior angles of a regular polygon, especcially octogon, nonogon, and hexagon.--User:Rat235478683--

The general formula is already there. I just added a subheading so you can find it. -- Steelpillow (talk) 15:28, 8 May 2008 (UTC)

## Request for suggestions

I have made my first 'noticeable' on Wikipedia. That is, not simply running a spell check. Could someone please give me some constructive criticism on how to better improve this article? 69.14.16.239 (talk) 23:52, 7 May 2008 (UTC)

Sorry, I just realized that I was logged out. I am now logged in with my real screen name.Thestarsprism (talk) 23:54, 7 May 2008 (UTC)

## rules

I want to know the rules for a regular polygon —Preceding unsigned comment added by 207.118.3.236 (talk) 23:03, 6 April 2009 (UTC)

Tell us what sort of rules you need, and we will be happy to supply them. Do you mean the methods of construction? Dbfirs 14:48, 7 April 2009 (UTC)

## convex or star

The article says that all angles of a polygon must be equal, however, it goes on to say that regular polygons can be convex or star. This is not possible, because star shapes have reflex angles as well as acute angles. Therefore, a star polygon cannot be regular because it has reflex angles. The information stating that a regular polygon can be a star should be removed. —Preceding unsigned comment added by 24.113.98.113 (talk) 02:04, 27 April 2009 (UTC)

You have a very good point there. I've tried to clarify, but there is probably a better way to do this. Perhaps we should split "regular star polygon" away from the standard definition of regular (convex) polygon. What does anyone else think? Dbfirs 18:17, 27 April 2009 (UTC)
Regular star polygons are very well defined. All angles are acute. Poinsot defined them in the same paper he described the four regular star polyhedra (the fact that both the regular star polygons and polyhedra have only acute angles even led him to think of them as convex figures, an idea which did not lose favour for over a hundred years). The requirement for equal angles, as stated at the beginning of the lead, is sufficient to ensure that the given equal-sided polygon is regular, whether it is convex or star. No other explanation is needed in the lead, so I have undone the addition.
A polygon which has equal sides and alternating reflex and acute angles is sometimes called a star polygon, but this usage is not well defined. Certainly such a figure cannot be regular. This article is specifically about Regular polygons. A star which is not regular is irrelevant, and consequently is excluded by the definition given in the lead.
As for splitting off the regular stars, that is quite unhelpful. We would then have to split off Regular convex polygon too, and use this article to stitch the subject back together. Urrrgh. You might like to take some time out to contemplate Schläfli symbols - light may then begin to dawn.
-- Cheers, Steelpillow (Talk) 20:36, 27 April 2009 (UTC)

Yes, point taken, and I am quite happy with your viewpoint from a mathematical point of view, but could we clarify somehow (not in the way that I attempted) to avoid misunderstanding by non-mathematicians? Perhaps we could just say that, in star polygons, only the acute angles count as internal angles. Dbfirs 13:33, 28 April 2009 (UTC)
The article begins, "A regular polygon is a polygon which is equiangular (all angles are equal in measure) and equilateral (all sides have the same length). Regular polygons may be convex or star." I do not know how it can be made clearer than that - I can see no room for misunderstanding, provided I suppose that one has read what is written. Whether a polygon is convex or star is neither here nor there in considering whether it is regular, it just happens that all (geometrically) regular polygons may be divided into these two classes. The rest of the article goes on to enlarge on this in detail. (Irregular polygons of any kind, whether star or convex or anything else, are utterly irrelevant to this article.) -- Cheers, Steelpillow (Talk) 15:59, 28 April 2009 (UTC)
For someone who has never encountered star polygons before I think there is a chance of confusion. With the pentagram there are 10 visible angles, the novice really needs pointing out that only five of these are important. A simple diagram indicating the important angles and edges could clear this up. --Salix (talk): 18:27, 28 April 2009 (UTC)
For someone who has never encountered star polygons before, the article already links to star polygon in both the lead and in the section on regular stars. The article does not explain what a polygon is, nor does it explain regularity or convexity. Sh why on Earth should one expect it to explain starriness either? It links to the relevant articles for all of these, and that is how it should be. -- Cheers, Steelpillow (Talk) 08:35, 29 April 2009 (UTC)
What it does not explain is that a two-dimensional representation of a star polygon is misleading because it creates extra angles. For those of us who can imagine a Klein bottle, this is no problem, but the statement that all angles are equal is obviously false for the drawn representation of a star pentagon seen as a pentagram (this was the point raised by 24.113.98.113). Dbfirs 21:14, 29 April 2009 (UTC)

## Area

I disagree with the equation stated as the area for a regular polygon:

$A = \frac{1}{2} n r^2 \sin \frac{2 \pi}{n}$

I believe it is:

$A = \frac{1}{2} n r^2 \sin \frac{\pi}{n}$

I have proved this years ago, tested it many times, found the limit, and even took it to my college professor who agrees with my proof and equation. I can even present my proof if anyone would like. Jeanlovecomputers (talk) 22:22, 29 December 2009 (UTC)

A quick check of the first equation with the unit square (n = 4, r = √2 / 2) gives
A = 1/2 * 4 * (√2 / 2)^2 * sin ( π / 2)
= 1/2 * 4 * 1/2 * 1 = 1
A = 1/2 * 4 * (√2 / 2)^2 * sin ( π / 4)
= 1/2 * 4 * 1/2 * (√2 / 2) = √2 / 2
I.e. wrong. You could post your proof if you want but I think the numerical accuracy and correctness of the current formula is pretty clear.--JohnBlackburne (talk) 22:32, 29 December 2009 (UTC)
I already replied with the same example on my talk page. It resulted in silence, so I guess Jean got the message. Cheers, DVdm (talk) 23:12, 29 December 2009 (UTC)
Yes I saw after posting as I was sure this had come up before. Can't really think of an easier one to do though than the unit square.
To Jeanlovecomputers have a look at Triangle#Using trigonometry, the current formula can be easily derived from that, a standard (i.e. I learned it at high school) formula, and the picture makes it clearer. --JohnBlackburne (talk) 23:18, 29 December 2009 (UTC)

Okay, I stand corrected as I look back at my proof. However, I do have a different equation than what is on the page:

$A = n \cdot r^2 \cdot \sin \frac{\pi}{n} \cdot \cos \frac{\pi}{n}$

Each regular polygon can be divided into triangles, where one side s is a side of the polygon, giving n triangles. This leaves the center angle L and corner angle R:

$L = \frac{2 \pi}{n}$

$R = \frac{\pi \cdot (n - 2)}{2n} = \frac{\pi}{2} - \frac{\pi}{n}$

Because each of these triangles are symmetrical, they can be divided by 2, which will be easier to find their area B:

$B = \frac{1}{2} \cdot \left(\frac{1}{2} \cdot s\right) \cdot a = \frac{1}{4} \cdot s \cdot a$

Using trigonometric functions, we can solve for s and a:

$\cos R = \frac{s}{2r}$

$s = 2r \cdot \cos R$

$s = 2r \cdot \sin \frac{\pi}{n}$

$\sin R = \frac{a}{r}$

$a = r \cdot \sin R$

$a = r \cdot \cos \frac{\pi}{n}$

We now substitute s and a into B:

$B = \frac{1}{4} \cdot s \cdot a = \frac{1}{4} \cdot 2r \cdot \sin \frac{\pi}{n} \cdot r \cdot \cos \frac{\pi}{n} = \frac{1}{2} \cdot r^2 \cdot \sin \frac{\pi}{n} \cdot \cos \frac{\pi}{n}$

Because we divided by 2 to get the smaller area, we now multiply by 2 and n to get the total area of the regular polygon A:

$A = \frac{1}{2} \cdot r^2 \cdot \sin \frac{\pi}{n} \cdot \cos \frac{\pi}{n} \cdot 2n = n \cdot r^2 \cdot \sin \frac{\pi}{n} \cdot \cos \frac{\pi}{n}$

If r remains the same as n increases, we know that s will eventually lead to 0. In other words, as the number of sides increases to infinity, the regular polygon becomes a circle. We know the area of a circle is:

$A = \pi \cdot r^2$

Stating this, we can take the limit of a regular polygon:

$\lim_{n \to \infty}n \cdot r^2 \cdot \sin \frac{\pi}{n} \cdot \cos \frac{\pi}{n} = \pi \cdot r^2$

Do you agree with this? I am sorry for my previous false equation. My mistake. Jeanlovecomputers (talk) 00:04, 30 December 2009 (UTC)

This
$A = n \cdot r^2 \cdot \sin \frac{\pi}{n} \cdot \cos \frac{\pi}{n}$
Is equivalent to the formula on the page. See Double-angle formula, where there's
$\sin 2\theta = 2 \sin \theta \cos \theta\$
which connects them --JohnBlackburne (talk) 00:13, 30 December 2009 (UTC)
So are we good now? I feel stupid for putting up the wrong equation and not even looking at my own proof beforehand. Jeanlovecomputers (talk) 00:17, 30 December 2009 (UTC)
We all get it wrong sometimes, so I wouldn't worry about it. Posting here instead of on the page was definitely the right thing to do if you're not sure about it.--JohnBlackburne (talk) 00:24, 30 December 2009 (UTC)
Minor quibble ;-) - DVdm (talk) 09:26, 30 December 2009 (UTC)

## Question

I'm asking this question after an argument in real life: how is the area of a star polygon defined? In the pentagram, is the central pentagon part of the pentagram? Or should it be counted twice in the area? It gets worse for heptagrams (7/3). 4 T C 12:24, 31 December 2009 (UTC)

I don't know there is a standard definition. Looking at Area it mentions a "region bounded by a closed curve", and there are various curves you could use here. I would go with the area of the pentagon + 5 triangles, but you can make a case for counting the pentagon twice or not at all. Really up to you. --JohnBlackburne (talk) 12:38, 31 December 2009 (UTC)
It depends on one's area (sorry) of interest. For example if I were cutting one out of plywood, I would count the centre once. But the nearest thing to a standard definition in mathematics is the generalisation of what we now call simplicial decomposition (i.e. dividing into triangles) that Cayley made about 150 years ago when he first figured out how density works: this method ends up counting it twice. But even this method runs into trouble with the volumes of non-orientable polyhedra. With respect to a "region bounded by a closed curve", the real difficulty is to understand what we mean by "bounded by" in such circumstances. BTW, if anybody can provide a respectable reference for not counting the central pentagon at all (effectively treating the figure as an embedding of a Möbius band), I would be most interested. -- Cheers, Steelpillow (Talk) 21:35, 31 December 2009 (UTC)
The only references I've seen (I just Googled 'pentagram area') treat the pentagram as a whole, i.e. the area is everything inside the outer boundary. But if you were to argue that the internal boundary was also an edge, and edges have to separate filled and unfilled regions, then you could say the centre must be unfilled and so not part of an area. Alternately use the winding number to argue it must be counted twice. As the original poster noted start polygons with 7 or more vertices can have higher winding numbers which makes both calculations far more complicated. --John Blackburne (wordsdeeds) 22:15, 31 December 2009 (UTC)
Cayley's (density) method is simple to use for any regular polygon, whether convex or star. For any {n} or {n/q}, mark the centre P. Draw rays from P to the two ends A, B of an edge. The area of the polygon is then just n times the area of the triangle ABP (where the angle APB = 360 deg * q / n). -- Cheers, Steelpillow (Talk) 20:27, 1 January 2010 (UTC)

## Area Formula

This method has been moved here from the article octagon. It needs some re-writing, then it could possibly go in this article? Dbfirs 21:34, 24 January 2010 (UTC)

Any regular polygon can be divided into numbers of equal right-angled triangles, so the area can also be calculated more simply, in the case of a polygon with an even number of sides, by taking the distance between any two opposite sides (A), dividing by two and then multiplying by the length of one side(B), divided by four and then multiplying by twice the total of the number of sides (N) as follows:

((A/2) * (B/4)) * 2N or just ABN/4

This works for any regular polygon by just changing the value of N provided that n is even. (For regular polygons with an uneven number of sides A is calculated as the distance between the point of one angle to the mid point of the side opposite.) diameter of the incircle.

The problem is not only do you need to know N, you need to know A and B too. And if you don't know them you have to calculate them. They are related by trigonometry to each other and the circumradius through the right angled triangles you mention, so it's not too difficult, but you end up with the current formula, i.e. it's no more simple.--JohnBlackburnewordsdeeds 21:46, 24 January 2010 (UTC)
Yes, I agree as a mathematician that it needs two measurements and so is "less elegant", but for finding the area of a regular polygon in the real world (outside our mathematical world of roots and trigonometry), this method is so much simpler that it is worth recording in any regular polygon article that might be read by a non-mathematician. Dbfirs 21:52, 24 January 2010 (UTC)
One flaw in your formula: it only works for polygons with an even number of sides, so it's also not as general. But even without this concern I think the current formula, which gives the area in terms of one easily measured quantity, is preferable.--JohnBlackburnewordsdeeds 22:55, 24 January 2010 (UTC)
Yes, I'm checking the version claimed for odd n before I post it anywhere. Of course the standard formula is preferable for mathematicians, but have you tried asking engineers, or the general public, which they prefer? I'm not proposing to give it status equal to that of the standard formula, but, for many readers, this simple "measure and multiply" formula is the only one that they will feel confident to use. Dbfirs 00:21, 25 January 2010 (UTC)
It's not my formula, by the way, it was posted by anon editor 88.105.90.141, and is the only contribution to Wikipedia from that IP address. I wonder who it was. I think I've seen something similar before (many years ago), but I hadn't remembered the details. I wonder if it is published anywhere. It would be good to find a general proof - I've only proved it so far for specific even values of n. Now that I'm awake, I realise it is trivial! Dbfirs 00:25, 25 January 2010 (UTC)
...(later) ... Now that I've checked, I realise that you are right (of course). This is really just another way of expressing the perimeter formula, and the anon's adjustment for odd n is not correct. I've altered the statememt above, but this means that this is not a useful general formula for odd n, and we already have the perimeter version, so it is probably not worth adding as a method for a general polygon. (Possibly we could express the formula in terms of the "in-radius", but I now share your doubts over whether it is worth the effort for any odd n.) The simplified formulae for even n are worth adding to the specific articles (as I've already done with hexagon, octagon, decagon & dodecagon. Dbfirs 01:33, 25 January 2010 (UTC)
Indeed, this ABN/4 for an octagon (and for any even-numbered regular polygon) is just a trivial special case of the general formula in the section Regular polygon#Area: A=1/2 Nsa with s=B and A=2a. I don't think these a need for this in this article. DVdm (talk) 08:31, 25 January 2010 (UTC)
Agreed. I hadn't realised that we already have it in this article in an alternative form. I moved it from Octagon because it was inappropriate there ( - too general). I've just adapted it for specific even-n polygons where it provides a simple calculation of area from two easy measurements, but this is the limit of its usefulness. Dbfirs 09:13, 25 January 2010 (UTC)

## Area table

I have reverted the table to this version for the following reasons:

1. The general formula should come first. Ssgxnh moved it back to the bottom. I'd like to insist having first.
2. The reason for the empty cells is WP:UNSOURCED: is it true that "For these polygons the exact area does not simplify out of trig functions"? Perhaps some do simplify, but we just don't know. One of the non-empty cells had the marker as well. I.m.o. an empty cell is really ok. We don't have to put reasons for being empty in there.
3. There is no reason why we should fear the cot-function and replace it with cos/sin. The latter formula does not belong in the approximated values column, not even with parenteses.

Ssgxnh, please propose and discuss on talk page before you make further changes? - Thanks - DVdm (talk) 21:12, 4 August 2010 (UTC)

I have also removed the formula -7 cos(2 Pi/7) / ( 4 sin(2 Pi/7) - 4 ) for the heptagon. It doesn't bring anything new and is more tedious to calculate than the general cot-formula. Furthermore that expression should haven been simplified to 7/4 cos(2 Pi/7) / ( 1 - sin(2 Pi/7) ) and then to 7/4 cot( Pi/7 ), but we already have that in the general formula on top.

Afterall, the idea of the exact values column was to have alternative expressions that do not involve trig functions. DVdm (talk) 07:43, 5 August 2010 (UTC)

The paragraph after the table says: "The amounts that the areas are less than those of circles with the same perimeter, are (rounded) equal to . . . ." As a non-mathematician Math Olympiad coach, I'd love to see a table or graph here showing the increasing area of regular polygons of equal perimeter, approaching but never reaching the area of a circle with the same circumference (last entry could be a circle). The existing chart keeps the side lengths at 1 unit, which gives astronomical areas for higher-number n-gons but doesn't make it easy to directly compare the areas of the various polygons. Thanks. JayBeeEye (talk) 17:54, 1 November 2011 (UTC)

Also, shouldn't the "exact area" for the square be the formula s^2 and the "approximate area" be "1"? Seems like the "1" is in the wrong column, even though it is exact, not approximate. Thanks. JayBeeEye (talk) 18:09, 1 November 2011 (UTC)

Done. That's a good idea. I have expanded the table with 3 colums for r=1 and ditto for a=1. The calculations are straightforward and were calculated with Maple 13. I guess we can assume that we have a consensus that this conforms to the spirit of wp:CALC. In any case, I have added a reference to the defined MAPLE function. DVdm (talk) 17:00, 4 November 2011 (UTC)

## 100,000

What is a 100,000 sided polygon?--90.217.236.77 (talk) 21:00, 31 May 2011 (UTC)

An almostcirclegon. DVdm (talk) 17:18, 4 November 2011 (UTC)
I learned it as a closenufegon but maybe it's different in Europe. JayBeeEye (talk) 04:06, 5 November 2011 (UTC)