Talk:Polygon

Archives

This page has archives. Topics inactive for 365 days are automatically archived 1 or more at a time by if there are more than 4.

Can We Add Names for Fake Fractional Polygons?

Latest comment: 3 years ago10 comments4 people in discussion

What's a $\frac{1}{2}$ -gon's name? Or a $7 \frac{13}{19}$ -gon too? We need the fractional names! Mariomaker-4 (talk) 22:23, 13 March 2023 (UTC)Reply

Wikipedia is not the place for publishing or asking about neologisms and original research. And this talk page is not the place for anything other than possible improvements to the polygon article based on published references. You need to look elsewhere for your speculation. —David Eppstein (talk) 22:33, 13 March 2023 (UTC)Reply

Maybe somewhere else? I didn't mean just on Wikipedia. Mariomaker-4 (talk) 22:49, 14 March 2023 (UTC)Reply

To repeat: this talk page is not the place for anything other than possible improvements to the polygon article. Other discussion does not belong here. —David Eppstein (talk) 22:55, 14 March 2023 (UTC)Reply

A 713⁄19-gon, Schläfli symbol

{\frac{146}{19}}

, is one of many 146-grams (the denominator can be any odd number 1<n<73). So what? —Tamfang (talk) 07:02, 17 March 2023 (UTC)Reply

And your { 1⁄2 } is a double cover of a henagon. —Tamfang (talk) 04:33, 18 March 2023 (UTC)Reply

Thank You! Also, can you make the page for a 146-gon? I don't want it to be a redirect. It's true name, I think, is a Hecatontetracontakaihexagon. I'll get work on the Regular Polygon db of it. It sadly won't be completely finished, but it's okay, right? Mariomaker-4 (talk) 15:16, 18 March 2023 (UTC)Reply

NO. There is nothing to say about this polygon, specifically, that is not true of polygons in general. It is not notable, in the sense that Wikipedia demands for articles: there are not multiple publications that cover it in-depth, independent of other polygons. —David Eppstein (talk) 15:38, 18 March 2023 (UTC)Reply

What if I make a... never mind. I can't even make websites! I'll ask someone else to make the page! Yeah, that's a good idea! Also, for now, it can be a redirect. Okay? I'm very sorry. :) Mariomaker-4 (talk) 13:24, 25 March 2023 (UTC)Reply

Didn't see my talk edit yet?I would like you to check before long. 219.104.224.225 (talk) 07:51, 19 March 2023 (UTC)Reply

A polygon does not have to be a plane figure; it can also be three-dimensional (since nothing about a shape made from line segments with endpoints each shared by exactly two segments restricts the shape to two dimensions).

"Enneacontahexagon" listed at Redirects for discussion

Latest comment: 6 months ago2 comments2 people in discussion

The redirect Enneacontahexagon has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2026 April 4 § Enneacontahexagon until a consensus is reached. 1234qwer 1234qwer 4 11:21, 4 April 2026 (UTC)Reply

96-gon is also listed with the same discussion. Jay 💬 19:41, 26 April 2026 (UTC)Reply

"All the 2d shapes" listed at Redirects for discussion

Latest comment: 7 months ago1 comment1 person in discussion

The redirect All the 2d shapes has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2026 April 4 § All the 2d shapes until a consensus is reached. 1234qwer 1234qwer 4 11:26, 4 April 2026 (UTC)Reply

Semi-protected edit request on 10 November 2026

Latest comment: 1 day ago2 comments2 people in discussion

This edit request has been answered. Set the |answered= parameter to no to reactivate your request.

ADD A NEW SECTION AFTER "Etymology": "

==Application==

polygons are widely applicable, not just because common shapes like rectangles and triangles fall exactly within this category. Any enclosed two-dimensional shapes (even circles and shapes with curved sides) can be approximated by polygons. The accuracy of this approximation can always be increased by adding more vertices. Transformations are also easily applicable. The polygon-approximation of a transformed shape can be formed by applying the transformation to each individual vertex-point of the polygon-approximation of the original shape.

The amount of calculation scales to the amount of vertices but the calculation methods stay the same for every polygon. This makes it ideal for general enclosed shapes inside computersoftware. "

ADD EXPLANATION TO "Area" SECTION: replace "If the polygon is non-self-intersecting (that is, simple), the signed area is " with: "

Every polygon is within its own x-domain. Within these x-positions, there are edges on the upper side and on the lower side. Assuming that the polygon doesn't intersect itself, all the edges on the upper side will have a different direction between left and right than the lower edges. The area can be determined by stretching all the upper parts towards the x-axis and determine its area. Also do this with the lower part and subtract its area from the higher area.

This area is determined for every edge and gets added and subtracted. The area of the trapezoid (created by stretching the edge to the x-axis) is calculated by multiplying the average y position of both points to the x-domain size:

$A_{1}_{,}_{2} = \frac{1}{2} (y_{1} + y_{2}) (x 1 - x 2)$

This results in negative areas underneath the y axis. Also, by defining the width of the trapezoid by the difference between start- and endpoint, the area (above the x-axis) becomes positive when the edge moves to the left and negative when it moves to the right. This automatically results in the areas in both directions being subtracted from each other. The resulting formula for the signed area is: "

Add after "...This is commonly called the shoelace formula or surveyor's formula.": " If the polygon is not a simple polygon and intersects itself, it is partially clockwise and counterclockwise. In this case, the area of the clockwise part gets subtracted by the counterclockwise part. "

ADD A NEW SECTION AFTER "Area": "

=== Center of mass and second area moment ===

Traversing edges to trapezoids to determine the area, can also be applied to other surface-superposition properties like the. second moment of area or the center of mass-area product. First the edge is defined as y as a function of x:

$y = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) + y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x - \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} + y_{1}$

The trapezoid in question is the edge stretched to the x-axis.

CM(x-position)-area product of the trapezoid:

$x_{z} \cdot A = \int_{x_{1}}^{x_{2}} x y d x$

$= \int_{x_{1}}^{x_{2}} x (\frac{y_{2} - y_{1}}{x_{2} - x_{1}} x - \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} + y_{1}) d x$

$= \int_{x_{1}}^{x_{2}} (\frac{y_{2} - y_{1}}{x_{2} - x_{1}} x^{2} - \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} x + y_{1} x) d x$

$=_{x_{1}}^{x_{2}} [\frac{1}{3} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x^{3} - \frac{1}{2} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} x^{2} + \frac{1}{2} y_{1} x^{2}]$

$= \frac{1}{3} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{2}^{3} - \frac{1}{2} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} x_{2}^{2} + \frac{1}{2} y_{1} x_{2}^{2} - \frac{1}{3} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1}^{3} + \frac{1}{2} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} x_{1}^{2} - \frac{1}{2} y_{1} x_{1}^{2}$

$= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (\frac{1}{3} x_{2}^{3} + \frac{1}{2} x_{1} x_{2}^{2} - \frac{1}{3} x_{1}^{3} + \frac{1}{2} x_{1} x_{1}^{2}) - \frac{1}{2} y_{1} x_{2}^{2} - \frac{1}{2} y_{1} x_{1}^{2}$

Final formula:

$x_{z} \cdot A = \frac{y_{2} - y_{1}}{6 (x_{2} - x_{1})} (2 x_{2}^{3} + 3 x_{1} x_{2}^{2} - 2 x_{1}^{3} + 3 x_{1} x_{1}^{2}) - \frac{1}{2} y_{1} (x_{2}^{2} + x_{1}^{2})$

The second area moment of the trapezoid around the y-axis

$I_{y} = \int_{x_{1}}^{x_{2}} x^{2} y d x$

$= \int_{x_{1}}^{x_{2}} x^{2} (\frac{y_{2} - y_{1}}{x_{2} - x_{1}} x - \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} + y_{1}) d x$

$= \int_{x_{1}}^{x_{2}} (\frac{y_{2} - y_{1}}{x_{2} - x_{1}} x^{3} - \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} x^{2} + y_{1} x^{2}) d x$

$=_{x_{1}}^{x_{2}} [\frac{1}{4} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x^{4} - \frac{1}{3} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} x^{3} + \frac{1}{3} y_{1} x^{3}]$

$= \frac{1}{4} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{2}^{4} - \frac{1}{3} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} x_{2}^{3} + \frac{1}{3} y_{1} x_{2}^{3} - \frac{1}{4} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1}^{4} + \frac{1}{3} \frac{y_{2} - y_{1}}{x_{2} - x_{1}} x_{1} x_{1}^{3} - \frac{1}{3} y_{1} x_{1}^{3}$

$= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (\frac{1}{4} x_{2}^{4} - \frac{1}{3} x_{1} x_{2}^{3} - \frac{1}{4} x_{1}^{4} + \frac{1}{3} x_{1} x_{1}^{3}) + \frac{1}{3} y_{1} x_{2}^{3} - \frac{1}{3} y_{1} x_{1}^{3})$

Final formula:

$I_{y} = \frac{y_{2} - y_{1}}{12 (x_{2} - x_{1})} (3 x_{2}^{4} - 4 x_{1} x_{2}^{3} - 3 x_{1}^{4} + 4 x_{1} x_{1}^{3}) + \frac{1}{3} y_{1} (x_{2}^{3} - x_{1}^{3})$

As with area, the properties of these trapezoids can be added to the total. Te determine the center of mass y-position or the second area moment, you can use the formulas with x and y switched. "

ADD A NEW SECTION BEFORE "Generalisations" (after "Properties and formulas"): "

==Point within a polygon==

Determining whether a point lies within a polygon can be valuable for various reasons. It can be used to make the polygon clickable or it can be used to render the polygon inside a bitmap structure. These processes need to be done in large quantities, which is why it is important make is as fast and easily as possible.

By drawing a ray from a point, the line will intersect the edges of the polygon a certain amount of times. At each intersection, the line will in turn either go inward or outward. Because the polygon can't take infinite space, the line will eventually end up outside the polygon, meaning the last intersection is always outward.

Following this logic, the whether the point lies within the polygon can be determined easily. If the first intersection of the ray from point in question is outward, then the point is inside the polygon. If the first intersection is inward, then the point is outside the polygon. Because inward and outward intersections take turns, the following can be stated:

Draw a straight line from the point in one direction and count the amount of intersections with the edges. If the amount of intersections is an even number (including 0), then the point lies outside the polygon. If the amount is an odd number, then the point lies inside the polygon.

This can be programmed as follows: Look at each edge of the polygon. Determine if the y position of both vertices are above or below the y position of the point. If one is higher and the other is lower, then the edge intersects with the vertical line from the point. Determine the x position of the intersection through interpolation and see if this x position is higher that that of the point (only look at the intersections at one side). Count them and see if the amount is even (n % 2 == 0) or odd (n % 2 == 1). "

Thank you StanHennink (talk) 11:07, 10 November 2026 (UTC)Reply

Edit requests are for simple changes, hopefully non-controversial. This request consists of the suggestion to add several sections. As presented, these sections are unsourced, and it is unclear whether they are relevant.

So the right procedure is to open sections in the talk page, one per proposed new section, which contain the proposed text and the reasons for including them in the article. D.Lazard (talk) 14:10, 10 November 2026 (UTC)Reply