Is there an equation to describe regular polygons?


Is there an equation to describe regular polygons?

So a blog reader, Michael Gmirkin, sent me an in-depth email about the possibility of the existence of a super equation that can describe any regular polygon. I wasn’t sure. For reference, you might want to check out these 2 blog posts about the equation for a square: question, answer.

I was going to just ask you here. Then I remembered there’s a Stack Overflow equivalent for maths. So I went there and asked the question. So if you know the answer, you can comment here, or go to the maths StackExchange site and earn yourself some points.

You can assume that the centre of the regular polygon in at the origin (0,0). Researching a little on the topic, I also learnt about the apothem, which is also the shortest distance from the centre to a polygon’s side. The “normal” radius is the distance from the centre to one of the regular polygon’s vertex.

If you trace 2 circles, one with the apothem and one with the radius, you get an inscribed circle and circumscribed circle respectively.

4 thoughts on “Is there an equation to describe regular polygons?

  1. Yes, the apothem of a polygon is the distance to the center of a side of a regular equilateral polygon. It is also the shortest distance to an intersection with the polygon and will be perpendicular to the side. It is the radius of an inscribed circle.

    Also, the circumradius is basically the same thing except it’s the distance to a corner or vertex of the same polygon. The line segment bisects the vertex angle and the distance is the radius of a circumscribed circle.

    with the help of a friend, and my dad, I now have a “Holy Grail” type equation that, describes any regular equilateral polygon in terms of angle theta and distance (radius). When graphed in polar coordinates, a perfect polygon is drawn and discrete values of all points thereon can be computed.

    In fact, I’ve got 3 versions (6 if you consider rotation a factor; to either align a vertex or the midpoint of a side with theta=0). One with circumradius = 1 (as vertices–> infinity, polygons expand outward toward the circumscribed circle), one with apothem = 1 (as vertices–> infinity, polygons collapse inward toward the inscribed circle) and one with the midpoint between circumradius & apothem = 1 (as vertices–> infinity, both the maxima and minima, thus the circumscribed and inscribed circles, collapse toward that ‘midpoint radius’).

    I’d be interested to know whether this approach, describing the radius of a polygon as a periodic function, has any precedent (has anyone else done this, or am I the first)? I’ve been working on this idea for some time (on and off for years), but just recently overcame some stumbling blocks with a little help from a friend and my dad. Most of the legwork was my own, though.

    The relatively final form(s) appear to be:

    (n-gon, circumradius=1, unrotated)

    (n-gon, circumradius=1, rotated -Pi/4)

    (n-gon, function centered around unit circle, unrotated)

    (n-gon, function centered around unit circle, rotated -Pi/4)

    (n-gon, apothem=1, unrotated)

    (n-gon, apothem=1, rotated -Pi/4)

    Don’t know whether they simplify at all to something less complicated… Even if not, they’re beauties!





    I submit these as the Gmirkin Polygon Radius Functions (or some suitably nifty sounding name that’s not too cumbersome). *Smile*

    I may write them up formally for publication at some point, once a few previous engagements clear up, assuming they’ve not previously been published or some directly correlated function has already been published elsewhere. (If so, I’d like to know when, where and by whom; for academic curiosity’s sake.)

    ~Michael Gmirkin

  2. I’ve also cross-posted to Stack Exchange. Feel free to check it out.

    ~Michael Gmirkin


    As to your statement “My friend wanted to know if there’s an equation that creates a square in any rotational orientation. I don’t believe there’s one, so please don’t bother to try. It’ll just waste your brain cells. But you’re welcome to try it as an intellectual exercise.”

    Belief refuted with maths! & my brain cells are still working just fine, thanks! =o]

    In the above formulas, the primary terms are Abs[Cos[(v*x)/4]] & Abs[Cos[((v*x)/4)-(Pi/2)]], added together and the whole thing inverted. Or Abs[Sin[]] + Abs[Cos[]] (the second Cos is just shifted Pi/2 to approximate Sin but get everything in terms of the same function). Anyway, the entire polygon is freely rotatable. All you have to do is add the same value to both primary functions, for instance: Abs[Cos[(v*x)/4 – (Pi/4)]] & Abs[Cos[((v*x)/4) – (Pi/2) – (Pi/4)]] will rotate the polygon by -Pi/4. This is handy for either aligning the center of a side or the vertex of an outer angle of the polygon with theta=0.

    So, yes, a complete description of all regular equilateral polygons is possible and it *is* freely rotatable. Granted it’s in polar coordinates. On wonders whether it translates into something where the polygons themselves are graphable in Cartesian coordinates.?

    Anyway, the numerator controls the length of the circumradius (and by extension one can control the length of the apothem by equivalently scaling the circumradius). So, one can arbitrarily enlarge or shrink the polygon in addition to arbitrarily rotating the polygon. Just multiply the numerator of the given equation you’re working with by the length you want (for the circumradius or the apothem or the radius half way between circumradius & apothem [I call it the ‘proper radius,’ but that’s just me] respectively depending on which formula you’re using), and it’ll be that length.

    It’s all pretty darned cool! Yay for maths!

  4. I retract the above equations. At the behest of someone on another site, I checked in Wolfram Alpha at a few data points. While it appears to work for the Square case (where the coefficients and corrective term basically cancel out), it doesn’t work for other cases, but is slightly off. I think I’ve got the coefficients wrong. Will have to poke around a bit more in the maths to see if it’s possible to get a technically correct exact solution.

    Sorry. Jumped the gun slightly.

    The graphs were so close as to fool me into thinking they were exact for all cases. Will get back to you if/when I get a technically correct solution. ‘Til then…

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