I recently found myself creating a set of regular polygons for a worksheet. I used GeoGebra to create them, and then free-handed the zoom in order to get them consistently sized. This led me to wonder what “consistently sized” would mean when it comes to polygons.
There are six basic values of a regular n-gon: Side length, radius length, apothem length, area, height, and width (the last two of which only making sense if the orientation of the polygons is fixed). Any of these could be set to a constant. Assume that the first two points will be on the x-axis. Which is best for creating a set of polygons that look like they’re all about the same size?
To explore this in GeoGebra, I created variables called sides (for the number of sides of the polygon), constant (for the specific value I’d be controlling), and displace that would determine the anchor points of the polygons. I created two points, (-displace, 0) and (displace, 0), then created a regular polygon through these points with sides sides.
The easiest value, side length, is also the worst choice. In this case, displace = constant/2. The polygons grow in size, and eventually become unmanageable (I’m showing the first six polygons overlapping here):
Using the radius length limits the polygons to a circumcircle with a constant radius (displace = constant sin(π / sides)). The polygons fairly quickly converge to the size of this circle, but the triangle and square look smaller than the rest:
Likewise, using the apothem length leads to polygons slightly bigger than an incircle with a constant radius (displace = constant tan(π / sides)). As with the radius, the polygons converge fairly quickly, but the triangle stands out more, this time being larger. Also, the even sides polygons look smaller than their odd sided neighbors to the right:
Of these three options, the second one looks best. Before proceeding with the other values, let’s consider two averages of the radius and the apothem, to see if either of those smooth out the problems at the lower sizes. First, the arithmetic mean (\(\frac{a + b}{2}\), so displace = constant (tan(π / sides) + sin(π / sides)) / 2):
This helps the triangle a little bit, but doesn’t really affect the square. Next is the geometric mean (\(\sqrt{a \times b}\), so displace = constant sqrt(tan(π / sides) sin(π / sides))):
I think this is the best of the options so far. However, there are three other values to consider.
At this point, the mathematics does get a bit trickier. Using area isn’t too bad, but the relative size of the polygons shrinks, so I tripled the constant value to get polygons sized comparably to the previous examples.
To get a polygon with n sides that has an area of a, the side length is \(\sqrt{\frac{a \tan{\frac{\pi}{n}}}{n}}\) (displace = sqrt(constant tan(π / sides) / sides)):
This is comparable to the geometric mean, but not identical. The triangle is slightly larger, for instance.
The height and width both require conditional statement. In the case of the height, polygons with an even number of sides have a height that is twice the apothem, yielding a side length of \(\frac{\tan{\frac{\pi}{n}}}{2}\) times the constant value. Meanwhile, polygons with an odd number of sides have a height that is the sum of the apothem and the radius. A bit of trigonometry shows that the side length is \(\tan{\frac{\pi}{2n}}\) times the constant value. This is used for the variable: displace = constant If[Mod[sides, 2] ≟ 0, tan(π / sides) / 2, tan(π / (2sides))], with the constant kept tripled:
This is the most satisfying set so far, in my o
pinion.
Using a constant width involves the trickiest mathematics of all. Polygons with doubly even (that is, multiples of four) numbers of sides have a width that is the same as their height, so we can use the same formula. Polygons with singly even (that is, even but not multiples of four) numbers of sides have a width that is twice their radius. In this case, the side length for a width of one will be \(\frac{\sin{\frac{\pi}{n}}}{2}\).
In the case that a polygon has an odd number of sides, its width will be the longest diagonal. I plan to demonstrate this in the next post, but the result of the relevant trigonometry is that the side length for a width of one will be \(\sin{\frac{\pi}{2n}}\).
Overall, then, the variable will be: displace = constant If[Mod[sides, 4] ≟ 0, tan(π / sides) / 2, If[Mod[sides, 4] ≟ 2, sin(π / sides) / 2, sin(π / (2sides))]], with the relevant set being:
In this case, the doubly even polygons stand out as largest.
It can be demonstrated that the width is always greater than or equal to the height. For an even number of sides, the proof is trivial, as it follows from the argument above. For an odd number of sides, the longest diagonal is the hypotenuse of a triangle formed by that segment, the diameter formed by an apothem and a radius, and half a side length. Since the height is the diameter formed by an apothem and a radius, the width must always be greater than the height.
In general, then, I would say that the height is the best value to use to create a consistent looking set of polygons. Of course, this demonstration is partly skewed by the presentation above, showing the polygons in a line, but I think even that aside, the height does the best job of bringing the size of the triangle and square in line with the others.
1 thought on “Polygon Sets”