Polygons: formality and intuition

A polygon is a closed geometric figure that consists of points - vertices - connected by straight line segments - the sides (or edges) of the polygon. Vertices are sequenced in a cyclic order, and sides only connect pairs of adjacent vertices. The terminology varies. In some sources (The Harper Collins Dictionary of Mathematics, Harper Perennial, 1991), the term polygon only applies to the cases where the sides do not intersect. Elsewhere, especially when star polygons form an object of study, sides are allowed to intersect. In the latter case, side intersections are not considered as vertices of the polygon.

Assume a polygon has n vertices and m sides. Each side connects two vertices, and each vertex belongs to two edges. (In the cyclic order of vertices, one of the edges may be called incoming while the other is naturally outgoing.) We thus have n = 2m/2. In other words, n = m. This argument shows that a polygon can be unambiguously referred to as an n-gon. There is no need to indicate whether n is the number of vertices or the sides.

The above argument is not, however, flawless. Imagine an 8-shaped polygon with a vertex shared by the two loops. Is it a polygon at all? Why not? Unless the definition explicitly precludes overlapping vertices, an 8-shaped figure fits the definition perfectly. Should the definition be restricted or not?

If it is, we can, as an argument above shows, freely talk of n-gons. Which is a short and a desirable notation. If overlapping vertices are allowed, the notational convenience seems to be lost; for we no longer may expect that the identity #vertices = #sides will necessarily hold. The situation can be saved, however, with the following device. The key here is the word overlapping. Let's agree that two vertices of a polygon may overlap without being one and the same. In other words, the points that physically represent the vertices may coincide, yet we agree to look at the vertices as different. Does this seem counterintuitive or farfetched? Will this threaten our geometric intuition?

Not necessarily. Once you warm up to the idea, it becomes quite natural. It even adds to our intuition somewhat. Here's an example.

In the discussion on the Isoperimetric Theorem I mention the fact that among all n-gons with the same perimeter, the regular n-gon has the largest area. According to the Isoperimetric Theorem, the largest area of all is enclosed by the circle. The circle can be approximated by regular n-gons. The larger is n, the closer is the approximation. Does it follow then that the area enclosed by regular n-gons grows along with n?

Yes, it does. And the proof becomes quite trivial and well-nigh intuitive if we stick with the definition of n-gon that permits vertices to overlap. Assume we know how to prove that indeed, among all n-gons, the regular n-gon encloses the largest area. Can we now show that a regular n-gon encloses a larger area than a regular (n-1)-gon with the same perimeter? What is here to show? An (n-1)-gon is, by definition, an n-gon with some two adjacent vertices overlapping. Looked at as an n-gon, the regular (n-1)-gon ceases to be regular, and, therefore, by our assumption, has an area less than that enclosed by the regular n-gon. Q.E.D.

The argument seems to push our logic even farther. Not only the vertices may overlap, but the sides should be permitted to have zero length. This is weird. What kind of line segments are those that have zero length? However, the argument is fully supported by the geometric intuition. It would be silly to consider only those polygons whose sides have length that is bounded below by a fixed quantity. So, if sides of a polygon may have arbitrary small lengths, why not to apply the "by continuity" argument, and let them vanish altogether? The idea, as we just saw, is quite fruitful.

{n} is the customary notation for a regular n-gon. To include the star-shaped polygons n is permitted to be a positive rational number above 2, although n = 2 is often included. As we know, rational numbers have infinitely many representations as common fractions (5/3 = 10/6 = 15/9 = ...) In general, to construct a star-shaped regular polygon of type {n/d} the fraction is reduced to lowest terms such that n and d become mutually prime. n equidistant points on a circle split the circle into n equals arcs. To obtain an {n/d} polygon, connect n points one after another skipping exactly d arcs at a time.

The renown geometer Branco Grünbaum argues that to carry out this construction, n and d must not be required to be coprime. If they are not, the polygon thus obtained will have its vertices and sides overlap and will appear as a polygon with a smaller n. It is usually said that the construction, in this case, leads to gcd(n,d) overlapping but distinct n/gcd(n,d)-gons. Against the prevailing norm, Grünbaum suggests that the construction yields a single polygon with overlapping vertices. To prove his point, Grünbaum continuously transforms vertices of the polygon by the same amount. Say, the even-numbered vertices slide by an angle t, whereas the odd-numbered vertices slide in the opposite direction, i.e. by the angle -t. Letting t change continuously, it is possible to transform a {14/2}-gon into a polygon of type {14/5}. Which points to the legitimacy of the {14/2}-gon as a single entity as opposed to its being a combination of two distinct 7-gons.

A generalization of Napoleon's Theorem appears to lend support to Grünbaum's point of view as it leads in the same breath to the "conventional" as well as "unconventional" star-shaped polygons.

The applet below illustrates Grünbaum's transformation. It works best when n equals twice a prime. It does not make a lot of sense for odd n because of the loss of symmetry. For n divisible by 4, the transformation may seem to have the effect opposite from the intended one. One consistently observes a polygon with a smaller number of sides. However, uncheck the "Auto update" box and drag the vertices manually. Even in this case there is a single polygon with multiple vertex overlaps.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

What if applet does not run?

Another problem is pertinent to the discussion. Consider a billiard table in the shape of an equilateral triangle. From a point on the base of the triangle, shoot the ball at 60° angle to the base. After covering a symmetric polygonal itinerary, the ball will return to the starting point. It's interesting that the length of the itinerary - the perimeter of the polygon - does not depend on the starting point. (Prove this.) There is one exception though. If the starting point is located in the middle of the base, the itinerary is less intricate and, in fact, reduces to a plain equilateral triangle whose perimeter is twice as small as the perimeter of all other polygons. Grünbaum's continuity argument suggests that we should talk of a family of 6-gons of the same perimeter of which one (corresponding to the middle point of the base) has vertices overlapping in pairs.

31 August 2015, Created with GeoGebra

This might be good from a geometric perspective. But it does not make sense from the point of view of the billiard ball. If shot from the midpoint, the ball travels along an equilateral triangle and, if not stopped upon its return, will do this indefinitely: going over and over the same triangle. What reason is there to combine 2 triangles into 1 hexagon? Why not to combine 3 or more of them as well? (For the names of (regular) n-gons see a page at the Math Forum.)

So, who is right? And what is the right answer? Are vertex overlaps allowed or not? I do not know about you, but, personally, I do not care. Mathematics is flexible. In one context one definition is more appropriate and/or propitious, in another context one may be better off leaning on another definition. I am often asked whether 0 is a natural number. While the existing ambiguity may be deplorable, I always feel uneasy having to answer this question. What in the world do you care? Pick the answer that best suits your goals. It's all the same to me.

In passing, I want to make one more remark. In popular Calculus texts, there is a tendency to restrict function definitions to functions given by algebraic formulas. Until Weierstrass came up with his example of a nowhere differentiable function some 150 years ago, this perception of function was quite common among mathematicians as well. Nowadays, elementary Calculus texts avoid the subject by insisting that functions that appear in practice (whatever this may mean) are all defined by algebraic expressions; all others belong to a higher mathematics.

I do not know. If we measure the distance traversed by the billiard ball in the above problem, how do we represent this distance as a function of a point on the base of the triangle? If we are only concerned with the distance travelled up to the first return to the starting point, what algebraic formula gives you a quantity constant everywhere except for one point, where it is twice as small? (The question is rhetorical. No such expression exists. The function is an example of a "naturally" occurring discontinuity.)

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