Recently I was very taken by a Mathologer video on polygons. Clearly I was not the only one because it already has been viewed about 150,00 times. The video is about the Petr-Douglas-Neumann theorem and I have already been inspired to blog about the special case of triangles (Napoleon's theorem). Today I would like to give a bird's eye view of the theorem and its proof. Then I shall give some thoughts to do with polygons in general which have been suggested by the proof of the general theorem.
First, for completeness, let me briefly describe what the theorem says. It begins with an arbitrary n-gon and then transforms it into a succession of n-2 further n-gons. A typical step Sr takes the current polygon and erects isosceles triangles on each of its sides; each side is the base of an isosceles triangle, and the apex angle of the triangle is an angle chosen from the set A = {360r/n | r =1,...,n-1}. The next polygon is this sequence of triangle apexes (or ears as Mathologer calls them). These ears appear on the left of each side (travelling clockwise) but, when the apex angle is greater than 180, they are on the right of the edge. The theorem's conclusion is that, after a series of n-2 steps with distinct angles chosen from the set A, the final n-gon is regular.
I encourage you to watch the video if only for the compelling animations used to describe the sequence of polygons. You will see (as Mathologer observes) that the polygons become increasing regular looking (whatever that means - but see below!).
Mathologer explains that one must consider every polygon to be an ordered sequence of complex numbers (each complex number representing a point in the Argand diagram). Then there is a special set of n regular polygons Pr (r = 0, 1, 2,..., n-1) where the corresponding sequences of complex numbers has the form (1,wr,w2r,...,w(n-1)r) with w representing a complex nth root of unity. He notes that this set polygons is a set of linearly independent vectors in n-dimensional complex space.
Therefore the initial arbitrary polygon is a linear combination of the n special polygons.
The next observation (which Mathologer only sketches) is that each step Sr which transforms a set of points (considered as a vector) into another set of points is actually a linear transformation. This fact is left to the reader to verify but, actually, it is not very difficult: each component in the new set of points depends on two of the points in the original vector and, when one calculates what it is from the geometrical construction, it turns out to be a linear formula.
The actual linear transformation effected by Sr is a bit too complicated to express in the typography provided by this blogging software (do I hear echoes of Fermat's last theorem?). You can find it in the video by John Harnad; and, in fact, two crucial facts flow from this formula. The first is that, if Sr is applied to Pr, the result is zero. The second is that, if Sr is applied to Pt with r≠t, the result is a polygon similar to Pt.
The proof of the theorem is now clear. The starting polygon is a linear combination of the n special polygons Pr. When a step of the form Sr is applied, the coefficient of Pr in this linear combination is reduced to zero. Once n-2 steps have been applied only one summand in the linear combination is non-zero; but that is a regular polygon.
We can now quantify the meaning of "becoming increasingly regular looking". At the beginning there is nothing special about the chosen polygon: it can be as irregular as possible. But after one step it is restricted: it is now in the linear span of n-1 of the Pr. And after each successive step it is in the linear span of fewer and fewer of the Pr, until at the end it is in the linear span of just one Pr.
Another way of putting this is that there is a hierarchy of increasingly restrictive types of polygon, depending on the increasingly restrictive linear spaces that the polygon belongs to. It is interesting to compare this with the conditions that classical types of polygons satisfy and this is particularly the case for triangles and quadrilaterals.
For example, what about equilateral triangles? Not all equilateral triangles are the same of course. They differ in their position in the plane, their size, and their rotation and so we have to think about the class of all such triangles, and we should also allow the triangle with all side lengths equal to zero. But that is not all! It is crucial whether we allow two equilateral triangles which are reflections of one another to be the same. Depending on whether we allow this or not we shall have one or two classes of equilateral triangles.
In the first case the set of all equilateral triangles is not a linear space (indeed it spans the entire set of triangles). In the second case it is a two-dimensional subspace (spanned, for example, by the one point triangle (0,0,0) and either the triangle (1,w,w2) or the triangle (1,w2,w) where w is a cube root of 1).
This suggests that when we consider other classical polygons it will be more interesting if we consider classes closed under displacement, inflation, and rotation but not reflection. We'll do that from now on.
What about the set of all triangles with the same angles (taken, say clockwise), i.e. all similar to one another? This is also a two-dimensional space (because two triangles S and T are similar if and only if S = wT +z for some complex numbers w and z).
Quadrilaterals provide richer pickings. Let's start with the simplest quadrilaterals - squares. Initially let's consider squares oriented clockwise: (a,b,c,d). To be a square, the side from a to b has to have the same length as the side from b to c and be perpendicular to it. This can be expressed by the equation
b-a = i(c-b)
Also the side from b to c has the same length as the side from c to d and is perpendicular to it
c-b = i(d-c)
These two linear conditions ensure that (a,b,c,d) is a square so the set of all clockwise oriented squares is a two-dimensional subspace.
What about the set of all squares (both orientations). This is not a subspace. However it is contained within the set of all parallelograms which is a subspace. Indeed, for (a,b,c,d) to be a parallelogram it is necessary and sufficient that the midpoints of the diagonals coincide. Thus the set of all parallelograms is defined by the single equation
a + c = b + d
and therefore the space of all parallelograms is 3-dimensional ("all" implies that reflections are also allowed). This implies the rather curious fact that every parallelogram is the sum of two squares and all parallelograms arise in this way.
For rectangles the conclusions are similar: the set of all rectangles is not a subspace but if we restrict to rectangles with the same aspect ratio we get a two-dimensional subspace.
For rhombi with one interior angle fixed we also get a two-dimensional subspace (otherwise they span the full space of parallelograms).
For trapezia (a,b,c,d) with the side from a to b parallel to the side from d to c we have the equation
b-a = -k(d-c)
where k is the ratio of the parallel sides (the aspect ratio). If all aspect ratios are allowed we don't get a subspace whereas if k is fixed we get a three-dimensional subspace. For trapezia, k will be real; but there are more general classes where k is complex number representing polygons with opposite sides inclined at a fixed angle and with fixed aspect ratio.
I will consider one more class of quadrilaterals that is not so well known - indeed I don't know if they even have a name. I shall call them DOEs because they are quadrilaterals whose Diagonals are Orthogonal and Equal. Of course there are many other generalisations but DOEs are interesting because, as we shall see, they are a sort of companion to parallelograms.
Suppose that (a,b,c,d) is a quadrilateral. The necessary and sufficient condition for it to be a DOE is
c - a = -i(d - b)
This constraint is linear so the set of DOEs is a three-dimensional subspace. Notice that quadrilaterals which satisfy c-a = +i(d - b) are another family of DOE-like quadrilaterals; the subtle difference between them and the original DOE family is that the angle from ac to bd is 270 degrees rather than 90 degrees.
We now examine some of these quadrilaterals in the language we used to describe the Petr-Douglas-Neumann theorem. The 4 special quadrilaterals are
P0 = (1, 1, 1, 1), the degenerate quadrilateral
P1 = (1, i, -1, -i), an anticlockwise square
P2 = (1, -1, 1, -1), a digon described twice
P3 = (1, -i, -1, i), a clockwise square
It follows from the equations defining parallelograms and DOEs that the linear span of P0, P1, P3 is the space of parallelograms and that the linear span of P0, P1, P2 is the space of DOEs.
We also note that the operators S1, S2, S3 are associated with apex ear angles of 90, 180, and 270 degrees; so the ears are apexes of right-angled triangles on the left of an edge, midpoints of edges, or apexes of right-angled triangles on the right of an edge.
There are essentially only two different ways to apply the Petr-Douglas-Neumann theorem to a quadrilateral. We can first apply the operator S2 and then the operator S3, or we can apply S3 and then S2; in either case the result will be a square. The proof of the Petr-Douglas-Neumann theorem however also shows that the first case (applying S2 first) gives a quadrilateral in the span of P0, P1 and P3 (so a parallelogram) and the second case (applying S3 first) gives a quadrilateral in the span of P0, P1 and P2 (so a DOE).
The last two statements are actually well-known geometrical theorems (Varignon's and Von Aubel's theorems respectively). The discussion also proves that, in any parallelogram, the apexes of right-angled triangles erected on its sides form a square; and also, in any DOE, the midpoints of its sides form a square.
All of this suggests that there may be similarly interesting theorems about pentagons and other n-gons. For example, there are 4 different regular pentagons: two normal pentagons oppositely oriented, and two pentagrams. What special properties are enjoyed by pentagons which are in the linear span of two of these? So far as I know (which is not very far) the study of sums of polygons is not at all well-developed.
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