Rational Equiangular Polygons

The main purpose of this note is to investigate equiangular polygons with rational edges. When the number of edges is the power of a prime, we determine simple, necessary and sufficient conditions for the existence of such polygons. As special cases of our investigations, we settle two conjectures involving arithmetic polygons.


Introduction
A simple way of extending the class of regular polygons is to maintain the congruence of vertex angles while no longer requiring that the edges be congruent.In this generality, the newly obtained equiangular polygons are not all that interesting given one can find plenty of such (nonsimilar) polygons with a given number of edges.Indeed, drawing a parallel line to one of the edges of a regular polygon through an arbitrary point on an adjacent edge yields a trapezoid and a new equiangular polygon with the same number of edges as the initial one (see Figure 1).However, if we also require that all edge lengths be rational numbers and that at least two of these numbers be different (thus excluding regular polygons), in general, such equiangular polygons may not even exist.For example, if we start with the regular pentagon 1 2 3 4 5 and draw the parallel P P P P P Q Q to 1 2 P P as in Figure 1 and if and 1 1 are rational numbers then, except for all edge lengths of the equiangular pentagon are rational.However, is irrational.While this, by no means, proves that equiangular pentagons with rational edges must be regular, it gives some credibility to the non-existence claim above.
An interesting investigation of equiangular polygons with integer sides is provided in [1], where the author considers the problem of tiling these polygons with either regular polygons or other pattern blocks of integer sides.In particular, he points out that every equiangular hexagon with integer sides can be tiled by a set of congruent equilateral triangles, also of integer sides, and also proposes a general tiling conjecture with an extended tiling set.On the other hand, if one no longer requires integer edges but asks that the vertices be integer lattice points, the only equiangular polygons that will do are squares and octagons (see [2,3]).
Further restricting the class of equiangular polygons with integer sides, in [4], R. Dawson considers the class of arithmetic polygons, i.e., equiangular polygons whose edge lengths form an arithmetic sequence (upon a suitable rearrangement) and shows that the existence of arithmetic n-gons is equivalent to that of equiangular n-gons whose side lengths form a permutation of the set In addition, some interesting existence as well as non-existence results are obtained, but the classification problem for arithmetic polygons with an arbitrary number of edges is left open.
In this note, we address the more general problem of determining all equiangular polygons with rational edges and, as a special case, we settle the classification problem above.
the existence of closed polygonal paths in terms of edge lengths and angle measures.
Based on the same type of argument, regardless of the orientation of triangles we have 1 2 ,1 , Combining these relations with .with n n all edge lengths and angle measures are as needed.We will prove that the closed polygonal path satisfies the requirements.To do so, if we let 1 n and denote the measure of the angle formed by     By applying the direct implication to our polygonal path (with edge lengths and angle measures 1 2 , , , ), we have By factoring out e n i  and applying the modulus on both sides of the equality above, we have However, the same type of operations can also be applied to the relation in our hypothesis (involving But then n n l l  based on the two formulas above.Now, factoring out e n i  and replacing by in (1.3), we have But we also have * All angles are measured counterclockwise from the first to the second referenced ray.
for some integer By hypothesis, .k Combining the two relations above finishes the proof.

Equiangular Polygons
If we consider a convex equiangular gon, then, with notations as in the previous section, we have then, based on Proposition 1, we obtain Theorem 1 Given 1 2 there exists a convex equiangular n-gon with side lengths Definition 1 A rational polygon is a polygon all of whose edge lengths are rational number.
Observation 1 The edges of a non-convex equiangular polygon can be rearranged to form a convex equiangular polygon, so we will only concentrate on the latter.
As a consequence of Theorem 1, we obtain

 .
n X  There exists a convex, rational, equiangular n-gon with edge lengths 1 2 (ordered counterclockwise) iff the following equalities are satisfied:  in the equality from Theorem 1 by By reorganizing the terms, the formula above becomes 2 ,1 0.
But then we get a polynomial of degree 1 N  with rational coefficients having n  as a root.This is only possible iff all the coefficients are zero, thus proving the proposition.Observation 2 By fixing 1 2 1 the conditions in the proposition above generate a system of equations


with N equations and  variables 1 , , .
N n l   l Comparing the number of equations and the number of variables, we obtain three cases depending on whether To better understand the three cases above, we have Lemma 1 For any positive integer we have the following for some odd prime and some positive integer To show that n has the desired form, let us assume by contradiction that But then, since and 2 p 3, 

we have or equivalently This implies
Together with the second inequality above yields which contradicts the hypothesis.Thus But then we must also have p 1 > 2 since otherwise Conversely, it is easy to see that if then For by considerations similar to the ones above we must have Since, by (1), we cannot have it must be that Also, it is clear that if then Next, we consider convex, rational, equiangular polygons in each of the three cases given by the lemma.For the overdetermined case we have the following: are the lengths of the edges of a convex, rational, -gon with p > 2 prime, then polygon is equiangular iff (see [5], page 31).In order to apply Proposition 2, we need to write ,  for all as a linear combination with integer coefficients of Starting with the equation in n , , .

  
given by its minimal polynomial and multiplying by we have 2 1 , , Thus, if and The consequence above proves conjecture 6 from [4].
For the fully determined case we have the following characterization: Proposition 4 Given a convex, rational polygon whose number of edges is a power of two, the polygon is equiangular iff opposite edges are congruent.
Proof Let 2 k n  be the number of edges of the polygon.Since   , , Thus, the relation from Theorem 1 becomes  is a root of a rational polynomial of degree less than that of .n  This is only possible if the polynomial is identically zero, which implies the conclusion.
As a consequence of the proposition above, we obtain a different proof of Theorem 3 from [4].
Consequence 2 There does not exist an equiangular -gon with integer edge lengths, all distinct.
For the underdetermined case , given the lack of a simple formula for in this case, we will only consider the following example.
Lemma 2 Based on these relations and Proposition 2, we must have

Arithmetic Polygons
Following the terminology from [4], a polygon is said to be arithmetic if it is equiangular and its edge lengths (in some order) form a nontrivial arithmetic sequence.As shown in the same paper, an arithmetic -gon exists iff there exists an equiangular polygon with edge lengths (in some order) In this section we find a necessary and sufficient condition for the existence of arithmetic polygons in terms of the number of edges.First, we have the following: n  Consequence 3 There are no arithmetic polygons whose number of edges is the power of a prime.
Proof This follows as a consequence of Propositions 3, 4, and Observation 1.
One case when arithmetic polygons do exist is provided by the example below.
Example 1 There exists a (convex) arithmetic 15-gon.Proof If we select then the conditions in Example 2 are satisfied since The proposition above provides a counterexample to conjecture 7 from [4]   Let us now observe that every integer between 1 and appears exactly once as an exponent in both (1.8) and (1.9) due to the fact that p and q are relatively prime.If we add all equations (1.8) and all equations (1.9), we obtain

Conclusions
In this note we determined all rational equiangular polygons whose number of sides a prime power.Although we also determined all rational equiangular 15-gons, the general problem remains open.In addition, we provided a complete characterization of arithmetic polygons.
As an interesting application, we note that, as mentioned in [6], there is a nice correspondence arising from the Schwarz-Christoffel transformations between equiangular n-gons and certain areas determined by binary forms of degree n with complete factorizations over It would be interesting to investigate the consequences of our results in the language of binary forms.
degree of the cyclotomic polynomial

p
is an integer between 1 and pq Moreover, differen a and b with 1 a p