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 and draw the parallel to as in Figure 1 and if and 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.

First, we derive a necessary and sufficient condition for the existence of closed polygonal paths in terms of edge lengths and angle measures.

Proposition 1 Let and be positive real numbers with There exists a closed polygonal path (with oriented counterclockwise) having edge lengths and the measures of the angles^* formed by with with with equal to , respectively, if and only if

and

for some integer

Proof Assuming that such a polygon exists, let be the complex number associated to. As the vector is the multiple of the rotation of in through (see Figure 2), we have

Based on the same type of argument, regardless of the orientation of triangles we have

Combining these relations with

yields

thus proving relation (1.1) from the conclusion. Relation (1.2) follows easily as a consequence of the last relation in the set of relations above.

Conversely, to prove the existence of a closed polygonal path with given satisfying (1.1), observe that, starting with an arbitrary point we can always consider the points such that, with the exception of and the measure of the angles formed by with and with 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 and denote the measure of the angle formed by with by and with by then, we need to show that and By applying the direct implication to our polygonal path (with edge lengths and angle measures ), we have

Equivalently,

By factoring out 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 ) to obtain

But then based on the two formulas above. Now, factoring out and replacing by in (1.3), we have

But we also have

Comparing relations (1.4) and (1.5), it follows that To show that let’s note that relation (1.2) applied to implies

for some integer By hypothesis,

Combining the two relations above finishes the proof.

If we consider a convex equiangular gon, then, with notations as in the previous section, we have

In addition, if we let

then, based on Proposition 1, we obtain Theorem 1 Given there exists a convex equiangular n-gon with side lengths (listed counterclockwise) iff

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 Proposition 2 Let and let be the degree of the cyclotomic polynomial There exists a convex, rational, equiangular n-gon with edge lengths (ordered counterclockwise) iff the following equalities are satisfied:

where are defined by

for all

Proof Let us first note that the definition of makes sense. Indeed, since forms a basis of for a fixed we can define to be the coefficients of in this basis.

For each if we replace in the equality from Theorem 1 by, we obtain

By reorganizing the terms, the formula above becomes

But then we get a polynomial of degree with rational coefficients having as a root. This is only possible iff all the coefficients are zero, thus proving the proposition.

Observation 2 By fixing the conditions in the proposition above generate a system of equations

with N equations and variables Comparing the number of equations and the number of variables, we obtain three cases depending on whether or

To better understand the three cases above, we have Lemma 1 For any positive integer we have the following 1) iff for some odd prime and some positive integer

2) iff for some positive integer

3) iff where the nonnegative integer and the odd integer are such that either or, if then is the product of the powers of at least two distinct primes.

Proof Since the third case is the complement of the first two, it is enough to prove the first two cases. So let, where are distinct primes and are nonnegative integers.

To prove observe that the inequality is equivalent to

or To show that n has the desired form, let us assume by contradiction that But then, since and we have or equivalently This implies Together with the second inequality above yields which contradicts the hypothesis. Thus But then we must also have p₁ > 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:

Proposition 3 If are the lengths of the edges of a convex, rational, -gon with p > 2 prime, then the polygon is equiangular iff

Proof Let

be the minimal polynomial of over (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 given by its minimal polynomial and multiplying by we have

Thus, if and otherwise. With these values of the conclusion follows.

Consequence 1 Any rational equiangular polygon with a prime number of edges is regular.

Proof This follows based on Observation 1 and the case in Proposition 3.

Observation 3 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 be the number of edges of the polygon. Since it follows that

Thus, the relation from Theorem 1 becomes

or

But then is a root of a rational polynomial of degree less than that of 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 are the edge lengths of a convex equiangular 15-gon, with the edges ordered counterclockwise, iff

and

Proof In this case, By letting we have

Based on these relations and Proposition 2, we must have

If we let

and

the relations above become

Clearly, these relations are equivalent to and thus proving the lemma.

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:

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

and

Observation 4 The proposition above provides a counterexample to conjecture 7 from [4] claiming that no arithmetic n-gons exist if is odd.

The example above suggests the following:

Theorem 2 There exists an arithmetic n-gon if and only if is not the power of a prime, i.e., has at least two distinct primes factors.

Proof By Consequence 3, it is enough to prove the converse. So, let’s consider for some positive integers Since is not the power of a prime, and q can be chosen to be relatively prime. If denotes a primitive th root of unity, then

and

Multiplying relations (1.6) by and (1.7) by we have

and

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

Whenever the sum of the corresponding coefficients is an integer between 1 and pq Moreover, different and with, generate different values for because and are relatively prime. Since there are exactly pq pairs the values of will represent a permutation of the set

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.