1. 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 
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
Figure 1. New equiangular pentagons from old.
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.
2. Preliminaries
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
(1.1)
and
(1.2)
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,
(1.3)
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
(1.4)
But we also have
(1.5)
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.
3. Equiangular Polygons
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.
4. 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:
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
(1.6)
and
(1.7)
Multiplying relations (1.6) by 
and (1.7) by 
we have
(1.8)
and
(1.9)
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 
5. 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.
NOTES