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 ![](https://www.scirp.org/html/14-7401789\5a0ea371-faad-40e4-a56d-576006edc3b2.jpg)
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
![](https://www.scirp.org/html/14-7401789\7a80ab9f-467e-4fe7-a740-e62243ef7250.jpg)
Based on the same type of argument, regardless of the orientation of triangles
we have
![](https://www.scirp.org/html/14-7401789\fe6418e4-1cce-4e5f-93af-9da5674d4cdb.jpg)
Combining these relations with
![](https://www.scirp.org/html/14-7401789\cd3f0315-f4df-44f4-997f-40012a0fdeba.jpg)
yields
![](https://www.scirp.org/html/14-7401789\f899de06-3e7a-46b4-abd2-03c691fbc368.jpg)
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
![](https://www.scirp.org/html/14-7401789\1ecd148b-c45c-43ca-b48f-c3fb53d1d5e1.jpg)
Equivalently,
(1.3)
By factoring out
and applying the modulus on both sides of the equality above, we have
![](https://www.scirp.org/html/14-7401789\e04e9bbe-7c92-4080-bb8a-df8608abe541.jpg)
However, the same type of operations can also be applied to the relation in our hypothesis (involving
) to obtain
![](https://www.scirp.org/html/14-7401789\b263ac00-5ba7-47a1-b990-cb94263b4c46.jpg)
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
![](https://www.scirp.org/html/14-7401789\a35d9085-c569-4e71-920b-b31dc67ae929.jpg)
for some integer
By hypothesis,
![](https://www.scirp.org/html/14-7401789\4676213c-15da-470b-b9ca-fd922ee2c698.jpg)
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
![](https://www.scirp.org/html/14-7401789\7ceb55e9-6458-4cf0-9562-f31b7dee819c.jpg)
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:
![](https://www.scirp.org/html/14-7401789\18f39a41-debf-47ee-b43f-b687b46c3830.jpg)
![](https://www.scirp.org/html/14-7401789\7f78a174-c24b-4e08-b0c0-b727e79e5a2d.jpg)
![](https://www.scirp.org/html/14-7401789\c1d06a35-258c-472d-aded-aa56ff86c8b3.jpg)
where
are defined by
![](https://www.scirp.org/html/14-7401789\3522bc47-8bdf-49a7-aa23-5a0f8ad2ba7d.jpg)
for all ![](https://www.scirp.org/html/14-7401789\c646f1ca-e90e-4eba-a84f-c1ea04b04f88.jpg)
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
![](https://www.scirp.org/html/14-7401789\a088fbd9-9532-40e4-9c10-d8baafec819c.jpg)
By reorganizing the terms, the formula above becomes
![](https://www.scirp.org/html/14-7401789\ff1fdf05-8c85-43bf-b686-7248d5a675fe.jpg)
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
![](https://www.scirp.org/html/14-7401789\a994be56-ef03-4949-b065-004aa840f44c.jpg)
with N equations and
variables
Comparing the number of equations and the number of variables, we obtain three cases depending on whether
or ![](https://www.scirp.org/html/14-7401789\dd244d10-1226-43b7-8080-623e6f9d8038.jpg)
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 ![](https://www.scirp.org/html/14-7401789\0dce67f7-05dd-4d46-a28e-84490af78284.jpg)
2)
iff
for some positive integer ![](https://www.scirp.org/html/14-7401789\db464eaa-07d4-4f1a-b32f-9fc6c7b31542.jpg)
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
![](https://www.scirp.org/html/14-7401789\548ae598-0e79-4f5c-9ee9-b0b95f62e1a2.jpg)
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 ![](https://www.scirp.org/html/14-7401789\ffd70349-9860-49ee-9296-1c5161ca7c48.jpg)
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 ![](https://www.scirp.org/html/14-7401789\a6abf30f-4996-4fb4-8332-ce11fcb146ed.jpg)
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
![](https://www.scirp.org/html/14-7401789\f5b063a1-950b-4bc6-9a1a-bf2cf2c8d867.jpg)
Proof Let
![](https://www.scirp.org/html/14-7401789\1512e4d7-4a9e-4d34-9e75-388d489b89b2.jpg)
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
![](https://www.scirp.org/html/14-7401789\ed73c3dd-f502-491c-aa73-6c2e5314950e.jpg)
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
![](https://www.scirp.org/html/14-7401789\64827993-5466-4dc6-8c04-17a3fa15c6d2.jpg)
or
![](https://www.scirp.org/html/14-7401789\cd256a2c-dc2c-4268-879d-5f13ecca6d36.jpg)
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
![](https://www.scirp.org/html/14-7401789\c13a95a7-3ce3-4fb8-8990-c9cdf3f34855.jpg)
and
![](https://www.scirp.org/html/14-7401789\bd365b7a-7a87-43f4-aa7a-2fc84465a5b7.jpg)
Proof In this case,
By letting
we have
![](https://www.scirp.org/html/14-7401789\882b8fa8-6ae2-4a9c-984f-bb562daeb65b.jpg)
![](https://www.scirp.org/html/14-7401789\e005b19a-9b6c-463f-b38e-d44f7510e984.jpg)
![](https://www.scirp.org/html/14-7401789\ce2e3df7-07cb-4103-937a-d56366935d33.jpg)
Based on these relations and Proposition 2, we must have
![](https://www.scirp.org/html/14-7401789\dbdd3580-eb4b-4903-b624-19534c74b375.jpg)
![](https://www.scirp.org/html/14-7401789\b7ba6f2a-6e53-490f-994e-f165300e8eed.jpg)
![](https://www.scirp.org/html/14-7401789\c0fca6fe-0598-47e9-a6e2-0782bf2a768b.jpg)
![](https://www.scirp.org/html/14-7401789\3f00ec7a-9604-45ed-a294-68de90dc03d8.jpg)
If we let
![](https://www.scirp.org/html/14-7401789\8bc38d12-17f1-4605-85ce-6082d8539231.jpg)
![](https://www.scirp.org/html/14-7401789\367ff3d7-7edb-40b1-9f73-587b9c4bc884.jpg)
and
![](https://www.scirp.org/html/14-7401789\626c6a24-97a7-4eea-bb62-f34610e14934.jpg)
![](https://www.scirp.org/html/14-7401789\6021e5bb-8c14-4d81-b6ee-756c41e5f57a.jpg)
the relations above become
![](https://www.scirp.org/html/14-7401789\52b4d2ab-9334-4105-af03-676c984b6d35.jpg)
![](https://www.scirp.org/html/14-7401789\fbe55c5c-d8bf-42be-8b8c-0926ce1cdf92.jpg)
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
![](https://www.scirp.org/html/14-7401789\06712041-380c-4a7c-842a-c2474c8b2ba5.jpg)
then the conditions in Example 2 are satisfied since
![](https://www.scirp.org/html/14-7401789\8920767a-432e-45dd-b732-41d9dd48b845.jpg)
and
![](https://www.scirp.org/html/14-7401789\b0f3a4a8-c3b4-42a1-8485-7c0b1e235c01.jpg)
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
![](https://www.scirp.org/html/14-7401789\28f8ae2b-c3d9-4c9a-82c2-03f37dc5f043.jpg)
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 ![](https://www.scirp.org/html/14-7401789\ae120229-d0d2-4c30-bc59-4128ffb2596a.jpg)
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