On Equations and Properties Concerning Some Classes of Chordal Polygons

M. Radić, T. K. Pogány, V. Kadum

Dedicated to the Memory of Grigorios TSAGAS (1935-2003), President of Balkan Society of Geometers (1997-2003)

Abstract

In the paper k-chordal (or k-inscribed) polygons of first and second kind with given index are considered. Existence result is proved for equilateral chordal polygon which side lengths are already known. The convex and nonconvex cases are discussed depending on the orientation of the polygon. Secondly, the number of different radii of circumcircles of equilateral k-inscribed n-gon cannot be greater then

s[n] = [\frac{n − 1}{2}] + [\frac{n − 3}{2}] + [\frac{n − 5}{2}] + ⋯ + 2 + 1.

A very natural conjecture is formulated on the existence of side lengths of kchordal n-gons when the minimal number of different circumcircle radii is

σ[n] := [\frac{n − 1}{2}]+( \begin{matrix}n \\ 1 \end{matrix} ) [\frac{n − 3}{2}]+( \begin{matrix}n \\ 2 \end{matrix} ) [\frac{n − 5}{2}]+⋯+( \begin{matrix}n \\ μ \end{matrix} ) [\frac{n − 2μ + 1}{2}] ,

where n−2μ = 3(4) for n odd (even). Thirdly, the so-called main equation (kind of related characteristic algebraic equation for a polygon) is introduced for the class C_{n}(a_{1}, ⋯, a_{n}) of k-chordal related polygons. In few illustrative examples we obtain the number and the numerical values of different radii of quadrangle, pentagon, octagon and enneagon, solving the related main equations, when only the side lengths of initial polygons are known. In the final section certain interesting properties of the so-called main equations are discussed, proving that the positive roots of the main equations are the radii of the circumcircles of the chordal n-gons whose sides have the lengths a_{1}, ⋯, a_{n}. The equilateral pentagon is presented in detail with three different positive solutions of its main equation which is an eigth degree algebraic equation. In the same section the main equation of λn-gons is characterized, when the initial n-gon is λ times continued on the same circumcircle, λ positive integer.

Mathematics Subject Classification: 51E12, 12D10; 51M04, 26C10. Key words: Algebraic equations, k-chordal polygon, k-inscribed chordal polygon, main equation, circumcircle, polygon of first kind, polygon of second kind, index of chordal polygon.

∗ Balkan Journal of Geometry and Its Applications, Vol.8, No.2, 2003, pp. 57-80.

\begin{matrix}c \\ ◯ \end{matrix} Balkan Society of Geometers, Geometry Balkan Press 2003.

1 Introduction

The subject and the main purposes of this article are very closely connected to the current interests of the first two authors in elementary geometry, more precisely in properties of generalized plane polygons. The main tools and definitions of the frequently mentioned basic geometrical object as k-chordal-, k-tangential-, (k, λ, l)- chordal-, (k, f, l)-chordal polygon, polygon of first/second kind are introduced, treated and discussed in ([6], [7], [8], [9], [10], [11]). There existence results are proved for chordal polygons under necessary and sufficient conditions upon the side lengths, while for (k, f, l)-chordal polygons upon the function f, i.e. upon the lengths a_{1}, ⋯, a_{n} and f(a_{1}), ⋯, f(a_{n}) being side lengths of a k-chordal and l-chordal polygons respectively in the same time ([6]). The approach in investigations by Góźdź in ([1]) and Pech in ([5]) is more or less different then in previously cited articles, namely, these authors uses complex methods and harmonic/Fourier series methods in getting characteristic inequalities and equations for different polygonal plane structures, such that are convex. The classical works by Kürschák concerns to the isoperimetric questions on the chordal and tangential n-gons to the given circle, where by their original method it was shown that the equilateral case is the extremal ([3], [4]). Some comments and explanations can be found in depending Horváth’s essay ([2]). Finally ([12]) is containing many known inequalities relating circumradius (and further characteristics) of planar convex sets, such that could be treated and generalized to our nonconvex, k-inscribed setting and similarly Temesvári’s optimization paper could be found interesting in further investigations for the maxima of the power sums of side lengths of some classes of k-chordal polygons, compare ([13]).

In this paper we follow the previous investigations focusing mainly ourselves to the so-called Main Equation of the k-chordal n-gon, and to computing all different radii of depending circumcircles, when the polygons side lenghts are already known.

A polygon with vertices A_{1}, ⋯, A_{n} (in this order) will be denoted by A ≡ A_{1} ⋯ A_{n} and the lengths of its sides by a_{1}, ⋯, a_{n}. The interior angle at the vertex A_{i} will be denoted by α_{i} or ^{}A̸_{i}. Thus

A̸_{i} = ^{}A̸_{i−1}A_{i}A_{i+1}, i = \overline{1, n},

where A_{0} ≡ A_{n}, A_{n+1} ≡ A_{1}.

A polygon A is called chordal if there exists a circle CA such that ∀A_{i} ∈ CA. Whenever A is chordal, then C, ρ and CA stand for the centre, radius and the circumcircle of A respectively. Throughout this paper very important roles are playing by (oriented) angles

(1)  β_{i} = ^{}C̸A_{i}A_{i+1},
(2)  φ_{i} = ^{}A̸_{i}CA_{i+1}, i = \overline{1, n}.

Also it is important to emphasize that β_{i} and φ_{i} are in opposite orientations, compare the following figure.

Figure 1. Angles of chordal polygons

Notice 1 We consider chordal polygons with property that no two of their consecutive vertices are the same.

Of course, the measure |ψ| of an oriented angle ψ we take depending on the orientation of ψ, in radians. So, by Notice 1. it is

0 ≤ |β_{i}| < \frac{π}{2} , 0 < |φ_{i}| ≤ π,

Notice 2 It will be no confusion there writing β_{i}, φ_{i} the measures of oriented angles β_{i}, φ_{i} given by (1),(2).

Notice 3 In the following we shall suppose that no β_{i} is zero.

Let us remark that in the case when some β_{i} is equal to zero, then we have

2ρ = max{a_{1}, ⋯, a_{n}}.

Accordingly, in the following when we speak about a chordal polygon A it will be meant (by Notice 1 and Notice 3) that A has no two the same consecutive vertices and no one of its sides is its diameter.

Definition 1 Let A = A_{1} ⋯ A_{n} be a chordal polygon. We say that A is of the first kind if inside CA there exists a point O that all oriented angles ^{}A̸_{i}OA_{i+1} have the same orientation. If such a point O does not exist, i.e. not all ^{}A̸_{i}OA_{i+1} have the same orientation, we say that A is of second kind.

Definition 2 Let A be a chordal polygon and let X be a point inside CA such that

|∑_{j=1}^{n} ψ_{j}| = 2ω(X)π,

where ψ_{j} = measure^{}A̸_{j}XA_{j+1} and ω(X) is a positive integer. Then we say that A is k-inscribed polygon of the first (second) kind when

k = max_{X∈int(CA)} ω(X).

Here int(S) stays for the interior of the set S.

We say that j is the index of A if |φ_{1} + ⋯ + φ_{n}| = 2jπ, j ∈ {0, 1, ⋯, k}.

Definition 3 The polygon A is said to be k-chordal polygon if it is of first kind and if j = k, where j is the index of A and k is given by Definition 2.

It is easy to see that A is k-chordal iff

|β_{1} + ⋯ + β_{n}| = (n − 2k)\frac{π}{2} ,

where β_{i} > 0 i = \overline{1, n} or β_{i} < 0, i = \overline{1, n}. For example, if ∀β_{i} > 0, then φ_{i} < 0, i = \overline{1, n}, and it is valid

φ_{1} + ⋯ + φ_{n} = −2kπ,

or in other words 2β_{1} + ⋯ + 2β_{n} = nπ − 2kπ, since φ_{i} = −π + 2β_{i}. Thus, if β_{i} > 0 , i = \overline{1, n}, then

β_{1} + ⋯ + β_{n} = (n − 2k)\frac{π}{2} .

The sign of the sum β_{1} + ⋯ + β_{n} depends of the orientation of the polygon, we discuss this in brief. Let A be a chordal k-inscribed polygon and let B = B_{1} ⋯ B_{n} be a polygon with vertices B_{j} = A_{n−1+j}, j = \overline{1, n}. Then A ≡ B but their orientations are opposite (orientation of A is positive (negative) depending on the circumscription of CA to A ”counter-clockwise” (”clockwise”)).

If A is k-chordal, then β_{j}, j = \overline{1, n} are negative if A is positively oriented and vice versa. But in the case when A is a chordal polygon of second kind, then there are β_{j}’s of opposite signes.

Notice 4 In the following we shall assume that polygons are negatively oriented. Then φ_{1} + ⋯ + φ_{n} ≤ 0 but β_{1} + ⋯ + β_{n} ≥ 0.

So, for example, the case Fig. 2.(a) gives β_{1} + ⋯ + β_{5} < 0, and the case Fig. 2.(b) one concludes β_{1} + ⋯ + β_{5} > 0.

Lemma 1 If A = A_{1} ⋯ A_{n} is a k-inscribed chordal polygon whose index is j, then

β_{1} + ⋯ + β_{n} = (n − 2(ν + j))\frac{π}{2} , (3)

where ν = ♯{m| β_{m} < 0} and

|β_{1}| + ⋯ + |β_{n}| = (n − 2(ν + j))\frac{π}{2} + 2τ, (4)

where τ = − ∑_{m: β_{m}<0} β_{m}.

Figure 2. Chordal pentagons of opposite orientations Proof. Since φ_{j} = −π + 2β_{j} if β_{j} > 0, and φ_{j} = π + 2β_{j} for β_{j} < 0, the equality φ_{1} + ⋯ + φ_{n} = −2jπ can be written as

2β_{1} + ⋯ + 2β_{n} + νπ − (n − ν)π = −2jπ,

from which follows (3). Now, by (3) we get (4) easily.

At this point we introduce certain symbols such that we will use frequently in the sequel.

1. [a] denotes the largest integer contained in a. Obviously, if A = A_{1} ⋯ A_{n} is k-inscribed chordal polygon, then

k ≤ [\frac{n − 1}{2}] , (5)

because it has to be n − 2k > 0. Of course, there are extremal cases, when the equality holds in (5), e.g. when A is equilateral, i.e. a_{1} = ⋯ = a_{n}.

2. P (a_{1}, ⋯, a_{n}; β_{1}, ⋯, β_{n}). Let A = A_{1} ⋯ A_{n} be a chordal n-gon. Then this ngon will also be written as

P (a_{1}, ⋯, a_{n}; β_{1}, ⋯, β_{n}). (6)

Sometimes instead of (6) we write

P (a_{1}, ⋯, a_{n}; β_{1}, ⋯, β_{n}; ρ). (7)

In the equilateral case (a_{j} = a) it stays P (a; β_{1}, ⋯, β_{n}) or P (a; β_{1}, ⋯, β_{n}; ρ).

3. P (a_{1}, ⋯, a_{n}; i_{1}, ⋯, i_{ν}). If β_{i_{j}}, j = \overline{1, ν} are negative, then this situation we note as P (a_{1}, ⋯, a_{n}; i_{1}, ⋯, i_{ν}), (8)

or appropriately P (a; i_{1}, ⋯, i_{ν}).

4. S_{p}^{n}. Let p be an integer such that 0 ≤ p ≤ n. If p = 0, then

S_{0}^{n} := cos β_{1} ⋯ cos β_{n}.

If p > 0, then S_{p}^{n} is the sum of ( \begin{matrix}n \\ p \end{matrix} ) products of the form

sin β_{i_{1}} ⋯ sin β_{i_{p}} cos β_{i_{p+1}} ⋯ cos β_{i_{n}},

where (i_{1}, ⋯, i_{n}) is a permutation of the set {1, ⋯, n}, e.g.

S_{2}^{3} = sin β_{1} sin β_{2} cos β_{3} + sin β_{1} cos β_{2} sin β_{3} + cos β_{1} sin β_{2} sin β_{3}.

2 Existence of k-inscribed chordal polygon

Let a_{1}, ⋯, a_{n} be given lengths such that satisfies the constraint

∑_{j=1}^{n} a_{j} > 2 max{a_{1}, ⋯, a_{n}}. (9)

Then there exists (at least) n-gon P (a_{1}, ⋯, a_{n}; β_{1}, ⋯, β_{n}), where ∀β_{j} > 0 and β_{1} + ⋯ + β_{n} = (n − 2)π/2. Namely, then there exists a positive real ρ (which is in fact certain length), such that satisfies

∑_{j=1}^{n} arccos \frac{a_{j}}{2ρ} = (n − 2)\frac{π}{2} . (10)

Intuitively it is easy to see this result. For example, observe the situation on Fig. 3.

Figure 3. Tightening of circumscribed circle one gets closed chordal polygon

Taking a sequence of circles with respect to decreasing radii we achieve the case M_{1} ≡ M_{2}; see the strong proof of (10) in ([7, proof of Theorem 2]).

Let us remark that (10) exists iff there are lengths r, R; r ≤ R, such that

∑_{j=1}^{n} arccos \frac{a_{j}}{2r} ≤ (n − 2)\frac{π}{2} ≤ ∑_{j=1}^{n} arccos \frac{a_{j}}{2R}.

Generally speaking, if ∑_{j=1}^{n} β_{j} = κπ/2, for κ integer, and if β_{j} = (−1)^{ϵ_{j}}|β_{j}|, where ϵ_{j} ∈ {0, 1}, then there exists P (a_{1}, ⋯, a_{n}; β_{1}, ⋯, β_{n}) iff there are lengths (positive reals) r, R; r ≤ R such that

∑_{j=1}^{n}(−1)^{ϵ_{j}} arccos \frac{a_{j}}{2r} ≤ κ\frac{π}{2} ≤ ∑_{j=1}^{n}(−1)^{ϵ_{j}} arccos \frac{a_{j}}{2R}.

Example 1 Put a_{j} = 2 + .1j, j = \overline{1, 5} and ∑_{j=1}^{5} β_{j} = π/2, where β_{1} < 0, then there is a pentagon P (2.1, ⋯, 2.5; β_{1}, ⋯, β_{5}), see Figure 4. below. In this case may be taken 2r = 2.806, 2R = 2.807, namely then the radius ρ of the circumcircle CA satisfies 2.806 < 2ρ < 2.807. But for β_{5} < 0 no pentagon is there, since it is

arccos \frac{2.1}{2ρ} + arccos \frac{2.2}{2ρ} + arccos \frac{2.3}{2ρ} + arccos \frac{2.4}{2ρ} − arccos \frac{2.5}{2ρ} > \frac{π}{2} , 2ρ ≥ a_{5}.

Figure 4. P (2.1, ⋯, 2.5; β_{1}, ⋯, β_{5}) with ∑_{j=1}^{5} β_{j} = \frac{π}{2} ; β_{1} < 0.

Theorem 1 If a_{j} = a, j = \overline{1, n}, then for each angle

β(k) = (n − 2k)\frac{π}{2n}, k = \overline{1, [(n − 1)/2]}

there is a k-chordal equilateral n-gon P (a; β(k)).

Proof. It is easy to see that for 2ρ_{k} = a/ cos(n − 2k)\frac{π}{2n} it is

n arccos \frac{a}{2ρ_{k}} = (n − 2k)\frac{π}{2} .

As an ilustration of this result we give the case of equilateral 2-chordal pentagon P (1; β(2)) presented on Figure 5.:

Figure 5. 2 - chordal pentagon P (1, β(2))

3 Number of k-inscribed chordal polygons

Let s[n] be defined by

s[n] := [\frac{n − 1}{2}] + [\frac{n − 3}{2}] + [\frac{n − 5}{2}] + ⋯ + 2 + 1. (11)

Theorem 2 If a_{1} = ⋯ = a_{n} = a, then the number of k-inscribed chordal n-gons wich have not equal radii cannot overgrow s[n].

Proof. From (n − 2ν)β(k, ν) = (n − 2(k + ν))π/2 it follows that

β(k, ν) = (1 − \frac{2k}{n − 2ν} ) \frac{π}{2} , (12)

where

k ∈ \{1, 2, ⋯, [\frac{n − 1}{2}]\} , ν ∈ \{0, 1, ⋯, [\frac{n − 3}{2}]\} , k + ν = [\frac{n − 1}{2}] .

Then by the Theorem 1. using the notation ν = ♯{m| β_{m} < 0} introduced in Lemma 1. we deduce that there are

[\frac{n − 2ν − 1}{2}] polygons P (a; β(k, ν)), k ∈ \{1, 2, ⋯, [\frac{n − 2ν − 1}{2}]\} .

Let us remark that here k refers to the term k-inscribed polygon in Definition 2. Now obvious transformations lead to the assertion of the theorem.

As an example we give the heptagon with parameters a = 1, k = 1, ν = 2. If i_{1} = 1, i_{2} = 5 (compare (8)), then the heptagon P (1; i_{1} = 1, i_{2} = 5) is presented on Figure 6 a). On Figure 6 b) the heptagon P (1; i_{1} = 1, i_{2} = 2) is shown. Although these heptagons are not equal, they have equal radii.

Figure 6. a) Heptagon P (1; i_{1} = 1, i_{2} = 5); b) Heptagon P (1; i_{1} = 1, i_{2} = 2)

Remark 1 If n is even then there is one more equilateral n-gon A = A_{1} ⋯ A_{n}, where A_{1} ≡ A_{3} ≡ ⋯ ≡ A_{n−1}, A_{2} ≡ A_{4} ≡ ⋯ ≡ A_{n}. But in the Notice 3 terminology speaking it is not included in s[n].

In the following considerations it is very important to see that the number of different radii of corresponding circumcircles of the chordal polygons is less then s[n] if n > 7. So, for example when n = 9 then the 3-chordal enneagon and the 1-inscribed chordal enneagon with three negative angles have equal radii since β(3, 0) = π/6 = β(1, 3).

Generally speaking if n − 2i and n − 2j are different entries of the sequences

n, n − 2, n − 4, ⋯, 3 where n is odd
n, n − 2, n − 4, ⋯, 4 where n is even

and GCD(n − 2i, n − 2j) ≥ 3(4) for n odd (even) respectively, then the number of different radii is less then s[n].

In the continuation we introduce the symbol

σ[n] := [\frac{n − 1}{2}]+( \begin{matrix}n \\ 1 \end{matrix} ) [\frac{n − 3}{2}]+( \begin{matrix}n \\ 2 \end{matrix} ) [\frac{n − 5}{2}]+⋯+( \begin{matrix}n \\ μ \end{matrix} ) [\frac{n − 2μ + 1}{2}] , (13)

where

(14) n − 2μ = \{ \begin{matrix}3 n in odd \\ 4 n in even. \end{matrix}

Having in mind Theorem 2 and the dicussion about Figure 3, we give the following hypothesis. Conjecture. There are the lengths a_{1}, ⋯, a_{n} such that the number of different radii of the circumcircles is at least σ[n].

Here we point out that we have proved the assertion of the Conjecture for n = 3, 4, 5, 6, 7. Something about this will be exposed in following examples.

Intuitively, this conjecture is very reasonable having in mind the following fact (in connection to Figure 3): if ϵ_{1}, ⋯, ϵ_{n} are different positive numbers, a_{j} = a + ϵ_{j}, j = \overline{1, n} and

p = na + ∑_{j=1}^{n} ϵ_{j},

then

p − (a + ϵ_{j}) ≠ p − (a + ϵ_{l})

whenever j ≠ l.

Example 2 As in Example 1 consider pentagon with the given side lengths a_{j} = 2 + .1j, j = \overline{1, 5}. (There is shown that β_{5} has to be positive!). Then using the equations

β_{1} + ⋯ + β_{5} = (5 − 2(j + ν))\frac{π}{2} , (15)

we find that

j ν negative angle 2ρ ∈

1 0  none  (3.9, 4)
2 0  none  (3.50, 2.51)
1 1  β_{1}  (2.806, 2.807)
1 1  β_{2}  (2.750, 2.760)
1 1  β_{3}  (2.680, 2.690)
1 1  β_{4}  (2.604, 2.605)

But, if we put e.g. a_{j} = 3 + .1j, j = \overline{1, 5}, then β_{5} < 0 is acceptable as well. Namely, then there is 3.6 < 2ρ < 3.7 such that

∑_{j=1}^{5} arccos \frac{a_{i}}{3.6} < \frac{π}{2} , ∑_{j=1}^{5} arccos \frac{a_{i}}{3.7} > \frac{π}{2} .

Thus by (15) we deduce that there are at least σ[5] = 7 chordal pentagons whose sides have the lengths a_{j} = 3 + .1j, j = \overline{1, 5}, and the corresponding radii ρ_{j} of the circumcircles are different.

In the case n = 7; a_{j} = 5 + .1j, j = \overline{1, 7}, it can be shown that there are σ[7] = 38 chordal heptagons with different corresponding radii of its circumcircles. As we can see by (13) the number σ[n] increases with the n growing: σ[5] = 7, σ[6] = 8, σ[7] = 38, σ[8] = 47, σ[9] = 187, etc.

4 Classes of related polygons and their main equations

In this section we consider certain relationships between polygons which possess the sides with same lengths. Firstly we introduce the term related polygons.

Definition 4 Let a_{1}, ⋯, a_{n} be given lengths and let X = X_{1} ⋯ X_{n}, Y = Y_{1} ⋯ Y_{n} be chordal polygons with the property that the lengths of their sides satisfy

x_{i} = y_{i} = a_{i}, i = \overline{1, n}. (16)

Then we say that X and Y are related polygons (with respect to their sides); the set consisting from all related polygons with respect to given a_{1}, ⋯, a_{n} we denote with the symbol C_{n}(a_{1}, ⋯, a_{n}).

In our next considerations we deal according to the Notice 1; also whenever we consider a polygon from C_{n}(a_{1}, ⋯, a_{n}) we will assume that no β_{j} vanishes (Notice 3).

Let a_{1}, ⋯, a_{n} be already known and let X ∈ C_{n}(a_{1}, ⋯, a_{n}) and choose the angles β_{j}, j = \overline{1, n} so that X ≡ P (a_{1}, ⋯, a_{n}; β_{1}, ⋯, β_{n}). If j is the index of X and ν = ♯{m| β_{m} < 0}, then by (3) we get

β_{1} + ⋯ + β_{n} = (n − 2(j + ν))\frac{π}{2} .

Hence we have the following equalities

(17)  cos(β_{1} + ⋯ + β_{n}) = 0, n odd,
(18)  sin(β_{1} + ⋯ + β_{n}) = 0, n even.

Using the symbol 𝔖_{p}^{n} we can transform the above two equalities into

(19)  𝔖_{0}^{n} − 𝔖_{2}^{n} + 𝔖_{4}^{n} − ⋯ + (−1)^{θ_{1}}𝔖_{n−1}^{n} = 0, n odd
(20)  𝔖_{1}^{n} − 𝔖_{3}^{n} + 𝔖_{5}^{n} − ⋯ + (−1)^{θ_{2}}𝔖_{n−1}^{n} = 0, n even

where θ_{1} = (1 + 3 + 5 + ⋯ + n) + 1, θ_{2} = (1 + 3 + 5 + ⋯ + (n − 1)) + 1. Now, the following steps will be done in (19) and (20). Replace sin β_{j} with

\sqrt{1 − (\frac{a_{j}}{2ρ})^{2}},

and put a_{j}/(2ρ) instead of cos β_{j}. Then rationalizing and simplifying (19) and (20) in ρ these equations become

(21)  F_{1}(a_{1}, ⋯, a_{n}; ρ) = 0, n odd
(22)  F_{2}(a_{1}, ⋯, a_{n}; ρ) = 0, n even,

where F_{m}(a_{1}, ⋯, a_{n}; ρ) are polynomials in ρ, m = 1, 2.

Definition 5 The equation (21) or (22) is said to be the Main Equation concerning the k - inscribed polygons in C_{n}(a_{1}, ⋯, a_{n}) for each k admissible in the sense of Definition 2.

Example 3 In this example we consider the main equation of chordal n-gons in C_{n}(1, ⋯, 1), when n is odd. Take n = 9. Let P (1; β_{1}(k, ν), ⋯, β_{9}(k, ν)) be k - inscribed enneagon with ν negative angles. Of course, it could be ν ∈ {0, 1, 2, 3} and it is unessential which ν angles are negative, since the depeneding radii of corresponding circumcircles equals in length.

As |β_{1}(k, ν)| = ⋯ = |β_{9}(k, ν)|, let β(k, ν) = |β_{1}(k, ν)| > 0. Then

\begin{aligned}β_{1}(k, ν) + ⋯ + β_{9}(k, ν) & = & (9 − 2ν)β(k, ν) \\ & = & ((9 − 2ν) − 2k)\frac{π}{2} = (9 − 2(k + ν))\frac{π}{2} . \end{aligned}

Consequently

\begin{aligned}β(k, 0) & = & (9 − 2k)\frac{π}{18}, k = 1, 2, 3, 4, \\ β(k, 1) & = & (7 − 2k)\frac{π}{14}, k = 1, 2, 3, \\ β(k, 2) & = & (5 − 2k)\frac{π}{10}, k = 1, 2, \\ β(1, 3) & = & \frac{π}{6} . \end{aligned}

Hence we have

(23)  cos 9β(k, 0) = 0, k = 1, 2, 3, 4,
(24)  cos 7β(k, 1) = 0, k = 1, 2, 3,
(25)  cos 5β(k, 2) = 0, k = 1, 2,
(26)  cos 3β(1, 3) = 0.

Now, from(23-26) using the well - known trigonometric equality

cos nα = cos^{n} α − ( \begin{matrix}n \\ 2 \end{matrix} ) cos^{n−2} α sin^{2} α + ( \begin{matrix}n \\ 4 \end{matrix} ) cos^{n−4} α sin^{4} α − ⋯,

it is easy to see that 1. x_{k}^{0} = cos β(k, 0), k = 1, 2, 3, 4 are the positive roots of the equation

x^{9} − 36x^{7}(1 − x^{2}) + 126x^{5}(1 − x^{2})^{2} − 42x^{3}(1 − x^{2})^{3} + 9x(1 − x^{2})^{4} = 0, (27)

2. x_{k}^{1} = cos β(k, 1), k = 1, 2, 3 are the positive roots of the equation x^{7} − 21x^{5}(1 − x^{2}) + 35x^{3}(1 − x^{2})^{2} − 7x(1 − x^{2})^{3} = 0, (28)

3. x_{k}^{2} = cos β(k, 2), k = 1, 2 are the positive roots of the equation x^{5} − 10x^{3}(1 − x^{2}) + 5x(1 − x^{2})^{2} = 0, (29)

4. x_{1}^{3} = cos β(1, 3) is the unique positive root of

x^{3} − 3x(1 − x^{2}) = 0. (30) Let the left hand sides of the equations (27-30) be denoted by f_{j}(x), j = 9, 7, 5, 3 respectively. Then the main equation of the chordal enneagons from C_{9}(1, ⋯, 1) of the form F (x) = f_{9}(x)f_{7}(x)f_{5}(x)f_{3}(x), (31) where x = 1/(2ρ). Its positive roots are

x_{kν} = \frac{1}{2ρ_{kν}} ,  ν = 0 , k = 1, 2, 3, 4,
ν = 1 , k = 1, 2, 3,
ν = 2 , k = 1, 2,
ν = 3 , k = 1,

where ρ_{kν} is the circumcircle radius of P (1; β_{1}(k, ν, ⋯, β_{9}(k, ν)).

Analogous results hold for all n odd. With respect to this question we can remark that the coefficients of the partial polynomials f_{n}(x) could be expressed as

c_{n−2j} = (−1)^{j} ∑_{i=0}^{(n−2j−1)/2} ( \begin{matrix}n \\ 2(i + j) \end{matrix} ) ( \begin{matrix}i + j \\ i \end{matrix} ) , j = \overline{0, (n − 1)/2}. (32)

Now, when n = 9, it is

\begin{aligned}f_{9}(x) & = & c_{9}x^{9} + c_{7}x^{7} + c_{5}x^{5} + c_{3}x^{3} + c_{1}x where \\ c_{9} & = & 1 + ( \begin{matrix}9 \\ 2 \end{matrix} ) + ( \begin{matrix}9 \\ 4 \end{matrix} ) + ( \begin{matrix}9 \\ 6 \end{matrix} ) + ( \begin{matrix}9 \\ 8 \end{matrix} ) = 2^{8} = 256, \\ −c_{7} & = & ( \begin{matrix}9 \\ 2 \end{matrix} ) + ( \begin{matrix}9 \\ 4 \end{matrix} ) ( \begin{matrix}2 \\ 1 \end{matrix} ) + ( \begin{matrix}9 \\ 6 \end{matrix} ) ( \begin{matrix}3 \\ 2 \end{matrix} ) + ( \begin{matrix}9 \\ 8 \end{matrix} ) ( \begin{matrix}4 \\ 3 \end{matrix} ) = 576, \\ c_{5} & = & ( \begin{matrix}9 \\ 4 \end{matrix} ) + ( \begin{matrix}9 \\ 6 \end{matrix} ) ( \begin{matrix}3 \\ 1 \end{matrix} ) + ( \begin{matrix}9 \\ 8 \end{matrix} ) ( \begin{matrix}4 \\ 2 \end{matrix} ) = 432, \\ −c_{3} & = & ( \begin{matrix}9 \\ 6 \end{matrix} ) + ( \begin{matrix}9 \\ 8 \end{matrix} ) ( \begin{matrix}4 \\ 1 \end{matrix} ) = 120, \\ c_{1} & = & ( \begin{matrix}9 \\ 8 \end{matrix} ) = 9. \end{aligned}

Generally, if c_{n} is the leading coefficient of f_{n}(x), n odd, then c_{n} = 2^{n−1}.

So the main equation of chordal n-gons in C_{n}(1, ⋯, 1), n odd may be written in the form f_{n}(x)f_{n−2}(x) ⋯ f_{5}(x)f_{3}(x) = 0. (33)

Of course, f_{j}(x) in (31) are the same as in (33). The number of positive roots in the equation (33) is at most s[n] = \frac{n^{2}−1}{8}. Also, each factor f_{j}(x) in (33) has a root x = 0. It is true because of cos n\frac{π}{2} = 0.

Finally, as the interesting consequence of the Example 1, we get some combinatorial/trigonometrical formulæ. From f_{n}(x)x^{−1} = 0, using Viète’s formulæ, it follows that

cos^{2} \frac{π}{2n} + cos^{2} \frac{3π}{2n} + ⋯ + cos^{2} \frac{(n − 2)π}{2n} = −\frac{c_{n−2}}{2^{n−1}} ,
cos^{2} \frac{π}{2n} cos^{2} \frac{3π}{2n} ⋯ cos^{2} \frac{(n − 2)π}{2n}  = \frac{n}{2^{n−1}} ,

where c_{n−2} is given by (32).

Example 4 Here we consider the main equation of chordal n-gons in C_{n}(1, ⋯, 1), n even. At first concentrate to the octagon, i.e. n = 8. In the same way as in Example 3 it can be found that

β(k, 0)  = (4 − k)\frac{π}{8} , k = 1, 2, 3,
β(k, 1)  = (3 − k)\frac{π}{6} , k = 1, 2,
β(1, 2)  = \frac{π}{4} ,
sin 8β(k, 0) = 0,  k = 1, 2, 3,
sin 6β(k, 1) = 0,  k = 1, 2,
sin 4β(1, 2) = 0.

70 M. Radić, T. K. Pogány, V. Kadum Now, using the identity

sin nα = ( \begin{matrix}n \\ 1 \end{matrix} ) cos^{n−1} α sin α − ( \begin{matrix}n \\ 3 \end{matrix} ) cos^{n−3} α sin^{3} α + ⋯,

we clearly get

1. x_{k}^{0} = cos β(k, 0), k = 1, 2, 3 are the positive roots of the equation x^{6} − 7x^{4}(1 − x^{2}) + 7x^{2}(1 − x^{2})^{2} − (1 − x^{2})^{3} = 0, (34)

2. x_{k}^{1} = cos β(k, 1), k = 1, 2 are the positive roots of the equation 3x^{4} − 10x^{2}(1 − x^{2}) + 3x^{4}(1 − x^{2})^{3} = 0, (35)

3. x_{1}^{2} = cos β(1, 2) is the unique positive root of

2x^{2} − 1 = 0. (36) Let the left hand side of equations (34-36) be denoted by f_{j}(x), j = 8, 6, 4 respectively.

Figure 7. Degenerated C_{8}(1, ⋯, 1) octagon with β(k, ν) = 0 Then the main equation of the chordal octagons from C_{8}(1, ⋯, 1) can be written

as

f_{8}(x)f_{6}(x)f_{4}(x) = 0, x = \frac{1}{2ρ}, (37)

excluding the polygon with circumcircle which possesses radius equal to 1/2. The positive roots of (37) are now given by

x_{k}^{ν} = \frac{1}{2ρ_{k}^{ν}} ,  ν = 0 , k = 1, 2, 3,
ν = 1 , k = 1, 2
ν = 2 , k = 1.

At this point we have to discuss the case of the circumcircle CA which possesses radius ρ = 1/2. As

sin nα = cos α sin α \{( \begin{matrix}n \\ 1 \end{matrix} ) cos^{n−2} α − ( \begin{matrix}n \\ 3 \end{matrix} ) cos^{n−4} α sin^{2} α + ⋯\} ,

there is a chordal octagon with x = 1, that means 2ρ = 1 (compare Figure 7). Indeed, by

(8 − 2ν)β(k, ν) = (4 − k − ν)π; k + ν = 4,

it is sin β(k, ν) = 0, cos β(k, ν) = 1.

So we don’t need Notice 3 in this example. Hence, instead of (37) the main equation of the considered octagon becomes (x − 1)f_{4}(x)f_{6}(x)f_{8}(x) = 0. On the other hand cos β(k, ν) = 0 results with β(k, ν) = π/2, see Figure 8.

Figure 8. Degenerated C_{8}(1, ⋯, 1) octagon with β(k, ν) = \frac{π}{2}

In general, analogous holds for all n-gons from the class C_{n}(1, ⋯, 1), n ≥ 4, n even. Therefore the coefficients γ_{j} of the factors f_{n}(x) in the main equation can be expressed in the following form

c_{n−2(j+1)} = (−1)^{j} ∑_{i=0}^{(n−2j−2)/2} ( \begin{matrix}n \\ 2(i + j) + 1 \end{matrix} ) ( \begin{matrix}i + j \\ i \end{matrix} ) , j = \overline{0, n/2 − 1}. (38)

Now, in our case we get

\begin{aligned}f_{8}(x) & = & c_{6}x^{6} + c_{4}x^{4} + c_{2}x^{2} + c_{0} where \\ c_{6} & = & ( \begin{matrix}8 \\ 1 \end{matrix} ) + + ( \begin{matrix}8 \\ 3 \end{matrix} ) + ( \begin{matrix}8 \\ 5 \end{matrix} ) + ( \begin{matrix}8 \\ 7 \end{matrix} ) = 2^{7} = 128, \\ −c_{4} & = & ( \begin{matrix}8 \\ 3 \end{matrix} ) + ( \begin{matrix}8 \\ 5 \end{matrix} ) ( \begin{matrix}2 \\ 1 \end{matrix} ) + ( \begin{matrix}8 \\ 7 \end{matrix} ) ( \begin{matrix}3 \\ 2 \end{matrix} ) = 192, \\ c_{2} & = & ( \begin{matrix}8 \\ 5 \end{matrix} ) + ( \begin{matrix}8 \\ 7 \end{matrix} ) ( \begin{matrix}3 \\ 1 \end{matrix} ) = 80, \\ −c_{0} & = & ( \begin{matrix}8 \\ 7 \end{matrix} ) = 8. \end{aligned}

We see that the leading coefficient c_{n−2} of f_{n}(x) is equal to 2^{n−1} similarly to the odd n case. Therefore the main equation of the chordal n-gons in the class C_{n}(1, ⋯, 1), n even could be written in the form

(x − 1)f_{4}(x)f_{6}(x) ⋯ f_{n−2}(x)f_{n}(x) = 0.

Finally as an application of (38) and the the Viète’s formulæ we derive the identity

cos^{2} (\frac{n}{2} − 1) \frac{π}{n} · cos^{2} (\frac{n}{2} − 2) \frac{π}{n} · cos^{2} (\frac{n}{2} − 3) \frac{π}{n} · . . . · cos^{2} \frac{π}{n} = \frac{n}{2^{n−1}} .

Example 5 Let a_{1}, a_{2}, a_{3}, a_{4} be given lengths. The main equation of the chordal quadrangles in C_{4}(a_{1}, a_{2}, a_{3}, a_{4}) will be considered. Since

β_{1} + β_{2} + β_{3} + β_{4} = (4 − 2(j + ν))\frac{π}{2} ,

we recognize three different cases; namely (j, ν) ∈ {(1, 0), (1, 1), (0, 2)}. For j = 1, ν = 0 it is β_{1} + β_{2} + β_{3} + β_{4} = π. So, by some heavy but straightforward trigonometry we deduce

((cos β_{1} cos β_{2} + cos β_{3} cos β_{4})^{2} − sin^{2} β_{1} sin^{2} β_{2}
− sin^{2} β_{3} sin^{2} β_{4})^{2} = 4 sin^{2} β_{1} sin^{2} β_{2} sin^{2} β_{3} sin^{2} β_{4}.

Putting a_{j}/(2ρ) instead of cos β_{j} and \sqrt{1 − (a_{j}/(2ρ))^{2}} instead of sin β_{j}, by rationalizing we get the equation in ρ reads as follows

R_{1}ρ^{2} − Q_{1} = 0, (39)

with

R_{1} = −a_{1}^{4} − a_{2}^{4} − a_{3}^{4} − a_{4}^{4} + 2(a_{1}^{2}a_{2}^{2} + a_{1}^{2}a_{3}^{2} + a_{1}^{2}a_{4}^{2} + a_{2}^{2}a_{3}^{2}
+ a_{2}^{2}a_{4}^{2} + a_{3}^{2}a_{4}^{2}) + 8a_{1}a_{2}a_{3}a_{4},
Q_{1} = a_{1}a_{2}a_{3}a_{4}(a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2}) + a_{1}^{2}a_{2}^{2}(a_{3}^{2} + a_{4}^{2}) + a_{3}^{2}a_{4}^{2}(a_{1}^{2} + a_{2}^{2}).

Figure 9. Three possible cases of C_{4}(a_{1}, a_{2}, a_{3}, a_{4}) - quadrangles

Similarly for (j, ν) ∈ {(1, 1), (0, 2)} we have the main equation in the form

R_{2}ρ^{2} − Q_{2} = 0, (40)

where

R_{2} = a_{1}^{4} + a_{2}^{4} + a_{3}^{4} + a_{4}^{4} − 2(a_{1}^{2}a_{2}^{2} + a_{1}^{2}a_{3}^{2} + a_{1}^{2}a_{4}^{2}
+ a_{2}^{2}a_{3}^{2} + a_{2}^{2}a_{4}^{2} + a_{3}^{2}a_{4}^{2}) + 8a_{1}a_{2}a_{3}a_{4},
Q_{2} = a_{1}a_{2}a_{3}a_{4}(a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2}) − a_{1}^{2}a_{2}^{2}(a_{3}^{2} + a_{4}^{2}) − a_{3}^{2}a_{4}^{2}(a_{1}^{2} + a_{2}^{2}).

Let us remark that R_{2} = Q_{2} = 0 when the quadrangle is equilateral. In this case (40) has infinitely many solutions, compare the Figure 10.

The cases (j, ν) = (1, 1), (0, 2) are mutually exclusive, cannot be satisfied simultaneously. Therefore just two different circumcircles exists, consult Figure 9 where the second circle realizes in the case j = 1, ν = 1, while the third circle happens for j = 0, ν = 2.

By β_{1} +β_{2} +β_{3} +β_{4} = (2−j −ν)π the main equation concerning chordal polygons living in C_{4}(a_{1}, a_{2}, a_{3}, a_{4}) we obtain using sin(β_{1} + β_{2} + β_{3} + β_{4}) = 0. Repeating the procedure explained about the equation (39), we get

(R_{1}ρ^{2} − Q_{1})(R_{2}ρ^{2} − Q_{2}) = 0. (41)

Figure 10. Trivial degenerated C_{4}(a, a, a, a) - quadrangles

Example 6 Let a_{1}, a_{2}, a_{3}, a_{4}, a_{5} be given. Then

β_{1} + β_{2} + β_{3} + β_{4} + β_{5} = (5 − 2(j + ν))π/2,

so we begin the main equation derivation transforming e.g.

cos(β_{1} + β_{2} + β_{3}) = ± sin(β_{4} + β_{5}).

After hard, but obvious computation we deduce

(42) G sin β_{1} sin β_{2} + H sin β_{2} sin β_{3} + K sin β_{3} sin β_{1} = L

where

G = −64a_{1}^{3}a_{2}^{3}a_{3}^{4}x^{10} + 16(4a_{1}^{3}a_{2}^{3}a_{3}^{2} + 2a_{1}^{3}a_{2}a_{3}^{4} + 2a_{1}a_{2}^{3}a_{3}^{4} − a_{1}a_{2}a_{3}^{2}a_{4}^{2}a_{5}^{2})x^{8}
+ 8(−4a_{1}^{3}a_{2}a_{3}^{2} − 4a_{1}a_{2}^{3}a_{3}^{2} − 2a_{1}a_{2}a_{3}^{4} − a_{1}^{3}a_{2}^{3} + a_{1}a_{2}a_{3}^{2}a_{4}^{2} + a_{1}a_{2}a_{3}^{2}a_{5}^{2}
+ a_{1}a_{2}a_{4}^{2}a_{5}^{2})x^{6} + 4(a_{1}a_{2}^{3} + a_{1}^{3}a_{2} + 3a_{1}a_{2}a_{3}^{2} − a_{1}a_{2}a_{4}^{2} − a_{1}a_{2}a_{5}^{2})x^{4},
H = −64a_{1}^{4}a_{2}^{3}a_{3}^{3}x^{10} + 16(4a_{1}^{2}a_{2}^{3}a_{3}^{3} + 2a_{1}^{4}a_{2}a_{3}^{3} + 2a_{1}^{4}a_{2}^{3}a_{3} − a_{1}^{2}a_{2}a_{3}a_{4}^{2}a_{5}^{2})x^{8}
+ 8(−4a_{1}^{2}a_{2}a_{3}^{3} − 4a_{1}^{2}a_{2}^{3}a_{3} − 2a_{1}^{4}a_{2}a_{3} − a_{2}^{3}a_{3}^{3} + a_{1}^{2}a_{2}a_{3}a_{4}^{2} + a_{1}^{2}a_{2}a_{3}a_{5}^{2}
+ a_{2}a_{3}a_{4}^{2}a_{5}^{2})x^{6} + 4(a_{2}a_{3}^{3} + a_{2}^{3}a_{3} + 3a_{1}^{2}a_{2}a_{3} − a_{2}a_{3}a_{4}^{2} − a_{2}a_{3}a_{5}^{2})x^{4},
K = −64a_{1}^{3}a_{2}^{4}a_{3}^{3}x^{10} + 16(4a_{1}^{3}a_{2}^{2}a_{3}^{3} + 2a_{1}a_{2}^{4}a_{3}^{3} + 2a_{1}^{3}a_{2}^{4}a_{3} − a_{1}a_{2}^{2}a_{3}a_{4}^{2}a_{5}^{2})x^{8}
+ 8(−4a_{1}a_{2}^{2}a_{3}^{3} − 4a_{1}^{3}a_{2}^{2}a_{3} − 2a_{1}a_{2}^{4}a_{3} − a_{1}^{3}a_{2}^{3} + a_{1}a_{2}^{2}a_{3}a_{4}^{2} + a_{1}a_{2}^{2}a_{3}a_{5}^{2}
+ a_{1}a_{3}a_{4}^{2}a_{5}^{2})x^{6} + 4(a_{1}a_{3}^{3} + a_{1}^{3}a_{3} + 3a_{1}a_{2}^{2}a_{3} − a_{1}a_{3}a_{4}^{2} − a_{1}a_{3}a_{5}^{2})x^{4},

L = −64a_{1}^{4}a_{2}^{4}a_{3}^{4}x^{12} + 16(4a_{1}^{2}a_{2}^{4}a_{3}^{4} + 4a_{1}^{4}a_{2}^{2}a_{3}^{4} + 4a_{1}^{4}a_{2}^{4}a_{3}^{2} − a_{1}^{2}a_{2}^{2}a_{3}^{2}a_{4}^{2}a_{5}^{2})x^{10}
+ 8(−8a_{1}^{2}a_{2}^{2}a_{3}^{4} − 8a_{1}^{2}a_{2}^{4}a_{3}^{2} − 8a_{1}^{4}a_{2}^{2}a_{3}^{2} − a_{1}^{4}a_{2}^{4} − a_{1}^{4}a_{3}^{4} − a_{2}^{4}a_{3}^{4}
+ a_{1}^{2}a_{2}^{2}a_{3}^{2}a_{4}^{2} + a_{1}^{2}a_{2}^{2}a_{3}^{2}a_{5}^{2} + a_{1}^{2}a_{2}^{2}a_{4}^{2}a_{5}^{2} + a_{1}^{2}a_{3}^{2}a_{4}^{2}a_{5}^{2} + a_{2}^{2}a_{3}^{2}a_{4}^{2}a_{5}^{2})x^{8}
+ 4(2a_{1}^{2}a_{2}^{4} + 2a_{1}^{4}a_{2}^{2} + 2a_{1}^{2}a_{3}^{4} + 2a_{1}^{4}a_{3}^{2} + 2a_{2}^{2}a_{3}^{4} + 2a_{2}^{4}a_{3}^{2} + 15a_{1}^{2}a_{2}^{2}a_{3}^{2}
− a_{1}^{2}a_{2}^{2}a_{4}^{2} − a_{1}^{2}a_{2}^{2}a_{5}^{2} − a_{1}^{2}a_{3}^{2}a_{4}^{2} − a_{1}^{2}a_{3}^{2}a_{5}^{2} − a_{1}^{2}a_{4}^{2}a_{5}^{2} − a_{2}^{2}a_{3}^{2}a_{4}^{2} − a_{2}^{2}a_{3}^{2}a_{5}^{2}
− a_{2}^{2}a_{4}^{2}a_{5}^{2} − a_{3}^{2}a_{4}^{2}a_{5}^{2})x^{6} + (−a_{1}^{4} − a_{2}^{4} − a_{3}^{4} − a_{4}^{4} − a_{5}^{2} − 6a_{1}^{2}a_{2}^{2} − 6a_{2}^{2}a_{3}^{2}
− 6a_{1}^{2}a_{3}^{2} + 2a_{1}^{2}a_{4}^{2} + 2a_{1}^{2}a_{5}^{2} + 2a_{2}^{2}a_{4}^{2} + 2a_{2}^{2}a_{5}^{2} + 2a_{3}^{2}a_{4}^{2} + 2a_{3}^{2}a_{5}^{2} + 2a_{4}^{2}a_{5}^{2})x^{4},

and the abreviation x = 1/(2ρ) is used.

Now, transforming once more (42), writing s_{j} = sin β_{j}, we have

4s_{1}^{2}s_{2}^{2}(HKs_{3}^{2} + GL)^{2} = (L^{2} + G^{2}s_{1}^{2}s_{2}^{2} − H^{2}s_{2}^{2}s_{3}^{2} − K^{2}s_{1}^{2}s_{3}^{2})^{2}. (43)

The equation (43) can be written in the form f(x) = 0, where f is a polynomial in x. To write this polynomial explicitely we need few pages therefore it is omitted. We shall here restrict ourselves to the use of the form (43) and consider the following special cases.

(i) Let a_{1} = a_{2} = a_{3} = 1, a_{4} = a_{5} = \sqrt{2}. In this case there are pentagons where sin β_{j} sin β_{k}, j, k ∈ {1, 2, 3}, j ≠ k are all positive. Then, (43) becomes

i  256x^{8} − 512x^{6} + 352x^{4} − 92x^{2} + 9 = 0,

where x_{1,2} = cos \frac{π}{3} is its double root (compare the first and the second pentagon on the Figure 11). Let us remark that x_{3} = cos β_{1} = \frac{\sqrt{7}}{4} is not the root of (i)

but the root of (x^{2} − 1)G = L which becomes 16x^{2} − 7 = 0. Thus 2ρ_{3} = \frac{4}{\sqrt{7}}

is the diameter of the circumcircle of the third pentagon. Finally, it is not hard to see that if three consecutive sides of the chordal pentagon have the same lengths, then at most three different circumcircles could arise.

(ii) Let a_{1} = a_{2} = a_{3} = 1, a_{4} = 2, a_{5} = 3. Then there is only one pentagon whose sides have given lengths (Figure 11). Since

\begin{aligned}G = H = K & = & −64x^{10} − 448x^{8} + 304x^{6} − 32x^{4}, \\ L & = & −64x^{12} − 384x^{10} + 752x^{8} − 480x^{6} + 32x^{4}, \end{aligned}

the equation (43) becomes now

16x^{8} − 96x^{6} + 188x^{4} − 93x^{2} + 8 = 0.

we find that 3.0364 < 2ρ < 3.0365 since

3 arccos \frac{1}{3.0364} + arccos \frac{2}{3.0364} + arccos \frac{3}{3.0364} = 269.99^{0} < \frac{3π}{2} ,
3 arccos \frac{1}{3.0365} + arccos \frac{2}{3.0365} + arccos \frac{3}{3.0365} = 270.01^{0} > \frac{3π}{2} .

Let us remark that using the equation (x^{2} − 1)G = M we deduce x = 0 or ρ = ∞, compare the second pentagon on Figure 11.

Figure 11. P (1, 1, 1, 2, 3) pentagons with 2ρ_{1} ≈ 3.0364, 2ρ_{2} = ∞

(iii) Now, let a_{1} = a_{2} = 1, a_{3} = 2, a_{4} = 3, a_{5} = 4. Then two pentagons are there whose sides have these given lengths, see Figure 12. Since

\begin{aligned}G & = & −1024x^{10} − 7936x^{8} + 1432x^{6} − 44x^{4}, \\ H & = & −512x^{10} − 3776x^{8} + 2288x^{6} − 136x^{4} = K, \\ L & = & −1024x^{12} − 6912x^{10} + 9368x^{8} − 3780x^{6} + 179x^{4}, \end{aligned}

then (43) can be written as

\begin{aligned}±2(1 − x^{2})(HK(1 − 4x^{2}) + GL) & = & L^{2} + G^{2}(1 − x^{2})^{2} \\ & − & (H^{2} + K^{2})(1 − x^{2})(1 − 4x^{2}), \end{aligned}

which means

iii  4(1 − x^{2})(1 − 4x^{2})H^{2} = (L − G(1 − x^{2}))^{2}, L + (1 − x^{2})G = 0,

where the first equation in (iii) corresponds to the sign +, and the second one to sign − in its initial equation.

Figure 12. Pentagons P (1, 1, 2, 3, 4) with 2ρ_{1} ≈ 4.12004, 2ρ_{2} ≈ 4.13118

From the first equation we obtain 4.12004 < 2ρ_{1} < 4.12005, and for the second one it is 2ρ_{2} ≈ 4.13118 (it is the diameter of the circumcircle of the the triangle A_{3}A_{4}A_{5} on the Figure 12).

(iv) Let a_{j} = j, j = \overline{1, 5}. Then there are only two different radii - the second and the third pentagon have equal diameters, compare Figure 13.

Figure 13. Pentagons P (1, 2, 3, 4, 5) with 2ρ_{1} ≈ 5.43513, 2ρ_{2} ≈ 507762

The diameters of these are 5.43513 < 2ρ_{1} < 5.43514 and 5.07762 < 2ρ_{2} <

5.07763.

5 On properties of main equation

It is clear that the main equations

21  F_{1}(a_{1}, ⋯, a_{n}; ρ) = 0, n odd

22  F_{2}(a_{1}, ⋯, a_{n}; ρ) = 0, n even

have the following properties.

1. The radius of the circumcircle of k - inscribed chordal polygon whose sides have the lengths a_{1}, ⋯, a_{n} is a positive root of the equation (21) or (22).

2. The equations (21), (22) depending on a_{1}, ⋯, a_{n} may have no one positive root, may have exactly one positive root or may have at least σ[n] positive roots if our Conjecture is true.

Here we give an another property of the main equation.

Theorem 3 The positive roots of the main equations (21), (22) are the radii of the circumcircles of the chordal n-gons whose sides have the lengths a_{1}, ⋯, a_{n}.

Proof. Let n be odd. We transform cos(β_{1} + ⋯ + β_{n}) = 0 into the form

𝔖_{0}^{n} − 𝔖_{2}^{n} + 𝔖_{4}^{n} − ⋯ + (−1)^{θ_{1}}𝔖_{n−1}^{n} = 0,

where θ_{1} = (1 + 3 + 5 + ⋯ + (n − 1)) + 1, compare (19). Expressing sin β_{1} from the previous equality, we have U_{1} sin β_{1} = V_{1}; (44)

no integer ζ_{1} exists that sin^{2ζ_{1}+1} β_{1} is a factor of addends in U_{1} or V_{1} unless sin^{2ζ_{1}+2} β_{1} is a factor there as well.

By squaring (44) we get U_{1}^{2} sin^{2} β_{1} = V_{1}^{2}. Writing this equality as

U_{2} sin β_{2} = V_{2}, (45)

On Equations and Properties Concerning Some Classes of Chordal Polygons 77 we deduce that no integer ζ_{2} is there that sin^{2ζ_{2}+1} β_{2} is a factor of an addend in U_{2} or of V_{2} unless sin^{2ζ_{2}+2} β_{2} is a factor as well.

Repeating this procedure we finish with U_{n}^{2} sin^{2} β_{n} = V_{n}^{2}. (46) Now we replace

\frac{a_{j}}{2ρ} ⟶ cos β_{j}, (1 − (\frac{a_{j}}{2ρ})^{2})^{ζ} ⟶ sin^{2ζ} β_{j} (47)

in all n equalities

U_{l}^{2} sin^{2} β_{l} = V_{l}^{2}, l = \overline{1, n}.

This iterative procedure ends with (46), which gives us equation (21).

If ρ_{m} is a positive root of (21), let p be the first integer with the property that ρ_{m} is a root of algebaric equation obtained by the replacement procedure (47) from U_{p}^{2} sin^{2} β_{p} = V_{p}^{2}. (48) Then the following cases arise

• ρ_{m} is a root of the equation which originates back to U_{p} sin β_{p} = V_{p},

• ρ_{m} is a root of the equation which originates back to U_{p} sin β_{p} = −V_{p}. In both cases ρ_{m} is the radius of the circumcircle CA of the chordal n-gon A with side lengths a_{1}, ⋯, a_{n}, whose angles are not all of the same sign.

Similar holds for n even. Remark 2 We point out that

U_{p}^{2} sin^{2} β_{p} = V_{p}^{2}

may generate the main equation of the chordal polygon for p < n. Example 7 So, consider e.g. a equilateral pentagon with unit side lengths. Then cos(4β + (−1)^{ϵ_{j}}β) = 0, (49) since in this case only one angle may be negative. Transforming (49) we derive the equation

64 cos^{6} β − 128 cos^{4} β + 80 cos^{2} β − 15 = 0

or 15ρ_{k}^{6} − 20ρ_{k}^{4} + 80ρ_{k}^{2} − 1 = 0, (50) where ρ_{k} cos β(k) = 1, k = 1, 2, 3 and

β(1) = \frac{3π}{10} , β(2) = \frac{π}{10}, β(3) = \frac{π}{6} .

Figure 14. P (1) pentagons with ρ_{1} = \frac{1}{2 cos \frac{3π}{10}} , ρ_{2} = \frac{1}{2 cos \frac{π}{10}} , ρ_{3} = \frac{1}{2 cos \frac{π}{6}}

Adopting the expression (43) in the Example 4 to our case, it becomes

256x^{8} − 704x^{6} + 704x^{4} − 300x^{2} + 45 = 0, (51)

with positive solutions

x_{1} = cos \frac{3π}{10} , x_{2} = cos \frac{π}{10}, x_{3} = cos \frac{π}{6} .

Thus 2ρ_{j}x_{j} = 1, j = 1, 2, 3, see Figure 14.

Also, let us remark that the equation (50) can be rewritten into

(5ρ_{k}^{4} − 5ρ_{k}^{2} + 1)(3ρ_{k}^{2} − 1) = 0,

and then we have at most

s[5] = [\frac{5 − 1}{2}] + [\frac{5 − 3}{2}] = 2 + 1 = 3

positive roots.

Consider at the end a new kind of n-gons. Denote A(1) ≡ A = A_{1} ⋯ A_{n} as above, and let λ ≥ 2 positive integer. Then we are interested in the λn-gon

A(λ) = \underbrace{A(1) ⋯ A(1)}_{λ},

which circumcircle C_{A(λ)} coincides with C_{A(1)}. Now, the new class of λn-gons C_{λn}((a_{1}, ⋯, a_{n})_{λ}), say, we build with the polygons A(λ) whose initial side lengths are a_{1}, ⋯, a_{n} using it λ times.

Theorem 4 Let a_{1}, ⋯, a_{n} be given lengths and let s and t be positive integers such that s|t. Then

F ((a_{1}, ⋯, a_{n})_{s}; ρ)|F ((a_{1}, ⋯, a_{n})_{t}; ρ),

where F ((a_{1}, ⋯, a_{n})_{s}; ρ) = 0 and F ((a_{1}, ⋯, a_{n})_{t}; ρ) = 0 are the main equations of the polygons in C_{sn}((a_{1}, ⋯, a_{n})_{s}) and C_{tn}((a_{1}, ⋯, a_{n})_{t}) respectively.

Proof. Firstly it could be say that the assertion of the Theorem is evident, since the polygons A(s), A(t) have equal diameters. Of course, it can be deduced the same using the properties of sine and cosine too. Namely, noting τ = β_{1} + ⋯ + β_{n}, we have (52) sin(sτ)| sin q(sτ) q integer,

(53) cos(sτ)| cos q(sτ) q odd,

(54) sin(sτ)| sin q(sτ) q even.

Starting with sin(sτ) = 0 we get F ((a_{1}, ⋯, a_{n})_{s}; ρ) = 0, and starting with sin q(sτ)/ sin(sτ) we finish with the equation T ((a_{1}, ⋯, a_{n})_{q}; ρ) = 0, that

F ((a_{1}, ⋯, a_{n})_{s}; ρ)T ((a_{1}, ⋯, a_{n})_{q}; ρ) = F ((a_{1}, ⋯, a_{n})_{t}; ρ).

Similarly we get the assertions for (53) and (54). Example. The circumcircles of triangle A(1) = A_{1}A_{2}A_{3} and the hexagon A(2) have the same diameters. Using

β_{1} + β_{2} + β_{3} + β_{1} + β_{2} + β_{3} = (6 − 2(j + ν))\frac{π}{2}

we have sin 2(β_{1} + β_{2} + β_{3}) = 0, i.e. sin(β_{1} + β_{2} + β_{3}) = 0, cos(β_{1} + β_{2} + β_{3}) = 0. Starting with cos(β_{1} + β_{2} + β_{3}) = 0, we obtain the equation F (a_{1}, a_{2}, a_{3}; ρ) = 0 and transforming sin(β_{1} +β_{2} +β_{3}) = 0 we finish the computation with T (a_{1}, a_{2}, a_{3}; ρ) = 0, so

F ((a_{1}, a_{2}, a_{3})_{2}; ρ) = F (a_{1}, a_{2}, a_{3}; ρ)T (a_{1}, a_{2}, a_{3}; ρ) = 0.

The diameter of C_{A(1)} and C_{A(2)} is a root of the equation F (a_{1}, a_{2}, a_{3}; ρ) = 0. The diameters of the circumcircles of all other hexagons (whose sides have prescribed lengths) are the roots of T (a_{1}, a_{2}, a_{3}; ρ) = 0. In this case the equations are:

F (a_{1}, a_{2}, a_{3}; ρ) = (4a_{1}^{2}a_{2}^{2}a_{3}^{2}x^{2} − 2a_{1}^{2}a_{2}^{2} − 2a_{1}^{2}a_{3}^{2} − 2a_{2}^{2}a_{3}^{2} + a_{1}^{4} + a_{2}^{4} + a_{3}^{4})x^{4} = 0,
T (a_{1}, a_{2}, a_{3}; ρ) = 4a_{1}^{2}a_{2}^{2}a_{3}^{2}x^{6} − (2a_{1}^{2}a_{2}^{2} + 2a_{1}^{2}a_{3}^{2} + 2a_{2}^{2}a_{3}^{2} + a_{1}^{4} + a_{2}^{4} + a_{3}^{4})x^{4}
+ 2(a_{1}^{2} + a_{2}^{2} + a_{3}^{2})x^{2} − 1 = 0,

where x = 1/(2ρ). Acknowledgement. It is the authors pleasure to thank Nataša Antonini from the Faculty of Maritime Studies, University of Rijeka for the her help in preparing the figures in the article.

References

[1] S. Góźdź, A convex polygon as a discrete plane curve, Balkan J. Geom. Appl. 4, No.2.(1999), 29-48.

[2] J. Horváth, On a Kürschák method, Proceedings of the II. Sopron Conference for Math Teachers, 1994, 21-24.

[3] J. Kürschák, On inscribed and circumscribed polygons to circle, Mat. Term. Tud.Értesítő 5(1886-1887), 153-160. (Hungarian)

[4] J. Kürschák, Über dem Kreis ein- und umgeschriebene, Math. Ann. 30(1887), 578-581.

[5] P. Pech, The harmonic analysis of the polygons and Napoleon’s theorem, J. Geom. Graph. 5, No.1.(2001), 13-22.

[6] T. Pogány and M. Radić, On (k, f, l)-chordal polygons, Balkan J. Geom. Appl. 5, No.2.(2000), 91-96.

[7] M. Radić, Some inequalities and properties concerning chordal polygons, Math. Inequal. Appl. 2, No.1.(1999), 141-150.

[8] M. Radić, Some inequalities and properties concerning chordal semi - polygons, Math. Inequal. Appl. 4, No.2.(2001), 301-322.

[9] M. Radić, On algerbaic equations conserning semi-tangential polygons, Math. Commun. 6, No.1.(2001), 73-81.

[10] M. Radić, Some relations concerning k-chordal and k-tangential polygons, Math. Commun. 7(2002), 21-34.

[11] M. Radić and T. Pogány, Some inequalities concerning to the existence of (k, λ, l)- chordal polygons, Acta Math. Acad. Paedagog. Nyházi (N.S.) (to appear).

[12] P.R. Scott and P.W. Yong, Inequalities for convex sets, JIPAM, J. Inequal. Pure Appl. Math. 1, No.1, A.6 (2000).

[13] Á. Temesvári, The power sum of the sides of a polygon, Mat.Lapok 24, No.3-4.(1973), 391-396. (Hungarian)

Mirko Radić, University of Rijeka, Faculty of Philosophy, Department of Mathematics, 51000 Rijeka, Omladinska 14, Croatia

Tibor K. Pogány, University of Rijeka, Faculty of Maritime Studies, Department of Sciences, 51000 Rijeka, Studentska 2, Croatia email: poganj@brod.pfri.hr

Vladimir Kadum, University of Rijeka, College of Education in Pula, 52100 Pula, Ivana Matetića Ronjgova 1, Croatia email: vladimirkadum@pu.hinet.hr