Applied Mathematics E-Notes, 1(2001), 118-123 \begin{matrix}c \\ ◯ \end{matrix} Available free at mirror sites of http://math2.math.nthu.edu.tw/∼amen/

Algebraic Equations Connected with Tangential Polygons and Their Solvability by Radicals ^{∗}

Mirko Radić^{†}, Tibor K. Pogány^{‡}

Received 1 March 2001

Abstract

Certain algebraic equations connected with tangential polygons are considered. By means of their geometrical interpretations, we prove that these equations are solvable by radicals.

1 Introduction

A polygon with vertices A_{1}, ⋯, A_{n} will be denoted by A ≡ A_{1} ⋯ A_{n}. The lengths of its sides will be denoted by a_{j} = |A_{j}A_{j+1}| and for the interior angle at vertex A_{j} we write

A̸_{j} = ^{}A̸_{j−1}A_{j}A_{j+1}, A_{0} ≡ A_{n}, j = \overline{1, n}.  (1)

We say that A is tangential if there exists a circle C_{A} such that each side of A is a tangential line of C_{A}. We will assume throughout the article that no two of the consecutive vertices are the same in A. The centre and the radius of C_{A} will be denoted by C and r respectively. Also, the integral part of a real number x will be denoted by [a].

Consider the angles

β_{j} = μ(^{}C̸A_{j}A_{j+1}), j = \overline{1, n},  (2)

where μ(x) means the “measure of x”. Thus, 2β_{j} = μ(^{}A̸_{j}).

Let A be a tangential n-gon, and let k be a positive integer such that k ≤ [(n − 1)/2]. Then the polygon A will be called k-tangential if any two of its consecutive sides have only one point in common, and if there holds

β_{1} + ⋯ + β_{n} = (n − 2k)\frac{π}{2} .  (3)

Consequently, a tangential polygon A is k-tangential if ∑_{j=1}^{n} φ_{j} = 2kπ, where φ_{j} = μ(^{}A̸_{j}CA_{j+1}). Indeed, from 2β_{j} = μ(^{}A̸_{j}) it follows that β_{j} + β_{j+1} = π − φ_{j}, and using (3) twice, our assertion follows.

∗Mathematics Subject Classifications: 11S05, 12E05, 51E12.

†Division of Mathematics, Faculty of Philosophy, University of Rijeka, Rijeka 51000, Croatia

‡Department of Maritime Studies, University of Rijeka, Rijeka 51000, Crotia

It is easy to see that k cannot be greater that (n − 1)/2 for n odd and cannot be greater than n/2 − 1 when n is even.

If A is k-tangential and if t_{1}, ⋯, t_{n} are the lengths of its tangents, then

t_{j} = r cot β_{j}, j = \overline{1, n}.  (4)

Let t_{1}, ⋯, t_{n} be any given lengths of some line segments (in fact any positive numbers),

and fix j between 1 and n. We will let S_{j}^{n} denote the sum of all ( \begin{matrix}n \\ j \end{matrix} ) products of

the form t_{i_{1}} ⋯ t_{i_{j}}, where i_{1}, ⋯ i_{j} are different elements of {1, ⋯, n}, i.e.

S_{j}^{n} = ∑_{1≤i_{1}<···<i_{j}≤n} t_{i_{j}} ⋯ t_{i_{j}}.  (5)

Finally we employ the symbol P (t_{1}, ⋯, t_{n}; β_{1}, ⋯, β_{n}; r) to denote the tangential n-gon A with given lengths of its tangents t_{1}, ⋯, t_{n} (the angles β_{j} are specified already and r is the radius of C_{A}).

2 Tangential Polygon and Corresponding Equation

Corollary 3 in the article [1] asserts that the radius r_{k} of C_{A} inscribed in P (t_{1}, ⋯, t_{n}; β_{1}, ⋯, β_{n}; r_{k}) is a root of the equation

S_{1}^{n}x^{n−1} − S_{3}^{n}x^{n−3} + ⋯ + (−1)^{ν}S_{n}^{n} = 0  (6)

if n is odd and a root of

S_{1}^{n}x^{n−2} − S_{3}^{n}x^{n−4} + ⋯ + (−1)^{η}S_{n−1}^{n} = 0  (7)

if n is even, where ν = (1 + 3 + 5 + ⋯ + n) + 1 and η = (1 + 3 + 5 + ⋯ + (n − 1)) + 1. Here we shall improve this result by the following theorem.

THEOREM 2.1. Let t_{1}, ⋯, t_{n} any given lengths (in fact any positive real numbers). Then there are tangential polygons

P (t_{1}, ⋯, t_{n}; β_{1}^{(k)}, ⋯, β_{n}^{(k)}; r_{k}), k = \overline{1, [(n − 1)/2]},

where β_{1}^{(k)} + ⋯ + β_{n}^{(k)} = (n − 2k)π/2.

PROOF. Since arctan x is continuous it is clear that for each fixed k = \overline{1, [(n − 1)/2]}, there is r_{k} which satisfies

∑_{j=1}^{n} arctan (\frac{r_{k}}{t_{j}} ) = (n − 2k)\frac{π}{2} .

Thus β_{j}^{(k)} = arctan(r_{k}/t_{j}) for j = \overline{1, n}.

From this result it follows that the equations (6), (7) possess real roots only.

COROLLARY 2.1. The roots x_{k} of the equations

∑_{j=1}^{(n+1)/2}(−1)^{j+1} ( \begin{matrix}n \\ 2j − 1 \end{matrix} ) x^{(n−2j+1)/2} = 0, n odd,  (8)

∑_{j=1}^{n/2}(−1)^{j+1} ( \begin{matrix}n \\ 2j − 1 \end{matrix} ) x^{(n−2j)/2} = 0, n even,  (9)

are of the form

x_{k} = r_{k}^{2} = tan^{2}(n − 2k)\frac{π}{2n}, k = \overline{1, [(n − 1)/2]}.  (10)

Indeed, this follows by putting t_{1} = ⋯ = t_{n} = 1 into (6) and (7) respectively.

COROLLARY 2.2. When n is a prime number, then equation (8) is normal over the field Q (of rational numbers).

PROOF. If n is a prime, and if ν = (1 + 3 + ⋯ + n) + 1, we may assert that the equation

( \begin{matrix}n \\ 1 \end{matrix} ) x^{(n−1)/2} − ( \begin{matrix}n \\ 3 \end{matrix} ) x^{(n−3)/2} + ⋯ + (−1)^{ν} = 0  (11)

is irreducible over the field Q because the equation

( \begin{matrix}n \\ 1 \end{matrix} ) − ( \begin{matrix}n \\ 3 \end{matrix} ) y + ⋯ + (−1)^{ν}y^{(n−1)/2} = 0

is irreducible over Q by the Eisenstein criterion. Using (10), and the straightforward formula

tan(mα) = \frac{( \begin{matrix}m \\ 1 \end{matrix} ) tan α − ( \begin{matrix}m \\ 3 \end{matrix} ) tan^{3} α + ⋯}{1 − ( \begin{matrix}m \\ 2 \end{matrix} ) tan^{2} α + ( \begin{matrix}m \\ 4 \end{matrix} ) tan^{4} α − ⋯} ,

it is easy to see that the rootfield of the equation (11) can be generated by only one root, that is,

Q (tan^{2} \frac{π}{2n}) = Q (tan^{2} \frac{π}{2n}, ⋯, tan^{2}(n − 2)\frac{π}{2n}) .

The proof is complete.

COROLLARY 2.3. The order of the Galois group of equation (8) is (n − 1)/2 when n is a prime, and this group is solvable.

PROOF. Since Q (tan^{2} \frac{π}{2n}) has the degree (n−1)/2 over the field Q, the order of the Galois group of (8) is (n − 1)/2 too. If Q_{1} = Q (tan^{2} \frac{π}{2n}) and Q_{2} = Q (exp\{iπ/(2n)\}), then Q_{1} is a subfield of Q_{2}. The subfield Q_{1} is the rootfield of the equation (8) and Q_{2} is the rootfield of the equation x^{2n} − 1 = 0. As is well known the Galois group of the equation x^{2n} − 1 = 0 is solvable and every subgroup of a solvable group is solvable.

COROLLARY 2.4. Equations (8) and (9) are solvable by radicals for any positive integer n.

PROOF. If m is a positive integer, then tan π/(2n) can be expressed by radicals over Q since exp{iπ/(2n)} can be expressed by radicals over Q.

THEOREM 2.2. Let λ and n be given positive integers and let t_{1}, ⋯, t_{λn} be any given lengths (positive real numbers), such that

t_{i+jn} = t_{i}, i = \overline{1, n}, j = \overline{1, λ − 1}.  (12)

Now, assume κ = λn, and let f_{1}(x), f_{2}(x), g_{1}(x), g_{2}(x) be the following polynomials:

f_{1}(x) = S_{1}^{κ}x^{κ−1} − S_{3}^{κ}x^{κ−3} + ⋯ + (−1)^{ν_{κ}}S_{κ}^{κ} κ odd
f_{2}(x) = S_{1}^{κ}x^{κ−2} − S_{3}^{κ}x^{κ−4} + ⋯ + (−1)^{η_{κ}}S_{κ−1}^{κ} κ even
g_{1}(x) = S_{1}^{n}x^{n−1} − S_{3}^{n}x^{n−3} + ⋯ + (−1)^{ν}S_{n}^{n} n odd
g_{2}(x) = S_{1}^{n}x^{n−2} − S_{3}^{n}x^{n−4} + ⋯ + (−1)^{η}S_{n−1}^{n} n even,

where ν_{κ} = (1 + 3 + ⋯ + κ) + 1, η_{κ} = (1 + 3 + ⋯ + (κ − 1)) + 1, and ν, η are defined already. Then

(g_{1}(x) |f_{1}(x) or g_{1}(x) |f_{2}(x)) and g_{2}(x)| f_{2}(x).  (13)

PROOF. As in the proof of Theorem 2.1, this theorem is almost evident since the circle C_{A(1)} inscribed in the n-gon A(1) = A_{1} ⋯ A_{n} has the same radius as the circle C_{A(λ)} inscribed in the λn-gon

A(λ) = \underbrace{A(1) ⋯ A(1)}_{λ} .

From

λ(β_{1}^{(k)} + ⋯ + β_{n}^{(k)}) = (λn − 2k)π/2,  k = \overline{1, [(λn − 1)/2]},  (14)

we have

β_{1}^{(k)} + ⋯ + β_{n}^{(k)} = (n − \frac{2k}{λ} ) \frac{π}{2} ,  k = \overline{1, [(λn − 1)/2]},  (15)

and it follows that the set {1, ⋯, [(λn − 1)/2]} contains m = [(n − 1)/2] numbers k_{1}, ⋯, k_{m}, say, that k_{j} = jλ, j = \overline{1, m}. Thus there are m tangential polygons

P (t_{1}, ⋯, t_{n}; β_{1}^{(k_{j})}, ⋯, β_{n}^{(k_{j})}; r_{k_{j}}), j = \overline{1, [(n − 1)/2]},

such that their radii are the roots of the equations

S_{1}^{n}x^{n−1} − S_{3}^{n}x^{n−3} + ⋯ + (−1)^{ν}S_{n}^{n} = 0  n odd,
S_{1}^{n}x^{n−2} − S_{3}^{n}x^{n−4} + ⋯ + (−1)^{η}S_{n−1}^{n} = 0  n even.

The proof is complete.

We remark that the polynomials which correspond to the angles (n − \frac{2k}{λ} ) \frac{π}{2} for different k may have different degrees with different parities. The following is an example.

122 Algebraic Equations

EXAMPLE. When λ = 4 and n = 5, we have

β_{1}^{(k)} + ⋯ + β_{5}^{(k)} = (5 − \frac{k}{2} ) \frac{π}{2} , k = \overline{1, 9}.

It follows that

tan (β_{1}^{(k)} + ⋯ + β_{5}^{(k)}) = 0,  k = 2, 6
cot (β_{1}^{(k)} + ⋯ + β_{5}^{(k)}) = 0,  k = 4, 8
tan (β_{1}^{(k)} + ⋯ + β_{5}^{(k)}) = 1,  k = 1, 5, 9
cot (β_{1}^{(k)} + ⋯ + β_{5}^{(k)}) = −1,  k = 3, 7.

Denote the corresponding polynomials by p_{j}(x), j = 1, 2, 3, 4. Then we have

p_{1}(x) = x^{4} − S_{2}^{5}x^{2} + S_{4}^{5},
p_{2}(x) = S_{1}^{5}x^{4} − S_{3}^{5}x^{2} + S_{5}^{5},
p_{3}(x) = x^{5} − S_{1}^{5}x^{4} − S_{2}^{5}x^{3} + S_{3}^{5}x^{2} + S_{4}^{5}x − S_{5}^{5},
p_{4}(x) = x^{5} + S_{1}^{5}x^{4} − S_{2}^{5}x^{3} − S_{3}^{5}x^{2} + S_{4}^{5}x + S_{5}^{5},

and consequently

p_{3}(x)p_{4}(x) = x^{10} − ((S_{1}^{5})^{2} + 2S_{2}^{5})x^{8} + (2S_{4}^{5} + 2S_{1}^{5}S_{3}^{5} + (S_{2}^{5})^{2})x^{6}
−(2S_{1}^{5}S_{5}^{5} + 2S_{2}^{5}S_{4}^{5} + (S_{3}^{5})^{2})x^{4} + (2S_{3}^{5}S_{5}^{5} + (S_{4}^{5})^{2})x^{2} − (S_{5}^{5})^{2},

and

4 ∏_{j=1}^{4} p_{j}(x) = S_{1}^{20}x^{18} − S_{3}^{20}x^{16} + S_{5}^{20}x^{14} − ⋯ − S_{19}^{20}.

We remark that

2p_{1}(x)p_{2}(x) = S_{1}^{10}x^{8} − S_{3}^{10}x^{6} + S_{5}^{10}x^{4} − S_{7}^{10}x^{2} + S_{9}^{10}.

Therefore the following relations

p_{2}(x)|S_{1}^{20}x^{18} − S_{3}^{20}x^{16} + ⋯ − S_{19}^{20}

and

2p_{1}(x)p_{2}(x)|S_{1}^{20}x^{18} − S_{3}^{20}x^{16} + ⋯ − S_{19}^{20},

are easily seen.

COROLLARY 2.5. If n ≤ 10, then both equations

S_{1}^{κ}x^{κ−1} − S_{3}^{κ}x^{κ−3} + ⋯ + (−1)^{ν_{κ}}S_{κ}^{κ} = 0, κ odd, (16)

and

S_{1}^{κ}x^{κ−2} − S_{3}^{κ}x^{κ−4} + ⋯ + (−1)^{η_{κ}}S_{κ−1}^{κ} = 0,  κ even,  (17)

are solvable by radicals for all positive integers λ. (Here κ = λn).

PROOF. From (15) it follows that

cot(λψ_{k}) = 0,  κ odd,
tan(λψ_{k}) = 0,  κ even,

where ψ_{k} = β_{1}^{(k)} + ⋯ + β_{n}^{(k)}. Transforming these two equations into sums of terms of the form tan(mβ_{j}^{(k)}) where m is a positive integer, then by means of

tan(mα) = \frac{1}{cot(mα)} = \frac{( \begin{matrix}m \\ 1 \end{matrix} ) tan α − ( \begin{matrix}m \\ 3 \end{matrix} ) tan^{3} α + ⋯}{1 − ( \begin{matrix}m \\ 2 \end{matrix} ) tan^{2} α + ( \begin{matrix}m \\ 4 \end{matrix} ) tan^{4} α − ⋯}

and by replacing tan β_{j}^{(k)} with x/t_{j}, one obtains algebraic equations in x which are solvable by radicals (here x denotes the radius of C_{A}). Indeed, when n ≤ 10, each of the equations

S_{1}^{n}x^{n−1} − S_{3}^{n}x^{n−3} + ⋯ + (−1)^{ν}S_{n}^{n} = 0,  n odd,  (18)

or

S_{1}^{n}x^{n−2} − S_{3}^{n}x^{n−4} + ⋯ + (−1)^{η}S_{n−1}^{n} = 0,  n even.  (19)

is solvable by radicals, and tan (n − \frac{2k}{λ} ) \frac{π}{2} can be explicitly expressed by radicals over Q(t_{1}, ⋯, t_{n}) according to Theorem 2.1 and its Corollaries.

COROLLARY 2.6. If equations (18) and (19) are solvable by radicals, then equations (16) and (17) are solvable by radicals as well.

We remark that it is quite possible that (16) and (17) are solvable by radicals for any n positive integer. But this remains to be proved.

References

[1] M. Radić, Some relations and properties concerning tangential polygons, Math. Commun. 4(1999), 197-206.

[2] M. Radić, On algebraic equations concerning semi-tangential polygons, Math. Commun., 6(2001), 73-81.