A Convex Polygon as a Discrete Plane Curve

Stanisław Góźdź

Abstract

In this paper we examine a convex polygon as a discrete substitute of a plane curve. We introduce a polygon with constant length of a diagonal as a counterpart of an oval with constant width. Moreover we define a convex polygon with constant perimeter of a special class circumscribed polygons.

Mathematics Subject Classification: 52C10, 42A32 Key words: n-polygon, diagonal, b-circumscribed, discrete Fourier sum

1 Introduction

In papers [1,3,4,6] applications of Fourier series to plane curves are presented . Plane curves examined in these papers are expressed by the following formulas:

t ⟼ z(t) = ∫_{0}^{t} f(s)e^{is}ds or z(t) = ∫_{0}^{t} k(s)f(s)e^{iK(s)}ds, (1.1)

where f is a periodic function. The representation of considered curves is associated with the integral. Therefore we search for a geometrical domain associated with a finite sum instead of an integral. The geometrical domain is included in the class of all convex polygons in the plane. To define a representation of a convex polygon we imitate formula (1.1). Therefore we consider a periodic sequence instead of a periodic function. Next we introduce a discrete Fourier series for a periodic sequence as follows:

Let x_{1}, x_{2}, x_{3}, … be a periodic sequence of real numbers with the period n,i.e.:

x_{v+n} = x_{v}, v = 0, 1, 2, ….

Then we apply a known trigonometrical interpolative polynomial

y(t) = a_{0} + ∑_{j=1}^{n−1}[a_{j} cos \frac{2πjt}{n} + b_{j}sin\frac{2πjt}{n}],

where

Balkan Journal of Geometry and Its Applications, Vol.4, No.2, 1999, pp. 29-48 \begin{matrix}c \\ ◯ \end{matrix}Balkan Society of Geometers, Geometry Balkan Press

a_{0} = \frac{1}{n} ∑_{μ=0}^{n−1} x_{μ}, a_{j} = \frac{1}{n} ∑_{μ=0}^{n−1} x_{μ} cos \frac{2π}{n} jμ, b_{j} = \frac{1}{n} ∑_{μ=0}^{n−1} x_{μ} sin \frac{2π}{n} jμ.

The trigonometrical interpolative polynomial satisfies the condition:

y(v) = x_{v}, v = 0, 1, … n − 1.

If we substitute instead the continuous variable t ∈ (−∞, +∞) the dicsrete variable v = 0, 1, 2, …, then we obtain

x_{v} = a_{0} + ∑_{j=1}^{n−1}[a_{j} cos \frac{2πjv}{n} + b_{j}sin\frac{2πjv}{n}]. (1.2)

In the sequel we call the formula (1.2) a discrete Fourier series of a periodic sequence {x_{v}}. We apply the discrete Fourier series to an n-polygon in the plane. The n-polygon

is a polygon with n sides having the same interior angles equal to 2π − \frac{2π}{n}, see [2,1]. In the paper n-polygon is ”a discrete curve”.

With reference to formula (1.1) we recall the following relations. There exists a strict correspondence between a property of curve (1.1) and a property of the function f. For example the following are known: Lemma A.(see [5,1]). A curve (1.1) is closed iff Fourier coefficients A_{1}, B_{1} of f vanish, i.e.: A_{1} = B_{1} = 0. Lemma B.(see [6]). If a closed curve represented by (1.1) is a curve with constant width ,then Fourier coefficients A_{2n}, B_{2n} of f vanish, i.e.:A_{2n} = B_{2n} = 0. Lemma C.(see [2,1]). If a closed curve represented by (1.1) is a curve with constant perimeter of a circumscribed m-polygon, then the Fourier coefficients A_{mj}, B_{mj}, j = 1, 2, 3, … vanish, i.e.:A_{mj} = B_{mj} = 0. Remark A. If f is a constant function (different from zero), then equation (1.1) forms a circle. This means that in this case all Fourier coefficients of f vanish with the exception of A_{0}.

In a discrete domain, n-polygon is represented as the sum

k ⟼ z_{k} = ∑_{v=0}^{k} x_{v}e^{i \frac{2πv}{n}} , (1.3)

where {x_{v}} is a sequence and k = 0, 1, … n − 1.

There exists a correspondence between a property of an n-polygon and a property of a sequence {x_{v}}. At the discrete domain a counterpart of a curve with constant width is a 2n-polygon with constant diagonal (see p.7). For 2n-polygon with constant diagonal the following counterpart of the Barbier theorem is satisfied:

L = π \frac{sin \frac{π}{2n}}{\frac{π}{2n}} d, (1.4)

where L denotes the perimeter of 2n-polygon and d is the length of a constant diagonal.

A counterpart of a curve with constant perimeter of a circumscribed m-polygon is an mn-polygon with constant perimeter of a b-circumscribed m-polygon defined as follows:

Let P be a convex polygon with vertices w_{1}, w_{2}, …, w_{n}, n > 2. To circumscribe a polygon with k sides (3 ≤ k ≤ n) on polygon P, we arbitrary choose vertices

w_{i_{1}}, w_{i_{2}}, …, w_{i_{k}}.

Next we draw a straight line l_{i_{1}}, l_{i_{2}}, …, l_{i_{k}} through vertices w_{i_{1}}, w_{i_{2}}, …, w_{i_{k}}. We consider only straight lines l_{i_{1}}, l_{i_{2}}, …, l_{i_{k}} passing through the outside angles of polygon P. The point of intersection of successive straight lines l_{i_{s}}, l_{i_{s+1}}, s = 1, 2, …, k − 1 is a vertex of the circumscribed polygon. We call the circumscribed polygon bcircumscribed on polygon P if and only if all straight lines l_{i_{1}}, l_{i_{2}}, …, l_{i_{k}} are bisectrices of the outside angles of the polygon P.

A property of a plane curve (represented by (1.1)) and ”a discrete theory plane curve ” are connected with the main result of the paper.

Perimeter 2πr of a circle with radius r can be obtained as the limit of perimeters of well-shaped regular polygons circumscribed on the circle. The above-mentioned idea and the counterpart of the Barbier theorem suggest that perimeter πd for an oval with constant width d can be obtained in the similarly way. To reach this aim we prove the following:

Theorem 1.1 Every 2n-polygon circumscribed on an oval with constant width δ is a 2n-polygon with constant diagonal equal to

\frac{δ}{cos \frac{π}{2n}} .

Theorem 1.2 Every mn-polygon circumscribed on an oval with constant perimeter l of a circumscribed m-polygon is mn-polygon with constant perimeter _{\frac{l}{cos \frac{π}{mn}}} of b-

circumscribed m-polygon.

2 Properties of a periodic sequence

A periodic sequence has some properties similar to a property of a periodic function. Therefore we recall (see [6,1]) those properties of a periodic function concerning of the discrete domain. Let f be 2π-periodic function having uniformly convergent Fourier series,

f(t) = \frac{1}{2}A_{0} + ∑_{n=0}^{∞}[A_{n} cos(nt) + B_{n} sin(nt)].

Theorem A. Expression f(t)+f(t+π) is a constant function iff Fourier coefficients A_{2j}, B_{2j}, j = 1, 2, … vanish. Theorem B. Expression

f(t) + f(t + \frac{2π}{m}) + f(t + 2\frac{2π}{m}) + … + f (t + (m − 1)\frac{2π}{m} )

is a constant function iff Fourier coefficients A_{mj}, B_{mj}, j = 1, 2, … vanish.

Theorems A and B have the following counterparts at the discrete domain:

Let x_{v} be a 2n-periodic sequence,i.e.: x_{v+2n} = x_{v}, v = 0, 1, 2, …. In this case sequence x_{v} has the discrete Fourier sum in the form

x_{v} = a_{0} + ∑_{j=1}^{2n−1}[a_{j} cos \frac{πjv}{n} + b_{j} sin \frac{πjv}{n}].

We prove two following lemmas:

Lemma 2.1 If {x_{v}} is a 2n-periodic sequence, then the discrete Fourier sum of x_{v} + x_{v+n} has the form

x_{v} + x_{v+n} = 2a_{0} + 2 ∑_{l=1}^{n−1}[a_{2l} cos \frac{2πlv}{n} + b_{2l} sin \frac{2πlv}{n}].

Moreover

Lemma 2.2 If

x_{v} = a_{0} + ∑_{j=1}^{2n−1}[a_{j} cos \frac{πjv}{n} + b_{j} sin \frac{πjv}{n}]

is the discrete Fourier sum of 2n-periodic sequence {x_{v}}, then the sequence {x_{v}+x_{v+n}} is a constant function if and only if a_{2l} = b_{2l} = 0 for l = 1, 2, …, n − 1.

Proof. To prove the lemma we verify that if x_{v} + x_{v+n} = c, v = 0, 1, …, then a_{2l} = b_{2l} = 0, l = 1, 2, …, n − 1. Indeed we have

a_{2l} = \frac{1}{2n} ∑_{μ=0}^{2n−1} x_{μ} cos(2l \frac{π}{n}μ) =

= \frac{1}{2n}(x_{0} + x_{1} cos \frac{2πl}{n} + x_{2} cos \frac{4πl}{n} + … + x_{n} cos \frac{2πl}{n} n + x_{n+1} cos \frac{4πl}{n} (n + 1) + …) =

= \frac{1}{2n}((x_{0} + x_{n}) + (x_{1} + x_{1+n}) cos \frac{2πl}{n} + (x_{2}+_{2+n}) cos \frac{4πl}{n} + …) =

= \frac{c}{2n}(1 + cos \frac{2πl}{n} + cos \frac{4πl}{n} + …) = 0,

because

1 + e^{i \frac{π}{n}} + … + e^{i \frac{(n−1)π}{n}} = 0

hence sum

∑_{v=0}^{n−1} cos \frac{vπ}{n} = 0

vanishes. Similarly we compute that b_{2l=0}, l = 1, 2, …, n − 1. So lemmas 1 and 2 are strict counterparts of the relation between Fourier coefficients of a 2π-periodic function f(t) and the function f(t) + f(t + π) ≡ C. Theorem B has the following discrete counterpart:

Lemma 2.3 If {x_{v}} is a m · n-periodic sequence and the discrete Fourier sum

x_{v} = a_{0} + ∑_{j=1}^{m·n−1}[a_{j} cos \frac{2πjv}{m · n} + b_{j} sin \frac{2πjv}{m · n}]

is given, then the sequence x_{v} + x_{v+n} + x_{v+2n} + … + x_{v+(m−1)·n} has the discrete Fourier sum in the form

x_{v} + x_{v+n} + x_{v+2n} + … + x_{v+(m−1)·n} =

= ma_{0} + m ∑_{l=1}^{n−1}[a_{ml} cos \frac{2πlv}{n} + b_{ml} sin \frac{2πlv}{n}].

Moreover

Lemma 2.4 If the discrete Fourier sum of m · n-periodic sequence is given, then the sequence x_{v} + x_{v+n} + x_{v+2n} + … + x_{v+(m−1)·n} is constant if and only if

a_{ml} = b_{ml} = 0, l = 1, 2, …, n − 1.

3 Convex polygons

Let x_{v} be an n-periodic sequence of real numbers . In this section we consider a polygon line represented by (1.3):

k ⟼ z_{k} = ∑_{v=0}^{k} x_{v}e^{i \frac{2πv}{n}} .

The correspondence (1.3) describes the polygon line whenever n > 2. This means that for each fixed sequence x_{v} points z_{k}, k = 0, 1, 2, … are vertices of the polygon line in the Euclidean plane, see Fig.1.

[figure: z_{0} = x_{0}; z_{1} = x_{0} + x_{1}e^{i \frac{2π}{n}}; z_{2} = x_{0} + x_{1}e^{i \frac{2π}{n}} + x_{2}e^{i2 \frac{2π}{n}}; …; Fig.1 Polygon line z_{k}.]

Obviously the value x_{k}, k = 0, 1, … is equal to the distance between points z_{k−1} and z_{k}. Polygon line with vertices z_{k} becomes a convex polygon if we assume that:

(a) x_{v} > 0,

(b) a_{1} = b_{1} = 0. Indeed, applying assumptions (a) and (b), we easy compute that

z_{k+n} = ∑_{v=0}^{k+n} x_{v}e^{i \frac{2πv}{n}} = z_{k} + ∑_{v=k+1}^{k+n} x_{v}e^{i \frac{2πv}{n}} .

Next we analyse the sum

S = ∑_{v=k+1}^{k+n} x_{v}e^{i \frac{2πv}{n}} = ∑_{v=k+1}^{k+n} x_{v} cos \frac{2πv}{n} + i ∑_{v=k+1}^{k+n} x_{v} sin \frac{2πv}{n} .

By the periodicity of sequence x_{v} we have:

∑_{v=k+1}^{k+n} x_{v} cos \frac{2πv}{n} =

= x_{k+1} cos \frac{2π(k + 1)}{n} + x_{k+2} cos \frac{2π(k + 2)}{n} + … +

+x_{n} cos \frac{2πn}{n} + x_{n+1} cos \frac{2π(n + 1)}{n} + … + x_{k+n} cos \frac{2π(k + n)}{n} =

= x_{0}+x_{1} cos \frac{2π}{n} +… x_{k} cos \frac{2πk}{n} +x_{k+1} cos \frac{2π(k + 1)}{n} +…+x_{n−1} cos \frac{2π(n − 1)}{n} = na_{1}.

Similarly we compute that

∑_{v=k+1}^{k+n} x_{v} sin \frac{2πv}{n} = nb_{1}.

Finally we obtain z_{k+n} = z_{k} + n(a_{1} + ib_{1}). (3.5)

Equality (3.5) implies the following:

Lemma 3.1 The polygon line with vertices z_{k} is closed if and only if the coefficients a_{1} and b_{1} of the discrete Fourier sum of sequence x_{v} vanish.

The above-mentioned lemma is the strict counterpart of Lm.A.

A polygon line with vertices z_{k} becomes a well-shaped regular polygon with n sides whenever x_{v} is a constant sequence. Therefore, comparing Remark A with formula (1.3), we state that a well-shaped regular polygon is a discrete counterpart of a circle.

4 On n-polygons with constant diagonal

In this section we examine 2n-polygons represented by formula

z_{k} = ∑_{v=0}^{k} x_{v}e^{i \frac{πv}{n}} , (4.6)

where n ≥ 2 and the sequence x_{v} satisfies the following conditions:

(i) x_{v} > 0,

(ii) x_{v+2n} = x_{v} v = 0, 1, …,

(iii) x_{v} + x_{v+n} = c, v = 0, 1, …,

(iv) a_{1} = b_{1} = 0.

Then the sector between points z_{k} and z_{k+n} is a diagonal of the polygon. Such a diagonal is called \frac{1}{2}-diagonal of 2n-polygon because the number all vertices of the polygon between z_{k} and z_{k+n} is equal the number all vertices of the polygon between z_{k+n} and z_{k}.

Now we prove the main result of the paper.

Theorem 4.1 If vertices of a 2n-polygon are determined by formula (4.6) and the

sequence x_{v} satisfies conditions (i)-(iv), then all \frac{1}{2}-diagonals of the polygon have the same length.

Proof. We consider a 2n-polygon represented by equation (4.6). Let p_{k} denote

a \frac{1}{2}-diagonal of the polygon. We put

T_{k} = e^{i \frac{πk}{n}} e^{i \frac{π}{2n}} = e^{i \frac{(2k+1)π}{2n}} and N_{k} = iT_{k}.

The vectors T_{k}, N_{k} are parallel to bisectrices of outside and inside angle at vertices z_{k}, k = 0, 1, … of 2n-polygon, respectively. Vectors T_{k}, N_{k} establish a basis for k = 0, 1, . . .. Therefore

p_{k} = D_{k}T_{k} − d_{k}N_{k},

where d_{k} = [p_{k}, T_{k}] and D_{k} = [p_{k}, N_{k}] are determinants of two pairs of vectors p_{k}, T_{k} and p_{k}, N_{k}, respectively. Now we obtain the discrete Fourier sum of d_{k} and D_{k}. First

d_{k} = [p_{k}, T_{k}] = ∑_{v=k+1}^{k+n} x_{v}[e^{i \frac{πv}{n}} , e^{i \frac{(2k+1)π}{2n}}] =

= ∑_{v=k+1}^{k+n} x_{v} sin(\frac{(2k + 1)π}{2n} − \frac{πv}{n}) =

= ∑_{v=k+1}^{k+n} x_{v} sin \frac{(2k − 2v + 1)π}{2n} .

On the other hand we have the following formula

x_{v} = a_{0} + ∑_{l=1}^{n−1}[a_{2l+1} cos \frac{(2l + 1)vπ}{n} + b_{2l+1} sin \frac{(2l + 1)vπ}{n}]

and we insert it into the formula d_{k} . Hence we obtain

d_{k} = a_{0} ∑_{v=k+1}^{k+n} sin \frac{(2k − 2v + 1)π}{2n} +

+ ∑_{l=1}^{n−1} (a_{2l+1} ∑_{v=k+1}^{k+n} cos \frac{(2l + 1)vπ}{n} sin \frac{(2k − 2v + 1)π}{2n} +

+ b_{2l+1} ∑_{v=k+1}^{k+n} sin \frac{(2l + 1)vπ}{n} sin \frac{(2k − 2v + 1)π}{2n} ) =

= −\frac{a_{0}}{sin \frac{π}{2n}} .

Similarly we compute that D_{k} = 0. So we have

p_{k} = \frac{a_{0}}{sin \frac{π}{2n}} T_{k}.

This means that every \frac{1}{2}-diagonal of the 2n-polygon has the same length equal to

|p_{k}| = \frac{a_{0}}{sin \frac{π}{2n}} .

□

Let L be a perimeter of 2n-polygon with constant diagonal and let d be a length of an \frac{1}{2}-diagonal. Then we have the following counterpart of the Barbier’s formula:

L = π \frac{sin \frac{π}{2n}}{\frac{π}{2n}} d (4.7)

because

a_{0} = \frac{1}{2n}(x_{0} + x_{1} + … + x_{2n−1}) = \frac{1}{2n}L.

The relation D_{k} = 0 means that

Corollary 4.1 Each \frac{1}{2}-diagonal of 2n-polygon determined by a sequence x_{v} satisfying conditions (i)-(iv) is a bisectrix of an inside angle of the polygon.

To define a 2n-polygon with constant diagonal by formula (4.6) we need a sequence x_{v} satisfying conditions (i)-(iv). This means that we solve the linear system n + 1 equations with 2n unknown quantities. We solve these equations for n = 3, 4. The sequence x_{0} = d, x_{1} = m − d, x_{2} = d, x_{3} = m − d, x_{4} = d, x_{5} = m − d determines 6-polygon with constant diagonal by formula (4.6) for fixed numbers d > 0 and m − d > 0. Then

\begin{aligned}z_{0} & = & d, \\ z_{1} & = & d + (m − d)e^{i \frac{π}{3}} , \\ z_{2} & = & z_{1} + de^{i2 \frac{π}{3}} , \\ z_{3} & = & z_{2} + (m − d)e^{iπ}, \\ z_{4} & = & z_{3} + de^{i4 \frac{π}{3}} , \\ z_{5} & = & z_{4} + (m − d)e^{i5 \frac{π}{3}} . \end{aligned}

To define a 8-polygon with constant diagonal we apply the following sequence.

\begin{aligned}x_{0} & = & a, \\ x_{1} & = & \frac{1}{2}(a + e) + e\frac{1}{\sqrt{2}} − c\frac{1}{\sqrt{2}}, \\ x_{2} & = & c, \\ x_{3} & = & \frac{1}{2}(a + e) + a\frac{1}{\sqrt{2}} − c\frac{1}{\sqrt{2}}, \\ x_{4} & = & e, \\ x_{5} & = & \frac{1}{2}(a + e) − e\frac{1}{\sqrt{2}} + c\frac{1}{\sqrt{2}}, \\ x_{6} & = & a + e − c, \\ x_{7} & = & \frac{1}{2}(a + e) − a\frac{1}{\sqrt{2}} + c\frac{1}{\sqrt{2}}, \end{aligned}

where a, c, e are arbitrary numbers changed such that x_{v} > 0, v = 0, 1, …, 7.

Now we present a simple method of defining a 2n-polygon with constant diagonal. Let f be a 2π-periodic real positive function such that

f(t) + f(t + π) = C for all t.

Then the Fourier series of f has the form (see[8]):

f(t) = \frac{1}{2}A_{0} + ∑_{j=0}^{∞}[A_{2j+1} cos((2j + 1)t) + B_{2j+1} sin((2j + 1)t)].

Moreover we assume that the series is uniformly convergent to f. Keeping the abovementioned notions we prove the following lemma:

38 S.Góźdź Lemma 4.1 For each fixed t the sequence

x_{v} = f(t + v \frac{π}{n}), v = 0, 1, …

determines the 2n-polygon with constant diagonal by formula (4.6) . Proof. Conditions (i) and (ii) are obvious. We verify the remaining relations. (iii)

x_{v} + x_{v+n} = f(t + v \frac{π}{n}) + f(t + (v + n)\frac{π}{n}) = C,

(iv)

2na_{1} = ∑_{v=0}^{n−1} f(t + \frac{vπ}{n}) cos \frac{vπ}{n} =

= ∑_{v=0}^{n−1} (\frac{1}{2}A_{0} + ∑_{j=1}^{∞}[A_{2j+1} cos((2j + 1)(t + \frac{vπ}{n})) + B_{2j+1} sin((2j + 1)(t + \frac{vπ}{n}))) cos \frac{vπ}{n} =

= \frac{1}{2}A_{0} ∑_{v=0}^{n−1} cos \frac{vπ}{n} + ∑_{j=1}^{∞} (A_{2j+1} ∑_{v=0}^{n−1} cos((2j + 1)(t + \frac{vπ}{n})) cos \frac{vπ}{n} +

+B_{2j+1} ∑_{v=0}^{n−1} sin((2j + 1)(t + \frac{vπ}{n})) cos \frac{vπ}{n} ) = 0.

To verify that the sums:

∑_{v=0}^{n−1} cos((2j + 1)(t + \frac{vπ}{n})) cos \frac{vπ}{n} ,

∑_{v=0}^{n−1} sin((2j + 1)(t + \frac{vπ}{n})) cos \frac{vπ}{n} ,

vanish, we apply simple trigonometric relations and we successively compute that

∑_{v=0}^{n−1} cos((2j + 1)(t + v \frac{π}{n}) + \frac{vπ}{n}) =

= \frac{sin(\frac{(j+1)π}{n} − 2jt − t)}{2 sin \frac{(j+1)π}{n}} − \frac{sin((j(\frac{1}{n} − 4) + \frac{1}{n} − 4)π − 2jt − t)}{2 sin((\frac{j}{n} + \frac{1}{n})π)} = 0

∑_{v=0}^{n−1} cos((2j + 1)(t + v \frac{π}{n}) − \frac{vπ}{n}) =

= \frac{sin(\frac{jπ}{n} − 2jt − t)}{2 sin(\frac{jπ}{n})} − \frac{sin(j(\frac{1}{n} − 4)π − 2jt − t)}{2 sin(\frac{jπ}{n}) sin(\frac{vπ}{n} + (2j + 1)(t + \frac{vπ}{n}))} = 0

∑_{v=0}^{n−1} sin((2j + 1)(t + v \frac{π}{n}) + \frac{vπ}{n}) =

= \frac{cos((\frac{j}{n} + \frac{1}{n})π − 2jt − t)}{2 sin((\frac{j}{n} + \frac{1}{n})π))} − \frac{cos((j(\frac{1}{n} − 4) + \frac{1}{n} − 4)π − 2jt − t)}{2 sin((\frac{j}{n} + \frac{1}{n})π)} = 0

∑_{v=0}^{n−1} sin((2j + 1)(t + v \frac{π}{n}) − \frac{vπ}{n}) =

= \frac{cos(\frac{jπ}{n} − 2jt − t)}{2 sin(\frac{jπ}{n})} − \frac{cos(j(\frac{1}{n} − 4)π − 2jt − t)}{2 sin(\frac{jπ}{n} j)} = 0.

To verify the above-mentioned equalities the computer programm ”Derive” was used. Therefore a_{1} = 0. Similarly we compute that b_{1} = 0.

4.1 On 2n-polygons circumscribed on an oval with constant width

In this subsection we prove the Th.1.1. i.e.: All 2n-polygons circumscribed on an oval with constant width δ are 2n-polygons with constant diagonal equal to

\frac{δ}{cos \frac{π}{2n}} .

Proof. Let an oval in arc length parametrization be represented by equation

s ⟼ z(s) = x(s) + iy(s).

We will denote a curvature,tangent and normal vectors at point z(s) by k(s), T_{s}, N_{s}, respectively. Moreover we define K(s) = ∫_{0}^{s} k(t)dt. Now we apply function φ(s) = K^{−1}(K(s)+\frac{π}{n}), where K^{−1} is an inverse function of K. Denoting φ^{v}(s) = \underbrace{φ(φ … φ}_{v−times}(s) …))

we easy observe that φ^{n} = K^{−1}(K(s) + π). Obviously

|z(s) − z(φ^{n}(s))| = |z(φ(s) − z(φ^{n+1}(s))| = δ, see F ig.2.

[figure: s; A; φ(s); φ^{n}(s); φ^{n+1}(s); Fig.2. 2n-polygon circumscribed on an oval.; B]

Next we consider the following expressions

d_{v} = [x(φ^{v}(s)) − x(φ^{v+1}(s)), T_{φ^{v}(s)}], v = 0, 1, …, 2n − 1,
D_{v} = [x(φ^{v}(s)) − x(φ^{v+1}(s)), N_{φ^{v}(s)}], v = 0, 1, … 2n − 1.

Applying the same considerations as in [6 p.373] we solve the following system of equations:

z(φ^{v}(s)) + ξ_{v}T_{φ^{v}(s)} = z(φ^{v+1}(s)) + η_{v}T_{φ^{v+1}(s)}, v = 0, 1, ….

Hence we obtain the points

A : z(s) + [−D_{0} − d_{0} cot \frac{π}{n}]T_{s},

B : z(φ^{n}(s) + [−D_{n} − d_{n} cot \frac{π}{n}](−T_{s}).

Now we compute the length of the diagonal AB:

|AB| = |z(s) − z(φ^{n}(s) + [(−D_{n} − D_{0}) + (−d_{n} − d_{0}) cot \frac{π}{n}]T_{s}|,

but

z(s) − z(φ^{n}(s)) = −δN_{s}

D_{n} + D_{0} = −δ sin \frac{π}{n}

d_{n} + d_{0} = δ(1 − cos \frac{π}{n}).

Inserting these relations we express the length |AB| as follows

|AB| = | − δN_{s} − δ[sin \frac{π}{n} + (1 − cos \frac{π}{n}) cot \frac{π}{n}]T_{s}| =

= δ| − N_{s} + tan \frac{π}{2n} T_{s}| = \frac{δ}{cos \frac{π}{2n}} .

This implies that a perimeter of 2n-polygons ( circumscribed on the oval) tends to πδ. Indeed we have

π (\frac{sin \frac{π}{2n}}{\frac{π}{2n}} ) \frac{δ}{cos \frac{π}{2n}} ⟼ πδ as n ⟼ ∞.

5 On m-polygons circumscribed on an m·n-polygon

The results of this section are discrete counterparts of theorems presented in papers [1,3,4].

In the section we examine m · n-polygons represented by the formula

z_{k} = ∑_{v=0}^{k} x_{v}e^{i \frac{2πv}{m·n}} , (5.8)

where n ≥ 2, m ≥ 3 and a sequence x_{v} satisfies the following conditions:

1^{o} x_{v} > 0,

2^{o} x_{v+m·n} = x_{v}, v = 0, 1, …,

3^{o} x_{v} + x_{v+n} + x_{v+2n} + … + x_{v+(m−1)n} = c, v = 0, 1, …,

4^{o} a_{1} = b_{1} = 0.

We consider an m-polygon b-circumscribed on an m·n-polygon. For a fixed integer k we draw bisectrices of outside angles in vertices

z_{k}, z_{k+n}, z_{k+2n}, …, z_{k+(m−1)n}.

This m-polygon b-circumscribed on m · n-polygon has vertices defined as a point of intersection of two successive bisectrices passing through vertices z_{k+jn}, z_{k+(j+1)n}. Keeping notions as before we show

Theorem 5.1 All m-polygons b-circumscribed on an m · n-polygon have the same perimeter whenever the sequence x_{v} satisfies conditions 1^{o} − 4^{o}.

Proof. To prove the theorem we denote vectors parallel to bisectrices of inside and outside angels at vertex z_{k} of the polygon by N_{k} and T_{k},respectively. Applying Fig.3 we easy observe that

T_{k} = e^{i \frac{2πk}{m·n}} e^{i \frac{π}{m·n}} = e^{i \frac{(2k+1)π}{m·n}} and N_{k} = iT_{k}.

To compute the perimeter of b-circumscribed m-polygon we use the following vectors

T_{k+jn} = ε^{j}T_{k}, and N_{k+jn} = iT_{k+jn}, j = 1, 2, … m − 1,

where ε = cos \frac{2π}{m} + i sin \frac{2π}{m}. Next we solve the following system of equations

z_{k+jn} + ξ_{k+jn}ε^{j}T_{k} = z_{k+(j+1)n} + η_{k+jn}ε^{j+1}T_{k}, j = 0, 1, 2, … m − 1.

The geometrical meaning of the above-mentioned equations is illustrated in Fig.3

z_{k+jn}  z_{k+(j+1)n} …  b_{k+jn} b_{k+(j+1)n}  A_{j} Fig.3.|z_{k+jn}A_{j}| = ξ_{k+jn}, |A_{j}z_{k+(j+1)n}| = η_{k+jn} ξ_{k+jn}  η_{k+jn}

Solving these equations we obtain

η_{k+jn} = \frac{[z_{k+(j+1)n} − z_{k+jn}, ε^{j}T_{k}]}{sin \frac{2π}{m}}, ξ_{k+jn} = \frac{[z_{k+(j+1)n} − z_{k+jn}, ε^{j+1}T_{k}]}{sin \frac{2π}{m}}.

Let L_{k} denote a perimeter of b-circumscribed m-polygon . The Fig.3 suggests that

L_{k} = ∑_{v}^{m−1}(ξ_{k+vn} − η_{k+vn}).

Inserting all formulas on ξ_{p}, η_{q} we obtain

L_{k} = \frac{1}{sin \frac{2π}{m}} ∑_{v=0}^{m−1}([z_{k+(v+1)n} − z_{k+vn}, ε^{v+1}T_{k}] − [z_{k+(v+1)n} − z_{k+vn}, ε^{v}T_{k}]) =

= \frac{1}{sin \frac{2π}{m}} ∑_{v=0}^{m−1}([ε^{m−v−1}(z_{k+(v+1)n} − z_{k+vn}), T_{k}] − [ε^{m−v}(z_{k+(v+1)n} − z_{k+vn}), T_{k}]) =

= \frac{1}{sin \frac{2π}{m}} [(2 − ε − \frac{1}{ε}) ∑_{v=0}^{m−1} ε^{m−v}z_{k+vn}, T_{k}],

but 2 − ε − \frac{1}{ε} = 2 − 2Re(ε) = 4 sin^{2} \frac{π}{m}. Hence putting

p_{k} = ∑_{v=0}^{m−1} ε^{m−v}z_{k+vn}

we express L_{k} as follows

L_{k} = 2[p_{k}, T_{k}] tan \frac{π}{m}.

Now introducing notions d_{k} = [p_{k}, T_{k}] and D_{k} = [p_{k}, N_{k}] we express vector p_{k} as follows

p_{k} = D_{k}T_{k} − d_{k}N_{k}.

In conclusion we show that discrete Fourier sums of sequences d_{k} and D_{k} have the form

d_{k} = [p_{k}, T_{k}] = [ ∑_{j=0}^{m−1} ε^{m−j} ∑_{v=0}^{k+jn} x_{v}e^{i \frac{2πv}{mn}} , e^{i \frac{(2k+1)π}{mn}} ] =

= ∑_{j=0}^{m−1} ∑_{v=0}^{k+jn} x_{v} sin \frac{(2k + 1 − 2v + 2jn)π}{mn} =

= ∑_{j=0}^{m−1} ∑_{v=0}^{k+jn} (a_{0} + ∑_{s=1}^{m−1} ∑_{l=1}^{n−1} [a_{ml+s} cos \frac{2π(ml + s)v}{mn} +

+b_{ml+s} sin \frac{2π(ml + s)v}{mn} ]) sin \frac{(2k − 2v + 1 + jn)π}{mn} =

= a_{0} ∑_{j=0}^{m−1} ∑_{v=0}^{k+jn} sin \frac{(2k − 2v + 2nj + 1)π}{mn} +

+ ∑_{s=1}^{m−1} ∑_{l=1}^{n−1} a_{ml+s} ∑_{j=0}^{m−1} ∑_{v=0}^{k+jn} cos \frac{2π(ml + s)v}{mn} sin \frac{(2k − 2v + 2nj + 1)π}{mn} +

+ ∑_{s=1}^{m−1} ∑_{l=1}^{n−1} b_{ml+s} ∑_{j=0}^{m−1} ∑_{v=0}^{k+jn} sin \frac{2π(ml + s)v}{mn} sin \frac{(2k − 2v + 2nj + 1)π}{mn} = \frac{ma_{0}}{2 sin \frac{π}{mn}} .

Similarly we compute that D_{k} = 0. Hence we finally obtain

L_{k} = 2[p_{k}, T_{k}] tan \frac{π}{m} = 2\frac{ma_{0}}{2 sin \frac{π}{mn}} tan \frac{π}{m}.

This means that all b-circumscribed m-polygons have the same perimeter independent of index k.

□

Moreover we observe that every vector p_{k} = −d_{k}N_{k} has the same length equal to \frac{ma_{0}}{2 sin \frac{π}{mn}} . We put d = |p_{k}|, then we express perimeter L of mn-polygon with a constant

perimeter of b-circumscribed m-polygon by the following relation

L = \frac{2π}{m} (\frac{sin \frac{π}{mn}}{\frac{π}{mn}} ) d, (5.9)

because mnL = x_{0} + x_{1} + … + x_{mn−1}. Tending to infinity with n we obtain

L = \frac{2πd}{m}.

Formula (5.9) is a discrete counterpart of Th.1.[3] and becomes formula (1.4) for m = 2. To define an mn-polygon with a constant perimeter of a b-circumscribed m-polygon we apply a 2π-periodic positive function f such that

∑_{v=0}^{m−1} f(t + v \frac{2π}{m}) ≡ C.

We assume that function f has uniformly convergent Fourier series and this series has a form

f(t) = \frac{1}{2}A_{0} + ∑_{l=1}^{∞}[A_{l} cos(lt) + B_{l} sin(lt)],

where A_{mj} = B_{mj} = 0, j = 1, 2, …,see [4,1]. Putting

x_{v} = f(t + \frac{2πv}{mn}), v = 0, 1, …

we obtain (for a fixed variable t) a sequence which satisfies conditions 1^{o}−4^{o}. Therefore a mn-polygon represented by equation

z_{k} = ∑_{v=0}^{k} x_{v}e^{i \frac{2πv}{mn}}

has the constant perimeter of each m-polygon b-circumscribed on it.

5.1 On an oval with constant perimeter of a circumscribed m-polygon and on an mn-polygon

In subsection 4.1 we proved that every 2n-polygon circumscribed on an oval with

constant width d is the polygon with a constant diagonal equal to \frac{d}{cos \frac{π}{2n}} . In this

subsection we prove Theorem 1.2. i.e.:

All m-polygons b-circumscribed on an mn-polygon which is circumscribed on an oval with a constant perimeter of a circumscribed m-polygon have the same perimeter equal to

\frac{l}{cos \frac{π}{mn}} ,

where l denotes the length of m-polygon circumscribed on this oval. Proof. We keep notion as before. Let z(s) be an oval with a constant perimeter of a circumscribed m-polygon. Putting φ(s) = K^{−1}(K(s) + \frac{2π}{mn}) we easy observe that φ^{mn}(s) = s+L and that mn-polygon circumscribed on the oval is tangent (to the oval) at points z(φ^{v}(s)), v = 0, 1, 2, …. Then vertices of an mn-polygon circumscribed on the oval are expressed as follows:

z(φ^{v}(s)) + ξ_{v}T_{φ^{v}(s))}, v = 0, 1, 2, …,

where

ξ = −D_{v} − d_{v} cot \frac{2π}{mn}, D_{v} = [z(φ^{v}(s)) − z(φ^{v+1}(s)), N_{φ^{v}(s))}]

and d_{v} = [z(φ^{v}(s))−z(φ^{v+1}(s)), T_{φ^{v}(s))}]. Now we consider m-polygon b-circumscribed on mn-polygon, see Fig.4.

[figure: …; A_{k+jn}; A_{k+(j+1)n}; φ^{k+jn}(s); φ^{k+(j+1)n}; Fig.4. m-polygon b-circumscribed on mn-polygon which is circumscribed on the oval.; B_{k+jn}]

We denote by T_{k+jn} the tangent vector at point z(φ^{k+jn}(s)) for fixed k and j = 0, 1, …, m − 1. To compute the perimeter of m-polygon b-circumscribed on mnpolygon we solve the following system of equations (Fig.3 and Fig.4):

z(φ^{k+jn}(s)) + ξ_{k+jn}T_{k+jn} + ξ_{k+jn}^{1}T_{k+jn}e^{i \frac{π}{mn}} =
= z(φ^{k+(j+1)n}(s)) + ξ_{k+(j+1)n}T_{k+(j+1)n} + η_{k+jn}^{1}T_{k+(j+1)n}e^{i \frac{π}{mn}} ,

where j = 0, 1, … m − 1. Moreover the length of sectors |z(φ^{k+jn}(s))A_{k+jn}| and |z(φ^{k+(j+1)n}(s)A_{k+(j+1)n}| is denoted by ξ_{k+jn} and ξ_{k+(j+1)n},respectively. The length of sectors |a_{k+jn}B_{k+jn}| and |B_{k+jn}A_{k+(j+1)n}| is denoted by ξ_{k+jn}^{1} and η_{k+(j+1)n}^{1}, respectively. Moreover vectors T_{k+jn}e^{i \frac{π}{mn}} and T_{k+(j+1)n}e^{i \frac{π}{mn}} are parallel to bisectrices of outside angles of mnpolygon. Obviously perimeter l_{k} of m-polygon b-circumscribed on mn-polygon is equal to 46 S.Góźdź

l_{k} = ∑_{j=0}^{m−1}(ξ_{k+jn}^{1} − η_{k+jn}^{1}).

At first we compute ξ_{k+jn}^{1}

ξ_{k+jn}^{1} = \frac{1}{−[T_{k+jn}e^{i \frac{π}{mn}} , T_{k+(j+1)n}e^{i \frac{π}{mn}}]} ([z(φ^{k+jn}(s)) − z(φ^{k+(j+1)n}(s)), T_{k+(j+1)n}e^{i \frac{π}{mn}}]+

+[ξ_{k+jn}T_{k+jn} − ξ_{k+(j+1)n}T_{k+(j+1)n}, T_{k+(j+1)n}e^{i \frac{π}{mn}}]) =

= \frac{1}{− sin \frac{2π}{mn}} ([z(φ^{k+jn}(s)) − z(φ^{k+(j+1)n}(s)), T_{k+(j+1)n}e^{i \frac{π}{mn}}]+

+ξ_{k+jn} sin \frac{(2n + 1)π}{mn} − ξ_{k+(j+1)n} sin \frac{π}{mn})

and

η_{k+jn}^{1} = \frac{1}{sin \frac{2π}{mn}} ([z(φ^{k+jn}(s)) − z(φ^{k+(j+1)n}(s)), T_{k+(j+1)n}e^{i \frac{π}{mn}}]+

+ξ_{k+jn} sin \frac{π}{mn} + ξ_{k+(j+1)n} sin \frac{(2n − 1)π}{mn} ) .

Inserting relation T_{k+jn} = ε^{j}T_{k}, ε^{m} = 1, ε ≠ 1 we obtain

W = ∑_{j=0}^{m−1}([z(φ^{k+jn}(s)) − z(φ^{k+(j+1)n}(s)), T_{k+(j+1)n}e^{i \frac{π}{mn}}]+

+ − [z(φ^{k+jn}(s)) − z(φ^{k+(j+1)n}(s)), T_{k+(j+1)n}e^{i \frac{π}{mn}}]) =

= [\overline{ε}p_{k} − p_{k} − p_{k} + εp_{k}, T_{k}e^{i \frac{π}{mn}}],

where

p_{k} = ∑_{j=0}^{m−1} ε^{m−j}z(φ^{k+jn}(s)).

By (3.1)[4,p.374] p_{k} = −dN_{k} ( where d denotes n-width of the oval) we obtain

W = (\overline{ε} + ε − 2)[p_{k}, T_{k}e^{i \frac{π}{mn}}] = −4d sin^{2} \frac{π}{m} cos \frac{π}{mn}.

It is easy to observe that

∑_{j=0}^{m−1} ξ_{k+jn} = ∑_{j=0}^{m−1} ξ_{k+(j+1)n}.

The sum

∑_{j=0}^{m−1} ξ_{k+jn}

is equal to

∑_{j=0}^{m−1} ξ_{k+jn} = − ∑_{j=0}^{m−1}([z(φ^{k+jn}(s)) − z(φ^{k+jn+1}(s)), N_{k+jn}]+

+[z(φ^{k+jn}(s)) − z(φ^{k+jn+1}(s)), T_{k+jn}] cot \frac{2π}{mn}) =

= − ∑_{j=0}^{m−1}([ε^{m−j}z(φ^{k+jn}(s)) − ε^{m−j}z(φ^{k+jn+1}(s)), N_{k}]+

+[ε^{m−j}z(φ^{k+jn}(s)) − ε^{m−j}z(φ^{k+jn+1}(s)), T_{k}] cot \frac{2π}{mn}) =

= −([p_{k} − p_{k+1}, N_{k}] + [p_{k} − p_{k+1}, T_{k}] cot \frac{2π}{mn} =

= d(sin \frac{2π}{mn} − tan \frac{π}{mn} cos \frac{2π}{mn}).

Finally we obtain perimeter l_{k} as the following expression

l_{k} = \frac{−4d sin^{2} \frac{π}{m} cos \frac{π}{mn}}{− sin \frac{2π}{m}} +

+\frac{1}{− sin \frac{2π}{m}} ( ∑_{j=0}^{m−1} ξ_{k+jn}(sin \frac{(2n + 1)π}{mn} − sin \frac{π}{mn})+

+ ∑_{j=0}^{m−1} ξ_{k+(j+1)n}(− sin \frac{π}{mn} − sin \frac{(2n − 1)}{mn})) =

= 2d tan \frac{π}{m}(cos \frac{π}{mn} + sin \frac{π}{mn}(sin \frac{2π}{mn} − tan \frac{π}{mn} cos \frac{2π}{mn})) =

= 2d tan \frac{π}{m} \frac{1}{cos \frac{π}{mn}} .

But by [4,p.373] l = 2d tan \frac{π}{m} is equal to the perimeter of m-polygon circumscribed on an oval . Therefore

l_{k} = \frac{l}{cos \frac{π}{mn}} .

This means that all m-polygons b-circumscribed on mn-polygon have the same perimeter. Moreover if n tends to infinity then perimeter l_{k} tends to the perimeter of m-polygon circumscribed on an oval. □

References

[1] W. Cieślak and S. Góźdź, On curves which bound special convex sets, Serdica Bulg. Math.Pub. Vol.13, 1987, 281-286.

[2] H. Groemer, Geometric Applications of Fourier Series and Spherical Harmonics, Cambridge University Press 1996.

[3] S. Góźdź, On curves with a constant perimeter of the described polygon lines, Serdica Bulg. Math.Pub. Vol.16, 1990, 66-74.

[4] S. Góźdź, Barbiers type theorem, Le Matematiche Vol.XLV(1990)-Fasc.II, 369-377.

[5] M.A. Hurwitz, Sur quelques applications geometriques des series de Fourier, Ann.Sc. de l’Ecole Normale superieure,3 serie,t.19. 1902, 357-408.

[6] R.L. Tennison ;Smooth curves of constant width, The. Math.Gaz.,60, 1976, 20-272.

Stanisław Góźdź UMCS Institute of Mathematics pl.M.Curie-Skłodowskiej 1 20-031 Lublin, Poland gozds@golem.umcs.lublin.pl