Ranking Small Regular Polygons by Area and by Perimeter C. Audet

Les Cahiers du GERAD

ISSN: 0711–2440

Ranking Small Regular Polygons by Area and by Perimeter C. Audet, P. Hansen, F. Messine G–2005–92 December 2005 Revised: November 2006

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Ranking Small Regular Polygons by Area and by Perimeter Charles Audet GERAD and Département de mathématiques et de génie industriel ´ Ecole Polytechnique de Montréal C.P. 6079, Succ. Centre-ville Montréal (Québec) Canada, H3C 3A7 [email protected]

Pierre Hansen GERAD and Méthodes quantitatives de gestion HEC Montréal 3000, chemin de la Cˆ ote-Sainte-Catherine Montréal (Québec) Canada, H3T 2A7 [email protected]

Fr´ ed´ eric Messine ENSEEIHT-IRIT, UMR-CNRS 5828 2, rue Charles Camichel 31071 Toulouse Cedex, France [email protected]

December 2005 Revised: November 2006

Les Cahiers du GERAD G–2005–92 c 2006 GERAD Copyright

Abstract From the pentagon onwards, the area of the regular convex polygon with n sides and unit diameter is greater for each odd number n than for the next even number n + 1. Moreover, from the heptagon onwards, the difference in areas decreases when n increases. Similar properties hold for the perimeter. A new proof of a result of Reinhardt follows. Key Words: Polygon, diameter, area, perimeter.

R´ esum´ e ` partir du pentagone, l’aire du polygone convexe régulier avec n côtés et un A diamètre unité est plus grande pour tout nombre n impair que pour le nombre pair suivant n + 1. De plus, à partir de l’heptagone, la différence des aires décroˆıt quand n croˆıt. Des propriétés semblables sont vérifiées pour le périmètre. Une nouvelle preuve d’un résultat de Reinhardt s’ensuit. Mots cl´ es :

polygone, périmètre, diamètre, aire.

Acknowledgments: Work of the first author was supported by NSERC grant 239436-01, AFOSR F4962-1-0013, and ExxonMobil. Work of the second author was supported by NSERC grant 239436-01.


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G–2005–92 – Revised

1

Introduction

Extremal problems on convex polygons have been studied since the ancient Greeks, see [5, 6, 9, 15, 16, 3] for surveys. The three following problems are central to this stream of research: P1 : What is the maximum area of an n-sided polygon with fixed perimeter; P2 : What is the maximum area of a n-sided polygon with fixed diameter; P3 : What is the maximum perimeter of a convex n-sided polygon with fixed diameter. Problem P1 was solved by Zenodorus in the second century b.c.e. (implicitly assuming existence of a solution): the regular n−sided polygon has maximal area for all n. His proof, and numerous further ones, using a variety of tools, are presented in a beautiful paper of Bl˚ asj¨ o [5]. For problem P2 , a fundamental result was obtained by Reinhardt in 1922 [17]: for odd n, the regular n-gon has maximal area among all isodiametric n−gons. Moreover, for any even n at least 6 the regular n−gon does not have maximum area. Graham [13] determined the hexagon with fixed diameter and maximum area, which is about 3.92% above the area of the regular hexagon. Combining geometric reasoning with extensive use of an algorithm for non-convex quadratic programming [1], the extremal octagon, which has an area about 2.79% above that of the regular octagon, was found in [4]. Several authors obtained, with non-linear programming codes, e.g. LANCELOT [8] and SNOPT [12], heuristic solutions for n even and at least 10. Numerical experiments from [7, 10, 11] are summarized in [3]. To bound the error, Mossinghoff [15, 16] focussed on n−gons with a particular diameter configuration introduced in [13], i.e., a n − 1 star polygon with a pending edge. Then, assuming axial symmetry and equality of some angles between diameters, n−gons with an area very close to be optimal are obtained. Indeed, it is shown that the areas obtained cannot be improved for large n by more than nc13 where c1 is a constant. For problem P3 , Reinhardt [17] proved that regular n-gons have extremal perimeter in the same cases than when they have extremal area: For odd n, the regular n-gon has maximum perimeter among all isodiametric n-gons. Moreover, if n has an odd factor m ≥ 3, a n-gon with maximum perimeter can be obtained as follows. (a) Construct a regular m-gon; (b) Transform it into a Reuleaux polygon by drawing an circle’s arc n centered at each vertex and through the end vertices of the opposite side; (c) Add m −1 vertices at equal interval on each of the m arcs; (d) take the convex hull of the vertices obtained in steps (a) and (c). The remaining cases are when n is a power of 2. Tamvakis [18] determined the quadrilateral with fixed diameter and maximum perimeter, which is about 7.31% above the perimeter of the square. Combining geometric reasoning with the use of an interval based global optimization algorithm [14], the extremal octagon, which has a perimeter about

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1.96% above that of the regular octagon was found in [2]. For larger powers of 2, Mossinghoff [15, 16] considers again polygons with a specific diameter configuration, from [18] and assuming once more axial symmetry and equality of some angles between diameters, obtained n−gons with a perimeter very close to be optimal. Indeed, the perimeters obtained cannot be improved for large n by more than nc25 where c2 is a constant. Areas and perimeters of regular and ⊛ polygons with n ≥ 5 are represented in Figure 1. Maximal values for areas of unit-diameter polygons, according to [17] for odd n, [13] for n = 6, [4] for n = 8 and (without proof) [15] for n = 10, 12, 14, 16, 18, 20 are represented by ’×’. Maximal values for perimeters, according to [17] for n containing an odd factor, [2] for n = 8 and [15] for n = 16 are represented by ’+’. Values for regular polygons are represented by ’◦’. For clarity, the regular polygons that maximize both the area and the perimeter (i.e., the polygons that would be represented by all three symbols ’◦’, ’×’ and ’+’) are represented by ’•’. q All these polygons satisfy the known inequality P ≥ 2 nA tan πn , which bounds the perimeter P of an n-sided polygon of area A, as represented by the dotted curves in that figure. This figure led by serendipity to the following observations: (i) From the pentagon onwards, the area of the regular n-sided convex polygon with unit diameter is greater for each odd n than the area for the next even n + 1; (ii) From the heptagon onwards, the difference in the areas of those successive polygons decreases when n increases. It is the purpose of this paper to prove these results, together with similar ones concerning the perimeter. They jointly lead to a new and very simple proof of a theorem of Reinhardt [17].

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Areas of small regular polygons

Graham [13] calls small a polygon with unit diameter; i.e., its longest diagonal equals one. Let Arn denote the area of the regular small n−gon; i.e., one for which all sides have equal length and all inner angles are identical. It is well known that, for even n = 2k with k ≥ 2, n 2π r An = sin ≤ Arn+2 8 n

(1)

and for odd n = 2k + 1 with k ≥ 2, Arn

n sin = 8 cos2

2π n π 2n

≤ Arn+2 .

(2)


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6

A

◦ × + •

Regular polygons Maximal area Maximal perimeter Maximal area and maximal perimeter

ccscs cccscss c c c cc ss c15 s×16 c 14 16c 13 s×+ 16 × 12 14 c s 11 +14 12 c

0.75

×10

s9 10 c 8×

+10

s 7 8c

+8

0.70 + 12 6 ×

s 0.65

5

6c

+ 6

3.00

3.05

3.10

Figure 1: Area versus perimeter of regular and optimal polygons

3.15 P

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Table 1 displays the area and perimeter of the small regular n-gons for values of n ranging from 3 to 14. Table 1: Area and perimeter of small regular n-gons for n = 3, 4, . . . 14. n 3 5 7 9 11 13

Arn 0.4330127020 0.6571638900 0.7197409265 0.7456192318 0.7587484457 0.7663089267

Pnr 3. 3.090169944 3.115293076 3.125667199 3.130926443 3.133953688

n 4 6 8 10 12 14

Arn 0.5 0.6495190530 0.7071067810 0.7347315655 0.75 0.7592965438

Pnr 2.828427124 3. 3.061467460 3.090169944 3.105828541 3.115293076

The sequence of areas is not monotonous, but follows a pattern. It turns out that for n ≥ 5 the values Arn alternately increase and decrease with n, as next shown. Theorem 2.1 Ar2k−1 > Ar2k for all integer k ≥ 3. Proof. Let us extend the area functions to the reals : x 2π x sin e A (x) = sin and Ao (x) = 8 x 8 cos2

2π x π . 2x

(3)

We now define lower and upper bounds on sin and cos2 based on Taylor expansions. For any a > 0, elementary calculus ensures that a3 a5 + , 6 120 a4 cℓ (a) = 1 − a2 < cos2 (a) < cu (a) = 1 − a2 + . 3 This allows us to derive valid bounds on the area functions. For any x > 0, xsℓ 2π xsu 2π e e x x e Aℓ (x) = < A (x) < Au (x) = , 8 8 π π xsℓ 2x xsu 2x o (x) < Ao (x) = Aoℓ (x) = < A u π π . 8cu 2x 8cℓ 2x sℓ (a) = a −

a3 < 6

sin(a)

< su (a) = a −

(4)

(5)

Consider the function f (x) = Ao (x) − Ae (x + 1),

and define

fℓ (x) = Aoℓ (x) − Aeu (x + 1)

(6)

to be a valid lower bound on f for any x > 1. Table 1 confirms that the result is true up to k = 7. Therefore, it suffices to show that fℓ (x) is positive whenever x > 14. Our proof does more than that, it shows it to be true for x ≥ 6.



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π x4 (x + 1)4 cu ( 2x ) is positive whenever x > 1. It Observe that the function µ(x) = 2880 π3 follows that we only need to show that p(x) = µ(x)fℓ (x) > 0 for all x ≥ 6. Elementary, but tedious, simplifications lead to

p(x) = 180x6 − 240x5 − (1320 − 231π 2 )x4 − (1200 + 300π 2 )x3 −(300 + 210π 2 − 34π 4 )x2 − (60π 2 − 20π 4 )x − (15π 2 − 10π 4 + 2π 6 ), a degree six polynomial. The largest positive real root is approximately 5.6029. The facts that it is less than 6, and that the coefficient of x6 is 180 > 0 imply that p(x) > 0 for all x ≥ 6. Considering differences between areas of consecutive small regular polygons (with n and n + 1 vertices) leads to a further result. As both small regular polygons with n even and with n odd converge to the circle when n goes to infinity, this suggests that these differences will decrease when n increases. We show that this is indeed the case, from the heptagon onwards. The proof is similar to that of Theorem 2.1. Theorem 2.2 Ar2k−1 − Ar2k > Ar2k+1 − Ar2k+2 , for all integer k ≥ 4. Proof. We use the functions introduced in equations (3-6) from the proof of Theorem 2.1. Table 1 implies that the result is true for k = 4, 5, 6. It suffices to show that fℓ (x − 2) − fu (x) > 0 for any x ≥ 14, where fu (x) = Aou (x) − Aeℓ (x + 1) is a valid upper bound on f for any x > 1. π π 4 2 4 4 Observe that the function ν(x) = 2880 π 3 x (x+1) (x−1) (x−2) cu ( 2x−4 )cℓ ( 2x ) is positive whenever x > 1. It follows that we only need to show that q(x) = ν(x)(fℓ (x−2)−fu (x)) > 0 for all x > 14. Elementary simplifications lead to q(x) = 720x11 + −10800 − 207π 2 x10 + 58320 + 606π 2 x9 + −155760 + 3087π 2 + 82π 4 x8 + 224880 − 20220π 2 − 196π 4 x7 + −169680 + 48111π 2 − 402π 4 − 10π 6 x6 + 46320 − 61602π 2 + 1928π 4 + 2π 6 x5 + 20400 + 43665π 2 − 2434π 4 + 38π 6 + 12 π 8 x4 + −19200 − 11520π 2 + 892π 4 − 42π 6 + π 8 x3 + 4800 − 6912π 2 + 514π 4 − 22π 6 + 12 π 8 x2 + 6144π 2 − 288π 4 + 4π 6 x − 1536π 2 + 96π 4 − 2π 6

a degree eleven polynomial. The largest positive real root is approximately 11.0747. The facts that it is less than 14, and that the coefficient of x11 is 720 > 0 imply that q(x) > 0 for all x ≥ 14.

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Perimeters of small regular polygons

Let Pnr denote the perimeter of the small regular n−gon. It is well known that, for even n = 2k with k ≥ 2, π r ≤ Pn+2 Pnr = n sin (7) n and for odd n = 2k + 1 with k ≥ 2, π r ≤ Pn+2 Pnr = 2n sin . (8) 2n The sequence Pnr is not monotone, and it follows directly from equations (7) and (8) that r r r r < P12 < P7r = P14 < P16 < . . . < π. P4r < P3r = P6r < P8r < P5r = P10 r for all odd k ≥ 3, and P r r More concisely, Pkr = P2k 2k−j > P2k for all k ≥ 2 and odd j < k. This immediately implies a result similar to Theorem 2.1. r r , for all integer k ≥ 2. Proposition 3.1 P2k−1 > P2k

Considering differences between perimeters of consecutive small regular polygons (with n and n+1 vertices) leads to a result similar to Theorem 2.2. Here, differences in perimeter of successive polygons decrease from the pentagon onwards. r r > Pr r − P2k Theorem 3.2 P2k−1 2k+1 − P2k+2 , for all integer k ≥ 3.

Proof. We use again the bounds on the sine function introduced in equations (4) from the proof of Theorem 2.1. This allows us to derive valid bounds on the perimeter functions. For any x > 0, π π Pℓe (x) = x sℓ < P e (x) < Pue (x) = x su , πx x π Pℓo (x) = 2x sℓ < P o (x) < Puo (x) = 2x su . 2x 2x Consider the function g(x) = P o (x) − P e (x + 1),

and define and

gℓ (x) = Pℓo (x) − Pue (x + 1) gu (x) = Puo (x) − Pℓe (x + 1).

It follows that gℓ (x − 2) − gu (x) for any x ≥ 4, is a valid upper bound on g(x − 2) − g(x). Observe that the function η(x) = 1920 x4 (x + 1)2 (x − 1)4 (x − 2)2 is always positive. π3 Therefore, it suffices to show that r(x) = η(x)(gℓ (x − 2) − gu (x)) > 0 for all x ≥ 4. Elementary simplifications lead to



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r(x) = 960x9 + −6720 − 17π 2 x8 + 16320 + 38π 2 x7 + −16960 + 37π 2 x6 + 6720 − 64π 2 x5 + 320 − 43π 2 x4 + −960 − 18π 2 x3 + 320 − 5π 2 x2 + 12π 2 x − 4π 2

a degree nine polynomial. There is a single real root approximately equal to 3.3876. The facts that it is is less than 4, and that the coefficient of x9 is 960 > 0 imply that r(x) > 0 for all x ≥ 4.

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A new proof of a theorem of Reinhardt

Theorem 2.1 and Proposition 3.1 lead to a short new proof of an often rediscovered result of Reinhardt [15][17]: Theorem 4.1 (Reinhardt) The regular unit diameter convex polygon does not have maximum area for all even n ≥ 6, nor maximum perimeter for all even n ≥ 4. Proof. Add to the regular unit diameter convex polygon with an odd number n − 1 of vertices a vertex outside of it but respecting the constraint on the diameter. Then take the convex hull of the vertices as n−vertex polygon. Its area and perimeter are larger than r , and from Theorem 2.1 and Proposition 3.1, than Ar for n ≥ 6, Arn−1 , respectively Pn−1 n r respectively Pn , for n ≥ 4. Note that the proofs of Reinhardt [17] and several subsequent authors apply adjustments to the regular polygon with n even sides. In the proofs of Mossinghoff [15, 16], adjustments are applied, as done here, beginning from the regular n − 1-gon, but in a more quantitative way.

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