ON THE ASYMMETRY FOR REULEAUX POLYGONS

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Let △abc and △a b c be triangles in R2. If |ac| = |bc|, |ab| = |a b |, Zc > π. 2. , Za = Zb, |a c |≤|ac|, |b c |≤|bc|, then Za + Zb ≤ 2Za = 2Zb. Proof. Taking x = a − c, ...
ON THE ASYMMETRY FOR REULEAUX POLYGONS Hai-lin Jin



Qi Guo

The Department of Mathematics Suzhou University of Technology and Science, Suzhou, 215009 China Abstract. In a previous paper, we showed that for regular Reuleaux polyπ gons Rn , as∞ (Rn ) = 1/(2 cos 2n − 1), where as∞ (·) denotes the Minkowski √ measure of asymmetry for convex bodies, and as∞ (K) ≤ 12 ( 3 + 1) for all convex bodies K of constant width and with equality holds iff K is a Reuleaux triangle. In this paper, we investigate the Minkowski measures of asymmetry among all Reuleaux polygons of order n and show that regular Reuleaux polygons of order n (n ≥ 3 and odd) have the minimal Minkowski measure of asymmetry. Keywords: Asymmetry measures, constant width, Reuleaux polygons MSC (2000): 52A38

1. Introduction The asymmetry (or symmetry) of convex bodies has been studied by mathematicians for a long time, and several kinds of asymmetry measure are introduced and studied (see [5∼6], [8], [10] and the references there). Among such researches, one popular topic is to find the extremal bodies, i.e., the bodies with maximal or minimal value for a given asymmetry measure, among some class of convex bodies. In 2-dimensional space R2 , the convex domains of constant width, in particular the Reuleaux polygons, got probably the most attentions of mathematicians in studying asymmetries (see [1], [3∼4], [7], [9], [11∼13]). It was known that, for most other kinds of measures of asymmetry, the Reuleaux triangles are the most asymmetric one among all convex domains. Recently, we proved that this is true also for the well-known Minkowski measure ([9]). In this paper, we continue to investigate the asymmetry of Reuleaux polygons and prove that, for the Minkowski measure of asymmetry, the regular Reuleaux polygons of order n are the extremal bodies among all Reuleaux polygons of order n. Precisely, we get the following Main Theorem. If n ≥ 3 is odd, then as∞ (RPn ) ≥ as∞ (Rn ) where RPn (Rn ) denotes the (regular) Reuleaux polygons of order n and as∞ (·) denotes the Minkowski measure of asymmetry for convex bodies. Remark 1. All regular Reuleaux polygons of order n have the same measure of asymmetry since they are affine equivalent. †

Project supported by The NSF of Jiangsu Higher Education (08KJD110016). 1

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H.L. Jin, Q. Guo

2. Preliminary Rn denotes the usual n-dimensional Euclidean space with the canonical inner product h·, ·i. A compact convex set C ⊂ Rn is called a convex body (convex domain for n = 2) if it has non-empty interior (int for brievity). The family of all convex bodies in Rn is denoted by Kn . A convex body C is said to be of constant width if its width function, i.e., the support function of C + (−C), is a constant. Equivalently, one can show that C is of constant width iff each boundary point of C is incident with (at least) one diameter (= chord of maximal length) of C. Assume that C ∈ Kn is of constant width and that V ⊂ bd(C) , the boundary of C. The set V is called a pinching set if each diameter of C is incident with (at least) one point of V . A convex body C of constant width is called a Reuleaux polygon if it admits a finite pinching set. Given a convex body C ∈ Kn and x ∈ int(C), for a hyperplane H through x and the pair H1 , H2 of support hyperplanes of C parallel to H, let r(H, x) be the ratio, not less than 1, in which H divides the distance between H1 and H2 . Denote r(C, x) = max{r(H, x) : H 3 x}. Then the Minkowski measure as∞ (C) of asymmetry of C is defined as (see [5],[6],[8]) as∞ (C) =

min r(C, x).

x∈int(C)

A point x ∈ int(C) satisfying r(C, x) = as∞ (C) is called a critical point (of C). The set of all critical points of C is denoted by C ∗ . The following is an equivalent definition of Minkowski measure (see [8], [10]). Let C ∈ Kn and x ∈ int(C). For a chord L through x with end points p, q of C, let |xq| r1 (L, x) = max{ |xp| |xq| , |xp| }, where |xp| denotes the length of xp, and denote r1 (C, x) = max{r1 (L, x) : L 3 x}. Then the Minkowski measure as∞ (C) =

min r1 (C, x).

x∈int(C)

It is known that if x ∈ int(C) is a critical point, then r1 (C, x) = as∞ (C) ([10]). A chord L satisfying r1 (L, x) = as∞ (C) is called a critical chord (of C). Denote SC (x) = {p ∈ bd(C) : x ∈ pq,

|xp| = r1 (C, x)}. |xq|

3. Proof of Main Theorem We state some known results as lemmas here first (see references for proofs). Lemma 1. ([9]) Let Rn be a regular Reuleaux polygon of order n. Then π as∞ (Rn ) = 1/(2 cos − 1). 2n Lemma 2. ([9]) Let K be a convex domain of constant width. Then 1 √ 1 ≤ as∞ (K) ≤ ( 3 + 1). 2

On the Asymmetry for Reuleaux Polygons

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Moreover, equality holds on the left-hand side precisely iff K is a circular disc, and on the right-hand side precisely iff K is a Reuleaux triangle. Lemma 3. ([10]) Suppose x ∈ ri(C ∗ ) (the relative interior) and y ∈ SC (x). Then ∞ (C)+1 y + asas (C ∗ − y) ⊂ bd(C) and y ∈ SC (p) for each p ∈ C ∗ . ∞ (C) Remark 2. This lemma shows that the set SC (p) does not vary as p ranges over ri(C ∗ ). We denote this set by C † . Lemma 4. ([10]) For C ∈ Kn , as∞ (C) + dimC ∗ ≤ n. If as∞ (C) + dimC ∗ ≥ n − 1, then ri(C ∗ ) ⊂ int(conv(C ∗ ∪ C † )), where dim denotes dimensions and conv denotes convex hulls. Lemma 5. ([10]) C † contains at least as∞ (C) + 1 points. Next, we prove two propositions which will be needed in the proof of Main Theorem. Proposition 1. If RPn is a Reuleaux polygon of order n with vertices e1 , ..., en , the (unique) critical point o and the width ω, then there are three vertices ei1 , ei2 , ei3 a , where a := maxi {|oei |}. such that |oei1 | = |oei2 | = |oei3 | = a and as∞ (RPn ) = ω−a Proof. By Lemma 5, there are three critical chords of RPn (notice as∞ (RPn ) > 1), i| denoted by pi qi , i = 1, 2, 3, such that as∞ (RPn ) = |op |oqi | . By Theorem 3.2 of [10], there are three pairs of line supporting RPn at pi , qi respectively. Since |opi | RPn is of constant width, |pi qi | = ω. Thus as∞ (RPn ) = ω−|op , which leads i| to |opi | =

as∞ (RPn ) as∞ (RPn )+1 ω,

i = 1, 2, 3.

Now we shall show that each pi is a vertex of RPn and |opi | = maxi {|oei |}. → By the defiLet x ∈ bd(RPn ), x0 be the intersection of bd(RPn ) and the ray − xo. |ox| ∞ (RPn ) |xx0 |. nition of r1 (RPn , o), |ox0 | ≤ as∞ (RPn ), which implies that |ox| ≤ asas ∞ (RPn )+1 0 Notice that |xx | ≤ ω implies |ox| ≤ |opi |, so RPn ⊂ B, where B := B(o, |op1 |) is the disc with radius |op1 | and centered at o. If one of these points pi , say p1 , is not a vertex, then q1 must be a vertex by Theorem 3.2 of [10]. Let ej1 , ej2 be the vertices such that |q1 ej1 | = |q1 ej2 | = |q1 p1 | = ω. Thus, if set B1 = B(q1 , ω), then ej1 , ej2 ∈ / B and B1 ⊃ B ⊃ RPn , a contradiction. ¤ Lemma 6. Under the constraint s2 + t2 = d2 , s, t > 0, where d ≥ 0 is a constant, the function f (s, t) := s2 +t2 −2cst, where c < 0 is a constant, attains its maximum (1 − c)d2 when s = t = √d2 . In particular, the maximum is increasing with respect to d. Proof. The proof is a routine calculation.

¤

Proposition 2. Let x, y, x0 , y 0 ∈ Rn , viewed as vectors. If |x−y| = |x0 −y 0 |, |x| = |y| and |x| ≥ |x0 |, |y| ≥ |y 0 |, θxy > π2 , then cos θxy ≥ cos θx0 y0 and in turn θxy ≤ θx0 y0 , where θxy denotes the angle between (vectors) x and y. Proof. It’s clear that θx0 y0 >

π 2.

By |x − y| = |x0 − y 0 |, we get

hx − y, x − yi = hx0 − y 0 , x0 − y 0 i.

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H.L. Jin, Q. Guo

Thus, if setting x00 , y 00 such that |x00 | = |y 00 | and |x00 |2 + |y 00 |2 = |x0 |2 + |y 0 |2 , then by Lemma 6, |x|2 + |y|2 − 2|x||y| cos θxy = |x0 |2 + |y 0 |2 − 2|x0 ||y 0 | cos θx0 y0 ≤ |x00 |2 + |y 00 |2 − 2|x00 ||y 00 | cos θx0 y0 ≤ |x|2 + |y|2 − 2|x||y| cos θx0 y0 which leads clearly to cos θxy ≥ cos θx0 y0 and in turn θxy ≤ θx0 y0 .

¤

Corollary 1. Let 4abc and 4a0 b0 c0 be triangles in R2 . If |ac| = |bc|, |ab| = |a0 b0 |, ∠c > π2 , ∠a = ∠b, |a0 c0 | ≤ |ac|, |b0 c0 | ≤ |bc|, then ∠a0 + ∠b0 ≤ 2∠a = 2∠b. Proof. Taking x = a − c, y = b − c, x0 = a0 − c0 , y 0 = b0 − c0 in Proposition 2, we get the desired conclusion. ¤ Now, we present the proof for our Main Theorem. Since all Reuleaux triangles are regular, we need only to prove it for n ≥ 5 . Proof. of Main Theorem. For simplicity, we start with the Reuleaux polygons of order 5. By the definition of Reuleaux polygons, |e1 e3 | = |e1 e4 | = |e2 e4 | = |e2 e5 | = |e3 e5 | = ω (see Fig.1). By Proposition 1 and Lemma 4, we may assume |oe | = √1 3+1 ∞ (RP5 ) |oe3 | = |oe4 | = a = asas ω, where a = max {|oe |}. Since as (RP ) ≤ i i ∞ 5 2 , ∞ (RP5 )+1 √

we get a = |oe1 | ≤ 33 ω and ∠e1 oe4 > π2 . Denote α := ∠oe1 e4 = ∠oe4 e1 , and denote by αi (i = 1, 2..., 10) the ten base angles (e.g. ∠oe1 e3 ) of these five triangles formed by taking the convex hull of e1 e3 , e1 e4 , e2 e4 , e2 e5 or e3 e5 in turn with the point o. By Corollary 1, we have Σ10 i=1 αi ≤ 10α. Thus, since each ∠ei could be expressed as the sum or difference of two of these αi0 s uniquely, we get π = Σ5i=1 ∠ei ≤ π Σ10 i=1 αi ≤ 10α, which leads to α ≥ 10 . Hence, by the fact that ω = 2|oe1 | cos α and Lemma 1, as∞ (RP5 ) =

|oe1 | 1 1 = as∞ (R5 ). = ≥ π −1 ω − |oe1 | 2 cos α − 1 2 cos 10

As for the general n, let ei , i = 1, ..., n be the vertices of RPn , then there exist at least three vertices, denoted by ei1 , ei2 , ei3 , such that |oei1 | = |oei2 | = |oei3 | = ∞ (RPn ) a = asas ω, where a := maxi {|oei |}. We consider the triangle conv{x, y, z}, ∞ (RPn )+1 where |xy| = |xz| = a, |yz| = ω and denote ∠xyz by α (see Fig. 1). By the same argument as used above, we obtain that ∠yxz > π2 . Similarly, denoting by αi (i = 1, 2..., n) the 2n base angles of these n triangles formed by taking in turn the convex hull of ei ej and o for these n chords with |ei ej | = ω, then by applying Corollary 1 repeatly, we have Σ2n i=1 αi ≤ 2nα. Since in this case each ∠ei could also be expressed as the sum or difference of two of these αi0 s uniquely, we have π = Σni=1 ∠ei ≤ Σ2n i=1 αi ≤ 2nα, which leads to α ≥

π 2n

and.

as∞ (RPn ) =

a 1 1 = as∞ (Rn ). = ≥ π −1 ω−a 2 cos α − 1 2 cos 2n ¤

On the Asymmetry for Reuleaux Polygons

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References [1] A. S. Besicovitch, Measures of asymmetry for convex curves II. Curves of constant width. J. London Math. Soc., no. 26, 81-93 (1951). [2] G. D. Charkerian, H. Gromer, Convex bodies of constant width. In: Convexity and its applications, Birkh¨ auser Verlag, Basel-Boston-Stuttgart , 49-96 (1983). [3] H. G. Eggleston, A proof of Blaschke’s theorem on the Reuleaux triangle. Quart. J. Math., vol. 2, no. 3, 296-297 (1952). [4] W. J. Firey, Isoperimetric ratios of Reuleaux polygons. Pacific J. Math. no. 10, 823-829 (1960). [5] Q. Guo, Stability of Minkowski measure of asymmetry of convex bodies. Discrete Comput. Geom., no. 34, 351-362 (2005). [6] Q. Guo, S. Kaijser, On asymmetry of some convex bodies. Discrete Comput. Geom., no. 27, 239-247 (2002). [7] H. Groemer, L. J. Wallen, A measure of asymmetry for domains of constant width. Beitr¨ age zur Algebra und Geom., no. 42, 517-521 (2001). [8] B. Gr¨ unbaum, Measure of convex sets. Convexity, Proceedings of symposia in Pure Mathematics 7 (American Math. Society, providence), 233-270 (1963). [9] H.L.Jin, Q.Guo, On the asymmetry for convex domains of constant width. Communications in Mathematics Research, vol. 26, no. 2, 176-182 (2010). [10] V. L. Klee, The critical set of a convex set. Amer. J. Math., no. 75, 178-188 (1953). [11] Y. K. Kupitz, H. Martini, On the isoperimetric inequalities for Reuleaux polygons. J. of Geom., no. 68, 171-191 (2000). [12] Y. K. Kupitz, H. Martini, B. Wegner, A linear-time construction of Reuleaux polygons. Beitr¨ age zur algebra und Geometrie, vol. 37, 415-427 (1996). [13] G. T. Sallee, Maximal area of Reuleaux polygons. Canad. Math. Bull., no. 13, 175-179 (1970). [14] R. Schneider, Convex bodies: the Brunn-Minkowski theory. Cambridge University Press (1993).