Simple Polygons of Maximum Perimeter Contained in a Unit Disk

Similar documents
Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

Analytic Continuation

Solutions for May. 3 x + 7 = 4 x x +

RADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify

Math 155 (Lecture 3)

THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART II Calculators are NOT permitted Time allowed: 2 hours

Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get

TR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT

Coffee Hour Problems of the Week (solutions)

Mathematical Foundations -1- Sets and Sequences. Sets and Sequences

Infinite Sequences and Series

INEQUALITIES BJORN POONEN

4755 Mark Scheme June Question Answer Marks Guidance M1* Attempt to find M or 108M -1 M 108 M1 A1 [6] M1 A1

SEQUENCE AND SERIES NCERT

2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.

Bertrand s Postulate

Math 220A Fall 2007 Homework #2. Will Garner A

Solutions. tan 2 θ(tan 2 θ + 1) = cot6 θ,

A class of spectral bounds for Max k-cut

USA Mathematical Talent Search Round 3 Solutions Year 27 Academic Year

CALCULUS BASIC SUMMER REVIEW

Optimization Methods MIT 2.098/6.255/ Final exam

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007

Sequences and Series of Functions

4.1 Sigma Notation and Riemann Sums

Week 5-6: The Binomial Coefficients

APPENDIX F Complex Numbers

CHAPTER 10 INFINITE SEQUENCES AND SERIES

NICK DUFRESNE. 1 1 p(x). To determine some formulas for the generating function of the Schröder numbers, r(x) = a(x) =

Complex Analysis Spring 2001 Homework I Solution

REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

Chapter 1. Complex Numbers. Dr. Pulak Sahoo

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations

Sequences and Limits

Chapter 4. Fourier Series

JEE ADVANCED 2013 PAPER 1 MATHEMATICS

A NOTE ON INVARIANT SETS OF ITERATED FUNCTION SYSTEMS

1. By using truth tables prove that, for all statements P and Q, the statement

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1

6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.

Notes #3 Sequences Limit Theorems Monotone and Subsequences Bolzano-WeierstraßTheorem Limsup & Liminf of Sequences Cauchy Sequences and Completeness

3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials

18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016

w (1) ˆx w (1) x (1) /ρ and w (2) ˆx w (2) x (2) /ρ.

Machine Learning for Data Science (CS 4786)

PRACTICE FINAL/STUDY GUIDE SOLUTIONS

Complex Numbers Solutions

Seunghee Ye Ma 8: Week 5 Oct 28

MAT1026 Calculus II Basic Convergence Tests for Series

Chapter 6 Infinite Series

End-of-Year Contest. ERHS Math Club. May 5, 2009

Lecture Notes for Analysis Class

Stanford Math Circle January 21, Complex Numbers

Solution. 1 Solutions of Homework 1. Sangchul Lee. October 27, Problem 1.1

Optimally Sparse SVMs

MA131 - Analysis 1. Workbook 2 Sequences I

SNAP Centre Workshop. Basic Algebraic Manipulation

6.3 Testing Series With Positive Terms

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j

Math 140A Elementary Analysis Homework Questions 3-1

SEQUENCES AND SERIES

ACO Comprehensive Exam 9 October 2007 Student code A. 1. Graph Theory

PUTNAM TRAINING INEQUALITIES

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled

MATH 10550, EXAM 3 SOLUTIONS

1 Introduction. 1.1 Notation and Terminology

Lecture 10 October Minimaxity and least favorable prior sequences

Solutions to Math 347 Practice Problems for the final

The Boolean Ring of Intervals

If a subset E of R contains no open interval, is it of zero measure? For instance, is the set of irrationals in [0, 1] is of measure zero?

MATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is

Solutions to Tutorial 3 (Week 4)

Sequences I. Chapter Introduction

4.1 SIGMA NOTATION AND RIEMANN SUMS

10-701/ Machine Learning Mid-term Exam Solution


International Contest-Game MATH KANGAROO Canada, Grade 11 and 12

Polynomials with Rational Roots that Differ by a Non-zero Constant. Generalities

Randomized Algorithms I, Spring 2018, Department of Computer Science, University of Helsinki Homework 1: Solutions (Discussed January 25, 2018)

Riemann Sums y = f (x)

PROBLEM SET 5 SOLUTIONS 126 = , 37 = , 15 = , 7 = 7 1.

Alternating Series. 1 n 0 2 n n THEOREM 9.14 Alternating Series Test Let a n > 0. The alternating series. 1 n a n.

REVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains.

September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1

SOME GENERALIZATIONS OF OLIVIER S THEOREM

LONG SNAKES IN POWERS OF THE COMPLETE GRAPH WITH AN ODD NUMBER OF VERTICES

Appendix F: Complex Numbers

THE INTEGRAL TEST AND ESTIMATES OF SUMS

Definitions and Theorems. where x are the decision variables. c, b, and a are constant coefficients.

Sequences and Series

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

MTH Assignment 1 : Real Numbers, Sequences

16 Riemann Sums and Integrals

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6

REGRESSION WITH QUADRATIC LOSS

Transcription:

Discrete Comput Geom (009) 1: 08 15 DOI 10.1007/s005-008-9093-7 Simple Polygos of Maximum Perimeter Cotaied i a Uit Disk Charles Audet Pierre Hase Frédéric Messie Received: 18 September 007 / Revised: May 008 / Accepted: 6 May 008 / Published olie: 0 Jue 008 Spriger Sciece+Busiess Media, LLC 008 Abstract A polygo is said to be simple if the oly poits of the plae belogig to two of its edges are its vertices. We aswer the questio of fidig, for a give iteger,asimple-sided polygo cotaied i a disk of radius 1 that has the logest perimeter. Whe is eve, the optimal solutio is arbitrarily close to a lie segmet of legth. Whe is odd, the optimal solutio is arbitrarily close to a isosceles triagle. Keywords Simple polygo Perimeter 1 Itroductio I this paper, we give a complete aswer to the questio posed by Brass, Moser ad Pach [1, p. 37]: for a give iteger, fid a simple -sided plaar polygo cotaied i a disk of radius 1 that has the logest perimeter. A polygo is said to be simple, or oitersectig, if the oly poits of the plae belogig to two of its edges are its vertices. These authors metio that the optimal solutio for the case where is eve is arbitrarily close to the trivial upper boud. It suffices to cosider a simple Work of the first author was supported by NSERC grat 3936-05, AFOSR FA9550-07-1-030, ad ExxoMobil. Work of the secod author was supported by NSERC grat 10557-0. C. Audet ( ) P. Hase GERAD ad Départemet de Mathématiques et de Géie Idustriel, École Polytechique de Motréal, C.P. 6079, Succ. Cetre-ville, Motréal, Québec, H3C 3A7 Caada e-mail: Charles.Audet@gerad.ca P. Hase e-mail: Pierre.Hase@gerad.ca F. Messie ENSEEIHT-IRIT, UMR-CNRS 5505, rue Camichel, BP 71, 31071 Toulouse Cedex 7, Frace e-mail: Frederic.Messie@7.fr

Discrete Comput Geom (009) 1: 08 15 09 polygo with all vertices with odd idices beig arbitrarily close to ( 1, 0) ad all vertices with eve idices beig arbitrarily close to (1, 0). We show that for = 3, the optimal solutio is the equilateral triagle ad that for odd values of 5, the optimal solutio is arbitrarily close to a isosceles triagle. A Simple Polygo with Logest Perimeter Let 3 be a odd iteger, ad v 1,v,...,v deote the cosecutive vertices of a simple polygo cotaied i a disk of radius 1. Defie N ={1,,...,}, ad for ay i N, letc i = v i+1 v i be the Euclidea distace betwee the two cosecutive vertices v i = (x i,y i ) ad v i+1 = (x i+1,y i+1 ) (throughout this ote, sums of idices are take modulo ). The perimeter of the polygo may the be writte as p = i=1 c i. Defie p to be the supremum of the perimeter i value over all simple polygos cotaied i a disk of radius 1. We use the term supremum istead of maximum sice we will see that there are o optimal simple polygos whe >3. Ideed, we already kow that the optimal solutio for a eve umber of vertices is arbitrarily close to a lie segmet, a figure with oly vertices. For = 3, a equilateral triagle yields a lower boud of p 3 = 3 3op3. For a odd umber of vertices 5, oe ca easily show that p = ( ) + = ( )> 3 is a lower boud o p. To do so, it suffices to costruct a polygo with all of its vertices with eve idex arbitrarily close to (1, 0), the vertex v at (0, 1), ad the remaiig oes arbitrarily close to ( 1, 0). These lower bouds o p are ot tight sice a small displacemet of all vertices at (±1, 0) o the circle (ad away from (0, 1)) produces a ew isosceles triagle. The the derivative of the legth of the edges betwee ( 1, 0) ad (1, 0) is zero, ad the derivative of the legth of the edges betwee (0, 1) ad (±1, 0) is strictly positive. Therefore, a small displacemet will icrease the perimeter, ad we will show that the optimal figure is obtaied by such a displacemet. By relabelig the idices if ecessary, we suppose that ad 1 are the idices of the cosecutive vertices that are the farthest apart, i.e., c c i for all i N. Furthermore, sice we are iterested i maximizig the perimeter, we will cosider that c p. I particular, this implies that c 3 3 ad that c ( )> 3 for odd 5. Furthermore, we will assume that the disk cotaiig the polygo is cetered at (0, 0) ad that the vertices v 1 ad v are vertically aliged: x = x 1 ad y = y 1 + c. Defie U = { (a, b) : a + b 1,b c 1 } ad L = { (a, b) : a + b } 1,b 1 c. The ext result shows that ay two cosecutive vertices of a optimal polygo caot both lie i the same zoe L or U. Lemma 1 If 5 ad if two cosecutive vertices v l ad v l+1 belog to same set U or L, the p<p.

10 Discrete Comput Geom (009) 1: 08 15 Fig. 1 Separatio of the disk ito zoes Proof If v l ad v l+1 areithesamesetl or U (represeted by the shaded regios of Fig. 1), the c l = v l+1 v l 1 (c 1) = c c. It follows that p = i=1 c i ( 1)c + c c. The maximum of the right part of the previous iequality is attaied either whe c is at oe of its boud or 3 or whe the first derivative with respect to c equals zero, i.e., whe ( 1) + (1 c )/ c c = 0. Oe ca easily verify that the maximum occurs whe c = + 1, ad therefore 1 +5 p + 5 + 1. Furthermore, oe ca show that if 5, the this upper boud o the perimeter is strictly iferior to the lower boud p = + sice + 5 < ( 3 + ). The ext lemma shows that there are three cosecutive vertices that satisfy a orderig property. Lemma There exist three cosecutive vertices v l 1,v l, ad v l+1 for some l {, 3,..., 1} such that either y l 1 y l y l+1 or y l 1 y l y l+1. Proof The previous lemma esures that y > 1 c y 1 ad y 1 <c 1 y.

Discrete Comput Geom (009) 1: 08 15 11 Fig. Maximizatio of c l 1 + c l without itersectig v 1 v Suppose ow that the result is false. The, y >y 1 implies that y 3 <y. Similarly y 3 <y implies that y >y 3. More geerally, if i is eve, the y i+1 <y i ad y i >y i 1. It follows that, sice is odd, y 1 >y, which is a cotradictio. Without loss of geerality ad by takig vertical ad horizotal symmetrical images if ecessary, we ca assume that the vertex v l of Lemma is such that x l x 1 ad y 1 y > 0 (i.e., that y 1 y 1 + c > 0). Defie u [ 1 c /, 0] to be the smallest scalar such that u + y1 1. The the segmet joiig (u, y 1) ad (u, y 1 + c ) does ot itersect the segmets v l 1 v l ad v l v l+1. These quatities are illustrated o Fig.. The objective of the ext results is to derive a upper boud o the sum c l 1 + c l = v l v l 1 + v l v l+1. The two previous lemmas esure that the vertex v l belogs to the set Ω ={(a, b) : u a 1 b, b 1 c }. Defie the fuctio f : R R as f(x)= dist ( ) ( ) (x, y l ), v l 1 + dist (x, yl ), v l+1 for (x, y l ) Ω. It is a covex fuctio, ad its maximum occurs whe (x, y l ) belogs to the boudary of Ω, i.e., whe x = u or whe x = 1 yl. The ext pair of lemmas cosiders these cases. Lemma 3 If 5 ad v l = (u, y l ), the p<p. Proof If v l = (u, y l ), the c l 1 + c l u + 1 yl + u + yl i Fig. 3. It follows that max u + 1 y + u + y + 1 y s.t. u + y 1 + 1, as illustrated

1 Discrete Comput Geom (009) 1: 08 15 Fig. 3 The vertex v l satisfies x l = u 0 is a valid upper boud o c l 1 + c l. The maximum is attaied whe y = 1 (1 u ), ad the optimal value is z(u) = u + + u + 1. Usig the facts that u 1 c / ad c > 3 whe 5, oe ca show that z(u) + c < 1 3/ + + (1 3/) + 1 + 3 = 3/ + 5/ + 3. Therefore p ( )c + z(u) = ( 3)c + z(u) + c < ( 3) + 3/ + 5/ + 3 = 9/ + 5/ + 3 < + = p. Lemma If 5 ad v l = ( 1 yl,y l) or if = 3, the c l 1 + c l + 1 c. Proof Cosider the case where v l = ( 1 yl,y l). Defie v 1 = (u, y 1), ad let v = (x,y ) be the poit o the boudary of the disk such that v v 1 =c ad x u. For ay 0 θ π, defie p(θ) to be the poit o the boudary of the circle that makes a agle of θ with v l, as illustrated i the left part of Fig.. The p(0) = v l. Defie 0 θ π θ 1 π to be the agles that satisfy v 1 = p(θ 1 ) ad v = p(θ ).

Discrete Comput Geom (009) 1: 08 15 13 Fig. The poit o the disk at maximum distace from ( c, 0) ad ( c, 0) The distace fuctio, p(θ) v l is icreasig for θ [0,π] ad decreasig for θ [π,π]. However, if θ <θ<θ 1, the the segmet joiig p(θ) ad v l itersects the segmet v 1 v. It follows that the sum c l 1 + c l is bouded above by v l v + v l v 1. To simplify the presetatio, let us move the disk so that the segmet v 1 v is horizotal with ed poits (±c /, 0), ad the coordiates of v l become (a, b) with b 0, as illustrated i the right part of Fig.. Observe that the case where = 3is represeted by the right side of that figure, as all three vertices of the optimal triagle ecessarily lie o the boudary of the disk. Equatio (1) provides a upper boud o the sum: ( ) ( ) (a, b) + c, 0 + (a, b) c, 0 max p + q a,b,p,q ( s.t. p = a + c ) + b, ( q = a c ) (1) + b, ( a + b 1 c ) = 1. The Lagrage multipliers system is ( 0 = a + c ) ( λ 1 + a c ) λ + aλ 3, ( ) 0 = bλ 1 + bλ + b 1 c λ 3,

1 Discrete Comput Geom (009) 1: 08 15 1 = pλ 1, 1 = qλ. Solvig the two last equatios for the Lagrage multipliers ad substitutig them ito the two first oes gives ( q a + c ) ( + p a c ) = apqλ 3, ( ) (q + p)b = pq b 1 c After elimiatig λ 3 ad makig some simplificatios, oe gets ( ) (q p) c b 1 c = (q + p)a 1 c. () Multiplyig both sides by (q p) ad usig the costraits of (1) to substitute (q + p)(q p) = q p = ac leads to ( ) (q p) c b 1 c = a c 1 c. Recall that we are lookig for a solutio with b 1 c 0 ad c > 0. Therefore, the left-had-side term is oegative, ad the right-had-side term is opositive. To get a equality, we eed that a = 0 or that a 0 ad c =. If a = 0, the p = q = + b = c c + ( 1 + 1 c λ 3. ) = + 1 c. If a 0 ad c =, the () becomes b(p q) = 0. The case where b = 0 correspods to a miimizer, ad p = q implies that (a + 1) = (a 1), i.e., that a = 0, which is a cotradictio. Therefore, we have show that c l 1 + c l + 1 c. Combiig the above lemmas leads to our mai result. Theorem 1 If is a odd iteger, the the supremum of the perimeter i value over all simple polygos cotaied i a disk of radius oe is p = ( 1 + 8( ) 1) 1 ( 1 + 8( ) + 3) 3 ( ) ad the polygo is arbitrarily close to a isosceles triagle.,

Discrete Comput Geom (009) 1: 08 15 15 Proof Combiig Lemmas 3 ad ad settig v = 1 c leads to p ( )c + + 1 c = ( ) 1 v + + v. I order to miimize this upper boud, we take the derivative ad set it equal to zero. This gives v ( ) + = 0 1 v 1 + v 1 + v ad may be rewritte as ( ) v + v 1 = 0. The uique oegative solutio is v = 1+ 1+8( ). Substitutig for v ad simplifyig leads to ( ) p ( 1 + 8( ) 1) 1 ( 1 + 8( ) + 3) 3. ( ) A simple polygo with a perimeter arbitrarily close to that value ca be costructed. Cosider a simple polygo i which ( 1) cosecutive vertices alterate back ad forth arbitrarily close to ( c, 0) ad ( c, 0), ad the remaiig vertex is located at (0, 1 + v), where c = 1 v ad v = 1+ 1+8( ). The resultig ( ) figure is arbitrarily close to a isosceles triagle, ad i the limit, the perimeter satisfies p = ( 1 + 8( ) 1) 1 ( 1 + 8( ) + 3) 3. ( ) I the case where = 3, the optimal triagle is equilateral. We coclude with the followig corollary. Corollary 1 For odd values of, the sequece { p } is mootoe icreasig. Proof For ay fixed value of, maximizig the perimeter or maximizig the average edge legth yields the same figure. Furthermore, for ay fixed odd value of, oe ca easily costruct a ( + )-sided polygo whose average edge legth strictly exceeds p. Ideed, oe simply has to cosider a optimal -sided polygo ad add a vertex ear ( a,0) ad aother oe ear (a, 0). This will add two edges of legth arbitrarily close to the value c > p. Refereces 1. Brass, P., Moser, W., Pach, J.: Research Problems i Discrete Geometry. Spriger, New York (005)