DOCSLIB.ORG

Bulging Triangles: Generalization of Reuleaux Triangles

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Bulging Triangles: Generalization of Reuleaux Triangles Bulging Triangles: Generalization of Reuleaux Triangles Norihiro Someyama∗ ∗Shin-yo-ji Buddhist Temple, Tokyo, Japan [email protected] ORCID iD: https://orcid.org/0000-0001-7579-5352 Abstract We introduce a bulging triangle like the generalization of the Reuleaux triangle. We may be able to propose various ways to bulge a triangle, but this paper presents the way so that its vertices are the same as them of the original triangle. We find some properties and theorems of our bulging triangles. In particular, we investigate, via calculus, whether basic facts such as triangle inequalities and Pythagorean theorem hold for bulging triangles. Keywords: Reuleaux triangle, Covexity, Triangle inequality, Pythagorean theorem. 2010 Mathematics Subject Classification: 51M04, 51M05, 51N20. 1 Introduction • the \triangle inequality" holds, i.e., the sum of the lengths of any two \round sides" is There is the Reuleaux triangle (also called the greater than the length of the remaining one spherical triangle) as one of famous and notable \round side", figures in geometry. It is named after F. Reuleaux (1829-1905), a German mechanical engineer. He • the three \round sides" are equal and the considered a figure in which an equilateral triangle common length is πa=3 ; 1:047a if the length is bulged in a natural way so that it has a constant of one side of 4ABC is a, p p width. To be more precise, it is defined as follows: 2 2 • its area is (π − 3)a =2 ; 1:628 × ( 3a =4) Let 4ABC be an equilateral triangle consisting (i.e., about 1.628 times the original area) if of vertices A, B and C. It namely holds that the length of one side of 4ABC is a. jABj = jBCj = jCAj Moreover, Reuleaux triangles are widely applied in various fields and there is no end to the list. where AB denotes the side whose end points are A See, e.g., [2] for the benefits and applications of and B, jABj denotes its length, and the other edges Reuleaux triangles. It is also well known that the are similar. We have a bulging equilateral triangle Reuleaux triangle is one of solutions for the Kakeya arXiv:2108.07725v1 [math.GM] 11 Aug 2021 like Figure 1 by drawing arcs with center C and problem (see, e.g., [1] for more information). radius BC, with center A and radius CA and with Our aim in this paper is to introduce the \gen- center B and radius AB. eralized" Reuleaux triangle with different lengths The most notable property of the Reuleaux tri- on the three \round sides" and study the proper- angle is that it can rotate completely while being ties. within the square and touching all four sides of the square. This property has been applied to archi- 2 Main Topics tectural and engineering technologies. Any Reuleaux triangle has some trivial mathe- 2.1 Definition of Bulging Triangles matical properties, e.g., Let 4ABC be a triangle. For convenience, think • it is (strictly) convex, of this as an acute triangle. We would like to make 1 a figure that 4ABC is bulged by the following 2.2 Basic Properties of Bulging Trian- way. Such a figure is called the bulging trian- gles gle (stemmed) from 4ABC in this paper. Proposition 2.1. The AB-center P of the bulging Consider the perpendicular bisector ` of side ( AB, and find out the intersection point P of an- triangle 4ABC is on the longer of the remaining other side and `. Then, remark that P lies on the two sides BC; CA. In other words, P is the in- longer (BC in the case of Figure 2) of the remain- ternally dividing point of either BC or CA. The ing two sides BC and CA. This fact is almost clear, BC-center and CA-center are similar. but we will prove it later (Proposition 2.1). Deter- Proof. Assume \ABC < \BAC and denote by P mine intersection points Q and R in the same way the point such that jAPj = jBPj. Then, for sides BC and CA. Here each P, Q, R is, of course, uniquely determined. Note that \BAP = \ABC < \BAC and so P is on BC. Since P is the AB-center of jAPj = jBPj; ( 4ABC, this completes the proof. jBQj = jCQj; The bulging triangle that dents at a vertex is jCRj = jARj: not ideal. We need to check the following fact. We obtain the \round side" of 4ABC by draw- ( Proposition 2.2. Let 4ABC be any bulging tri- ing the circular arc whose center and radius are P angle from 4ABC. and jAPj (= jBPj), respectively. The other \round ( sides" are similarly obtained. From the above, we 1) If 4ABC is acute, then 4ABC is convex; have been able to define the \bulging triangle" ( (Figure 3) that is naturally extended the Reuleaux 2) If 4ABC is right, then 4ABC is convex; triangle to the general case. We can treat vertices ( 3) If 4ABC is obtuse, then 4ABC is concave. of the original triangle as vertices of the bulging triangle. Proof. Without loss of generality, we can visualize ( In what follows, we set the following promises: 4ABC from 4ABC with \BCA = \R. Note that ( the BC-center and the CA-center match and the • We write 4 ABC for the bulging triangle common point is the middle point of AB. In other whose vertices are A, B and C. words, BC+f CAg (i.e., the arc formed by connecting ( ( • Its \round sides" are called edges of 4ABC BCf and CA)g coincides with semi-circular arc BA and are denoted by AB,g BCf and CA.g Their containing point C. By virtue of the above, the lengths are denoted by jABgj, jBCf j and jCAgj, proof is easily obtained. respectively. In response to the result of Proposition 2.2, we • Three points P, Q and R defined above are exclusively consider bulging triangles from acute or called centers associated with edges of right-angled triangles in this paper. ( 4ABC. We often abbreviate such P as the For convenience, we put the following term. AB-center and the others are similar. Definition 2.1. Let 4ABC be a right-angled tri- angle with BCA = R. The bulging triangle from • The radius (i.e., AP, BP, and so on) of the \ \ ( sector consisting of A, B and P is called the this 4ABC is denoted by 4RBCA. The others are AB-radius. The others are similar. similar. Such a bulging triangle is called the right- angled bulging triangle in this paper. In partic- ( • The \length of the radius" is simply abbrevi- ular, we call 4RBCA the isosceles right-angled ated as \the radius" throughout this paper. bulging triangle if jBCj = jCAj. Remark 2.1. If 4ABC is an equilateral triangle, We study the edges of a bulging triangle from the AB-center is C, the BC-center is A and the the famous right triangle. In particular, the result CA-center is B. Hence, the bulging triangle defined for an isosceles right-angled bulging triangle can be above is certainly an extension of the Reuleaux tri- shown to be exactly the same as the conventional angle. one. 2 ( Proposition 2.3. For 4RBCA with jBCj = a, it Proof. Denote by P and Q the AB-center and CA- ( follows that center of 4 ABC, respectively (see Figure 5). It is obviously that QP k BC. Considering the per- 1) one has pendicular bisector L, passing through point A, of p BC, it follows that Q is the point of symmetry of jABgj : jBCf j : jCAgj = 2 : 1 : 1 P with respect to L. B is also, of course, the point if \ABC = π=4, of symmetry of C with respect to L. Hence, the claim clearly holds. This completes the proof. 2) one has p p Proposition 2.5. Suppose that 4ABC obeys jABj : jBCj : jCAj = 4 : 3 : 2 3 g f g π 0 < \ABC < \BCA ≤ : (2.4) if \ABC = π=3. 2 ( Proof. We leave the proof of 1) to the reader and Then, 4ABC from such 4ABC satisfies that prove only 2). jCAj < jABj: (2.5) Since jBCj = a and \ABC = π=3, we have g g p π jABj = 2a; jCAj = 3a; BAC = : Proof. Put \ 6 a := jBCj; b := jCAj; c := jABj (2.6) Denoting by P thep AB-center, P lies on CA, the AB-radius is 2a= 3 and \APB = 2π=3. Hence we and have p α := BAC; β := ABC; γ := BCA: (2.7) 4 3 \ \ \ jABgj = πa: (2.1) 9 We prove tha claim by dividing into two patterns. (See Figure 4.) Denoting by Q the BC-center, Q i) The case of α ≤ β: lies on AB, the BC-radius is a and \BPC = π=3. Hence we have By noting that β < γ, we have 1 b jBCf j = πa: (2.2) jCAgj = (π − 2α) 3 2 cos α Denoting by R the CA-center, R = Q, the CA- and radius is a and \CPA = 2π=3. Hence we have c jABgj = (π − 2α): 2 2 cos α jCAgj = πa: (2.3) 3 From β < γ again, we obtain b < c. This (2.1)-(2.3) imply the claim, so the proof is com- implies (2.5). plete. ii) The case of β ≤ α: We investigate the relationship between the By noting that β < γ, we have edge of the bulging triangle and the angle of the b original triangle. Take note that we can meet ex- jCAgj = (π − 2α) actly the same results for a usual triangle. 2 cos α ( and Proposition 2.4.
Load more

Recommended publications
  • Extremal Problems for Convex Polygons ∗
    Extremal Problems for Convex Polygons ∗ Charles Audet Ecole´ Polytechnique de Montreal´ Pierre Hansen HEC Montreal´ Fred´ eric´ Messine ENSEEIHT-IRIT Abstract. Consider a convex polygon Vn with n sides, perimeter Pn, diameter Dn, area An, sum of distances between vertices Sn and width Wn. Minimizing or maximizing any of these quantities while fixing another defines ten pairs of extremal polygon problems (one of which usually has a trivial solution or no solution at all). We survey research on these problems, which uses geometrical reasoning increasingly complemented by global optimization meth- ods. Numerous open problems are mentioned, as well as series of test problems for global optimization and nonlinear programming codes. Keywords: polygon, perimeter, diameter, area, sum of distances, width, isoperimeter problem, isodiametric problem. 1. Introduction Plane geometry is replete with extremal problems, many of which are de- scribed in the book of Croft, Falconer and Guy [12] on Unsolved problems in geometry. Traditionally, such problems have been solved, some since the Greeks, by geometrical reasoning. In the last four decades, this approach has been increasingly complemented by global optimization methods. This allowed solution of larger instances than could be solved by any one of these two approaches alone. Probably the best known type of such problems are circle packing ones: given a geometrical form such as a unit square, a unit-side triangle or a unit- diameter circle, find the maximum radius and configuration of n circles which can be packed in its interior (see [46] for a recent survey and the site [44] for a census of exact and approximate results with up to 300 circles).
    [Show full text]
  • Network Editorial Rsula Martin’S Article in the Last Issue Was Particularly Topical in Celebrating the Impact of Mathematics Upon U Society
    network Editorial rsula Martin’s article in the last issue was particularly topical in celebrating the impact of mathematics upon U society. Although I have yet to see Hidden Figures, I have read the book by Margot Lee Shetterly that inspired this film. It is astonishing that the brilliant mathematicians – who performed the extensive analyses and laborious calculations that enabled the USA to fly supersonically, journey through space and visit the moon – were not properly recognised for 50 years. © Drake2uk | Dreamstime.com © Drake2uk These forgotten geniuses were known simply as computers and included Dorothy Vaughan, Katherine Johnson and Mary Jackson. At least their dedication and achievements have finally been acknowledged. The accomplishments of many other gifted mathematicians are similarly hidden from public view due to their sensitive nature, particularly in defence as evidenced by vending machines, supermarket trolleys and other automated James Moffat’s remarkable recollections that also appeared in payment devices. Instead, it approximates a regular convex April’s issue of Mathematics Today. curvilinear dodecagon, similar to an old threepenny bit, and its I hope that you have visited the IMA’s fabulous new website. diameter varies from 23.03 mm to 23.43 mm. There is plenty to explore, including puzzles, careers, jobs, In contrast, the 20p and 50p coins rather surprisingly have policies, newsletters, articles, publications, awards, conferences, constant widths despite being non-circular, as they approximate branches, workshops and access to the Reuleaux heptagons. The width-pre- newly upgraded database via myIMA. … surprisingly have constant serving property of Reuleaux polygons Among these webpages, you will find is quite impressive.
    [Show full text]
  • Geometry by Construction
    GEOMETRY BY CONSTRUCTION GEOMETRY BY CONSTRUCTION Object Creation and Problem-solving in Euclidean and Non-Euclidean Geometries MICHAEL MCDANIEL Universal-Publishers Boca Raton Geometry by Construction: Object Creation and Problem-solving in Euclidean and Non-Euclidean Geometries Copyright © 2015 Michael McDaniel All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the publisher Universal-Publishers Boca Raton, Florida USA • 2015 ISBN-10: 1-62734-028-9 ISBN-13: 978-1-62734-028-1 www.universal-publishers.com Cover image: Kamira/Bigstock.com Publisher's Cataloging-in-Publication Data McDaniel, Michael. Geometry by construction : object creation and problem-solving in euclidean and non- euclidean geometries / Michael McDaniel. pages cm Includes bibliographical references and index. ISBN: 978-1-62734-028-1 (pbk.) 1. Euclid's Elements. 2. Geometry, Non-Euclidean. 3. Geometry, Modern. 4. Geometry—Foundations. 5. Geometry—Problems, Exercises, etc. I. Title. QA445 .M34 2015 516—dc23 2015930041 January 7, 2015 10:13 World Scientific Book - 9.75in x 6.5in swp0000 Contents Michael McDaniel Aquinas College Preface ix 1. Euclidean geometry rules and constructions 3 1.1 Euclidean Geometry Vocabulary and Definitions . 4 1.2 Constructions to know ....................... 10 1.3 Tangent construction ........................ 15 1.4 Cyclic quadrilaterals ........................ 16 1.5 Similar triangles ........................... 18 1.6 The theorem of Menelaus ..................... 19 1.7 sAs for similar triangles ...................... 22 1.8 Ceva’sTheorem ........................... 24 1.9 Incircles, excircles and circumcircles . 25 1.10 The 9-point circle .........................
    [Show full text]
  • Exercise Sheet 6
    TECHNISCHE UNIVERSITAT¨ BERLIN Institut fur¨ Mathematik Prof. Dr. John M. Sullivan Geometry II Dott. Matteo Petrera SS 10 http://www.math.tu-berlin.de/∼sullivan/L/10S/Geo2/ Exercise Sheet 6 Exercise 1: Reuleaux triangles. (4 pts) A Reuleaux triangle is the planar body of constant width made from three circular arcs centered at the corners of an equilateral triangle. Glue four Reuleaux triangles along their edges in tetrahedral fashion. (This can be done in 3-space, say with paper models, or we can think about the resulting piecewise-smooth surface intrinsically.) What is the Guass curvature? Exercise 2: CMC surfaces. (4 pts) 3 If p is a vertex on a polyhedral surface M in R , we defined the area vector Ap and the mean curvature vector Hp as the gradients of volume and surface area. We said that M is a discrete CMC surface if there is a constant λ such that Hp = λAp for all p. Consider a triangular bipyramid M with three vertices equally spaced around the unit xy-circle and two more at heights ±h along the z-axis. For what value(s) of h is M a discrete CMC surface? Exercise 3: Convex polytopes. (4 pts) Let P ⊂ R3 be a convex polytope containing the origin O. For a facet F of P , denote by α(F ) the sum of the angles of F and by β(F ) the sum of the angles of the projection of F onto a unit sphere centered at O. Finally, let ω(F ) = β(F ) − α(F ). Prove that X ω(F ) = 4π.
    [Show full text]
  • Individual Round 7 November 2015
    BmMT 2015 Individual Round 7 November 2015 1. What is the units digit of 1 + 9 + 92 + ::: + 92015 ? Solution: The even powers have units digit 1 and the odd powers have units digit 9, so the sum has units digit 0. 2. In Fourtown, every person must have a car and therefore a license plate. Every license plate must be a 4-digit number where each digit is a value between 0 and 9 inclusive. However 0000 is not a valid license plate. What is the minimum population of Fourtown to guarantee that at least two people who have the same license plate? Solution: Since there are 9999 possible license plates, there would need to be 10; 000 people in Fourtown to guarantee a duplicate plate. 3. Two sides of an isosceles triangle 4ABC have lengths 9 and 4. What is the area of 4ABC? Solution: The third side must be 9, since a 4; 4; 9 triangle can't exist. If we drop a perpendicular from the vertex opposite the base to the side of length 4, thenp we have a right triangle with leg 2 p 2 2 andp hypotenusep 9, so the altitude has length 9 − 2 = 77. Thus, the area of the triangle is 4 77=2 = 2 77. 1 3 2x 4. Let x be a real number such that 10 x = x. Find (x ) Solution: (x3)2x = (xx)6 = 106 = 1; 000; 000. 5. A Berkeley student and a Stanford student are going to visit each others campus and go back to their own campuses immediately after they arrive by riding bikes.
    [Show full text]
  • Meissner's Mysterious Bodies
    Meissner's Mysterious Bodies Bernd Kawohl & Christof Weber June 19, 2011 Introduction Certain three-dimensional convex bodies have a counterintuitive property; they are of constant width. In this particular respect they resemble a sphere without being one. Discovered a century ago, Meissner's bodies have often been conjectured to minimize volume among bodies of given constant width. However, this conjecture is still open. We draw attention to this challenging and beautiful open problem by presenting some of its history and its recent development. One Century of Bodies of Constant Width In §32 of their famous book \Geometry and the Imagination", Hilbert and Cohn-Vossen list eleven properties of the sphere and discuss which of these suffices to uniquely determine the shape of the sphere [22]. One of those properties is called constant width: if a sphere is squeezed between two parallel (supporting) planes, it can rotate in any direction and always touches both planes. As the reader may suspect, there are many other con- vex sets with this property of constant width. To indicate this property in common with spheres, such three-dimensional objects are sometimes called spheroforms ([8, p. 135], [36], [7, p. 33]). Some of the three-dimensional convex sets of constant width have a rota- tional symmetry. They can be generated by rotating plane sets of constant width with a reflection symmetry about their symmetry line. The drawing in Figure 1 is taken from a catalogue of mathematical models produced by the publisher Martin Schilling in 1911 [34, p. 149]. Influenced by mathematicians like Felix Klein, such models were produced for educational purposes, many of which were made by plaster.
    [Show full text]
  • Reuleaux Triangle
    What Makes a Circle a Circle? Loyola University Chicago Feb 7, 2017 Steven L. Jordan Adjunct Professor, Loyola University Chicago Professor Emeritus, University of Illinois at Chicago Circles – features Always curved = 3 Never ends Rolling a circle on its≈ “center” and “radius” edge Angles Area = Diameter Area = 2 Tangent 2 Area 4 3 2 ≈ 4 How many squares make Area of square a circle? and area of circle G K H C A B J F I More properties, please! Circle and areas of plane figures Circle is the most fat of all! Isoperimetric inequality If a plane figure has perimeter C and area A, then 4 If this is equal,2 then we have a circle! ≤ Maybe someone mentions AB = 3.00 in. Chords in a circle C Angles? D Lengths? A I B E Rolling out a rope Circular section? If a plane figure has constant diameter, is it a circle? What do we mean by “diameter” or “width?” Support Line For finite set, or compact planar figure. Line defines a half-plane. Figure is on one side of the line. Figure touches the line. Try with our pictures. Now draw a 2nd line, parallel Move it until it becomes a support line on the other side. C E A B H “Width” – distance between support lines m CE = 3.00 in. C E M A B m MN = 4.13 in. H N What happens? Minimum “width”? Maximum width? Width as a function of the angle? (trigonometry if you like it) Graph of width as a function of angle? Other figures? E.g., See Warm-up figures.
    [Show full text]
  • MYSTERIES of the EQUILATERAL TRIANGLE, First Published 2010
    MYSTERIES OF THE EQUILATERAL TRIANGLE Brian J. McCartin Applied Mathematics Kettering University HIKARI LT D HIKARI LTD Hikari Ltd is a publisher of international scientific journals and books. www.m-hikari.com Brian J. McCartin, MYSTERIES OF THE EQUILATERAL TRIANGLE, First published 2010. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission of the publisher Hikari Ltd. ISBN 978-954-91999-5-6 Copyright c 2010 by Brian J. McCartin Typeset using LATEX. Mathematics Subject Classification: 00A08, 00A09, 00A69, 01A05, 01A70, 51M04, 97U40 Keywords: equilateral triangle, history of mathematics, mathematical bi- ography, recreational mathematics, mathematics competitions, applied math- ematics Published by Hikari Ltd Dedicated to our beloved Beta Katzenteufel for completing our equilateral triangle. Euclid and the Equilateral Triangle (Elements: Book I, Proposition 1) Preface v PREFACE Welcome to Mysteries of the Equilateral Triangle (MOTET), my collection of equilateral triangular arcana. While at first sight this might seem an id- iosyncratic choice of subject matter for such a detailed and elaborate study, a moment’s reflection reveals the worthiness of its selection. Human beings, “being as they be”, tend to take for granted some of their greatest discoveries (witness the wheel, fire, language, music,...). In Mathe- matics, the once flourishing topic of Triangle Geometry has turned fallow and fallen out of vogue (although Phil Davis offers us hope that it may be resusci- tated by The Computer [70]). A regrettable casualty of this general decline in prominence has been the Equilateral Triangle. Yet, the facts remain that Mathematics resides at the very core of human civilization, Geometry lies at the structural heart of Mathematics and the Equilateral Triangle provides one of the marble pillars of Geometry.
    [Show full text]
  • A Novel Omnidirectional Wheel Based on Reuleaux-Triangles
    A novel omnidirectional wheel based on Reuleaux-triangles Jochen Brunhorn, Oliver Tenchio, Ra´ulRojas Institut f¨ur Informatik Freie Universit¨atBerlin Takustr. 9, 14195 Berlin, Germany Abstract. This paper discusses the mechanical design and simulation of a novel omnidirectional wheel based on Reuleaux-triangles. The main feature of our omniwheel is that the point of contact of the wheel with the floor is always kept at the same distance from the center of rotation by mechanical means. This produces smooth translational movement on a flat surface, even when the profile of the complete wheel assembly has gaps between the passive rollers. The grip of the wheel with the floor is also improved. The design described in this paper is ideal for hard surfaces, and can be scaled to fit small or large vehicles. This is the first design for an omnidirectional wheel without circular profile, yet capable of rolling smoothly on a hard surface. 1 Motivation and Reuleaux triangles It has been thought for many years, that the only way of providing a smooth rolling effect when using omnidirectional wheels1 with gaps between rollers is: a) by stacking two wheels on the same axis, producing a combined circular profile; b) by using several synchronized wheels which combine in order to support the vehicle keeping the distance to the floor constant; or c) by using spheres or quasi- spheres as wheels. There is a fourth alternative, which is to design omniwheels with a circular profile, in which the gaps are almost closed by using two kinds of rollers alternatively, as in [3].
    [Show full text]
  • Sets of Constant Width
    Pacific Journal of Mathematics SETS OF CONSTANT WIDTH GULBANK D. CHAKERIAN Vol. 19, No. 1 May 1966 PACIFIC JOURNAL OF MATHEMATICS Vol. 19, No. 1. 1966 SETS OF CONSTANT WIDTH G. D. CHAKERIAN A lower bound, better than those previously known, is given for the volume of a 3-dimensional body of constant width 1. Bounds are also given in the case of n-dimensional bodies of constant width 1, n ^ 4. Short proofs of the known sharp bounds for such bodies in the Euclidean and Minkowskian planes are given using properties of mixed areas. An appli- cation is made to a measure of outer symmetry of sets of constant width in 2 and 3 dimensions. Let K be a convex body in ^-dimensional Euclidean space En. For each point u on the unit sphere S centered at the origin, let b(u) be the distance between the two parallel supporting hyperplanes of K orthogo- nal to the direction. The function b(u) is the "width function" of K. If b(u) is constant on S, then we say K is a body of constant width. If jKi and K2 are convex bodies, then Kx + K2 is the "Minkowski sum" or "vector sum" of K, and K2 [5, p. 79]. The following useful theorem is well-known. THEOREM 1. A convex body K has constant width b if and only if K + (-K) is a spherical ball of radius δ. In the case of E2, a number of special properties of sets of constant width are known-for example, the following theorem of Pal (see [5, p.
    [Show full text]
  • Get on Board with Geogebra! Suzanne Harper and Dana Cox Miami University [email protected] [email protected]
    Get on Board with GeoGebra! Suzanne Harper and Dana Cox Miami University [email protected] [email protected] http://padlet.com/harpersr/414 Geometry Perspective: Explore each of the tool menus along the top bar. Each has a drop-down menu that enables additional tools. Structured Play: Drawing Play with the toolbars and menus until you can create the following objects (in any order): a. A point b. A circle c. A line, line segment, and a ray d. A point at the intersection of objects e. Parallel lines f. Perpendicular lines g. The midpoint of a line segment h. An arc on the circle Structured Play: Measurement Play with the toolbars and menus until you can measure/Calculate the following objects (in any order): a. The length of a line segment b. The distance between two points c. The ratio of the circumference of a circle to its diameter. d. The ratio of the lengths of two line segments e. The area of a circle f. The sum of any two angles of a triangle. (What happens when you drag the vertices?) g. The angles made at the intersection of two diagonals of a quadrilateral. Drag-Proof Shapes Create a drag-proof version of each shape. A drag-proof shape is one that remains true to the definition even if the vertices and edges are dragged. a. Triangles: Right, Isosceles, and Equilateral. b. Square c. Parallelogram d. Rectangle e. Kite f. Rhombus g. Two circles with the same radius (as you move one, the other changes to match.) h.
    [Show full text]
  • Asymmetry of Convex Bodies of Constant Width
    Discrete Comput Geom (2012) 47:415–423 DOI 10.1007/s00454-011-9370-8 Asymmetry of Convex Bodies of Constant Width HaiLin Jin · Qi Guo Received: 19 November 2010 / Revised: 4 May 2011 / Accepted: 10 June 2011 / Published online: 13 July 2011 © Springer Science+Business Media, LLC 2011 Abstract The symmetry of convex bodies of constant width is discussed in this pa- ⊂ Rn ≤ ≤ per.√ We proved that for any convex body K of constant width, 1 as∞(K) n+ 2n(n+1) · n+2 , where as∞( ) denotes the Minkowski measure of asymmetry for convex bodies. Moreover, the equality holds on the left-hand side precisely iff K is an Eu- clidean ball and the upper bounds are attainable, in particular, if n = 3, the equality holds on the right-hand side if K is a Meissner body. Keywords Convex body of constant width · Asymmetry measure · Meissner’s body 1 Introduction Measures of (central) symmetry or, as we prefer, asymmetry for convex bodies have been extensively investigated (see [11, 13]). Among these researches, for a given asymmetry measure, it is a popular topic to determine the extremal bodies, i.e., the bodies with maximal asymmetry measure, in some class of convex bodies. For instance, for many known asymmetry measures, circles are most symmetric and Reuleaux triangles are most asymmetric among convex bodies of constant width in R2 (see [4, 5, 7, 8, 10]). Recently, in [12] we showed that this is also valid for the well known Minkowski measure (for the definition see below) (see [12]). Con- cretely, for any 2-dimensional convex body K of constant width, √ 3 + 1 1 ≤ as∞(K) ≤ 2 H.L.
    [Show full text]