DOCSLIB.ORG

Some Relations Concerning K-Chordal and K-Tangential Polygons

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Some Relations Concerning K-Chordal and K-Tangential Polygons Mathematical Communications 7(2002), 21-34 21 Some relations concerning k-chordal and k-tangential polygons Mirko Radic´∗ Abstract. In papers [6] and [7] the k-chordal and the k-tangential polygons are defined and some of their properties are proved. In this paper we shall consider some of their other properties. Theorems 1-4 are proved. Key words: geometrical inequalities, k-chordal polygon, k-tangential polygon AMS subject classifications: 51E12, 52A40 Received October 8, 2001 Accepted April 4, 2002 1. Preliminaries A polygon with vertices A1 ...An (in this order) will be denoted by A1 ...An and the lengths ofthe sides of A1 ...An will be denoted by a1,... ,an,whereai = | AiAi+1 |, i =1, 2,... ,n. For the interior angle at the vertex Ai we write αi or ∠Ai, i.e. ∠Ai = ∠An−1+iAiAi+1,i=1,... ,n. Ofcourse, indices are calculated modulo n. For convenience we list some definitions given in [6] and [7]. Definition 1. Let A = A1 ...An be a chordal polygon and let C be its circum- circle. By SAi and SAi we denote the semicircles of C such that SAi ∪ SAi = C, Ai ∈ SAi ∩ SAi . The polygon A is said to be of the first kind if the following is fulfilled: 1. all vertices A1 ...An do not lie on the same semicircle, 2. for every three consecutive vertices Ai,Ai+1,Ai+2 it holds Ai ∈ SAi+1 ⇒ Ai+2 ∈ SAi+1 3. any two consecutive vertices Ai,Ai+1 do not lie on the same diameter. Definition 2. Let A = A1 ...An be a chordal polygon and let k be a positive integer. The polygon A is said to be a k-chordal polygon if it is of the first kind and if there holds n ∠AiCAi+1 =2kπ, (1) i=1 ∗Department of Mathematics, University of Rijeka, Omladinska 14, HR-51 000 Rijeka, Croatia 22 M. Radic´ where C is the centre of the circumcircle of the polygon A. Using (1) it is easy to see that the angles ofa k-chordal polygon A1 ...An satisfy the relation: n ∠Ai =(n − 2k)π. (2) i=1 Definition 3. Let A = A1 ...An be a tangential polygon and let k be a positive n−1 n−1 n−2 integer so that k ≤ 2 , that is, k ≤ 2 if n is odd, and k ≤ 2 if n is even. The polygon A will be called a k-tangential polygon if any two of its consecutive sides have only one common point, and if there holds π β1 + ···+ βn =(n − 2k) , (3) 2 where 2βi = ∠Ai,i=1, ..., n. Consequently, a tangential polygon A is k-tangential if ϕ1 + ···+ ϕn =2kπ, (4) where ϕi = ∠AiCAi+1 and C is the centre ofthe circle inscribed into the polygon A. n−1 n−2 The integer k in relations (1)-(4) can be at most 2 if n is odd and 2 if n is even. Remark 1. In the following considerations we shall denote the angles β1,... ,βn such that βi = ∠CAiAi+1, if it is a question of a chordal polygon, 1 βi = ∠Ai, if it is a question of a tangential polygon. 2 2. Some inequalities concerning the radius of k-chordal and k-tangential polygons At first we prove some results concerning a k-chordal polygon. Theorem 1. Let a1,... ,an be the lengths of the sides of a k-chordal polygon A = A1 ...An and let a1 = min{a1 ...an}. If there exist angles γ1,... ,γn such that π π γ1 + ···+ γn =(n − 2k) , 0 <γi < , 1=1,... ,n, (5) 2 2 a1 sin γ1 = a2 sin γ2 = ...= an sin γn. (6) k-chordal and k-tangential polygons 23 Then n 2 ai i=1 2r>n n , (7) 1 1 2 2 π ai − a1 sin (n − 2k) i=1 2 i=1 ai 2n where r is the radius of the circumcircle of the polygon A. Proof. Since βi = ∠CAiAi+1,i=1,... ,n, we have the following relations π π β1 + ···+ βn =(n − 2k) , 0 <βi < ,i=1,... ,n (8) 2 2 2r cos βi = ai,i=1,... ,n (9) from which it follows 2 2rai cos βi = ai ,i=1,... ,n n 2 ai i=1 2r = n . (10) ai cos βi i=1 In addition to the angles β1,... ,βn there are infinitely many angles γ1,... ,γn such that π π γ1 + ···+ γn =(n − 2k) , 0 <γi < ,i=1,... ,n. 2 2 n We shall prove that ai cos γi = maximum ifthe angles γ1,... ,γn satisfy i=1 a1 sin γ1 = a2 sin γ2 = ...= an sin γn. First we shall prove the following lemma. Lemma 1. If a1 and a2 are positive numbers and π γ1 + γ2 = s, 0 <s<π, 0 <γi < ,i=1, 2 2 then the function f(γ1,γ2)=a1 cos γ1 + a2 cos γ2 assumes maximum if a1 sin γ1 = a2 sin γ2. Proof. Let g(γ1)=a1 cos γ1 + a2 cos(s − γ1), then g (γ1)=−a1 sin γ1 + a2 sin(s − γ1), g (γ1)=−a1 cos γ1 − a2 cos(s − γ1) < 0, 24 M. Radic´ a1 sin γ1 + a2 sin(s − γ1)=0 ⇒ a1 sin γ1 = a2 sin(s − γ2). ✷ n From the above lemma it is clear that the sum ai cos γi assumes maximum i=1 iffor each sum ai cos γi + aj cos γj,i,j∈{1,... ,n} there holds ai sin γi = aj sin γj, since we can put γi + γj = s. Now, we are going to prove that the inequality (7) is valid if(6) is fulfiled. Based on the assumption that equations (6) exist, we can write ai sin γi = λ, i =1,... ,n from which it follows 2 2 λ 1 λ cos γi = 1 − < 1 − ,i=1,... ,n, ai 2 ai n 2 n n 1 λ ai 1 − > ai cos γi ≥ ai cos βi i=1 2 ai i=1 i=1 so that instead of(10) we can write n 2 ai i=1 2r>n n . (11) 1 1 2 ai − λ i=1 2 i=1 ai λ i Since γ =arcsinai ,i=1,... ,n we have the equation n λ π arcsin =(n − 2k) , (12) i=1 ai 2 or 3 3 λ λ 1 λ λ π + ···+ + + ···+ + ···=(n − 2k) . (13) a1 an 6 a1 an 2 Since by assumption a1 =min{a1,... ,an}, from (13) it follows that 3 3 λ λ 1 λ λ π + ···+ + + ···+ + ···≥(n − 2k) a1 a1 6 a1 a1 2 k-chordal and k-tangential polygons 25 or λ π arcsin ≥ (n − 2k) . a1 2n Hence π λ ≥ a1 sin(n − 2k) . (14) 2n Now using (11) and (14) we readily get (7). So, Theorem 1 is proved. ✷ Before stating some of its corollaries here is an example. If A1 ...A5 is a l-chordal pentagon as shown in Figure 1, then there are angles γ1,... ,γ5 such that π γ1 + ···+ γ5 =(5− 2) ,a1 sin γ1 = ...= a5 sin γ5 2 ifinstead ofthe drawn circles these can be drawn greater such that the above 3π equalities are valid. (For these drawn ones it is γ1 + ···+ γ5 < 2 . Let us remark that in the case when a side is small enough, then there are no angles γ1,... ,γ5 3π such that γ1 + ...+ γ5 = 2 .) A3 a2 A2 γ2 γ3 a1 γ1 a3 C A1 A4 γ4 γ5 a5 A5 Figure 1. Now we state some ofthe corollaries of Theorem 1. Corollary 1. There are angles γ1,... ,γn such that (5) and (6) hold if and only if 3 a1 1 a1 ···≥ − π + 3 3 + (n 2k) (15) H(a1,... ,an) 6 H(a1,... ,an) 2n 26 M. Radic´ i i i i where H(a1,... ,an) is the harmonic mean of a1,... ,an. Proof. It is clear from (13) since λ may be at most a1. ✷ Corollary 2. A sufficient condition for the existence of the angles γ1,... ,γn such that (5) and (6) hold is the inequality π a1 ≥ H(a1,... ,an)sin(n − 2k) (16) 2n Proof. If(16) holds, then obviously (15) holds, too. Namely, if 3 a1 1 a1 π + + ···≥(n − 2k) , H(a1,... ,an) 6 H(a1,... ,an) 2n then certainly (15) is valid because ofthe property ofthe arithmetics mean. ✷ Corollary 3. If there exists a k-chordal polygon whose sides have the lengths 1 1 2k π ≥ − 1 n a1 ,... , an and n sin(n 2k) 2n , then there exist angles γ ,... ,γ such that (5) and (6) hold. Proof. We shall use Corollary 2 in [6]. If a1,... ,an are the lengths ofthe sides ofthe k-chordal polygon A,then n ai > 2kaj,j=1,... ,n. (17) i=1 1 1 If ,... , are also the lengths ofthe sides ofa k-chordal polygon, then a1 an 1 1 2k + ···+ > a1 an a1 or 2k a1 > H(a1,... ,an). (18) n 2k π Accordingly, if ≥ sin(n − 2k) then (16) is valid. ✷ n 2n n − 1 Corollary 4. If n is odd and k is maximal, i.e. k = , then there exist the 2 angles γ1,... ,γn such that (5) and (6) hold. n − 1 Proof. If k = , then equation (5) can be written as 2 π γ1 + ...+ γn = , 2 n λ π and obviously there is λ such that arcsin = . ✷ i=1 ai 2 Corollary 5. If n =3and a, b, c are the lengths of the sides of an acute triangle, then a2 + b2 + c2 2r> (19) 3 a2 a + b + c − 8 H(a, b, c) k-chordal and k-tangential polygons 27 where a =min{a, b, c}.
Load more

Recommended publications
  • Downloaded from Bookstore.Ams.Org 30-60-90 Triangle, 190, 233 36-72
    Index 30-60-90 triangle, 190, 233 intersects interior of a side, 144 36-72-72 triangle, 226 to the base of an isosceles triangle, 145 360 theorem, 96, 97 to the hypotenuse, 144 45-45-90 triangle, 190, 233 to the longest side, 144 60-60-60 triangle, 189 Amtrak model, 29 and (logical conjunction), 385 AA congruence theorem for asymptotic angle, 83 triangles, 353 acute, 88 AA similarity theorem, 216 included between two sides, 104 AAA congruence theorem in hyperbolic inscribed in a semicircle, 257 geometry, 338 inscribed in an arc, 257 AAA construction theorem, 191 obtuse, 88 AAASA congruence, 197, 354 of a polygon, 156 AAS congruence theorem, 119 of a triangle, 103 AASAS congruence, 179 of an asymptotic triangle, 351 ABCD property of rigid motions, 441 on a side of a line, 149 absolute value, 434 opposite a side, 104 acute angle, 88 proper, 84 acute triangle, 105 right, 88 adapted coordinate function, 72 straight, 84 adjacency lemma, 98 zero, 84 adjacent angles, 90, 91 angle addition theorem, 90 adjacent edges of a polygon, 156 angle bisector, 100, 147 adjacent interior angle, 113 angle bisector concurrence theorem, 268 admissible decomposition, 201 angle bisector proportion theorem, 219 algebraic number, 317 angle bisector theorem, 147 all-or-nothing theorem, 333 converse, 149 alternate interior angles, 150 angle construction theorem, 88 alternate interior angles postulate, 323 angle criterion for convexity, 160 alternate interior angles theorem, 150 angle measure, 54, 85 converse, 185, 323 between two lines, 357 altitude concurrence theorem,
    [Show full text]
  • An Innovative Analysis to Develop New Theorems on Irregular Polygon
    International Journal of Physics and Mathematical Sciences ISSN: 2277-2111 (Online) An Online International Journal Available at http://www.cibtech.org/jpms.htm 2013 Vol. 3 (1) January-March, pp.73-81/Kalaimaran Research Article AN INNOVATIVE ANALYSIS TO DEVELOP NEW THEOREMS ON IRREGULAR POLYGON *Kalaimaran Ara Construction & Civil Maintenance Unit, Central Food Technological Research Institute, Mysore-20, Karnataka, India *Author for Correspondence ABSTRACT The irregular Polygon is a four sided polygon of two dimensional geometrical figures. The triangle, square, rectangle, tetragon, pentagon, hexagon, heptagon, octagon, nonagon, dodecagon, parallelogram, rhombus, rhomboid, trapezium or trapezoidal, kite and dart are the members of the irregular polygon family. A polygon is a two dimensional example of the more general prototype in any number of dimensions. However the properties are varied from one to another. The author has attempted to develop two new theorems for the property of irregular polygon for a point anywhere inside of the polygon with necessary illustrations, appropriate examples and derivation of equations for better understanding. Key Words: Irregular Polygon, Triangle, Right-angled triangle, Perpendicular and Vertex INTRODUCTION Polygon (Weisstein, 2003) is a closed two dimensional figure formed by connecting three or more straight line segments, where each line segment end connects to only one end of two other line segments. Polygon is one of the most all-encompassing shapes in two- dimensional geometry. The sum of the interior angles is equal to 180 degree multiplied by number of sides minus two. The sum of the exterior angles is equal to 360 degree. From the simple triangle up through square, rectangle, tetragon, pentagon, hexagon, heptagon, octagon, nonagon, dodecagon (Weisstein, 2003,) and beyond is called n-gon.
    [Show full text]
  • Some Relations and Properties Concerning Tangential Polygons
    View metadata, citation and similar papers at core.ac.uk brought to you by CORE Mathematical Communications 4(1999), 197-206 197 Some relations and properties concerning tangential polygons Mirko Radic´∗ Abstract. The k-tangential polygon is defined, and some of its properties are proved. Key words: k-tangential polygon AMS subject classifications: 51E12 Received October 10, 1998 Accepted June 7, 1999 1. Preliminaries A polygon with the vertices A1, ..., An (in this order) will be denoted by A1...An. The lengths of the sides of the polygon A1...An will be denoted by |A1A2|, ..., |AnA1| or a1, ..., an. The interior angle at the vertex Ai will be denoted by αi or ∠Ai, i.e. ∠Ai = ∠An−1+iAiAi+1,i=1, ..., n (0 <αi <π). (1) Of course, indices are calculated modulo n. A polygon A = A1...An is a tangential polygon if there exists a circle C such that each side of A is on a tangent line of C. Definition 1. Let A = A1...An be a tangential polygon, and let k be a positive ≤bn−1 c ≤ n−1 ≤ n−2 integer such that k 2 , that is, k 2 if n is odd and k 2 if n is even. Then the polygon A will be called a k-tangential polygon if any two of its consecutive sides have only one point in common, and if there holds π β + ···+ β =(n − 2k) , (2) 1 n 2 where 2βi = ∠Ai, i =1, ..., n. Consequently, a tangential polygon A is k-tangential if n X ϕi =2kπ, i=1 where ϕi = ∠AiCAi+1 and C is the centre of the circle inscribed into the polygon A.
    [Show full text]
  • Arxiv:1712.00299V1 [Math.GT]
    POLYGONS WITH PRESCRIBED EDGE SLOPES: CONFIGURATION SPACE AND EXTREMAL POINTS OF PERIMETER JOSEPH GORDON, GAIANE PANINA, YANA TEPLITSKAYA Abstract. We describe the configuration space S of polygons with pre- scribed edge slopes, and study the perimeter as a Morse function on S. We characterize critical points of (these areP tangential polygons) and compute their Morse indices. ThisP setup is motivated by a number of re- sults about critical points and Morse indices of the oriented area function defined on the configuration space of polygons with prescribed edge lengths (flexible polygons). As a by-product, we present an independent computa- tion of the Morse index of the area function (obtained earlier by G. Panina and A. Zhukova). 1. Introduction Consider the space L of planar polygons with prescribed edge lengths1 and the oriented area as a Morse function defined on it. It is known that gener- ically: A L is a smooth closed manifold whose diffeomorphic type depends on • the edge lengths [1, 2]. The oriented area is a Morse function whose critical points are cyclic • configurations (thatA is, polygons with all the vertices lying on a cir- cle), whose Morse indices are known, see Theorem 2, [3, 5, 6]. The Morse index depends not only on the combinatorics of a cyclic poly- arXiv:1712.00299v1 [math.GT] 1 Dec 2017 gon, but also on some metric data. Direct computations of the Morse index proved to be quite involved, so the existing proof comes from bifurcation analysis combined with a number of combinatorial tricks. Bifurcations of are captured by cyclic polygons P whose dual poly- • gons P ∗ have zeroA perimeter [5]; see also Lemma 4.
    [Show full text]
  • (Theorem 1) Is Proved
    Rad HAZU Volume 503 (2009), 41–54 ABOUT ONE RELATION CONCERNING TWO CIRCLES MIRKO RADIC´ AND ZORAN KALIMAN Abstract. This article can be considered as an appendix to the article [1]. Here the article [1] is extended to the cases when one circle is outside of the other and when circles are intersecting. 1. Preliminaries In [1] the following theorem (Theorem 1) is proved: Let C1 and C2 be any given two circles such that C1 is inside of the C2 and let A1 , A2 , A3 be any given three different points on C2 such that there are points T1 and T2 on C1 with properties |A1A2| = t1 +t2, |A2A3| = t2 +t3, (1) where t1 = |A1T1|, t2 = |T1A2|, t3 = |T2A3|.Then 2rR |A A | =(t +t ) , (2) 1 3 1 3 R2 − d2 where r = radius of C1 , R = radius of C2 , d = |IO|, I is the center of C1 , O is center of C2 .(SeeFigure1.) In short about the proof of this theorem. First the following lemma is proved. It t1 is given then t2 can be calculated using the expression √ t (R2 − d2) ± D (t ) = 1 1 (3a) 2 1,2 2 + 2 r t1 where = 2( 2 − 2)2 +( 2 + 2) 2 2 − 2 2 − ( 2 + 2 − 2)2 . D1 t1 R d r t1 4R d r t1 R d r (3b) The values (t2)1,2 given by (3) are solutions of the equation ( 2 + 2) 2 − ( 2 − 2)+ 2 2 − 2 2 +( 2 + 2 − 2)2 = . r t1 t2 2t1t2 R d r t1 4R d R d r 0 (4) Mathematics subject classification (2000): 51M04.
    [Show full text]
  • Topics in Elementary Geometry
    Topics in Elementary Geometry Second Edition O. Bottema (deceased) Topics in Elementary Geometry Second Edition With a Foreword by Robin Hartshorne Translated from the Dutch by Reinie Erne´ 123 O. Bottema Translator: (deceased) Reinie Ern´e Leiden, The Netherlands [email protected] ISBN: 978-0-387-78130-3 e-ISBN: 978-0-387-78131-0 DOI: 10.1007/978-0-387-78131-0 Library of Congress Control Number: 2008931335 Mathematics Subject Classification (2000): 51-xx This current edition is a translation of the Second Dutch Edition of, Hoofdstukken uit de Elementaire Meetkunde, published by Epsilon-Uitgaven, 1987. c 2008 Springer Science+Business Media, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connec- tion with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper 987654321 springer.com At school I was good in mathematics; now I discovered that I found the so-called higher mathematics – differential and integral calculus – easier than the complicated (but elementary) plane geometry.
    [Show full text]
  • 3Doodler Geometry: Pyramids
    3Doodler EDU 3Doodler Geometry: Pyramids Suitable Ages: Suitable for ages 14+ Skill Level: Basic Materials Required: • 3Doodler 2.0: At a minimum, one 3Doodler per group (six groups per class); if more 3Doodlers are available, one for every one to two students is advisable • 3Doodler PLA or ABS strands: 3 strands per 3Doodler • Pyramid Stencils (N.B.: If PLA is used, it is advisable to apply a layer of masking tape over the stencil to ensure that the shapes can be peeled off the stencil with ease) • The Pyramid Worksheet (pages 16-19) Duration: This classroom activity will take between two and three hours, which can be divided between a number of classroom sessions Instructional Video: Instructional videos showing how to build the Pyramids 1 & 4 of this activity are available at: YouTube: https://youtu.be/f0wQQKNY51w and https://youtu.be/coKS6OJh8WU Vimeo: https://vimeo.com/140039345 and https://vimeo.com/140039344 Dropbox: https://www.dropbox.com/s/hvpr8mcv6f135yi/Example%20Pyramid%201.mpeg?dl=0 and https://www.dropbox.com/s/xf004dw7o4lzcwq/Example%20Pyramid%204.mpeg?dl=0 Written by: Linda Giampieretti Part of the “Fascinating World of Geometric Forms” Project Page 1 of 23 Contents Objective 3 Activity 4 Sequence & Pacing 6 Striving Students 6 Accelerated Students 6 Evaluation Strategies 6 Evaluation Rubric 7 Reference Materials 8 Stencil: Pyramid 1 10 Stencil: Pyramid 2 11 Stencil: Pyramid 3 12 Stencil: Pyramid 4 13 Stencil: Pyramid 5 14 Stencil: Pyramid 6 15 Pyramid Worksheet 16 Worksheet Answer Key 19 Additional Resources 22 Page 2 of 23 Objective Triangles, and in turn, pyramids, form the basis for much of the structural strength in modern architecture and engineering.
    [Show full text]
  • Some Relations and Properties Concerning Tangential Polygons
    Mathematical Communications 4(1999), 197-206 197 Some relations and properties concerning tangential polygons Mirko Radic´∗ Abstract. The k-tangential polygon is defined, and some of its properties are proved. Key words: k-tangential polygon AMS subject classifications: 51E12 Received October 10, 1998 Accepted June 7, 1999 1. Preliminaries A polygon with the vertices A1, ..., An (in this order) will be denoted by A1...An. The lengths of the sides of the polygon A1...An will be denoted by |A1A2|, ..., |AnA1| or a1, ..., an. The interior angle at the vertex Ai will be denoted by αi or ∠Ai, i.e. ∠Ai = ∠An−1+iAiAi+1,i=1, ..., n (0 <αi <π). (1) Of course, indices are calculated modulo n. A polygon A = A1...An is a tangential polygon if there exists a circle C such that each side of A is on a tangent line of C. Definition 1. Let A = A1...An be a tangential polygon, and let k be a positive ≤bn−1 c ≤ n−1 ≤ n−2 integer such that k 2 , that is, k 2 if n is odd and k 2 if n is even. Then the polygon A will be called a k-tangential polygon if any two of its consecutive sides have only one point in common, and if there holds π β + ···+ β =(n − 2k) , (2) 1 n 2 where 2βi = ∠Ai, i =1, ..., n. Consequently, a tangential polygon A is k-tangential if n X ϕi =2kπ, i=1 where ϕi = ∠AiCAi+1 and C is the centre of the circle inscribed into the polygon A.
    [Show full text]
  • American Mathematical Monthly Geometry Problems 1894 –
    YIU : Problems in Elementary Geometry 1 American Mathematical Monthly Geometry Problems 1894 – Vis171. (Marcus Baker) In a traingle ABC, the center of the circumscribed circle is O, the center of the inscribed circle is I, and the orthocenter is H. Knowing the sides of the triangle OIH, determine the sides of triangle ABC. Solution by W.P. Casey: Let N be the nine-point center. From IN = R − 2 − 2 r and OI = R(R 2r), R and r can be determined. The circumcircle and the incircle, and the nine-point circle, can all be constructed. Then, Casey wrote, “[i]t only remains to find a point C in the circumcircle of ABC so that the tangent CA CA to circle I may be bisected by the circle N in the point S, which is easily done”. [sic] The construction of ABC from OIH cannot be effected by ruler and compass in general. Casey continued to derive the cubic equation with roots cos α,cosβ,cosγ, and coefficients in terms of R, r,and := OH,namely, r 4(R + r)2 − (2 +3R2) R2 − 2 x3 − (1 + )x2 + x − =0. R 8R2 8R2 Given triangle OIH,letN be the midpoint of OH. Construct the circle through N tangent to OI at O.ExtendIN to intersect this circle again at M. The diameter of the circumcircle is equal to the length of IM.From this, the circumcircle, the nine-point circle, and the incircle can be con- structed. Now, it remains to select a point X on the nine point circle so that the perpendicular to OX is tangent to the incircle.
    [Show full text]
  • Robbie Robinson, Professor of Mathematics, George Washington University
    Robbie Robinson, Professor of Mathematics, George Washington University Georgetown Math Club March 1, 2016 Leonardo da Vinci, illustrations for Luca Pacioli’s 1509 book The Divine Proportion. From: http://www.georgehart.com/virtual-polyhedra/leonardo.html Melencolia I Albrecht Dürer, 1514 Anatoly Fomenko 1945- Nicholas Neufchatel, Johannes Neudorfer and son 1561 M. C. Escher Stars Salvador Dali, The Last Supper, 1955 Tony Robbin, 2000-10 http://tonyrobbin.home.att.net/work.htm The 2-dimensional case… Ehrhard Bohne, 2000 http://www.math-inf.uni-greifswald.de/mathematik+kunst/mathematiker_bohne.html\ Pas de Deux, Nat Friedman, Fractal Stone Print http://www.albany.edu/~artmath/cetl/am97/ Theorem 1. Let A1, A2, …, An be the interior angles of a polygon. Let o Sn=A1+A2+…+An. Then Sn=180 (n-2). A2 A3 A4 A1 Example. For n=4 we have o o o o. A1+A2+A3+A4 = 180 (n-2)=180 (4-2)=180 (2)=360 Lemma. Let B1, B2, …, Bn be the exterior o angles of a polygon. Then B1+B2+…+Bn= 360 B2 B3 B1 B4 Note: this does not depend on the number of sides! n Sn Rn Sn = 180(n-2) 3 180 60 � � − 2 � = " = 180 4 360 90 " � � 5 540 108 Rn is the angle of a regular n-gon 6 720 120 �" → 180 as � → ∞ 7 900 128.5… 8 1080 135 9 1260 140 10 1440 144 Schlflaly symbol {n,k}={n/k}={9,3} � − 2� � = 180 " � Vertex e.g. Edge Dodecahedron Face Vertices Dihedral angle (plural) Vertex: interior angle/angle deficit wire frame or stick figure solid (perspective) Schlegel web diagram diagram Convex polyhedra projected to sphere=tiling of sphere.
    [Show full text]
  • Inscribed Angles and Quadrilaterals Worksheet Answers Nano
    Inscribed Angles And Quadrilaterals Worksheet Answers Sapient and heterodyne John-Patrick smoke-dry her emigrants tabulating or suds aliunde. Gilded and oftenmetaphysical recalls some Bennie Brunei covings deservedly her mangrove or oversteer hewings readably. or photoengraved whence. Fibroblastic and octantal Abbot Right angles for an inscribed angles quadrilaterals worksheet will allow the fourth set of Correct measure on and inscribed quadrilaterals answers for firefox because its sides are an inscribed angles are coming to spare to operate this problem. As a bus, inscribed angles quadrilaterals worksheet answers using the printing of a question is a circle and the document. Incorrect answer keys, inscribed angles answers for a unit on the google drive traffic and examples of notes are the instructions. Software please see the inscribed angles and worksheet answers for remote learning and its event handler order is your identity as possible after exploring various square. Online tutoring and angles worksheet answers using the congruent parts, you to use with it of a unique website? Quadrilateral theorem in finding angles and worksheet you the issue. Numbers and angles quadrilaterals answers by composing into triangles are supplementary is: the geometric shape and inside inscribed triangles and a set your traffic? Third set of worksheets encloses charts, inscribed quadrilaterals inscribed in our teacher answer is. Fully annotated typed answer, angles quadrilaterals and link to find the parallelograms worksheets. Easily digestible for the angles and quadrilaterals worksheet answers are of this is a header and link. Riddle practice and inscribed angles quadrilaterals answers for your workforce and triangle. Binder geometry notes is inscribed angles quadrilaterals worksheet answers for free information about central angles with your choices at properties; employing the parallelograms with circles.
    [Show full text]
  • Semidefinite Programming for Optimizing Convex Bodies Under
    Semidefinite programming for optimizing convex bodies under width constraints T´erence Bayen1, Didier Henrion2,3 October 22, 2010 Abstract We consider the problem of minimizing a functional (like the area, perimeter, surface) within the class of convex bodies whose support functions are trigonometric polynomials. The convexity constraint is transformed via the Fej´er-Riesz theorem on positive trigono- metric polynomials into a semidefinite programming problem. Several problems such as the minimization of the area in the class of constant width planar bodies, rotors and space bodies of revolution are revisited. The approach seems promising to investigate more difficult optimization problems in the class of three-dimensional convex bodies. Keywords: convexity, optimization, semidefinite programming, support function. AMS Subject Classification: 49M27,90C22,52C05 1 Introduction Many optimization problems involve convexity constraints. Such constraints naturally appear in the framework of geometric optimization problems, where the underlying admissible class is the set of convex bodies in Euclidean space. This paper aims to tackle the issue of a certain class of planar geometric optimization problems via semidefinite programming. A convex body can be parametrized by its support function defined by homogeneity on the unit hypersphere. The convexity constraint amounts to the non-negativity of the principal radii of curvature of the boundary of the body (if they are defined). For the planar case, the convexity constraint is equivalent to the non-negativity of the radius of curvature of the boundary of the body, easily expressed via the support function. A reformulation of the con- vexity constraint allows the use of the calculus of variation and optimal control theory to treat 1Institut de Math´ematiques et de Mod´elisation de Montpellier, Universit´e Montpellier 2, Case Courrier 051, Place Eug`eneBataillon, 34095 Montpellier Cedex, France.
    [Show full text]