Introduction to Geometry" ©2014 Aops Inc
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Combinatorics: the Fine Art of Counting
Combinatorics: The Fine Art of Counting Week Three Solutions Note: in these notes multiplication is assumed to take precedence over division, so 4!/2!2! = 4!/(2!*2!), and binomial coefficients are written horizontally: (4 2) denotes “4 choose 2” or 4!/2!2! Appetizers 1. How many 5 letter words have at least one double letter, i.e. two consecutive letters that are the same? 265 – 26*254 = 1,725,126 2. A Venn diagram is drawn using three circular regions of radius 1 with their centers all distance 1 from each other. What is the area of the intersection of all three regions? What is the area of their union? Let A, B, and C be the three circular regions. A∪B∪C can be seen to be equal to the union of three 60° sectors of radius 1, call them X, Y, and Z. The intersection of any 2 or 3 of these is an equilateral triangle with side length 1. Applying PIE, , we have |XUYUZ| = |X| + |Y| + |Z| - (|X∪Y| + |Y∪Z| + |Z∪X|) + |X∪Y∪Z| which is equal to 3*3*π/6 – 3*√3/4 + √3/4 = π/2 - √3/2. The intersection of any two of A, B, or C has area 2π/3 - √3/2, as shown in class. Applying PIE, |AUBUC| = |A| + |B| + |C| - (|A∪B| + |B∪C| + |C∪A|) + |A∪B∪B| equal to 3π - 3*(2π/3 - √3/2) + (π/2 - √3/2) = 3π/2 + √3 3. A diagonal of polygon is any line segment between vertices which is not an edge of the polygon. -
Almost-Magic Stars a Magic Pentagram (5-Pointed Star), We Now Know, Must Have 5 Lines Summing to an Equal Value
MAGIC SQUARE LEXICON: ILLUSTRATED 192 224 97 1 33 65 256 160 96 64 129 225 193 161 32 128 2 98 223 191 8 104 217 185 190 222 99 3 159 255 66 34 153 249 72 40 35 67 254 158 226 130 63 95 232 136 57 89 94 62 131 227 127 31 162 194 121 25 168 200 195 163 30 126 4 100 221 189 9 105 216 184 10 106 215 183 11 107 214 182 188 220 101 5 157 253 68 36 152 248 73 41 151 247 74 42 150 246 75 43 37 69 252 156 228 132 61 93 233 137 56 88 234 138 55 87 235 139 54 86 92 60 133 229 125 29 164 196 120 24 169 201 119 23 170 202 118 22 171 203 197 165 28 124 6 102 219 187 12 108 213 181 13 109 212 180 14 110 211 179 15 111 210 178 16 112 209 177 186 218 103 7 155 251 70 38 149 245 76 44 148 244 77 45 147 243 78 46 146 242 79 47 145 241 80 48 39 71 250 154 230 134 59 91 236 140 53 85 237 141 52 84 238 142 51 83 239 143 50 82 240 144 49 81 90 58 135 231 123 27 166 198 117 21 172 204 116 20 173 205 115 19 174 206 114 18 175 207 113 17 176 208 199 167 26 122 All rows, columns, and 14 main diagonals sum correctly in proportion to length M AGIC SQUAR E LEX ICON : Illustrated 1 1 4 8 512 4 18 11 20 12 1024 16 1 1 24 21 1 9 10 2 128 256 32 2048 7 3 13 23 19 1 1 4096 64 1 64 4096 16 17 25 5 2 14 28 81 1 1 1 14 6 15 8 22 2 32 52 69 2048 256 128 2 57 26 40 36 77 10 1 1 1 65 7 51 16 1024 22 39 62 512 8 473 18 32 6 47 70 44 58 21 48 71 4 59 45 19 74 67 16 3 33 53 1 27 41 55 8 29 49 79 66 15 10 15 37 63 23 27 78 11 34 9 2 61 24 38 14 23 25 12 35 76 8 26 20 50 64 9 22 12 3 13 43 60 31 75 17 7 21 72 5 46 11 16 5 4 42 56 25 24 17 80 13 30 20 18 1 54 68 6 19 H. -
On the Standard Lengths of Angle Bisectors and the Angle Bisector Theorem
Global Journal of Advanced Research on Classical and Modern Geometries ISSN: 2284-5569, pp.15-27 ON THE STANDARD LENGTHS OF ANGLE BISECTORS AND THE ANGLE BISECTOR THEOREM G.W INDIKA SHAMEERA AMARASINGHE ABSTRACT. In this paper the author unveils several alternative proofs for the standard lengths of Angle Bisectors and Angle Bisector Theorem in any triangle, along with some new useful derivatives of them. 2010 Mathematical Subject Classification: 97G40 Keywords and phrases: Angle Bisector theorem, Parallel lines, Pythagoras Theorem, Similar triangles. 1. INTRODUCTION In this paper the author introduces alternative proofs for the standard length of An- gle Bisectors and the Angle Bisector Theorem in classical Euclidean Plane Geometry, on a concise elementary format while promoting the significance of them by acquainting some prominent generalized side length ratios within any two distinct triangles existed with some certain correlations of their corresponding angles, as new lemmas. Within this paper 8 new alternative proofs are exposed by the author on the angle bisection, 3 new proofs each for the lengths of the Angle Bisectors by various perspectives with also 5 new proofs for the Angle Bisector Theorem. 1.1. The Standard Length of the Angle Bisector Date: 1 February 2012 . 15 G.W Indika Shameera Amarasinghe The length of the angle bisector of a standard triangle such as AD in figure 1.1 is AD2 = AB · AC − BD · DC, or AD2 = bc 1 − (a2/(b + c)2) according to the standard notation of a triangle as it was initially proved by an extension of the angle bisector up to the circumcircle of the triangle. -
THE ZEN of MAGIC SQUARES, CIRCLES, and STARS Also by Clifford A
THE ZEN OF MAGIC SQUARES, CIRCLES, AND STARS Also by Clifford A. Pickover The Alien IQ Test Black Holes: A Traveler’s Guide Chaos and Fractals Chaos in Wonderland Computers and the Imagination Computers, Pattern, Chaos, and Beauty Cryptorunes Dreaming the Future Fractal Horizons: The Future Use of Fractals Frontiers of Scientific Visualization (with Stuart Tewksbury) Future Health: Computers and Medicine in the 21st Century The Girl Who Gave Birth t o Rabbits Keys t o Infinity The Loom of God Mazes for the Mind: Computers and the Unexpected The Paradox of God and the Science of Omniscience The Pattern Book: Fractals, Art, and Nature The Science of Aliens Spider Legs (with Piers Anthony) Spiral Symmetry (with Istvan Hargittai) The Stars of Heaven Strange Brains and Genius Surfing Through Hyperspace Time: A Traveler’s Guide Visions of the Future Visualizing Biological Information Wonders of Numbers THE ZEN OF MAGIC SQUARES, CIRCLES, AND STARS An Exhibition of Surprising Structures across Dimensions Clifford A. Pickover Princeton University Press Princeton and Oxford Copyright © 2002 by Clifford A. Pickover Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, 3 Market Place, Woodstock, Oxfordshire OX20 1SY All Rights Reserved Library of Congress Cataloging-in-Publication Data Pickover, Clifford A. The zen of magic squares, circles, and stars : an exhibition of surprising structures across dimensions / Clifford A. Pickover. p. cm Includes bibliographical references and index. ISBN 0-691-07041-5 (acid-free paper) 1. Magic squares. 2. Mathematical recreations. I. Title. QA165.P53 2002 511'.64—dc21 2001027848 British Library Cataloging-in-Publication Data is available This book has been composed in Baskerville BE and Gill Sans. -
Geometrygeometry
Math League SCASD 2019-20 Meet #5 GeometryGeometry Self-study Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Solid Geometry (Volume and Surface Area) 3. Number Theory: Set Theory and Venn Diagrams 4. Arithmetic: Combinatorics and Probability 5. Algebra: Solving Quadratics with Rational Solutions, including word problems Important Information you need to know about GEOMETRY: Solid Geometry (Volume and Surface Area) Know these formulas! SURFACE SHAPE VOLUME AREA Rect. 2(LW + LH + LWH prism WH) sum of areas Any prism H(Area of Base) of all surfaces Cylinder 2 R2 + 2 RH R2H sum of areas Pyramid 1/3 H(Base area) of all surfaces Cone R2 + RS 1/3 R2H Sphere 4 R2 4/3 R3 Surface Diagonal: any diagonal (NOT an edge) that connects two vertices of a solid while lying on the surface of that solid. Space Diagonal: an imaginary line that connects any two vertices of a solid and passes through the interior of a solid (does not lie on the surface). Category 2 Geometry Calculator Meet Meet #5 - April, 2018 1) A cube has a volume of 512 cubic feet. How many feet are in the length of one edge? 2) A pyramid has a rectangular base with an area of 119 square inches. The pyramid has the same base as the rectangular solid and is half as tall as the rectangular solid. The altitude of the rectangular solid is 10 feet. How many cubic inches are in the volume of the pyramid? 3) Quaykah wants to paint the inside of a cylindrical storage silo that is 63 feet high and whose circular floor has a diameter of 19 feet. -
Advanced Euclidean Geometry
Advanced Euclidean Geometry Paul Yiu Summer 2016 Department of Mathematics Florida Atlantic University July 18, 2016 Summer 2016 Contents 1 Some Basic Theorems 101 1.1 The Pythagorean Theorem . ............................ 101 1.2 Constructions of geometric mean . ........................ 104 1.3 The golden ratio . .......................... 106 1.3.1 The regular pentagon . ............................ 106 1.4 Basic construction principles ............................ 108 1.4.1 Perpendicular bisector locus . ....................... 108 1.4.2 Angle bisector locus . ............................ 109 1.4.3 Tangency of circles . ......................... 110 1.4.4 Construction of tangents of a circle . ............... 110 1.5 The intersecting chords theorem ........................... 112 1.6 Ptolemy’s theorem . ................................. 114 2 The laws of sines and cosines 115 2.1 The law of sines . ................................ 115 2.2 The orthocenter ................................... 116 2.3 The law of cosines .................................. 117 2.4 The centroid ..................................... 120 2.5 The angle bisector theorem . ............................ 121 2.5.1 The lengths of the bisectors . ........................ 121 2.6 The circle of Apollonius . ............................ 123 3 The tritangent circles 125 3.1 The incircle ..................................... 125 3.2 Euler’s formula . ................................ 128 3.3 Steiner’s porism ................................... 129 3.4 The excircles .................................... -
4 3 E B a C F 2
See-Saw Geometry and the Method of Mass-Points1 Bobby Hanson February 27, 2008 Give me a place to stand on, and I can move the Earth. — Archimedes. 1. Motivating Problems Today we are going to explore a kind of geometry similar to regular Euclidean Geometry, involving points and lines and triangles, etc. The main difference that we will see is that we are going to give mass to the points in our geometry. We will call such a geometry, See-Saw Geometry (we’ll understand why, shortly). Before going into the details of See-Saw Geometry, let’s look at some problems that might be solved using this different geometry. Note that these problems are perfectly solvable using regular Euclidean Geometry, but we will find See-Saw Geometry to be very fast and effective. Problem 1. Below is the triangle △ABC. Side BC is divided by D in a ratio of 5 : 2, and AB is divided by E in a ration of 3 : 4. Find the ratios in which F divides the line segments AD and CE; i.e., find AF : F D and CF : F E. (Note: in Figure 1, below, only the ratios are shown; the actual lengths are unknown). B 3 E 5 4 F D 2 A C Figure 1 1My notes are shamelessly stolen from notes by Tom Rike, of the Berkeley Math Circle available at http://mathcircle.berkeley.edu/archivedocs/2007 2008/lectures/0708lecturespdf/MassPointsBMC07.pdf . 1 2 Problem 2. In Figure 2, below, D and E divide sides BC and AB, respectively, as before. -
MAΘ Theta CPAV Solutions 2018
MAΘ Theta CPAV Solutions 2018 Since the perimeter of the square is 40, one side of the square is 10 & its area is 100. Since the 1. B diameter of the circle is 8 its radius is 4. One quarter of the circle is inside the square; its area is 1 1 퐴 = 휋(42) = 휋(16) = 4휋. The area inside the square but outside the circle is 100 − 4휋. 4 4 Since the diameter of the sphere is 20 feet, its radius is 10 feet. 4 4 4000휋 1200휋 푉 = 휋푟3 = 휋103 = ; 퐴 = 4휋푟2 = 4휋102 = 400휋 = ; 3 3 3 3 2. C 1200휋 4000휋 5200휋 퐴 + 푉 = + = 3 3 3 3. B Since the diameter is 18 퐶 = 휋푑 = 18휋 4. A 3 3 288 2 푉 = 푎 ⁄ = 12 ⁄ = 1728⁄ = 288⁄ = √ ⁄ = 144√2 6√2 6√2 6√2 √2 2 20 Use the given circumference to find the radius of the circle. 퐶 = 2휋푟, 40 = 2휋푟, 20 = 휋푟, ⁄휋 = 푟 . If the inscribed angle is 45∘ its arc is 90∘ and has length 10, one quarter of the circumference. The chord of the 5. D segment is the hypotenuse of an isosceles right triangle whose legs are radii of the circle. 20 20√2 ℎ = 푙푒√2 = ⁄휋 √2 = ⁄휋 20√2 Perimeter of the segment is arc length plus chord: 10 + ⁄휋 Since the diameter of the sphere is the space diagonal of the inscribed cube, we must first find the diameter of the sphere. 6. B 4 125휋 3 4 125휋(3√3) 125 27 5 3 푉 = 휋푟3; √ = 휋푟3; = 푟3; √ = 푟3; √ = 푟 and diameter is 5√3 3 2 3 8휋 8 2 Space diagonal of a cube is 푎√3 = 5√3; 푎 = 5 and surface area is 6푎2 = 6 ∙ 52 = 150 Radius found using the circular cross-section area 퐴 = 휋푟2; 9휋 = 휋푟2; 푟 = 3; ℎ = 3푟 == 9 Volume is equal to volume of cylinder minus volume of hemisphere. -
Homework 2: Vectors and Dot Product
Math 21a: Multivariable calculus Fall 2017 Homework 2: Vectors and Dot product This homework is due Monday, 9/11 respectively Tuesday 9/12 at the beginning of class. 1 A kite surfer gets pulled with a force F~ = h7; 1; 4i. She moves with velocity ~v = h4; −2; 1i. The dot product of F~ with ~v is power. a) Find the angle between the F~ and ~v. b) Find the vector projection of the F~ onto ~v. Solution: (a) To find the angle between the force and velocity, we use the formula: F · v cos θ = : jF jjvj p p 2 2 2 The magnitude of the forcep is 7 + 1 + 4 = 66. The mag- nitude of the velocity is 21. The dot product F · v = 30. Thus, F · v 30 cos θ = = p p jF jjvj 66 · 21 Hence, θ = arccos(p30 ). 1386 (b) The projection of F~ onto ~v is given by F · v 30 10 projvF = v = p 2 h4; −2; 1i = h4; −2; 1i: jvj2 21 7 2 Light shines long the vector ~a = ha1; a2; a3i and reflects at the three coordinate planes where the angle of incidence equals the 1 angle of reflection. Verify that the reflected ray is −~a. Hint. Reflect first at the xy-plane. Solution: If we reflect at the xy-plane, then the vector ~a = ha1; a2; a3i gets changed to ha1; a2; −a3i. You can see this by watching the reflection from above. Notice that the first two components stay the same. Do the same process with the other planes. -
Icons of Mathematics an EXPLORATION of TWENTY KEY IMAGES Claudi Alsina and Roger B
AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 45 Icons of Mathematics AN EXPLORATION OF TWENTY KEY IMAGES Claudi Alsina and Roger B. Nelsen i i “MABK018-FM” — 2011/5/16 — 19:53 — page i — #1 i i 10.1090/dol/045 Icons of Mathematics An Exploration of Twenty Key Images i i i i i i “MABK018-FM” — 2011/5/16 — 19:53 — page ii — #2 i i c 2011 by The Mathematical Association of America (Incorporated) Library of Congress Catalog Card Number 2011923441 Print ISBN 978-0-88385-352-8 Electronic ISBN 978-0-88385-986-5 Printed in the United States of America Current Printing (last digit): 10987654321 i i i i i i “MABK018-FM” — 2011/5/16 — 19:53 — page iii — #3 i i The Dolciani Mathematical Expositions NUMBER FORTY-FIVE Icons of Mathematics An Exploration of Twenty Key Images Claudi Alsina Universitat Politecnica` de Catalunya Roger B. Nelsen Lewis & Clark College Published and Distributed by The Mathematical Association of America i i i i i i “MABK018-FM” — 2011/5/16 — 19:53 — page iv — #4 i i DOLCIANI MATHEMATICAL EXPOSITIONS Committee on Books Frank Farris, Chair Dolciani Mathematical Expositions Editorial Board Underwood Dudley, Editor Jeremy S. Case Rosalie A. Dance Tevian Dray Thomas M. Halverson Patricia B. Humphrey Michael J. McAsey Michael J. Mossinghoff Jonathan Rogness Thomas Q. Sibley i i i i i i “MABK018-FM” — 2011/5/16 — 19:53 — page v — #5 i i The DOLCIANI MATHEMATICAL EXPOSITIONS series of the Mathematical As- sociation of America was established through a generous gift to the Association from Mary P. -
Foundations of Euclidean Constructive Geometry
FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY MICHAEL BEESON Abstract. Euclidean geometry, as presented by Euclid, consists of straightedge-and- compass constructions and rigorous reasoning about the results of those constructions. A consideration of the relation of the Euclidean “constructions” to “constructive mathe- matics” leads to the development of a first-order theory ECG of “Euclidean Constructive Geometry”, which can serve as an axiomatization of Euclid rather close in spirit to the Elements of Euclid. Using Gentzen’s cut-elimination theorem, we show that when ECG proves an existential theorem, then the things proved to exist can be constructed by Eu- clidean ruler-and-compass constructions. In the second part of the paper we take up the formal relationships between three versions of Euclid’s parallel postulate: Euclid’s own formulation in his Postulate 5, Playfair’s 1795 version, which is the one usually used in modern axiomatizations, and the version used in ECG. We completely settle the questions about which versions imply which others using only constructive logic: ECG’s version im- plies Euclid 5, which implies Playfair, and none of the reverse implications are provable. The proofs use Kripke models based on carefully constructed rings of real-valued functions. “Points” in these models are real-valued functions. We also characterize these theories in terms of different constructive versions of the axioms for Euclidean fields.1 Contents 1. Introduction 5 1.1. Euclid 5 1.2. The collapsible vs. the rigid compass 7 1.3. Postulates vs. axioms in Euclid 9 1.4. The parallel postulate 9 1.5. Polygons in Euclid 10 2. -
3-D (Intermediate UKMT)
3-D (Intermediate UKMT) These questions must be attempted without a calculator Topics covered in the questions below may not necessarily be from the topic of the title. 1. A solid ‘star’ shape is created by gluing a square-based pyramid, in which each edge is of length 1 unit, precisely onto each face of a cube of edge 1 unit. How many faces does this ‘star’ have? A 18 B 24 C 30 D 36 E 48 2. Four of these shapes can be placed together to make a cube. Which is the odd one out? A B C D E 3. The figure shows a net for a cube with a number on each face. When the cube is made, three faces meet at each vertex. The numbers on the three faces which meet at each vertex are multiplied together. What is the largest product obtained? 1 4 2 5 6 3 A 40 B 60 C 72 D 90 E 120 4. Each edge of a cube is coloured either red or black. If every face of the cube has at least one black edge, what is the smallest possible number of black edges? A 2 B 3 C 4 D 5 E 6 5. The total length of all the edges of a cube is L cm. If the surface area of the cube has the same numerical value Lcm2 what is its volume in cm3? A 1 B L C 2 D L3 E 8 6. Three rectangular-shaped holes have been drilled passing all the way through a solid 3 × 4 × 5 cuboid.