Real Life Example of a Regular Polygon Quarts
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Islamic Geometric Ornaments in Istanbul
►SKETCH 2 CONSTRUCTIONS OF REGULAR POLYGONS Regular polygons are the base elements for constructing the majority of Islamic geometric ornaments. Of course, in Islamic art there are geometric ornaments that may have different genesis, but those that can be created from regular polygons are the most frequently seen in Istanbul. We can also notice that many of the Islamic geometric ornaments can be recreated using rectangular grids like the ornament in our first example. Sometimes methods using rectangular grids are much simpler than those based or regular polygons. Therefore, we should not omit these methods. However, because methods for constructing geometric ornaments based on regular polygons are the most popular, we will spend relatively more time explor- ing them. Before, we start developing some concrete constructions it would be worthwhile to look into a few issues of a general nature. As we have no- ticed while developing construction of the ornament from the floor in the Sultan Ahmed Mosque, these constructions are not always simple, and in order to create them we need some knowledge of elementary geometry. On the other hand, computer programs for geometry or for computer graphics can give us a number of simpler ways to develop geometric fig- ures. Some of them may not require any knowledge of geometry. For ex- ample, we can create a regular polygon with any number of sides by rotat- ing a point around another point by using rotations 360/n degrees. This is a very simple task if we use a computer program and the only knowledge of geometry we need here is that the full angle is 360 degrees. -
Vertex-Transplants on a Convex Polyhedron
CCCG 2020, Saskatoon, Canada, August 5{7, 2020 Vertex-Transplants on a Convex Polyhedron Joseph O'Rourke Abstract Regular Tetrahedron. Let the four vertices of a regu- lar tetrahedron of unit edge length be v1; v2; v3 forming Given any convex polyhedron P of sufficiently many ver- the base, and apex v0. Place a point x on the v3v0 edge, tices n, and with no vertex's curvature greater than π, close to v3. Then one can form a digon starting from x it is possible to cut out a vertex, and paste the excised and surrounding v0 with geodesics γ1 and γ2 to a point y portion elsewhere along a vertex-to-vertex geodesic, cre- on 4v1v2v0, with jγ1j = jγ2j = 1. See Fig. 1(a,b). This ating a new convex polyhedron P0 of n + 2 vertices. digon can then be cut out and its hole sutured closed. The removed digon surface can be folded to a doubly covered triangle, and pasted into edge v1v2. The re- 1 Introduction sulting convex polyhedron guaranteed by Alexandrov's Theorem is a 6-vertex irregular octahedron P0. The goal of this paper is to prove the following theorem: Theorem 1 For any convex polyhedron P of n > N vertices, none of which have curvature greater than π, there is a vertex v0 that can be cut out along a digon of geodesics, and the excised surface glued to a geodesic on P connecting two vertices v1; v2. The result is a new convex polyhedron P0 with n + 2 vertices. N = 16 suffices. -
Framing Cyclic Revolutionary Emergence of Opposing Symbols of Identity Eppur Si Muove: Biomimetic Embedding of N-Tuple Helices in Spherical Polyhedra - /
Alternative view of segmented documents via Kairos 23 October 2017 | Draft Framing Cyclic Revolutionary Emergence of Opposing Symbols of Identity Eppur si muove: Biomimetic embedding of N-tuple helices in spherical polyhedra - / - Introduction Symbolic stars vs Strategic pillars; Polyhedra vs Helices; Logic vs Comprehension? Dynamic bonding patterns in n-tuple helices engendering n-fold rotating symbols Embedding the triple helix in a spherical octahedron Embedding the quadruple helix in a spherical cube Embedding the quintuple helix in a spherical dodecahedron and a Pentagramma Mirificum Embedding six-fold, eight-fold and ten-fold helices in appropriately encircled polyhedra Embedding twelve-fold, eleven-fold, nine-fold and seven-fold helices in appropriately encircled polyhedra Neglected recognition of logical patterns -- especially of opposition Dynamic relationship between polyhedra engendered by circles -- variously implying forms of unity Symbol rotation as dynamic essential to engaging with value-inversion References Introduction The contrast to the geocentric model of the solar system was framed by the Italian mathematician, physicist and philosopher Galileo Galilei (1564-1642). His much-cited phrase, " And yet it moves" (E pur si muove or Eppur si muove) was allegedly pronounced in 1633 when he was forced to recant his claims that the Earth moves around the immovable Sun rather than the converse -- known as the Galileo affair. Such a shift in perspective might usefully inspire the recognition that the stasis attributed so widely to logos and other much-valued cultural and heraldic symbols obscures the manner in which they imply a fundamental cognitive dynamic. Cultural symbols fundamental to the identity of a group might then be understood as variously moving and transforming in ways which currently elude comprehension. -
Polygon Review and Puzzlers in the Above, Those Are Names to the Polygons: Fill in the Blank Parts. Names: Number of Sides
Polygon review and puzzlers ÆReview to the classification of polygons: Is it a Polygon? Polygons are 2-dimensional shapes. They are made of straight lines, and the shape is "closed" (all the lines connect up). Polygon Not a Polygon Not a Polygon (straight sides) (has a curve) (open, not closed) Regular polygons have equal length sides and equal interior angles. Polygons are named according to their number of sides. Name of Degree of Degree of triangle total angles regular angles Triangle 180 60 In the above, those are names to the polygons: Quadrilateral 360 90 fill in the blank parts. Pentagon Hexagon Heptagon 900 129 Names: number of sides: Octagon Nonagon hendecagon, 11 dodecagon, _____________ Decagon 1440 144 tetradecagon, 13 hexadecagon, 15 Do you see a pattern in the calculation of the heptadecagon, _____________ total degree of angles of the polygon? octadecagon, _____________ --- (n -2) x 180° enneadecagon, _____________ icosagon 20 pentadecagon, _____________ These summation of angles rules, also apply to the irregular polygons, try it out yourself !!! A point where two or more straight lines meet. Corner. Example: a corner of a polygon (2D) or of a polyhedron (3D) as shown. The plural of vertex is "vertices” Test them out yourself, by drawing diagonals on the polygons. Here are some fun polygon riddles; could you come up with the answer? Geometry polygon riddles I: My first is in shape and also in space; My second is in line and also in place; My third is in point and also in line; My fourth in operation but not in sign; My fifth is in angle but not in degree; My sixth is in glide but not symmetry; Geometry polygon riddles II: I am a polygon all my angles have the same measure all my five sides have the same measure, what general shape am I? Geometry polygon riddles III: I am a polygon. -
The Construction, by Euclid, of the Regular Pentagon
THE CONSTRUCTION, BY EUCLID, OF THE REGULAR PENTAGON Jo˜ao Bosco Pitombeira de CARVALHO Instituto de Matem´atica, Universidade Federal do Rio de Janeiro, Cidade Universit´aria, Ilha do Fund˜ao, Rio de Janeiro, Brazil. e-mail: [email protected] ABSTRACT We present a modern account of Ptolemy’s construction of the regular pentagon, as found in a well-known book on the history of ancient mathematics (Aaboe [1]), and discuss how anachronistic it is from a historical point of view. We then carefully present Euclid’s original construction of the regular pentagon, which shows the power of the method of equivalence of areas. We also propose how to use the ideas of this paper in several contexts. Key-words: Regular pentagon, regular constructible polygons, history of Greek mathe- matics, equivalence of areas in Greek mathematics. 1 Introduction This paper presents Euclid’s construction of the regular pentagon, a highlight of the Elements, comparing it with the widely known construction of Ptolemy, as presented by Aaboe [1]. This gives rise to a discussion on how to view Greek mathematics and shows the care on must have when adopting adapting ancient mathematics to modern styles of presentation, in order to preserve not only content but the very way ancient mathematicians thought and viewed mathematics. 1 The material here presented can be used for several purposes. First of all, in courses for prospective teachers interested in using historical sources in their classrooms. In several places, for example Brazil, the history of mathematics is becoming commonplace in the curricula of courses for prospective teachers, and so one needs materials that will awaken awareness of the need to approach ancient mathematics as much as possible in its own terms, and not in some pasteurized downgraded versions. -
Refer to the Figure. 1. If Name Two Congruent Angles. SOLUTION: Isosceles Triangle Theorem States That If Two Sides of T
4-6 Isosceles and Equilateral Triangles Refer to the figure. 1. If name two congruent angles. SOLUTION: Isosceles Triangle Theorem states that if two sides of the triangle are congruent, then the angles opposite those sides are congruent. Therefore, in triangle ABC, ANSWER: BAC and BCA 2. If EAC ECA, name two congruent segments. SOLUTION: Converse of Isosceles Triangle Theorem states that if two angles of a triangle are congruent, then the sides opposite those angles are congruent. Therefore, in triangle EAC, ANSWER: Find each measure. 3. FH SOLUTION: By the Triangle Angle-Sum Theorem, Since the measures of all the three angles are 60°; the triangle must be equiangular. All the equiangular triangles are equilateral. Therefore, FH = GH = 12. ANSWER: 12 eSolutions4. m ManualMRP - Powered by Cognero Page 1 SOLUTION: Since all the sides are congruent, is an equilateral triangle. Each angle of an equilateral triangle measures 60°. Therefore, m MRP = 60°. ANSWER: 60 SENSE-MAKING Find the value of each variable. 5. SOLUTION: In the figure, . Therefore, triangle RST is an isosceles triangle. By the Converse of Isosceles Triangle Theorem, That is, . ANSWER: 12 6. SOLUTION: In the figure, Therefore, triangle WXY is an isosceles triangle. By the Isosceles Triangle Theorem, . ANSWER: 16 7. PROOF Write a two-column proof. Given: is isosceles; bisects ABC. Prove: SOLUTION: ANSWER: 8. ROLLER COASTERS The roller coaster track appears to be composed of congruent triangles. A portion of the track is shown. a. If and are perpendicular to is isosceles with base , and prove that b. If VR = 2.5 meters and QR = 2 meters, find the distance between and Explain your reasoning. -
( ) Methods for Construction of Odd Number Pointed
Daniel DOBRE METHODS FOR CONSTRUCTION OF ODD NUMBER POINTED POLYGONS Abstract: The purpose of this paper is to present methods for constructing of polygons with an odd number of sides, although some of them may not be built only with compass and straightedge. Odd pointed polygons are difficult to construct accurately, though there are relatively simple ways of making a good approximation which are accurate within the normal working tolerances of design practitioners. The paper illustrates rather complicated constructions to produce unconstructible polygons with an odd number of sides, constructions that are particularly helpful for engineers, architects and designers. All methods presented in this paper provide practice in geometric representations. Key words: regular polygon, pentagon, heptagon, nonagon, hendecagon, pentadecagon, heptadecagon. 1. INTRODUCTION n – number of sides; Ln – length of a side; For a specialist inured to engineering graphics, plane an – apothem (radius of inscribed circle); and solid geometries exert a special fascination. Relying R – radius of circumscribed circle. on theorems and relations between linear and angular sizes, geometry develops students' spatial imagination so 2. THREE - POINT GEOMETRY necessary for understanding other graphic discipline, descriptive geometry, underlying representations in The construction for three point geometry is shown engineering graphics. in fig. 1. Given a circle with center O, draw the diameter Construction of regular polygons with odd number of AD. From the point D as center and, with a radius in sides, just by straightedge and compass, has preoccupied compass equal to the radius of circle, describe an arc that many mathematicians who have found ingenious intersects the circle at points B and C. -
Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 1
Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 1 Similar Quadrilaterals Authors Guangqi Cui, Akshaj Kadaveru, Joshua Lee, Sagar Maheshwari Special thanks to Cosmin Pohoata and the AMSP Cornell 2014 Geometric Proofs Class B0 C0 B A A0 D0 C D Additional thanks to Justin Stevens and David Altizio for the LATEX Template Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 2 Contents 1 Introduction 3 2 Interesting Property 4 3 Example Problems 5 4 Practice Problems 11 Similar Quadrilaterals Cui, Kadaveru, Lee, Maheshwari Page 3 1 Introduction Similar quadrilaterals are a very useful but relatively unknown tool used to solve olympiad geometry problems. It usually goes unnoticed due to the confinement of geometric education to the geometry of the triangle and other conventional methods of problem solving. Also, it is only in very special cases where pairs of similar quadrilaterals exist, and proofs using these qualities usually shorten what would have otherwise been an unnecessarily long proof. The most common method of finding such quadrilaterals involves finding one pair of adjacent sides with identical ratios, and three pairs of congruent angles. We will call this SSAAA Similarity. 0 0 0 0 Example 1.1. (SSAAA Similarity) Two quadrilaterals ABCD and A B C D satisfy \A = AB BC A0, B = B0, C = C0, and = . Show that ABCD and A0B0C0D0 are similar. \ \ \ \ \ A0B0 B0C0 B0 C0 B A A0 D0 C D 0 0 0 0 0 0 Solution. Notice 4ABC and 4A B C are similar from SAS similarity. Therefore \C A D = 0 0 0 0 0 0 0 0 0 0 \A − \B A C = \A − \BAC = \CAD. -
Polygons and Convexity
Geometry Week 4 Sec 2.5 to ch. 2 test section 2.5 Polygons and Convexity Definitions: convex set – has the property that any two of its points determine a segment contained in the set concave set – a set that is not convex concave concave convex convex concave Definitions: polygon – a simple closed curve that consists only of segments side of a polygon – one of the segments that defines the polygon vertex – the endpoint of the side of a polygon 1 angle of a polygon – an angle with two properties: 1) its vertex is a vertex of the polygon 2) each side of the angle contains a side of the polygon polygon not a not a polygon (called a polygonal curve) polygon Definitions: polygonal region – a polygon together with its interior equilateral polygon – all sides have the same length equiangular polygon – all angels have the same measure regular polygon – both equilateral and equiangular Example: A square is equilateral, equiangular, and regular. 2 diagonal – a segment that connects 2 vertices but is not a side of the polygon C B C B D A D A E AC is a diagonal AC is a diagonal AB is not a diagonal AD is a diagonal AB is not a diagonal Notation: It does not matter which vertex you start with, but the vertices must be listed in order. Above, we have square ABCD and pentagon ABCDE. interior of a convex polygon – the intersection of the interiors of is angles exterior of a convex polygon – union of the exteriors of its angles 3 Polygon Classification Number of sides Name of polygon 3 triangle 4 quadrilateral 5 pentagon 6 hexagon 7 heptagon 8 octagon -
Day 7 Classwork for Constructing a Regular Hexagon
Classwork for Constructing a Regular Hexagon, Inscribed Shapes, and Isosceles Triangle I. Review 1) Construct a perpendicular bisector to 퐴퐵̅̅̅̅. 2) Construct a perpendicular line through point P. P • A B 3) Construct an equilateral triangle with length CD. (LT 1c) C D II. Construct a Regular Hexagon (LT 1c) Hexagon A six-sided polygon Regular Hexagon A six-sided polygon with all sides the same length and all interior angles the same measure Begin by constructing an equilateral with the segment below. We can construct a regular hexagon by observing that it is made up of Then construct 5 more equilateral triangles with anchor point on point R 6 adjacent (sharing a side) equilateral for one arc and anchor point on the most recent point of intersection for the second ( intersecting) arc. triangles. R III. Construct a Regular Hexagon Inscribed in a Circle Inscribed (in a Circle) All vertices of the inscribed shape are points on the circle Inscribed Hexagon Hexagon with all 6 vertex points on a circle Construct a regular hexagon with side length 퐴퐵̅̅̅̅: A B 1) Copy the length of 퐴퐵̅̅̅̅ onto the compass. 2) Place metal tip of compass on point C and construct a circle. 3) Keep the same length on the compass. 4) Mark any point (randomly) on the circle. C 5) Place the metal tip on the randomly marked • point and mark an arc on the circle. 6) Lift the compass and place the metal tip on the last arc mark, and mark new arc on the circle. 7) Repeat until the arc mark lands on the original point. -
Locating Diametral Points
Results Math (2020) 75:68 Online First c 2020 Springer Nature Switzerland AG Results in Mathematics https://doi.org/10.1007/s00025-020-01193-5 Locating Diametral Points Jin-ichi Itoh, Costin Vˆılcu, Liping Yuan , and Tudor Zamfirescu Abstract. Let K be a convex body in Rd,withd =2, 3. We determine sharp sufficient conditions for a set E composed of 1, 2, or 3 points of bdK, to contain at least one endpoint of a diameter of K.Weextend this also to convex surfaces, with their intrinsic metric. Our conditions are upper bounds on the sum of the complete angles at the points in E. We also show that such criteria do not exist for n ≥ 4points. Mathematics Subject Classification. 52A10, 52A15, 53C45. Keywords. Convex body, diameter, geodesic diameter, diametral point. 1. Introduction The tangent cone at a point x in the boundary bdK of a convex body K can be defined using only neighborhoods of x in bdK. So, one doesn’t normally expect to get global information about K from the size of the tangent cones at one, two or three points. Nevertheless, in some cases this is what happens! A convex body K in Rd is a compact convex set with interior points; we shall consider only the cases d =2, 3. A convex surface in R3 is the boundary of a convex body in R3. Let S be a convex surface and x apointinS. Consider homothetic dilations of S with the centre at x and coefficients of homothety tending to infinity. The limit surface is called the tangent cone at x (see [1]), and is denoted by Tx. -
Tessellations: Lessons for Every Age
TESSELLATIONS: LESSONS FOR EVERY AGE A Thesis Presented to The Graduate Faculty of The University of Akron In Partial Fulfillment of the Requirements for the Degree Master of Science Kathryn L. Cerrone August, 2006 TESSELLATIONS: LESSONS FOR EVERY AGE Kathryn L. Cerrone Thesis Approved: Accepted: Advisor Dean of the College Dr. Linda Saliga Dr. Ronald F. Levant Faculty Reader Dean of the Graduate School Dr. Antonio Quesada Dr. George R. Newkome Department Chair Date Dr. Kevin Kreider ii ABSTRACT Tessellations are a mathematical concept which many elementary teachers use for interdisciplinary lessons between math and art. Since the tilings are used by many artists and masons many of the lessons in circulation tend to focus primarily on the artistic part, while overlooking some of the deeper mathematical concepts such as symmetry and spatial sense. The inquiry-based lessons included in this paper utilize the subject of tessellations to lead students in developing a relationship between geometry, spatial sense, symmetry, and abstract algebra for older students. Lesson topics include fundamental principles of tessellations using regular polygons as well as those that can be made from irregular shapes, symmetry of polygons and tessellations, angle measurements of polygons, polyhedra, three-dimensional tessellations, and the wallpaper symmetry groups to which the regular tessellations belong. Background information is given prior to the lessons, so that teachers have adequate resources for teaching the concepts. The concluding chapter details results of testing at various age levels. iii ACKNOWLEDGEMENTS First and foremost, I would like to thank my family for their support and encourage- ment. I would especially like thank Chris for his patience and understanding.