Physical Proof of Only Five Regular Solids
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Volumes of Prisms and Cylinders 625
11-4 11-4 Volumes of Prisms and 11-4 Cylinders 1. Plan Objectives What You’ll Learn Check Skills You’ll Need GO for Help Lessons 1-9 and 10-1 1 To find the volume of a prism 2 To find the volume of • To find the volume of a Find the area of each figure. For answers that are not whole numbers, round to prism a cylinder the nearest tenth. • To find the volume of a 2 Examples cylinder 1. a square with side length 7 cm 49 cm 1 Finding Volume of a 2. a circle with diameter 15 in. 176.7 in.2 . And Why Rectangular Prism 3. a circle with radius 10 mm 314.2 mm2 2 Finding Volume of a To estimate the volume of a 4. a rectangle with length 3 ft and width 1 ft 3 ft2 Triangular Prism backpack, as in Example 4 2 3 Finding Volume of a Cylinder 5. a rectangle with base 14 in. and height 11 in. 154 in. 4 Finding Volume of a 6. a triangle with base 11 cm and height 5 cm 27.5 cm2 Composite Figure 7. an equilateral triangle that is 8 in. on each side 27.7 in.2 New Vocabulary • volume • composite space figure Math Background Integral calculus considers the area under a curve, which leads to computation of volumes of 1 Finding Volume of a Prism solids of revolution. Cavalieri’s Principle is a forerunner of ideas formalized by Newton and Leibniz in calculus. Hands-On Activity: Finding Volume Explore the volume of a prism with unit cubes. -
Build a Tetrahedral Kite
Aeronautics Research Mission Directorate Build a Tetrahedral Kite Suggested Grades: 8-12 Activity Overview Time: 90-120 minutes In this activity, you will build a tetrahedral kite from Materials household supplies. • 24 straws (8 inches or less) - NOTE: The straws need to be Steps straight and the same length. If only flexible straws are available, 1. Cut a length of yarn/string 4 feet long. then cut off the flexible portion. • Two or three large spools of 2. Take six straws and place them on a flat surface. cotton string or yarn (approximately 100 feet total) 3. Use your piece of string to join three straws • Scissors together in a triangular shape. On the side where • Hot glue gun and hot glue sticks the two strings are extending from it, one end • Ruler or dowel rod for kite bridle should be approximately 20 inches long, and the • Four pieces of tissue paper (24 x other should be approximately 4 inches long. 18 inches or larger) See Figure 1. • All-purpose glue stick Figure 1 4. Tie these two ends of the string tightly together. Make sure there is no room for the triangle to wiggle. 5. The three straws should form a tight triangle. 6. Cut another 4-inch piece of string. 7. Take one end of the 4-inch string, and tie that end to a corner of the triangle that does not have the string ends extending from it. Figure 2. 8. Add two more straws onto the longest piece of string. 9. Next, take the string that holds the two additional straws and tie it to the end of one of the 4-inch strings to make another tight triangle. -
On the Archimedean Or Semiregular Polyhedra
ON THE ARCHIMEDEAN OR SEMIREGULAR POLYHEDRA Mark B. Villarino Depto. de Matem´atica, Universidad de Costa Rica, 2060 San Jos´e, Costa Rica May 11, 2005 Abstract We prove that there are thirteen Archimedean/semiregular polyhedra by using Euler’s polyhedral formula. Contents 1 Introduction 2 1.1 RegularPolyhedra .............................. 2 1.2 Archimedean/semiregular polyhedra . ..... 2 2 Proof techniques 3 2.1 Euclid’s proof for regular polyhedra . ..... 3 2.2 Euler’s polyhedral formula for regular polyhedra . ......... 4 2.3 ProofsofArchimedes’theorem. .. 4 3 Three lemmas 5 3.1 Lemma1.................................... 5 3.2 Lemma2.................................... 6 3.3 Lemma3.................................... 7 4 Topological Proof of Archimedes’ theorem 8 arXiv:math/0505488v1 [math.GT] 24 May 2005 4.1 Case1: fivefacesmeetatavertex: r=5. .. 8 4.1.1 At least one face is a triangle: p1 =3................ 8 4.1.2 All faces have at least four sides: p1 > 4 .............. 9 4.2 Case2: fourfacesmeetatavertex: r=4 . .. 10 4.2.1 At least one face is a triangle: p1 =3................ 10 4.2.2 All faces have at least four sides: p1 > 4 .............. 11 4.3 Case3: threefacesmeetatavertes: r=3 . ... 11 4.3.1 At least one face is a triangle: p1 =3................ 11 4.3.2 All faces have at least four sides and one exactly four sides: p1 =4 6 p2 6 p3. 12 4.3.3 All faces have at least five sides and one exactly five sides: p1 =5 6 p2 6 p3 13 1 5 Summary of our results 13 6 Final remarks 14 1 Introduction 1.1 Regular Polyhedra A polyhedron may be intuitively conceived as a “solid figure” bounded by plane faces and straight line edges so arranged that every edge joins exactly two (no more, no less) vertices and is a common side of two faces. -
THE GEOMETRY of PYRAMIDS One of the More Interesting Solid
THE GEOMETRY OF PYRAMIDS One of the more interesting solid structures which has fascinated individuals for thousands of years going all the way back to the ancient Egyptians is the pyramid. It is a structure in which one takes a closed curve in the x-y plane and connects straight lines between every point on this curve and a fixed point P above the centroid of the curve. Classical pyramids such as the structures at Giza have square bases and lateral sides close in form to equilateral triangles. When the closed curve becomes a circle one obtains a cone and this cone becomes a cylindrical rod when point P is moved to infinity. It is our purpose here to discuss the properties of all N sided pyramids including their volume and surface area using only elementary calculus and geometry. Our starting point will be the following sketch- The base represents a regular N sided polygon with side length ‘a’ . The angle between neighboring radial lines r (shown in red) connecting the polygon vertices with its centroid is θ=2π/N. From this it follows, by the law of cosines, that the length r=a/sqrt[2(1- cos(θ))] . The area of the iscosolis triangle of sides r-a-r is- a a 2 a 2 1 cos( ) A r 2 T 2 4 4 (1 cos( ) From this we have that the area of the N sided polygon and hence the pyramid base will be- 2 2 1 cos( ) Na A N base 2 4 1 cos( ) N 2 It readily follows from this result that a square base N=4 has area Abase=a and a hexagon 2 base N=6 yields Abase= 3sqrt(3)a /2. -
Applying the Polygon Angle
POLYGONS 8.1.1 – 8.1.5 After studying triangles and quadrilaterals, students now extend their study to all polygons. A polygon is a closed, two-dimensional figure made of three or more non- intersecting straight line segments connected end-to-end. Using the fact that the sum of the measures of the angles in a triangle is 180°, students learn a method to determine the sum of the measures of the interior angles of any polygon. Next they explore the sum of the measures of the exterior angles of a polygon. Finally they use the information about the angles of polygons along with their Triangle Toolkits to find the areas of regular polygons. See the Math Notes boxes in Lessons 8.1.1, 8.1.5, and 8.3.1. Example 1 4x + 7 3x + 1 x + 1 The figure at right is a hexagon. What is the sum of the measures of the interior angles of a hexagon? Explain how you know. Then write an equation and solve for x. 2x 3x – 5 5x – 4 One way to find the sum of the interior angles of the 9 hexagon is to divide the figure into triangles. There are 11 several different ways to do this, but keep in mind that we 8 are trying to add the interior angles at the vertices. One 6 12 way to divide the hexagon into triangles is to draw in all of 10 the diagonals from a single vertex, as shown at right. 7 Doing this forms four triangles, each with angle measures 5 4 3 1 summing to 180°. -
Squaring the Circle a Case Study in the History of Mathematics the Problem
Squaring the Circle A Case Study in the History of Mathematics The Problem Using only a compass and straightedge, construct for any given circle, a square with the same area as the circle. The general problem of constructing a square with the same area as a given figure is known as the Quadrature of that figure. So, we seek a quadrature of the circle. The Answer It has been known since 1822 that the quadrature of a circle with straightedge and compass is impossible. Notes: First of all we are not saying that a square of equal area does not exist. If the circle has area A, then a square with side √A clearly has the same area. Secondly, we are not saying that a quadrature of a circle is impossible, since it is possible, but not under the restriction of using only a straightedge and compass. Precursors It has been written, in many places, that the quadrature problem appears in one of the earliest extant mathematical sources, the Rhind Papyrus (~ 1650 B.C.). This is not really an accurate statement. If one means by the “quadrature of the circle” simply a quadrature by any means, then one is just asking for the determination of the area of a circle. This problem does appear in the Rhind Papyrus, but I consider it as just a precursor to the construction problem we are examining. The Rhind Papyrus The papyrus was found in Thebes (Luxor) in the ruins of a small building near the Ramesseum.1 It was purchased in 1858 in Egypt by the Scottish Egyptologist A. -
Archimedean Solids
University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln MAT Exam Expository Papers Math in the Middle Institute Partnership 7-2008 Archimedean Solids Anna Anderson University of Nebraska-Lincoln Follow this and additional works at: https://digitalcommons.unl.edu/mathmidexppap Part of the Science and Mathematics Education Commons Anderson, Anna, "Archimedean Solids" (2008). MAT Exam Expository Papers. 4. https://digitalcommons.unl.edu/mathmidexppap/4 This Article is brought to you for free and open access by the Math in the Middle Institute Partnership at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in MAT Exam Expository Papers by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. Archimedean Solids Anna Anderson In partial fulfillment of the requirements for the Master of Arts in Teaching with a Specialization in the Teaching of Middle Level Mathematics in the Department of Mathematics. Jim Lewis, Advisor July 2008 2 Archimedean Solids A polygon is a simple, closed, planar figure with sides formed by joining line segments, where each line segment intersects exactly two others. If all of the sides have the same length and all of the angles are congruent, the polygon is called regular. The sum of the angles of a regular polygon with n sides, where n is 3 or more, is 180° x (n – 2) degrees. If a regular polygon were connected with other regular polygons in three dimensional space, a polyhedron could be created. In geometry, a polyhedron is a three- dimensional solid which consists of a collection of polygons joined at their edges. The word polyhedron is derived from the Greek word poly (many) and the Indo-European term hedron (seat). -
Regular Polyhedra of Index 2
REGULAR POLYHEDRA OF INDEX 2 A dissertation presented by Anthony Cutler to The Department of Mathematics In partial fulfillment of the requirements for the degree of Doctor of Philosophy in the field of Mathematics Northeastern University Boston, Massachusetts April, 2009 1 REGULAR POLYHEDRA OF INDEX 2 by Anthony Cutler ABSTRACT OF DISSERTATION Submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics in the Graduate School of Arts and Sciences of Northeastern University, April 2009 2 We classify all finite regular polyhedra of index 2, as defined in Section 2 herein. The definition requires the polyhedra to be combinatorially flag transitive, but does not require them to have planar or convex faces or vertex-figures, and neither does it require the polyhedra to be orientable. We find there are 10 combinatorially regular polyhedra of index 2 with vertices on one orbit, and 22 infinite families of combinatorially regular polyhedra of index 2 with vertices on two orbits, where polyhedra in the same family differ only in the relative diameters of their vertex orbits. For each such polyhedron, or family of polyhedra, we provide the underlying map, as well as a geometric diagram showing a representative face for each face orbit, and a verification of the polyhedron’s combinatorial regularity. A self-contained completeness proof is given. Exactly five of the polyhedra have planar faces, which is consistent with a previously known result. We conclude by describing a non-Petrie duality relation among regular polyhedra of index 2, and suggest how it can be extended to other combinatorially regular polyhedra. -
Math 366 Lecture Notes Section 11.4 – Geometry in Three Dimensions
Section 11-4 Math 366 Lecture Notes Section 11.4 – Geometry in Three Dimensions Simple Closed Surfaces A simple closed surface has exactly one interior, no holes, and is hollow. A sphere is the set of all points at a given distance from a given point, the center . A sphere is a simple closed surface. A solid is a simple closed surface with all interior points. (see p. 726) A polyhedron is a simple closed surface made up of polygonal regions, or faces . The vertices of the polygonal regions are the vertices of the polyhedron, and the sides of each polygonal region are the edges of the polyhedron. (see p. 726-727) A prism is a polyhedron in which two congruent faces lie in parallel planes and the other faces are bounded by parallelograms. The parallel faces of a prism are the bases of the prism. A prism is usually names after its bases. The faces other than the bases are the lateral faces of a prism. A right prism is one in which the lateral faces are all bounded by rectangles. An oblique prism is one in which some of the lateral faces are not bounded by rectangles. To draw a prism: 1) Draw one of the bases. 2) Draw vertical segments of equal length from each vertex. 3) Connect the bottom endpoints to form the second base. Use dashed segments for edges that cannot be seen. A pyramid is a polyhedron determined by a polygon and a point not in the plane of the polygon. The pyramid consists of the triangular regions determined by the point and each pair of consecutive vertices of the polygon and the polygonal region determined by the polygon. -
10-4 Surface and Lateral Area of Prism and Cylinders.Pdf
Aim: How do we evaluate and solve problems involving surface area of prisms an cylinders? Objective: Learn and apply the formula for the surface area of a prism. Learn and apply the formula for the surface area of a cylinder. Vocabulary: lateral face, lateral edge, right prism, oblique prism, altitude, surface area, lateral surface, axis of a cylinder, right cylinder, oblique cylinder. Prisms and cylinders have 2 congruent parallel bases. A lateral face is not a base. The edges of the base are called base edges. A lateral edge is not an edge of a base. The lateral faces of a right prism are all rectangles. An oblique prism has at least one nonrectangular lateral face. An altitude of a prism or cylinder is a perpendicular segment joining the planes of the bases. The height of a three-dimensional figure is the length of an altitude. Surface area is the total area of all faces and curved surfaces of a three-dimensional figure. The lateral area of a prism is the sum of the areas of the lateral faces. Holt Geometry The net of a right prism can be drawn so that the lateral faces form a rectangle with the same height as the prism. The base of the rectangle is equal to the perimeter of the base of the prism. The surface area of a right rectangular prism with length ℓ, width w, and height h can be written as S = 2ℓw + 2wh + 2ℓh. The surface area formula is only true for right prisms. To find the surface area of an oblique prism, add the areas of the faces. -
Binary Icosahedral Group and 600-Cell
Article Binary Icosahedral Group and 600-Cell Jihyun Choi and Jae-Hyouk Lee * Department of Mathematics, Ewha Womans University 52, Ewhayeodae-gil, Seodaemun-gu, Seoul 03760, Korea; [email protected] * Correspondence: [email protected]; Tel.: +82-2-3277-3346 Received: 10 July 2018; Accepted: 26 July 2018; Published: 7 August 2018 Abstract: In this article, we have an explicit description of the binary isosahedral group as a 600-cell. We introduce a method to construct binary polyhedral groups as a subset of quaternions H via spin map of SO(3). In addition, we show that the binary icosahedral group in H is the set of vertices of a 600-cell by applying the Coxeter–Dynkin diagram of H4. Keywords: binary polyhedral group; icosahedron; dodecahedron; 600-cell MSC: 52B10, 52B11, 52B15 1. Introduction The classification of finite subgroups in SLn(C) derives attention from various research areas in mathematics. Especially when n = 2, it is related to McKay correspondence and ADE singularity theory [1]. The list of finite subgroups of SL2(C) consists of cyclic groups (Zn), binary dihedral groups corresponded to the symmetry group of regular 2n-gons, and binary polyhedral groups related to regular polyhedra. These are related to the classification of regular polyhedrons known as Platonic solids. There are five platonic solids (tetrahedron, cubic, octahedron, dodecahedron, icosahedron), but, as a regular polyhedron and its dual polyhedron are associated with the same symmetry groups, there are only three binary polyhedral groups(binary tetrahedral group 2T, binary octahedral group 2O, binary icosahedral group 2I) related to regular polyhedrons. -
Can Every Face of a Polyhedron Have Many Sides ?
Can Every Face of a Polyhedron Have Many Sides ? Branko Grünbaum Dedicated to Joe Malkevitch, an old friend and colleague, who was always partial to polyhedra Abstract. The simple question of the title has many different answers, depending on the kinds of faces we are willing to consider, on the types of polyhedra we admit, and on the symmetries we require. Known results and open problems about this topic are presented. The main classes of objects considered here are the following, listed in increasing generality: Faces: convex n-gons, starshaped n-gons, simple n-gons –– for n ≥ 3. Polyhedra (in Euclidean 3-dimensional space): convex polyhedra, starshaped polyhedra, acoptic polyhedra, polyhedra with selfintersections. Symmetry properties of polyhedra P: Isohedron –– all faces of P in one orbit under the group of symmetries of P; monohedron –– all faces of P are mutually congru- ent; ekahedron –– all faces have of P the same number of sides (eka –– Sanskrit for "one"). If the number of sides is k, we shall use (k)-isohedron, (k)-monohedron, and (k)- ekahedron, as appropriate. We shall first describe the results that either can be found in the literature, or ob- tained by slight modifications of these. Then we shall show how two systematic ap- proaches can be used to obtain results that are better –– although in some cases less visu- ally attractive than the old ones. There are many possible combinations of these classes of faces, polyhedra and symmetries, but considerable reductions in their number are possible; we start with one of these, which is well known even if it is hard to give specific references for precisely the assertion of Theorem 1.