PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS


 Lilian Ford
 2 years ago
 Views:
Transcription
1 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS JORGE REZENDE. Introduction Consider a polyhedron. For example, a platonic, an arquemidean, or a dual of an arquemidean polyhedron. Construct flat polygonal plates in the same number, shape and size as the faces of the referred polyhedron. Adjacent to each side of each plate draw a number like it is shown in figure. Some of the plates, or all, can have numbers on both faces. We call these plates, twofaced plates. In this case, they have the same number adjacent to the same side. In figure, c and d are the two faces of a pentagonal twofaced plate example. Now the game is to put the plates over the polyhedron faces in such a way that the two numbers near each polyhedron edge are equal. If there is at least one solution for this puzzle one says that we have a polyhedron puzzle with numbers. These puzzles are a tool in teaching and learning mathematics, and a source of examples and exercises. They may be used in various fields such as elementary group theory and computational geometry. It is obvious that not all puzzles are interesting. But, amazingly or not, there are natural ways of constructing interesting puzzles. From now on, assume that the numbers belong to the set {,,..., n}, and that all the numbers are used. If we have plate faces which have the shape of a regular polygon with j sides, one can ask how many possible ways ν are there to draw the numbers,,..., n, without repeating them on each plate face. The answer is ( ) n ν = (j )! = j n! (n j)!j. For n = and j = (equilateral triangle), this gives ν =. The Mathematical Physics Group is supported by the Portuguese Ministry for Science and Technology. 000 Mathematics Subject Classification. B0, B, 0B0, 0B0.
2 JORGE REZENDE For n = and j =, this gives ν = 8. Note that 8 is precisely the number of the octahedron faces. We naturally call the related puzzle, the octahedron puzzle. a b c d Figure. For n = and j =, this gives ν = 0. Note that 0 is precisely the number of the icosahedron faces. We call the related puzzle, the icosahedron first puzzle (or icosahedron ()). For n = and j =, this gives ν = 0. Note that 0 is the double of the number of the icosahedron faces. Construct different plates with the numbers written on both faces. This gives 0 plates. We call the related puzzle, the icosahedron second puzzle (or icosahedron ()). Consider again n = and j =. Construct different plates with the numbers written only on one face, but in such a way that the numbers grow if we read them, beginning with the minimum, counter clockwise. See an example in figure b. This gives 0 plates. We call the related puzzle, the icosahedron third puzzle (or icosahedron ()). For n = and j = (square), this gives ν =. Note that is precisely the number of the cube faces. We naturally call the related puzzle, the cube puzzle. For n = and j = (regular pentagon), this gives ν =. Note that is precisely the double of the number of the dodecahedron faces. Construct different plates with the numbers written on both faces. This gives plates. In figure, c and d show the two faces of a plate of this type. We call the related puzzle, the dodecahedron first puzzle (or dodecahedron ()).
3 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS Let again n = and j =. Construct different plates with the numbers written only on one face, but in such a way that the numbers read counter clockwise, abcd, are such that abcd form an even permutation. This gives plates. In figure, c and d show two plates of this type. We call the related puzzle, the dodecahedron second puzzle (or dodecahedron ()). Consider n =. With j =, one has ν = 8. With j =, one has ν =. Note that 8 is precisely the number of the cuboctahedron triangular faces and is precisely the number of its square faces. This is an example of an interesting puzzle using an arquemidean polyhedron. Take now a deltoidal icositetrahedron. It has deltoidal faces. If we have plates which have the deltoidal shape the number of possible different ways to draw the numbers,,,, without repeating them on each plate is precisely. This an example of an interesting puzzle using a dual of an arquemidean polyhedron. These are simple examples of polyhedron puzzles with numbers, which are enough in order to understand the following sections. There are, obviously, others. For more examples see Reference [], which is a development of Reference []. Reference [] is a collection of some of these puzzles paper models.. Polyhedron symmetries and permutation groups Consider a puzzle with numbers,,..., n drawn on the plates. From now on P denotes the set of its plate faces which have numbers drawn, and call it the plate set. S n denotes the group of all permutations of {,,..., n}; σ S n means that σ is a onetoone function σ : {,,..., n} {,,..., n}. The identity is σ 0 : σ 0 () =, σ 0 () =,..., σ 0 (n) = n. The alternating group, the S n subgroup of the even permutations, is denoted by A n. We shall use also the group {, } S n denoted by S ± n. If δ, δ {, } and σ, σ S n, then (δ, σ ) (δ, σ ) = (δ δ, σ σ ). We denote S + n = {} S n S n. If Λ is a set, then Λ denotes its cardinal. Hence, if G is group, G denotes its order. From now on E denotes the set of the polyhedron edges and F denotes the set of the polyhedron faces. A solution of the puzzle defines a function ε : E {,,..., n}. Denote E the set of these functions. One can say that E is the set of the puzzle solutions. Denote Ω the group of the polyhedron symmetries. Every symmetry ω Ω induces two bijections, that we shall also denote ω, whenever there is no confusion possible: ω : E E and ω : F F. Denote also
4 JORGE REZENDE Ω {ω : E E} {ω : F F }, the two sets of these functions. One can say that each one of these two sets Ω is the set of the polyhedron symmetries. With the composition of functions each one of these two sets Ω forms a group that is isomorphic to the group of the polyhedron symmetries. If ω, ω Ω, we shall denote ω ω ω ω. If Ω is a subgroup of Ω, then Ω acts naturally on the face set, F : for ω Ω and ϕ F, one defines the action ωϕ = ω (ϕ). In the following Ω + denotes the subgroup of Ω of the symmetries with determinant. We shall also consider the group S n Ω. If (σ, ω ), (σ, ω ) S n Ω, one defines the product (σ, ω ) (σ, ω ) = (σ σ, ω ω ). On group theory we follow essentially references [] and []... The plate group. Some S n subgroups act naturally on P. Let π P and σ S n. Assume that a, b, c,... are drawn on π, by this order. Then σπ is the plate face where the numbers σ (a), σ (b), σ (c),... are drawn replacing a, b, c,.... Let s S n ± and π P. If s = (, σ) σ, then sπ = σπ. If s = (, σ) σ, then sπ is a reflection of σπ. In this last case, if the numbers a, b, c,... are drawn on π, by this order, then sπ is a plate face where the numbers..., σ (c), σ (b), σ (a) are drawn by this order. The plate group, G P, is the greatest subgroup of S n ± that acts on P. If s S n ± and sπ P, for every π P, then s G P... The solution group. Let ε : E {,,..., n} be a solution of the puzzle. The group of this solution, G ε, is a subgroup of S n Ω; (σ, ω) G ε if and only if σ ε ω = ε. Denote Ω ε the following subgroup of Ω: ω Ω ε if and only if there exists σ S n such that (σ, ω) G ε. Note that if ω Ω ε there exists only one σ S n such that (σ, ω) G ε. From this one concludes that ω (σ, ω) defines an isomorphism between Ω ε and G ε. On the other hand (det ω, σ) G P. This defines g ε : Ω ε G P, g ε (ω) = (det ω, σ), which is an homomorphism of groups. If all the plates are different, as is the case in the given examples and as we shall assume from now on, (det ω, σ) defines completely ω. Hence, g ε establishes an isomorphism between Ω ε and g ε (Ω ε ) G P. Denote G P ε g ε (Ω ε ). Finally, G ε and G P ε are isomorphic. We can identify (σ, ω) with (det ω, σ), and G ε with the subgroup G P ε of G P. One can define a function u ε : F P, that associates to every face ϕ the plate face π that ε puts in ϕ. Define also P ε = u ε (F ). Note that
5 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS P ε = F and that G P ε acts on P ε. Then one has for every ϕ F. (det ω, σ) (u ε ω (ϕ)) = u ε (ϕ),.. Equivalent solutions. Let ε, ε : E {,,..., n} be solutions of the puzzle. One says that these solutions are equivalent, ε ε, if there are ω Ω and σ S n, such that Note that (det ω, σ ) G P. Then, σ ε ω = ε. (σ, ω) ( σ σσ, ω ωω ) = (σ, ω ) (σ, ω) (σ, ω ) defines an isomorphism between G ε one has that s σ sσ and G ε. As det ( ω ωω ) = det ω, defines an isomorphism between G P ε and G P ε. If σ = σ 0 and det ω =, what distinguishes the solutions ε and ε is only a rotational symmetry. In this case ε ω = ε expresses another equivalence relation, ε ε. When we make a puzzle, in practice, we do not recognize the difference between ε and ε. We shall say that they represent the same natural solution, an equivalence class of the relation. Let ε, ε, ε E. As ε ε and ε ε, implies ε ε, one can say that the natural solution represented by ε is equivalent to ε. For ε E, represent by [ε] the set of natural solutions equivalent to ε. Choose now ω Ω, such that det ω =. For ε E and s = (δ, σ) G P, denote ε s = σ ε ω, where ω is the identity if δ = and ω = ω if δ =. The set {ε s : s G P } includes representatives of all natural solutions equivalent to ε. Then [ε] = G P G P ε. The cardinal of all the natural solutions is then given by [ε] G P G P ε, where the sum is extended to all different equivalence classes [ε].
6 JORGE REZENDE.. Equivalent actions. Consider two groups H and H that act on two sets K and K. For ρ j H j and x j K j, ρ j x j ( K j ) represents the action of ρ j on x j. One says that the action of H on K is equivalent to the action of H on K if there exists an isomorphism ξ : H H and a onetoone function ζ : K K, such that for every ρ H and every x K one has ξ (ρ) ζ (x) = ζ (ρx). Note that if the action of H on K is equivalent to the action of H on K, and H is a subgroup of H, then the action of H on K is equivalent to the action of H = ξ (H ) on K. Assume that the action of H on K is equivalent to the action of H on K. Let K and K be the sets of the equivalence classes of such actions. Then there exists a bijection Z : K K, such that for every ρ H and every X K one has and ξ (ρ) Z (X) = Z (ρx) X = Z (X). These two conditions are necessary in order to have equivalent actions. Example.. If ε and ε are equivalent solutions of the polyhedron puzzle, the actions of G P ε and G P ε on P are equivalent. Let ξ (s) = σ sσ and ζ (π) = (det ω, σ ) π. Then ξ (s) ζ (π) = σ sσ (det ω, σ ) π = (det ω, σ ) sπ = ζ (sπ). Example.. If ε and ε are equivalent solutions of the polyhedron puzzle, the actions of Ω ε and Ω ε on F are equivalent. Let ξ (ω) = ω ωω and ζ (ϕ) = ω ϕ. Then ξ (ω) ζ (ϕ) = ω ωω ω ϕ = ω ωϕ = ζ (ωϕ). Example.. If ε is a solution of the polyhedron puzzle, the actions of Ω ε on F and of G P ε on P ε = u ε (P ) are equivalent. Let ξ (ω) = g ε (ω ) = (det ω, σ ) and ζ (ϕ) = u ε (ϕ). Then ξ (ω) ζ (ϕ) = ( det ω, σ ) u ε (ϕ) = (u ε ω (ϕ)) = ζ (ωϕ). Denote F = {F, F,..., F k } and P = {P, P,..., P k } be the sets of the equivalence classes of such actions. Then u ε induces a onetoone function U : F P, U (F j ) = P j, F j = P j, and ( det ω, σ ) U (F j ) = U (ωf j ), for every j =,,..., k, ω Ω ε.
7 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS 7 From now on we shall deal only with the action of a subgroup Ω of Ω on F, or the action of a subgroup G of G P on P. Hence, when we say that two groups H and H are equivalent (H H ), we are talking about these actions.. Examples Some of the examples we give in this section can easily be studied directly. It is the case of the cube, the octahedron and the dodecahedron () puzzles. We give also results for the dodecahedron (), icosahedron () and () puzzles. These results were obtained mostly with a computer. Similar results for the icosahedron () puzzle are left to the reader. There are good reasons for presenting results for these two icosahedron puzzles. The icosahedron () puzzle is mathematically rich, has a lot of natural solutions (over one million!). On the other hand, the icosahedron () puzzle is also very instructive because it has few natural solutions ( only ), which means that it is difficult to do it without the help of a computer... The cube puzzle. In this case there is only one equivalence class. G ε Ω + S. G ε =. G P = 8. If one puts a plate on a face then one has different possibilities for the other plates ( natural solutions): 8/. These two different possibilities and the equivalences G ε Ω + S can be used to translate the cube symmetries into permutations, in the same way as we shall do later with the icosahedron () puzzle and its canonical solution... The octahedron puzzle. There three equivalence classes. One can distinguish them in the following way. Take a solution. On every vertex of the octahedron, note the numbers that correspond to its four edges. There are two possibilities for a vertex: a) four distinct numbers; b) three distinct numbers, with one of them repeated. In the solution, count the number of vertices where the situation a) happens. They can be, or 0, that distinguish the three equivalence classes. The first class group is of order. The second class group is of order 8. The third class group is of order. As G P = 8, one has that if one puts a plate on a face then one has different possibilities for the other plates ( natural solutions): Although the cube and the octahedron are dual polyhedra, puzzles are not. Solutions can be dual. The first class of the octahedron puzzle is dual of the cube puzzle class.
8 8 JORGE REZENDE.. The dodecahedron () puzzle. This puzzle has three equivalence classes that we can distinguish in the following way. Consider a solution. On every dodecahedron vertex note the numbers that are on the edges around the vertex. There are 0 possibilities, but not all of them belong to the solution. There are some repetitions: 7 or. The solutions that have 7 repetitions on the vertices are equivalent. The solutions that have repetitions on the vertices belong to different equivalence classes. One of these classes has the repetitions on opposite vertices. The other has the repetitions on vertices that belong to the same edges. The first class group is of order 8. The second class group is of order. The third class group is of order. As G P = 0, one has that if one puts a plate on a face then one has ( 0 different possibilities for the other plates (0 natural solutions): ) The dodecahedron () puzzle. This puzzle has equivalence classes that can be distinguished in the following form. Consider two opposite dodecahedron edges. There are other four that are orthogonal to these two. The six edges are over the faces of a virtual cube where the dodecahedron is inscribed. There are five such cubes. The first equivalence class has the same number associated to the edges that belong to the faces of each cube. The second equivalence class has the same number associated to the edges that belong to the faces of one of the fives cubes. The first class group is of order 0 and the second class group is of order. As G P = 0, one has that if one puts a plate on a face then one has different possibilities for the other plates ( natural solutions): The icosahedron () puzzle. The icosahedron () puzzle has 9 equivalence classes: have groups of order, have groups of order, have groups of order, has a group of order, have groups of order, 0 have groups of order, has a group of order 8, have groups of order 0, have groups of order, have groups of order, has a group of order 0. As G P = 0, one has that once one puts a plate over a face there are 0 different possibilities (0 natural solutions): 0 ( There are two possible natural solutions with a group of order 0. One of them, that we call canonical solution, is dual of the dodecahedron () solution with a group of order 0. The group of these two solutions ).
9 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS 9 z x y Figure. is equivalent to the icosahedron group: G ε Ω {, } A. The canonical solution is represented in figure... The icosahedron () puzzle. This puzzle has 97 equivalence classes: 90 have groups of order, and 7 have groups of order. As G P =, one has that, once one puts a plate over a face there are different possibilities ( natural solutions): Note that the actions of all these 7 groups of order are equivalent. Figure shows representatives of all these 7 equivalence classes.. More on icosahedron puzzles As we have already seen there is a natural solution of the icosahedron () puzzle which is dual of the dodecahedron () natural solution with a group of order 0 (see figure ). This group is equivalent to the icosahedron group Ω: G ε Ω {, } A. As this solution is very easy to construct, one can use it in order to translate in terms of
10 0 JORGE REZENDE Figure. {, } A everything that happens in Ω. For example, all subgroups of Ω are equivalent to the corresponding subgroups of {, } A... Subgroups of the icosahedron group. The icosahedron group Ω has different equivalence classes of subgroups: of order ({σ 0 }); of order ; of order ; of order ; of order ; of order ; of order 8; of order 0; of order ; of order 0; of order ; of order 0; of order 0 (Ω {, } A ). In the following a, b, c, d, e {,,,, } are different numbers, and in angle brackets we give the number of equivalence classes of solutions in the icosahedron () puzzle.... Groups of order. The equivalence classes are generated by (ab) (cd) 8, (, (ab) (cd)) and (, σ 0 ). Look at figure. The rotation around the zaxis by an angle of 80, corresponds to the permutation () (). This rotation belongs to the first equivalence class. The reflection using the xyplane (orthogonal to the zaxis) as a mirror, corresponds to (, () ()). This reflection belongs to the second equivalence class. The third equivalence class corresponds to the central symmetry (x, y, z) (x, y, z).
11 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS... Groups of order. The unique equivalence class is generated by (abc). In figure, the rotations around the axis x = y = z by angles multiple of 0 correspond to group generated by ().... Groups of order. Let σ = (ab) (cd) and σ = (ac) (bd). Note that σ σ = σ σ = (ad) (bc). The equivalence classes are generated by σ and σ 0, σ and (, σ 0 ), σ and (, σ ) 0, respectively.... Groups of order. The unique equivalence class is generated by (abcde). In figure, the rotations around the axis x = 0, z = + y (this is the z axis in figure ), by angles multiple of 7, c orrespond to group generated by ().... Groups of order. Let σ = (ab) (cd) and σ = (ae) (cd). The equivalence classes are generated by σ and σ ( S ), (, σ ) and (, σ ) 0, (abc) and (, σ 0 ) 7, respectively.... Groups of order 8. Let σ = (ab) (cd) and σ = (ac) (bd). As before, note that σ σ = σ σ = (ad) (bc). The unique equivalence class is generated by σ and σ and (, σ 0 )...7. Groups of order 0. Let σ = (ab) (cd) and σ = (ac) (be). The equivalence classes are generated by σ and σ ( D, the dihedral group of order 0), (, σ ) and (, σ ) 0, (abcde) and (, σ 0 ), respectively...8. Groups of order. Let σ = (ab) (cd) and σ = (acd). The equivalence classes are generated by σ and σ ( A ), S and (, σ 0 ) ( {, } S ) 0, respectively...9. Groups of order 0. The unique equivalence class is generated by D and (, σ 0 ) ( {, } D ) Groups of order. The unique equivalence class is generated by A and (, σ 0 ) ( {, } A ).... Groups of order 0. The unique equivalence class is A Groups of order 0. The unique equivalence class is {, } A Ω.
12 JORGE REZENDE.. Subgroups of plate groups. All the subgroups of the icosahedron group have cyclic groups of order, or as generators. Let us look at the actions on the faces of such cyclic groups. The first equivalence class of the order groups has 0 orbits with elements and the determinant of the generator is : (, 0 ). The second equivalence class has 8 orbits with elements, orbits with element and the determinant of the generator is : (, 8 + ). The third equivalence class has 0 orbits with elements and the determinant of the generator is : (, 0 ). The equivalence class of the order groups has orbits with elements, orbits with element and the determinant of the generator is : (, + ). The equivalence class of the order groups has orbits with elements and the determinant of the generator is : (, ).... The icosahedron () puzzle. The plate group is G P = {, } S. The only cyclic groups of G P whose actions on the plates have orbits of the type (, 0 ), (, 8 + ), (, 0 ), (, + ) and (, ) are those generated, precisely, by (, (ab) (cd)), (, (ab) (cd)), (, σ 0 ), (, (abc)) and (, (abcde)).... The icosahedron () puzzle. Let σ = () and σ = () () (). The plate group, G P, is generated by (, σ ) and (, σ ). The only cyclic groups of G P whose actions on the plates have orbits of the type (, 0 ), (, 8 + ), (, 0 ) and (, + ) are generated by (, σ), ( ) ( ), σσ j,, σ k σ, (, σ ), j =,,, k = 0,,. There are no subgroups of order. It is easy to see directly that there are no solutions corresponding to the subgroups of order. The solutions corresponding to subgroups of order, are only those related to the generators ( ), σσ k, k = 0,,... Finding solutions of a puzzle with a prescribed group. A way to find a solution for a puzzle is to use the group relations between the polyhedron and the plate groups. Take, for example, the icosahedron () puzzle and the third equivalence class for groups of order 0. This class is generated by an element σ of order and (, σ 0 ). The numbers on the edges must respect the central symmetry and a rotational symmetry group of order. Let σ = (). In figure, that illustrates this example, a, b {,,,, }, and a j = σ j (a), b j = σ j (b), j =,,,. The numbers a and b must be chosen so that a b, a σ (b) = b, b σ (a) = a. These three conditions give 9 possibilities, but only of them correspond to solutions of the puzzle: (a, b) = (, ), (, ), (, ), (, ), (, ). The third one is the canonical
13 PUZZLES WITH POLYHEDRA AND PERMUTATION GROUPS z a a b b a b b b b a a Figure. solution. The first and the fourth ones belong to the same equivalence class. The same happens with the second and the fifth ones. These last two classes are the ones already listed. References [] Jorge Rezende: Puzzles numéricos poliédricos. Diário da República, III Série, n o 00, p. 7 (988). [] Jorge Rezende: Jogos com poliedros e permutações. Bol. Soc. Port. Mat., 0 (000). PT.html [] Jorge Rezende: Puzzles com poliedros e números. SPM: Lisboa 00. [] M. A. Armstrong: Groups and Symmetry. Berlin: SpringerVerlag 988. [] R. A. Wilson, R. A. Parker and J. N. Bray: ATLAS of Finite Group Representations. Grupo de FísicaMatemática da Universidade de Lisboa, Av. Prof. Gama Pinto, 900 Lisboa, Portugal, e Departamento de Matemática, Faculdade de Ciências da Universidade de Lisboa address: URL: PT.html
1 Symmetries of regular polyhedra
1230, notes 5 1 Symmetries of regular polyhedra Symmetry groups Recall: Group axioms: Suppose that (G, ) is a group and a, b, c are elements of G. Then (i) a b G (ii) (a b) c = a (b c) (iii) There is an
More informationCLASSIFYING FINITE SUBGROUPS OF SO 3
CLASSIFYING FINITE SUBGROUPS OF SO 3 HANNAH MARK Abstract. The goal of this paper is to prove that all finite subgroups of SO 3 are isomorphic to either a cyclic group, a dihedral group, or the rotational
More informationConstruction and Properties of the Icosahedron
Course Project (Introduction to Reflection Groups) Construction and Properties of the Icosahedron Shreejit Bandyopadhyay April 19, 2013 Abstract The icosahedron is one of the most important platonic solids
More informationThe 5 Platonic solids:
The 5 Platonic solids: The Tetrahedron (3 equilateral triangles at each vertex) The Hexahedron (3 squares at each vertex, cube) The Octahedron (4 equilateral triangles at each vertex) The Dodecahedron
More informationSOLIDS, NETS, AND CROSS SECTIONS
SOLIDS, NETS, AND CROSS SECTIONS Polyhedra In this section, we will examine various threedimensional figures, known as solids. We begin with a discussion of polyhedra. Polyhedron A polyhedron is a threedimensional
More informationStar and convex regular polyhedra by Origami.
Star and convex regular polyhedra by Origami. Build polyhedra by Origami.] Marcel Morales Alice Morales 2009 E D I T I O N M O R A L E S Polyhedron by Origami I) Table of convex regular Polyhedra... 4
More information7. The GaussBonnet theorem
7. The GaussBonnet theorem 7. Hyperbolic polygons In Euclidean geometry, an nsided polygon is a subset of the Euclidean plane bounded by n straight lines. Thus the edges of a Euclidean polygon are formed
More informationNotes on polyhedra and 3dimensional geometry
Notes on polyhedra and 3dimensional geometry Judith Roitman / Jeremy Martin April 23, 2013 1 Polyhedra Threedimensional geometry is a very rich field; this is just a little taste of it. Our main protagonist
More informationCK12 Geometry: Exploring Solids
CK12 Geometry: Exploring Solids Learning Objectives Identify different types of solids and their parts. Use Euler s formula to solve problems. Draw and identify different views of solids. Draw and identify
More informationBASIC GEOMETRY GLOSSARY
BASIC GEOMETRY GLOSSARY Acute angle An angle that measures between 0 and 90. Examples: Acute triangle A triangle in which each angle is an acute angle. Adjacent angles Two angles next to each other that
More information(Q, ), (R, ), (C, ), where the star means without 0, (Q +, ), (R +, ), where the plussign means just positive numbers, and (U, ),
2 Examples of Groups 21 Some infinite abelian groups It is easy to see that the following are infinite abelian groups: Z, +), Q, +), R, +), C, +), where R is the set of real numbers and C is the set of
More informationChapter 7. Permutation Groups
Chapter 7 Permutation Groups () We started the study of groups by considering planar isometries In the previous chapter, we learnt that finite groups of planar isometries can only be cyclic or dihedral
More informationS on n elements. A good way to think about permutations is the following. Consider the A = 1,2,3, 4 whose elements we permute with the P =
Section 6. 1 Section 6. Groups of Permutations: : The Symmetric Group Purpose of Section: To introduce the idea of a permutation and show how the set of all permutations of a set of n elements, equipped
More informationENGINEERING DRAWING. UNIT I  Part A
ENGINEERING DRAWING UNIT I  Part A 1. Solid Geometry is the study of graphic representation of solids of  dimensions on plane surfaces of  dimensions. 2. In the orthographic projection,
More information13 Solutions for Section 6
13 Solutions for Section 6 Exercise 6.2 Draw up the group table for S 3. List, giving each as a product of disjoint cycles, all the permutations in S 4. Determine the order of each element of S 4. Solution
More informationGeometrical symmetry and the fine structure of regular polyhedra
Geometrical symmetry and the fine structure of regular polyhedra Bill Casselman Department of Mathematics University of B.C. cass@math.ubc.ca We shall be concerned with geometrical figures with a high
More informationChapter 1 Symmetry of Molecules p. 1  Symmetry operation: Operation that transforms a molecule to an equivalent position
Chapter 1 Symmetry of Molecules p. 11. Symmetry of Molecules 1.1 Symmetry Elements Symmetry operation: Operation that transforms a molecule to an equivalent position and orientation, i.e. after the operation
More informationShape Dictionary YR to Y6
Shape Dictionary YR to Y6 Guidance Notes The terms in this dictionary are taken from the booklet Mathematical Vocabulary produced by the National Numeracy Strategy. Children need to understand and use
More information3D shapes. Level A. 1. Which of the following is a 3D shape? A) Cylinder B) Octagon C) Kite. 2. What is another name for 3D shapes?
Level A 1. Which of the following is a 3D shape? A) Cylinder B) Octagon C) Kite 2. What is another name for 3D shapes? A) Polygon B) Polyhedron C) Point 3. A 3D shape has four sides and a triangular
More informationTeachers should read through the following activity ideas and make their own risk assessment for them before proceeding with them in the classroom.
Platonic solids investigation Teacher notes Teachers should read through the following activity ideas and make their own risk assessment for them before proceeding with them in the classroom. Platonic
More information2. If C is the midpoint of AB and B is the midpoint of AE, can you say that the measure of AC is 1/4 the measure of AE?
MATH 206  Midterm Exam 2 Practice Exam Solutions 1. Show two rays in the same plane that intersect at more than one point. Rays AB and BA intersect at all points from A to B. 2. If C is the midpoint of
More informationTopological Treatment of Platonic, Archimedean, and Related Polyhedra
Forum Geometricorum Volume 15 (015) 43 51. FORUM GEOM ISSN 15341178 Topological Treatment of Platonic, Archimedean, and Related Polyhedra Tom M. Apostol and Mamikon A. Mnatsakanian Abstract. Platonic
More informationA. 32 cu ft B. 49 cu ft C. 57 cu ft D. 1,145 cu ft. F. 96 sq in. G. 136 sq in. H. 192 sq in. J. 272 sq in. 5 in
7.5 The student will a) describe volume and surface area of cylinders; b) solve practical problems involving the volume and surface area of rectangular prisms and cylinders; and c) describe how changing
More informationThreedimensional finite point groups and the symmetry of beaded beads
Journal for Mathematics and the Arts, Vol. X, No. X, Month 2007, xxxxxx Threedimensional finite point groups and the symmetry of beaded beads G. L. FISHER* AND B. MELLOR California Polytechnic State
More informationBegin recognition in EYFS Age related expectation at Y1 (secure use of language)
For more information  http://www.mathsisfun.com/geometry Begin recognition in EYFS Age related expectation at Y1 (secure use of language) shape, flat, curved, straight, round, hollow, solid, vertexvertices
More informationUndergraduate Research Opportunity Programme in Science
Undergraduate Research Opportunity Programme in Science Polyhedra Name: Kavitha d/o Krishnan Supervisor: Associate Professor Helmer Aslaksen Department of Mathematics National University of Singapore 2001/2002
More informationChapter 18 Symmetry. Symmetry of Shapes in a Plane 18.1. then unfold
Chapter 18 Symmetry Symmetry is of interest in many areas, for example, art, design in general, and even the study of molecules. This chapter begins with a look at two types of symmetry of twodimensional
More informationPRACTICAL GEOMETRY SYMMETRY AND VISUALISING SOLID SHAPES NCERT
UNIT 12 PRACTICAL GEOMETRY SYMMETRY AND VISUALISING SOLID SHAPES (A) Main Concepts and Results Let a line l and a point P not lying on it be given. By using properties of a transversal and parallel lines,
More informationGeometry Vocabulary Booklet
Geometry Vocabulary Booklet Geometry Vocabulary Word Everyday Expression Example Acute An angle less than 90 degrees. Adjacent Lying next to each other. Array Numbers, letter or shapes arranged in a rectangular
More informationEvery 3dimensional shape has three measurements to describe it: height, length and width.
Every 3dimensional shape has three measurements to describe it: height, length and width. Height Width Length Faces A face is one of the flat sides of a threedimensional shape. Faces A cuboid has 6 flat
More informationAngles that are between parallel lines, but on opposite sides of a transversal.
GLOSSARY Appendix A Appendix A: Glossary Acute Angle An angle that measures less than 90. Acute Triangle Alternate Angles A triangle that has three acute angles. Angles that are between parallel lines,
More informationJunior Math Circles March 10, D Geometry II
1 University of Waterloo Faculty of Mathematics Junior Math Circles March 10, 2010 3D Geometry II Centre for Education in Mathematics and Computing Opening Problem Three tennis ball are packed in a cylinder.
More informationGeometry A Solutions. Written by Ante Qu
Geometry A Solutions Written by Ante Qu 1. [3] Three circles, with radii of 1, 1, and, are externally tangent to each other. The minimum possible area of a quadrilateral that contains and is tangent to
More informationSession 9 Solids. congruent regular polygon vertex. cross section edge face net Platonic solid polyhedron
Key Terms for This Session Session 9 Solids Previously Introduced congruent regular polygon vertex New in This Session cross section edge face net Platonic solid polyhedron Introduction In this session,
More informationGeometry Progress Ladder
Geometry Progress Ladder Maths Makes Sense Foundation Endofyear objectives page 2 Maths Makes Sense 1 2 Endofblock objectives page 3 Maths Makes Sense 3 4 Endofblock objectives page 4 Maths Makes
More informationof Nebraska  Lincoln
University of Nebraska  Lincoln DigitalCommons@University of Nebraska  Lincoln MAT Exam Expository Papers Math in the Middle Institute Partnership 712008 Archimedean Solids Anna Anderson University
More informationLine Segments, Rays, and Lines
HOME LINK Line Segments, Rays, and Lines Family Note Help your child match each name below with the correct drawing of a line, ray, or line segment. Then observe as your child uses a straightedge to draw
More informationDMATH Algebra I HS 2013 Prof. Brent Doran. Solution 5
DMATH Algebra I HS 2013 Prof. Brent Doran Solution 5 Dihedral groups, permutation groups, discrete subgroups of M 2, group actions 1. Write an explicit embedding of the dihedral group D n into the symmetric
More informationSection 12.1 Translations and Rotations
Section 12.1 Translations and Rotations Any rigid motion that preserves length or distance is an isometry (meaning equal measure ). In this section, we will investigate two types of isometries: translations
More informationand their Rotational Groups
Adam Siege1 Senior Capstone Dr. Papantonopoulou Spring 2007 and their Rotational Groups I. Introduction The symmetry group for a physical object is the set of ways that object can be repositioned so that
More informationMA.7.G.4.2 Predict the results of transformations and draw transformed figures with and without the coordinate plane.
MA.7.G.4.2 Predict the results of transformations and draw transformed figures with and without the coordinate plane. Symmetry When you can fold a figure in half, with both sides congruent, the fold line
More informationTransformations Packet Geometry
Transformations Packet Geometry 1 Name: Geometry Chapter 14: Transformations Date Due Section Topics Assignment 14.1 14.2 14.3 Preimage Image Isometry Mapping Reflections Note: Notation is different in
More informationHOMEWORK #3 SOLUTIONS  MATH 3260
HOMEWORK #3 SOLUTIONS  MATH 3260 ASSIGNED: FEBRUARY 26, 2003 DUE: MARCH 12, 2003 AT 2:30PM (1) Show either that each of the following graphs are planar by drawing them in a way that the vertices do not
More informationHyperbolic Islamic Patterns A Beginning
Hyperbolic Islamic Patterns A Beginning Douglas Dunham Department of Computer Science University of Minnesota, Duluth Duluth, MN 558122496, USA Email: ddunham@d.umn.edu Web Site: http://www.d.umn.edu/
More informationTeaching Guidelines. Knowledge and Skills: Can specify defining characteristics of common polygons
CIRCLE FOLDING Teaching Guidelines Subject: Mathematics Topics: Geometry (Circles, Polygons) Grades: 46 Concepts: Property Diameter Radius Chord Perimeter Area Knowledge and Skills: Can specify defining
More informationChapter 17. Mensuration of Pyramid
44 Chapter 7 7. Pyramid: A pyramid is a solid whose base is a plane polygon and sides are triangles that meet in a common vertex. The triangular sides are called lateral faces. The common vertex is also
More information12 Surface Area and Volume
12 Surface Area and Volume 12.1 ThreeDimensional Figures 12.2 Surface Areas of Prisms and Cylinders 12.3 Surface Areas of Pyramids and Cones 12.4 Volumes of Prisms and Cylinders 12.5 Volumes of Pyramids
More informationVertex : is the point at which two sides of a polygon meet.
POLYGONS A polygon is a closed plane figure made up of several line segments that are joined together. The sides do not cross one another. Exactly two sides meet at every vertex. Vertex : is the point
More informationNets of a cube puzzle
Nets of a cube puzzle Here are all eleven different nets for a cube, together with four that cannot be folded to make a cube. Can you find the four impostors? Pyramids Here is a squarebased pyramid. A
More information1. Determine all real numbers a, b, c, d that satisfy the following system of equations.
altic Way 1999 Reykjavík, November 6, 1999 Problems 1. etermine all real numbers a, b, c, d that satisfy the following system of equations. abc + ab + bc + ca + a + b + c = 1 bcd + bc + cd + db + b + c
More informationAcute triangulations of the double triangle
Acute triangulations of the double triangle Carol Zamfirescu Abstract. We prove that every doubly covered triangle can be triangulated with 12 acute triangles, and this number is best possible. Keywords:
More informationPROPERTIES OF TRIANGLES AND QUADRILATERALS
Mathematics Revision Guides Properties of Triangles, Quadrilaterals and Polygons Page 1 of 21 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier PROPERTIES OF TRIANGLES AND QUADRILATERALS
More informationLevel 1. AfL Questions for assessing Strand Five : Understanding Shape. NC Level Objectives AfL questions from RF
AfL Questions for assessing Strand Five : Understanding Shape NC Level Objectives AfL questions from RF Level 1 I can use language such as circle or bigger to describe the shape and size of solids and
More information1. By how much does 1 3 of 5 2 exceed 1 2 of 1 3? 2. What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3.
1 By how much does 1 3 of 5 exceed 1 of 1 3? What fraction of the area of a circle of radius 5 lies between radius 3 and radius 4? 3 A ticket fee was $10, but then it was reduced The number of customers
More informationBaltic Way 1995. Västerås (Sweden), November 12, 1995. Problems and solutions
Baltic Way 995 Västerås (Sweden), November, 995 Problems and solutions. Find all triples (x, y, z) of positive integers satisfying the system of equations { x = (y + z) x 6 = y 6 + z 6 + 3(y + z ). Solution.
More informationSYMMETRY MATHEMATICAL CONCEPTS AND APPLICATIONS IN TECHNOLOGY AND ENGINEERING
Daniel DOBRE, Ionel SIMION SYMMETRY MATHEMATICAL CONCEPTS AND APPLICATIONS IN TECHNOLOGY AND ENGINEERING Abstract: This article discusses the relation between the concept of symmetry and its applications
More informationSOLID SHAPES M.K. HOME TUITION. Mathematics Revision Guides Level: GCSE Higher Tier
Mathematics Revision Guides Solid Shapes Page 1 of 19 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier SOLID SHAPES Version: 2.1 Date: 10112015 Mathematics Revision Guides Solid
More information2 feet Opposite sides of a rectangle are equal. All sides of a square are equal. 2 X 3 = 6 meters = 18 meters
GEOMETRY Vocabulary 1. Adjacent: Next to each other. Side by side. 2. Angle: A figure formed by two straight line sides that have a common end point. A. Acute angle: Angle that is less than 90 degree.
More informationPlanarity Planarity
Planarity 8.1 71 Planarity Up until now, graphs have been completely abstract. In Topological Graph Theory, it matters how the graphs are drawn. Do the edges cross? Are there knots in the graph structure?
More informationLecture 4 DISCRETE SUBGROUPS OF THE ISOMETRY GROUP OF THE PLANE AND TILINGS
1 Lecture 4 DISCRETE SUBGROUPS OF THE ISOMETRY GROUP OF THE PLANE AND TILINGS This lecture, just as the previous one, deals with a classification of objects, the original interest in which was perhaps
More informationGeometry Chapter 1 Vocabulary. coordinate  The real number that corresponds to a point on a line.
Chapter 1 Vocabulary coordinate  The real number that corresponds to a point on a line. point  Has no dimension. It is usually represented by a small dot. bisect  To divide into two congruent parts.
More informationVisualize and Create 3D Geometry
Visualize and Create 3D Geometry A Creative Maker Project This project introduces you to base concepts of building 3D geometry with examples and walkthroughs. Big Idea: Learn to visualize and design 3D
More information56 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 224 points.
6.1.1 Review: Semester Review Study Sheet Geometry Core Sem 2 (S2495808) Semester Exam Preparation Look back at the unit quizzes and diagnostics. Use the unit quizzes and diagnostics to determine which
More informationABEL S THEOREM IN PROBLEMS AND SOLUTIONS
TeAM YYePG Digitally signed by TeAM YYePG DN: cn=team YYePG, c=us, o=team YYePG, ou=team YYePG, email=yyepg@msn.com Reason: I attest to the accuracy and integrity of this document Date: 2005.01.23 16:28:19
More informationA FAMILY OF BUTTERFLY PATTERNS INSPIRED BY ESCHER
15th INTERNATIONAL CONFERENCE ON GEOMETRY AND GRAPHICS 1 5 AUGUST, 2012, MONTREAL, CANADA A FAMILY OF BUTTERFLY PATTERNS INSPIRED BY ESCHER Douglas DUNHAM ABSTRACT: M.C. Escher is noted for his repeating
More informationThe Sepak Takraw Ball Puzzle
The Sepak Takraw Ball Puzzle Yutaka Nishiyama Department of Business Information, Faculty of Information Management, Osaka University of Economics, 2, Osumi Higashiyodogawa Osaka, 5338533, Japan nishiyama@osakaue.ac.jp
More information24HourAnswers.com. Online Homework. Focused Exercises for Math SAT. Skill Set 12: Geometry  2D Figures and 3D Solids
24HourAnsers.com Online Homeork Focused Exercises for Math SAT Skill Set 12: Geometry  2D Figures and 3D Solids Many of the problems in this exercise set came from The College Board, riters of the SAT
More informationPlatonic Solids. Some solids have curved surfaces or a mix of curved and flat surfaces (so they aren't polyhedra). Examples:
Solid Geometry Solid Geometry is the geometry of threedimensional space, the kind of space we live in. Three Dimensions It is called threedimensional or 3D because there are three dimensions: width,
More informationComputing the Symmetry Groups of the Platonic Solids With the Help of Maple
Computing the Symmetry Groups of the Platonic Solids With the Help of Maple Patrick J. Morandi Department of Mathematical Sciences New Mexico State University Las Cruces NM 88003 USA pmorandi@nmsu.edu
More informationSolutions to Exercises, Section 6.1
Instructor s Solutions Manual, Section 6.1 Exercise 1 Solutions to Exercises, Section 6.1 1. Find the area of a triangle that has sides of length 3 and 4, with an angle of 37 between those sides. 3 4 sin
More informationDate: Period: Symmetry
Name: Date: Period: Symmetry 1) Line Symmetry: A line of symmetry not only cuts a figure in, it creates a mirror image. In order to determine if a figure has line symmetry, a figure can be divided into
More informationChapter 8 Geometry We will discuss following concepts in this chapter.
Mat College Mathematics Updated on Nov 5, 009 Chapter 8 Geometry We will discuss following concepts in this chapter. Two Dimensional Geometry: Straight lines (parallel and perpendicular), Rays, Angles
More informationCharacteristics of Solid Figures Face Edge Vertex (Vertices) A shape is characterized by its number of... Faces, Edges, and Vertices Faces  6 Edges  12 A CUBE has... Vertices  8 Faces of a Prism The
More informationColorings, monodromy, and impossible triangulations
Colorings, monodromy, and impossible triangulations Ivan Izmestiev FU Berlin Computational geometry in noneuclidean spaces Nancy, August 6 8, 015 Impossible triangulations Theorem If a triangulation of
More informationChapter 5. Molecular Symmetry
Chapter 5. Molecular Symmetry Symmetry is present in nature and in human culture An introduction to symmetry analysis Symmetry perations and Elements Definitions Symmetry peration = a movement of a body
More informationBinary Operations. Discussion. Power Set Operation. Binary Operation Properties. Whole Number Subsets. Let be a binary operation defined on the set A.
Binary Operations Let S be any given set. A binary operation on S is a correspondence that associates with each ordered pair (a, b) of elements of S a uniquely determined element a b = c where c S Discussion
More informationActivity Set 4. Trainer Guide
Geometry and Measurement of Solid Figures Activity Set 4 Trainer Guide Mid_SGe_04_TG Copyright by the McGrawHill Companies McGrawHill Professional Development GEOMETRY AND MEASUREMENT OF SOLID FIGURES
More informationTopics Covered on Geometry Placement Exam
Topics Covered on Geometry Placement Exam  Use segments and congruence  Use midpoint and distance formulas  Measure and classify angles  Describe angle pair relationships  Use parallel lines and transversals
More informationWinter 2016 Math 213 Final Exam. Points Possible. Subtotal 100. Total 100
Winter 2016 Math 213 Final Exam Name Instructions: Show ALL work. Simplify wherever possible. Clearly indicate your final answer. Problem Number Points Possible Score 1 25 2 25 3 25 4 25 Subtotal 100 Extra
More information(a) 5 square units. (b) 12 square units. (c) 5 3 square units. 3 square units. (d) 6. (e) 16 square units
1. Find the area of parallelogram ACD shown below if the measures of segments A, C, and DE are 6 units, 2 units, and 1 unit respectively and AED is a right angle. (a) 5 square units (b) 12 square units
More informationLesson 21: Line of Symmetry and Rotational Symmetry
Lesson 21: Line of Symmetry and Rotational Symmetry Warmup 1. A(1, 6), B(4, 7), C(1, 3) R O,90 r x axis ( ABC) A B C A B C 2. Using the rules, determine the coordinates of the missing point. a) O,90 R
More informationSelected practice exam solutions (part 5, item 2) (MAT 360)
Selected practice exam solutions (part 5, item ) (MAT 360) Harder 8,91,9,94(smaller should be replaced by greater )95,103,109,140,160,(178,179,180,181 this is really one problem),188,193,194,195 8. On
More informationCSU Fresno Problem Solving Session. Geometry, 17 March 2012
CSU Fresno Problem Solving Session Problem Solving Sessions website: http://zimmer.csufresno.edu/ mnogin/mfdprep.html Math Field Day date: Saturday, April 21, 2012 Math Field Day website: http://www.csufresno.edu/math/news
More informationENUMERATION BY ALGEBRAIC COMBINATORICS. 1. Introduction
ENUMERATION BY ALGEBRAIC COMBINATORICS CAROLYN ATWOOD Abstract. Pólya s theorem can be used to enumerate objects under permutation groups. Using group theory, combinatorics, and many examples, Burnside
More informationM243. Fall Homework 2. Solutions.
M43. Fall 011. Homework. s. H.1 Given a cube ABCDA 1 B 1 C 1 D 1, with sides AA 1, BB 1, CC 1 and DD 1 being parallel (can think of them as vertical ). (i) Find the angle between diagonal AC 1 of a cube
More informationGeometry B Solutions. Written by Ante Qu
Geometry B Solutions Written by Ante Qu 1. [3] During chemistry labs, we oftentimes fold a diskshaped filter paper twice, and then open up a flap of the quartercircle to form a cone shape, as in the diagram.
More informationMaximizing Angle Counts for n Points in a Plane
Maximizing Angle Counts for n Points in a Plane By Brian Heisler A SENIOR RESEARCH PAPER PRESENTED TO THE DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE OF STETSON UNIVERSITY IN PARTIAL FULFILLMENT OF
More informationQuadrilaterals Properties of a parallelogram, a rectangle, a rhombus, a square, and a trapezoid
Quadrilaterals Properties of a parallelogram, a rectangle, a rhombus, a square, and a trapezoid Grade level: 10 Prerequisite knowledge: Students have studied triangle congruences, perpendicular lines,
More informationPolyhedra: Plato, Archimedes, Euler
Polyhedra: Plato, Archimedes, Euler Robert L. Benedetto Amherst College MathPath at Mount Holyoke College Tuesday, July 15, 2014 Regular Polygons Definition A polygon is a planar region R bounded by a
More informationUNIT H1 Angles and Symmetry Activities
UNIT H1 Angles and Symmetry Activities Activities H1.1 Lines of Symmetry H1.2 Rotational and Line Symmetry H1.3 Symmetry of Regular Polygons H1.4 Interior Angles in Polygons Notes and Solutions (1 page)
More informationMath 475, Problem Set #13: Answers. A. A baton is divided into five cylindrical bands of equal length, as shown (crudely) below.
Math 475, Problem Set #13: Answers A. A baton is divided into five cylindrical bands of equal length, as shown (crudely) below. () ) ) ) ) ) In how many different ways can the five bands be colored if
More informationGeometric Transformations
Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted
More informationSolutions to Problems for Mathematics 2.5 DV Date: April 14,
Solutions to Problems for Mathematics 25 DV Date: April 14, 2010 1 Solution to Problem 1 The equation is 75 = 12 6 + 3 Solution to Problem 2 The remainder is 1 This is because 11 evenly divides the first
More informationIf a question asks you to find all or list all and you think there are none, write None.
If a question asks you to find all or list all and you think there are none, write None 1 Simplify 1/( 1 3 1 4 ) 2 The price of an item increases by 10% and then by another 10% What is the overall price
More informationName Date Period. 3D Geometry Project
Name 3D Geometry Project Part I: Exploring ThreeDimensional Shapes In the first part of this WebQuest, you will be exploring what threedimensional (3D) objects are, how to classify them, and several
More informationConstructing Geometric Solids
Constructing Geometric Solids www.everydaymathonline.com Objectives To provide practice identifying geometric solids given their properties; and to guide the construction of polyhedrons. epresentations
More informationVector Notation: AB represents the vector from point A to point B on a graph. The vector can be computed by B A.
1 Linear Transformations Prepared by: Robin Michelle King A transformation of an object is a change in position or dimension (or both) of the object. The resulting object after the transformation is called
More informationGeometry Module 4 Unit 2 Practice Exam
Name: Class: Date: ID: A Geometry Module 4 Unit 2 Practice Exam Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which diagram shows the most useful positioning
More informationWorking in 2 & 3 dimensions Revision Guide
Tips for Revising Working in 2 & 3 dimensions Make sure you know what you will be tested on. The main topics are listed below. The examples show you what to do. List the topics and plan a revision timetable.
More informationalternate interior angles
alternate interior angles two nonadjacent angles that lie on the opposite sides of a transversal between two lines that the transversal intersects (a description of the location of the angles); alternate
More information