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# Class Note for MATH 409 with Professor Martin at KU (3)

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Date Created: 02/06/15

Notes on polyhedra and 3 dimensional geometry Judith Roitman Jeremy Martin April 30 2009 1 Polyhedra Threedimensional geometry is a very rich eld Which we Will only taste slightly We Will focus on a particular kind of solid known as a polyhedron plural polyhedral lts characteristics are 0 it is made up of polygons glued together along their edges 0 it separates R3 into itself the space inside and the space outside the polygons it is made of are called facesi the edges of the faces are called the edges of the polyhedron the vertices of the faces are called the vertices of the polyhedroni The most familiar example of a polyhedron is a cube lts faces are squares and it has 6 of themi It also has 12 edges and 8 verticesi Another familiar example is a pyramid A pyramid has a bottom face Which can be any polygon you are probably most famiiar With pyramids that have square bottoms and the rest of its faces meet in one point Cube Square pyramid Pentagonal pyramid Problem 1 If the bottom face of a pyramid has n sides how many faces edges and vertices Will it have Another familiar example is a prism which is a polyhedron with two congruent parallel faces in which the other faces are rectangles The two congruent faces can be triangles quadrilaterals or anything else Another perhaps slightly less familiar example is a bipyramid which is built by taking two pyramids with congruent bases and gluing the bases together so that only the triangular faces are left Triangular prism Hexagonal prism Pentagonal bipyramid Problem 2 If one of the parallel faces of a prism has n sides how many faces edges and vertices will the prism have Problem 3 If the base or equator of a bipyramid has n sides how many faces edges and vertices will the bipyramid have While there are lots of different polyhedra they all have some common features just by Virtue of being polyhedra Theorem 1 In any polyhedron 0 Every vertex must lie in at least three faces Otherwise the polyhedron collapses to have no volume 0 Every face has at least three vertices It s a polygon so it better have at least three sides 0 Every edge must lie in exactly two faces Otherwise the polyhedron wouldn t have an inside and an outside As usual you can learn a lot by playing with nonstandard examples For example Problem 4 Can you construct a polyhedron with two parallel faces one a triangle the other a rectangle Problem 5 Can you construct a polyhedron so that exactly one face is not a quadrilateral Problem 6 Can you construct a polyhedron so that exactly one face is not a rectangle Problem 7 Can you construct a polyhedron in which every face is a hexagon 2 Euler s formula Let 1 e and f be the numbers of vertices edges and faces of a polyhedron For example if the polyhedron isacubethenv8e12 andf6 Problem 8 Make a table of the values for the polyhedra shown above as well as the ones you have built What do you notice You should observe that v 7 e f 2 for all these polyhedra This relationship is called Euler7s formula and it is vitally important in geometry topology and many other areas of mathematics The idea of adding up things in different dimensions counting even dimensions as positive and odd dimensions as negative just comes up everywhere Here is a cute proof of Euler7s formula from p 198 of George E Martin no relation Transformation Geometry An Introduction to Symmetry SpringerVerlag 1982 New York To prove the famous formula imagine that all the edges of a convex polyhedron are dikes exactly one face contains the raging sea and all other faces are dry We break dikes one at a time until all the faces are inundated following the rule that a dike is broken only if this results in the ooding of a face Now after this rampage we have ooded f 7 1 faces and so destroyed exactly f 7 1 dikes Noticing that we can walk with dry feet along the remaining dikes from any vertex p to any other vertex along exactly one path we conclude there is a onetoone correspondence between the remaining dikes and the vertices excluding 11 Hence there remain exactly 1 7 1 unbroken dikes So e 7 1 v 7 1 and we have proved Euler7s formula This is a vivid dramatic proof But is it correct If we take away the colorful imagery the proof boils down to four assertions about what happens after all the dikes are broken Assertion 1 We have ooded f 7 1 faces Assertion 2 We can walk with dry feet along the unbroken edges from any vertex to any other vertex Assertion 3 There is a unique path consisting of unbroken edges connecting any two vertices Assertion 4 There are exactly 1 7 1 unbroken edges Our task is to show that these assertions are correct and ii they imply Euler7s formula Proof of assertion 1 The only face we have not ooded is the one which originally contained the raging sea Proof of assertion 2 What if there is some pair of vertices that are cut off from each other Then there must have been some bridge B such that the dike network was connected just before B was broken and disconnected just after B was broken But that could only happen if the raging sea had already ooded both sides of B 7 and in that case we wouldn7t have broken B in the first place So it s impossible for there to be two mutually unwalkable between vertices Proof of assertion 3 Suppose by way of contradiction that there were two different paths between some pair of vertices But then those paths would enclose some un ooded region and that means that we didn t break enough bridges Proof of assertion 4 Pick a vertex 11 There are exactly 1 7 1 vertices other than p and for each other vertex 4 there is a unique edge that is the rst step from L to p lf 4 and q are different vertices then it cannot be the case that fq otherwise one or both of the paths to p involves doubling back On the other hand every edge is for w 4 because the network is connected so it s possible to walk from p towards that edge and eventually cross it That is the function f vertices other than p A edges is a bijection and so 6 v 7 1 21 Inequalities from Euler7s formula When combined with other observations about the numberse v e and f Eulerls formula has other conse quences Remember that the degree of a vertex is the number of edges attached to it For instance all vertices in a cube have degree 3 while all vertices in an octahedron have degree 4 The degrees don7t have to be the same in an arbitrary polhedron In a pentagonal pyramid see p 1 the apex of the pyramid has degree 5 while each of the base vertices has degree 3 If you add up all the degrees of vertices in a polyhedron in fact in any graph each edge will be counted twicei That is degree of vertex 1 degree of vertex 2 degree of vertex 1 26 1 If you add up the numbers of edges in all the faces of a polyhedra you will again count each edge twice because each edge lies in exactly two adjacent faces That is number of edges in face lnumber of edges in face 2 number of edges in face f 26 2 For example a cube has 8 vertices of degree 3 each so the sum of all degrees is 24 and 6 quadrilateral faces so the sum in 2 is also 24 and 12 242 edgesi Formula 1 goes by the name of the Handshaking Theorem in graph theory 7 if you think of the vertices as people and each edge as a handshake between two people then the degree of v is the number of people with whom v shakes hands so the theorem says that adding up those numbers for all people then dividing by 2 gives the total number of handshakes Problem 9 Verify that these general rules are true for your favorite polyhedral Now every face has to have at least 3 sides So the sum in 2 has to be at least Sfi this observation together with 2 it tells us that 2 3f S 26 or equivalently f S Eel Now substitute this inequality into Eulerls formula to get rid of the f 2vievaie2 e e v 7 g or 6 S 31 7 e or eS3v76i Also every vertex has to have degree at least 3 so the same calculation says that e S 3 f 7 6 or equivalently 2e 12 7lt677lt6i f f So what Well 2ef is the average number of edges in a face just because there are f faces in total and the sum of their numbers of edges is 2e Therefore we have proved Theorem 2 In every polyhedron the average number of sides in a face is less than 6 In particular its impossible to build a polyhedron all of whose faces are hexagonsl 3 Cubes cubes and more cubes What does a 4dimensional cube look like This is a scarysounding question 7 how can you see anything in 4dimensional space But if we work by analogy and use what we can see about ordinary ire visible 1 2 and 3 dimensional spaces then we can get a pretty good idea We know what a 3dimensional cube looks like What s a 2dimensional cube Its a square 7 the thing you get by pressing a cube at And if you press the squuare at you get a line segment a ldimensional cube and if you press the line segment at you get a single point an Odimensional cube Q3 Q2 Q1 1 Q0 squash gt lt T squash squash Of course in the preceding paragraph the word cube is being used more generally than it usually is its shorthand for thing that is analogous to a cube but happens to live in a differentdimensional space77 A common notation for the n dimensional cube is Q so Q0 is a point Q1 is a line segment Q2 is a square and Q3 is the familiar threedimensional cube We7ve seen how to make smallerdimensional cubes from bigger ones What about the reverse process To make a square out of a line segment Q1 you can make two copies of the line segment and attach each corresponding pair of vertices with a new line segment This same procedure lets you build a cube starting with a square Q2 or even a line segment from a point Q0 So what about Q4 By analogy you can build Q4 by starting with two copies of Q3 and attaching each corresponding pair of vertices with a line segment Its not so easy to see the symmetry in this picture7 but miraculously there is a beautiful and surprising other way to draw Q4i Make a 7 X 7 chessboard and place dots in the middle squares of the bottom and top rowsi Then draw in all the possible ways that a knight can move from one dot to the other in four moves The result is Q4 f6 ltltgt k gt ltgtgt 3 y XOR How many copies of Q3 are there inside Q4 Since 0 there are two Qols inside Q1 iiei7 a line segment has two points 0 there are four Q17s inside Q2 iiei7 a square has four sides 0 there are six Q27s inside Q3 iiei7 a cube has six faces if this pattern continues then the answer should be eight You can see this by coloring the edges of Q4 like this Now each set of three colors and there are four such sets can be used to make two Q37s For example here are the two Q37s with red green and blue but no purple edges In general how many copies of Qk sit inside Q If we call this number fn k and make a table of values we notice a variety of wonderful patterns lk0 k1k2 k3 k4 k5 n0 1 0 0 0 0 0 711 2 1 0 0 0 0 n2 4 4 1 0 0 0 n73 8 12 6 1 0 0 n4 16 32 24 8 1 0 n5 32 80 80 10 1 Here are some of the patterns 0 fn n 11 This is pretty simple everything contains one copy of itself 0 fn0 27h A point has 1 vertex a line segment has two a square has four a 3D cube has eight 1 1 ln general since you can build Qn out of two copies of Qn1 it makes sense that the number of vertices doubles each time o fn n 7 l 2n1 We already noticed this 0 What about fn 1 ie the number of edges iiei line segments This is the k 1 column of the table The pattern is not as obvious but it turns out to be fn l n 2 1 11 Here s why Each vertex in Qn has degree n and there are 2 vertices so adding up all the degrees gives n 27K This we know is twice the number of edges Therefore the number of edges re are n 2 2 n 27 1 edges 0 What about fn2 ie the number of squares in Q This is the k 2 column of the table and the pattern is even less obvious but fortunately we can use a similar counting technique First of all every vertex belongs to n edges and each of those pairs of edges forms a square So every vertex belongs to nn7l2 squares and multiplying this by the number of vertices gives 7 l2 2 nn 7 l 2 1 11 On the other hand we have now counted each square four times 7 once for each of its corner verticesi Therefore the actual number of squares in Q7 is nn 7 l 27 1 100172 4 nn 71 2 31 Using these observations we can ll out almost all of that table except for the question mark for f531 What about Eulerls formula In this notation the relation 1 7 e f 2 for the cube Q3 becomes 370 f371 f372 2 This is true 8 7 12 6 21 What if we look at the alternating sum of numbers for each row ignoring the 17s nl fl0 2 2 n2 f20 7 f21 474 0 n73 370 7 f371 7 1152 7 87126 7 2 0 n74 470 7 f471 f572 7 f573 7 16732247s n75 f570 7 f571 f572 7 f573 11574 327soso7v1o 7 If this pattern continues and it doesl then the alternating sum for n 5 should be 2 which means that the question mark should be f53 401 In fact this is a general rule about polyhedra in all dimensions For any n dimensional polyhedron the alternating sum number of vertices 0dimensional pieces 7 number of edges ldimensional pieces number of faces 2dimensional pieces 7 number of 3 dimensional pieces whatever they7re called i number of n 7 l dimensional pieces comes out to either 2 when n is odd or 0 when n is even 4 The Platonic solids De nition 1 A polyhedron is called regular or a Platonic solid iff a all of its faces are congruent b all of its faces are regular polygons and c each of its vertices meets the same number of edges as every other vertex So cubes are Platonic7 but most prisms and pyramids are not You may have heard of the following Platonic solids besides the cube 0 tetrahedron the faces are 4 equilateral triangles o octahedron 8 equilateral triangles o dodecahedron 12 regular pentagons o icosahedron 20 equilateral triangles Cube Tetmhedron Octahedron D odecahedron Icosahedron You havenlt heard of any others because i i Theorem 3 There are exactly ve Platonic solids This is a surprising theoremi Frequently when we de ne a nice class of objects in mathematics it is large in fact usually in nite think of the set of primes7 the set of isometries of the plane the collection of all regular polygons But this class is not only nite it has only ve things in it One part of the proof 7 that there are at least ve Platonic solids 7 is already done We know what they are In fact you7ve built them out of Zometooli So the interesting part is proving that there aren7t any more In order to do this7 you need to think carefully about how you t polygons together to make polyhedra 7 both the numerical observations that we made above see Theorem 1 and some geometric facts for example thinking about the angles that faces meet at Problem 10 Fold a piece of paper Cut it so that you have two congruent polygons joined at the fold Let p be a point at the edge of the fold and let Lm be the sides through p which do not coincide with the fold Fold and unfold the paper so that you narrow and expand the angle between E and m at p When is this angle the biggest possible If the two faces were a piece of a polyhedron would you need at least one more face at p Could you have more than one more face at p Problem 11 Fit some polygons together at a vertex p to start a polyhedron What is the sum of the angles touching p Now atten the part touching p you will have to cut at least one edge to do this No matter what your polyhedron is can the sum of the angles touching p sum up to more than 360 Can they sum up to exactly 360 Explain brie yi Platonic solids are very special because each vertex must belong to the same number of faces say n and each face must be a polygon with the same number of sides say 8 What can we say about these numbers We already know that s 2 3 and n 2 3 see Theorem 1 Each face F of 73 is a regular polygon with 8 sides and s angles The sum of the angles of F is 1803 7 2 so each single angle measures 1803 7 2amp Therefore if we consider the n faces that t together at a single vertex we see that their angles add up to 1803 7 2n 8 On the other hand we know that this quantity has to be lt 3600 If n and s are too large then this condition will fail So we can gure out all the possibilities just by brute force n 3 n 4 n 5 n 6 s 3 180 240 300 360 4 270 360 450 324 432 s 6 360 l l on There are evidently only ve possibilities o the faces are equilateral triangles s 3 the number n of faces meeting at a vertex may be 3 the tetrahedron 4 the octahedron or 5 the icosahedroni o the faces are squares s 4 the number of faces meeting at a vertex must be 3 the cube 0 the faces are regular pentagons s 5 the number of faces meeting at a vertex must be 3 the dodecahedroni This almost proves the theoremi Except how do you know that the parameters 8 and n determine the polytope Problem 12 Suppose you have a bunch of equilateral triangles squares or regular pentagonsi Suppose you are told how many of such faces each vertex of a regular polyhedron must meet Suppose your constraints are consistent with conclusion 2 Then there is exactly one way to construct the regular polyhedron and it must be one of those listed in conclusion 2 10

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