# Three applications of Euler s formula. Chapter 12

Save this PDF as:

Size: px
Start display at page:

Download "Three applications of Euler s formula. Chapter 12"

## Transcription

1 Three applications of Euler s formula Chapter 1 From: Aigner, Martin; Ziegler, Günter (009). Proofs from THE BOOK (4th ed.). Berlin, New York: Springer-Verlag. A graph is planar if it can be drawn in the plane R without crossing edges (or, equivalently, on the -dimensional sphere S ). We talk of a plane graph if such a drawing is already given and fixed. Any such drawing decomposes the plane or sphere into a finite number of connected regions, including the outer (unbounded) region, which are referred to as faces. Euler s formula exhibits a beautiful relation between the number of vertices, edges and faces that is valid for any plane graph. Euler mentioned this result for the first time in a letter to his friend Goldbach in 1750, but he did not have a complete proof at the time. Among the many proofs of Euler s formula, we present a pretty and self-dual one that gets by without induction. It can be traced back to von Staudt s book Geometrie der Lage from Euler s formula. If G is a connected plane graph with n vertices, e edges and f faces, then n e + f =. Proof. Let T E be the edge set of a spanning tree for G, that is, of a minimal subgraph that connects all the vertices of G. This graph does not contain a cycle because of the minimality assumption. We now need the dual graph G of G: To construct it, put a vertex into the interior of each face of G, and connect two such vertices of G by edges that correspond to common boundary edges between the corresponding faces. If there are several common boundary edges, then we draw several connecting edges in the dual graph. (Thus G may have multiple edges even if the original graph G is simple.) Consider the collection T E of edges in the dual graph that corresponds to edges in E\T. The edges in T connect all the faces, since T does not have a cycle; but also T does not contain a cycle, since otherwise it would separate some vertices of G inside the cycle from vertices outside (and this cannot be, since T is a spanning subgraph, and the edges of T and of T do not intersect). Thus T is a spanning tree for G. For every tree the number of vertices is one larger than the number of edges. To see this, choose one vertex as the root, and direct all edges away from the root : this yields a bijection between the non-root vertices and the edges, by matching each edge with the vertex it points at. Applied to the tree T this yields n = e T + 1, while for the tree T it yields f = e T + 1. Adding both equations we get n+f = (e T +1)+(e T +1) = e +. Leonhard Euler A plane graph G: n = 6, e = 10, f = 6 Dual spanning trees in G and in G

2 76 Three applications of Euler s formula The five platonic solids Euler s formula thus produces a strong numerical conclusion from a geometric-topological situation: the numbers of vertices, edges, and faces of a finite graph G satisfy n e+f = whenever the graph is or can be drawn in the plane or on a sphere. Many well-known and classical consequences can be derived from Euler s formula. Among them are the classification of the regular convex polyhedra (the platonic solids), the fact that K 5 and K 3,3 are not planar (see below), and the five-color theorem that every planar map can be colored with at most five colors such that no two adjacent countries have the same color. But for this we have a much better proof, which does not even need Euler s formula see Chapter 34. This chapter collects three other beautiful proofs that have Euler s formula at their core. The first two a proof of the Sylvester Gallai theorem, and a theorem on two-colored point configurations use Euler s formula in clever combination with other arithmetic relationships between basic graph parameters. Let us first look at these parameters. 5 4 The degree of a vertex is the number of edges that end in the vertex, where loops count double. Let n i denote the number of vertices of degree i in G. Counting the vertices according to their degrees, we obtain n = n 0 + n 1 + n + n (1) 5 Here the degree is written next to each vertex. Counting the vertices of given degree yields n = 3, n 3 = 0, n 4 = 1, n 5 =. 1 9 The number of sides is written into each region. Counting the faces with a given number of sides yields f 1 = 1, f = 3, f 4 = 1, f 9 = 1, and f i = 0 otherwise. 4 On the other hand, every edge has two ends, so it contributes to the sum of all degrees, and we obtain e = n 1 + n + 3n 3 + 4n () You may interpret this identity as counting in two ways the ends of the edges, that is, the edge-vertex incidences. The average degree d of the vertices is therefore d = e n. Next we count the faces of a plane graph according to their number of sides: a k-face is a face that is bounded by k edges (where an edge that on both sides borders the same region has to be counted twice!). Let f k be the number of k-faces. Counting all faces we find f = f 1 + f + f 3 + f (3) Counting the edges according to the faces of which they are sides, we get e = f 1 + f + 3f 3 + 4f (4) As before, we can interpret this as double-counting of edge-face incidences. Note that the average number of sides of faces is given by f = e f.

3 Three applications of Euler s formula 77 Let us deduce from this together with Euler s formula quickly that the complete graph K 5 and the complete bipartite graph K 3,3 are not planar. For a hypothetical plane drawing of K 5 we calculate n = 5, e = ( 5 ) = 10, thus f = e + n = 7 and f = e < 3. But if the average number of sides is smaller than 3, then the embedding would have a face with at most two sides, which cannot be. f = 0 7 Similarly for K 3,3 we get n = 6, e = 9, and f = e + n = 5, and thus f = e f = 18 5 < 4, which cannot be since K 3,3 is simple and bipartite, so all its cycles have length at least 4. It is no coincidence, of course, that the equations (3) and (4) for the f i s look so similar to the equations (1) and () for the n i s. They are transformed into each other by the dual graph construction G G explained above. From the double counting identities, we get the following important local consequences of Euler s formula. K 5 drawn with one crossing K 3,3 drawn with one crossing Proposition. Let G be any simple plane graph with n > vertices. Then (A) G has at most 3n 6 edges. (B) G has a vertex of degree at most 5. (C) If the edges of G are two-colored, then there is a vertex of G with at most two color-changes in the cyclic order of the edges around the vertex. Proof. For each of the three statements, we may assume that G is connected. (A) Every face has at least 3 sides (since G is simple), so (3) and (4) yield and f = f 3 + f 4 + f e = 3f 3 + 4f 4 + 5f and thus e 3f 0. Euler s formula now gives 3n 6 = 3e 3f e. (B) By part (A), the average degree d satisfies d = e n 6n 1 n < 6. So there must be a vertex of degree at most 5.

4 78 Three applications of Euler s formula Arrows point to the corners with color changes. (C) Let c be the number of corners where color changes occur. Suppose the statement is false, then we have c 4n corners with color changes, since at every vertex there is an even number of changes. Now every face with k or k + 1 sides has at most k such corners, so we conclude that 4n c f 3 + 4f 4 + 4f 5 + 6f 6 + 6f 7 + 8f f 3 + 4f 4 + 6f 5 + 8f f = (3f 3 + 4f 4 + 5f 5 + 6f 6 + 7f ) 4(f 3 + f 4 + f 5 + f 6 + f ) = 4e 4f using again (3) and (4). So we have e n + f, again contradicting Euler s formula. 1. The Sylvester Gallai theorem, revisited It was first noted by Norman Steenrod, it seems, that part (B) of the proposition yields a strikingly simple proof of the Sylvester Gallai theorem (see Chapter 10). The Sylvester Gallai theorem. Given any set of n 3 points in the plane, not all on one line, there is always a line that contains exactly two of the points. Proof. (Sylvester Gallai via Euler) If we embed the plane R in R 3 near the unit sphere S as indicated in our figure, then every point in R corresponds to a pair of antipodal points on S, and the lines in R correspond to great circles on S. Thus the Sylvester Gallai theorem amounts to the following: Given any set of n 3 pairs of antipodal points on the sphere, not all on one great circle, there is always a great circle that contains exactly two of the antipodal pairs. v Now we dualize, replacing each pair of antipodal points by the corresponding great circle on the sphere. That is, instead of points ±v S we consider the orthogonal circles given by C v := {x S : x, v = 0}. (This C v is the equator if we consider v as the north pole of the sphere.) Then the Sylvester Gallai problem asks us to prove: v C v Given any collection of n 3 great circles on S, not all of them passing through one point, there is always a point that is on exactly two of the great circles. But the arrangement of great circles yields a simple plane graph on S, whose vertices are the intersection points of two of the great circles, which divide the great circles into edges. All the vertex degrees are even, and they are at least 4 by construction. Thus part (B) of the proposition yields the existence of a vertex of degree 4. That s it!

5 Three applications of Euler s formula 79. Monochromatic lines The following proof of a colorful relative of the Sylvester Gallai theorem is due to Don Chakerian. Theorem. Given any finite configuration of black and white points in the plane, not all on one line, there is always a monochromatic line: a line that contains at least two points of one color and none of the other. Proof. As for the Sylvester Gallai problem, we transfer the problem to the unit sphere and dualize it there. So we must prove: Given any finite collection of black and white great circles on the unit sphere, not all passing through one point, there is always an intersection point that lies either only on white great circles, or only on black great circles. Now the (positive) answer is clear from part (C) of the proposition, since in every vertex where great circles of different colors intersect, we always have at least 4 corners with sign changes. 3. Pick s theorem Pick s theorem from 1899 is a beautiful and surprising result in itself, but it is also a classical consequence of Euler s formula. For the following, call a convex polygon P R elementary if its vertices are integral (that is, they lie in the lattice Z ), but if it does not contain any further lattice points. Lemma. Every elementary triangle = conv{p 0, p 1, p } R has area A( ) = 1. Proof. Both the parallelogram P with corners p 0, p 1, p, p 1 + p p 0 and the lattice Z are symmetric with respect to the map σ : x p 1 + p x, p p 1 + p p 0 which is the reflection with respect to the center of the segment from p 1 to p. Thus the parallelogram P = σ( ) is elementary as well, and its integral translates tile the plane. Hence {p 1 p 0, p p 0 } is a basis of the lattice Z, it has determinant ±1, P is a parallelogram of area 1, and has area 1. (For an explanation of these terms see the box on the next page.) Theorem. The area of any (not necessarily convex) polygon Q R with integral vertices is given by A(Q) = n int + 1 n bd 1, p 0 p 1 where n int and n bd are the numbers of integral points in the interior respectively on the boundary of Q. n int = 11, n bd = 8, so A = 14

6 80 Three applications of Euler s formula Lattice bases A basis of Z is a pair of linearly independent vectors e 1, e such that Z = {λ 1 e 1 + λ e : λ 1, λ Z}. Let e 1 = ( ) a b and e = ( c d), then the area of the parallelogram spanned by e 1 and e is given by A(e 1, e ) = det(e 1, e ) = det ( ) a c b d. If f1 = ( ) r s and f = ( t u) is another basis, then there exists an invertible Z-matrix Q with ( ) ( r t s u = a c b d) Q. Since QQ 1 = ( ), and the determinants are integers, it follows that detq = 1, and hence det(f 1, f ) = det(e 1, e ). Therefore all basis parallelograms have the same area 1, since A (( ( 1 0), 0 1)) = 1. Proof. Every such polygon can be triangulated using all the n int lattice points in the interior, and all the n bd lattice points on the boundary of Q. (This is not quite obvious, in particular if Q is not required to be convex, but the argument given in Chapter 35 on the art gallery problem proves this.) Now we interpret the triangulation as a plane graph, which subdivides the plane into one unbounded face plus f 1 triangles of area 1, so A(Q) = 1 (f 1). Every triangle has three sides, where each of the e int interior edges bounds two triangles, while the e bd boundary edges appear in one single triangle each. So 3(f 1) = e int +e bd and thus f = (e f) e bd +3. Also, there is the same number of boundary edges and vertices, e bd = n bd. These two facts together with Euler s formula yield and thus References f = (e f) e bd + 3 = (n ) n bd + 3 = n int + n bd 1, A(Q) = 1 (f 1) = n int + 1 n bd 1. [1] G. D. CHAKERIAN: Sylvester s problem on collinear points and a relative, Amer. Math. Monthly 77 (1970), [] D. EPPSTEIN: Nineteen proofs of Euler s formula: V E + F =, in: The Geometry Junkyard, [3] G. PICK: Geometrisches zur Zahlenlehre, Sitzungsberichte Lotos (Prag), Natur-med. Verein für Böhmen 19 (1899), [4] K. G. C. VON STAUDT: Geometrie der Lage, Verlag der Fr. Korn schen Buchhandlung, Nürnberg [5] N. E. STEENROD: Solution 4065/Editorial Note, Amer. Math. Monthly 51 (1944),

### Topological Treatment of Platonic, Archimedean, and Related Polyhedra

Forum Geometricorum Volume 15 (015) 43 51. FORUM GEOM ISSN 1534-1178 Topological Treatment of Platonic, Archimedean, and Related Polyhedra Tom M. Apostol and Mamikon A. Mnatsakanian Abstract. Platonic

### Arrangements And Duality

Arrangements And Duality 3.1 Introduction 3 Point configurations are tbe most basic structure we study in computational geometry. But what about configurations of more complicated shapes? For example,

### IMO Training 2010 Russian-style Problems Alexander Remorov

Solutions: Combinatorial Geometry 1. No. Call a lattice point even if the sum of its coordinates is even, and call it odd otherwise. Call one of the legs first, and other one second. Then every time a

### INCIDENCE-BETWEENNESS GEOMETRY

INCIDENCE-BETWEENNESS GEOMETRY MATH 410, CSUSM. SPRING 2008. PROFESSOR AITKEN This document covers the geometry that can be developed with just the axioms related to incidence and betweenness. The full

### Introduction to Graph Theory

Introduction to Graph Theory Allen Dickson October 2006 1 The Königsberg Bridge Problem The city of Königsberg was located on the Pregel river in Prussia. The river divided the city into four separate

### Induction Problems. Tom Davis November 7, 2005

Induction Problems Tom Davis tomrdavis@earthlin.net http://www.geometer.org/mathcircles November 7, 2005 All of the following problems should be proved by mathematical induction. The problems are not necessarily

### Mathematical Induction. Mary Barnes Sue Gordon

Mathematics Learning Centre Mathematical Induction Mary Barnes Sue Gordon c 1987 University of Sydney Contents 1 Mathematical Induction 1 1.1 Why do we need proof by induction?.... 1 1. What is proof by

### Homework Exam 1, Geometric Algorithms, 2016

Homework Exam 1, Geometric Algorithms, 2016 1. (3 points) Let P be a convex polyhedron in 3-dimensional space. The boundary of P is represented as a DCEL, storing the incidence relationships between the

### Actually Doing It! 6. Prove that the regular unit cube (say 1cm=unit) of sufficiently high dimension can fit inside it the whole city of New York.

1: 1. Compute a random 4-dimensional polytope P as the convex hull of 10 random points using rand sphere(4,10). Run VISUAL to see a Schlegel diagram. How many 3-dimensional polytopes do you see? How many

### 1. 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

### Sum of Degrees of Vertices Theorem

Sum of Degrees of Vertices Theorem Theorem (Sum of Degrees of Vertices Theorem) Suppose a graph has n vertices with degrees d 1, d 2, d 3,...,d n. Add together all degrees to get a new number d 1 + d 2

### Midterm Practice Problems

6.042/8.062J Mathematics for Computer Science October 2, 200 Tom Leighton, Marten van Dijk, and Brooke Cowan Midterm Practice Problems Problem. [0 points] In problem set you showed that the nand operator

### Graph Theory Lecture 3: Sum of Degrees Formulas, Planar Graphs, and Euler s Theorem Spring 2014 Morgan Schreffler Office: POT 902

Graph Theory Lecture 3: Sum of Degrees Formulas, Planar Graphs, and Euler s Theorem Spring 2014 Morgan Schreffler Office: POT 902 http://www.ms.uky.edu/~mschreffler Different Graphs, Similar Properties

### Divide-and-Conquer. Three main steps : 1. divide; 2. conquer; 3. merge.

Divide-and-Conquer Three main steps : 1. divide; 2. conquer; 3. merge. 1 Let I denote the (sub)problem instance and S be its solution. The divide-and-conquer strategy can be described as follows. Procedure

### Triangulation by Ear Clipping

Triangulation by Ear Clipping David Eberly Geometric Tools, LLC http://www.geometrictools.com/ Copyright c 1998-2016. All Rights Reserved. Created: November 18, 2002 Last Modified: August 16, 2015 Contents

### We call this set an n-dimensional parallelogram (with one vertex 0). We also refer to the vectors x 1,..., x n as the edges of P.

Volumes of parallelograms 1 Chapter 8 Volumes of parallelograms In the present short chapter we are going to discuss the elementary geometrical objects which we call parallelograms. These are going to

### Euclidean Geometry. We start with the idea of an axiomatic system. An axiomatic system has four parts:

Euclidean Geometry Students are often so challenged by the details of Euclidean geometry that they miss the rich structure of the subject. We give an overview of a piece of this structure below. We start

### Induction. Margaret M. Fleck. 10 October These notes cover mathematical induction and recursive definition

Induction Margaret M. Fleck 10 October 011 These notes cover mathematical induction and recursive definition 1 Introduction to induction At the start of the term, we saw the following formula for computing

### Geometry 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.

### 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

### 5.3 The Cross Product in R 3

53 The Cross Product in R 3 Definition 531 Let u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ] Then the vector given by [u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] is called the cross product (or

### Elementary triangle geometry

Elementary triangle geometry Dennis Westra March 26, 2010 bstract In this short note we discuss some fundamental properties of triangles up to the construction of the Euler line. ontents ngle bisectors

### Removing Even Crossings

EuroComb 2005 DMTCS proc. AE, 2005, 105 110 Removing Even Crossings Michael J. Pelsmajer 1, Marcus Schaefer 2 and Daniel Štefankovič 2 1 Department of Applied Mathematics, Illinois Institute of Technology,

### Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis

Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis 2. Polar coordinates A point P in a polar coordinate system is represented by an ordered pair of numbers (r, θ). If r >

### ON SELF-INTERSECTIONS OF IMMERSED SURFACES

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 12, December 1998, Pages 3721 3726 S 0002-9939(98)04456-6 ON SELF-INTERSECTIONS OF IMMERSED SURFACES GUI-SONG LI (Communicated by Ronald

### When is a graph planar?

When is a graph planar? Theorem(Euler, 1758) If a plane multigraph G with k components has n vertices, e edges, and f faces, then n e + f = 1 + k. Corollary If G is a simple, planar graph with n(g) 3,

### COMBINATORIAL PROPERTIES OF THE HIGMAN-SIMS GRAPH. 1. Introduction

COMBINATORIAL PROPERTIES OF THE HIGMAN-SIMS GRAPH ZACHARY ABEL 1. Introduction In this survey we discuss properties of the Higman-Sims graph, which has 100 vertices, 1100 edges, and is 22 regular. In fact

### Selected 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

### Pick s Theorem. Tom Davis Oct 27, 2003

Part I Examples Pick s Theorem Tom Davis tomrdavis@earthlink.net Oct 27, 2003 Pick s Theorem provides a method to calculate the area of simple polygons whose vertices lie on lattice points points with

### Tilings of the sphere with right triangles III: the asymptotically obtuse families

Tilings of the sphere with right triangles III: the asymptotically obtuse families Robert J. MacG. Dawson Department of Mathematics and Computing Science Saint Mary s University Halifax, Nova Scotia, Canada

### Centroid: The point of intersection of the three medians of a triangle. Centroid

Vocabulary Words Acute Triangles: A triangle with all acute angles. Examples 80 50 50 Angle: A figure formed by two noncollinear rays that have a common endpoint and are not opposite rays. Angle Bisector:

### The Orthogonal Art Gallery Theorem with Constrained Guards

The Orthogonal Art Gallery Theorem with Constrained Guards T. S. Michael 1 Mathematics Department United States Naval Academy Annapolis, MD, U.S.A. Val Pinciu 2 Department of Mathematics Southern Connecticut

### Adding vectors We can do arithmetic with vectors. We ll start with vector addition and related operations. Suppose you have two vectors

1 Chapter 13. VECTORS IN THREE DIMENSIONAL SPACE Let s begin with some names and notation for things: R is the set (collection) of real numbers. We write x R to mean that x is a real number. A real number

### ALGEBRA. sequence, term, nth term, consecutive, rule, relationship, generate, predict, continue increase, decrease finite, infinite

ALGEBRA Pupils should be taught to: Generate and describe sequences As outcomes, Year 7 pupils should, for example: Use, read and write, spelling correctly: sequence, term, nth term, consecutive, rule,

### arxiv:1409.4299v1 [cs.cg] 15 Sep 2014

Planar Embeddings with Small and Uniform Faces Giordano Da Lozzo, Vít Jelínek, Jan Kratochvíl 3, and Ignaz Rutter 3,4 arxiv:409.499v [cs.cg] 5 Sep 04 Department of Engineering, Roma Tre University, Italy

### Graphs without proper subgraphs of minimum degree 3 and short cycles

Graphs without proper subgraphs of minimum degree 3 and short cycles Lothar Narins, Alexey Pokrovskiy, Tibor Szabó Department of Mathematics, Freie Universität, Berlin, Germany. August 22, 2014 Abstract

### A convex polygon is a polygon such that no line containing a side of the polygon will contain a point in the interior of the polygon.

hapter 7 Polygons A polygon can be described by two conditions: 1. No two segments with a common endpoint are collinear. 2. Each segment intersects exactly two other segments, but only on the endpoints.

### Lecture 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

### Foundations of Geometry 1: Points, Lines, Segments, Angles

Chapter 3 Foundations of Geometry 1: Points, Lines, Segments, Angles 3.1 An Introduction to Proof Syllogism: The abstract form is: 1. All A is B. 2. X is A 3. X is B Example: Let s think about an example.

### Planar Graphs. Complement to Chapter 2, The Villas of the Bellevue

Planar Graphs Complement to Chapter 2, The Villas of the Bellevue In the chapter The Villas of the Bellevue, Manori gives Courtel the following definition. Definition A graph is planar if it can be drawn

### alternate interior angles

alternate interior angles two non-adjacent 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

### The minimum number of distinct areas of triangles determined by a set of n points in the plane

The minimum number of distinct areas of triangles determined by a set of n points in the plane Rom Pinchasi Israel Institute of Technology, Technion 1 August 6, 007 Abstract We prove a conjecture of Erdős,

### HOLES 5.1. INTRODUCTION

HOLES 5.1. INTRODUCTION One of the major open problems in the field of art gallery theorems is to establish a theorem for polygons with holes. A polygon with holes is a polygon P enclosing several other

### Zachary Monaco Georgia College Olympic Coloring: Go For The Gold

Zachary Monaco Georgia College Olympic Coloring: Go For The Gold Coloring the vertices or edges of a graph leads to a variety of interesting applications in graph theory These applications include various

### Chapter 1: Essentials of Geometry

Section Section Title 1.1 Identify Points, Lines, and Planes 1.2 Use Segments and Congruence 1.3 Use Midpoint and Distance Formulas Chapter 1: Essentials of Geometry Learning Targets I Can 1. Identify,

### HOW MANY HOLES DOES MY CAT HAVE?

HOW MANY HOLES DOES MY CAT HAVE? DAVID ROSENTHAL In memory of Professor Bill Watson Abstract. Euler s famous formula V E+F = 2 is one of the most elegant and powerful formulas in mathematics, and for this

### Synthetic Projective Treatment of Cevian Nests and Graves Triangles

Synthetic Projective Treatment of Cevian Nests and Graves Triangles Igor Minevich 1 Introduction Several proofs of the cevian nest theorem (given below) are known, including one using ratios along sides

### Graph Theory Problems and Solutions

raph Theory Problems and Solutions Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November, 005 Problems. Prove that the sum of the degrees of the vertices of any finite graph is

### SPERNER S LEMMA AND BROUWER S FIXED POINT THEOREM

SPERNER S LEMMA AND BROUWER S FIXED POINT THEOREM ALEX WRIGHT 1. Intoduction A fixed point of a function f from a set X into itself is a point x 0 satisfying f(x 0 ) = x 0. Theorems which establish the

### Examination paper for MA0301 Elementær diskret matematikk

Department of Mathematical Sciences Examination paper for MA0301 Elementær diskret matematikk Academic contact during examination: Iris Marjan Smit a, Sverre Olaf Smalø b Phone: a 9285 0781, b 7359 1750

### Three Lemmas in Geometry

Winter amp 2010 Three Lemmas in Geometry Yufei Zhao Three Lemmas in Geometry Yufei Zhao Massachusetts Institute of Technology yufei.zhao@gmail.com 1 iameter of incircle T Lemma 1. Let the incircle of triangle

### On planar regular graphs degree three without Hamiltonian cycles 1

On planar regular graphs degree three without Hamiltonian cycles 1 E. Grinbergs Computing Centre of Latvian State University Abstract. Necessary condition to have Hamiltonian cycle in planar graph is given.

### Chapter 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

### Max-Min Representation of Piecewise Linear Functions

Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 43 (2002), No. 1, 297-302. Max-Min Representation of Piecewise Linear Functions Sergei Ovchinnikov Mathematics Department,

### Classical theorems on hyperbolic triangles from a projective point of view

tmcs-szilasi 2012/3/1 0:14 page 175 #1 10/1 (2012), 175 181 Classical theorems on hyperbolic triangles from a projective point of view Zoltán Szilasi Abstract. Using the Cayley-Klein model of hyperbolic

### Labeling outerplanar graphs with maximum degree three

Labeling outerplanar graphs with maximum degree three Xiangwen Li 1 and Sanming Zhou 2 1 Department of Mathematics Huazhong Normal University, Wuhan 430079, China 2 Department of Mathematics and Statistics

### Solutions to Homework 10

Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x

### Orthic Quadrilaterals of a Convex Quadrilateral

Forum Geometricorum Volume 10 (2010) 79 91. FORUM GEOM ISSN 1534-1178 Orthic Quadrilaterals of a Convex Quadrilateral Maria Flavia Mammana, Biagio Micale, and Mario Pennisi Abstract. We introduce the orthic

### Euler s Theorem. Tom Davis July 14, 2009

Euler s Theorem Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles July 14, 2009 Abstract Euler s theorem is a nice result that is easy to investigate with simple models from Euclidean

### Solutions to Math 51 First Exam January 29, 2015

Solutions to Math 5 First Exam January 29, 25. ( points) (a) Complete the following sentence: A set of vectors {v,..., v k } is defined to be linearly dependent if (2 points) there exist c,... c k R, not

### Optimal 3D Angular Resolution for Low-Degree Graphs

Journal of Graph Algorithms and Applications http://jgaa.info/ vol. 17, no. 3, pp. 173 200 (2013) DOI: 10.7155/jgaa.00290 Optimal 3D Angular Resolution for Low-Degree Graphs David Eppstein 1 Maarten Löffler

### Surface Area of Rectangular & Right Prisms Surface Area of Pyramids. Geometry

Surface Area of Rectangular & Right Prisms Surface Area of Pyramids Geometry Finding the surface area of a prism A prism is a rectangular solid with two congruent faces, called bases, that lie in parallel

### PYTHAGOREAN TRIPLES KEITH CONRAD

PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient

### 28 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE. v x. u y v z u z v y u y u z. v y v z

28 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE 1.4 Cross Product 1.4.1 Definitions The cross product is the second multiplication operation between vectors we will study. The goal behind the definition

### Final Review Geometry A Fall Semester

Final Review Geometry Fall Semester Multiple Response Identify one or more choices that best complete the statement or answer the question. 1. Which graph shows a triangle and its reflection image over

### Solving Geometric Problems with the Rotating Calipers *

Solving Geometric Problems with the Rotating Calipers * Godfried Toussaint School of Computer Science McGill University Montreal, Quebec, Canada ABSTRACT Shamos [1] recently showed that the diameter of

### MATH 131 SOLUTION SET, WEEK 12

MATH 131 SOLUTION SET, WEEK 12 ARPON RAKSIT AND ALEKSANDAR MAKELOV 1. Normalisers We first claim H N G (H). Let h H. Since H is a subgroup, for all k H we have hkh 1 H and h 1 kh H. Since h(h 1 kh)h 1

### Topics 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

### a = bq + r where 0 r < b.

Lecture 5: Euclid s algorithm Introduction The fundamental arithmetic operations are addition, subtraction, multiplication and division. But there is a fifth operation which I would argue is just as fundamental

### Geometry Course Summary Department: Math. Semester 1

Geometry Course Summary Department: Math Semester 1 Learning Objective #1 Geometry Basics Targets to Meet Learning Objective #1 Use inductive reasoning to make conclusions about mathematical patterns Give

### A FIBONACCI TILING OF THE PLANE

A FIBONACCI TILING OF THE PLANE Charles W. Huegy Polygonics, 2 Mann St., Irvine, CA 92612-2709 Douglas B. West University of Illinois, Urbana, IL 61801-2975, west@math.uiuc.edu Abstract. We describe a

### 55 questions (multiple choice, check all that apply, and fill in the blank) The exam is worth 220 points.

Geometry Core Semester 1 Semester Exam Preparation Look back at the unit quizzes and diagnostics. Use the unit quizzes and diagnostics to determine which topics you need to review most carefully. The unit

### On end degrees and infinite cycles in locally finite graphs

On end degrees and infinite cycles in locally finite graphs Henning Bruhn Maya Stein Abstract We introduce a natural extension of the vertex degree to ends. For the cycle space C(G) as proposed by Diestel

### n 2 + 4n + 3. The answer in decimal form (for the Blitz): 0, 75. Solution. (n + 1)(n + 3) = n + 3 2 lim m 2 1

. Calculate the sum of the series Answer: 3 4. n 2 + 4n + 3. The answer in decimal form (for the Blitz):, 75. Solution. n 2 + 4n + 3 = (n + )(n + 3) = (n + 3) (n + ) = 2 (n + )(n + 3) ( 2 n + ) = m ( n

### Unknown Angle Problems with Inscribed Angles in Circles

: Unknown Angle Problems with Inscribed Angles in Circles Student Outcomes Use the inscribed angle theorem to find the measures of unknown angles. Prove relationships between inscribed angles and central

### THE NUMBER OF GRAPHS AND A RANDOM GRAPH WITH A GIVEN DEGREE SEQUENCE. Alexander Barvinok

THE NUMBER OF GRAPHS AND A RANDOM GRAPH WITH A GIVEN DEGREE SEQUENCE Alexer Barvinok Papers are available at http://www.math.lsa.umich.edu/ barvinok/papers.html This is a joint work with J.A. Hartigan

### Disjoint Compatible Geometric Matchings

Disjoint Compatible Geometric Matchings Mashhood Ishaque Diane L. Souvaine Csaba D. Tóth Abstract We prove that for every even set of n pairwise disjoint line segments in the plane in general position,

### Common Unfolding of Regular Tetrahedron and Johnson-Zalgaller Solid

Journal of Graph Algorithms and Applications http://jgaa.info/ vol. 20, no. 1, pp. 101 114 (2016) DOI: 10.7155/jgaa.0086 Common Unfolding of Regular Tetrahedron and Johnson-Zalgaller Solid Yoshiaki Araki

### Math 181 Handout 16. Rich Schwartz. March 9, 2010

Math 8 Handout 6 Rich Schwartz March 9, 200 The purpose of this handout is to describe continued fractions and their connection to hyperbolic geometry. The Gauss Map Given any x (0, ) we define γ(x) =

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### Incenter Circumcenter

TRIANGLE: Centers: Incenter Incenter is the center of the inscribed circle (incircle) of the triangle, it is the point of intersection of the angle bisectors of the triangle. The radius of incircle is

### 39 Symmetry of Plane Figures

39 Symmetry of Plane Figures In this section, we are interested in the symmetric properties of plane figures. By a symmetry of a plane figure we mean a motion of the plane that moves the figure so that

### THE DIMENSION OF A VECTOR SPACE

THE DIMENSION OF A VECTOR SPACE KEITH CONRAD This handout is a supplementary discussion leading up to the definition of dimension and some of its basic properties. Let V be a vector space over a field

### Just the Factors, Ma am

1 Introduction Just the Factors, Ma am The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive

### Mathematical Induction

Mathematical Induction (Handout March 8, 01) The Principle of Mathematical Induction provides a means to prove infinitely many statements all at once The principle is logical rather than strictly mathematical,

### Similarity and Diagonalization. Similar Matrices

MATH022 Linear Algebra Brief lecture notes 48 Similarity and Diagonalization Similar Matrices Let A and B be n n matrices. We say that A is similar to B if there is an invertible n n matrix P such that

### Unit 6 Coordinate Geometry

Mathematics I Frameworks Student Edition Unit 6 Coordinate Geometry 2 nd Edition Table of Contents Introduction:... 3 Video Game Learning Task... 6 New York Learning Task... 11 Quadrilaterals Revisited

Quadrilaterals / Mathematics Unit: 11 Lesson: 01 Duration: 7 days Lesson Synopsis: In this lesson students explore properties of quadrilaterals in a variety of ways including concrete modeling, patty paper

### Curriculum Map by Block Geometry Mapping for Math Block Testing 2007-2008. August 20 to August 24 Review concepts from previous grades.

Curriculum Map by Geometry Mapping for Math Testing 2007-2008 Pre- s 1 August 20 to August 24 Review concepts from previous grades. August 27 to September 28 (Assessment to be completed by September 28)

### 8 Square matrices continued: Determinants

8 Square matrices continued: Determinants 8. Introduction Determinants give us important information about square matrices, and, as we ll soon see, are essential for the computation of eigenvalues. You

### V. Adamchik 1. Graph Theory. Victor Adamchik. Fall of 2005

V. Adamchik 1 Graph Theory Victor Adamchik Fall of 2005 Plan 1. Basic Vocabulary 2. Regular graph 3. Connectivity 4. Representing Graphs Introduction A.Aho and J.Ulman acknowledge that Fundamentally, computer

### THREE DIMENSIONAL GEOMETRY

Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,

### E XPLORING QUADRILATERALS

E XPLORING QUADRILATERALS E 1 Geometry State Goal 9: Use geometric methods to analyze, categorize and draw conclusions about points, lines, planes and space. Statement of Purpose: The activities in this

### SOLVED PROBLEMS REVIEW COORDINATE GEOMETRY. 2.1 Use the slopes, distances, line equations to verify your guesses

CHAPTER SOLVED PROBLEMS REVIEW COORDINATE GEOMETRY For the review sessions, I will try to post some of the solved homework since I find that at this age both taking notes and proofs are still a burgeoning

### . 0 1 10 2 100 11 1000 3 20 1 2 3 4 5 6 7 8 9

Introduction The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive integer We say d is a

### CSU Fresno Problem Solving Session. Geometry, 17 March 2012

CSU Fresno Problem Solving Session Problem Solving Sessions website: http://zimmer.csufresno.edu/ mnogin/mfd-prep.html Math Field Day date: Saturday, April 21, 2012 Math Field Day website: http://www.csufresno.edu/math/news

### Geometric 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

### Inscribed Angle Theorem and Its Applications

: Student Outcomes Prove the inscribed angle theorem: The measure of a central angle is twice the measure of any inscribed angle that intercepts the same arc as the central angle. Recognize and use different