geometry of convex sets · 2015-10-12 · give a proof of borsuk’s problem in the plane using...

GEOMETRY OF CONVEX SETS

GEOMETRY OFCONVEX SETS

I. E. LEONARDDepartment of Mathematical and Statistical SciencesUniversity of Alberta, Edmonton, AB, Canada

J. E. LEWISDepartment of Mathematical and Statistical SciencesUniversity of Alberta, Edmonton, AB, Canada

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Leonard, I. Ed., 1938-Geometry of convex sets / I.E. Leonard, J.E. Lewis.

pages cmIncludes bibliographical references and index.ISBN 978-1-119-02266-4 (cloth)1. Convex sets. 2. Geometry. I. Lewis, J. E. (James Edward) II. Title.QA640.L46 2016516′.08–dc23

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Contents

Preface ix

1 Introduction to N-Dimensional Geometry 11.1 Figures in N-Dimensions 11.2 Points, Vectors, and Parallel Lines 2

1.2.1 Points and Vectors 21.2.2 Lines 41.2.3 Segments 111.2.4 Examples 121.2.5 Problems 18

1.3 Distance in N-Space 191.3.1 Metrics 191.3.2 Norms 201.3.3 Balls and Spheres 23

1.4 Inner Product and Orthogonality 291.4.1 Nearest Points 321.4.2 Cauchy–Schwarz Inequality 361.4.3 Problems 41

1.5 Convex Sets 411.6 Hyperplanes and Linear Functionals 45

1.6.1 Linear Functionals 451.6.2 Hyperplanes 521.6.3 Problems 66

2 Topology 692.1 Introduction 692.2 Interior Points and Open Sets 72

2.2.1 Properties of Open Sets 80

vi Contents

2.3 Accumulation Points and Closed Sets 832.3.1 Properties of Closed Sets 882.3.2 Boundary Points and Closed Sets 882.3.3 Closure of a Set 902.3.4 Problems 92

2.4 Compact Sets in R 942.4.1 Basic Properties of Compact Sets 992.4.2 Sequences and Compact Sets in R 1042.4.3 Completeness 106

2.5 Compact Sets in Rn 108

2.5.1 Sequences and Compact Sets in Rn 112

2.5.2 Completeness 1152.6 Applications of Compactness 117

2.6.1 Continuous Functions 1172.6.2 Equivalent Norms on R

n 1192.6.3 Distance between Sets in R

n 1212.6.4 Support Hyperplanes for Compact Sets in R

n 1272.6.5 Problems 130

3 Convexity 1353.1 Introduction 1353.2 Basic Properties of Convex Sets 137

3.2.1 Problems 1443.3 Convex Hulls 146

3.3.1 Problems 1553.4 Interior and Closure of Convex Sets 157

3.4.1 The Closed Convex Hull 1613.4.2 Accessibility Lemma 1623.4.3 Regularity of Convex Sets 1643.4.4 Problems 169

3.5 Affine Hulls 1703.5.1 Flats or Affine Subspaces 1703.5.2 Properties of Flats 1723.5.3 Affine Basis 1733.5.4 Problems 178

3.6 Separation Theorems 1803.6.1 Applications of the Separation Theorem 1923.6.2 Problems 196

Contents vii

3.7 Extreme Points of Convex Sets 1993.7.1 Supporting Hyperplanes and Extreme Points 1993.7.2 Existence of Extreme Points 2033.7.3 The Krein–Milman Theorem 2053.7.4 Examples 2073.7.5 Polyhedral Sets and Polytopes 2103.7.6 Birkhoff’s Theorem 2203.7.7 Problems 224

4 Helly’s Theorem 2274.1 Finite Intersection Property 227

4.1.1 The Finite Intersection Property 2274.1.2 Problems 229

4.2 Helly’s Theorem 2304.3 Applications of Helly’s Theorem 235

4.3.1 The Art Gallery Theorem 2354.3.2 Vincensini’s Problem 2424.3.3 Hadwiger’s Theorem 2494.3.4 Theorems of Radon and Carathéodory 2574.3.5 Kirchberger’s Theorem 2604.3.6 Helly-type Theorems for Circles 2624.3.7 Covering Problems 2664.3.8 Piercing Problems 2744.3.9 Problems 276

4.4 Sets of Constant Width 2774.4.1 Reuleaux Triangles 2774.4.2 Properties of Sets of Constant Width 2794.4.3 Adjunction Complete Convex Sets 2854.4.4 Sets of Constant Width in the Plane 2934.4.5 Barbier’s Theorem 2944.4.6 Constructing Sets of Constant Width 2974.4.7 Borsuk’s Problem 3044.4.8 Problems 309

Bibliography 311

Index 317

Preface

This book is designed for a one-semester course studying the geometry ofconvex sets in n-dimensional space. It is meant for students in education, arts,science, and engineering and is usually given in the third or fourth year ofuniversity. Prerequisites are courses in elementary geometry, linear algebra,and some familiarity with coordinate geometry.

The geometry of convex sets in n-dimensional space usually starts with thebasic definitions of the linear concepts of addition and scalar multiplicationof vectors and then defines the notion of convexity for subsets of then-dimensional space. Many properties of convex sets can be discoveredusing just the linear structure. However, in order to prove more interestingresults, we need the notion of a metric space, that is, the notion of distance, sothat we can talk about open and closed sets, bounded sets, compact sets, etc.It is the interplay between the linear and topological concepts that makes thenotion of convexity so interesting. Convexity can be defined using only thelinear structure, but it is the marriage of the linear structure and the topologicalstructure that allows us to see the really beautiful results.

In the first chapter, we introduce the linear or vector space notions of additionand scalar multiplication, linear subspaces, linear functionals, and hyperplanes.We also get a start on the convexity and topology by defining different distancesin n-space and studying the geometric properties of subsets, subspaces, andhyperplanes.

The second chapter discusses the notion of topology in the setting of metricsderived from a norm on the n-dimensional space. First, we define the basictopological notions of open sets and interior points, closed sets and accumu-lation points, boundary points, the interior of a set, the closure of a set, andthe boundary of a set. Next we discuss the basic properties of compact setsand give some geometric examples. We also give examples to show that inthe setting of an n-dimensional linear space which is a metric space, all thesenotions can be given in terms of convergence of sequences and subsequences.

x Preface

In the third chapter, we introduce the notion of convexity and discuss the basicproperties of convex sets. Here we define the convex hull of a set as well asthe interior and closure of convex sets. We prove Carathéodory’s theorem andshow that the convex hull of a compact set is compact. We discuss flats oraffine subspaces and their properties and go on to discuss the various separationtheorems for convex sets. Finally, we discuss the notions of extreme pointsand exposed points of convex sets, and polyhedral sets and polytopes, andprove the existence of extreme points for compact convex sets. We prove theKrein-Milman theorem and Birkhoff’s theorem on the extreme points of theset of n× n doubly stochastic matrices.

In the fourth and final chapter, we prove Helly’s theorem and give some appli-cations involving transversals of families of pairwise disjoint compact convexsubsets of the plane. We include a proof of Vicensini’s theorem and Klee’sreduction of the Helly number and give the results of Santalo and Hadwigeron finite families. We also discuss covering problems, piercing or stabbingproblems, and construction of sets of constant width in the plane. Finally, wegive a proof of Borsuk’s problem in the plane using Pál’s theorem and alsoMelzak’s proof of Borsuk’s problem for smooth sets of constant width in R

n.

The book is based on classroom notes for a course on the geometry of convexsets, a course in a sequence of four undergraduate geometry courses given atthe University of Alberta for the past 30 years. The choice of topics is basedon years of experience in teaching these courses and covers typical materialcovered in most introductory geometry courses on convex sets.

Both the authors were influenced by the texts by Hadwiger, Debrunner, andKlee [55], Eggleston [36], and Valentine [117] and later by the graduate textsof Holmes [61] and Wilansky [120]. We do not cover all of the material inthis book in a one-semester undergraduate course; however, the topics that wehave selected have all been included over the years.

Ed and Ted

Edmonton, Alberta, CanadaAugust, 2015

1 INTRODUCTION TON-DIMENSIONAL GEOMETRY

1.1 FIGURES IN N-DIMENSIONS

There is much to be said for a hands-on approach in geometry. We enhanceour understanding in two dimensions by drawing plane figures or by experi-menting with polygons cut out of paper. In three dimensions, we can constructpolyhedrons out of cardboard, wire, or plastic straws. In some cases, we mayuse a computer program to provide a visual representation. Although every-one knows that a picture does not constitute a “proof,” there is no doubt thata decent diagram can be utterly convincing.

However, we cannot make a full-dimensional model of an object that has morethan three dimensions. We cannot visualize a cube of four or five dimensions,at least not in the usual sense. Nevertheless, if you can imagine that “hy-percubes” of four or five dimensions exist, you might suspect that they are alot like three-dimensional cubes. You would be correct. Geometry in four orhigher dimensions is quite similar to geometry in two and three dimensions,and although we cannot visualize space of any more than three dimensions,we can build up a fairly reliable intuition about what such a space is like.

In this section, we examine some n-dimensional objects and learn how towork with them. We begin with the higher dimensional analogs of points,lines, planes, and spheres. We find out that these can be described in adimension-free way and that their geometric properties are as they should be.

We also examine how geometric notions such as perpendicular and parallellines extend to higher dimensions. The emphasis will be on how n-dimensionalspace is like two- or three-dimensional space, but we also examine some ofthe ways in which they differ.

Geometry of Convex Sets, First Edition. I. E. Leonard and J. E. Lewis.© 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc.

2 Introduction to N-Dimensional Geometry

Geometry at this level necessarily involves proofs. Even if you have alreadyhad a geometry course, you will find that some of the techniques are new. Inthis first section, we spend some time explaining how a proof proceeds beforeactually carrying it out. We apologize to anyone who is already familiar withthe methods.

1.2 POINTS, VECTORS, AND PARALLEL LINES

1.2.1 Points and Vectors

We denote the usual two-dimensional coordinate space by R2, three-dimen-

sional coordinate space by R3, four-dimensional coordinate space by R

4, andso on. In general, Rn denotes n-dimensional space (the n is not a variable—itstands for some fixed nonnegative integer) and is the set of all n-tuples of realnumbers, that is,

Rn = { (x1,x2, . . . ,xn) : xi ∈ R for i = 1, 2, . . . ,n }.

Points in Rn will be denoted by italic letters such as x, y, and z. Numbers are

usually (but not always) denoted with lowercase Greek letters, α, β, γ, and soon.∗

A point in Rn is just an n-tuple and can also be described by giving its

coordinates. In R4, for example, the ordered 4-tuple (α1,α2,α3,α4) denotes

the point whose ith coordinate is αi. The origin (0, 0, 0, 0) is denoted by 0(the bar is used to avoid confusion with the real number 0). The symbol 0 isalso used to denote the origin in any space R

n.

Points from the same space can be added together, subtracted from each other,and multiplied by scalars (that is, real numbers), and these operations areperformed coordinatewise.

∗ The Greek alphabet is as follows:

A α Alpha I ι Iota P ρ, � RhoB β Beta K κ Kappa Σ σ, ς SigmaΓ γ Gamma Λ λ Lambda T τ TauΔ δ Delta M μ Mu Υ υ UpsilonE ε, ε Epsilon N ν Nu Φ φ,ϕ PhiZ ζ Zeta Ξ ξ Xi X χ ChiH η Eta O o Omicron Ψ ψ PsiΘ θ Theta Π π,� Pi Ω ω Omega

The alternate pi (�), sigma (ς), and upsilon (υ) are very seldom used.

1.2 Points, Vectors, and Parallel Lines 3

If x = (α1,α2, . . . ,αn) and y = (β1,β2, . . . ,βn) are two points in Rn and if

ρ is any number, then we define

• x = y if and only if αi = βi for i = 1, 2, . . . ,n (equality)• x+ y = (α1 + β1,α2 + β2, . . . ,αn + βn) (addition)• ρx = (ρα1, ρα2, . . . , ραn) (scalar multiplication).

With these definitions of addition and scalar multiplication, Rn becomes ann-dimensional vector space.∗

In other words, points behave algebraically as though they were vectors. As aconsequence, a notation such as (1, 2) can be interpreted as a vector as well asa point. The two interpretations may be tied together geometrically by thinkingof the vector as an arrow whose tail is at the origin and whose tip is at thepoint (1, 2), as in the figure below.

or

x1

x2

x1

x2(1, 2) (1, 2)

A more abstract notion of a vector is as a class of arrows (an equivalence class,to be more exact), each arrow in the class having the same length and pointingin the same direction. Any particular arrow from the class of a given vector iscalled a representative of the vector. The arrow, in the example above, whosetail is at the point (0, 0) and whose tip is at the point (1, 2), is but one ofinfinitely many representatives of the vector, as in the figure below.

(0, 0) x1

x2(1, 2)

Note. Representatives of a given vector are called free vectors, since apartfrom their length and direction, there is no restriction as to their position.

Although a given n-tuple (α1,α2, . . . ,αn) has infinitely many free vectorsassociated with it, the n-tuple always represents one and only one point in R

n.When we refer to the point associated with a given vector, we mean that point.

∗ Unlike the situation in synthetic geometry, the notions of point and line are no longer primitive orundefined terms but are defined in terms of n-tuples (coordinatized space).


1.2.2 Lines

If we were to multiply the point v = (1, 2) by the numbers

−1, 12 , and 3,

we would get the points

(−1,−2), (12 , 1), and (3, 6),

respectively. The three points all lie on the straight line through 0 and v, asin the figure below.

(3, 6)

(−1,−2)

(0, 0)

(1, 2)

If μ is any number, then the point μv also lies on the straight line through 0and v. This is not just a property of R2: if v is any point other than 0 in R

1,R

2, or R3, then the point μv also lies on the straight line through 0 and v.

In fact, we can describe the line L through 0 and v algebraically as the set ofall multiples of the point v:

L = {μv : −∞ < μ < ∞}.

The line L passes through the origin.

To obtain a line M parallel to L, but passing through a given point p, wesimply add p to every point of L:

M = { p+ μv : −∞ < μ < ∞},

since p is on M (take μ = 0) and the vector v is parallel to L.

The above equations for L and M are also used to describe lines in spaces ofdimension n > 3. These equations, in fact, define what is meant by straightlines in higher dimensions.


Often a line is described by a vector equation. For example, a point x is onM if and only if

x = p+ μv,

for some real number μ. In this vector equation, p and v are fixed, and x is avariable point; thus, the line passing through the point p in the direction ofv can be written as

M = { x ∈ Rn : x = p+ μv, −∞ < μ < ∞}.

To give a concrete example, suppose that p = (1,−2, 4, 1) and v = (7, 8,−6, 5)are points in R

4. Denoting x by (x1,x2,x3,x4), we can describe the vectorequation for the line through p parallel to the vector v as the set of all pointsx ∈ R

4 such that

(x1,x2,x3,x4) = (1,−2, 4, 1) + μ(7, 8,−6, 5), −∞ < μ < ∞,

which is the same as

(x1,x2,x3,x4) = (1 + 7μ,−2 + 8μ, 4− 6μ, 1 + 5μ), −∞ < μ < ∞.

This is sometimes written in parametric form:⎧⎪⎪⎪⎨⎪⎪⎪⎩

x1 = 1 + 7μ

x2 = −2 + 8μ

x3 = 4− 6μ

x4 = 1 + 5μ,

where −∞ < μ < ∞.

Perhaps some words of caution are worthwhile at this point. Giving a nameto something does not endow it with any special properties. Simply becausewe have called an object in R

n by the same name as something from R2 or

R3, it does not follow that the n-dimensional object has the same properties

as its namesake in R2 or R

3. We cannot assume that what we are callingstraight lines in R

n automatically have the same properties as do straight linesin two or three dimensions. This has to be proved, which is what the next fewtheorems do.

Theorem 1.2.1. In the definition of the line

M = { x ∈ Rn : x = p+ μv, −∞ < μ < ∞},

the point p may be replaced by any point on M .


Proof. It is often worthwhile to rephrase a statement to make sure that weunderstand what it means. This theorem is saying that if p′ is any point onthe line M and if we replace p with p′ in the equation of M , then we will getexactly the same line.

In a situation similar to this, it helps to give different names to the potentiallydifferent lines. We will use M ′ to denote the line that we get when we replacep by p′:

M ′ = { x ∈ Rn : x = p′ + μv, −∞ < μ < ∞}.

Our task is to show that M and M ′ must be the same line.

How do we show that two lines are the same? Well, lines are just special typesof sets, and we will use the standard strategy for showing that two sets areequal—namely, we will show that every point in the first set also belongs tothe second set and vice versa.

(i) Note that a typical point x on M ′ can be written as

x = p′ + μv

for some number μ. We claim that x is also on M . To show that this isthe case, we have to show that there is some number λ such that x canbe written as p+ λv, which is a point on M .Now, since p′ is on M , there must be some number β such that

p′ = p+ βv,

and therefore,

x = p′ + μv

= (p+ βv) + μv

= p+ (β + μ)v.

Setting λ = β + μ shows that x is on M , and since x was an arbitrarypoint of M ′, then M ′ ⊆ M .

(ii) Conversely, we will show that every point on M is also on M ′. If welet y be a typical point on M , then

y = p+ μv

for some number μ. Using the fact that p′ is on M , we have

p′ = p+ βv

for some number β, so that p = p′ − βv. Therefore,

y = p+ μv

= (p′ − βv) + μv

= p′ + (−β + μ)v,


which shows that y is on M ′, and since y was an arbitrary point of M ,then M ⊆ M ′.

Thus, M = M ′, which completes the proof.

The proof of the next theorem is somewhat similar and is left as an exercise.

Theorem 1.2.2. In the definition of the line

M = { x ∈ Rn : x = p+ μv, −∞ < μ < ∞},

the vector v may be replaced by any nonzero multiple of v.

Two nonzero vectors are said to be parallel if one is a multiple of the other.With this terminology, the previous theorems can be combined as follows.

Corollary 1.2.3. In the definition of the line

M = { x ∈ Rn : x = p+ μv, −∞ < μ < ∞},

the point p may be replaced by any point on the line and the vector v may bereplaced by any vector parallel to v.

In the next theorem, we would like to show that if one line is a subset ofanother line, then the two lines must be the same. Before proving this, weshould convince ourselves that we actually have something to prove, since itlooks like this might be just another way of stating that a point and vectordetermine a unique straight line. Perhaps the two statements

(i) “A point and a vector determine a unique straight line”(ii) “If one line is contained in another, then the two lines must coincide”

are logically equivalent in the sense that one follows from the other withoutreally invoking any geometry.

To see that this is not the case, try replacing the word “line” by the words“solid ball.” It is true that a point and a vector determine a unique solid ball,namely, the ball with the point as its center and with a radius equal to thelength of the vector. It is clearly not true that two solid balls must coincide ifone is a subset of the other.


Having talked ourselves into believing that there is something to prove, thenext problem is to find a way to carry it out. This is how we will do it:

We will first suppose that M1 passes through the point p1 and is parallel to thenonzero vector v1 and that M2 passes through the point p2 and is parallel tothe nonzero vector v2. The assumption is that every point on M1 is containedin M2. We will show that this means that v1 and v2 are parallel, which iswhat our intuition suggests should be the case. We will then use the previoustheorems to finish the proof.

Theorem 1.2.4. If M1 and M2 are straight lines in Rn, and if M1 ⊂ M2,

then M1 = M2.

Proof. We may suppose that M1 and M2 are the lines

M1 = { x ∈ Rn : x = p1 + μv1, −∞ < μ < ∞}

M2 = { x ∈ Rn : x = p2 + μv2, −∞ < μ < ∞}.

Since M1 is a subset of M2, then p1 ∈ M1 ⊂ M2, and we can replace p2 inM2 with the point p1. We can then write M2 as

M2 = { x ∈ Rn : x = p1 + μv2, −∞ < μ < ∞}.

Again, using the fact that M1 ⊂ M2, the point p1+v1 of M1 must also belongto M2, and from the previous equation, there must be a number μ0 such that

p1 + v1 = p1 + μ0v2,

so that v1 = μ0v2. But this means that in the last equation for M2, we canreplace the vector v2 with the parallel vector v1, that is,

M2 = { x ∈ Rn : x = p1 + μv1, −∞ < μ < ∞},

and the right-hand side of this equation is precisely M1.

Theorem 1.2.5. Let M be the straight line given by

M = { x ∈ Rn : x = p+ μv, −∞ < μ < ∞}.

If x1 and x2 are distinct points on M , then the vector x1−x2 is parallel to v.

Proof. To prove this, one only needs to write down what x1 and x2 are interms of p and v and then perform the subtraction. Since each of the twopoints is on M , there must be numbers μ1 and μ2 such that

x1 = p+ μ1v and x2 = p+ μ2v,


and, therefore,

x1 − x2 = (p+ μ1v)− (p+ μ2v)

= (μ1 − μ2)v.

Since x1 and x2 are distinct, (μ1 − μ2)v �= 0, and if μ1 = μ2, then

(μ1 − μ2)v = 0 · v = 0.

Therefore, μ1 �= μ2, so that x1−x2 is a nonzero multiple of v, that is, x1−x2

is parallel to v.

Another way to describe a line is to specify points on it.

Theorem 1.2.6. A straight line in Rn is completely determined by any two

distinct points on the line.

Proof. This is proved using Theorem 1.2.5 and Corollary 1.2.3.

If p and q are distinct points on M , then M must be parallel to the vectorq − p by Theorem 1.2.5. Since p is on M , Corollary 1.2.3 now implies thatM must be the line

{ x ∈ Rn : x = p+ μ(q − p), −∞ < μ < ∞}.

In other words, p and q completely determine M .

In the proof of the previous theorem, we showed that if p and q are two distinctpoints in R

n, then the line determined by p and q will have the equation

x = p+ μ(q − p), −∞ < μ < ∞.

p

0

q

x


This equation may be rearranged to get the more usual form

x = (1− μ)p+ μq, −∞ < μ < ∞.

Note that if μ = 0, then x = p, while if μ = 1, then x = q. Since p and q areon the line, then the vector q − p must be parallel to the line.

Other equations that also describe the same line are

x = ηp+ μq, η + μ = 1

andx = μp+ (1− μ)q, −∞ < μ < ∞.

Example 1.2.7. If p = (−1,−1) and q = (1, 1), find the equation of the linepassing through p and q.

Solution. The equation of the line M passing through p and q can be writtenas

(x1,x2) = (1− μ)(−1,−1) + μ(1, 1)

= (−1 + 2μ,−1 + 2μ)

or parametrically as

x1 = −1 + λ

x2 = −1 + λ

for −∞ < λ < ∞.

As in R2 or R3, two distinct lines in R

n either do not intersect or they meetin exactly one point.

Theorem 1.2.8. Two distinct lines in Rn meet in at most one point.

Proof. Suppose that

M1 = { p+ μv : −∞ < μ < ∞}

andM2 = { q + μw : −∞ < μ < ∞}

are distinct lines in Rn and M1 ∩M2 �= ∅.


We may assume that the vectors v and w are not parallel. If they are paralleland M1 ∩ M2 �= ∅, then by Theorem 1.2.2, M1 and M2 coincide, whichcontradicts the fact that they are distinct.

Suppose that x0 and x1 are in M1 ∩M2, then

x0 − p = λ0v

x0 − q = μ0w

for some real numbers λ0 and μ0. Also,

x1 − p = λ1v

x1 − q = μ1w

for some numbers λ1 and μ1.

Therefore,p− q = −λ0v + μ0w = μ1w − λ1v,

so that(μ1 − μ0)w = (λ1 − λ0)v.

If μ0 �= μ1, then

w =

(λ1 − λ0

μ1 − μ0

)v,

which is a contradiction. Thus, μ0 = μ1.

Similarly, if λ0 �= λ1, then

v =

(μ1 − μ0

λ1 − λ0

)w,

which is a contradiction. Thus, λ0 = λ1.

Therefore,x1 = p+ λ1v = p+ λ0v = x0.

1.2.3 Segments

The part of a straight line between two distinct points p and q in Rn is called

a straight line segment, or simply a line segment.


x

0

p

q

• [p, q] denotes the closed line segment joining p and q

[p, q] = { x ∈ Rn : x = (1− μ)p+ μq : 0 ≤ μ ≤ 1 }.

• (p, q) denotes the open line segment joining p and q

(p, q) = { x ∈ Rn : x = (1 − μ)p+ μq : 0 < μ < 1 }.

• [p, q) and (p, q] denote the half open line segments joining p and q

[p, q) = { x ∈ Rn : x = (1− μ)p+ μq : 0 ≤ μ < 1 },

(p, q] = { x ∈ Rn : x = (1− μ)p+ μq : 0 < μ ≤ 1 }.

Note that as the scalar increases from μ = 0 to μ = 1, the point x movesalong the line segment from x = p to x = q.

1.2.4 Examples

Example 1.2.9. Show that the points

a = (−2,−2), b = (−1, 1), c = (1, 7)

are collinear.

Solution. Note that

b− a = (−1, 1)− (−2,−2) = (1, 3)

c− a = (1, 7)− (−2,−2) = (3, 9) = 3(1, 3)

so that c− a = 3(b− a) and b − a and c− a are parallel.

Therefore,x = a+ λ(b − a), −∞ < λ < ∞


andx = a+ λ(c− a), −∞ < λ < ∞

are the same line.

For λ = 0, x = a is on the line from the first equation, while for λ = 1, x = band x = c are on the line from the first and second equation, respectively.Therefore, a, b, and c are collinear.

Example 1.2.10. Let a and b be distinct points in Rn and let μ and ν be

scalars such that μ + ν = 1. Show that the point c = μa + νb is on the linethrough a and b.

Solution. Note that

c− a = μa+ νb− a

= νb− (1− μ)a

= νb− νa

= ν(b− a).

Therefore, c− a is parallel to b − a, and the points a, b, and c are collinear.

Example 1.2.11. Given distinct point a1 and a2 in Rn, the midpoint of the

segment [a1, a2] is given by12a1 +

12a2.

Solution. Since 12 + 1

2 = 1, then the point

12a1 +

12a2

is on the line joining a1 and a2.

Also,

12a1 +

12a2 − a1 = 1

2a1 −12a2

= 12 (a2 − a1)

so that12a1 +

12a2 = a1 +

12 (a2 − a1).


Also,

a2 −(12a1 +

12a2

)= 1

2a2 −12a1

= 12 (a2 − a1).

0

12a1 + 1

2a2

a2

a1

Therefore,12a1 +

12a2

is the midpoint of the segment [a1, a2].

Exercise 1.2.12. Given three noncollinear points a, b, and c in Rn, show that

the medians of the triangle with vertices a, b, and c intersect at a point G, thefamiliar centroid from synthetic geometry, and that

G = 13a+

13b+

13c.

In general, if a1, a2, . . . , ak are k points in Rn, where n > 2, we may define

the point1ka1 +

1ka2 + · · ·+ 1

kak

to be the centroid of the set { a1, a2, . . . , ak }. It is then obvious from theprevious examples that this definition agrees with the notion of the centroid ofa segment or a triangle, that is, for n = 1 or n = 2.

Example 1.2.13. Show that the point of intersection of the lines joining themidpoints of the opposite sides of a plane quadrilateral is the centroid of thevertices of the quadrilateral.

Solution. Let a, b, c, and d be the vertices of the quadrilateral.


.

a

w

b

y z

d

x

c

The centroid of the vertices is14(a + b + c + d)

The midpoints of the sides, in opposite pairs, are

w = 12a+ 1

2b, x = 12c+

12d

y = 12a+

12d, z = 1

2b+12c.

The midpoint of the segment [w,x] is

12w + 1

2x = 14 (a+ b + c+ d),

and the midpoint of the segment [y, z] is

12y +

12z = 1

4 (a+ b+ c+ d).

Thus, the segments [w,x] and [y, z] intersect at 14 (a+ b+ c+ d), the centroid

of the vertices of the quadrilateral.

Example 1.2.14. Given a triangle with vertices 0, a, b, as shown in the figurebelow,

12(a + b)

12b

a

12a

0 b

show that there exists a triangle whose sides are equal in length and parallelto the medians of the triangle.


Solution. The medians of the triangle are the following line segments:

from 0 to 12 (a+ b), from b to 1

2a, from a to 12b

and the (free) vectors corresponding to these sides are

u = 12 (a+ b), v = 1

2a− b, w = 12b− a

and we need only show that these vectors can be positioned to form atriangle.

Let one vertex of the triangle be 0, let a second vertex be the point u, and letthe third vertex be the point u+ v.

v

u

0

w

• Clearly, the edge from 0 to u is parallel to and equal in length to u, thefirst median.

• The second edge from u to u+ v is parallel to and equal in length to v,the second median.

• The third edge from u+ v to 0 is parallel to and equal in length to thevector u+ v, but

u+ v = 12 (a+ b) + 1

2a− b = a− 12b = −w,

so that the third side of the triangle is parallel to and equal in length tow, the third median.

Example 1.2.15. Given the quadrilateral [a, b, c, d] in the plane, the sides[a, b] and [c, d], when extended, meet at the point p. The sides [b, c] and [a, d],when extended, meet at the point q. On the rays from p through b and care points u and v so that [p,u] and [p, v] are of the same lengths as [a, b]and [c, d], respectively. On the rays from q through a and b are points xand y so that [q,x] and [q, y] are of the same lengths as [a, d] and [b, c],respectively.


x

d c

y

vp

q

ub

a

Show that [u, v] is parallel to [x, y].

Solution. In the figure, we have

u = p+ (a− b)

v = p+ (d− c)

x = q + (a− d)

y = q + (b − c),

so that

u− v = (a− b)− (d− c) = a+ c− (b + d)

x− y = (a− d)− (b − c) = a+ c− (b + d),

and [u, v] is parallel to [x, y].

Example 1.2.16. Let a, b, c, and d be the vertices of a tetrahedron in R3.

a

b

c

d

Show that the three lines through the midpoints of the opposite sides areconcurrent.


Solution. Note that in each case, the centroid of the tetrahedron

14 (a+ b+ c+ d)

is the midpoint of the segment.

For example, consider the opposite edges [a, b] and [c, d], their midpoints are12 (a+b) and 1

2 (c+d), and the midpoint of the segment[12 (a+ b), 1

2 (c+ d)]is

12

(12 (a+ b) + 1

2 (c+ d))= 1

4 (a+ b+ c+ d).

1.2.5 Problems

A remark about the exercises is necessary. Certain questions are phrased asstatements to avoid the incessant use of “prove that.” See Problem 1, forexample. Such statements are supposed to be proved. Other questions have a“true–false” or “yes–no” quality. The point of such questions is not to guess,but to justify your answer. Questions marked with ∗ are considered to bemore challenging. Hints are given for some problems. Of course, a hint maycontain statements that must be proved.

1. Let S be a nonempty set in Rn. If every three points of S are collinear,

then S is collinear.2. In R

2, there are two different types of equations that describe a straightline:(a) A vector equation: (x1,x2) = (α1,α2) + μ(β1,β2).(b) A linear equation: μ1x1 + μ2x2 = δ.Given that the line L has the vector equation

(x1,x2) = (4, 5) + μ(−3, 2),

find a linear equation for L.3. Given that the line L has the linear equation

μ1x1 + μ2x2 = δ,

show that the point (μ1δ

μ21 + μ2

2

,μ2δ

μ21 + μ2

2

)is on the line and that the vector (−μ2,μ1) is parallel to the line.Hint. If p is on the line and if p+ v is also on the line, then v must beparallel to the line.

geometry of convex sets · 2015-10-12 · give a proof of borsuk’s problem in the plane using...

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