Ideas in Geometry Distributing This Document

Ideas in Geometry Distributing this Document

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This document was typeset on December 16, 2009. Preface

The aim of these notes is to convey the spirit of mathematical thinking as demonstrated through topics mainly from geometry. The reader must be careful not to forget this emphasis on deduction and visual reasoning. To this end, many questions are asked in the text that follows. Sometimes these questions are answered, other times the questions are left for the reader to ponder. To let the reader know which questions are left for cogitation, a large question mark is displayed: ? The instructor of the course will address some of these questions. If a question is not discussed to the reader’s satisfaction, then I encourage the reader to put on a thinking-cap and think, think, think! If the question is still unresolved, go to the World Wide Web and search, search, search! This document is licensed under the Creative Commons Attribution-ShareAlike (CC BY-SA) License. Loosely speaking, this means that this document is available for free. Anyone can get a free copy of this document (source or PDF) from the following site:

http://www.math.uiuc.edu/Courses/math119/

Please report corrections, suggestions, gripes, complaints, and criticisms to: [email protected] or [email protected]

Thanks and Acknowledgments

This document is based on a set of lectures originally given by Bart Snapp at the University of Illinois at Urbana-Champaign in the Fall of 2005 and Spring of 2006. Each semester since, these notes have been revised and modiﬁed. A growing number of instructors have made contributions, including Tom Cooney, Melissa Dennison, Jesse Miller, and Bart Snapp. Thanks to Alison Ahlgren, the Quantitative Reasoning Coordinator at the University of Illinois at Urbana-Champaign, for developing this course and for working with me and all the other instructors during the continual development of this document. Also thanks to Harry Calkins for help with Mathematica graphics and acting as a sounding-board for some of the ideas expressed in this document. In 2009, Greg Williams, a Master of Arts in Teaching student at Coastal Car- olina University, worked with Bart Snapp to produce the chapter on geometric transformations. A number of students have also contributed to this document by either typing or suggesting problems. They are: Camille Brooks, Michelle Bruno, Marissa Colatosti, Katie Colby, Anthony ‘Tino’ Forneris, Amanda Genovise, Melissa Peterson, Nicole Petschenko, Jason Reczek, Christina Reincke, David Seo, Adam Shalzi, Allice Son, Katie Strle, Beth Vaughn. Contents

1 Beginnings, Axioms, and Viewpoints 1 1.1 Euclid and Beyond ...... 1 1.1.1 The Most Successful Textbook Ever Written ...... 1 1.1.2 The Parallel Postulate ...... 5 1.2 PointsofView ...... 12 1.2.1 Synthetic Geometry ...... 13 1.2.2 Algebraic Geometry ...... 14 1.2.3 Analytic Geometry ...... 17 1.3 CityGeometry ...... 23 1.3.1 GettingWorkDone ...... 25 1.3.2 (Un)Common Structures ...... 28

2 Proof by Picture 38 2.1 BasicSetTheory ...... 38 2.1.1 Union ...... 39 2.1.2 Intersection ...... 39 2.1.3 Complement ...... 40 2.1.4 Putting Things Together ...... 41 2.2 Logic...... 45 2.3 Tessellations...... 51 2.3.1 Tessellations and Art ...... 52 2.4 ProofbyPicture ...... 56 2.4.1 Proofs Involving Right Triangles ...... 56 2.4.2 Proofs Involving Boxy Things ...... 60 2.4.3 Proofs Involving Sums ...... 62 2.4.4 Proofs Involving Sequences ...... 66 2.4.5 Thinking Outside the Box ...... 67

3 Topics in Plane Geometry 81 3.1 Triangles ...... 81 3.1.1 Centers in Triangles ...... 81 3.1.2 Theorems about Triangles ...... 85 3.2 Numbers...... 91 3.2.1 Areas ...... 91 3.2.2 Ratios...... 92 3.2.3 Combining Areas and Ratios—Probability ...... 100

4 Compass and Straightedge Constructions 113 4.1 Constructions...... 113 4.2 Trickier Constructions ...... 122 4.2.1 Challenge Constructions ...... 123 4.2.2 Problem Solving Strategies ...... 127 4.3 Constructible Numbers ...... 130 4.4 Impossibilities...... 140 4.4.1 Doubling the Cube ...... 140 4.4.2 Squaring the Circle ...... 140 4.4.3 Trisecting the Angle ...... 141

5 Transformations 144 5.1 Basic Transformations ...... 144 5.1.1 Translations...... 145 5.1.2 Reﬂections ...... 147 5.1.3 Rotations ...... 149 5.2 The Algebra of Transformations ...... 156 5.2.1 Matrix Multiplication ...... 156 5.2.2 Compositions of Transformations ...... 157 5.2.3 Mixing and Matching ...... 160 5.3 The Theory of Groups ...... 164 5.3.1 Groups of Reﬂections ...... 164 5.3.2 Groups of Rotations ...... 165 5.3.3 SymmetryGroups ...... 166

6 Convex Sets 170 6.1 BasicDeﬁnitions ...... 170 6.1.1 An Application ...... 173 6.2 Convex Sets in Three Dimensions ...... 177 6.2.1 Analogies to Two Dimensions ...... 177 6.2.2 Platonic Solids ...... 177 6.3 Ideas Related to Convexity ...... 182 6.3.1 The Convex Hull ...... 182 6.3.2 Sets of Constant Width ...... 182 6.4 Advanced Theorems ...... 187

References and Further Reading 189

Index 191 Chapter 1

Beginnings, Axioms, and Viewpoints

Since you are now studying geometry and trigonometry, I will give you a problem. A ship sails the ocean. It left Boston with a cargo of wool. It grosses 200 tons. It is bound for Le Havre. The mainmast is broken, the cabin boy is on deck, there are 12 passengers aboard, the wind is blowing East-North-East, the clock points to a quarter past three in the afternoon. It is the month of May. How old is the captain? —Gustave Flaubert

1.1 Euclid and Beyond 1.1.1 The Most Successful Textbook Ever Written Question Think of all the books that were ever written. What are some of the most inﬂuential of these? ? The Elements by the Greek mathematician Euclid of Alexandria should be high on this list. Euclid lived in Alexandria, Egypt, around 300 BC. His book, The Elements is an attempt to compile and write down everything that was known about geometry. This book is perhaps the most successful textbook ever written, having been used in nearly all universities up until the 20th century. Even today its heritage can be seen in scientiﬁc thought and writing. Here are three reasons this book is so important: (1) The Elements is of practical use. (2) The Elements contains powerful ideas.

1 1.1. EUCLID AND BEYOND

(3) The Elements provides a playground for the development of logical thought.

We’ll address each of these in turn.

The Elements is of practical use. Any time something large is built, some geometry must be used. The roads we drive on every day, the buildings we live in, the malls we shop at, the stadiums our favorite sports teams compete in; in short, if it is bigger than a shack, then geometry must have been used at some point. Moreover, geometry is crucial to modern transportation—in fact, any large scale transportation. An airplane could never make it to its destination without geometry, nor could any ship at sea. People of the past were faced with the diﬃculties of geometry on a continuing basis. Euclid’s The Elements was their handbook to solve everyday problems.

The Elements contains powerful ideas. Around 200 BC, the head librar- ian at the Great Library of Alexandria was a man by the name of Eratosthenes. Not only was Eratosthenes a great athlete, he was a scholar of astronomy, ethics, music, philosophy, poetry, theater, and important to this discussion, mathematics. His nickname was Beta, the second letter in the Greek alphabet. This is because with so many interests and accomplishments, he seemed to be second best at everything in the world. It came to Eratosthenes that every year, on the longest day of the year, at noon, sunlight would shine down to the bottom of a deep well located in present day city of Aswan, Egypt. Eratosthenes reasoned that this meant that the sun was directly overhead Aswan at this time. However, Eratosthenes knew that the sun was not directly overhead in Alexandria. He realized that the situation must be something like this:

Alexandria

Sun’s rays Aswan center of the Earth

Using ideas found in The Elements, Eratosthenes realized that if he drew imaginary lines from Alexandria and Aswan down to the center of the Earth, and if he could compute the angle between these lines, then to compute the circumference

2 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS of the Earth he would need only to solve the equation: total degrees in a circle circumference of the Earth = angle between the cities distance between the cities Thus Eratosthenes hired a man to pace the distance from Alexandria to Aswan. It was found to be about 5000 stadia. A stadia is an ancient unit of measurement which is the the length of a stadium. To measure the angle, he measured the angle of the shadow of a perpendicular stick in Alexandria, and found:

Sun’s rays stick

ground

Where the lower left angle is about 83◦ and the upper right angle is about 7◦. Thus we have 360 x 360 = 5000 = x. 7 5000 ⇒ 7 · So we see that x, the circumference of the Earth is about 250000 stadia. Unfortunately, as you may realize, the length of a stadium can vary. If the length of the stadium is deﬁned to be 157 meters, the length of an ancient Egyptian stadium, then it can be calculated that the circumference of the Earth is about 39250 kilometers. Considering the true amount is 40075 kilometers, Eratos- thenes made a truly remarkable measurement. However, the most important part is not that his measurement was close to being exactly right, but that his logic was correct. Question What did Eratosthenes assume when he made his measurement of the circumference of the Earth? ?

The Elements provides a playground for the development of logical thought. By answering the question raised above, we see it is necessary to understand what one assumes when doing science. When Euclid wrote The Elements, he started by stating his assumptions. By stating his assumptions, he gave rigor to his arguments. By focusing on the logical reasoning that goes into problem solving, Euclid put the method of solving a problem, and not merely the solution, into the spotlight. Euclid’s assumptions were stated as ﬁve axioms1. 1Actually, in Euclid’s time the word axiom was reserved for something obvious, a common notion, while postulate meant something to be assumed. However, in present day language we use the word axiom to mean something that is assumed. Henceforth we will always use the modern terminology.

3 1.1. EUCLID AND BEYOND

Definition An axiom is a statement that is accepted without proof. Euclid’s five axioms can be paraphrased as: (1) A line can be drawn from a point to any other point. (2) A finite line can be extended indefinitely. (3) A circle can be drawn, given a center and a radius. (4) All right angles are ninety degrees. (5) If a line intersects two other lines such that the sum of the interior angles on one side of the intersecting line is less than the sum of two right angles, then the lines meet on that side and not on the other side. The first four axioms are easy to understand, but the fifth is more complex. We should draw a picture describing the situation. Here is an example of how to draw pictures describing mathematical statements: Example Here is a picture describing the fifth axiom above:

α δ the lines don’t meet on this side the lines meet yonder

β γ

The fifth axiom says that if α + β is less than 180 degrees, the sum of two right angles, then the lines will meet on that side. Likewise the axiom says that if δ + γ is less than than 180 degrees, then the angles will meet on that side. The latter looks to be the case in the diagram above. One may wonder, what if we just ignore the Euclid’s 5th Axiom? By remov- ing or changing the fifth axiom (or any independent axiom) a different geometry is created. The sort of geometry that Euclid wrote about takes place on a plane. We call this sort of geometry Euclidean Geometry in honor of Euclid. By changing Euclid’s 5th Axiom, we stop doing geometry on the plane and start doing it on other types of surfaces, say spheres or other beasts. While The Elements may be the most successful textbook ever written, with over one thousand editions and over two thousand years of usage, there is still room for improvement. In the early 20th century, mathematicians pointed out that there are some logical flaws in the proofs that Euclid gives. David Hilbert, one of the great mathematicians of the 20th century, required around 20 axioms to prove all the theorems in The Elements. Nevertheless most of the theorems in The Elements are proved more-or-less correctly, and the text continues to have influence to this day.

4 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

1.1.2 The Parallel Postulate

Euclidean geometry seems to be a wonderful description of the universe in which we live. Is it really? How useful is it if you want to travel all the way around the world? What if you are standing in the middle of a city with large buildings obstructing your path? In each of these situations, a different sort of geometry is needed. Let’s look at new geometries that are different from but closely related to Euclidean geometry. The first four of Euclid’s axioms discussed in the previous section were always widely accepted. The fifth attracted more attention. For convenience’s sake, here is the fifth axiom again:

(5) If a line intersects two other lines such that the sum of the interior angles on one side of the line is less than the sum of two right angles, then the lines meet on that side and not on the other side.

Here are some other statements closely related to Euclid’s ﬁfth axiom:

(5A) Exactly one line can be drawn through any point not on a given line parallel to the given line.

(5B) The sum of the angles in every triangle is equal to 180◦.

(5C) If two lines ℓ1 and ℓ2 are both perpendicular to some third line, then ℓ1 and ℓ2 do not meet.

Question Can you draw pictures depicting these statements? Can you explain why Euclid’s ﬁfth axiom is sometimes called the parallel postulate? ?

Let’s replace Euclid’s ﬁfth axiom by the following:

(⋆) Given a point and a line, there does not exist a line through that point parallel to the given line.

There is a very natural geometry where this new axiom holds and the essences of the ﬁrst four also still hold. Instead of working with a plane, we now work on a sphere. We call this sort of geometry Spherical Geometry. Points, circles, angles, and distances are exactly what we would expect them to be. But what do we mean by lines on a sphere? Lines are supposed to be extended indeﬁnitely. In Spherical Geometry, the lines are the great circles.

Deﬁnition A great circle is a circle on the sphere with the same center as the sphere.

5 1.1. EUCLID AND BEYOND

Here is a picture to help you out. On the left, we have great circles drawn. On the right, we have regular old circles drawn.

A great circle cuts the sphere into two equal hemispheres. Great circles of the planet Earth include the equator and the lines of longitude. A great circle through a point P also goes through the point directly opposite to P on the sphere. This point is the called the antipodal point for P . For example, any great circle through the North Pole also goes through the South Pole. Question Why should we choose great circles to be the lines in Spherical Geometry? I’ll take this one. It is a theorem of Euclidean Geometry that the shortest path between any two points on a plane is given by a line segment. We have a similar theorem in Spherical Geometry. Theorem 1 The shortest path between any two points on a sphere is given by an arc of a great circle. This theorem is really handy! For one thing, it explains why an airplane ﬂies over Alaska when it is ﬂying from Chicago to Tokyo.

Many of the results from Euclid’s Elements still hold once we make suitable changes. For example, Euclid’s second axiom says that a ﬁnite line segment

6 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS can be extended. This idea still holds: Given a line segment (an arc of a great circle) we can extend it to a line (a great circle). However this line is no longer inﬁnite in length. It will loop around and meet itself after traveling around the circumference of the sphere. However, not all of our results will still hold. Question Are statements (5A), (5B), and (5C) true in Spherical Geometry? ? Let’s take a closer look at (5B). Question What is a triangle in Spherical Geometry? What is a polygon? ?

The picture above shows a triangle in Spherical Geometry. Here a region bounded by three line segments that meet at their endpoints. The sum of the angles in this triangle is clearly greater than 180◦, contradicting (5B). In Spherical Geometry, the sum of the angles in a triangle can be any number between 180◦ and 900◦. Question How is it that in Spherical Geometry the angles of a triangle can sum to any number between 180◦ and 900◦? ? You can go further and develop a whole theory of spherical trigonometry. This proved to be very important in cartography and navigation with huge rewards (more than $5,000,000 in today’s money) being offered to anyone who could devise a practical, accurate way of determining a ship’s location when it is in the middle of the ocean, a problem that was not deemed fully solved until 1828. Unless your journey is very short, the fact that the Earth is not flat makes a big difference.

7 1.1. EUCLID AND BEYOND

Question Suppose you replaced Euclid’s ﬁfth axiom by the statement: Given a line and a point not on that line, there exists multiple lines through that point parallel to the given line. What kind of geometry would this lead to? ?

8 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

Problems for Section 1.1 (1) Brieﬂy explain what Eratosthenes assumed when he computed the circumference of the Earth.

(2) Doug drove from Columbus, Ohio to Urbana, Illinois in 5 hours. The drive is almost exactly 300 miles. Deena says, “Doug, it looks like you were speeding.” Doug replies, “No, I was driving 60 miles per hour.”

(a) How did Doug come to his conclusion? (b) How did Deena come to her conclusion? (c) What assumptions were made? (d) Whose statement is correct? Explain your answer.

(3) Consider the following proposition of Euclid:

Given a line segment, one can construct an equilateral triangle with the line segment as its side.

Draw a picture depicting this statement and give a short explanation of how your picture depicts the above statement.

(4) Consider the following proposition of Euclid:

If two lines intersect, then the opposite angles at the intersection point are equal.

Draw a picture depicting this statement and give a short explanation of how your picture depicts the above statement.

(5) Consider the following proposition of Euclid:

In any triangle, the sum of the lengths of any two sides is greater than the length of the third.

Draw a picture depicting this statement and give a short explanation of how your picture depicts the above statement.

(6) Euclid’s fourth axiom states: “All right angles are ninety degrees.” This is not quite what Euclid said. Euclid said that a right angle is formed when two lines intersect and adjacent angles on either side of one of the lines are equal. In particular, Euclid asserted that the angles in every such case will be equal. Draw a picture depicting this statement and give a short explanation of how your picture depicts the above statement.

(7) Consider the following axiom of Hilbert:

9 1.1. EUCLID AND BEYOND

Let AB and BC be two segments of a line ℓ that have no points in common aside from the point B, and, furthermore, let A′B′ and B′C′ be two segments of another line ℓ′ having, likewise, no point other than B′ in common. If AB is the same length as A′B′ and BC is the same length as B′C′, then AC is the same length as A′C′. Draw a picture depicting this statement and give a short explanation of how your picture depicts the above statement.

(8) Consider the following axiom of Hilbert:

Let A, B, and C be three points not lying on the same line, and let ℓ be a straight line lying in the plane ABC and not passing through any of the points A, B, or C. Then, if the line ℓ passes through a point of the segment AB, it will also pass through either a point of the segment BC or a point of the segment AC.

Draw a picture depicting this statement and give a short explanation of how your picture depicts the above statement.

(9) Consider the following proposition:

If two lines ℓ1 and ℓ2 are both perpendicular to some third line, then ℓ1 and ℓ2 do not meet. Draw a picture depicting this statement and give a short explanation of how your picture depicts the above statement.

(10) State the deﬁnition of a great circle and compare/contrast it to a line in Euclidean geometry.

(11) In Spherical Geometry, what is the diﬀerence between a great circle and a regular Spherical Geometry circle?

(12) One way of writing Euclid’s ﬁrst axiom is “Any two distinct points determine a unique line.” Explain how you would alter this so that it holds in Spherical Geometry.

(13) One way of writing Euclid’s second axiom is “A ﬁnite line segment can be extended to an inﬁnite line.” Explain how you would alter this so that it holds in Spherical Geometry.

(14) One way of writing Euclid’s third axiom is “Given any point and any radius, a circle can be drawn with this center and this radius.” Explain how you would alter this so that it holds in Spherical Geometry.

(15) Explain why the following proposition from Euclidean geometry does not hold in Spherical Geometry: If two lines ℓ1 and ℓ2 are both perpendicular to some third line, then ℓ1 and ℓ2 do not meet.

10 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

(16) Explain why the following proposition from Euclidean geometry does not hold in Spherical Geometry: A triangle has at most one right angle. (Can you ﬁnd a triangle in Spherical Geometry with three right angles?)

(17) Explain why the following result from Euclidean geometry does not hold in Spherical Geometry: When the radius of a circle increases, its circumference also increases. (18) Deﬁne distinct lines to be parallel if they do not intersect. Can you have parallel lines in Spherical Geometry? Explain why or why not.

(19) Come up with a deﬁnition of a circle that will be true in both Euclidean and Spherical Geometry.

(20) Come up with a deﬁnition of a polygon that will be true in both Euclidean and Spherical Geometry.

(21) True or False. Explain your conclusions.

(a) Any two distinct lines in Spherical Geometry have at most one point of intersection. (b) All polygons in Spherical Geometry have at least three sides. (c) In Spherical Geometry there exist points arbitrarily far apart. (d) In Spherical Geometry all triangles have ﬁnite area. (e) In Spherical Geometry any two points can be connected by more than one line.

(22) A mathematician goes camping. She leaves her tent, walks one mile due south, then one mile due east. She then sees a bear before walking one mile north back to her tent. What color was the bear?

(23) The great German mathematician Gauss measured the angles of the triangle formed by the mountain peaks of Hohenhagen, Inselberg, and Brocken. What reasons might one have for doing this?

11 1.2. POINTS OF VIEW

1.2 Points of View

By studying geometry from different viewpoints we gain insight. Consider something very simple: U-shaped curves. U-shaped curves appear all the time in nature. Pick something up and toss it into the air. The object should follow a U-shaped path, we hope an upside down U-shape! Question Where else do U-shaped curves appear in nature? ? U-shaped curves often appear in mathematics. There are many different U- shaped curves, including catenary curves, hyperbolas, and most famous of all, parabolas. Question What is a parabola? We will answer the above question multiple times in the discussion that follows. Here is our first definition of a parabola: Definition A parabola is a set of points such that each of those points is the same distance from a given point as it is from a given line.

Question Why study parabolas? Well, the mirror in your makeup kit or reﬂecting telescope has parabolic cross-sections. The cables in suspension bridges approximate parabolas. But probably the most important application of parabolas is how they describe projectile motion:

Of course, if we are actually interested in projectile motion, we are probably most interested in two speciﬁc questions: (1) At a given time, how fast is the object moving? (2) At a given time, what direction is the object moving in?

12 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

Question Why are we interested in the above questions? ? The two questions above are directly related to the idea of a tangent line. So if we are interested in the two above questions, then we are interested in tangent lines. Question What is a tangent line? ? So we want to know about parabolas and tangent lines of parabolas. We will now look at these ideas in diﬀerent ways using: (1) Synthetic Geometry. (2) Algebraic Geometry. (3) Analytic Geometry.

1.2.1 Synthetic Geometry When you study geometry without the use of a coordinate system (that is, an (x,y)-plane) you are studying synthetic geometry. Good examples of this are when you study properties of triangles, circles, compass and straightedge constructions, or any other idea that goes back to classical Greek geometry. Deﬁnition When studying synthetic geometry, the classical way to deﬁne a parabola is as a special slice of a cone:

Hence people often refer to a parabola as a conic-section.

13 1.2. POINTS OF VIEW

Question What curves do you get if you cut the cone some other way? ? Now how do we study lines that are tangent to some slice of a cone? I don’t know! But here is another way to think about a parabola using synthetic geometry that makes the job easy. Check this out:

How do we know that the above picture is a parabola? Well, we would need to prove this, but we will not do that here. So if you accept that the above picture is a parabola, then you can just see those tangent lines. But this interpretation really doesn’t help us solve problems. We need a more sophisticated approach.

1.2.2 Algebraic Geometry In the 1600s, Rene Descartes revolutionized geometry. Descartes was a philoso- pher and a mathematician. Outside of mathematics, he is most famous for his phrase: Je pense, donc je suis. Which is often translated as: I think, therefore I am. With that statement, Descartes was laying the foundations for his future arguments on the nature of the universe around him, with his ﬁrst argument being that he, the arguer, actually exists. This rigor that Descartes employs is no doubt inherited from Euclid and other Greek mathematicians. However, Descartes’ connection to geometry does not stop there. Descartes is best known in mathematics as the inventor of the (x,y)-plane, also called the Cartesian plane in his honor. The (x,y)-plane was a brilliant breakthrough as it allowed geometry to be combined with algebra in ways that were not previously imagined. Question What is a parabola?

14 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

Deﬁnition Algebraically, a parabola is the graph of:

y = ax2 + bx + c

Question What is a tangent line?

To answer this question, we must ask ourselves, “how do lines go through parabolas?” Look at this:

Of the lines that intersect a parabola, most go through two points—ignore vertical lines. Take one of the good lines, one that intersects the parabola at two points. We can slide this line around, without changing the slope, until it intersects the parabola at only one point. This line is tangent to the parabola.

Question A tangent line will go through a parabola at exactly one point. Is this true for tangent lines of all curves? ?

Question If line(x) is a line that goes through the parabola, how many roots does the equation parabola(x) = line(x) have? How many roots does

parabola(x) line(x) = 0 − have? ?

15 1.2. POINTS OF VIEW

Example So suppose you wish to ﬁnd the line tangent to the parabola y = x2 at x = 2. To do this, write

x2 line(x) = (x 2)(x 2), − − − since x2 line(x) must have a double-root at x = 2. Now we see that − x2 line(x) = x2 4x + 4 − − line(x) = 4x 4. − So, the line tangent to the parabola y = x2 at x = 2 is line(x) = 4x 4. − What if you want to ﬁnd the line tangent to a higher degree polynomial? In that case, life gets a bit harder, but not impossible. Again we need to ask the question:

Question What is a tangent line?

I’ll take this one. Above we see that a tangent line to a parabola is a line that passes through the parabola in such a way that

parabola(x) = line(x) has a double-root. So we’ll make the following deﬁnition:

Deﬁnition Algebraically speaking, a tangent line is a line that passes through a curve such that curve(x) = line(x) has a double-root.

Example Suppose you wish to ﬁnd the line tangent to the curve y = x3 at x = 2. To do this write

x3 line(x) = (x 2)(x 2)(x c), − − − − where c is some unknown constant. We know that x3 line(x) factors with two (x 2) since 2 must be a double-root. Now write − − x3 line(x) = (x2 4x + 4)(x c) − − − x3 line(x) = x3 4x2 + 4x cx2 + 4cx 4c. − − − − Since there are no terms involving x2 on the left-hand side of the equals sign, we see that

4x2 cx2 = 0 − − 4x2 = cx2 − 4 = c. −

16 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

Plugging this back in, we ﬁnd x3 line(x) = x3 4x2 + 4x ( 4)x2 + 4( 4)x 4( 4) − − − − − − − x3 line(x) = x3 4x2 + 4x + 4x2 16x + 16 − − − x3 line(x) = x3 12x + 16 − − line(x) = 12x 16. − So we see that the line tangent to the curve y = x3 at x = 2 is line(x) = 12x 16. − Question What are the limitations of this method? ?

1.2.3 Analytic Geometry Let me tell you a story: A young man graduates from college at the age of 23. He graduated without honors and without distinction. He then traveled to the country to meditate on, among other things, the following question: Question If we know the position of an object at every instant of time, shouldn’t we know its velocity? This question helped lead this man to develop Calculus. The year was 1665 and the man was Isaac Newton. OK, but what does this have to do with what we’ve been talking about? It turns out that if you have a line that represents the position of an object, then the slope of that line is the velocity of the object. So now what we want to do is look at the “slope” of a curve. How do we do this? We look at the slope of the line tangent to the curve. Fine—but how do you do that? Here’s the idea: Suppose you want to ﬁnd the slope of the following curve at x = a. Look at:

a a+s

That s up there stands for a small (near zero) number. So if we look at the slope of that line we ﬁnd f(a + s) f(a) slope (a, s) = − . f(x) s

17 1.2. POINTS OF VIEW

If we want to ﬁnd the slope of the tangent line at x = a, all we do is plug in values for s that get closer and closer to zero. Example Suppose you need to ﬁnd the slope of the line tangent to the curve f(x) = x2 at the point x = 2. So you write (2 + s)2 22 slope (2,s) = − . f(x) s Now you plug in values for s that approach zero. Look at this: s = 0.1 slope (2, 0.1) = 4.1 ⇒ f(x) s = 0.01 slope (2, 0.01) = 4.01 ⇒ f(x) s = 0.001 slope (2, 0.001) = 4.001 ⇒ f(x) Ah! It looks like as s gets really close to zero that slopef(x)(2,s) = 4. Now we should check the slope when s is a small negative number.

Question What is slopef(x)(2,s) when s is a small negative number? ? Let’s see another example: Example Suppose you need to ﬁnd the slope of the line tangent to the curve g(x) = x3 at the point x = 2. So you write (2 + s)3 23 slope (2,s) = − . g(x) s Now you plug in values for s that approach zero. Look at this: s = 0.1 slope (2, 0.1) = 12.61 ⇒ g(x) s = 0.01 slope (2, 0.01) = 12.0601 ⇒ g(x) s = 0.001 slope (2, 0.001) = 12.006001 ⇒ g(x) Ah! It looks like as s gets really close to zero that slopeg(x)(2,s) = 12. Now we should check the slope when s is a small negative number.

Question What is slopeg(x)(2,s) when s is a small negative number? ? If you think that the above method is a bit sloppy and imprecise, then you are correct. How do you clean up this sloppiness? You must learn the martial art known as Calculus! Question Can you come up with a function f(x) (a sketch will suﬃce) where

slopef(x)(2, 0.1), slopef(x)(2, 0.01), slopef(x)(2, 0.001), do not approach the slope of f(x) at x = 2? ?

18 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

Old Enemies In a previous math course, you may have come upon the mysterious function:

f(x) = ex

What’s the deal with this? Some common answers are:

e is easy to work with. • e appears naturally in real world problems. • I don’t know about you, but I was never satisﬁed by answers like those above. Here’s the real deal: ex is a function such that

a slope x (a, s) e e ≈ as s gets smaller and smaller. In fact, you can get slopeex (a, s) to be as close to ea as you want!

Question Suppose that for some number a, ea = 0. What would you conclude about ex then? ? Question What does the graph of ex look like? ? Question Can you explain ex in terms of:

Driving. • Speed-limit signs. • Mile-marker signs. • Hint: What would happen if you got the mile-marker signs confused with the speed-limit signs? ?

19 1.2. POINTS OF VIEW

Problems for Section 1.2 (1) Explain the differences between the synthetic, algebraic, and analytic ap- proaches to geometry. (2) Explain how to define a parabola knowing a point and a line. (3) What is a tangent line? (4) Explain how to define a parabola using conic-sections. (5) Draw a parabola given two lines using tangent lines. (6) Give an algebraic definition of a tangent line. (7) Given:

3x7 x5+x4 16x3+27 = a x7+a x6+a x5+a x4+a x3+a x2+a x1+a − − 7 6 5 4 3 2 1 0 Find a0, a1, a2, a3, a4, a5, a6, a7. (8) Given:

6x5 + a x4 x2 + a = a x5 24x4 + a x3 + a x2 5 4 − 0 5 − 3 2 − Find a0, a1, a2, a3, a4, a5. (9) Algebraically find the line tangent to y = x2 at the point x = 2. Explain your work. (10) Algebraically find the line tangent to y = x2 3x + 1 at the point x = 3. Explain your work. − (11) Algebraically find the line tangent to y = x2 + 12x 4 at the point x = 0. Explain your work. − (12) Algebraically find the line tangent to y = x2 + 4x 2 at the point x = 0. Explain your work. − − (13) Algebraically find the line tangent to y = x2 at the point x = P , in terms of P . Explain your work. (14) Algebraically find the line tangent to y = x3 at the point x = 2. Explain your work. (15) Algebraically find the line tangent to y = x3 3x2 + 4x 1 at the point x = 0. Explain your work. − − (16) Algebraically find the line tangent to y = x3 + 5x2 + 2 at the point x = 1. Explain your work. (17) Algebraically find the line tangent to y = x20 23x+4 at the point x = 0. Explain your work. Hint: If you have trouble− with this one, do Problems 7 and 8 above.

20 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

(18) Algebraically ﬁnd the line tangent to y = x4 + 3x3 5x2 + 12 at the point x = 0. Explain your work. −

(19) Algebraically ﬁnd the line tangent to y = x14 at the point x = 0. Explain your work.

(20) Explain why f(a + s) f(a) slope (a, s) = − f(x) s gives you the slope of the tangent line that passes through the point (a, f(a)), when s is near zero.

(21) For a given function f(x), write out the formula for slopef(x)(a, 0).

(22) Approximate the slope of the line tangent to the function f(x) = x2 at x = 2 to 2 decimal places. Explain your work.

(23) Approximate the slope of the line tangent to the function f(x) = x3 at x = 2 to 2 decimal places. Explain your work.

(24) Approximate the slope of the line tangent to the function f(x) = x2+2x+1 at x = 1 to 2 decimal places. Explain your work.

(25) Approximate the slope of the line tangent to the function f(x) = x6 at x = 0 to 2 decimal places. Explain your work.

(26) Approximate the slope of the line tangent to the function f(x) = x3 + 5x2 + 2 at x = 1 to 2 decimal places. Explain your work.

(27) True or False: Explain your conclusions.

(a) A tangent line can intersect a curve at more than 1 point. (b) Any line which intersects a curve at exactly one point is a tangent line. (c) Some points on the graph of a function might not have a tangent line. (d) Any quadratic equation will always have 2 distinct roots. (e) If x10 x + 1 line(x) = x2g(x) where g(x) is a polynomial, then line(x)− = x +− 1. − (28) Give an example of a curve C and a line ℓ where ℓ is not a tangent line of C at any point and only intersects C at a single point. Clearly label your sketch.

(29) Give an example of a curve C and a line ℓ where ℓ is a tangent line of C at some point, but ℓ also intersects C in exactly 4 points. Clearly label your sketch.

21 1.2. POINTS OF VIEW

(30) Can you come up with a function f(x) (a sketch will suﬃce) where

slopef(x)(2, 0.1) = 0 and slopef(x)(2, 0.01) = 1?

Explain your answer.

(31) Can you come up with a function f(x) (a sketch will suﬃce) where

slopef(x)(2, 0.1) = 0, slopef(x)(2, 0.01) = 1, slopef(x)(2, 0.001) = 0?

Explain your answer.

(32) Can you come up with a function f(x) (a sketch will suﬃce) where

slopef(x)(2, 0.1) = 0, slopef(x)(2, 0.01) = 0, slopef(x)(2, 0.001) = 0,

but the slope of f(x) at x = 2 is 1? Explain your answer.

(33) Approximate the slope of the line tangent to the function f(x) = ex at x = 1 to 2 decimal places. Recall that e = 2.718281828459045 . . . . Explain your work. (34) Approximate the slope of the line tangent to the function f(x) = ex at x = 2 to 2 decimal places. Recall that e = 2.718281828459045 . . . . Explain your work.

(35) Approximate the slope of the line tangent to the function f(x) = ex at x = 3 to 2 decimal places. Recall that e = 2.718281828459045 . . . . Explain your work.

(36) Suppose that for some number a, ea = 0. In light of the previous three questions, what would you conclude about ex then? Explain your answer.

(37) What does the graph of ex look like?

(38) Explain ex in terms of a combination of the following:

Driving. • Speed-limit signs. • Mile-marker signs. •

22 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

1.3 City Geometry

One day I was walking through the city—that’s right, New York City. I had the most terrible feeling that I was lost. I had just passed a Starbucks Coﬀee on my left and a Sbarro Pizza on my right, when what did I see? Another Starbucks Coﬀee and Sbarro Pizza! Three options occurred to me:

(1) I was walking in circles.

(2) I was at the nexus of the universe.

(3) New York City had way too many Starbucks and Sbarro Pizzas!

Regardless, I was lost. My buddy Joe came to my rescue. He pointed out that the city is organized like a grid. “Ah! City Geometry!” I exclaimed. At this point all Joe could say was “Huh?”

Question What the heck was I talking about?

Most cities can be viewed as a grid of city blocks:

In City Geometry we have points and lines, just like in Euclidean Geometry. However, since we can only travel on city blocks, the distance between points is computed in a bit of a strange way. We don’t measure distance as the crow ﬂies. Instead we use the Taxicab distance:

Deﬁnition Given two points A = (ax, ay) and B = (bx, by), we deﬁne the Taxicab distance as:

dT (A, B) = ax bx + ay by | − | | − | The approach taken in this section was adapted from [14].

23 1.3. CITY GEOMETRY

Example Consider the following points:

Let A = (0, 0). Now we see that B = (7, 4). Hence

dT (A, B) = 0 7 + 0 4 | − | | − | =7+4 = 11.

Of course in real life, you would want to add in the appropriate units to your ﬁnal answer.

Question How do you compute the distance between A and B as the crow ﬂies? ?

Here’s the scoop: When we consider our points and lines to be like those in Euclidean Geometry, but when we use the Taxicab distance, we are working with City Geometry.

Question Compare and contrast the notion of a line in Euclidean Geometry and in City Geometry. In either geometry is a line the unique shortest path between any two points? ?

If you are interested in real-world types of problems, then maybe City Ge- ometry is the geometry for you. The concepts that arise in City Geometry are directly applicable to everyday life.

Question Will just bought himself a brand new gorilla suit. He wants to show it oﬀ at three parties this Saturday night. The parties are being held at his friends’ houses: the Antidisestablishment (A), Hausdorﬀ (H), and the

24 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

Wookie Loveshack (W ). If he travels from party A to party H to party W , how far does he travel this Saturday night?

Solution We need to compute

dT (A, H) + dT (H, W )

Let’s start by ﬁxing a coordinate system and making A the origin. Then H is (2, 5) and W is ( 10, 2). Then − − −

dT (A, H) = 0 2 + 0 ( 5) | − | | − − | =2+5 = 7 and

dT (H, W ) = 2 ( 10) + 5 ( 2) | − − | | − − − | = 12 + 3 = 15.

Will must trudge 7 + 15 = 22 blocks in his gorilla suit.

1.3.1 Getting Work Done

Okay, that’s enough monkey business. Time to get some work done.

Question Brad and Melissa are going to downtown Champaign, Illinois. Brad wants to go to Jupiter’s for pizza (J) while Melissa goes to Boardman’s Art Theater (B) to watch a movie. Where should they park to minimize the

25 1.3. CITY GEOMETRY total distance walked by both?

Solution Again, let’s set up a coordinate system so that we can say what points we are talking about. If J is (0, 0), then B is ( 5, 4). −

No matter where they park, Brad and Melissa’s two paths joined together must make a path from B to J. This combined path has to be at least 9 blocks long since dT (B,J) = 9. They should look for a parking spot in the rectangle formed by the points (0, 0), (0, 4), ( 5, 0), and ( 5, 4). Suppose they park within− this rectangle− and call this point C. Melissa now walks 4 blocks from C to B and Brad walks 5 blocks from C to J. The two paths joined together form a path from B to J of length 9. If they park outside the rectangle described above, for example at point D, then the corresponding path from B to J will be longer than 9 blocks. Any path from B to J going through D goes a block too far west and then has to backtrack a block to the east making it longer than 9 blocks.

Question If we consider the same question in Euclidean Geometry, what is the answer? ? Question Tom is looking for an apartment that is close to Altgeld Hall (H) but is also close to his favorite restaurant, Crane Alley (C). Where should Tom

26 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS live?

Solution If we ﬁx a coordinate system with its origin at Altgeld Hall, H, then C is at (8, 2). We see that dT (H, C) = 10. If Tom wants to live as close as possible to both of these, he should look for an apartment, A, such that dT (A, H) = dT (A, C) = 5. He would then be living halfway along one of the shortest paths from Altgeld to the restaurant. Mark all the points 5 blocks away from H. Now mark all the points 5 blocks away from C.

We now see that Tom should check out the apartments near (5, 0), (4, 1), and (3, 2).

Question Johann is starting up a new business, Cafe Battle Royale. He knows mathematicians drink a lot of coﬀee so he wants it to be near Altgeld Hall. Balancing this against how expensive rent is near campus, he decides the cafe should be 3 blocks from Altgeld Hall. Where should his cafe be located?

Solution What are the possibilities? The cafe could be 3 blocks due north or due south of Altgeld Hall, which is labeled A in the ﬁgure below. It could be also be 2 blocks north and 1 block west or 1 block north and 2 blocks west.

27 1.3. CITY GEOMETRY

Continuing in this fashion, we obtain the following ﬁgure:

Johann can have his coﬀee shop on any of the point above surrounding Altgeld Hall.

1.3.2 (Un)Common Structures How diﬀerent is life in City Geometry from life in Euclidean geometry? In this section we’ll try to ﬁnd out!

Triangles If we think back to Euclidean Geometry, we may recall some lengthy discussions on triangles. Yet so far, we have not really discussed triangles in City Geometry.

Question What does a triangle look like in City Geometry and how do you measure its angles?

I’ll take this one. Triangles look the same in City Geometry as they do in Eu- clidean Geometry. Also, you measure angles in exactly the same way. However, there is one minor hiccup. Consider these two triangles in City Geometry:

28 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

Question What are the lengths of the sides of each of these triangles? Why is this odd? ?

Hence we see that triangles are a bit funny in City Geometry.

Circles

Circles are also discussed in many geometry courses and this course is no different. However, in City Geometry the circles are a little less round. The ﬁrst question we must answer is the following:

Question What is a circle?

Well, a circle is the collection of all points equidistant from a given point. So in City Geometry, we must conclude that a circle of radius 2 would look like:

Question How many points are there at the intersection of two circles in Euclidean Geometry? How many points are there at the intersection of two circles in City Geometry? ?

Midsets

Deﬁnition Given two points A and B, their midset is the set of points that are an equal distance away from both A and B.

Question How do we ﬁnd the midset of two points in Euclidean Geometry? How do we ﬁnd the midset of two points in City Geometry?

29 1.3. CITY GEOMETRY

In Euclidean Geometry, we just take the the following line:

If we had no idea what the midset should look like in Euclidean Geometry, we could start as follows:

Draw circles of radius r1 centered at both A and B. If these circles inter- • sect, then their points of intersection will be in our midset. (Why?)

Draw circles of radius r2 centered at both A and B. If these circles inter- • sect, then their points of intersection will be in our midset.

We continue in this fashion until we have a clear idea of what the midset • looks like. It is now easy to check that the line in our picture is indeed the midset.

How do we do it in City Geometry? We do it basically the same way.

Example Suppose you wished to ﬁnd the midset of two points in City Geom- etry.

We start by ﬁxing coordinate axes. Considering the diagram below, if A = (0, 0), then B = (5, 3). We now use the same idea as in Euclidean Geometry. Drawing circles of radius 3 centered at A and B respectively, we see that there are no points 3 points away from both A and B. Since dT (A, B) = 8, this is to be expected. We will need to draw larger Taxicab circles before we will ﬁnd points in the midset. Drawing Taxicab circles of radius 5, we see that the points

30 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

(1, 4) and (4, 1) are both in our midset. −

Now it is time to sing along. You draw circles of radius 6, to get two more points (1, 5) and (4, 2). Drawing circles with larger radii yields more and more points “due north”− of (1, 5) and “due south” of (4, 2). However, if we draw circles of radius 4 centered at A and B respectively,− their intersection is the line segment between (1, 3) and (4, 0). Unlike Euclidean circles, distinct City Geometry circles can intersect in more than two points and City Geometry midsets can be more complicated than their Euclidean counterparts. Question How do you draw the City Geometry midset of A and B? What could the midsets look like? ?

Parabolas Recall that a parabola is a set of points such that each of those points is the same distance from a given point, A, as it is from a given line, L.

31 1.3. CITY GEOMETRY

This deﬁnition still makes sense when we work with Taxicab distance instead of Euclidean distance. Draw a line parallel to L at Taxicab distance r away from L. Now draw a City circle of radius r centered at A. The points of intersection of this line and this circle will be r away from L and r away from A and so will be points on our City parabola. Repeat this process for diﬀerent values of r.

Unlike the Euclidean case, the City parabola need not grow broader and broader as the distance from the line increases. In the picture above, as we go from B to C on the parabola, both the Taxicab and Euclidean distances to the line L increase by 1. The Taxicab distance from the point A also increases by 1 as we go from B to C but the Euclidean distance increases by less than 1. For the Euclidean distance from A to the parabola to keep increasing at the same rate as the distance to the line L, the Euclidean parabola has to keep spreading to the sides.

Question How do you draw the City Geometry parabolas? What do diﬀerent parabolas look like? ?

A Paradox To be completely clear on what a paradox is, here is the deﬁnition we will be using:

Deﬁnition A paradox is a statement that seems to be contradictory. This means it seems both true and false at the same time.

There are many paradoxes in mathematics. By studying them we gain insight—and also practice tying our brain into knots! Here is the ﬁrst paradox we will study in this course:

32 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

Paradox √2 = 2.

False-Proof Consider the following sequence of diagrams:

On the far right-hand side, we see a right-triangle. Suppose that the lengths of the legs of the right-triangle are one. Now by the Pythagorean Theorem, the length of the hypotenuse is √12 + 12 = √2. However, we see that the triangles coming from the left converge to the triangle on the right. In every case on the left, the stair-step side has length 2. Hence when our sequence of stair-step triangles converges, we see that the hypotenuse of the right-triangle will have length 2. Thus √2 = 2.

Question What is wrong with the proof above? ?

33 1.3. CITY GEOMETRY

Problems for Section 1.3

(1) Given two points A and B in City Geometry, does dT (A, B) = dT (B, A)? Explain your reasoning. (2) Explain how City Geometry shows that Euclid’s ﬁve axioms are not enough to determine all of the familiar properties of the plane. (3) Do Euclid’s axioms hold in City Geometry? How would you change these axioms so that they do hold in City Geometry? (4) Brad and Melissa are going to downtown Champaign. Brad wants to go to Jupiter’s for pizza while Melissa goes to Boardman’s to watch a movie. Where should they park to minimize the total distance walked by both and Brad insists that Melissa should not have to walk a longer distance than him?

(5) Brad and Melissa are going to downtown Champaign. Brad wants to go to Jupiter’s for pizza while Melissa goes to Boardman’s to watch a movie. Where should they park to minimize the total distance walked by both and Melissa insists that they should both walk the same distance?

(6) Lisa just bought a 3-wheeled zebra-striped electric car. It has a top speed of 40 mph and a maximum range of 40 miles. Suppose that there are 4

34 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

blocks to a mile and she wishes to drive 4 miles from her house. What points can she reach? (7) A group of hooligans think it would be hilarious to place a bucket on the Alma Mater’s head,2 point A. Moreover, these hooligans are currently at point S and wish to celebrate their accomplishment at Murphy’s Pub, point M. If there are campus police at points P and Q, what path should the hooligans take from S to A to M to best avoid detainment for their hijinks?

(8) Scott wants to live within 4 blocks of a cafe, C, within 5 blocks of a bar, B, and within 10 blocks of Altgeld Hall, A. Where should he go apartment hunting?

(9) The university is installing emergency phones across campus. Where should they place them so that their students are never more than a block away from an emergency phone? (10) Suppose that you have two triangles ABC and DEF in City Geometry such that △ △

(a) dT (A, B) = dT (D, E).

2The Alma Mater is a statue of a “Loving Mother” at the University of Illinois.

35 1.3. CITY GEOMETRY

(b) dT (B, C) = dT (E,F ).

Is it necessarily true that ABC DEF ? Explain your reasoning. △ ≡ △

(11) In City Geometry, if all the angles of ABC are 60◦, is ABC necessarily an equilateral triangle? Explain your△ reasoning. △

(12) In City Geometry, if two right triangles have legs of the same length, is it true that their hypotenuses will be the same length? Explain your reasoning.

(13) Considering that π is the ratio of the circumference of a circle to its diam- eter, what is the value of π in City Geometry? Explain your reasoning.

(14) Considering that the area of a circle of radius r is given by πr2, what is the value of π in City Geometry. Explain your reasoning.

(15) How many points are there at the intersection of two circles in Euclidean Geometry? How many points are there at the intersection of two circles in City Geometry?

(16) What would the City Geometry equivalent of a compass be?

(17) Cafe Battle Royale, Inc. is expanding. Johann wants his potential cus- tomers to always be within 4 blocks of one of his cafes. Where should his cafes be located?

(18) When is the Euclidean midset of two points equal to their City Geometry midset?

(19) Find the City Geometry midset of ( 2, 2) and (3, 2). − (20) Find the City Geometry midset of ( 2, 2) and (4, 1). − − (21) Find the City Geometry midset of ( 2, 2) and (2, 2). − (22) Draw the City Geometry parabola determined by the point (2, 0) and the line y = 0.

(23) Draw the City Geometry parabola determined by the point (2, 0) and the line y = x.

(24) Find the distance in City Geometry from the point (3, 4) to the line y = 1/3x. Explain your reasoning. − (25) Draw the City Geometry parabola determined by the point (3, 0) and the line y = 2x + 6. Explain your reasoning. This problem was suggested by Marissa− Colatosti.

36 CHAPTER 1. BEGINNINGS, AXIOMS, AND VIEWPOINTS

(26) There are hospitals located at A, B, and C. Ambulances should be sent to medical emergencies from whichever hospital is closest. Divide the city into regions in a way that will help the dispatcher decide which ambulance to send.

(27) Find all points P such that dT (P, A) + dT (P,B) = 8. Explain your work. (In Euclidean Geometry, this condition determines an ellipse. The solution to this problem could be called the City Geometry ellipse.)

(28) True/False: Three noncollinear points lie on a unique Euclidean circle. Explain your reasoning.

(29) True/False: Three noncollinear points lie on a unique Taxicab circle. Ex- plain your reasoning. (30) Explain why no Euclidean circle can contain three collinear points. Can a Taxicab circle contain three collinear points? Explain your conclusion.

(31) Can you ﬁnd a false-proof showing that π = 2?

37 Chapter 2

Proof by Picture

A picture is worth a thousand words. —Unknown

2.1 Basic Set Theory

The word set has more definitions in the dictionary than any other word. In our case we’ll use the following definition: Definition A set is any collection of elements for which we can always tell whether an element is in the set or not. Question What are some examples of sets? What are some examples of things that are not sets? ? If we have a set X and the element x is inside of X, we write:

x X ∈ This notation is said “x in X.” Pictorially we can imagine this as:

38 CHAPTER 2. PROOF BY PICTURE

Deﬁnition A subset Y of a set X is a set Y such that every element of Y is also an element of X. We denote this by:

Y X ⊆

If Y is contained in X, we will sometimes loosely say that X is bigger than Y .

Question Can you think of a set X and a subset Y where saying X is bigger than Y is a bit misleading? ?

Question How is the meaning of the symbol diﬀerent from the meaning of the symbol ? ∈ ⊆ ?

2.1.1 Union

Deﬁnition Given two sets X and Y , X union Y is the set of all the elements in X and all the elements in Y . We denote this by X Y . ∪ Pictorially, we can imagine this as:

2.1.2 Intersection

Deﬁnition Given two sets X and Y , X intersect Y is the set of all the elements that are simultaneously in X and in Y . We denote this by X Y . ∩

39 2.1. BASIC SET THEORY

Pictorially, we can imagine this as:

Question Consider the sets X and Y below:

What is X Y ? ∩ I’ll take this one: Nothing! We have a special notation for the set with no elements, it is called the empty set. We denote the empty set by the symbol ∅.

2.1.3 Complement Deﬁnition Given two sets X and Y , X complement Y is the set of all the elements that are in X and are not in Y . We denote this by X Y . − Pictorially, we can imagine this as:

40 CHAPTER 2. PROOF BY PICTURE

Question Check out the two sets below:

What is X Y ? What is Y X? − − ?

2.1.4 Putting Things Together

OK, let’s try something more complex:

Question Prove that:

X (Y Z) = (X Y ) (X Z) ∪ ∩ ∪ ∩ ∪

Proof Look at the left-hand side of the equation ﬁrst:

41 2.1. BASIC SET THEORY

And so we see:

Now look at the right-hand side of the equation:

And:

42 CHAPTER 2. PROOF BY PICTURE

So we see that:

Comparing the diagrams representing the left-hand and right-hand sides of the equation, we see that we are done.

43 2.1. BASIC SET THEORY

Problems for Section 2.1 (1) Given two sets X and Y , explain what is meant by X Y . ∪ (2) Given two sets X and Y , explain what is meant by X Y . ∩ (3) Given two sets X and Y , explain what is meant by X Y . − (4) Explain the diﬀerence between the symbols and . ∈ ⊆ (5) Prove that: X = (X Y ) (X Y ) ∩ ∪ − (6) Prove that: X (X Y ) = (X Y ) − − ∩ (7) Prove that: X (Y X) = (X Y ) ∪ − ∪ (8) Prove that: X (Y X) = ∅ ∩ − (9) Prove that: (X Y ) (Y X) = (X Y ) (X Y ) − ∪ − ∪ − ∩ (10) Prove that: X (Y Z) = (X Y ) (X Z) ∪ ∩ ∪ ∩ ∪ (11) Prove that: X (Y Z) = (X Y ) (X Z) ∩ ∪ ∩ ∪ ∩ (12) Prove that: X (Y Z) = (X Y ) (X Z) − ∩ − ∪ − (13) Prove that: X (Y Z) = (X Y ) (X Z) − ∪ − ∩ − (14) If X Y = X, what can we say about the relationship between the sets X and∪ Y ? Explain your reasoning. (15) If X Y = Y , what can we say about the relationship between the sets X and∪ Y ? Explain your reasoning. (16) If X Y = X, what can we say about the relationship between the sets X and∩ Y ? Explain your reasoning. (17) If X Y = Y , what can we say about the relationship between the sets X and∩ Y ? Explain your reasoning. (18) If X Y = ∅, what can we say about the relationship between the sets X and− Y ? Explain your reasoning. (19) If Y X = ∅, what can we say about the relationship between the sets X and− Y ? Explain your reasoning.

44 CHAPTER 2. PROOF BY PICTURE

2.2 Logic

Logic is a great tool to have around. It turns out that we can solve lots of logical problems using simple tables. Moreover, one can often look at logic using the ideas of Set Theory that we learned in the previous section. When working with logic, there are certain buzz words you need to be on the watch for: not—, —and—, —or—, if—, then—, —if and only if—. Any time you see the above buzz words you need to stop and think. We will address each of these words in turn. The ﬁrst buzz word above is not. Suppose you have a statement: P = I love math! We use symbol to mean not. To negate the above statement, you just put the in front of¬ the P : ¬ P = It is not the case that I love math. ¬ When is P true? Well, only when P is false. We can display this with a truth-table¬ : P P ¬ T F F T When you apply not to a statement, it simply swaps true for false in the truth- tables: Now consider the statement: I’m strong and I’m cool. Q P ∧ and let P = I’m strong, while| Q{z= I’m} | cool.{z } | When{z is} the above statement true? Well it is only true when both P and Q are true. We can display this with a truth-table. Note that we use the symbol to mean and: ∧ P Q P Q ∧ T T T T F F F T F F F F Now what if we want to look at the statement: I eat ice cream or I eat cookies. P Q ∨ When is the above statement| {z true?} It|{z} is| true{z when either} P or Q are true. In fact it is true even when they are both true. We can display this with a truth-table. Note that we use the symbol to mean or: ∨

45 2.2. LOGIC

P Q P Q ∨ T T T T F T F T T F F F If you look at the truth-tables for both and and or you see a sort of symmetry. This can best explained by the use of not. WARNING Applying not changes an and to an or and vice versa. So if we have the statement:

I’m strong and I’m cool. Q P ∧ Then | {z } | {z } | {z }

(P Q) = ( P ) ( Q) ¬ ∧ ¬ ∨ ¬ = I’m not strong or I’m not cool.

P Q ¬ ∨ ¬ We can see this best using a truth-table:| {z } |{z} | {z } P Q P Q (P Q) P Q ( P ) ( Q) ∧ ¬ ∧ ¬ ¬ ¬ ∨ ¬ T T T F F F F T F F T F T T F T F T T F T F F F T T T T Question What does the truth-table for (P Q) and ( P ) ( Q) look like? ¬ ∨ ¬ ∧ ¬ ? While and, or, and not really aren’t all that bad, if-then is much trickier. Allow me to demonstrate how tricky if-then can be. I will do this with the Wason selection task: Suppose I had a set of cards each with a number on one side and a letter on the other side, and I laid four of them on a table in front of you:

A 7 B 6

Consider the statement: If one side of the card shows an even number, then the other side of the card shows a vowel.

46 CHAPTER 2. PROOF BY PICTURE

Question Exactly which card(s) above do you need to flip over to see whether my statement is true? ? Now suppose you are a police officer at a local bar that has four tables. At the first table nobody is drinking alcohol, at the second table every customer looks quite old, at the third table there are many pitchers of beer, and at the fourth table everybody looks quite young:

no old young alcohol people beer people

Consider the law: If you are under 21, then you cannot drink alcohol. Question Exactly which table(s) do you need to check to see if the law is being upheld? ? Question Are the two questions above diﬀerent? ? Question Of the two questions above, which one was easier? ? I think that the two situations above involving if-then statements show that we need to be careful when dealing with them, especially when the situation is somewhat abstract. Let’s look at the truth-tables for if-then. Note that we use the symbol to mean if-then: ⇒ P Q P Q ⇒ T T T T F F F T T F F T

To make P Q easier to read it is sometimes helpful to read it P implies Q. A curious⇒ fact is that often the easiest way to negate an if-then statement is to rewrite it in terms of or and not:

47 2.2. LOGIC

P Q P P Q P Q ¬ ¬ ∨ ⇒ T T F T T T F F F F F T T T T F F T T T From the truth-table we see that

P Q = P Q. ⇒ ¬ ∨ Now we can negate this easily:

(P Q) = ( P Q) = P ( Q). ¬ ⇒ ¬ ¬ ∨ ∧ ¬ Finally if you see if-and-only-if, denoted by the symbol , this is nothing more than: ⇔ P Q = (P Q) (Q P ). ⇔ ⇒ ∧ ⇒ Question Can you connect the ideas in this section to ideas in Set Theory? Speciﬁcally, let a statement be a set. The “points” that make it true are what are inside the set. What do

not—, —and—, —or—, if—, then—, —if and only if—, look like? ?

48 CHAPTER 2. PROOF BY PICTURE

Problems for Section 2.2 (1) Knowing that P Q = (P Q) (Q P ), write a truth-table for P Q. ⇔ ⇒ ∧ ⇒ ⇔ (2) Use a truth-table to show that (P Q) = ( P ) ( Q). ¬ ∨ ¬ ∧ ¬ (3) Use a truth-table to show that (P Q) = (Q P ). ⇒ 6 ⇒ (4) Explain why P Q is not the same as Q P by giving a real-world sentence for P and⇒ a real world sentence for Q⇒and analyzing what P Q and Q P mean. ⇒ ⇒ (5) Use a truth-table to show that P Q = ( Q) ( P ). ⇒ ¬ ⇒ ¬ (6) Explain why P Q is the same as ( Q) ( P ) by giving a real-world sentence for P and⇒ a real world sentence¬ for⇒Q and¬ analyzing what P Q and ( Q) ( P ) mean. ⇒ ¬ ⇒ ¬ (7) Go out and find some friends. Set up the card example as explained above. See how many of them can get it right. Then set up the example of the tables at a bar as explained above. How many get it right now? (8) Suppose I give you the statement: If your name is Agatha, then you like to eat tomatoes. Which of the following don’t contradict the above statement: (a) Your name is Agatha and you like to eat tomatoes. (b) Your name is Agatha and you don’t like to eat tomatoes. (c) Your name is Joe and you like to eat tomatoes. (d) Your name is Joe and you don’t like to eat tomatoes. (9) Suppose I give you the statement: If your name is Jen, then you have a cat named Hypie. Which of the following don’t contradict the above statement: (a) Your have a cat named Hypie and your name is Jen. (b) Your have a cat named Hypie and your name is Joe. (c) Your have no cats and your name is Jen. (d) Your have no cats and your name is Joe. (10) Give an if-then statement involving traffic laws and use it in an example to explain why (false true) is a true statement. Explain your answer. ⇒ (11) Give an if-then statement involving traffic laws and use it in an example to explain why (false false) is a true statement. Explain your answer. ⇒

49 2.2. LOGIC

(12) Let P and Q be true statements, and let X and Y be false statements. Determine the truth value of the following statements:

(a) P Y . ∧ (b) X Q. ∨ (c) P Q. ⇒ (d) X Y . ⇒ (e) Y Q. ⇒ (f) P Y . ⇒ (g) P (X Y ). ⇒ ∨ (h) P P . ¬ ⇒ (i) (X Q) Y . ∧ ⇒ (j) (P Y ) Q. ∨ ⇒ (k) (P ( P )). ¬ ∧ ¬ (l) (X ( X)). ¬ ∨ ¬ (13) Here is the truth-table for neither-nor, denoted by the symbol : × P Q P Q × T T F T F F F T F F F T

(a) Make a truth-table for P P . × (b) Make a truth-table for (P Q) (P Q). × × × (c) Make a truth-table for (P P ) (Q Q). × × × (d) Make a truth-table for ((P P ) Q) ((P P ) Q). × × × × × Use your work above to express not, and, or, and if-then purely in terms of neither-nor.

(14) Draw pictures showing the connection between intersection and and, and union and or. What does not look like? What does if-then look like? What does if-and-only-if look like?

50 CHAPTER 2. PROOF BY PICTURE

2.3 Tessellations

Go to the internet and look up M.C. Escher. He was an artist. Look at some of his work. When you do your search be sure to include the word “tessellation” OK? Back already? Very good. With some of Escher’s work he started with a tessellation. What’s a tessellation? I’m glad you asked:

Deﬁnition A tessellation is a pattern of polygons ﬁtted together to cover the entire plane without overlapping. A tessellation is called a regular tessellation if the polygons are regular and they have common vertexes.

Example Here are some examples of regular tessellations:

Johannes Kepler was one of the ﬁrst people to study tessellations. He cer- tainly knew the next theorem:

Theorem 2 There are only 3 regular tessellations.

Since one can prove that there are only three regular tessellations, and we have shown three above, then that is all of them. On the other hand there are lots of nonregular tessellations. Here are two diﬀerent ways to tessellate the plane with a triangle:

Here is a way that you can tessellate the plane with any old quadrilateral:

51 2.3. TESSELLATIONS

2.3.1 Tessellations and Art

How does one make art with tessellations? To start, a little decoration goes a long way. Check this out: Decorate two squares as such:

Tessellate them randomly in the plane to get this lightning-like picture:

Question What sort of picture do you get if you tessellate these decorated squares randomly in a plane?

Another way to go is to start with your favorite tessellation:

52 CHAPTER 2. PROOF BY PICTURE

Then you modify it a bunch to get something diﬀerent:

Question What kind of art can you make with tessellations? ?

53 2.3. TESSELLATIONS

Problems for Section 2.3 (1) Show two diﬀerent ways of tessellating the plane with a given scalene triangle. Label your picture as necessary. (2) Show how to tessellate the plane with a given quadrilateral. Label your picture. (3) Show how to tessellate the plane with a nonregular hexagon. Label your picture. (4) Give an example of a polygon with 9 sides that tessellates the plane. (5) Give examples of polygons that tessellate and polygons that do not tessellate. (6) True or False: Explain your conclusions. (a) There are exactly 5 regular tessellations. (b) Any quadrilateral tessellates the plane. (c) Any triangle will tessellate the plane. (d) If a triangle is used to tessellate the plane, then exactly 4 angles will ﬁt around each vertex. (e) If a polygon has more than 6 sides, then it cannot tessellate the plane. (7) Fill in the following table:

Regular Does it Measure If it tessellates, how n-gon tessellate? of angles many surround each vertex? 3-gon 4-gon 5-gon 6-gon 7-gon 8-gon 9-gon 10-gon

Hint: A regular n-gon has interior angles of 180(n 2)/n degrees. − (a) What do the shapes that tessellate have in common? (b) Make a graph with the number of angles on the horizontal axis and the measure of the angles on the vertical axis. (c) What regular polygons could a bee use for building hives? Give some reasons that bees seem to use hexagons. (8) Given a regular tessellation, what is the sum of the angles around a given vertex?

54 CHAPTER 2. PROOF BY PICTURE

(9) Given that the regular octagon has 135 degree angles, explain why you cannot give a regular tessellation of the plane with a regular octagon.

(10) Considering that the regular n-gon has interior angles of 180(n 2)/n degrees, and Problem 7 above, prove that there are only 3 regular− tessellations of the plane.

55 2.4. PROOF BY PICTURE

2.4 Proof by Picture

2.4.1 Proofs Involving Right Triangles

We’ll start this oﬀ with a question:

Question What is the most famous theorem in mathematics?

Probably the Pythagorean Theorem comes to mind. Let’s recall the statement of the Pythagorean Theorem:

Theorem 3 (Pythagorean Theorem) Given a right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. Symbolically, if a and b represent the lengths of the legs and c is the length of the hypotenuse,

c a

b then

a2 + b2 = c2.

Question What is the converse to the Pythagorean Theorem? Is it true? How do you prove it?

While everyone may know the Pythagorean Theorem, not as many know how to prove it. Euclid’s proof goes kind of like this:

Nearly all of the pictures from this section are adapted from the wonderful source books: [17] and [18].

56 CHAPTER 2. PROOF BY PICTURE

Consider the following picture:

Now, cut up the squares a2 and b2 in such a way that they ﬁt into c2 perfectly. When you give a proof that involves cutting up the shapes and putting them back together, it is called a dissection proof. The trick to ensure that this is actually a proof is in making sure that your dissection will work no matter what right triangle you are given. Does it sound complicated? Well it can be. Is there an easier proof? Sure, look at:

Question How does the picture above “prove” the Pythagorean Theorem?

Solution Both of the large squares above are the same size. Moreover both the unshaded regions above must have the same area. The large white square

57 2.4. PROOF BY PICTURE on the left has an area of c2 and the two white squares on the right have a combined area of a2 + b2. Thus we see that:

c2 = a2 + b2

Let’s give another proof! This one looks at a tessellation involving 2 squares.

Question How does the picture above “prove” the Pythagorean Theorem?

Solution The striped triangle is our right triangle. The area of the overlaid square is c2, the area of the small squares is a2, and the area of the medium square is b2. Now label all the “parts” of the large overlaid square:

58 CHAPTER 2. PROOF BY PICTURE

From the picture we see that

a2 = 3 and 4 { } b2 = 1, 2, and 5 { } c2 = 1, 2, 3, 4, and 5 { } Hence c2 = a2 + b2 Since we can always put two squares together in this pattern, this proof will work for any right triangle.

Question Can you use the above tessellation to give a dissection proof of the Pythagorean Theorem? ? Now a paradox: Paradox What is wrong with this picture?

Question How does this happen? ?

59 2.4. PROOF BY PICTURE

2.4.2 Proofs Involving Boxy Things Consider the problem of Doubling the Cube. If a mathematician asks us to double a cube, he or she is asking us to double the volume of a given cube. One may be tempted to merely double each side, but this doesn’t double the volume! Question Why doesn’t doubling each side of the cube double the volume of the cube? ? Well, let’s answer an easier question ﬁrst. How do you double the area of a square? Does taking each side and doubling it work?

No! You now have four times the area. So you cannot double the area of a square merely by doubling each side. What about for the cube? Can you double the volume of a cube merely by doubling the length of every side? Check this out:

Ah, so the answer is again no. If you double each side of a cube you have 8 times the volume.

The Arithmetic-Geometric Mean Inequality The arithmetic mean of two numbers is just the average of those two numbers. However, the geometric mean is a bit more mysterious. Essentially with the geometric mean, you are “squaring” a rectangle. What do we mean by this? We mean that you are ﬁnding a square whose area is the same as the original rectangle.

60 CHAPTER 2. PROOF BY PICTURE

Question Suppose you have have a rectangle with sides a and b. What is the side length of the square whose area is the same as that rectangle? ?

Theorem 4 (Arithmetic-Geometric Mean Inequality) If a and b are positive numbers then: a + b √a b 6 · 2 and the inequality above is an equality when a = b.

Question Can you state the above theorem in English? Can you give some examples of how it is true? Can you show me a graph?

Now look at this picture:

a b

Question How does the picture above “prove” the Arithmetic-Geometric Mean Inequality?

Solution Consider the area of the large square. This is

(a + b)2.

On the other hand, the area of the four smaller rectangles is 4ab. Since we can see from the picture that this area is less than the area of the large square (unless of course a = b), we have

4ab 6 (a + b)2 (a + b)2 ab 6 4 a + b √a b 6 , · 2 which is what we wanted to show.

OK, but in mathematics we really want to know that we are correct. So to do this, we will often give as many proofs of the same theorem as possible. Look

61 2.4. PROOF BY PICTURE at this picture:

Question How does the picture above “prove” the Arithmetic-Geometric Mean Inequality?

Solution As usual, in the above right-triangles, let the short leg be of length a, and the longer leg—that is not the hypotenuse, be of length b. Now the area of the big square is (a + b)2. But the area of the all the triangles is 4ab. Hence

4ab 6 (a + b)2 (a + b)2 ab 6 4 a + b √a b 6 , · 2 which is what we wanted to show. Note that if a = b, then the triangles would ﬁll the square and we would have equality.

2.4.3 Proofs Involving Sums Finite Sums According to legend, when Gauss was in elementary school, his teacher gave the problem of summing all the integers from 1 to 100. Supposedly Gauss gave the answer in seconds, infuriating the teacher. How did he do it? Well he probably did something like this: Write the numbers in a funky array:

012345 95 96 97 98 99 100 + + + + + + · · · + + + + + + 100 99 98 97 96 95 · · · 543210 · · · 100k 100 k 100 k 100 k 100 k 100 k···k100 100 k 100 k 100 k 100 k 100 k · · · 101 terms

So| if we sum up all the integers from 1{z to 100 twice we get }

100 101. ·

62 CHAPTER 2. PROOF BY PICTURE

Thus, the sum of all the integers from 1 to 100 is 100 101 1 + 2 + 3 + 4 + 5 + + 95 + 96 + 97 + 98 + 99 + 100 = · = 5050. · · · 2 That’s nice. But it isn’t a very useful fact to know. I mean, suppose you now want to know what is: 1 + 2 + 3 + + 9 + 10 · · · Or: 1 + 2 + + 452 + 453 · · · What we really want is a formula that somehow encapsulates what Gauss did above. Question What is the formula for the following sum of integers?

1 + 2 + 3 + 4 + 5 + + n · · · ? Consider this picture:

Question Explain how the picture above “proves” that: n(n + 1) 1 + 2 + 3 + 4 + 5 + + n = · · · 2 Solution Since the light circles make up half of the rectangle above, and the rectangle has n(n + 1) circles in it, then the number of light circles is

n(n + 1) . 2 However, from the picture, we can see that there are

1 + 2 + 3 + 4 + + n · · ·

63 2.4. PROOF BY PICTURE light colored circles. Therefore n(n + 1) 1 + 2 + 3 + 4 + 5 + + n = . · · · 2

Now you may object to this proof because the speciﬁc picture shown, n is 7 and not any old value. However, to this objection one could retort, that the pattern is clear, and that one need only continue the pattern to the desired value of n.

Infinite Sums As is our style, we will start off with a question: Question Can you add up an infinite number of terms and still get a finite number? Consider 1/3. Actually, consider the decimal notation for 1/3: 1 = .333333333333333333333333333333 . . . 3 But this is merely the sum: .3 + .03 + .003 + .0003 + .00003 + .000003 + · · · It stays less than 1 because the terms get so small so quickly. Are there other infinite sums of this sort? You bet! In fact: 1 1 2 1 3 1 4 1 5 + + + + + = 1 2 2 2 2 2 · · · Wow! How can we visualize this? Consider this picture:

1 1/2

(1/2)2

Question Explain how the picture above “proves” that:

1 1 2 1 3 1 4 1 5 + + + + + = 1 2 2 2 2 2 · · ·

64 CHAPTER 2. PROOF BY PICTURE

Solution So we take a unit square and divide it in half. This half piece has an area of 1/2. Now look at the other half of the square, divide it in half. This piece has an area of (1/2)2. Look at the next half and so on. From the picture above we see that we will eventually ﬁll the entire square. Therefore, summing the areas we see: 1 1 2 1 3 1 4 1 5 + + + + + = 1 2 2 2 2 2 · · ·

Now look at:

Question Explain how the picture above “proves” that: 1 1 2 1 3 1 4 1 5 1 + + + + + = 4 4 4 4 4 · · · 3 Solution Let’s take it in steps. If the big triangle has area 1, the area of the shaded region below is 1/4.

We also see that the area of the shaded region below

65 2.4. PROOF BY PICTURE is: 1 1 2 + 4 4 Continuing on in this fashion we see that the area of all the shaded regions is:

1 1 2 1 3 1 4 1 5 + + + + + 4 4 4 4 4 · · · But look, the unshaded triangles have twice as much area as the shaded triangle. Thus the shaded triangles must have an area of 1/3.

2.4.4 Proofs Involving Sequences Get your calculator out and play along at home:

1 1 = 3 3 1 + 3 =? 5 + 7 1+3+5 =? 7 + 9 + 11 1+3+5+7 =? 9 + 11 + 13 + 15 1+3+5+7+9 =? 11 + 13 + 15 + 17 + 19 Question What is happening here? Does it always happen? ?

One of the ﬁrst people to study this phenomenon was the physicist Galileo Galilei. To help us understand it, look at this picture:

Question Explain how the picture above “proves” that:

1 + 3 + + (2n 1) 1 · · · − = (2n + 1) + (2n + 3) + + (4n 1) 3 · · · −

66 CHAPTER 2. PROOF BY PICTURE

Solution Looking at the rows of the pyramid, the numerator of the fraction is represented by the top part and the denominator is represented by the bottom part. While it is not completely general since it stops at 5 circles, it is obvious how to extend the picture to work for any number. Since we can see that the top part of the pyramid is 1/3 of the bottom part,

1 + 3 + + (2n 1) 1 · · · − = . (2n + 1) + (2n + 3) + + (4n 1) 3 · · · −

Now look at this:

1 1 = 1 · 11 11 = 121 · 111 111 = 12321 · 1111 1111 = 1234321 ·

Question What is happening here? Does it always happen? Hint: You don’t need a picture to ﬁgure this one out. ?

2.4.5 Thinking Outside the Box

A calisson is a French candy that sort of looks like two equilateral triangles stuck together. They usually come in a hexagon-shaped box.

Question How do the calissons ﬁt into their hexagon-shaped box?

If you start to put the calissons into a box, you quickly see that they can be placed in there with exactly three diﬀerent orientations:

Theorem 5 In any packing, the number of calissons with a given orientation is exactly one-third the total number of calissons in the box.

67 2.4. PROOF BY PICTURE

Look at this picture:

Question How does the picture above “prove” Theorem 5? Hint: Think outside the box! ?

68 CHAPTER 2. PROOF BY PICTURE

Problems for Section 2.4

(1) Explain how the following picture “proves” the Pythagorean Theorem.

(2) Explain how the following picture “proves” the Pythagorean Theorem.

(3) Explain how the following picture “proves” the Pythagorean Theorem.

Note: This proof is due to Leonardo da Vinci.

69 2.4. PROOF BY PICTURE

(4) Explain how the following picture “proves” the Pythagorean Theorem.

(5) Use the following tessellation to give a dissection proof of the Pythagorean Theorem.

(6) Explain how the following picture “proves” the Pythagorean Theorem.

c a ×b b ×c b2 ×a bc bc c2

ab a2

ac ac

70 CHAPTER 2. PROOF BY PICTURE

(7) Recall that a trapezoid is a quadrilateral with two parallel sides. Consider the following picture:

How does the above picture prove that the area of a trapezoid is

h(b + b ) area = 1 2 , 2

where h is the height of the trapezoid and b1, b2, are the lengths of the parallel sides?

(8) Explain how the following picture “proves” the Pythagorean Theorem.

Note: This proof is due to James A. Garﬁeld, the 20th President of the United States.

(9) Explain how the following picture “proves” that if a quadrilateral has two opposite angles that are equal, then the bisectors of the other two angles are parallel or on top of each other.

71 2.4. PROOF BY PICTURE

(10) Explain how the following picture “proves” that the area of a quadrilateral is equal to half of the area of the parallelogram whose sides are parallel to and equal in length to the diagonals of the original quadrilateral.

(11) Why might someone ﬁnd the following picture disturbing? How would you assure them that actually everything is good and well in the geometrical world?

72 CHAPTER 2. PROOF BY PICTURE

(12) Why might someone ﬁnd the following picture disturbing? How would you assure them that actually everything is good and well in the geometrical world?

(13) How could you explain to someone that doubling the lengths of each side of a cube does not double the volume of the cube? (14) Look back at Problem 7. Can you use a similar picture to prove that the area of a parallelogram

is the length of the base times the height? (15) Explain how the following picture “proves” that the area of a parallelogram is base times height.

73 2.4. PROOF BY PICTURE

(16) You probably have noticed Geometry Giorgio in your class. In an attempt to prove the formula for the area of a parallelogram, Geometry Giorgio draws the following picture:

What is he doing wrong? How could he ﬁx his “proof”?

(17) Which of the above “proofs” for the formula for the area of a parallelogram is your favorite? Explain why.

(18) Explain how the following picture “proves” that if a and b are positive numbers then: a + b √a b 6 · 2 and the inequality above is an equality when a = b.

a b

(19) Explain how the following picture “proves” that the sum of a number x and its inverse 1/x is at least 2.

1 x 1 1

x 1 1

1 x x

(20) Explain how the following picture “proves” that if a and b are positive numbers then: a + b √a b 6 · 2

74 CHAPTER 2. PROOF BY PICTURE

and the inequality above is an equality when a = b.

(21) Explain how the following picture “proves” that:

n(n + 1) 1 + 2 + 3 + 4 + 5 + + n = · · · 2

(22) Explain how the following picture “proves” that:

1 n(n + 1) 1 + 2 + 3 + 4 + 5 + + n = (n2 + n) = · · · 2 2

75 2.4. PROOF BY PICTURE

(23) Explain how the following picture “proves” that:

n2 n n(n + 1) 1 + 2 + 3 + 4 + 5 + + n = + = · · · 2 2 2

(24) Explain how the following picture “proves” that:

1 + 3 + 5 + + (2n 1) = n2 · · · −

76 CHAPTER 2. PROOF BY PICTURE

(25) Explain how the following picture “proves” that:

1 + 2 + 3 + + (n 1) + n + (n 1) + + 3 + 2 + 1 = n2 · · · − − · · ·

(26) Explain how the following picture “proves” that:

1 1 2 1 3 1 4 1 5 + + + + + = 1 2 2 2 2 2 · · ·

1 1/2

(1/2)2

(27) Explain how the following picture “proves” that if 0 < r < 1:

r + r(1 r) + r(1 r)2 + r(1 r)3 + = 1 − − − · · ·

77 2.4. PROOF BY PICTURE

r(1 − r)3

r(1 − r)2

1 r r(1 − r)

r(1 − r) r

(28) Explain how the following picture “proves” that:

1 1 2 1 3 1 4 1 5 1 + + + + + = 4 4 4 4 4 · · · 3

(29) Explain how the following picture “proves” that:

1 1 2 1 3 1 4 + + + + = 1 2 2 2 2 · · ·

78 CHAPTER 2. PROOF BY PICTURE

Hint: Add up the area of the shaded regions and the area of the unshaded regions. (30) Explain how the following picture “proves” that: 1 + 3 + + (2n 1) 1 · · · − = (2n + 1) + (2n + 3) + + (4n 1) 3 · · · −

(31) Explain how the following picture “proves” that: 1 + 3 + + (2n 1) 1 · · · − = (2n + 1) + (2n + 3) + + (4n 1) 3 · · · − 2n

1 3 5

2n − 1

(32) Look at: 1 1 = 1 · 11 11 = 121 · 111 111 = 12321 · 1111 1111 = 1234321 ·

79 2.4. PROOF BY PICTURE

Does the pattern continue? Explain your answer.

(33) Explain how the following picture “proves” that in any packing, the number of calissons with a given orientation is exactly one-third the total number of calissons in the box.

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Topics in Plane Geometry

Geometry is the science of correct reasoning on incorrect ﬁgures. —George Polya

3.1 Triangles

3.1.1 Centers in Triangles

The idea of a center for an equilateral triangle makes sense. However, for an arbitrary triangle, there can be several diﬀerent ideas for what the center is.

Question How could one deﬁne the “center” of an arbitrary triangle? ?

The Circumcenter

Theorem 6 The perpendicular bisector of the sides of a triangle meet at a point. This point is called the circumcenter.

Here is a picture illustrating the theorem above:

81 3.1. TRIANGLES

Paper-Folding Construction To construct the circumcenter using paper-folding, perform the following steps: (1) Fold the leg of the triangle over top of itself so that its endpoints meet. (2) Repeat Step 1 for each of the 3 legs of the triangle. (3) The creases made in Steps 1 and 2 above should meet at the circumcenter of the triangle.

What is it really? The circumcenter is the center of the circle that circumscribes the triangle:

This circle is sometimes called the circumcircle.

The Incenter Theorem 7 The interior bisectors of the angles of a triangle meet at a point. This point is called the incenter.

Paper-Folding Construction To construct the incenter using paper-folding, perform the following steps: (1) Choose a vertex of the triangle and fold the triangle over top of itself so that legs adjacent to the chosen vertex line up. (2) Repeat Step 1 for each of the 3 vertexes of the triangle.

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(3) The creases made in Steps 1 and 2 above should meet at the incenter of the triangle.

What is it really? If we draw a circle inside the triangle such that the circle touches each of the sides of the triangle, then the incenter is the center of this circle.

This circle is sometimes called the incircle.

The Orthocenter Recall the following deﬁnition: Deﬁnition An altitude of a triangle is a line segment originating at a vertex of the triangle that meets the line containing the opposite side at a right angle. Since the perpendicular bisectors of the large triangle form the altitudes of the smaller triangle, we now have the following theorem: Theorem 8 The altitudes of a triangle meet at a point. This point is called the orthocenter.

Paper-Folding Construction To construct the orthocenter using paper-folding, perform the following steps: (1) Fold the leg of the triangle over top of itself so that the vertex opposite the leg is on the crease. (2) Repeat Step 1 for each of the 3 legs of the triangle. (3) The creases made in Steps 1 and 2 above should meet at the orthocenter of the triangle.

83 3.1. TRIANGLES

The Centroid Theorem 9 The lines of a triangle that connect the vertexes to the midpoints of the opposite sides of a triangle meet at a point. This point is called the centroid.

The line segments used in ﬁnding the centroid have a special name:

Deﬁnition A median of a triangle is a line segment that connects a vertex to the midpoint of the opposite side.

Paper-Folding Construction To construct the centroid using paper-folding, perform the following steps:

(1) Fold the leg of the triangle over top of itself so that its endpoints meet.

(2) Take the vertex opposite the leg above, and make a crease that starts at the vertex and extends to where the crease made in Step 1 meets the leg opposite the vertex.

(3) Repeat Steps 1 and 2 for each of the 3 legs of the triangle.

(4) The creases made in the paper in Steps 2 and 3 above should meet at the centroid of the triangle.

What is it really? The centroid is the center of mass of the triangle. The center of mass is the balancing point of an object. That is, if we had a triangle made of say cardboard, then you could balance the cardboard triangle on its centroid.

Putting it all Together The next turn we should make on our path is to understand the relationship between the centers above.

Question Is it always/ever the case that the circumcenter, the orthocenter, the incenter, and the centroid are all actually the same point? ?

84 CHAPTER 3. TOPICS IN PLANE GEOMETRY

Question Can the circumcenter ever be outside the triangle? ? Question Can the incenter ever be outside the triangle? ? Question Can the orthocenter ever be outside the triangle? ? Question Can the centroid ever be outside the triangle? ? Keeping all these ideas straight can be tough, and unfortunately, I don’t know of an easy way to learn them. However, a ﬁrst step is knowing the names of each of the centers described above. Here is simple mnemonic that may help: c e C O I N i rn t r t c r c h e o u on i m c t d c e e e n r n t t e e r r Note that the centers that correspond to the IN part of COIN are always inside the triangle!

3.1.2 Theorems about Triangles Now consider some triangle and look at the orthocenter, the centroid, and the circumcenter. In the illustration below, O is the orthocenter, N is the centroid, and C is the circumcenter:

O N C

85 3.1. TRIANGLES

Do you notice anything?

Theorem 10 (Euler) The circumcenter, the centroid, and the orthocenter are on a line. The centroid lies a third of the distance from the circumcenter to the orthocenter. This line is called the Euler line.

What about all the other points on the above triangle? Well there are many many theorems about all kinds of points. Here is one that relates nine diﬀerent points to each other:

Theorem 11 Given any triangle, three sets of three points all lie on a circle. Those three sets are:

(1) The midpoints of the sides of the triangle.

(2) Where the altitudes meet the lines containing the sides of the triangle.

(3) Midpoints of the segments joining the orthocenter and the vertexes.

This circle is called the Nine-Point Circle.

Question How would one go about drawing this? ?

Theorem 12 The center of the Nine-Point Circle bisects the segment joining the orthocenter, point O, and the circumcenter, point C.

O C

While the previous theorems concerned themselves with points, the next theorem involves arcs.

86 CHAPTER 3. TOPICS IN PLANE GEOMETRY

Theorem 13 (Miquel) Consider three points, one on each side of a triangle, none of which is on a vertex of the triangle. Then the three circles determined by a vertex and the two points on adjacent sides meet at a point. This point is called a Miquel point.

Question How would one go about drawing this? ? Question Why do we want the points in Miquel’s Theorem not to be on the vertexes of the triangle? ? A ﬁnal theorem about triangles:

Theorem 14 (Morley) If you trisect the angles of any triangle with lines, then those lines form a new equilateral triangle inside the original triangle.

Paper-Folding Construction To construct the triangle described in Morley’s Theorem using paper-folding, perform the following steps:

(1) We must trisect an angle of the triangle. Choose a vertex of the triangle and fold the paper so that the crease leads up to the vertex, with the edge of the ﬂap being folded over bisecting the new angle of the crease and the edge that was not moved.

87 3.1. TRIANGLES

(2) Now fold the edge that was not moved on top of the flap that was just made. It should fit perfectly near the vertex. If done correctly, Steps 1 and 2 should trisect the chosen vertex. (3) Repeat Steps 1 and 2 for each vertex of the triangle. (4) Now mark the adjacent intersections of creases coming from different vertexes. (5) Connect the marks made in Step 4 above. This should form an equilateral triangle.

A Paradox Here is another devilish paradox! Paradox All triangles are isosceles.

False-Proof Consider

X Z

B C Y

The central point is the point where the bisector of ∠XAZ meets the perpendicular bisector of line BC, we’ll call this point P . We want to show that AB AC. To see this ﬁrst note that ≡ AXP AZP △ ≡ △ since they have equal angles and share a side. Next note that

BYP Y CP. △ ≡ △ Hence BXP ZCP as they are right triangles with two equal sides. Thus AB AC△ . ≡ △ ≡ This is a paradox as not all triangle are isosceles and yet we seem to have proved that all triangles are isosceles! There must be something wrong with the proof. Question What is wrong with the proof above? Hint: Use paper-folding to construct the ﬁgure in the above false-proof. ?

88 CHAPTER 3. TOPICS IN PLANE GEOMETRY

Problems for Section 3.1 (1) Use paper-folding to construct the circumcenter of a triangle.

(2) Use paper-folding to construct the orthocenter of a triangle.

(3) Use paper-folding to construct the incenter of a triangle.

(4) Use paper-folding to construct the centroid of a triangle.

(5) Explain how a perpendicular bisector is diﬀerent from an altitude. Draw an example to illustrate the diﬀerence.

(6) Explain how a median diﬀerent from an angle bisector. Draw an example to illustrate the diﬀerence.

(7) What is the name of the point that is the same distance from all three sides of a triangle? Explain your answer.

(8) What is the name of the point that is the same distance from all three vertexes of a triangle? Explain your answer.

(9) Could the circumcenter be outside the triangle? If so, draw a picture and explain. If not, explain why not using pictures as necessary.

(10) Could the orthocenter be outside the triangle? If so, draw a picture and explain. If not, explain why not using pictures as necessary.

(11) Could the incenter be outside the triangle? If so, draw a picture and explain. If not, explain why not using pictures as necessary.

(12) Could the centroid be outside the triangle? If so, draw a picture and explain. If not, explain why not using pictures as necessary.

(13) Are there shapes that do not contain their centroid? If so, draw a picture and explain. If not, explain why not using pictures as necessary.

(14) Draw a triangle. Now draw the altitudes of this triangle. How many orthocenters do you have as intersections of lines in your drawing? Hints:

(a) More than one. (b) How many triangles are in the picture you drew?

(15) Where is the circumcenter of a right triangle?

(16) Where is the orthocenter of a right triangle?

(17) Can you draw a triangle where the circumcenter, orthocenter, incenter, and centroid are all the same point? If so, draw a picture and explain. If not, explain why not using pictures as necessary.

(18) True or False: Explain your conclusions.

89 3.1. TRIANGLES

(a) An altitude of a triangle is always perpendicular to a line containing some side of the triangle. (b) An altitude of a triangle always bisects some side of the triangle. (c) The incenter is always inside the triangle. (d) The circumcenter, the centroid, and the orthocenter always lie in a line. (e) The circumcenter can be outside the triangle. (f) The orthocenter is always inside the triangle. (g) The orthocenter is always the center of the Nine-Point Circle. (19) How many Euler-Lines does a given triangle have? Explain your reasoning. (20) How many Nine-Point Circles does a given triangle have? Explain your reasoning. (21) How many Miquel points does a given triangle have? Explain your reasoning. (22) Sketch and label a Nine-Point Circle. (23) Sketch and label an Euler-Line. (24) True or False: Explain your conclusion. (a) The centroid of a triangle always lies on the Nine-Point Circle. (b) If a given altitude of a triangle is extended to an inﬁnite line, then the Nine-Point Circle will always intersect that line in two distinct points. (c) Given a triangle, the Nine-Point Circle for that triangle is the circumscribed circle for a triangle whose vertexes are the midpoints of the sides of original triangle. (d) A triangle does not ever share 9 points with its Nine-Point Circle. (e) There is a triangle that only shares 3 points with its Nine-Point Circle. (25) Does a Miquel point always lie inside the triangle? Explain your answer. (26) Use paper-folding to illustrate Morley’s Theorem. (27) Illustrate the statement of Morley’s Theorem using triangles with the following angles:

(a) 60◦, 60◦, and 60◦.

(b) 30◦, 60◦, and 90◦.

90 CHAPTER 3. TOPICS IN PLANE GEOMETRY

3.2 Numbers

Numbers are synonymous with mathematics. In this section we will discuss several ways that numbers come up in geometry.

3.2.1 Areas Heron’s Formula The standard formula for the area of a triangle is: area = (1/2)base height · Could we compute the area of a triangle just knowing the lengths of the sides? Well after a moments thought, and some sketches, one should decide that it is indeed possible, but the many different configurations of possible triangles might make it difficult. Enter Heron’s Formula: Theorem 15 (Heron’s Formula) Given a triangle whose sides have length a, b, and c, then the area of the triangle is given by

P P P P area = a b c 2 − 2 − 2 − 2 s where P = a + b + c, the perimeter of the triangle. Question Why does it make sense that a formula for the area of a triangle can be given knowing only the sides? Could a similar statement be made for all quadrilaterals? ? Giving away part of this question, we only have similar formulas for quadrilaterals in certain cases. Our next lemma will help us out. What is a lemma, you ask? A lemma is nothing but a little theorem that we have to help us solve another problem. Note that a lemma should not be confused with the more sour lemon, as that is something diﬀerent and unrelated to what we are discussing. Lemma 16 A quadrilateral can be circumscribed in a circle if and only if opposite angles sum to 180 degrees. If a quadrilateral can be circumscribed by a circle then we have another version of Heron’s Formula: Theorem 17 (Heron’s Formula) Given a quadrilateral that can be circumscribed by a circle, whose sides have length a, b, c, and d, then the area of the quadrilateral is given by

P P P P area = a b c d 2 − 2 − 2 − 2 − s where P = a + b + c + d, the perimeter of the quadrilateral.

91 3.2. NUMBERS

It should be noted that we can see that Heron’s Formula for the area of a quadrilateral reduces to Heron’s Formula for the triangle when d = 0.

Lattice points Lattice points are just points that are spaced 1 unit apart, horizontally and vertically, in the plane. The following picture shows some lattice points:

If a polygon has a vertexes on lattice points, then we can easily compute its area using Pick’s Theorem: Theorem 18 (Pick) If a polygon has its vertexes on a lattice, then

b area = + n 1 2 − where b is the number of lattice points on the border of the polygon and n is the number of lattice points inside our polygonal region.

Example Suppose you want to ﬁnd the area of the following polygon:

Since it has its vertexes on lattice points, we can use Pick’s Theorem: 8 area = + 2 1 = 5 square units. 2 − 3.2.2 Ratios Turning Tricks into Techniques We will show you three separate tricks, which are all quite similar. By considering these tricks, we will develop techniques for solving problems.

First Trick How is 0.999 . . . related to 1? I claim we have the following paradox: I intend to show that

0.999 . . . = 1.

92 CHAPTER 3. TOPICS IN PLANE GEOMETRY

To see this, set: x = 0.999 . . . Now we have: 10x = 9.999 . . . So we see that

10x x = (9.999 . . .) (0.999 . . .) − − 9x = 9, and we are forced to conclude that x = 1, but we started oﬀ with the assumption that x = 0.999 . . . , hence 1 = 0.999 . . . . This paradox challenges a common implicit (and false) notion that every number has exactly one decimal representation. We are forced to conclude that 0.999 . . . and 1 are representations for the same number! To be completely explicit, using a similar method as described above we see that

4.999 . . . = 5, 7.3999 . . . = 7.4, 23.745999 . . . = 23.746, and so on. Hence numbers can have multiple decimal representations.

Second Trick Keeping the ﬁrst trick used above in the back of your mind, consider this next similar trick. What is:

x = 1 + 1 + 1 + √1 + s · · · r q Now we have: x = √1 + x Squaring both sides we get: x2 = 1 + x. Now putting everything on the left-hand side:

x2 x 1 = 0. − − By the Quadratic Formula, 1 √5 x = ± , 2 and since we can see that x > 1, we must conclude that

1 + √5 x = . 2 Well this seems like a strange number. Oh well, let’s keep on going.

93 3.2. NUMBERS

Third Trick Now what about: 1 x = 1 + 1 1 + 1 1 + 1 1 + 1 + · · · Again using a similar trick as above,

1 x = 1 + . x Multiplying both sides by x we get:

x2 = x + 1.

Now putting everything on the left-hand side:

x2 x 1 = 0. − − By the Quadratic Formula, 1 √5 x = ± , 2 and since we can see that x > 1 we must conclude that

1 + √5 x = . 2 Wait a minute, this says that:

1 1 + 1 + 1 + √1 + = 1 + s · · · 1 r 1 + q 1 1 + 1 1 + 1 + · · · Wow! Who would have ever thought that? These two crazy looking formulas are equal, and despite the fact that the only number in them is a one, they are both equal to the messy number:

1 + √5 = 1.6180339887 . . . 2

Continued Fractions Can we use similar techniques (tricks) to study other numbers that have a nasty form? You bet! Before we do that, we’ll need a deﬁnition:

94 CHAPTER 3. TOPICS IN PLANE GEOMETRY

Deﬁnition A fraction of the form

a1 a0 + a b + 2 1 a3 b2 + a b + 4 3 b + 4 · · · is called a continued fraction. If a1, a2, a3,... are all 1, we will call this a simple continued fraction.

So what about the continued fraction: 1 x = 1 + 1 2 + 1 2 + 1 2 + 2 + · · · Using the technique as above,

1 x = 1 + . 1 + x

Multiplying both sides by 1 + x we get:

x + x2 = (1 + x) + 1.

Now putting all the x’s on the left-hand side and all the numbers on the right:

x2 = 2.

Ah! So x = √2. Using your calculator, you can see that:

√2 = 1.4142135623 . . .

This means that: 1 1.4142135623 . . . = 1 + 1 2 + 1 2 + 1 2 + 2 + · · · Note that on the left-hand side you don’t see much of a pattern. However, on the right-hand side a clear pattern is formed. This is part of the beauty of continued fractions. Now it turns out that you can do this with other numbers and and get lots of other cool patterns!

95 3.2. NUMBERS

Check out e = 2.718281828459045 . . . . It turns out that 1 e = 2 + 1 1 + 1 2 + 1 1 + 1 1 + 1 4 + 1 1 + 1 1 + 1 6 + 1 + · · · Also check out π = 3.14159265358 . . . . It turns out that we can get a nice continued fraction for π: 12 π = 3 + 32 6 + 52 6 + 72 6 + 92 6 + 112 6 + 6 + · · · Going the Other Way Given a number, can you find the continued fraction of the number? It turns out that if the number is sufficiently nice, then this is not so hard. Before we start with an example, let’s get some definitions and a lemma out of the way: Definition The whole-number part of a number is the largest whole number which is less than or equal to the given number. Example The whole-number part of 2 is 2, while the whole-number part of 5.32 is 5. Definition The fractional part of a number is the number minus its whole- number part. Example The fractional part of 2 is 0, while the fractional part of 5.32 is 0.32. Question Why don’t we just describe the fractional part of a number as the part that is to the right of the decimal point? Hint: Think about 0.99999 . . . . ? Given any number, we can write it as a simple continued fraction. Consider 13/5. To start note that 13 3 > > 2. 5

96 CHAPTER 3. TOPICS IN PLANE GEOMETRY

So this means that 13 3 = 2 + . 5 5 Here 2 is the whole-number part and 3/5 is the fractional part of 13/5. But in the simple continued fraction, our numerator is 1, not 3. How do we deal with this? Well, 13 3 1 = 2 + = 2 + 5 . 5 5 3 This is an improvement but we only want whole numbers in our simple continued fractions and not 5/3. So we write 5 2 = 1 + 3 3 which gives us 13 1 = 2 + . 5 2 1 + 3 Again, we want our numerator to be 1, not 2 so we will repeat the steps above to get 13 1 1 1 = 2 + = 2 + = 2 + 5 2 1 1 1 + 1 + 1 + 3 3 1 2 1 + 2 and this last expression is the simple continued fraction for 13/5. We could also list our steps as: 13 3 = 2 + 5 5 1 5 2 3 = = 1 + 5 3 3 1 3 1 2 = = 1 + 3 2 2 1 1 = 2 + 0 2 These boldface numbers tell us our continued fraction expansion. We can also ﬁnd the simple continued fraction of numbers which are not already fractions (otherwise this would all be a bit silly). Consider √2. To start note that 2 > √2 > 1. So this means that √2=1+(√2 1). − Where 1 is the whole-number part and (√2 1) is the fractional part of √2. Alright, now look at 1/(√2 1). Again we want− to separate the whole-number part and the fractional part.− With a little algebra we see that 1 √2 + 1 = = √2+1=2+(√2 + 1 2) = 2 + (√2 1). √2 1 2 1 − − − − 97 3.2. NUMBERS

Now don’t you get bogged down in the steps. Here it is in fast forward: √2 = 1 + (√2 1) − 1 = 2 + (√2 1) (√2 1) − − 1 = 2 + (√2 1) (√2 1) − − 1 = 2 + (√2 1), (√2 1) − − . . At each step we want: number = whole-number part + fractional part Now from the bold-faced numbers above we will make our continued fraction: 1 √2 = 1 + 1 2 + 1 2 + 1 2 + 2 + · · · Question Can you explain why this works? ?

The Golden Ratio It turns out that the number 1 + √5 1 φ = = 1 + 2 1 1 + 1 1 + 1 1 + 1 + · · · is a special number that we call the golden ratio. We denote the golden ratio by the symbol φ. Question What’s so special about the golden ratio? Given any rectangle you can divide it into a square and another smaller rectangle:

98 CHAPTER 3. TOPICS IN PLANE GEOMETRY

Suppose that you want the new smaller rectangle to have the same proportions as the original rectangle. This will only happen if the ratio of the sides of the rectangle are φ to 1. We can show this to be true algebraically. Start by making a sketch of a rectangle: x − 1

x So we want 1 x 1 = − x 1 and so: 1 = x2 x − Thus x2 x 1 = 0. − − But we have already solved this equation several times, its solution is: 1 + √5 x = = φ. 2 Deﬁnition A rectangle with the proportions of φ for one of its sides and 1 for the other is called a golden rectangle. Given a golden rectangle, we can put a spiral in side, by making a quarter of a circle in every square:

Moreover, we can make what is called a golden triangle, an isosceles triangle with two long sides being related to the shorter side by a ratio of φ to 1. We can place a similar spiral in this shape as before:

99 3.2. NUMBERS

Question How would you draw the above ﬁgures? ?

3.2.3 Combining Areas and Ratios—Probability Deﬁnition The probability of an event occurring is a number between 0 and 1 giving a linear scale of the likelihood of the event occurring, with 0 meaning impossible, and 1 meaning certain.

Question Could we build a machine that would take random numbers and from them produce a closer and closer approximation of π? ? To get at the above question, let’s talk about probability.

Question What is the probability that a point chosen at random in the square below will land in the shaded region?

? Question What is the probability that a point chosen at random in the square below will land in the shaded region?

100 CHAPTER 3. TOPICS IN PLANE GEOMETRY

? OK! Now a tricky question: Question What is the probability that a point chosen at random in the square below will land inside the shaded area of the arc?

I’ll help you solve this one. What if we expand our view a bit? Say the square above has area 1. Now consider

In this case the area of the big square is 4 and the area of the shaded region is π, as it is a circle of radius 1. Thus if a point is chosen at random in the big square, the probability that it lands in the shaded circle is π/4. However, the little square and the little quarter-circle above are only 1/4 of this each. So we have a probability of π/4 = π/4 4/4 of landing in the shaded region in the little square above. So the above example gives us a hint how we could take random numbers and turn them into an approximation for π. Here is an algorithm that should do the trick: (1) Take a random set of numbers all between 0 and 1. (2) Take pairs of these numbers (a, b). Let n be the total number of pairs that we have. (3) To see if a pair (a, b) lands inside the circle use the Pythagorean Theorem, that is if a2 + b2 6 1 then the point is inside the circle, otherwise the point is outside the circle.

101 3.2. NUMBERS

(4) Count how many pairs land inside the circle and divide by the total number of pairs.

(5) This ratio approximates π/4. To get an approximation for π, multiply your answer by 4.

The Monty Hall Problem

There used to be a TV show called Let’s Make a Deal. It was hosted by Monty Hall. At the end of the show something like this was presented to the leading contestant.

1 2 3

These are three doors. Behind two of the doors is something you don’t want—say a goat. But behind one of the doors is something you might like— say a brand new car! Here is how the game works: You point to a door and then Monty Hall opens another door revealing a goat. He then oﬀers to let you switch or stay. If you stay, then you open your door to reveal either the second goat or the fabulous car. If you decide to switch, you open the other remaining door to reveal either the second goat or the fabulous car.

Question Considering the Monty Hall problem, is it better to switch or is it better to stay? Hint: This problem is very tricky and has fooled many a good mathematician! ?

Bertrand’s Paradox

Here is an innocent looking question:

Question Given a circle, ﬁnd the probability that a chord chosen at random is longer than the side of an inscribed equilateral triangle.

Let’s use our new found skills in probability to hack this one to pieces.

Solution 1 How do we pick a random chord? Well imagine two spinners mounted at the center of the circle. When they get done spinning, just connect the points that they point at to get your chord. Let the one spinner ﬁnish anywhere. Now the second spinner will tell us whether the chord is longer than

102 CHAPTER 3. TOPICS IN PLANE GEOMETRY the edge of the triangle.

Since there are 60 degrees in a triangle, and there are 180 degrees in a straight line, the probability that the random chord will be longer than the edge of the triangle is 60 = 1/3. 180

Here is another solution:

Solution 2 Let’s consider another way of picking a random chord. By drawing some pictures, one can see that a chord can be determined completely by its midpoint, unless the midpoint is the center of the circle—but what is the probability of a random point landing in the exact center of a circle?

Thus a chord is longer than the edge of the triangle if and only if its midpoint lands inside the circle inscribed inside of the triangle. However, it can be shown that the radius of the circle inside the triangle is half of the radius of the large circle and hence the ratio of the areas of the circles is:

π(1/2)2 = 1/4, π12 and this must be the probability that the random chord will be longer than the edge of the triangle.

D’oh! 1/3 = 1/4! 6 Question What is wrong in the above discussion? ?

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Wacky Dice

Consider the game Rock-Paper-Scissors. In this game two players make hand gestures. Each player’s hand gesture represents one of the following: a rock, a piece of paper, or scissors. According to the rules of the game:

Rock beats scissors, by breaking them of course. • Scissors beats paper, by cutting it of course. • Paper beats rock, by covering it of course1. • Now consider these dice:

3 2 1 0 33 3 62 6 5 1 5 4 0 4 3 2 1 4 3 2 5 4

(3) (2) (1) (0)

Here is a new game. Considering the dice above, let your opponent pick a die. Then you pick one of the remaining three dice. Each player throws their die and the highest number wins a point. Play until someone reaches 10 points. You’ll have a much better chance of winning the game if you know these facts: Die (3) has a probability of 2/3 beating Die (2), Die (2) has a probability of 2/3 beating Die (1), Die (1) has a probability of 2/3 beating Die (0), and Die (0) has a probability of 2/3 beating Die (3). Just as in Rock-Paper-Scissors we have made a full circle. These crazy dice were invented by Bradley Efron. Die (3) has a 2/3 probability of beating Die (2) since no matter what Die (3) rolls, there are 4 losing squares out of a total of 6 squares on Die (2):

2 62 6 2 2

(2)

Before we can compute the next probability we need two lemmas.

1OK I admit it, the method through which paper actually beats rock has always been a mystery to me.

104 CHAPTER 3. TOPICS IN PLANE GEOMETRY

Lemma 19 If two events E1 and E2 are disjoint, meaning that they cannot both occur together, then the probability that E1 or E2 will happen is the sum of the probability that E1 happens with the probability that E2 happens. Symbolically, if P (Ei) is the probability that event Ei happens, then

P (E E ) = P (E ) + P (E ). 1 ∨ 2 1 2

Lemma 20 If two events E1 and E2 are independent, meaning that the occurrence of one has no eﬀect on the occurrence of the other, then the probability that E1 and E2 will happen is the product of the probability that E1 happens with the probability that E2 happens. Symbolically, if P (Ei) is the probability that event Ei happens, then

P (E E ) = P (E ) P (E ). 1 ∧ 2 1 · 2 Die (2) has a 2/3 probability of beating Die (1) since you roll a 6 with Die (2) and win 2 62 6 2 2

(2) with a 2/6 = 1/3 probability or you roll as follows:

2 1 62 6 and 5 1 5 2 1 2 5

(2) (1)

Since there is an and there, we must multiply the probabilities to get a probability of 4 1 1 = . 6 · 2 3 Since there was an or above, we must add

1 1 2 + = . 3 3 3 Thus Die (2) has a 2/3 probability of beating Die (1).

105 3.2. NUMBERS

Die (1) has a 2/3 probability of beating Die (0) since you roll a 5 with Die (1) and win 1 5 1 5 1 5

(1) with a 3/6 = 1/2 probability or you roll as follows:

1 0 5 1 5 and 4 0 4 1 4 5 4 (1) (0)

Since there is an and there, we must multiply the probabilities to get a probability of 1 2 1 = . 2 · 6 6 Since there was an or above, we must add 1 1 2 + = . 2 6 3 Thus Die (1) has a 2/3 probability of beating Die (0). Finally Die (0) has a 2/3 probability of beating Die (3) since you roll a 4 with Die (0) and win 0 4 0 4 4 4

(0) with a 4/6 = 2/3 probability.

106 CHAPTER 3. TOPICS IN PLANE GEOMETRY

Problems for Section 3.2 (1) Find the area of the triangle whose sides are of length 3, 4, and 5. Explain your work. (2) Find the area of the triangle whose sides are of length 10, 6, and 10. Explain your work. (3) Find the area of the triangle whose sides are of length 5, 5, and 5. Explain your work. (4) Find the area of the triangle whose sides are of length 4, 6, and 10. Explain your work. Does this answer make sense? Why or why not? (5) Find the area of the quadrilateral that can be inscribed in a circle whose sides are of lengths 9, 3, 5, and 3. Explain your work. (6) Find the area of the quadrilateral that can be inscribed in a circle whose sides are of lengths 6, 4, 5, and 9. Explain your work. (7) Find the area of the quadrilateral that can be inscribed in a circle whose sides are of lengths 2, 5, 6, and 3. Explain your work. (8) Find the area of the quadrilateral that can be inscribed in a circle whose sides are of lengths 2, 10, 5, and 3. Explain your work. Does your answer make sense? Why or why not? (9) Compute the area of:

Explain your work and include units with your answer. (10) Compute the area of:

Explain your work and include units with your answer. (11) Compute the area of:

107 3.2. NUMBERS

Explain your work and include units with your answer.

(12) Explain what φ is in terms of squares and rectangles.

(13) Explain what φ is in terms of a continued square-root.

(14) Explain what φ is in terms of a continued fraction.

(15) Courtney Gibbons is someone who has a rather unusual tattoo. She was kind enough to let an unusual person like me take a picture of it. What does her tattoo represent? Explain your reasoning.

(16) Find the exact value for x when:

x = 2 + 2 + 2 + √2 + s · · · r q Explain your work.

(17) Find the exact value for x when:

x = 6 + 6 + 6 + √6 + s · · · r q Explain your work.

(18) Find the exact value for x when:

x = 12 + 12 + 12 + √12 + s · · · r q Explain your work.

108 CHAPTER 3. TOPICS IN PLANE GEOMETRY

(19) Find the exact value for x when:

x = 20 + 20 + 20 + √20 + s · · · r q Explain your work. (20) Find the exact value for x when:

x = 30 + 30 + 30 + √30 + s · · · r q Explain your work. (21) Find the exact value for x when: 1 x = 2 + 1 4 + 1 4 + 1 4 + 4 + · · · Explain your work. (22) Find x when: 1 x = 4 + 1 6 + 1 6 + 1 6 + 6 + · · · Explain your work. (23) Find the exact value for x when: 1 x = 4 + 1 8 + 1 8 + 1 8 + 8 + · · · Explain your work. (24) Find the exact value for x when: 1 x = 3 + 1 10 + 1 10 + 1 10 + 10 + · · · Explain your work.

109 3.2. NUMBERS

(25) Explain what the whole-number part and what the fractional part of a number are. Give examples.

(26) Find the simple continued fraction expansion of 5/3. Explain your work.

(27) Find the simple continued fraction expansion of 15/11. Explain your work.

(28) Find the simple continued fraction expansion of 22/17. Explain your work.

(29) Using a calculator, find the first five terms in the simple continued fraction expansion of π. What number do you get by only considering the first term? The first four?

(30) Find the simple continued fraction expansion of √5. Explain your work.

(31) Find the simple continued fraction expansion of √10. Explain your work.

(32) Find the simple continued fraction expansion of √17. Explain your work.

(33) Find the simple continued fraction expansion of √26. Explain your work.

(34) Find the simple continued fraction expansion of 1/2. Explain your work.

(35) Find the simple continued fraction expansion of 11. Explain your work.

(36) What is it about the numbers 2, 5, 10, 17, 26 that makes it easy to compute the continued fraction expansion of the square-roots of these numbers? Explain your answer.

(37) What is the deﬁnition of probability?

(38) Aloof old Professor Rufus came into his Calculus class one day and said, “I have chosen a real number randomly and will give anyone an ‘A’ on the next exam who can guess it!” Aloof old Professor Rufus knew that the probability of someone finding a single real number with a finite number of guesses is 0, and so was quite confident that none could guess his number. It should be no surprise to you that aloof old Professor Rufus turned quite white when after 37 guesses, Smart Sally guessed his number which was 13. How is it that the students were able to guess aloof old Professor Rufus’ number? How was aloof old Professor Rufus wrong about the probability?

(39) What is the probability that a point chosen at random in the larger rectangle below will land in the shaded region? Explain your answer.

110 CHAPTER 3. TOPICS IN PLANE GEOMETRY

(40) What is the probability that a point chosen at random in the larger rectangle below will land in the shaded region? Explain your answer.

(41) What is the probability that a point chosen at random in the larger rectangle below will land in the shaded region? Explain your answer.

(42) What is the probability that a point chosen at random in the larger region below will land in the smaller shaded region? Explain your answer.

(43) What is the probability that a point chosen at random in the larger region below will land in the smaller shaded region? Explain your answer.

(44) What is the probability that a point chosen at random in the larger region below will land in the smaller shaded region? Explain your answer.

111 3.2. NUMBERS

(45) Explain how someone could start to approximate π using a dart-board and darts.

(46) Here are two dice. Which die has a better probability of rolling higher, and what is that probability? Explain your answer.

3 2 4 4 4 62 6 3 2 3 2

(a) (b)

(47) Here are two dice. Which die has a better probability of rolling higher, and what is that probability? Explain your answer.

3 5 4 4 4 2 5 2 3 5 3 2

(a) (b)

(48) Here are two dice. Which die has a better probability of rolling higher, and what is that probability? Explain your answer.

3 2 4 4 4 6 6 6 3 2 3 2

(a) (b)

112 Chapter 4

Compass and Straightedge Constructions

Mephistopheles: I must say there is an obstacle That prevents my leaving: It’s the pentagram on your threshold. Faust: The pentagram impedes you? Tell me then, you son of hell, If this stops you, how did you come in? Mephistopheles: Observe! The lines are poorly drawn; That one, the outer angle, Is open, the lines don’t meet. —G¨othe, Faust act I, scene III

4.1 Constructions

About a century before the time of Euclid, Plato—a student of Socrates— declared that the compass and straightedge should be the only tools of the geometer. Why would he do such a thing? For one thing, both the the compass and straightedge are fairly simple instruments. One draws circles, the other draws lines—what else could possibly be needed to study geometry? Moreover, rulers and protractors are far more complex in comparison and people back then couldn’t just walk to the campus bookstore and buy whatever they wanted. However, there are other reasons:

(1) Compass and straightedge constructions are independent of units.

(2) Compass and straightedge constructions are theoretically correct.

(3) Combined, the compass and straightedge seem like powerful tools.

113 4.1. CONSTRUCTIONS

Compass and straightedge constructions are independent of units. Whether you are working in centimeters or miles, compass and straightedge constructions work just as well. By not being locked to set of units, the constructions given by a compass and straightedge have certain generality that is appreciated even today.

Compass and straightedge constructions are theoretically correct. In mathematics, a correct method to solve a problem is more valuable than a correct solution. In this sense, the compass and straightedge are ideal tools for the mathematician. Easy enough to use that the rough drawings that they produce can be somewhat relied upon, yet simple enough that the tools themselves can be described theoretically. Hence it is usually not too diﬃcult to connect a given construction to a formal proof showing that the construction is correct.

Combined, the compass and straightedge seem like powerful tools. No tool is useful unless it can solve a lot of problems. Without a doubt, the compass and straightedge combined form a powerful tool. Using a compass and straightedge, we are able to solve many problems exactly. Of the problems that we cannot solve exactly, we can always produce an approximate solution. We’ll start by giving the rules of compass and straightedge constructions:

Rules for Compass and Straightedge Constructions (1) You may only use a compass and straightedge.

(2) You must have two points to draw a line.

(3) You must have a point and a line segment to draw a circle. The point is the center and the line segment gives the radius.

(4) Points can only be placed in two ways:

(a) As the intersection of lines and/or circles. (b) As a free point, meaning the location of the point is not important for the ﬁnal outcome of the construction.

Our ﬁrst construction is also Euclid’s ﬁrst construction:

Construction (Equilateral Triangle) We wish to construct an equilateral triangle given the length of one side.

(1) Open your compass to the width of the line segment.

(2) Draw two circles, one with the center being each end point of the line segment.

(3) The two circles intersect at two points. Choose one and connect it to both of the line segment’s endpoints.

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Euclid’s second construction will also be our second construction: Construction (Transferring a Segment) Given a segment, we wish to move it so that it starts on a given point. (1) Draw a line through the point in question. (2) Open your compass to the length of the line segment and draw a circle with the given point as its center. (3) The line segment consisting of the given point and the intersection of the circle and the line is the transferred segment. If you read The Elements, you’ll see that Euclid’s construction is much more complicated than ours. Apparently, Euclid felt the need to justify the ability to move a distance. Many sources say that Euclid used what is called a collapsing compass, that is a compass that collapsed when it was picked up. However, I do not believe that such an invention ever existed. Rather this is something that lives in the conservative geometer’s head. Regardless of whether the diﬃculty of transferring distances was theoretical or physical, we need not worry when we do it. In fact, Euclid’s proof of the above theorem proves that our modern way of using the compass to transfer distances is equivalent to using the so-called collapsing compass. Question Exactly how would one prove that the modern compass is equivalent to the collapsing compass? Hint: See Euclid’s proof. ? Construction (Bisecting a Segment) Given a segment, we wish to cut it in half. (1) Open your compass to the width of the segment. (2) Draw two circles, one with the center being at each end point of the line segment. (3) The circles intersect at two points. Draw a line through these two points.

115 4.1. CONSTRUCTIONS

(4) The new line bisects the original line segment.

Construction (Perpendicular through a Point) Given a point and a line, we wish to construct a line perpendicular to the original line that passes through the given point.

(1) Draw a circle centered at the point large enough to intersect the line in two distinct points.

(2) Bisect the line segment. The line used to do this will be the desired line.

Construction (Bisecting an Angle) We wish to divide an angle in half.

(1) Draw a circle with its center being the vertex of the angle.

(2) Draw a line segment where the circle intersects the lines.

(3) Bisect the new line segment. The bisector will bisect the angle.

116 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

We now come to a very important construction: Construction (Copying an Angle) Given a point on a line and some angle, we wish to copy the given angle so that the new angle has the point as its vertex and the line as one of its edges. (1) Open the compass to a ﬁxed width and make a circle centered at the vertex of the angle. (2) Make a circle of the same radius on the line with the point. (3) Open the compass so that one end touches the 1st circle where it hits an edge of the original angle, with the other end of the compass extended to where the 1st circle hits the other edge of the original angle. (4) Draw a circle with the radius found above with its center where the second circle hits the line. (5) Connect the point to where the circles meet. This is the other leg of the angle we are constructing.

117 4.1. CONSTRUCTIONS

Construction (Parallel through a Point) Given a line and a point, we wish to construct another line parallel to the ﬁrst that passes through the given point.

(1) Draw a circle around the given point that passes through the given line at two points.

(2) We now have an isosceles triangle, duplicate this triangle.

(3) Connect the top vertexes of the triangles and we get a parallel line.

Question Can you give another diﬀerent construction?

Construction (Tangent to a Circle) Given a circle and a point, we wish to construct a line tangent to the circle that goes through the point.

(1) Draw a line segment connecting the point to the center of the circle.

(2) Bisect the above segment.

(3) Draw a circle centered at the bisector whose radius is half the length of the above segment.

(4) Draw lines connecting the given point to where the two circles intersect.

118 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

Question What if the point is inside the circle? What if the point is on the circle? ?

119 4.1. CONSTRUCTIONS

Problems for Section 4.1 (1) What is a collapsing compass? Why don’t we use them or worry about them any more?

(2) Prove that the collapsing compass is equivalent to the modern compass.

(3) Given a line segment, construct an equilateral triangle whose edge has the length of the given segment. Explain the steps in your construction.

(4) Use a compass and straightedge to bisect a given line segment. Explain the steps in your construction.

(5) Given a line segment with a point on it, construct a line perpendicular to the segment that passes through the given point. Explain the steps in your construction.

(6) Use a compass and straightedge to bisect a given angle. Explain the steps in your construction.

(7) Given an angle and some point, use a compass and straightedge to copy the angle so that the new angle has as its vertex the given point. Explain the steps in your construction.

(8) Given a point and line, construct a line perpendicular to the given line that passes through the given point. Explain the steps in your construction.

(9) Given a point and line, construct a line parallel to the given line that passes through the given point. Explain the steps in your construction.

(10) Given a circle and a point, construct a line tangent to the given circle that passes through the given point.

(11) Given a triangle, construct the circumcenter. Explain the steps in your construction.

(12) Given a triangle, construct the orthocenter. Explain the steps in your construction.

(13) Given a triangle, construct the incenter. Explain the steps in your construction.

(14) Given a triangle, construct the centroid. Explain the steps in your construction.

(15) Given a triangle, construct the incircle. Explain the steps in your construction.

(16) Given a triangle, construct the circumcircle. Explain the steps in your construction.

120 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

(17) Given a triangle, construct the Euler line. Explain the steps in your construction.

(18) Given 3 distinct points not all in a line, construct a circle that passes through all three points. Explain the steps in your construction.

(19) Give 3 diﬀerent constructions of the Nine-Point Circle. Explain the steps in your constructions.

(20) Given a triangle with a point on each side, construct a Miquel point. Explain the steps in your construction.

121 4.2. TRICKIER CONSTRUCTIONS

4.2 Trickier Constructions

Question How do you construct regular polygons? In particular, how do you construct regular: 3-gons, 4-gons, 5-gons, 6-gons, 7-gons, 8-gons, 10-gons, 12-gons, 17-gons, 24-gons, and 144-gons? ? Well the equilateral triangle is easy. It was the first construction that we did. What about squares? What about regular hexagons? It turns out that they aren’t too difficult. What about pentagons? Or say n-gons? We’ll have to think about that. Let’s leave the difficult land of n-gons and go back to thinking about nice, three-sided triangles.

Construction (SAS Triangle) Given two sides with an angle between them, we wish to construct the triangle with that angle and two adjacent sides.

(1) Transfer the one side so that it starts at the vertex of the angle.

(2) Transfer the other side so that it starts at the vertex.

(3) Connect the end points of all moved line segments.

The “SAS” in this construction’s name spawns from the fact that it requires two sides with an angle between them. The SAS Theorem states that we can obtain a unique triangle given two sides and the angle between them.

Construction (SSS Triangle) Given three line segments we wish to construct the triangle that has those three sides if it exists.

(1) Choose a side and select one of its endpoints.

(2) Draw a circle of radius equal to the length of the second side around the chosen endpoint.

(3) Draw a circle of radius equal to the length of the third side around the other endpoint.

(4) Connect the end points of the ﬁrst side and the intersection of the circles. This is the desired triangle.

Question Can this construction fail to produce a triangle? If so, show how. If not, why not? ? Question Remember earlier when we asked about the converse to the Pythagorean Theorem? Can you use the construction above to prove the converse of the Pythagorean Theorem?

122 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

? Question Can you state the SSS Theorem? ? Construction (SAA Triangle) Given a side and two angles, where the given sided does not touch one of the angles, we wish to construct the triangle that has this side and these angles if it exists.

(1) Start with the given side and place the adjacent angle at one of its endpoints.

(2) Move the second angle so that it shares a leg with the leg of the ﬁrst angle–not the leg with the side.

(3) Extend the side past the ﬁrst angle, forming a new angle with the leg of the second angle.

(4) Move this new angle to the other endpoint of the side, extending the legs of this angle and the ﬁrst angle will produce the desired triangle.

Question Can this construction fail to produce a triangle? If so, show how. If not, why not? ? Question Can you state the SAA Theorem? ?

4.2.1 Challenge Constructions Question How can you construct a triangle given the length of one side, the length of the the median to that side, and the length of the altitude of the opposite angle? Hint: Recall that the median connects the vertex to the midpoint of the opposite side.

Solution

(1) Start with the given side. (2) Since the median hits our side at the center, bisect the given side. (3) Make a circle of radius equal to the length of the median centered at the bisector of the given side. (4) Construct a line parallel to our given line of distance equal to the length of the given altitude away.

123 4.2. TRICKIER CONSTRUCTIONS

(5) Where the line and the circle intersect is the third point of our triangle. Connect the endpoints of the given side and the new point to get the triangle we want.

Question How can you construct a triangle, given one angle, the length of an adjacent side and the altitude to that side?

Solution

(1) Start with a line containing the side.

(2) Put the angle at the end of the side.

(3) Draw a parallel line to the side of the length of the altitude away.

(4) Connect the angle to the parallel side. This is the third vertex. Connect the endpoints of the given side and the new point to get the triangle we want.

124 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

Question How can you construct a circle with a given radius tangent to two other circles? Solution

(1) Let r be the given radius, and let r1 and r2 be the radii of the given circles.

(2) Draw a circle of radius r1 + r around the center of the circle of radius r1.

(3) Draw a circle of radius r2 + r around the center of the circle of radius r2. (4) Where the two circles drawn above intersect is the center of the desired circle.

125 4.2. TRICKIER CONSTRUCTIONS

Question Place two tacks in a wall. Insert a sheet of paper so that the edges hit the tacks and the corner passes through the imaginary line between the tacks. Mark where the corner of the piece of paper touches the wall. Repeat this process, sliding the paper around. What curve do you end up drawing? ?

Question How can you construct a triangle given an angle and the length of the opposite side?

Solution We really can’t solve this problem completely because the infor- mation given doesn’t uniquely determine a triangle. However, we can still say something. Here is what we can do:

(1) Put the known angle at one end of the line segment. Note in the picture below, it is at the left end of the line segment and it is opening downwards.

(2) Bisect the segment.

(3) See where the bisector in Step 2 intersects the perpendicular of the other leg of the angle drawn from the vertex of the angle.

(4) Draw the circle centered at the point found in Step 3 that touches the endpoints of the original segment.

All points on the circle could be the vertex.

Question Why does the above method work? ?

126 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

Question You are on a boat at night. You can see three lighthouses, and you know their position on a map. Also you know the angles of the light rays between the lighthouses as measured from the boat. How do you ﬁgure out where you are? ?

4.2.2 Problem Solving Strategies The harder constructions discussed in this section can be diﬃcult to do. There is no rote method to solve these problems, hence you must rely on your brain. Here are some hints that you may ﬁnd helpful:

Construct what you can. You should start by constructing anything you can, even if you don’t see how it will help you with your final construction. In doing so you are “chipping away” at the problem just as a rock-cutter chips away at a large boulder. Here are some guidelines that may help when constructing triangles: (1) If a side is given, then you should draw it. (2) If an angle is given and you know where to put it, draw it. (3) If an altitude of length ℓ is given, then draw a line parallel to the side that the altitude is perpendicular to. This new line must be distance ℓ from the side. (4) If a median is given, then bisect the segment it connects to and draw a circle centered around the bisector, whose radius is the length of the median. (5) If you are working on a figure, construct any “mini-figures” inside the figure you are trying to construct. For example, many of the problems below ask you to construct a triangle. Some of these constructions have right-triangles inside of them, which are easier to construct than the final figure.

Sketch what you are trying to find. It is a good idea to try to sketch the figure that you are trying to construct. Sketch it accurately and label all pertinent parts. If there are special features in the figure, say two segments have the same length or there is a right-angle, make a note of it on your sketch. Also mark what is unknown in your sketch. We hope that doing this will help organize your thoughts and get your “brain juices” flowing. Question Why are the above strategies good? ?

127 4.2. TRICKIER CONSTRUCTIONS

Problems for Section 4.2 (1) Construct a square. Explain the steps in your construction. (2) Construct a regular hexagon. Explain the steps in your construction. (3) Your friend Margy is building a clock. She needs to know how to align the twelve numbers on her clock so that they are equally spread out on a circle. Explain how to use a compass and straightedge construction to help her out. Illustrate your answer with a construction. (4) Construct a triangle given two sides of a triangle and the angle between them. Explain the steps in your construction. (5) Construct a triangle given three sides of a triangle. Explain the steps in your construction. (6) Construct a triangle given two adjacent sides of a triangle and a median to one of the given sides. Explain the steps in your construction. (7) Construct a figure showing that a triangle cannot always be uniquely determined when given an angle, a side adjacent to that angle, and the side opposite the angle. Explain the steps in your construction and explain how your figure shows what is desired. Hint: Draw many pictures to help yourself out. (8) Give a construction showing that a triangle is uniquely determined if you are given a right-angle, a side touching that angle, and another side not touching the angle. Explain the steps in your construction and explain how your figure shows what is desired. (9) Construct a triangle given a side, the median to the side, and the angle opposite to the side. Explain the steps in your construction. (10) Construct a triangle given two sides and the altitude to the third side. Explain the steps in your construction. (11) Construct a triangle given two altitudes and an angle touching one of them. Explain the steps in your construction. (12) Construct a triangle given an altitude, and two angles not touching the altitude. Explain the steps in your construction. (13) Construct a triangle given the length of one side, the length of the the median to that side, and the length of the altitude of the opposite angle. Explain the steps in your construction. (14) Construct a triangle, given one angle, the length of an adjacent side and the altitude to that side. Explain the steps in your construction. (15) Construct a circle with a given radius tangent to two other given circles. Explain the steps in your construction.

128 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

(16) Does a given angle and a given opposite side uniquely determine a triangle? Explain your answer.

(17) You are on the bank of a river. There is a tree directly in front of you on the other side of the river. Directly left of you is a friend a known distance away. You friend knows the angle between you and the tree across from you. How wide is the river? Explain your work. (18) You are on a boat at night. You can see three lighthouses, and you know their position on a map. Also you know the angles of the light rays from the lighthouses. How do you ﬁgure out where you are? Explain your work.

(19) Construct a triangle given an angle, the length of the angle’s bisector to the opposite side, and the length of a side adjacent to the given angle. Explain the steps in your construction.

(20) Construct a triangle given an angle, the length of the opposite side and the length of the altitude of the given angle. Explain the steps in your construction.

(21) Construct a triangle given one side, the altitude of the opposite angle, and the radius of the circumcircle. Explain the steps in your construction.

(22) Construct a triangle given one side, the altitude of an adjacent angle, and the radius of the circumcircle. Explain the steps in your construction.

(23) Construct a triangle given one side, the length of the median connecting that side to the opposite angle, and the radius of the circumcircle. Explain the steps in your construction. Hint: Recall that the median connects the vertex to the midpoint.

(24) Given a circle and a line, construct another circle of a given radius which is tangent to both the original circle and line. Explain the steps in your construction.

(25) Construct a circle with a given radius tangent to two given intersecting lines. Explain the steps in your construction.

(26) Construct a triangle given one angle and the altitudes to the two other angles. Explain the steps in your construction.

(27) Construct a circle with three smaller circles of equal size inside such that each smaller circle is tangent to the other two and the larger outside circle. Explain the steps in your construction.

129 4.3. CONSTRUCTIBLE NUMBERS

4.3 Constructible Numbers

First of all, what do we mean by the words constructible numbers? Imagine a line with two points on it:

0 1

Label the left point 0 and the right point 1. If we think of this as a starting point for a number line, then a constructible number is nothing more than a point we can obtain on the above number line using only a compass and a straightedge starting with the points 0 and 1. Call the set of constructible numbers . C Question Exactly which numbers are constructible? ? How do we attack this question? Well ﬁrst let’s get a bit of notation. Recall that we use the symbol “ ” to mean is in. So we know that 0 and 1 are in the set of constructible numbers.∈ So we write 0 and 1 . ∈C ∈C If we could use constructions to make the operations +, , , and , then we would be able to say a lot more. In fact we will do just this.− · In the÷ following constructions, the segments of length 1, a, and b are as given below:

1 a b

Construction (Addition) Adding is simple, use the compass to extend the given line segment as necessary. a b a+b

Construction (Subtraction) Subtracting is easy too: a−b

Question What does our number line look like at this point? At this point we have all the whole numbers and their negatives. We have a special name for this set, we call it the integers and denote it by the letter Z: Z = ..., 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5,... . { − − − − − } But we still have some more operations:

130 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

Construction (Multiplication) The idea of multiplication is based on the idea of similar triangles. Start with given segments of length a, b, and 1:

(1) Make a small triangle with the segment of length 1 and segment of length b.

(2) Now place the segment of length a on top of the unit segment with one end at the vertex.

(3) Draw a line parallel to the segment connecting the unit to the segment of length b starting at the other end of segment of length a.

(4) The length from the vertex to the point that the line containing b intersects the line drawn in Step 3 is of length a b. ·

b 1 a

Construction (Division) The idea of division is also based on the idea of similar triangles. Again, you start with given segments of length a, b, and 1:

(1) Make a triangle with the segment of length a and the segment of length b.

(2) Put the unit along the segment of length a starting at the vertex where the segment of length a and the segment of length b meet.

(3) Make a line parallel to the third side of the triangle containing the segment of length a and the segment of length b starting at the end of the unit.

(4) The distance from where the line drawn in Step 3 meets the segment of length b to the vertex is of length b/a.

b b/a 1 a

Question What does our number line look like at this point?

Currently we have Z, the integers, and all of the fractions. In other words: a Q = such that a Z and b Z with b = 0 b ∈ ∈ 6 n o 131 4.3. CONSTRUCTIBLE NUMBERS

Fancy folks will replace the words such that with a colon “:” to get: a Q = : a Z and b Z with b = 0 b ∈ ∈ 6 n o We call this set the rational numbers. The letter Q stands for the word quotient, which should remind us of fractions. In mathematics we study sets of numbers. In any field of science, the first step to understanding something is to classify it. One sort of classification that we have is the notion of a field.

Deﬁnition A ﬁeld is a set of numbers, which we will call F , that is closed under two associative and commutative operations + and such that: · (1) (a) There exists an additive identity 0 F such that for all x F , ∈ ∈ x + 0 = x.

(b) For all x F , there is an additive inverse x F such that ∈ − ∈ x + ( x) = 0. − (2) (a) There exists a multiplicative identity 1 F such that for all x F , ∈ ∈ x 1 = x. · 1 (b) For all x F where x = 0, there is a multiplicative inverse x− such that ∈ 6 1 x x− = 1. · (3) Multiplication distributes over addition. That is, for all x,y,z F ∈ x (y + z) = x y + x z. · · · Now, a word is in order about three tricky words I threw in above: closed, associative, and commutative:

Deﬁnition A set F is closed under an operation ⋆ if for all x,y F , x⋆y F . ∈ ∈ Example The set of integers, Z, is closed under addition, but is not closed under division.

Deﬁnition An operation ⋆ is associative if for all x, y, and z

x ⋆ (y⋆z) = (x⋆y) ⋆ z.

Deﬁnition An operation ⋆ is commutative if for all x, y

x⋆y = y⋆x.

132 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

Question Is Z a field? Is Q a field? Can you think of other fields? What about the set of constructible number ? C ?

From all the constructions above we see that the set of constructible numbers is a ﬁeld. However, which ﬁeld is it? In fact, the set of constructible numbers isC bigger than Q!

Construction (Square-Roots) Start with given segments of length a and 1:

(1) Put the segment of length a immediately to the left of the unit segment on a line.

(2) Bisect the segment of length a + 1.

(3) Draw an arc centered at the bisector that starts at one end of the line segment of length a + 1 and ends at the other end.

(4) Construct the perpendicular at the point where the segment of length a meets the unit.

(5) The line segment connecting the meeting point of the segment of length a and the unit to the arc drawn in Step 3 is of length √a.

a 1

This tells us that square-roots are constructible. In particular, the square- root of two is constructible. But the square-root of two is not rational! That is, there is no fraction a = √2 such that a, b Z. b ∈ How do we know the above fact? Well there are several ways to do it. Why not use the Rational Roots Test?

133 4.3. CONSTRUCTIBLE NUMBERS

Theorem 21 (Rational Roots Test) If we have a polynomial with integer coeﬃcients, n n 1 anx + an 1x − + + a1x + a0 − · · · then given any rational root r/s of the above polynomial, r must be a factor of a0 and s must be a factor of an. Now what does it mean to be the square-root of 2? It means that you are a solution to the following equation:

x2 2 = 0. − By the Rational Roots Test, if r/s is a rational root then r is a factor of 2 and s is a factor of 1. So here are our choices: −

r = 1, 2 ± ± s = 1. ± But no combination of the above numbers form fractions r/s such that

(r/s)2 = 2.

Thus, the square root of two is not rational. OK, so how do we talk about a ﬁeld that contains both Q and √2? Simple, use this notation:

Q(√2) = the smallest ﬁeld containing both Q and √2 { } So the set of constructible numbers contains all of Q(√2). Does the set of constructible numbers contain even more numbers? Yes! In fact the √3 is also not rational, but is constructible. So here is our situation:

Z Q Q(√2) Q(√2, √3) ⊆ ⊆ ⊆ ⊆C So all the numbers in Q(√2, √3) are also in . But is this all of ? Hardly! We could keep on going, adding more and moreC square-roots ’til theC cows come home, and we still will not have our hands on all of the constructible numbers. But all is not lost. We can still say something: Theorem 22 The use of compass and straightedge alone on a ﬁeld F can at most produce numbers in a ﬁeld F (√α) where α F . ∈ The upshot of the above theorem is that the only numbers that are constructible are expressible as a combination of rational numbers and the symbols:

+ √ − · ÷ So what are examples of numbers that are not constructible? Well to start √3 2 is not constructible. Also π is not constructible. While both of these facts can be carefully explained, we will spare you gentle reader—for now.

134 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

Question Which of the following numbers are constructible?

16 3 6 3.1415926, √5, √27, √27. ?

Question Is the golden ratio constructible?

Well 1 + √5 φ = , 2 so of course it is. But, how do you actually construct it? Here is an easy way:

Construction (Golden Ratio)

(1) Consider a unit on a line.

(2) Construct a perpendicular of unit length at the right end point of the unit.

(3) Bisect the original unit.

(4) Draw an arc, centered at the point found in Step 3 that goes through the top of the perpendicular drawn in Step 2.

(5) The segment starting at the left end of the unit and ending at the point found in Step 4 is of length φ.

Question How how do you construct a regular pentagon?

135 4.3. CONSTRUCTIBLE NUMBERS

One way is to use golden triangles:

1 1

φ φ

1 1

What about other regular n-gons? Carl Friedrich Gauss, one of the greatest mathematicians of all time, solved this problem when he was 18. He did this around the year 1800, nearly 2000 years after the time of the Greeks. How did he do it? He thought of constructions algebraically as we have been doing. Using these methods, he discovered this theorem: Theorem 23 (Gauss) One can construct a regular n-gon if and only if n > 3 and i n = 2 p p pj · 1 · 2 · · · where each subscripted p is a distinct prime number of the form

k 2(2 ) + 1 where i, j, and k are nonnegative integers.

Question Find i, j, and k for a regular 3-gon, 4-gon, 5-gon, and 6-gon. ?

136 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

Problems for Section 4.3

(1) Explain what the set denoted by Z is.

(2) Explain what the set denoted by Q is.

(3) Explain what the set of constructible numbers is. C (4) Given two line segments a and b, construct a + b. Explain the steps in your construction.

(5) Given two line segments a and b, construct a b. Explain the steps in your construction. −

(6) Given three line segments 1, a, and b, construct a b. Explain the steps in your construction. ·

(7) Given three line segments 1, a, and b, construct a/b. Explain the steps in your construction.

(8) Given a unit, construct 4/3. Explain the steps in your construction.

(9) Given a unit, construct 3/4. Explain the steps in your construction.

(10) Given a unit, construct √2. Explain the steps in your construction.

(11) Use the construction for multiplication to explain why when multiplying two numbers between 0 and 1, the product is always still between 0 and 1.

(12) Use the construction for division to explain why when dividing a positive number by a number between 0 and 1, the quotient is always larger than the initial positive number.

(13) Fill in the following table:

+ − · ÷ ∧ Commutative yes Associative yes Closed in Z: yes Closed in Q: yes Closed in : yes C

137 4.3. CONSTRUCTIBLE NUMBERS

(14) Which of the following are constructible numbers? Explain your answers. (a) 3.141 (b) √3 5 (c) 3 + √17 8 (d) p√5 (e) 10√37 (f) 16√37 (g) √3 28 (h) 13 + √3 2 + √11 5 (i)p 3 + √4 (j) 3 + √19 + √10 (15) Supposep that you know that all the roots of

x4 10x3 + 35x2 50x + 24 − − are rational. Find them and explain your work. (16) Suppose that you know that all the roots of

x4 5x2 + 4 − are rational. Find them and explain your work. (17) Suppose that you know that all the roots of

x4 2x3 13x2 + 14x + 24 − − are rational. Find them and explain your work. (18) Is √7 a rational number? Is it a constructible number? Explain your answers. (19) Is √8 a rational number? Is it a constructible number? Explain your answers. (20) Is √9 a rational number? Is it a constructible number? Explain your answers. (21) Is √3 7 a rational number? Is it a constructible number? Explain your answers. (22) Is √3 8 a rational number? Is it a constructible number? Explain your answers. (23) Is √3 9 a rational number? Is it a constructible number? Explain your answers.

138 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

(24) The Rational Roots Test, Theorem 21, is actually more general than you really need to merely prove that a number is not rational. Can you state a more specialized theorem, based upon the Rational Roots Test, that would do everything we need to prove that a number is not rational? Give examples of your theorem in action. (25) Construct the golden ratio. Explain the steps in your construction. (26) Construct the golden ratio given a unit using only the basic constructions for addition, division, and square roots, and the fact that: 1 + √5 φ = 2 Explain the steps in your construction. (27) Construct a golden rectangle. Explain the steps in your construction. (28) Construct a golden triangle. Explain the steps in your construction. (29) Construct a golden spiral associated to a golden rectangle. Explain the steps in your construction. (30) Construct a golden spiral associated to a golden triangle. Explain the steps in your construction. (31) Construct a regular pentagon. Explain the steps in your construction. (32) Explain Gauss’ Theorem, Theorem 23 above. (33) Which of the following polygons can be constructed using a compass and straightedge? Explain your answers. (a) A regular 3-gon. (b) A regular 5-gon. (c) A regular 7-gon. (d) A regular 9-gon. (e) A regular 11-gon. (f) A regular 12-gon. (g) A regular 13-gon. (h) A regular 15-gon. (i) A regular 17-gon. (j) A regular 34-gon. (k) A regular 2-gon. (l) A regular 4-gon. (m) A regular 10-gon. (n) A regular 20-gon. (o) A regular 70-gon.

139 4.4. IMPOSSIBILITIES

4.4 Impossibilities

Oddly enough, the importance of compass and straightedge constructions is not so much what we can construct, but what we cannot construct. It turns out that classifying what we cannot construct is an interesting question. There are three classic problems which are impossible to solve with a compass and straightedge alone:

(1) Doubling the cube.

(2) Squaring the circle.

(3) Trisecting the angle.

4.4.1 Doubling the Cube The goal of this problem is to double the volume of a given cube. This boils down to trying to construct roots to the equation:

x3 2 = 0 − But we can see that the only root of the above equation is √3 2 and we already know that this number is not constructible.

Question Why does doubling the cube boil down to constructing a solution to the equation x3 2=0? − ?

4.4.2 Squaring the Circle Given a circle of radius r, we wish to construct a square that has the same area. Why would someone want to do such a thing? Well to answer this question you must ask yourself:

Question What is area? ?

So what is the deal with this problem? Well suppose you have a circle of radius 1. Its area is now π square units. How long should the edge of a square be if it has the same area? Well the square should have sides of length √π units. In 1882, it was proved that π is not the root of any polynomial equation, and hence √π is not constructible. Therefore, it is impossible to square the circle.

140 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

4.4.3 Trisecting the Angle This might sound like the easiest to understand, but it’s a bit subtle. Given any angle, the goal is to trisect that angle. It can be shown that this cannot be done using a compass and straightedge. However, we are not saying that you cannot trisect some angles with compass and straightedge alone, in fact there are special angles which can be trisected using a compass and straightedge. However the methods used to trisect those special angles will fail miserably in nearly all other cases.

Question Can you think of any angles that can be trisected using a compass and straightedge? ?

Just because it is impossible to trisect an arbitrary angle with compass and straightedge alone does not stop people from trying.

Question If you did not know that it was impossible to trisect an arbitrary angle with a compass and straightedge alone, how might you try to do it? ?

One common way that people try to trisect angles is to take an angle, make a triangle using the angle, and divide the line segment opposite the angle into three equal parts. While you can divide the opposite side into three equal parts, it in fact never trisects the angle. When you do this procedure to acute angles, it seems to work, though it doesn’t really. You can see that it doesn’t by looking at an obtuse angle:

Trisecting the line segment opposite the angle clearly leaves the middle angle much larger than the outer two angles. This happens regardless of the measure of the angle. This mistake is common among people who think that they can trisect an angle with compass and straightedge alone.

How to Trisect the Angle with a Marked Straightedge The ancient Greek mathematicians were a tenacious bunch. While they wanted very badly to trisect the angle with a compass and straightedge alone, this did not stop them from devising other methods of doing it. So they cheated and used a marked straightedge. Now to trisect an angle using a marked straightedge, we will use a method attributed to Archimedes:

141 4.4. IMPOSSIBILITIES

Construction (Cheating) We will show how to trisect an arbitrary angle using a marked straightedge.

(1) Take an angle and draw a circle centered at the vertex. (2) Draw a line extending one leg of the angle.

(3) Draw a line from the intersection of the circle and the other leg that intersects the line drawn in Step 2, such that the part of the line outside the circle is as long as the radius of the circle.

(4) The angle formed by the lines drawn in Step 2 and Step 3 is one third of the original angle.

WARNING The above method of trisecting an angle is not a legitimate compass and straightedge construction. If you are attempting a problem in this chapter and the problem says to “construct,” then you may not use the above method. You may only use the above method if the question explicitly says “use a marked straightedge and compass.”

142 CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS

Problems for Section 4.4 (1) Explain the three classic problems that cannot be solved with a compass and straightedge alone.

(2) Use a compass and straightedge construction to trisect an angle of 90◦. Explain the steps in your construction.

(3) Use a compass and straightedge construction to trisect an angle of 135◦. Explain the steps in your construction.

(4) Use a compass and straightedge construction to trisect an angle of 45◦. Explain the steps in your construction.

(5) Use a compass and straightedge construction to trisect an angle of 67.5◦. Explain the steps in your construction.

(6) Use a marked straightedge and compass to trisect a given angle. Explain the steps in your construction.

143 Chapter 5

Transformations

And since you know you cannot see yourself, so well as by reﬂection, I, your glass, will modestly discover to yourself, that of yourself which you yet know not of. —William Shakespeare

5.1 Basic Transformations

We’re going to be discussing some basic transformations in geometry. Specifi- cally, we are talking about translations, reflections, and rotations. To start us off, we need a little background on matrices. Question What is a matrix? You might think of a matrix as just a jumble of large brackets and numbers. However, we are going to think of matrices as functions. Just as we write f(x) for a function f acting on a number x, we’ll write:

Mp = q to represent a function M mapping point p to point q. To make things work out nice, we need to write our points all straight and narrow, with a little buddy at the end: x (x,y) y 1   Throughout this chapter, we will abuse notation slightly, freely interchanging several notations for a point: x p ! (x,y) ! y 1   144 CHAPTER 5. TRANSFORMATIONS

With this in mind, our transformations will be done via matrices and points that look like this:

a b c x M = d e f and p = y 0 0 1 1     Now recall the nitty gritty details of matrix multiplication:

a b c x Mp = d e f y 0 0 1 1  ax + by+ c1 = dx + ey + f · 1 0 x + 0 y + 1· 1 · · · ax + by + c  = dx + ey + f  1    Question Fine, but what does this have to do with geometry?

In this chapter we are going to study a special type of functions, called geometric transformations.

Deﬁnition A geometric transformation is a function M that maps points in the plane to other points in the plane such that if

Mp = Mq, then p = q.

We’re going to see that several ideas in geometry, specifically translations, reflections, and rotations which all seem very different, are actually all geometric transformations. Hence, we will be thinking of these concepts as matrices.

5.1.1 Translations

Of all the geometric transformations, translations are probably the easiest. With a translation, all we do is move our object in a straight line. Let’s see what

145 5.1. BASIC TRANSFORMATIONS happens to Louie Llama:

Pretty simple eh? We can give a more “mathematical” deﬁnition of a translation using our newly-found knowledge of matrices! Check it:

T Deﬁnition A translation, denoted by (u,v), is a geometric transformation which moves a point a given distance u in the x-direction and a given distance v in the y-direction. We will use the following type of matrix to represent translations:

1 0 u T (u,v) = 0 1 v 0 0 1   Example Suppose that you have a point p = ( 3, 2) and you want to translate it 5 units right and 4 units down. −

146 CHAPTER 5. TRANSFORMATIONS

Here is how you do it:

1 0 5 3 T − (5, 4)p = 0 1 4 2 − 0 0− 1   1   3+0+5   = −0 + 2 4  0+0+1−   2  = 2 −1    Hence, we end up with the point (2, 2). But you knew that already, didn’t you? −

Question What matrix will translate your point back to its original position? T 1 T 1 T Call it (4−, 5). What is (4−, 5) (4, 5)p equal to? − − − ?

Question We know how to translate individual points. How do we move entire ﬁgures and other funky shapes? ?

5.1.2 Reﬂections

The act of reflection has fascinated humanity for millennia. It has a strong effect on our perception of beauty and has a defined place in the field of art. We all think we know what reflections are, here is our definition:

Deﬁnition A reﬂection about a line ℓ, denoted by Fℓ, is a geometric transformation that moves a point p to a point Fℓp such that:

(1) If p is on ℓ, then Fℓp = p.

(2) If p is not on ℓ, then ℓ is the perpendicular bisector of the segment connecting p and Fℓp.

You might be saying, “Huh?” It’s not as hard as it looks. Check out this

147 5.1. BASIC TRANSFORMATIONS picture of the situation, again Louie Llama will help us out:

A Collection of Reﬂections

We are going to look at a trio of reﬂections. We’ll start with a horizontal reﬂection over the y-axis. Using our matrix notation, we write:

1 0 0 − Fx=0 = 0 1 0  0 0 1  

The next reﬂection in our collection is a vertical reﬂection over the x-axis. Using our matrix notation, we write:

1 0 0 Fy=0 = 0 1 0 0− 0 1   The final reflection to add to our collection is a diagonal reflection over the line y = x. Using our matrix notation, we write:

0 1 0 Fy=x = 1 0 0 0 0 1  

Example Consider the point p = (3, 1). What would p look like if we −

148 CHAPTER 5. TRANSFORMATIONS reﬂected it across the line y = x?

Here is how you do it:

0 1 0 3 Fy=xp = 1 0 0 1 0 0 1 −1  0 1 + 0   = 3+0+0− 0+0+1  1  − = 3  1    Hence we end up with the point ( 1, 3). − Question Let p be some point. It is not hard to see that Fy=xp lands in the first quadrant of the (x,y)-plane. What reflection will place this point in Quadrant II? What about Quadrant IV? What about Quadrant III? ? Question How do we deal with reflections which are not about the lines y = 0, x = 0, or y = x? How would you reflect points about the line y = 1? ?

5.1.3 Rotations Imagine that you are on a swing set, going higher and higher until you are actually able to make a full circle1. At the point where you are directly above

1Face it, I think we all dreamed of doing that when we were little—or in my case, last week.

149 5.1. BASIC TRANSFORMATIONS where you would be if the swing were at rest, where is your head, comparatively? Your feet? Your hands? Rotations should bring circles to mind. This is not a coincidence. Check out our deﬁnition of a rotation:

Deﬁnition A rotation of n degrees about the origin, denoted by Rn, is a geometric transformation that moves a point p to a point Rnp such that:

(1) The points p and Rnp are equidistant from the origin.

(2) An angle of n degrees is formed by p, the origin, and Rnp.

Louie Llama, can you do the honors?

WARNING We always measure our angles in a counterclockwise fashion.

Looking back on your trigonometry, there were a few special angle measure- ments, namely 90◦, 60◦, and 45◦. We’ll focus on these degrees as well.

√ 1 1 1 3 − 0 0 1 0 2 −2 0 √2 √2 R − R √3 1 R 1 1 90 = 1 0 0 60 = 0 45 = 0    2 2   √2 √2  0 0 1 0 0 1 0 0 1        

Example Suppose that you have a point p = (3, 2) and you want to rotate −

150 CHAPTER 5. TRANSFORMATIONS

it 60◦ about the origin.

Here is how you do it:

1 √3 2 −2 0 4 R √3 1 60p = 0 2  2 2  −  0 0 1 1      2 + √3 + 0  = 2√3 1 + 0   0+0+1−  2 + √3  = 2√3 1   1−   Hence, we end up with the point (2 + √3, 2√3 1). − Question Do the numbers in the matrices above look familiar? If so, why? ? Question How do you rotate a point 180 degrees? ?

151 5.1. BASIC TRANSFORMATIONS

Problems for Section 5.1 (1) What is a geometric transformation? (2) What is a translation? (3) What is a rotation? (4) What is a reﬂection? (5) In what direction does a positive rotation occur? (6) Consider the following matrix:

1 0 0 I = 0 1 0 0 0 1   Is I a geometric transformation? Explain your conclusion. (7) Consider the following matrix:

0 2 0 M = 3 0 0 −0 0 9   Is M a geometric transformation? Explain your conclusion. (8) Consider the following matrix:

0 2 0 M = 1 1 0 0− 0 1 −   Is M a geometric transformation? Explain your conclusion. (9) Use a matrix to translate the point ( 1, 6) three units right and two units up. Sketch this situation and explain− your work. T (10) The matrix ( 2,6) was used to translate the point p to ( 1, 3). What is p? Sketch this− situation and explain your work. − − T T (11) Given the point p = (0, 7), what would be the result of (4,2) (6, 5)p? Sketch this situation and− explain your work. − (12) Use a matrix to reflect the point (5, 2) across the x-axis. Sketch this situation and explain your work. (13) Use a matrix to reflect the point ( 3, 4) across the y-axis. Sketch this situation and explain your work. − (14) Use a matrix to reflect the point ( 1, 1) across the line y = x. Sketch this situation and explain your work. −

152 CHAPTER 5. TRANSFORMATIONS

(15) Use a matrix to reﬂect the point (1, 1) across the line y = x. Sketch this situation and explain your work.

(16) The matrix Fy=0 was used to reﬂect the point p to (4, 3). What is p? Explain your work.

(17) The matrix Fy=0 was used to reﬂect the point p to (0, 8). What is p? Explain your work. −

(18) The matrix Fx=0 was used to reﬂect the point p to ( 5, 1). What is p? Explain your work. − −

(19) The matrix Fy=x was used to reﬂect the point p to (9, 2). What is p? Explain your work. −

(20) The matrix Fy=x was used to reﬂect the point p to ( 3, 3). What is p? Explain your work. − −

(21) Considering the point (3, 2), use a matrix to rotate this point 60◦ about the origin. Sketch this situation and explain your work.

(22) Considering the point (√2, √2), use a matrix to rotate this point 45◦ about the origin. Sketch this− situation and explain your work.

(23) Considering the point ( 7, 6), use a matrix to rotate this point 90◦ about the origin. Sketch this situation− and explain your work.

(24) Considering the point ( 1, 3), use a matrix to rotate this point 0◦ about the origin. Sketch this situation− and explain your work.

(25) Considering the point (0, 0), use a matrix to rotate this point 120◦ about the origin. Sketch this situation and explain your work.

(26) Considering the point (1, 1), use a matrix to rotate this point 90◦ about the origin. Sketch this situation and explain your work. −

(27) The matrix R90 was used to reﬂect the point p to (2, 5). What is p? Explain your work. −

(28) The matrix R60 was used to reﬂect the point p to (0, 2). What is p? Explain your work.

R 1 5 (29) The matrix 45 was used to reﬂect the point p to ( 2 , 2 ). What is p? Explain your work. −

(30) The matrix R 90 was used to reﬂect the point p to (4, 3). What is p? Explain your work.−

(31) If someone wanted to plot y = 3x, they might start by ﬁlling in the following table:

153 5.1. BASIC TRANSFORMATIONS

x 3x 0 1 1 − 2 2 − 3 3 − Reﬂect each point you obtain from the table above about the line y = x. Give a plot of this situation. What curve do you obtain? What is its relationship to y = 3x? Explain your reasoning.

(32) Some translation T was used to map point p to point q. Given p = (1, 2) and q = (3, 4), ﬁnd T and explain your reasoning.

(33) Some translation T was used to map point p to point q. Given p = ( 2, 3) and q = (2, 3), ﬁnd T and explain your reasoning. −

(34) Some reflection F was used to map point p to point q. Given p = (1, 4) and q = (1, 4), find F and explain your reasoning. − (35) Some reflection F was used to map point p to point q. Given p = (5, 0) and q = (0, 5), find F and explain your reasoning.

(36) Some rotation R was used to map point p to point q. Given p = (3, 0) and q = (0, 3), ﬁnd R and explain your reasoning.

(37) Some rotation R was used to map point p to point q. Given p = (√2, √2) and q = (0, 2), ﬁnd R and explain your reasoning.

(38) Some geometric transformation M maps

(0, 0) (3, 2), 7→ (1, 0) (4, 2), 7→ (0, 1) (3, 3). 7→ Find M and explain your reasoning.

(39) Some geometric transformation M maps

(0, 0) ( 1, 1), 7→ − (1, 0) (0, 1), 7→ (0, 1) ( 1, 2). 7→ − Find M and explain your reasoning.

154 CHAPTER 5. TRANSFORMATIONS

(40) Some geometric transformation M maps

(0, 0) (0, 0), 7→ (1, 0) (1, 0), 7→ (0, 1) (0, 1). 7→ − Find M and explain your reasoning.

(41) Some geometric transformation M maps

(0, 0) (0, 0), 7→ (1, 0) (0, 1), 7→ (0, 1) (1, 0). 7→ Find M and explain your reasoning.

(42) Some geometric transformation M maps

(0, 0) (0, 0), 7→ (1, 0) (0, 1), 7→ (0, 1) ( 1, 0). 7→ − Find M and explain your reasoning.

(43) Some geometric transformation M maps

(0, 0) (0, 0), 7→ 1 √3 (1, 0) , , 7→ 2 2 ! √3 1 (0, 1) − , . 7→ 2 2!

Find M and explain your reasoning.

155 5.2. THE ALGEBRA OF TRANSFORMATIONS

5.2 The Algebra of Transformations

5.2.1 Matrix Multiplication

We know how to multiply a matrix and a column. Multiplying two matrices is a similar procedure:

a b c j k l aj + bm + cp ak + bn + cq al + bo + cr d e f m n o = dj + em + fp dk + en + fq dl + eo + fr g h i  p q r gj + hm + ip gk + hn + iq gl + ho + ir       Variables are all good and well, but let’s do this with actual numbers. Con- sider the following two matrices:

1 2 3 1 0 0 M = 4 5 6 and I = 0 1 0 7 8 9 0 0 1     Let’s multiply them together and see what we get:

1 2 3 1 0 0 MI = 4 5 6 0 1 0 7 8 9 0 0 1 1(1) + 2(0)  + 3(0) 1(0) + 2(1) + 3(0) 1(0) + 2(0) + 3(1) = 4(1) + 5(0) + 6(0) 4(0) + 5(1) + 6(0) 4(0) + 5(0) + 6(1) 7(1) + 8(0) + 9(0) 7(0) + 8(1) + 9(0) 7(0) + 8(0) + 9(1) 1 2 3  = 4 5 6 7 8 9 = M 

Question What is IM equal to? ?

It turns out that we have a special name for I. We call it the identity matrix.

WARNING Matrix multiplication is not generally commutative. Check it out: 1 0 0 0 1 0 − F = 0 1 0 and R = 1 0 0 0− 0 1 0 0 1     156 CHAPTER 5. TRANSFORMATIONS

When we multiply these matrices, we get: 1 0 0 0 1 0 FR = 0 1 0 1− 0 0 0− 0 1 0 0 1  1(0) + 0(1)  + 0(0) 1( 1) + 0(0) + 0(0) 1(0) + 0(0) + 0(1) = 0(0) + 1(1) + 0(0) 0( −1) + 1(0) + 0(0) 0(0) + 1(0) + 0(1)  0(0) +− 0(1) + 1(0) 0(− 1) +− 0(0) + 1(0) 0(0) +− 0(0) + 1(1)  −  0 1 0  − = 1 0 0 −0 0 1 On the other hand, we get: 0 1 0 RF = 1 0 0 0 0 1 Question Is it always the case that (LM)N = L(MN)? ? Question Could we take a matrix with 2 rows and 3 columns and multiply it by one with 3 rows and 3 columns? How about one with 2 rows and 2 columns? Why or why not? ?

5.2.2 Compositions of Transformations It is often the case that we wish to apply several geometric transformations successively to a point. Consider the following: a b 0 e f 0 x M = c d 0 N = g h 0 and p = y 0 0 1 0 0 1 1 Now let’s compute      a b 0 e f 0 x M(Np) = c d 0 g h 0 y 0 0 1 0 0 1 1 a b 0 ex + fy    = c d 0 gx + hy 0 0 1  1  aex + afy + bgx + bhy = cex + cfy + dgx + dhy  1  Now you compute (MN)p and compare what you get to what we got above.

157 5.2. THE ALGEBRA OF TRANSFORMATIONS

Compositions of Translations A composition of translations occurs when two or more successive translations are applied to the same point. Check it out:

1 0 5 1 0 3 T T − (5, 4) ( 3,2) = 0 1 4 0 1 2 − − 0 0− 1  0 0 1  1 0 2    = 0 1 2 0 0− 1  T  = (5+( 3),( 4)+2) − − T = (2, 2) − T T Theorem 24 The composition of two translations (u,v) and (s,t) is equiva- T lent to the translation (u+s,v+t). Question How do you prove the theorem above? ? Question Can you give a single translation which is equivalent to the following composition? T T T T ( 7,5) (0, 6) (2,8) (5, 4) − − − ? Question Are compositions of translations commutative? Are they associative? ?

Compositions of Reflections A composition of reflections occurs when two or more successive reflections are applied to the same point. Check it out:

1 0 0 0 1 0 Fy=0Fy=x = 0 1 0 1 0 0 0− 0 1 0 0 1  0 1 0   = 1 0 0 −0 0 1   Question Which line does the composition Fy=0Fy=x reﬂect points over?

158 CHAPTER 5. TRANSFORMATIONS

Question Are compositions of reﬂections commutative? Are they associative? ?

Compositions of Rotations

A composition of rotations occurs when two or more successive rotations are applied to the same point. Check it out:

1 √3 1 √3 2 −2 0 2 −2 0 √3 √3 R60R60 = 1 1  2 2 0  2 2 0 0 0 1 0 0 1      1 √3    −2 −2 0 = √3 1  2 −2 0 0 0 1     Theorem 25 The product of two rotations Ra and Rb with the same center is equivalent to the rotation Ra+b.

From this we see that:

1 √3 −2 −2 0 √3 R120 = 1  2 −2 0 0 0 1     Question What is the rotation matrix for a 360◦ rotation? What about a 405◦ rotation? ?

Question What makes a rotation diﬀerent from a reﬂection? ?

Question Are compositions of rotations commutative? Are they associative? ?

159 5.2. THE ALGEBRA OF TRANSFORMATIONS

5.2.3 Mixing and Matching Life gets interesting when we start composing translations, reflections, and rotations together. Let’s look at what happens when we show the matrix multiplication of different geometric translations. First we’ll take a look at a reflection mixed with a rotation:

1 √3 1 0 0 2 −2 0 F R √3 1 y=0 60 = 0 1 0 0  −   2 2  0 0 1 0 0 1     1 √3   2 −2 0 = √3 1  2 2 0 −0− 0 1   Question Does this result look familiar?  ? Now how about a rotation with a translation:

0 1 0 1 0 3 R T − 90 (3, 4) = 1 0 0 0 1 4 − 0 0 1 0 0− 1  0 1 4   = 1− 0 3 0 0 1   T R Question What do you think it would look like if instead we had (3, 4) 90? Would the result be the same? − ?

Question Find a matrix that represents the reﬂection Fy= x. − I’ll take this one. Note that

Fy= x = R180Fy=x − = R90R90Fy=x 0 1 0 0 1 0 0 1 0 = 1− 0 0 1− 0 0 1 0 0 0 0 1 0 0 1 0 0 1  0 1 0    − = 1 0 0 −0 0 1 OK looks good, but you, the reader, are going to have to check the above computation yourself.

160 CHAPTER 5. TRANSFORMATIONS

Question When we mix geometric transformations, will we retain commuta- tivity? Associativity? ?

161 5.2. THE ALGEBRA OF TRANSFORMATIONS

Problems for Section 5.2 T T (1) Give a single translation which is equivalent to ( 3,2) (5, 1). Explain your work. − − T T (2) Consider the two translations ( 4,8) and (4, 8). Do these commute? Explain your reasoning. − −

p T 1 T p (3) Given the point = (7, 4), compute ( 2 ,6) (2, 1) . Sketch this situation and explain your work. −

1 T 1 T (4) Given the point p = ( √3, ), compute , √ 1 p. Sketch this − − 4 (2 2 ) ( 3, 4 ) situation and explain your work. −

(5) Give a matrix representing Fy= x. Explain your work. −

(6) Give a matrix representing R 45. Explain your work. −

(7) Give a matrix representing R 60. Explain your work. −

(8) Give a single reﬂection which is equivalent to Fx=0Fy=0. Explain your work.

(9) Given the point p = ( 4, 2), compute Fy=0Fy=xp. Sketch this situation and explain your work.−

(10) Given the point p = (5, 0), compute Fy=xFy= xp. Sketch this situation and explain your work. −

(11) Give a single rotation which is equivalent to R45R60. Explain your work.

(12) Given the point p = (1, 3), compute R45R90p. Sketch this situation and explain your work.

(13) Given the point p = ( 7, 2), compute R45R 45p. Sketch this situation and explain your work.− −

(14) Given the point p = ( 2, 5), compute R90R 90R360p. Sketch this situation and explain your work.− − F T (15) Given the point p = (5, 4), compute y=0 (2, 4)p. Sketch this situation and explain your work. − R T (16) Given the point p = ( 1, 6), compute 45 (0,0)p. Sketch this situation and explain your work.− T R (17) Given the point p = (11, 13), compute ( 6, 3) 135p. Sketch this situation and explain your work. − −

(18) Given the point p = ( 7, 5), compute R540Fx=0p. Sketch this situation and explain your work.− −

162 CHAPTER 5. TRANSFORMATIONS

R F T (19) Given the point p = (8, 1), compute 90 y=x ( 2,3)p. Sketch this situation and explain your work. − F T (20) Give a single geometric transformation which is equivalent to y= x ( 3,10). −T − F Also give a single geometric transformation which is equivalent to ( 3,10) y= x. Explain your work. − −

(21) Give a single geometric transformation which is equivalent to R30Fy=0. Also give a single geometric transformation which is equivalent to Fy=0R30. Explain your work.

163 5.3. THE THEORY OF GROUPS

5.3 The Theory of Groups

One of the most fundamental notions in all of modern mathematics is that of a group. Sadly, many students never see a group in their education. Deﬁnition A group is a set of elements (in our case matrices) which we will call G that is closed under an associative operation (in our case matrix multiplication) such that: (1) There exists an identity I G such that for all M G, ∈ ∈ MI = M.

1 (2) For all M G there is an inverse M− such that ∈ 1 MM− = I.

5.3.1 Groups of Reﬂections To start, the identity I is a transformation which is equivalent to doing nothing at all. To this end, MI = IM = M where M is any other translation that can be applied to the ﬁgure. For our basic study, we’re going to use the simplest shape, the triangle, and center it at the origin. Question Consider the following equilateral triangle.

How many different ways can we reflect this triangle back on to itself? With a triangle there are only three reflections that preserve the location of the vertexes of the triangle. The easiest of these is the reflection over Fx=0.

Let’s use this as the basis for our ﬁrst group table. We’ll start with just two items: I and Fx=0.

I Fx ◦ =0 I I Fx=0 Fx=0 Fx=0 I

164 CHAPTER 5. TRANSFORMATIONS

Notice what happens when we apply the Fx=0 twice, we’re right back where we started. Hence, Fx=0Fx=0 = I. Since matrix multiplication is associative, we see that I, Fx { =0} forms a group as every element clearly has an inverse—we can see this from the table. Speciﬁcally this is a group of reﬂections of the triangle. Question What are the equations of the other two lines of symmetry for the triangle? ? Question Would these same equations work with a square? If not, what equations would result in symmetries? ?

5.3.2 Groups of Rotations How many degrees does it take to rotate and equilateral triangle so that the vertexes are still at the same positions? Well, we have 3 angles and 360◦ of rotation. That gives us 120◦ for each rotation. Remember the matrix for a 120◦ rotation?

1 √3 −2 −2 0 √3 R120 = 1  2 −2 0 0 0 1   Question Consider the following equilateral triangle.

How many diﬀerent ways can we rotate this triangle back on to itself?

Because R120R120 does not give us the identity, we need to include it in our R2 R group table. We will write this as 120. However, if we apply the 120 rotation R R2 one more time, we do get back to the identity. This shows that 120 and 120 are inverses. Everything now has an inverse and we can complete our rotation group table: I R R2 ◦ 120 120 I I R R2 120 120 R R R2 I 120 120 120 R2 R2 I R 120 120 120

165 5.3. THE THEORY OF GROUPS

Question What kind of rotations would we apply when working with a square? A pentagon? A hexagon? ?

5.3.3 Symmetry Groups What happens when we mix rotations and reﬂections? Take for instance if we go back to our triangle and do Fx=0R120:

What you may not immediately notice is that we obtain the same transformation by taking the original triangle and reﬂecting it over the line y = 1 x. √3

As it turns out, every possible arrangement of the vertexes of the equilateral triangle can be represented using compositions of reﬂections and rotations. Hence we’ll call such a composition a symmetry of the triangle. The collection of all symmetries forms a group called the symmetry group of the equilateral triangle. Let’s see the group table:

2 2 (D , ) I R R Fx Fx R Fx R 3 ◦ 120 120 =0 =0 120 =0 120 I I R R2 F F R F R2 120 120 x=0 x=0 120 x=0 120 R R R2 I F R F R2 F 120 120 120 x=0 120 x=0 120 x=0 R2 R2 I R F R2 F F R 120 120 120 x=0 120 x=0 x=0 120 F F F R F R2 I R R2 x=0 x=0 x=0 120 x=0 120 120 120 F R F R F R2 F R R2 I x=0 120 x=0 120 x=0 120 x=0 120 120 F R2 F R2 F F R R2 I R x=0 120 x=0 120 x=0 x=0 120 120 120

This table shows every symmetry of the triangle. By comparing the rows and columns of the group table, you can see that every element has an inverse and the identity is included. This combined with the fact that matrix multiplication is associative shows that the symmetries of the triangle form a group.

166 CHAPTER 5. TRANSFORMATIONS

Question Can you express the symmetries of the square in terms of reﬂec- tions and rotations? What does the group table look like for the symmetry group of the square? ?

167 5.3. THE THEORY OF GROUPS

Problems for Section 5.3 (1) How many lines of symmetry exist for a square? Provide a drawing to justify your answer. (2) How many lines of symmetry exist for a hexagon? Provide a drawing to justify your answer. (3) What are the equations for the lines of symmetry that exist for the square? Explain your answers. (4) What are the equations for the lines of symmetry for a hexagon? Explain your answers. (5) Give the group table for the reflections of a square. (6) Give the group table for the reflections of a hexagon. (7) Give the group table for the rotations of a square. (8) Give the group table for the rotations of a hexagon. (9) Give the group table for the symmetries of a square. (10) Give the group table for the symmetries of a hexagon. (11) How many consecutive rotations are needed to return the vertexes of a square to their original position? Provide a drawing to justify your answer, labeling the vertexes. (12) How many degrees are in one rotation of the square? Explain your answer. (13) How many rotations are needed to return to the identity in a hexagon? Provide a drawing to justify your answer, labeling the vertexes. (14) How many degrees are in one rotation of the hexagon? Explain your answer. (15) In this section, we’ve focused on a 3-sided figure, a 4-sided figure, and a 6-sided figure. Why do we not include the rotation group for the pentagon in this section? If we did, how many degrees would be in one rotation? (16) Find two symmetries of the equilateral triangle, neither of which is the identity, such that their composition is R120. Explain and illustrate your answer. (17) Find two symmetries of the equilateral triangle, neither of which is the identity, such that their composition is Fx=0. Explain and illustrate your answer. (18) Find two symmetries of the equilateral triangle, neither of which is the R2 identity, such that their composition is 120. Explain and illustrate your answer.

168 CHAPTER 5. TRANSFORMATIONS

(19) Find two symmetries of the square, neither of which is the identity, such that their composition is R180. Explain and illustrate your answer. (20) Find two symmetries of the square, neither of which is the identity, such that their composition is Fy=x. Explain and illustrate your answer. (21) Find two symmetries of the square, neither of which is the identity, such that their composition is R270. Explain and illustrate your answer. (22) Find two symmetries of the square, neither of which is the identity, such that their composition is Fx=0. Explain and illustrate your answer.

169 Chapter 6

Convex Sets

If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is. —John von Neumann

6.1 Basic Deﬁnitions

The following pictures are examples of convex sets:

The next set of pictures are not examples of convex sets:

Question What is the diﬀerence between a set which is convex and a set which is not convex?

Deﬁnition A set X is convex if for all points A, B in X the line segment AB is also in X.

Question Is a straight line a convex set? ?

170 CHAPTER 6. CONVEX SETS

From two separate convex sets, we can build new ones: Theorem 26 The intersection of two convex sets is a convex set.

Proof Let X1 and X2 be two convex sets. Let

Y = X X . 1 ∩ 2

If A, B are points in X1, then the line segment AB is a subset of X1. Likewise, if A, B are points in X2, then the line segment AB is a subset of X2. Thus we see that if A, B are points in Y , the line segment AB is a subset of Y . Thus Y = X X is convex. 1 ∩ 2 Question If X , X , and X are convex, is X X X convex? 1 2 3 1 ∩ 2 ∩ 3 ? Question Can you generalize the above question? Is the intersection of 4 convex sets convex? What about 5 convex sets? What about n convex sets? ? Imagine you are blind. How would you identify convex sets in real life? One way to identify convex sets would be with your hands. A mathematical analogue of this is the idea of supporting lines. A supporting line is like a rigid stick that one presses against the set in question. If you can “wedge” a point of the stick into the set, then it is not convex. To be more explicit let’s look at the next deﬁnition: Deﬁnition If X is a set with interior points, a supporting line for X is a line that goes through at least one boundary point of X and cuts the plane in half such that X is only on one side. Pictorially we have this as an example:

Theorem 27 A line is a supporting line for a convex set with interior points if and only if it goes through at least 1 boundary point of the set and no interior points. Question In the above theorem, why does the set need to have interior points? Why does the set need to be convex? ?

171 6.1. BASIC DEFINITIONS

Since we are studying mathematics, we should ask ourselves, “How do we prove the above theorem?” Well, the proof is not hard, but you need to remember the logic we did earlier. Proof Let P = a line is a supporting line for a convex set { } and a line goes through at least 1 boundary point Q = of a convex set and no interior points Since the statement of the theorem above is: P Q ⇔ we must prove: (P Q) (Q P ) ⇒ ∧ ⇒ We will do this in two steps, proving the left-hand side first, and then proving the right-hand side. ( ) Suppose that P Q is false. In other words, suppose that ⇒ ⇒ (P Q) = P ( Q) ¬ ⇒ ∧ ¬ is true. So we are supposing a line is a supporting line and it does not go through at least 1 boundary point or it goes through some interior points. Since it is a supporting line, it must touch the boundary. So could it go through some interior points? If so then there is a small zone around those points contained entirely inside the convex set. Thus we may pick points in our set on either side of the line. Thus our line cannot be a supporting line, a contradiction. ( ) Suppose that Q P is false. In other words, suppose that ⇐ ⇒ (Q P ) = Q ( P ) ¬ ⇒ ∧ ¬ is true. So we are supposing a line goes through at least 1 boundary point and no interior points and it is not a supporting line for a convex set. The points of the set must be on both sides of the line. Therefore, each line connecting the points on both sides of the line are in the set, as the set is convex. Hence, the line must touch interior points. Definition A supporting half-plane is the half-plane formed by a supporting line which contains the set. Here is a picture of a supporting half plane to help you understand the definition:

The supporting half-plane is shaded in the picture above.

172 CHAPTER 6. CONVEX SETS

Theorem 28 A convex set is the intersection of all its supporting half-planes.

Question How could you use the above theorem to deﬁne a triangular region? How could you use the above theorem to prove that a triangular region is convex? ?

6.1.1 An Application Convex sets actually have uses in the real world. One use that we can discuss easily is that they can help us maximize or minimize certain functions. Check out the next theorem:

Theorem 29 Consider a function f(x,y) = ax + by + c, and consider a convex polygonal region P in the (x,y)-plane. Then:

(1) The maximum value for f(x,y) occurs at a vertex of P .

(2) The minimum value for f(x,y) occurs at a vertex of P .

Question How do you think about functions of two variables? Hint: Have you ever seen a topographical map? ?

Example Suppose you’re taking a math exam, 60 problems total, 30 of which are worth 7 points and 30 of which are worth 6 points. You are instructed to do only 30 problems and you have 105 minutes to do it. It takes you 4 minutes to do the 7 point problems and 3 minutes to do the 6 point problems. Assuming you make no mistakes, how many of each problem should you do in order to maximize your score? To solve this problem ﬁrst set

x = number of 7 point problems done. y = number of 6 point problems done.

Now we have score(x,y) = 7x + 6y We want to maximize score(x,y). What sort of constraints do we have? Well, we can only do 30 problems so we have:

x + y 6 30 y 6 x + 30 ⇒ We only have 105 minutes to do our problems so:

4x + 3y 6 105 y 6 ( 4/3)x + 35 ⇒ −

173 6.1. BASIC DEFINITIONS

So what is the mystery point which maximizes the score? Well we have two equations:

y = x + 30 − y = ( 4/3)x + 35. − Setting them equal to each other we ﬁnd:

( 4/3)x + 35 = x + 30 − − Solving for x we get that x = 15 and by plugging 15 in for x in one of the above equations we see that y = 15 too. Now look at our region:

5 10 15 20 25

Since the shaded region is a convex set, the theorem above says that the maximum must occur at a vertex. Plugging the appropriate points in we ﬁnd:

score(0, 0) = 0, score(0, 30) = 180, score(26.25, 0) = 183.75, score(15, 15) = 195.

So we see that solving 15 of the 7 point problems and 15 of the 6 point problems is best.

174 CHAPTER 6. CONVEX SETS

Problems for Section 6.1 (1) What is a convex set?

(2) State which of the following are convex sets:

(a) A straight line. (b) Two lines intersecting at a point. (c) Two parallel lines. (d) A half-plane. (e) The entire plane. (f) A solid disk. (g) An empty box. (h) A solid 7 pointed star.

Explain your answers.

(3) Describe all possible 1-dimensional convex sets.

(4) Draw a series of diagrams illustrating the steps in the proof of Theorem 26.

(5) Draw a diagram showing how the conclusion of Theorem 26 is false if either X1 or X2 are not convex. (6) Prove that the intersection of two convex sets is a convex set.

(7) Is the union of two convex sets ever convex? Is the union of two convex sets always convex?

(8) Draw a diagram showing how the conclusion of Theorem 27 is false if the set in question is not convex.

(9) Draw a series of diagrams illustrating the steps in the proof of Theorem 27.

(10) Use the fact that a half-plane is a convex set to prove that a triangular region is a convex set.

(11) Use the fact that a half-plane is a convex set to prove that a square region is a convex set.

(12) Draw a diagram showing how the conclusion of Theorem 28 is false if the set in question is not convex.

(13) Draw three nonconvex sets and identify one point on the boundary of each through which no supporting line passes.

(14) Consider a bounded convex set. How many boundary points does a ray originating the from the interior of the convex set hit? Explain your answer.

175 6.1. BASIC DEFINITIONS

(15) Is the following picture a counterexample to Theorem 29? Why or why not?

(16) Dieter is a farmer who owns 45 acres of land on each acre of which he plans to grow either wheat or corn. Here are the facts: Each acre of wheat yields $200 profit and requires 2 tons of fertilizer. • Each acre of corn yields $300 profit and requires 4 tons of fertilizer. • Only 120 tons of fertilizer are available. How many acres of wheat and how many acres of corn should Dieter plant to maximize his profit? Explain your work. (17) Jennifer is a carpenter who makes desks and chairs. Here are the facts: Each desk sells for a profit of $40 and requires 4 units of wood. • Each chair sells for a profit of $25 and requires 2 units of wood. • If only 16 units of wood are available and she has to make at least twice as many chairs as desks, how many desks and how many chairs should Jennifer make to maximize her profit? Explain your work. (18) Brübaki Breweries produces two type of beer: David’s Death-From-Above Stout and Han’s Honey-Delight Ale. Here are the facts: David’s Death-From-Above Stout makes a profit of $5 a barrel, and • each barrel requires 6 lbs of barley and 3 oz of hops. Han’s Honey-Delight Ale makes a profit of $2 a barrel, and each • barrel requires 3 lbs of barley and 1 oz of hops. If 60 lbs of barley and 25 oz of hops are available, how many barrels of each type of beer should be made to maximize profit? Explain your work. (19) A clock-smith and her assistant produce two types of watches, a fancy watch and a utilitarian watch. Here are the facts: The fancy watch sells for a profit of $50, and requires 1 hour of work • from the clock-smith and 2 hours of detailing by the assistant. The utilitarian watch sells for a profit of $40, and requires 1 hour of • work from the clock-smith and 1/2 hour of detailing by the assistant. If both people will only work 12 hours a day, how many of each type of watch should be manufactured each day for a maximum profit? Explain your work.

176 CHAPTER 6. CONVEX SETS

6.2 Convex Sets in Three Dimensions

6.2.1 Analogies to Two Dimensions Before we were doing things in two dimensions. A central idea in all of mathematics is that of generalization. In this section we are going to generalize the ideas we had before to three dimensions. Let’s start with a question:

Question Does the idea of a supporting line make sense in three dimensions? ?

Deﬁnition A plane π is called a supporting plane for a three-dimensional set S if π contains a boundary point of S and S lies entirely in one of the half-spaces determined by π.

WARNING Be careful, we currently have two very diﬀerent ideas, the supporting half-plane and supporting plane. Make sure you know the diﬀerence!

Since the supporting plane is the analogue of the supporting line, we have a new idea for the analogue of the supporting half-plane:

Deﬁnition A supporting half-space is the half-space determined by a supporting plane containing the set.

Now that we have our new deﬁnitions, we also have analogous theorems:

Theorem 30 (See Theorem 27) A plane is a supporting plane for a convex set if and only if it goes through at least 1 boundary point of the set and no interior points.

Theorem 31 (See Theorem 28) A three-dimensional convex set is the intersection of all its supporting half-spaces.

6.2.2 Platonic Solids Between 500 BC and 300 BC, there was a group of mystics—today we might call it a cult—who called themselves the Pythagoreans. Being mystics, the Pythagoreans had some strange ideas, but on the other hand they were an enlightened group of people, because they believed that they could better understand the universe around them by studying mathematics. Being that the current physical sciences have strong roots in mathematics, you must agree that we side with the Pythagoreans on this issue. As part of the Pythagoreans’ numerological religion, they thought that some convex solid bodies had special powers. The Pythagoreans associated regular convex polyhedra to elements of nature as follows:

177 6.2. CONVEX SETS IN THREE DIMENSIONS

Fire: Tetrahedron Air: Octahedron

Earth: Cube Water: Icosahedron

However, the Pythagoreans knew of one more regular convex polyhedra, the dodecahedron: Æther: Dodecahedron

Since the ﬁrst four regular convex solids had elements associated to them, the Pythagoreans reasoned that this ﬁfth solid must also be associated to an element. However, all the elements were accounted for. So the Pythagoreans associated the dodecahedron to what we might call the æther, a mysterious non-earthly substance.

Question Look at the regular convex polyhedra above. What does the word regular mean in the phrase regular convex polyhedra? In particular why is a triangular dipyramid, the shape which is two tetrahedrons joined at a face, not a regular polyhedra. How does your deﬁnition of the word regular above discount the above polyhedra from being regular? ? Question How could we use Theorem 31 to deﬁne the Platonic Solids? ?

178 CHAPTER 6. CONVEX SETS

Question Are there other regular convex polygons other than the ones shown above?

We’ll give you the answer to this one:

Theorem 32 There are at most 5 regular convex polyhedra.

Proof To start, a corner of a three-dimensional object made of polygons must have at least 3 faces. Now start with the simplest regular polygon, an equilateral triangle. You can make a corner by placing:

(1) 3 triangles together.

(2) 4 triangles together.

(3) 5 triangles together.

Thus each of the above configuration of triangles could give rise to a regular convex polyhedra. However, you cannot make a corner out of 6 or more equilateral triangles, as 6 triangles all connected lie flat on the plane. Now we can make a corner with 3 square faces, but we cannot make a corner with 4 or more square faces as 4 or more squares lie flat on the plane. Finally we can make a corner with 3 faces each shaped like a regular pentagon, but we cannot make a corner with 4 or more regular pentagonal faces as they will be forced to overlap. Could we make a corner with 3 faces each shaped line a regular hexagon? No, because any number of hexagons will lie flat on the plane. A similar argument works to show that we cannot make a corner with 3 faces shaped like a regular n-gon where n > 6, except now instead of the shapes lying flat on the plane, they overlap. Thus we could at most have 5 regular convex solids.

Question Above we prove that there are at most 5 regular convex polyhedra. How do we know that these actually exist? ?

The ﬁve regular convex polyhedra came to be known as the Platonic Solids, after Plato discussed them in his work Timaeus. To help you on your way, here is a table of facts about the Platonic Solids: Solid Faces Edges Vertexes tetrahedron 4 6 4 octahedron 8 12 6 cube 6 12 8 icosahedron 20 30 12 dodecahedron 12 30 20

179 6.2. CONVEX SETS IN THREE DIMENSIONS

These solids have haunted men for centuries. People who were trying to understand the universe wondered what was the reason that there were only 5 regular convex polyhedra. Some even thought there was something special about certain numbers.

Question In what ways does the number 5 come up in life that might lead a person into believing it is a special number? Does this mean that the number 5 is more special than any other number? ?

180 CHAPTER 6. CONVEX SETS

Problems for Section 6.2 (1) Explain what a supporting half-plane and a supporting plane are, and discuss how they are diﬀerent. (2) State which of the following are convex sets: (a) A hollow sphere. (b) A half-space. (c) The intersection of two spherical solids. (d) A solid cube. (e) A solid cone. (f) The union of two spherical solids. Explain your answers. (3) Use words and pictures to describe the following objects: (a) A tetrahedron. (b) An octahedron. (c) A cube. (d) An icosahedron. (e) A dodecahedron. (f) A triangular dipyramid. (4) How is a regular convex polyhedra diﬀerent from any old convex polyhedra? (5) True or False: If true, explain why. If false explain why. (a) There are only 5 convex polyhedra. (b) The icosahedron has exactly 12 faces. (c) Every vertex of the tetrahedron touches exactly 3 faces. (d) The octahedron has exactly 6 vertexes. (e) Every vertex of the dodecahedron touches exactly 3 faces. (6) While there are only 5 regular convex polyhedra, there are also nonconvex regular polyhedra. Draw some examples. (7) Draw picture illustrating the steps of the proof of Theorem 32. (8) Where does the proof of Theorem 32 use the fact that the solids are convex? (9) A dual polyhedron is the polyhedron obtained when one connects the centers of all the pairs of adjacent faces of a given polyhedron. What are the dual polyhedra of the Platonic Solids?

181 6.3. IDEAS RELATED TO CONVEXITY

6.3 Ideas Related to Convexity 6.3.1 The Convex Hull So far we have talked a lot about convex sets, but what about nonconvex sets? For this occasion, we have the idea of a convex hull, which is sometimes called the convex envelope or convex cover:

Deﬁnition The convex hull of a set S is the intersection of all the convex sets containing S.

Here are some examples of sets and their convex hull:

Theorem 33 A set is convex if and only if it is its own convex hull.

Proof ( ) This is clear from the deﬁnition of convex hull. ( ) If⇒ a set is its own convex hull, then it is the intersection of convex sets and⇐ so it is convex.

Question Can you deﬁne the Platonic Solids using convex hulls? ?

6.3.2 Sets of Constant Width NASA When the space shuttle launches into space there are three sets of rocket engines used:

The main shuttle engines. • Two rocket boosters on either side of the shuttle. • Minutes after the shuttle is launched, the two rocket boosters separate from the shuttle and fall back to Earth, landing in the ocean. Later people retrieve the rocket boosters and they are reused in future missions.

Question What happens when you heat up metal and then cool it down rapidly?

182 CHAPTER 6. CONVEX SETS

? Since the rocket boosters are being heated and rapidly cooled, and also are being stressed as by the enormous forces of the launch, they become slightly warped. Being that the boosters are reused, they come apart, but if they are warped, then they do not go back together as well as they should. If the boosters do not go back together very well, then the boosters could start burning their fuel in unexpected ways, and cause an explosion.

Question If you worked at NASA, how would you check to see if the boosters are still round? ? Before the Challenger accident, here is how NASA did it: They measured the width of the booster every 60 degrees. However, this doesn’t work! There are many shapes with constant width besides the circle! To explain this, ﬁrst we should say what we mean by the width of a set:

Deﬁnition If we have a set, with two parallel supporting lines, then a width of the set is the distance between those supporting lines.

Most sets do not have constant width. However, there are sets with constant width. The easiest example is a circle. But another example is the Reuleaux triangle:

Thus the method NASA was using to see how round their boosters were was not a sound one, as there are many noncircular sets with constant width. To obtain more examples of sets of constant width, you can draw one yourself. We will give two methods of drawing sets of constant width.

Drawing a Set of Constant Width: Method 1 (1) Start with a regular polygon with a odd number of sides.

(2) Place the pointy end of a compass on a vertex of the polygon, and draw an arc spanning the edge opposite to the vertex you started with.

(3) Repeat the process for every vertex of the polygon.

Method 1 is how the Reuleaux triangle is drawn. Any set of constant width drawn using Method 1 is called a Reuleaux polygon. Here is another way of drawing sets of constant width:

183 6.3. IDEAS RELATED TO CONVEXITY

Drawing a Set of Constant Width: Method 2 (1) Draw some straight lines that all intersect each other. (2) Put the pointy end of your compass on an intersection of two of the lines.

(3) Draw an arc connecting the lines that intersect on one side of their intersection.

(4) Moving clockwise around, extend the arc that was drawn before connecting the next two lines with the pointy end of the compass at their intersection. You will need to pick up the compass and resize it to do this.

(5) Repeat step 5 until the curve closes.

Here is a picture which may help explain the above algorithm:

In the above picture, the ﬁrst three arcs are drawn and the point which each arc was drawn around is circled. If the reader continues around in the same fashion, the curve will close and the ﬁnal set will be of constant width.

184 CHAPTER 6. CONVEX SETS

Problems for Section 6.3 (1) Draw the convex hull of the following:

(a) A circle. (b) Two distinct points. (c) Three distinct points, not all on the same line. (d) Two parallel line segments. (e) Two intersecting Line segments. (f) Seven points in a line. (g) Two circles that intersect. (h) Two circles that do not intersect. (i) The shape of the letter “D”. (j) The shape of the letter “X”.

(2) True or False: If true, explain why. If false, give a counterexample.

(a) The convex hull of a set is always a convex set. (b) The convex hull of a finite set of points is a finite set of points. (c) A set of points could have two different convex hulls. (d) The convex hull of a half-plane is the whole plane. (e) The convex hull of a two-dimensional set is always two-dimensional. (f) A line is its own convex hull.

(3) True or False: If true, explain why. If false give a counterexample.

(a) Every width of the Reuleaux triangle is the same. (b) The intersection of a two Reuleaux triangles is always a Reuleaux triangle. (c) The intersection of a two Reuleaux triangles is never a Reuleaux triangle. (d) The Reuleaux triangle has 3 corner points. (e) The Reuleaux triangle is a polygon.

(4) Draw a set of constant width using one of the methods described above.

(5) Find the perimeter of a Reuleaux triangle in terms of its constant width. Explain your work.

(6) Find the perimeter of a Reuleaux pentagon in terms of its constant width. Explain your work. Hint: Recall that if an angle of measure d degrees has its vertex on a circle of circumference 1, then the arc spanned by the intersection of the edges of the angle with the circle is 2d/360.

185 6.3. IDEAS RELATED TO CONVEXITY

(7) Find the perimeter of a Reuleaux 7-gon in terms of its constant width. Explain your work. Hint: Recall that if an angle of measure d degrees has its vertex on a circle of circumference 1, then the arc spanned by the intersection of the edges of the angle with the circle is 2d/360.

(8) Considering the previous 3 exercises, make a conjecture about the perimeter of a set of constant width.

186 CHAPTER 6. CONVEX SETS

6.4 Advanced Theorems

In this section we simply state several advanced theorems related to the above concepts and then ask questions about their statements. Theorem 34 Let X and X be two convex sets. If X X , then the 1 2 2 ⊆ 1 perimeter of X1 is greater than or equal to the perimeter of X2. Question What does this this theorem look like? ? Question If you leave oﬀ or change the assumptions of the theorem, is it still true? ?

Theorem 35 (Helly) Let X1,...,Xn be n convex sets lying in d-space with n > d+1 such that any collection of d+1 of the sets has a nonempty intersection. Then the intersection of all the sets is nonempty.

Question What does this this theorem look like? ? Question If you leave oﬀ or change the assumptions of the theorem, is it still true? ?

Theorem 36 (Radon) Let S = P1,...,Pn be a set of points in d-space, if n > d + 2, then we can partition{ the set into} two disjoint sets whose convex hulls intersect.

Question What does this this theorem look like? ? Question If you leave oﬀ or change the assumptions of the theorem, is it still true? ?

Theorem 37 Let S = P1,...,Pn be a set of n points in the plane. Then there is a point A such that{ every half-plane} determined by a line through A contains at least n/4 points of S. Question What does this this theorem look like?

187 6.4. ADVANCED THEOREMS

? Question Will any point for A work? Or does A need to be chosen somewhat carefully? ? Theorem 38 If we have a set S of n points in a plane such that each set of 3 points can be enclosed in a circle of radius 1, then every point in S can be enclosed in a single circle of radius 1. Question What does this this theorem look like? ? Question If you leave oﬀ or change the assumptions of the theorem, is it still true? ? Theorem 39 Given n parallel line segments in a plane, if there exists a line that intersects all sets of 3 of them, then there is a line that intersects all of them.

Question What does this this theorem look like? ? Question If you leave oﬀ or change the assumptions of the theorem, is it still true? ?

188 CHAPTER 6. CONVEX SETS

Problems for Section 6.4 (1) Draw a picture depicting the statement of Theorem 34 and give a short explanation of how your picture depicts the statement.

(2) Give an example showing that the conclusion of Theorem 34 does not hold for nonconvex sets.

(3) Draw a picture depicting the statement of Helly’s Theorem and give a short explanation of how your picture depicts the statement.

(4) Give an example where d = 2 showing that the conclusion of Helly’s Theorem does not hold for nonconvex sets.

(5) Draw a picture depicting the statement of Radon’s Theorem and give a short explanation of how your picture depicts the statement.

(6) Give an example where n = 3 and d = 2 showing that the conclusion of Radon’s Theorem does not hold in this case.

(7) Draw a picture depicting the statement of Theorem 37 and give a short explanation of how your picture depicts the statement.

(8) Give an example showing that the conclusion Theorem 37 does not hold for all choices of the point A. (9) Draw a picture depicting the statement of Theorem 38 and give a short explanation of how your picture depicts the statement.

(10) Draw a picture depicting the statement of Theorem 39 and give a short explanation of how your picture depicts the statement.

(11) Give an example of four parallel line segments such that a line intersects 3 of them, yet no line intersects all the segments simultaneously.

(12) Give an example of three parallel line segments such that some line intersects every pair of line segments, yet no line intersects all three of the segments simultaneously.

189 References and Further Reading

[1] N. Bourbaki. Elements of the History of Mathematics. Springer-Verlag, 1984.

[2] C.B. Boyer and Uta C. Merzbach. A History of Mathematics. Wiley, 1991.

[3] R.G. Brown. Transformational Geometry. Ginn and Company, 1973.

[4] H. D¨orrie. 100 Great Problems of Elementary Mathematics: Their History and Solution. Dover, 1965.

[5] Euclid and T.L. Heath. The Thirteen Books of Euclid’s Elements: Volumes 1, 2, and 3. Dover, 1956.

[6] H.M. Enzensberger. The Number Devil: A Mathematical Adventure. Metropolitan Books, 1998.

[7] H. Eves. An Introduction to the History of Mathematics. Saunders College Publishing, 1990.

[8] R.P. Feynman. “What Do You Care What Other People Think?”. Norton, 2001.

[9] M. Gardner. The Colossal Book of Mathematics. Norton, 2001.

[10] . Mathematics, Magic and Mystery. Dover, 1956.

[11] J. Gullberg. Mathematics: From the Birth of Numbers. Norton, 1997.

[12] S. Hawking. God Created the Integers. Running Press, 2005.

[13] J.L. Heilbron. Geometry Civilized. Oxford, 1998.

[14] E.F. Krause. Taxicab Geometry: An Adventure in Non-Euclidean Geome- try. Dover, 1986.

[15] M. Livio. The Golden Ratio: The Story of Phi, the World’s Most Aston- ishing Number. Broadway Books, 2002.

190 REFERENCES AND FURTHER READING

[16] C.H. MacGillavry. Symmetry Aspects of M.C. Escher’s Periodic Drawings. A. Oosthoek’s Uitgeversmaatschappij, 1965.

[17] R.B. Nelsen. Proofs Without Words. Mathematical Association of America, 1993.

[18] . Proofs Without Words II. Mathematical Association of America, 2000.

[19] G. Polya. Mathematical Discovery: On Understanding, Learning, and Teaching Problem Solving . Wiley, 1981.

[20] C. Sagan. Cosmos. Random House, 2002.

[21] T.S. Row. Geometric Exercises in Paper Folding. The Open Court Pub- lishing Company, 1901.

[22] J.R. Smart. Modern Geometries. Brooks/Cole, 1998. [23] E.W. Weisstein. MathWorld—A Wolfram Web Resource. http://mathworld.wolfram.com/

[24] D. Wells. The Penguin Dictionary of Curious and Interesting Geometry. Penguin Books, 1991.

191 Index

æther, 178 Spherical Geometry, 6 algebraic geometry, 14 circumcenter, 81, 86, 120 altitude, 83, 86 circumcircle, 82, 120 analytic geometry, 17 circumscribe, 82, 91 and, 45 City Geometry, 23, 24 and probability, 105 circle, 29 antipodal point, 6 midset, 29 Archimedes, 141 parabola, 32 area triangle, 28 Heron’s Formula, 91 closed, 132 Pick’s Theorem, 92 collapsing compass, 115 arithmetic mean, 60 commutative, 132 Arithmetic-Geometric Mean Inequality, compass 61 collapsing, 115 associative, 132 compass and straightedge axiom, 4 addition, 130 bisecting a segment, 115 Battle Royale, 27 bisecting an angle, 116 beauty, see truth copying an angle, 117 beer, 47, 176 division, 131 bees, 54 equilateral triangle, 114 Bertrand’s paradox, 102 boat golden ratio, 135 lost at night, 127 impossible problems, 140 brain juices, 127 multiplication, 131 bucket parallel through a point, 118 Alma Mater, 35 pentagon, 135 perpendicular through a point, 116 , 130 SAA triangle, 123 calisson,C 67 SAS triangle, 122 Cartesian plane, 14 SSS triangle, 122 center of mass, 84 subtraction, 130 centroid, 84, 86, 120 tangent to a circle, 118 circle transferring a segment, 115 circumcircle, 82 complement, 40 City Geometry, 29 conic-section, 13 incircle, 83 constructible numbers, 130

192 INDEX constructions, see compass and straight- spirals, 99 edge or paper-folding triangle, 99, 136 continued fraction, 95 gorilla suit, 24 convex, 170 great circle, 5 convex hull, 182 group, 164 convex polyhedra regular, 178 Helly’s Theorem, 187 Crane Alley, 26 Heron’s Formula, 91 for quadrilaterals, 91 diagonal reflection, 148 for triangles, 91 dissection proof, 57 horizontal reflection, 148 doubling the cube, 60, 140 dual identity matrix, 156 polyhedron, 181 if-and-only-if, 48 if-then, 46, 47 , 38, 130 iff, 48 e∈, 96 incenter, 82, 120 The Elements, 1 incircle, 83, 120 empty set, 40 integers, 130 equilateral triangle, 87 intersection, 39 Eratosthenes, 2 Kepler, Johannes, 51 Escher, M.C., 51 Euclid, 1, 113 lemma, 91 Euclidean Geometry, 4 lemon, see lemma Euler line, 86, 121 Let’s Make a Deal, 102 logical symbols field, 132 , 45, 105 fractional part, 96 ∧ , 48 free point, 114 ⇔, 47 ⇒, 45 Galileo Galilei, 66 ¬, 45, 105 Gauss, Carl Friedrich, 62, 136 Louie∨ Llama, 146, 148 geometric mean, 60 Louie Llama, 150 geometric transformation, 145 geometry matrix, 144 algebraic, 14 multiplication, 145 analytic, 17 median, 84 City, 24 midset Spherical, 5 City Geometry, 29 synthetic, 13 Miquel point, 87, 121 geometry,City, 23 Miquel’s Theorem, 87 goat, 102 Monty Hall problem, 102 golden Morley’s Theorem, 87 ratio, 98, 135 compass and straightedge, 135 NASA, 182 rectangle, 99 nexus of the universe, 23

193 INDEX

Nine-Point Circle, 86, 121 rational numbers, 132 nontransitive dice, 104 rational roots test, 134 not, 46 reflection, 147 diagonal, 148 or, 45 horizontal, 148 and probability, 105 vertical, 148 orthocenter, 83, 86, 120 regular paper-folding convex polyhedra, 177 Morley’s Theorem, 87 tessellation, 51 the centroid, 84 Reuleaux the circumcenter, 82 polygon, 183 the incenter, 82 triangle, 183 the orthocenter, 83 Rock-Paper-Scissors, 104 parabola rotation, 150 algebraic definition, 15 City Geometry, 32 set, 38 conic-section definition, 13 set theory symbols intrinsic definition, 12 , 40 − paradox, 32 ∅, 40 √2 = 2, 33 , 38, 130 ∈ 1 = 0.999 . . . , 92 , 39 ∩ all triangles are isosceles, 88 , 39 ⊆ Bertrand’s paradox, 102 , 39 ∪ involving dice, 104 sets of constant width, 182 the Monty Hall problem, 102 drawing, 183, 184 triangle dissection, 59 slopef(x)(x,s), 17 parallel postulate, 5 Socrates, 113 pentagon, 135 Spherical Geometry, 5 φ, 98 great circle, 5 π, 96, 101 Spherical Geometry Pick’s Theorem, 92 circle, 5 Plato, 113 line, 5 Platonic Solids, 177, 179 triangle, 7 polyhedra spirals, 99 convex regular, 178 √2, 33, 95, 97 prime numbers, 136 squaring the circle, 140 probability, 100 subset, 39 Pythagorean Theorem, 56, 122 supporting The Pythagoreans, 177 half-plane, 172 Q, 131 half-space, 177 Q(α), 134 line, 171 quadrilateral plane, 177 tessellation of, 51 symmetry group group, 166 Radon’s Theorem, 187 synthetic geometry, 13

194 INDEX tables truth, 45 tangent line, 13, 15, 16 Taxicab distance, 23 tessellation, 51 any quadrilateral, 51 regular, 51 triangles, 51 three doors, 102 transformation, 144 reﬂection, 147 rotation, 150 translation, 146 translation, 146 triangle altitude, 83 centroid, 84 circumcenter, 81 circumcircle, 82 City Geometry, 28 incenter, 82 incircle, 83 orthocenter, 83 Spherical Geometry, 7 triangular dipyramid, 178 trisecting the angle, 87, 141 by cheating, 142 truth, see beauty truth-table, 45

, 39 union,∪ 39 vertical reﬂection, 148

Wason selection task, 46 whole-number part, 96 width of a set, 183

Z, 130

195