Ideas in Geometry
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c 2010 Alison Ahlgren and Bart Snapp.
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This document was typeset on December 15, 2010.
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 open-source. It 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 Carolina 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 Euclid and Beyond . . . . . . . . . . . . . . . . . . .
1.1.1 The Most Successful Textbook Ever Written
1.1.2 The Parallel Postulate . . . . . . . . . . . . .
1.2 Points of View . . . . . . . . . . . . . . . . . . . . .
1.2.1 Synthetic Geometry . . . . . . . . . . . . . .
1.2.2 Algebraic Geometry . . . . . . . . . . . . . .
1.2.3 Analytic Geometry . . . . . . . . . . . . . . .
1.3 City Geometry . . . . . . . . . . . . . . . . . . . . .
1.3.1 Getting Work Done . . . . . . . . . . . . . .
1.3.2 (Un)Common Structures . . . . . . . . . . . .
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1
1
1
5
12
13
14
17
23
25
28
2 Proof by Picture
2.1 Basic Set Theory . . . . . . . . . . . . .
2.1.1 Union . . . . . . . . . . . . . . .
2.1.2 Intersection . . . . . . . . . . . .
2.1.3 Complement . . . . . . . . . . .
2.1.4 Putting Things Together . . . . .
2.2 Logic . . . . . . . . . . . . . . . . . . . .
2.3 Tessellations . . . . . . . . . . . . . . . .
2.3.1 Tessellations and Art . . . . . . .
2.4 Proof by Picture . . . . . . . . . . . . .
2.4.1 Proofs Involving Right Triangles
2.4.2 Proofs Involving Boxy Things . .
2.4.3 Proofs Involving Sums . . . . . .
2.4.4 Proofs Involving Sequences . . .
2.4.5 Thinking Outside the Box . . . .
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38
38
39
39
40
41
45
51
52
56
56
60
63
67
68
3 Topics in Plane Geometry
3.1 Triangles . . . . . . . . . . . . .
3.1.1 Centers in Triangles . . .
3.1.2 Theorems about Triangles
3.2 Numbers . . . . . . . . . . . . . .
3.2.1 Areas . . . . . . . . . . .
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83
83
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87
93
93
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3.2.2
3.2.3
Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
Combining Areas and Ratios—Probability . . . . . . . . . 102
4 Compass and Straightedge Constructions
4.1 Constructions . . . . . . . . . . . . . . . .
4.2 Trickier Constructions . . . . . . . . . . .
4.2.1 Challenge Constructions . . . . . .
4.2.2 Problem Solving Strategies . . . .
4.3 Constructible Numbers . . . . . . . . . . .
4.4 Impossibilities . . . . . . . . . . . . . . . .
4.4.1 Doubling the Cube . . . . . . . . .
4.4.2 Squaring the Circle . . . . . . . . .
4.4.3 Trisecting the Angle . . . . . . . .
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115
115
124
125
129
132
142
142
142
143
5 Isometries
5.1 Matrices as Functions . . . . . .
5.1.1 Translations . . . . . . . .
5.1.2 Reflections . . . . . . . .
5.1.3 Rotations . . . . . . . . .
5.2 The Algebra of Matrices . . . . .
5.2.1 Matrix Multiplication . .
5.2.2 Compositions of Matrices
5.2.3 Mixing and Matching . .
5.3 The Theory of Groups . . . . . .
5.3.1 Groups of Reflections . .
5.3.2 Groups of Rotations . . .
5.3.3 Symmetry Groups . . . .
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146
146
147
149
152
159
159
160
163
167
167
168
169
6 Convex Sets
6.1 Basic Definitions . . . . . . . . . . .
6.1.1 An Application . . . . . . . .
6.2 Convex Sets in Three Dimensions . .
6.2.1 Analogies to Two Dimensions
6.2.2 Platonic Solids . . . . . . . .
6.3 Ideas Related to Convexity . . . . .
6.3.1 The Convex Hull . . . . . . .
6.3.2 Sets of Constant Width . . .
6.4 Advanced Theorems . . . . . . . . .
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173
173
176
180
180
180
185
185
185
190
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References and Further Reading
192
Index
194
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
1.1.1
Euclid and Beyond
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 librarian 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
x
360
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, Eratosthenes 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 .
1 Actually, 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 removing 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 Ushaped 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 Given a point and a line, a parabola is the set of points such that
each of these points is the same distance from the given point as it is from the
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 philosopher 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
slopef (x) (a, s) =
f (a + s) − f (a)
.
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
.
s
Now you plug in values for s that approach zero. Look at this:
slopef (x) (2, s) =
s = 0.1
⇒
s = 0.01
slopef (x) (2, 0.1) = 4.1
⇒
s = 0.001
slopef (x) (2, 0.01) = 4.01
⇒
slopef (x) (2, 0.001) = 4.001
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
.
s
Now you plug in values for s that approach zero. Look at this:
slopeg(x) (2, s) =
s = 0.1
s = 0.01
s = 0.001
⇒
⇒
⇒
slopeg(x) (2, 0.1) = 12.61
slopeg(x) (2, 0.01) = 12.0601
slopeg(x) (2, 0.001) = 12.006001
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),
do not approach the slope of f (x) at x = 2?
?
18
slopef (x) (2, 0.001),
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
slopeex (a, s) ≈ ea
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 approaches 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 = a7 x7 +a6 x6 +a5 x5 +a4 x4 +a3 x3 +a2 x2 +a1 x1 +a0
Find a0 , a1 , a2 , a3 , a4 , a5 , a6 , a7 .
(8) Given:
6x5 + a4 x4 − x2 + a0 = a5 x5 − 24x4 + a3 x3 + a2 x2 − 5
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)
s
gives you the slope of the tangent line that passes through the point
(a, f (a)), when s is near zero.
slopef (x) (a, s) =
(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) = x2 g(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 [15].
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
flies?
How do you compute the distance between A and B as the crow
?
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 Geometry 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 Euclidean 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 intersect, 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 intersect, 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
etry.
Suppose you wished to find the midset of two points in City Geom-
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 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
√
False-Proof
2 = 2.
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’s2 head, 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).
2 The
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 ).
(c) dT (C, A) = dT (F, D).
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 diameter, 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 customers 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. Explain 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
T
F
¬P
F
T
When you apply not to a statement, it simply swaps true for false in the truthtables:
Now consider the statement:
I’m strong |and
I’m cool.
| {z } {z } | {z }
Q
∧
P
and let P = I’m strong, while Q = I’m cool. When 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
T
T
F
F
Q
T
F
T
F
P ∧Q
T
F
F
F
Now what if we want to look at the statement:
I| eat ice
or I| eat cookies.
{z cream} |{z}
{z
}
P
∨
Q
When is the above statement true? It is true 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
T
T
F
F
Q
T
F
T
F
P ∨Q
T
T
T
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.
| {z } {z } | {z }
Then
Q
∧
P
¬(P ∧ Q) = (¬P ) ∨ (¬Q)
or I’m
= I’m not strong |{z}
| not
{z cool.}
{z
}
|
∨
¬P
¬Q
We can see this best using a truth-table:
P
T
T
F
F
Question
Q
T
F
T
F
P ∧Q
T
F
F
F
¬(P ∧ Q)
F
T
T
T
¬P
F
F
T
T
¬Q
F
T
F
T
(¬P ) ∨ (¬Q)
F
T
T
T
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
alcohol
old
people
beer
young
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
T
T
F
F
Q
T
F
T
F
P ⇒Q
T
F
T
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
T
T
F
F
Q
T
F
T
F
¬P
F
F
T
T
¬P ∨ Q
T
F
T
T
P ⇒Q
T
F
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—,
look like?
?
48
—if and only if—,
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) 6= (Q ⇒ P ).
(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
T
T
F
F
Q
T
F
T
F
P ×Q
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 certainly 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) Give an example of a triangle that tessellates the plane where both 4 and
8 angles fit around each vertex.
(7) 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 it is always the case
that exactly 6 angles will fit around each vertex.
(e) If a polygon has more than 6 sides, then it cannot tessellate the plane.
(8) Fill in the following table:
Regular
n-gon
3-gon
4-gon
5-gon
6-gon
7-gon
8-gon
9-gon
10-gon
Does it
tessellate?
Measure
of angles
If it tessellates, how
many surround each vertex?
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.
54
CHAPTER 2. PROOF BY PICTURE
(c) What regular polygons could a bee use for building hives? Give some
reasons that bees seem to use hexagons.
(9) Given a regular tessellation, what is the sum of the angles around a given
vertex?
(10) Given that the regular octagon has 135 degree angles, explain why you
cannot give a regular tessellation of the plane with a regular octagon.
(11) Considering that the regular n-gon has interior angles of 180(n − 2)/n
degrees, and Problem (8) above, prove that there are only 3 regular tessellations of the plane.
55
2.4. PROOF BY PICTURE
2.4
Proof by Picture
Pictures generally do not constitute a proof on their own. However, a good
picture can show insight and communicate concepts better than words alone.
In this section we will show you pictures giving the idea of a proof and then ask
you to supply the words to finish off the argument.
2.4.1
Proofs Involving Right Triangles
Let’s start with something easy:
Question Explain how the following picture “proves” that the area of a right
triangle is half the base times the height.
?
That wasn’t so bad was it? Now for a game of whose-who:
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
a 2 + b2 = c 2 .
Nearly all of the pictures from this section are adapted from the wonderful source books:
[18] and [19].
56
CHAPTER 2. PROOF BY PICTURE
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:
Consider the following picture:
c2
a2
b2
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.
57
2.4. PROOF BY PICTURE
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
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:
c 2 = a 2 + b2
Let’s give another proof! This one looks at a tessellation involving 2 squares.
1111
0000
0000
1111
0000
1111
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
58
CHAPTER 2. PROOF BY PICTURE
square is b2 . Now label all the “parts” of the large overlaid square:
5
1111
0000
0000
1111
0000
1111
3
4
2
1
From the picture we see that
a2 = {3 and 4}
b2 = {1, 2, and 5}
c2 = {1, 2, 3, 4, and 5}
Hence
c 2 = a 2 + 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:
59
2.4. PROOF BY PICTURE
Paradox
What is wrong with this picture?
Question
How does this happen1 ?
?
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
the cube?
Why doesn’t doubling each side of the cube double the volume of
?
1 See
[10] Chapter 8, for a wonderful discussion of puzzling pictures like this one.
60
CHAPTER 2. PROOF BY PICTURE
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.
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)
numbers then:
√
a+b
a·b6
2
and the inequality above is an equality when a = b.
61
If a and b are positive
2.4. PROOF BY PICTURE
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
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
4
√
a+b
a·b6
,
2
ab 6
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
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 .
62
CHAPTER 2. PROOF BY PICTURE
But the area of the all the triangles is 4ab. Hence
4ab 6 (a + b)2
(a + b)2
4
√
a+b
,
a·b6
2
ab 6
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:
0
+
100
k
100
1
+
99
k
100
|
2
+
98
k
100
3
+
97
k
100
4
+
96
k
100
5
+
95
k
100
···
···
···
···
···
{z
95
+
5
k
100
96
+
4
k
100
97
+
3
k
100
98
+
2
k
100
99
+
1
k
100
100
+
0
k
100
}
101 terms
So if we sum up all the integers from 1 to 100 twice we get
100 · 101.
Thus, the sum of all the integers from 1 to 100 is
1 + 2 + 3 + 4 + 5 + · · · + 95 + 96 + 97 + 98 + 99 + 100 =
100 · 101
= 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
63
2.4. PROOF BY PICTURE
?
Consider this picture:
Question
Explain how the picture above “proves” that:
1 + 2 + 3 + 4 + 5 + ··· + n =
n(n + 1)
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
light colored circles. Therefore
1 + 2 + 3 + 4 + 5 + ··· + n =
n(n + 1)
.
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
number?
Can you add up an infinite number of terms and still get a finite
64
CHAPTER 2. PROOF BY PICTURE
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
+
2
2 3 4 5
1
1
1
1
+
+
+
+ ··· = 1
2
2
2
2
Wow! How can we visualize this? Consider this picture:
1
1/2
(1/2)2
1
Question
Explain how the picture above “proves” that:
1
+
2
2 3 4 5
1
1
1
1
+
+
+
+ ··· = 1
2
2
2
2
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
+
2
2 3 4 5
1
1
1
1
+
+
+
+ ··· = 1
2
2
2
2
65
2.4. PROOF BY PICTURE
Now look at:
Question
Explain how the picture above “proves” that:
2 3 4 5
1
1
1
1
1
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
is:
2
1
1
+
4
4
Continuing on in this fashion we see that the area of all the shaded regions is:
2 3 4 5
1
1
1
1
1
+
+
+
+ ···
+
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.
66
CHAPTER 2. PROOF BY PICTURE
2.4.4
Proofs Involving Sequences
Get your calculator out and play along at home:
1
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
=
1
3
=?
=?
=?
=?
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
1 + 3 + · · · + (2n − 1)
=
(2n + 1) + (2n + 3) + · · · + (4n − 1)
3
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
67
2.4. PROOF BY PICTURE
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.
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” that the area of a right triangle
is half the base times the height.
(2) Building on Problem (1), explain how the following picture “proves” that
the area of any triangle is half the base times the height.
(3) You may have noticed Geometry Giorgio in your class. In an attempt to
prove the formula for the area of a triangle, Geometry Giorgio draws the
following picture:
What is he doing wrong? How could he fix his “proof”?
(4) Again building on Problem (1), explain how the following picture “proves”
that the area of any triangle is half the base times the height. Note, this
way of thinking is the basis for Cavalieri’s Principle.
1111
0000
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
1111
0000
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
69
2.4. PROOF BY PICTURE
(5) Explain how the following picture “proves” that the area of any parallelogram is base times height. Note, this way of thinking is the basis for
Cavalieri’s Principle.
11111111111
00000000000
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
00000000000
11111111111
111111
000000
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
(6) Explain how the following picture “proves” the Pythagorean Theorem.
(7) Explain how the following picture “proves” the Pythagorean Theorem.
(8) Explain how the following picture “proves” the Pythagorean Theorem.
70
CHAPTER 2. PROOF BY PICTURE
Note: This proof is due to Leonardo da Vinci.
(9) Explain how the following picture “proves” the Pythagorean Theorem.
111
000
000
111
(10) Use the following tessellation to give a dissection proof of the Pythagorean
Theorem.
111
000
000
111
71
2.4. PROOF BY PICTURE
(11) Explain how the following picture “proves” the Pythagorean Theorem.
a
c
b
×b
×c
b2
×a
bc
bc
c2
ab
a2
ac
ac
(12) 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
area =
h(b1 + b2 )
,
2
where h is the height of the trapezoid and b1 , b2 , are the lengths of the
parallel sides?
(13) 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.
(14) Look at Problem (12). Can you use a similar picture to prove that the
72
CHAPTER 2. PROOF BY PICTURE
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.
(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 quadrilateral has two
opposite angles that are equal, then the bisectors of the other two angles
73
2.4. PROOF BY PICTURE
are parallel or on top of each other.
(19) 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.
74
CHAPTER 2. PROOF BY PICTURE
(20) Why might someone find the following picture disturbing? How would you
assure them that actually everything is good and well in the geometrical
world?
75
2.4. PROOF BY PICTURE
(21) Why might someone find the following picture disturbing? How would you
assure them that actually everything is good and well in the geometrical
world?
(22) How could you explain to someone that doubling the lengths of each side
of a cube does not double the volume of the cube?
(23) Explain how the following picture “proves” that if a and b are positive
numbers then:
√
a+b
a·b6
2
and the inequality above is an equality when a = b.
a
b
a
b
(24) Explain how the following picture “proves” that the sum of a number x
76
CHAPTER 2. PROOF BY PICTURE
and its inverse 1/x is at least 2.
1
x
1
1
x
1
1
x
1
x
(25) Explain how the following picture “proves” that if a and b are positive
numbers then:
√
a+b
a·b6
2
and the inequality above is an equality when a = b.
(26) Explain how the following picture “proves” that:
1 + 2 + 3 + 4 + 5 + ··· + n =
n(n + 1)
2
(27) Explain how the following picture “proves” that:
1 + 2 + 3 + 4 + 5 + ··· + n =
77
1 2
n(n + 1)
(n + n) =
2
2
2.4. PROOF BY PICTURE
(28) Explain how the following picture “proves” that:
1 + 2 + 3 + 4 + 5 + ··· + n =
n
n(n + 1)
n2
+ =
2
2
2
(29) Explain how the following picture “proves” that:
1 + 3 + 5 + · · · + (2n − 1) = n2
78
CHAPTER 2. PROOF BY PICTURE
(30) Explain how the following picture “proves” that:
1 + 2 + 3 + · · · + (n − 1) + n + (n − 1) + · · · + 3 + 2 + 1 = n2
(31) Explain how the following picture “proves” that:
1
+
2
2 3 4 5
1
1
1
1
+
+
+
+ ··· = 1
2
2
2
2
1
1/2
(1/2)2
1
(32) Explain how the following picture “proves” that if 0 < r < 1:
r + r(1 − r) + r(1 − r)2 + r(1 − r)3 + · · · = 1
79
2.4. PROOF BY PICTURE
r
r(1 − r)3
r(1 − r)2
1
r
r(1 − r)
r(1 − r)
r
1
(33) Explain how the following picture “proves” that:
1
+
4
2 3 4 5
1
1
1
1
1
+
+
+
+ ··· =
4
4
4
4
3
(34) Explain how the following picture “proves” that:
1
+
2
2 3 4
1
1
1
+
+
+ ··· = 1
2
2
2
80
CHAPTER 2. PROOF BY PICTURE
Hint: Add up the area of the shaded regions and the area of the unshaded
regions.
(35) Explain how the following picture “proves” that:
1 + 3 + · · · + (2n − 1)
1
=
(2n + 1) + (2n + 3) + · · · + (4n − 1)
3
(36) Explain how the following picture “proves” that:
1
1 + 3 + · · · + (2n − 1)
=
(2n + 1) + (2n + 3) + · · · + (4n − 1)
3
2n
1
3
5
2n − 1
(37) Look at:
1·1=1
11 · 11 = 121
111 · 111 = 12321
1111 · 1111 = 1234321
81
2.4. PROOF BY PICTURE
Does the pattern continue? Explain your answer.
(38) 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.
82
Chapter 3
Topics in Plane Geometry
Geometry is the science of correct reasoning on incorrect figures.
—George Polya
3.1
3.1.1
Triangles
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:
83
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.
84
CHAPTER 3. TOPICS IN PLANE GEOMETRY
(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 lines containing 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.
85
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?
?
86
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
i
r
c
u
m
c
e
n
t
e
r
O
r
t
h
o
c
e
n
t
e
r
I
n
c
e
n
t
e
r
c
e
N
t
r
o
i
d
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
87
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.
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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.
89
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
A
X
B
Z
Y
C
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
△BY P ≡ △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.
?
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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 lines containing 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.
91
3.1. TRIANGLES
(18) True or False: Explain your conclusions.
(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◦ .
(c) 15◦ , 45◦ , and 120◦ .
Use a protractor as necessary. Give a short explanation of your illustrations.
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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
s
P
P
P
P
area =
−a
−b
−c
2
2
2
2
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
s
P
P
P
P
−a
−b
−c
−d
area =
2
2
2
2
where P = a + b + c + d, the perimeter of the quadrilateral.
93
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
area =
b
+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:
area =
3.2.2
8
+ 2 − 1 = 5 square units.
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.
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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:
s
r
q
√
x = 1 + 1 + 1 + 1 + ···
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.
95
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,
x=1+
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:
s
r
q
1+
1+
1+
√
1 + ··· = 1 +
1
1
1+
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:
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CHAPTER 3. TOPICS IN PLANE GEOMETRY
Definition
A fraction of the form
a0 +
a1
b1 +
b2 +
a2
a3
a4
b4 + · · ·
b3 +
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+
2+
1
2 + ···
Using the technique as above,
x=1+
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+
2+
1
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!
97
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+
6+
1
1 + ···
Also check out π = 3.14159265358 . . . . It turns out that we can get a nice
continued fraction for π:
π =3+
12
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
5.32 is 5.
The whole-number part of 2 is 2, while the whole-number part of
Definition The fractional part of a number is the number minus its wholenumber 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
> 2.
3>
5
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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,
3
1
13
=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
=2+
5
1
.
2
3
Again, we want our numerator to be 1, not 2 so we will repeat the steps above
to get
1
1
13
1
=2+
=2+
=2+
2
1
1
5
1+
1+ 3
1+
1
3
2
1+
2
and this last expression is the simple continued fraction for 13/5. We could also
list our steps as:
1+
13
3
=2+
5
5
2
5
1
3 = 3 =1+ 3
5
1
1
3
2 = 2 =1+ 2
3
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
2. To start
already fractions
(otherwise
this
would
all
be
a
bit
silly).
Consider
√
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
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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
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.
1+
Question
What’s so special about the golden ratio?
Given any rectangle you can divide it into a square and another smaller
rectangle:
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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
1
x
So we want
x−1
1
=
x
1
and so:
Thus
1 = x2 − x
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:
101
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?
?
Question What is the probability that a point chosen at random in the square
below will land in the shaded region?
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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
a 2 + b2 6 1
then the point is inside the circle, otherwise the point is outside the circle.
103
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
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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 6= 1/4!
Question
What is wrong in the above discussion?
?
105
3.2. NUMBERS
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
3
3
2
3
6
2
1
6
5
1
0
5
4
0
3
2
1
4
3
2
5
4
(3)
(2)
(1)
(0)
4
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
6
2
6
2
2
(2)
Before we can compute the next probability we need two lemmas.
1 OK I admit it, the method through which paper actually beats rock has always been a
mystery to me.
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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 (E1 ∨ E2 ) = P (E1 ) + P (E2 ).
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 (E1 ∧ E2 ) = P (E1 ) · P (E2 ).
Die (2) has a 2/3 probability of beating Die (1) since you roll a 6 with Die (2)
and win
2
6
2
6
2
2
(2)
with a 2/6 = 1/3 probability or you roll as follows:
2
6
2
1
6
and
5
1
2
1
2
5
(2)
(1)
5
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
2
1 1
+ = .
3 3
3
Thus Die (2) has a 2/3 probability of beating Die (1).
107
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:
5
1
1
1
5
5
and
4
(1)
0
0
4
4
4
(0)
Since there is an and there, we must multiply the probabilities to get a probability of
1
1 2
· = .
2 6
6
Since there was an or above, we must add
2
1 1
+ = .
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
(0)
with a 4/6 = 2/3 probability.
108
4
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:
109
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:
s
r
x=
2+
q
2+
√
6+
q
6+
√
12 +
q
12 +
2+
2 + ···
Explain your work.
(17) Find the exact value for x when:
s
r
x=
6+
6 + ···
Explain your work.
(18) Find the exact value for x when:
s
r
x=
12 +
Explain your work.
110
√
12 + · · ·
CHAPTER 3. TOPICS IN PLANE GEOMETRY
(19) Find the exact value for x when:
s
r
x=
20 +
20 +
q
20 +
√
20 + · · ·
q
30 +
√
30 + · · ·
Explain your work.
(20) Find the exact value for x when:
s
r
x=
30 +
30 +
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+
8+
Explain your work.
1
8 + ···
(24) Find the exact value for x when:
1
x=3+
1
10 +
1
10 +
10 +
Explain your work.
111
1
10 + · · ·
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.
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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.
113
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
4
4
2
4
6
2
3
2
3
2
(a)
(b)
6
(47) Here are two dice. Which die has a better probability of rolling higher,
and what is that probability? Explain your answer.
3
4
4
5
4
2
5
3
5
3
2
(a)
(b)
2
(48) Here are two dice. Which die has a better probability of rolling higher,
and what is that probability? Explain your answer.
3
4
4
2
4
6
6
3
2
3
2
(a)
(b)
114
6
Chapter 4
Compass and Straightedge
Constructions
Mephistopheles:
Faust:
Mephistopheles:
I must say there is an obstacle
That prevents my leaving:
It’s the pentagram on your threshold.
The pentagram impedes you?
Tell me then, you son of hell,
If this stops you, how did you come in?
Observe! The lines are poorly drawn;
That one, the outer angle,
Is open, the lines don’t meet.
—Göthe, 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.
115
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)
angle given the length of one side.
We wish to construct an equilateral tri-
(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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CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS
Euclid’s second construction will also be our second construction:
Construction (Transferring a Segment)
it so that it starts on a given point.
Given a segment, we wish to move
(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)
half.
Given a segment, we wish to cut it in
(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.
117
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.
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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.
119
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.
120
CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS
Question
circle?
What if the point is inside the circle? What if the point is on the
?
121
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.
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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.
123
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?
124
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.
125
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.
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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.
127
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 information 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?
?
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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?
?
129
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 spaced on a circle.
Explain how to use a compass and straightedge construction to help her
out. Illustrate your answer with a construction and explain the steps in
your 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.
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CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS
(15) Construct a circle with a given radius tangent to two other given circles.
Explain the steps in your construction.
(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. Your friend knows the angle starting with them, going to the tree,
and ending with 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 that
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.
131
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∈C
and
1 ∈ 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:
a
1
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:
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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.
ab
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:
o
na
such that a ∈ Z and b ∈ Z with b 6= 0
Q=
b
133
4.3. CONSTRUCTIBLE NUMBERS
Fancy folks will replace the words such that with a colon “:” to get:
na
o
Q=
: a ∈ Z and b ∈ Z with b 6= 0
b
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.
(b) For all x ∈ F where x 6= 0, there is a multiplicative inverse x−1 such
that
x · x−1 = 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.
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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 numbers C?
?
From all the constructions above we see that the set of constructible numbers
C is a field. However, which field is it? In fact, the set of constructible numbers
is 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 squareroot of two is constructible. But the square-root of two is not rational! That is,
there is no fraction
a √
such that a, b ∈ Z.
= 2
b
How do we know the above fact? Well there are several ways to do it. Why not
use the Rational Roots Test?
135
4.3. CONSTRUCTIBLE NUMBERS
Theorem 21 (Rational Roots Test) If we have a polynomial with integer
coefficients,
an xn + an−1 xn−1 + · · · + a1 x + 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 C. But is this all of C? Hardly!
We could keep on going, adding more and more square-roots ’til the 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.
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CHAPTER 4. COMPASS AND STRAIGHTEDGE CONSTRUCTIONS
Question
Which of the following numbers are constructible?
3.1415926,
√
5,
16
√
3
27,
√
6
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?
137
4.3. CONSTRUCTIBLE NUMBERS
One way is to use golden triangles:
1
1
φ
φ
1
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 n > 3 and
n = 2i · p1 · p2 · · · pj
where each subscripted p is a distinct prime number of the form
2(2
k
)
+1
where i, j, and k are nonnegative integers.
Around thirty years later, Wantzel proved that these were the only regular
polygons that could be constructed.
Question
Find i, j, and k for a regular 3-gon, 4-gon, 5-gon, and 6-gon.
?
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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 C of constructible numbers is.
(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
Associative
Closed in Z:
Closed in Q:
Closed in C:
+
yes
yes
yes
yes
yes
139
−
·
÷
∧
4.3. CONSTRUCTIBLE NUMBERS
(14) Which of the following are constructible numbers? Explain your answers.
(a) 3.141
√
3
5
p
√
3 + 17
(c)
√
(d) 8 5
√
(e) 10 37
√
(f) 16 37
√
(g) 3 28
p
√
√
13 + 3 2 + 11
(h)
√
(i) 3 + 5 4
p
√
√
3 + 19 + 10
(j)
(b)
(15) Suppose 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
(18)
(19)
(20)
(21)
(22)
(23)
are rational. Find them and explain your work.
√
Is 7 a rational number? Is it a constructible
answers.
√
Is 8 a rational number? Is it a constructible
answers.
√
Is 9 a rational number? Is it a constructible
answers.
√
Is 3 7 a rational number? Is it a constructible
answers.
√
Is 3 8 a rational number? Is it a constructible
answers.
√
Is 3 9 a rational number? Is it a constructible
answers.
140
number? Explain your
number? Explain your
number? Explain your
number? Explain your
number? Explain your
number? Explain your
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)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
regular
regular
regular
regular
regular
regular
regular
regular
regular
regular
regular
regular
regular
regular
regular
3-gon.
5-gon.
7-gon.
9-gon.
11-gon.
12-gon.
13-gon.
15-gon.
17-gon.
34-gon.
2-gon.
4-gon.
10-gon.
20-gon.
70-gon.
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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
know that this number is not constructible.
√
3
2 and we already
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.
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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
an isosceles 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:
143
4.4. IMPOSSIBILITIES
Construction (Cheating)
a marked straightedge.
We will show how to trisect an arbitrary angle using
(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.”
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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.
145
Chapter 5
Isometries
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
Matrices as Functions
We’re going to be discussing some basic functions in geometry. Specifically, 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 matrix 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
y 
(x, y)
1
Throughout this chapter, we will abuse notation slightly, freely interchanging
several notations for a point:
 
x
p ! (x, y) ! y 
1
146
CHAPTER 5. ISOMETRIES
With this in mind, our work
this:

a
M = d
0
will be done via matrices and points that look like

b c
e f
0 1
and
 
x
p = y 
1
Now recall the nitty gritty details of matrix multiplication:

a
Mp = d
0

 
b c
x
e f  y 
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
isometries.
Definition An isometry is a function M that maps points in the plane to
other points in the plane such that
d(p, q) = d(Mp, Mq),
where d is the distance function.
Question
How do you compute the distance between two points again?
?
We’re going to see that several ideas in geometry, specifically translations, reflections, and rotations which all seem very different, are actually all isometries.
Hence, we will be thinking of these concepts as matrices.
5.1.1
Translations
Of all the isometries, translations are probably the easiest. With a translation,
all we do is move our object in a straight line. Let’s see what happens to Louie
147
5.1. MATRICES AS FUNCTIONS
Llama:
Pretty simple eh? We can give a more “mathematical” definition of a translation using our newly-found knowledge of matrices! Check it:
Definition A translation, denoted by T(u,v) , is a function that moves every
point a given distance u in the x-direction and a given distance v in the ydirection. We will use the following type of matrix to represent translations:
T(u,v)


1 0 u
= 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.
148
CHAPTER 5. ISOMETRIES
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
Can you demonstrate with algebra why translations are isometries?
?
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 function that moves
every point p to a point Fℓ p such that:
(1) If p is on ℓ, then Fℓ p = p.
(2) If p is not on ℓ, then ℓ is the perpendicular bisector of the segment connecting p and Fℓ p.
You might be saying, “Huh?” It’s not as hard as it looks. Check out this
149
5.1. MATRICES AS FUNCTIONS
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:
Fx=0

−1 0
= 0 1
0 0

0
0
1
The next reflection in our collection is a vertical reflection over the x-axis.
Using our matrix notation, we write:
Fy=0

1
= 0
0

0 0
−1 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:
Fy=x
Example

0 1
= 1 0
0 0

0
0
1
Consider the point p = (3, −1). What would p look like if we
150
CHAPTER 5. ISOMETRIES
reflected it across the line y = x?
Here is how you do it:

 
0 1 0
3
Fy=x p = 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=x p 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 Can you demonstrate with algebra why our collection of reflections
above are isometries?
?
Question How do we deal with reflections that are not about the lines y = 0,
x = 0, or y = x? How would you reflect points about the line y = 1?
?
151
5.1. MATRICES AS FUNCTIONS
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
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
function that moves every point p to a point Rn p such that:
(1) The points p and Rn p are equidistant from the origin.
(2) An angle of n degrees is formed by p, the origin, and Rn p.
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 measurements, namely 90◦ , 60◦ , and 45◦ . We’ll focus on these degrees as well.
R90

0
= 1
0
Example

−1 0
0 0
0 1
R60 =

1
 √23
 2
0
√
− 3
2
1
2
0

0

0
1
R45 =
 √1
2
 √1
2
0
−1
√
2
√1
2
0

0
0
1
Suppose that you have a point p = (4, −2) and you want to rotate
1 Face it, I think we all dreamed of doing that when we were little—or in my case, last
week.
152
CHAPTER 5. ISOMETRIES
it 60◦ about the origin.
Here is how you do it:
R60 p =

√
− 3
2
1
2
1
 √23
 2
0
√

0
 
0
4
 
−2

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 +
Question
√
√
3, 2 3 − 1).
Do the numbers in the matrices above look familiar? If so, why?
?
Question
How do you rotate a point 180 degrees?
?
Question Can you demonstrate with algebra why our rotations above are
isometries?
?
153
5.1. MATRICES AS FUNCTIONS
Problems for Section 5.1
(1) How do you compute the distance between two points p and q in the
plane?
(2) What is an isometry?
(3) What is a translation?
(4) What is a rotation?
(5) What is a reflection?
(6) In what direction does a positive rotation occur?
(7) Devise a way to explain translations using compass and straightedge constructions.
(8) Devise a way to explain translations using origami constructions.
(9) Devise a way to explain reflections using origami constructions.
(10) Devise a way to explain rotations using compass and straightedge constructions.
(11) Consider the following matrix:

0 0
M = 2 1
0 0

8
0
1
Is M an isometry? Explain your reasoning.
(12) Consider the following matrix:

2 3
M = 2 3
0 0

2
2
1
Is M an isometry? Explain your reasoning.
(13) Consider the following matrix:

1
I = 0
0

0 0
1 0
0 1
Is I an isometry? Explain your reasoning.
(14) Consider the following matrix:


0 2 0
M = −3 0 0
0 0 1
Is M an isometry? Explain your reasoning.
154
CHAPTER 5. ISOMETRIES
(15) Consider the following matrix:

1 3
M = 3 9
0 0

2
6
1
Is M an isometry? Explain your reasoning.
(16) Consider the following matrix:


0 2 0
M = 1 −1 0
0 0 1
Is M an isometry? Explain your reasoning.
(17) Use a matrix to translate the point (−1, 6) three units right and two units
up. Sketch this situation and explain your reasoning.
(18) The matrix T(−2,6) was used to translate the point p to (−1, −3). What
is p? Sketch this situation and explain your reasoning.
(19) Given the point p = (0, −7), what would be the result of T(4,2) T(6,−5) p?
Sketch this situation and explain your reasoning.
(20) Use a matrix to reflect the point (5, 2) across the x-axis. Sketch this
situation and explain your reasoning.
(21) Use a matrix to reflect the point (−3, 4) across the y-axis. Sketch this
situation and explain your reasoning.
(22) Use a matrix to reflect the point (−1, 1) across the line y = x. Sketch this
situation and explain your reasoning.
(23) Use a matrix to reflect the point (1, 1) across the line y = x. Sketch this
situation and explain your reasoning.
(24) The matrix Fy=0 was used to reflect the point p to (4, 3). What is p?
Explain your reasoning.
(25) The matrix Fy=0 was used to reflect the point p to (0, −8). What is p?
Explain your reasoning.
(26) The matrix Fx=0 was used to reflect the point p to (−5, −1). What is p?
Explain your reasoning.
(27) The matrix Fy=x was used to reflect the point p to (9, −2). What is p?
Explain your reasoning.
(28) The matrix Fy=x was used to reflect the point p to (−3, −3). What is p?
Explain your reasoning.
155
5.1. MATRICES AS FUNCTIONS
(29) Considering the point (3, 2), use a matrix to rotate this point 60◦ about
the origin. Sketch this situation and explain your reasoning.
√
√
(30) Considering the point ( 2, − 2), use a matrix to rotate this point 45◦
about the origin. Sketch this situation and explain your reasoning.
(31) Considering the point (−7, 6), use a matrix to rotate this point 90◦ about
the origin. Sketch this situation and explain your reasoning.
(32) Considering the point (−1, 3), use a matrix to rotate this point 0◦ about
the origin. Sketch this situation and explain your reasoning.
(33) Considering the point (0, 0), use a matrix to rotate this point 120◦ about
the origin. Sketch this situation and explain your reasoning.
(34) Considering the point (1, 1), use a matrix to rotate this point −90◦ about
the origin. Sketch this situation and explain your reasoning.
(35) The matrix R90 was used to rotate the point p to (2, −5). What is p?
Explain your reasoning.
(36) The matrix R60 was used to rotate the point p to (0, 2). What is p?
Explain your reasoning.
(37) The matrix R45 was used to rotate the point p to (− 12 , 25 ). What is p?
Explain your reasoning.
(38) The matrix R−90 was used to rotate the point p to (4, 3). What is p?
Explain your reasoning.
(39) If someone wanted to plot y = 3x, they might start by filling in the
following table:
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.
(40) 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.
(41) 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.
156
CHAPTER 5. ISOMETRIES
(42) 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.
(43) 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.
(44) 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.
√ √
(45) 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.
(46) Some isometry M maps
(0, 0) 7→ (3, 2),
(1, 0) 7→ (4, 2),
(0, 1) 7→ (3, 3).
Find M and explain your reasoning.
(47) Some isometry M maps
(0, 0) 7→ (−1, 1),
(1, 0) 7→ (0, 1),
(0, 1) 7→ (−1, 2).
Find M and explain your reasoning.
(48) Some isometry M maps
(0, 0) 7→ (0, 0),
(1, 0) 7→ (1, 0),
(0, 1) 7→ (0, −1).
Find M and explain your reasoning.
(49) Some isometry M maps
(0, 0) 7→ (0, 0),
(1, 0) 7→ (0, 1),
(0, 1) 7→ (1, 0).
Find M and explain your reasoning.
(50) Some isometry M maps
(0, 0) 7→ (0, 0),
(1, 0) 7→ (0, 1),
(0, 1) 7→ (−1, 0).
Find M and explain your reasoning.
157
5.1. MATRICES AS FUNCTIONS
(51) Some isometry M maps
(0, 0) 7→ (0, 0),
√ !
1
3
,
,
(1, 0) 7→
2 2
!
√
− 3 1
(0, 1) 7→
.
,
2 2
Find M and explain your reasoning.
(52) Consider the line 3x + 4y = 2.
(a) If p is a point on the line, and the x-coordinate of p is 7, what is the
y coordinate?
(b) If p is a point on the line, and the x-coordinate of p is −2, what is
the y coordinate?
(c) If p is a point on the line, what are the coordinates of p in terms of
x and y?
(d) Now consider the line ax + by = c. If p is a point on the line, what
are the coordinates of p in terms of x and y?
(53) Consider T(u,v) and a line ℓ : ax + by = c. Use algebra to explain why
T(u,v) ℓ is a new line and not some other curve. Hint: See Problem (52).
(54) Consider T(u,v) and a line ℓ : ax + by = c. Use algebra to explain why
T(u,v) ℓ is a new line that is parallel to the original line. Hint: See Problem (52).
158
CHAPTER 5. ISOMETRIES
5.2
The Algebra of Matrices
5.2.1
Matrix Multiplication
We know how to multiply a matrix and a column. Multiplying two matrices is
a similar procedure:

a
d
g

c
j k
f  m n
i
p q
b
e
h
 
l
aj + bm + cp
o = dj + em + f p
r
gj + hm + ip

al + bo + cr
dl + eo + f r
gl + ho + ir
ak + bn + cq
dk + en + f q
gk + hn + iq
Variables are all good and well, but let’s do this with actual numbers. Consider the following two matrices:


1 2 3
M =  4 5 6
7 8 9
and

1 0
I = 0 1
0 0

0
0
1
Let’s multiply them together and see what we get:


1 2 3
1 0
MI = 4 5 6 0 1
7 8 9
0 0

1·1+2·0+3·0
= 4 · 1 + 5 · 0 + 6 · 0
7·1+8·0+9·0


1 2 3
=  4 5 6
7 8 9

0
0
1
1·0+2·1+3·0
4·0+5·1+6·0
7·0+8·1+9·0

1·0+2·0+3·1
4 · 0 + 5 · 0 + 6 · 1
7·0+8·0+9·1
=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
out:
Matrix multiplication is not generally commutative. Check it

1 0
F = 0 −1
0 0

0
0
1
and
159

0 −1
R = 1 0
0 0

0
0
1
5.2. THE ALGEBRA OF MATRICES
When we multiply these matrices, we get:


0 0
0
−1 0 1
0 1
0
1
FR = 0
0


−1 0
0 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
RF = 1 0
0 0
Question

0
0
1
Is it always the case that (LM)N = L(MN)?
?
5.2.2
Compositions of Matrices
It is often the case that we wish to apply several isometries successively to a
point. Consider the following:

a
M = d
0
b
e
0

c
f
1

g
N = j
0
h
k
0

i
l
1
and
 
x
p = y 
1
Now let’s compute

 
  
a b c
g h i
x
M(Np) = d e f  j k l  y 
0 0 1
0 0 1
1



a b c
gx + hy + i
= d e f   jx + ky + l 
0 0 1
1


agx + ahy + ai + bjx + bky + bl + c
= dgx + dhy + di + ejx + eky + el + f 
1
Now you compute (MN)p and compare what you get to what we got above.
160
CHAPTER 5. ISOMETRIES
Compositions of Translations
A composition of translations occurs when two or more
are applied to the same point. Check it out:


1 0 5
1 0
T(5,−4) T(−3,2) = 0 1 −4 0 1
0 0 1
0 0


1 0 2
= 0 1 −2
0 0 1
successive translations

−3
2
1
= T(5+(−3),(−4)+2)
= T(2,−2)
Theorem 24 The composition of two translations T(u,v) and T(s,t) is equivalent to the translation T(u+s,v+t) .
Question
How do you prove the theorem above?
?
Question Can you give a single translation that is equivalent to the following
composition?
T(−7,5) T(0,−6) T(2,8) T(5,−4)
?
Question
tive?
Are compositions of translations commutative? Are they associa-
?
Compositions of Reflections
A composition of reflections occurs when two or more successive reflections are
applied to the same point. Check it out:
Fy=0 Fy=x
Question

1 0
= 0 −1
0 0

0 1
= −1 0
0 0

0 0
0 1
1 0

0
0
1

1 0
0 0
0 1
Which line does the composition Fy=0 Fy=x reflect points over?
161
5.2. THE ALGEBRA OF MATRICES
?
Question
tive?
Are compositions of reflections commutative? Are they associa-
?
Compositions of Rotations
A composition of rotations occurs when two or more successive rotations are
applied to the same point. Check it out:
R60 R60 =

1
 √23
 2

0
−1
 √2
= 3
2
0
√
− 3
2
1
2
0
√
− 3
2
−1
2
0

1
0
  √23
0  2
1
0

0

0
1
√
− 3
2
1
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:
R120 =

−1
 √23
 2
0
√
− 3
2
−1
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?
?
162
CHAPTER 5. ISOMETRIES
5.2.3
Mixing and Matching
Life gets interesting when we start composing translations, reflections, and rotations together. First we’ll take a look at a reflection mixed with a rotation:
Fy=0 R60
Question
 1
1 0 0
 √2
= 0 −1 0  3
2
0 0 1
0


√
− 3
1
0
2

 2√
= − 3 − 1 0
2
2
0
0
1
√
− 3
2
1
2

0

0

0
1
Does this result look familiar?
?
Now how about a rotation with a translation:
R90 T(3,−4)

0 −1
= 1 0
0 0

0 −1
= 1 0
0 0

0
1 0
0 0 1
1
0 0

4
3
1

3
−4
1
Question What do you think it would look like if instead we had T(3,−4) R90 ?
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 = R180 Fy=x
= R90 R90 Fy=x


0 −1 0
0 −1
= 1 0 0 1 0
0 0 1
0 0


0 −1 0
= −1 0 0
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.
163
5.2. THE ALGEBRA OF MATRICES
Question
tivity?
When we mix isometries, will we retain commutativity? Associa-
?
164
CHAPTER 5. ISOMETRIES
Problems for Section 5.2
(1) Give a single translation that is equivalent to T(−3,2) T(5,−1) . Explain your
reasoning.
(2) Consider the two translations T(−4,8) and T(4,−8) . Do these commute?
Explain your reasoning.
(3) Given the point p = (7, 4), use matrices to compute T( 21 ,6) T(2,−1) p. Sketch
this situation and explain your reasoning.
√
(4) Given the point p = (− 3, − 41 ), use matrices to compute T(2, 12 ) T(√3,− 1 ) p.
4
Sketch this situation and explain your reasoning.
(5) Give a matrix representing Fy=−x . Explain your reasoning.
(6) Give a matrix representing R−45 . Explain your reasoning.
(7) Give a matrix representing R−60 . Explain your reasoning.
(8) Give a single reflection that is equivalent to Fx=0 Fy=0 . Explain your reasoning.
(9) Given the point p = (−4, 2), use matrices to compute Fy=0 Fy=x p. Sketch
this situation and explain your reasoning.
(10) Given the point p = (5, 0), use matrices to compute Fy=x Fy=−x p. Sketch
this situation and explain your reasoning.
(11) Give a single rotation that is equivalent to R45 R60 . Explain your reasoning.
(12) Given the point p = (1, 3), use matrices to compute R45 R90 p. Sketch this
situation and explain your reasoning.
(13) Given the point p = (−7, 2), use matrices to compute R45 R−45 p. Sketch
this situation and explain your reasoning.
(14) Given the point p = (−2, 5), use matrices to compute R90 R−90 R360 p.
Sketch this situation and explain your reasoning.
(15) Given the point p = (5, 4), use matrices to compute Fy=0 T(2,−4) p. Sketch
this situation and explain your reasoning.
(16) Given the point p = (−1, 6), use matrices to compute R45 T(0,0) p. Sketch
this situation and explain your reasoning.
(17) Given the point p = (11, 13), use matrices to compute T(−6,−3) R135 p.
Sketch this situation and explain your reasoning.
(18) Given the point p = (−7, −5), use matrices to compute R540 Fx=0 p. Sketch
this situation and explain your reasoning.
165
5.2. THE ALGEBRA OF MATRICES
(19) Given the point p = (8, 1), use matrices to compute R90 Fy=x T(−2,3) p.
Sketch this situation and explain your reasoning.
(20) Give a single isometry that is equivalent to Fy=−x T(−3,10) . Also give a
single isometry that is equivalent to T(−3,10) Fy=−x . Explain your reasoning.
(21) Give a single isometry that is equivalent to R30 Fy=0 . Also give a single
isometry that is equivalent to Fy=0 R30 . Explain your reasoning.
166
CHAPTER 5. ISOMETRIES
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.
(2) For all M ∈ G there is an inverse M−1 such that
MM−1 = I.
5.3.1
Groups of Reflections
Let’s see a basic group:
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
elements: I and Fx=0 .
◦
I
I
I
Fx=0
Fx=0
167
Fx=0
Fx=0
I
5.3. THE THEORY OF GROUPS
Notice what happens when we apply Fx=0 twice, we’re right back where we
started. Hence, Fx=0 Fx=0 = I. Since matrix multiplication is associative, we see
that
{I, Fx=0 }
forms a group. Specifically this is a group of reflections of the triangle.
Question Above, we reflected across the line x = 0. 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 they be?
?
5.3.2
Groups of Rotations
How many degrees does it take to rotate an 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?


√
− 3
−1
0
2

 √2
−1
R120 =  3
0
2
2
0
0
1
Question
Consider the following equilateral triangle.
How many different ways can we rotate this triangle back on to itself?
Because R120 R120 does not give us the identity, we need to include it in our
group table. We will write this as R2120 . However, if we apply the R120 rotation
one more time, we do get back to the identity. This shows that R120 and R2120
are inverses. Everything now has an inverse and we can complete our rotation
group table:
◦
I
R120 R2120
I
I
R120 R2120
R120 R120 R2120
I
R2120 R2120
I
R120
168
CHAPTER 5. ISOMETRIES
Question What kind of rotation matrices would we use 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=0 R120 :
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 = √13 x.
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:
(D3 , ◦)
I
R120
R2120
Fx=0
Fx=0 R120
Fx=0 R2120
I
I
R120
R2120
Fx=0
Fx=0 R120
Fx=0 R2120
R120
R120
R2120
I
Fx=0 R120
Fx=0 R2120
Fx=0
R2120
R2120
I
R120
Fx=0 R2120
Fx=0
Fx=0 R120
Fx=0
Fx=0
Fx=0 R2120
Fx=0 R120
I
2
R120
R120
Fx=0 R120
Fx=0 R120
Fx=0
Fx=0 R2120
R120
I
R2120
Fx=0 R2120
Fx=0 R2120
Fx=0 R120
Fx=0
R2120
R120
I
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.
169
5.3. THE THEORY OF GROUPS
Question Can you express the symmetries of the square in terms of reflections and rotations? What does the group table look like for the symmetry
group of the square?
?
170
CHAPTER 5. ISOMETRIES
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 consecutive rotations are needed to return the vertexes of a
hexagon to their original position? Provide a drawing to justify your
answer, labeling the vertexes.
(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.
171
5.3. THE THEORY OF GROUPS
(18) Find two symmetries of the equilateral triangle, neither of which is the
identity, such that their composition is R2120 . Explain and illustrate your
answer.
(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.
172
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?
?
173
6.1. BASIC DEFINITIONS
From two separate convex sets, we can build new ones:
Theorem 26
Proof
The intersection of two convex sets is a convex set.
Let X1 and X2 be two convex sets. Let
Y = X1 ∩ X2 .
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 = X1 ∩ X2 is convex.
Question
If X1 , X2 , and X3 are convex, is X1 ∩ X2 ∩ X3 convex?
?
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?
?
174
CHAPTER 6. CONVEX SETS
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.
175
6.1. BASIC DEFINITIONS
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
⇒
176
y 6 (−4/3)x + 35
CHAPTER 6. CONVEX SETS
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:
30
20
10
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.
177
6.1. BASIC DEFINITIONS
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.
178
CHAPTER 6. CONVEX SETS
(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.
179
6.2. CONVEX SETS IN THREE DIMENSIONS
6.2
6.2.1
Convex Sets in Three Dimensions
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:
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CHAPTER 6. CONVEX SETS
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?
?
181
6.2. CONVEX SETS IN THREE DIMENSIONS
Question
above?
Are there other regular convex polygons other than the ones shown
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
tetrahedron
octahedron
cube
icosahedron
dodecahedron
Faces
4
8
6
20
12
182
Edges
6
12
12
30
30
Vertexes
4
6
8
12
20
CHAPTER 6. CONVEX SETS
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?
?
183
6.2. CONVEX SETS IN THREE DIMENSIONS
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?
184
CHAPTER 6. CONVEX SETS
6.3
6.3.1
Ideas Related to Convexity
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
rapidly?
What happens when you heat up metal and then cool it down
185
6.3. IDEAS RELATED TO CONVEXITY
?
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:
186
CHAPTER 6. CONVEX SETS
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.
187
6.3. IDEAS RELATED TO CONVEXITY
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.
188
CHAPTER 6. CONVEX SETS
(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.
189
6.4. ADVANCED THEOREMS
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 X1 and X2 be two convex sets. If X2 ⊆ X1 , then the
perimeter of X1 is greater than or equal to the perimeter of X2 .
Question
What does this this theorem look like?
?
Question
true?
If you leave off or change the assumptions of the theorem, is it still
?
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
true?
If you leave off or change the assumptions of the theorem, is it still
?
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
true?
If you leave off or change the assumptions of the theorem, is it still
?
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?
190
CHAPTER 6. CONVEX SETS
?
Question
carefully?
Will any point for A work? Or does A need to be chosen somewhat
?
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
true?
If you leave off or change the assumptions of the theorem, is it still
?
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
true?
If you leave off or change the assumptions of the theorem, is it still
?
191
6.4. ADVANCED THEOREMS
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.
192
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. Dörrie. 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:
Metropolitan Books, 1998.
A Mathematical Adventure.
[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] M. Jeger. Transformation Geometry. John Wiley & Sons, 1966.
[15] E.F. Krause. Taxicab Geometry: An Adventure in Non-Euclidean Geometry. Dover, 1986.
193
REFERENCES AND FURTHER READING
[16] M. Livio. The Golden Ratio: The Story of Phi, the World’s Most Astonishing Number. Broadway Books, 2002.
[17] C.H. MacGillavry. Fantasy & Symmetry: The Periodic Drawings of M.C.
Escher. Harry N. Abrams Inc, 1965.
[18] R.B. Nelsen. Proofs Without Words. Mathematical Association of America,
1993.
[19]
. Proofs Without Words II. Mathematical Association of America,
2000.
[20] G. Polya. Mathematical Discovery: On Understanding, Learning, and
Teaching Problem Solving. Wiley, 1981.
[21] C. Sagan. Cosmos. Random House, 2002.
[22] T.S. Row. Geometric Exercises in Paper Folding. The Open Court Publishing Company, 1901.
[23] J.R. Smart. Modern Geometries. Brooks/Cole, 1998.
[24] E.W. Weisstein. MathWorld—A Wolfram Web Resource.
http://mathworld.wolfram.com/
[25] D. Wells. The Penguin Dictionary of Curious and Interesting Geometry.
Penguin Books, 1991.
194
Index
æther, 181
algebraic geometry, 14
altitude, 85, 88
analytic geometry, 17
and, 45
and probability, 107
antipodal point, 6
Archimedes, 143
area
Heron’s Formula, 93
Pick’s Theorem, 94
arithmetic mean, 61
Arithmetic-Geometric Mean Inequality,
61
associative, 134
axiom, 4
Battle Royale, 27
beauty, see truth
beer, 47, 179
bees, 55
Bertrand’s paradox, 104
boat
lost at night, 129
brain juices, 129
bucket
Alma Mater, 35
C, 132
calisson, 68
Cartesian plane, 14
Cavalieri’s Principle, 69, 70
center of mass, 86
centroid, 86, 88, 122
circle
circumcircle, 84
City Geometry, 29
incircle, 85
Spherical Geometry, 6
circumcenter, 83, 88, 122
circumcircle, 84, 122
circumscribe, 84, 93
City Geometry, 23, 24
circle, 29
midset, 29
parabola, 32
triangle, 28
closed, 134
collapsing compass, 117
commutative, 134
compass
collapsing, 117
compass and straightedge
addition, 132
bisecting a segment, 117
bisecting an angle, 118
copying an angle, 119
division, 133
equilateral triangle, 116
golden ratio, 137
impossible problems, 142
multiplication, 133
parallel through a point, 120
pentagon, 137
perpendicular through a point, 118
SAA triangle, 125
SAS triangle, 124
SSS triangle, 124
subtraction, 132
tangent to a circle, 120
transferring a segment, 117
complement, 40
conic-section, 13
constructible numbers, 132
195
INDEX
constructions, see compass and straightedge or paper-folding
continued fraction, 97
convex, 173
convex hull, 185
convex polyhedra
regular, 181
Crane Alley, 26
diagonal reflection, 150
dissection proof, 57
doubling the cube, 60, 142
dual
polyhedron, 184
∈, 38, 132
e, 98
The Elements, 1
empty set, 40
equilateral triangle, 89
Eratosthenes, 2
Escher, M.C., 51
Euclid, 1, 115
Euclidean Geometry, 4
Euler line, 88, 123
field, 134
fractional part, 98
free point, 116
Galileo Galilei, 67
Gauss, Carl Friedrich, 63, 138
geometric mean, 61
geometry
algebraic, 14
analytic, 17
City, 24
Spherical, 5
synthetic, 13
geometry,City, 23
goat, 104
golden
ratio, 100, 137
compass and straightedge, 137
rectangle, 101
spirals, 101
triangle, 101, 138
gorilla suit, 24
great circle, 5
group, 167
Helly’s Theorem, 190
Heron’s Formula, 93
for quadrilaterals, 93
for triangles, 93
horizontal reflection, 150
identity matrix, 159
if-and-only-if, 48
if-then, 46, 47
iff, 48
incenter, 84, 122
incircle, 85, 122
integers, 132
intersection, 39
isometry, 147
Kepler, Johannes, 51
lemma, 93
lemon, see lemma
Let’s Make a Deal, 104
logical symbols
∧, 45, 107
⇔, 48
⇒, 47
¬, 45
∨, 45, 107
Louie Llama, 148, 150, 152
matrix, 146
multiplication, 147
median, 86
midset
City Geometry, 29
Miquel point, 89, 123
Miquel’s Theorem, 89
Monty Hall problem, 104
Morley’s Theorem, 89
NASA, 185
nexus of the universe, 23
Nine-Point Circle, 88, 123
196
INDEX
nontransitive dice, 106
not, 46
or, 45
and probability, 107
orthocenter, 85, 88, 122
paper-folding
Morley’s Theorem, 89
the centroid, 86
the circumcenter, 84
the incenter, 84
the orthocenter, 85
parabola
algebraic definition, 15
City Geometry, 32
conic-section definition, 13
intrinsic definition, 12
paradox,
√ 32
2 = 2, 33
1 = 0.999 . . . , 94
all triangles are isosceles, 90
Bertrand’s paradox, 104
involving dice, 106
the Monty Hall problem, 104
triangle dissection, 59
parallel postulate, 5
pentagon, 137
φ, 100
π, 98, 103
Pick’s Theorem, 94
Plato, 115
Platonic Solids, 180, 182
polyhedra
convex regular, 181
prime numbers, 138
probability, 102
Pythagorean Theorem, 56, 124
The Pythagoreans, 180
Q, 133
Q(α), 136
quadrilateral
tessellation of, 51
Radon’s Theorem, 190
rational numbers, 134
rational roots test, 136
reflection, 149
diagonal, 150
horizontal, 150
vertical, 150
regular
convex polyhedra, 180
tessellation, 51
Reuleaux
polygon, 186
triangle, 186
Rock-Paper-Scissors, 106
rotation, 152
set, 38
set theory symbols
−, 40
∅, 40
∈, 38, 132
∩, 39
⊆, 39
∪, 39
sets of constant width, 185
drawing, 186, 187
slopef (x) (x, s), 17
Socrates, 115
Spherical Geometry, 5
great circle, 5
Spherical Geometry
circle, 5
line, 5
triangle, 7
spirals,
101
√
2, 33, 97, 99
squaring the circle, 142
subset, 39
supporting
half-plane, 175
half-space, 180
line, 174
plane, 180
symmetry group
group, 169
synthetic geometry, 13
197
INDEX
tables
truth, 45
tangent line, 13, 15, 16
Taxicab distance, 23
tessellation, 51
any quadrilateral, 51
regular, 51
triangles, 51
three doors, 104
translation, 148
triangle
altitude, 85
centroid, 86
circumcenter, 83
circumcircle, 84
City Geometry, 28
incenter, 84
incircle, 85
orthocenter, 85
Spherical Geometry, 7
triangular dipyramid, 181
trisecting the angle, 89, 143
by cheating, 144
truth, see beauty
truth-table, 45
∪, 39
union, 39
vertical reflection, 150
Wantzel, Pierre, 138
Wason selection task, 46
whole-number part, 98
width of a set, 186
Z, 132
198