1. science
2. mathematics
3. geometry

Some Geometry You Never Met
There is no royal road to geometry.
—Euclid, to Ptolemy I, ruler of Egypt
Out of nothing I have created a strange new universe.
—Janos Bolyai, referring to the creation
of a non-euclidean geometry
A
C
B
Figure 1. Outline of a Farmer’s Property
I begin this chapter with a challenge exercise. A farmer, who owns a triangular property
that has the shape of Figure 1, has the plot surveyed. The surveyor reports the angles
at the lot corners. Their measures are ∠A = 78○ 20′ 23.52”, ∠B = 42○ 51′ 3.13” and ∠C =
58○ 48′ 33.37”. From these measurements the farmer is able to determine the area of the
triangle. What is that area and how does he calculate it?
1
Triangle area formulas
Students who studied geometry in high school may recall several ways to calculate the area
of a triangle. If it is a right triangle or if you know the triangle’s base and height, you can
use the simple area formula
1
A = bh
(1.1)
2
If you know two sides of a triangle and the angle between them, you can use the formula
1
A = ab sin C
2
(1.2)
A less well known formula for area is the one for a triangle in which you know two angles,
A and C and the side between them, b. It is
A=
1 2 tan A tan C
b
2 tan A + tan C
1
(1.3)
And finally you can use Heron’s Formula if you know all three sides. First calculate the
semiperimeter s = (a + b + c)/2 and then use that value in the formula
√
A = s(s − a)(s − b)(s − c)
(1.4)
Of course, none of those methods addresses the farmer’s problem, because they all require
the length of at least one side of the triangle. In fact, you were probably cautioned when
you studied geometry in high school that the so-called AAA case, when you know only
the triangle’s angles, only allows for similarity — triangles with the same shape but not
necessarily the same size
Figure 1.5 The AAA Case: Similar triangles with different areas
so their area cannot be calculated.
There is, however, something different about the farmer’s problem and we will address
that difference later in this chapter.
2
Euclid’s Elements
The Greek mathematician, Euclid, lived about 2300 years ago in Alexandria, where he
compiled and organized the most successful and influential mathematics text of all times,
probably providing many of the proofs it contained himself. The 13 books1 of his Elements
served as the basis for understanding mathematics until the early 20th century.
Two Elements propositions remain famous, but for quite different reasons. The first is
Elements I, 5 (that is, Book I, Proposition 5) which we usually state today as “Base angles
of an isosceles triangle are equal.”2 Because Euclid included in his proof the fact that the
supplements of those angles would also be equal, his diagram took on several additional
line segments, making it appear as in Figure 2.
1
The books are more like the chapters of modern texts, but one modern Elements reissue is still in three
volumes.
2
In this book I am not adopting today’s formal usage that would require me to restate that theorem
either replacing “equal” with “congruent” or adding “The measure of” at the beginning. I can imagine no
reader who does not understand what I am talking about when I say that angles of a triangle are equal.
2
A
B
C
!
!
Figure 2. ! The Diagram
for Elements I, 5
A
That structure gave the diagram the appearance of a bridge with trusses and the proposition
became known to teachers and students alike as Pons asinorum, Latin for Bridge of Asses
(a.k.a. fools). In this sense, Elements
I, 5 was said C
to separate the students who were able
B
to continue from those who would not make it past this point in the course.
!
!
!
!
!
Figure 3. The Diagram for Elements I, 47
The other proposition is Elements I, 47, the Pythagorean Theorem, easily the most
famous theorem of all time. Euclid knew it in terms of areas: “The sum of the areas of the
squares on the legs of a right triangle is equal to the area of the square on the triangle’s
hypotenuse.” Today we usually think of this theorem in algebraic terms: for legs a and b
and hypotenuse c, a2 + b2 = c2 .
Although algebra and even our Hindu-Arabic numeration system would not be invented
for centuries after Euclid’s time, many other algebraic ideas are presented in his books in
geometric form. For example, using arguments involving rectangular areas, Elements II,
1 proves a form of the distributive law a(b + c + d) = ab + ac + ad and the theorems that
follow prove such complex identities as (2a + b)2 + b2 = 2(a2 + (a + b)2 ). Later books prove
theorems normally found in other areas of math. For example, another famous theorem
you will meet again, Elements IX, 20 proves that the number of primes is infinite.
Today, although Euclid is scarcely mentioned in school geometry texts, the secondary
school course follows much of the development of the early books in the series.
3
3
Problems with Euclid
Despite the power and influence of the Elements, it is easy today to identify problems with
some of the foundations from which Euclid developed his proofs. The whole idea of his
development was to list a series of definitions and axioms that are at once simple and easy
to understand and accept. He would then use these as the basis for proving what was to
follow.
Some of the definitions are strange and of little value: “A point is that which has no
part” and “A line is breadthless length.” Others are oddly stated: “A circle is a plane
figure contained by one line such that all the straight lines falling upon it from one point
among those lying within the figure are equal to one another,” but that one is at least
partly straightened out by the following definition: “And the point is called the center of
the circle.”
A more significant problem arises in Euclid’s approach to parallel lines. He defines them
in what seems a reasonable way, as “straight lines which, being in the same plane and being
produced indefinitely in both directions, do not meet one another in either direction.” But
then he lists his five basic postulates:3
1. To draw a straight line from any point to any point.
2. To produce a finite straight line continuously in a straight line.
3. To describe a circle with any center and distance.
4. That all right angles are equal to one another.
5. That, if a straight line falling on two straight lines make the interior angles on the
same side less than two right angles, the two straight lines, if produced indefinitely,
meet on that side on which are the angles less than the two right angles.
Now think about those postulates. The first three merely describe the basic constructions
done with compasses and straightedge. The fourth is evident. But the fifth is quite unlike
all the others. At 45 words, it is longer than all the others combined. And it is quite
complicated. Figure 4 shows what it is saying.
3
Postulates and axioms, like definitions, are accepted statements on which subsequent proofs are based.
They are supposed to be self-evident as the Postulate 5 clearly is not.
4
!
Figure 4. A diagram illustrating Elements Postulate 5
According to the postulate, when those two angles marked in red sum to less than 180○ , the
lines will meet on that side of the transverse line. The smaller figure illustrates a situation
when the two angles do add to 180○ .
Because of the strange characteristics of Postulate 5, mathematicians tried first to prove
it based on the other postulates and definitions. Failing this, they identified postulates
equivalent to it, that is, statements that can be proved on the basis of Postulate 5 and
that, at the same time, can be used to prove Postulate 5. The list of equivalent theorems
is quite remarkable. It includes:
The sum of the angles in every triangle is 180.
There exists a pair of similar, but not congruent, triangles.
Two lines that are parallel to the same line are also parallel to each other.
Even the Pythagorean Theorem
Still another equivalent to Proposition 5 is called the Playfair Postulate. It states that
a line can be drawn parallel to a given line through a given point outside that line. Thus
we have the situation of Figure 5.
C
A
B
!
Figure 5. The Playfair Postulate equivalent of Euclid’s Postulate 5
The Playfair Postulate or equivalently Elements Postulate 5 allows exactly one line to be
drawn through C parallel to AB. This should be quite familiar to you as you almost
certainly learned in school mathematics to construct that line.
5
A number of mathematicians, however, saw in that new formulation the possibility of
different answers. Among them were the 12th century Persian, Omar Khayy´am;4 the 17th
century Italian, Giovanni Girolamo Saccheri; the 18th century German, Carl Friedrich
Gauss; and in the 19th century the Russian Nikolai Ivanovich Lobachevsky and the Hungarian, J´
anos Bolyai.
What these famous mathematicians finally found was that the Playfair Postulate’s “one”
line was arbitrary and that there are, in fact, three alternatives to this postulate:
(1) Playfair: Through a given point one line can be drawn parallel to a given line.
(2) Through a given point less than one line (that is, no lines) can be drawn parallel
to a given line.
(3) Through a given point more than one line can be drawn parallel to a given line.
Your first response to (2) and (3) may be that they are simply nonsense. Surely (2) is
wrong because we learned to construct that line through C on Figure 5. And (3) would
require us to have additional lines through C. What you have to understand, however, is
that (2) and (3) define entirely different geometries, in a real sense different worlds. And,
believe it or not, one of those geometries is all around us. We will explore that geometry
now and in doing so find the answer to the challenge with which this chapter began.
4
Geometry on the Surface of a Sphere
The plane geometry of our high school studies makes perfect sense because locally —
within a few miles, that is — we can consider ourselves living on a plane. But our Earth
is a sphere and geometry on a sphere is different from geometry on a plane. A major
difference becomes immediately apparent. There are no straight lines like those of plane
geometry on a sphere. A line of sight looking toward the horizon, for example, would shoot
away from the Earth on a tangent.
We need to modify the concept of a straight line in the plane to give us a reasonable
substitute on a sphere. We still want to be able to travel “in a straight line” in any direction
along the earth’s surface and have our path along that line shorter than any alternate path.5
In order to do this, we substitute for a straight line on a sphere a line following what is
called a great circle. Indeed, great circles do represent the shortest distance between two
points on them.6
4
Omar Khayy´
am is better known as the poet who wrote the Rubi´
ay´
at, a series of quatrains translated
into English by Edward FitzGerald in 1859. Often quoted are the lines, “A Flask of Wine, a Book of Verse
— and Thou Beside me singing in the Wilderness — Oh, Wilderness were Paradise enow!”
5
Shortest paths are called geodesics.
6
There is an exception to this. If the points on the sphere are at opposite ends of a diameter there are
many equal paths.
6
You are familiar with one great circle, the equator. Lines of longitude (those north-south
lines measured east and west from Greenwich, England) also run along great circles. It is
important to note, however that lines of latitude (east-west lines), except for the equator
itself, do not follow great circles. They do not represent shortest paths. It is easy to show
this with a globe. You need only connect two points at some distance apart with a string.
When pulled taught, in seeking the minimal path, the string will lie along a great circle
and not along a line of latitude other than the equator.
Most maps and charts display geographic features with the lines of latitude horizontal.
For larger areas, especially those involving continents and oceans, this significantly distorts
both shapes and measurements. One consequence on those maps is that Greenland and
Antarctica appear much larger than they really are. Another is that minimum distances are
not along straight lines. Consider, for example, the shortest (great circle) route compared
with the route that follows a line of latitude between New York’s Kennedy Airport and the
Ciampino Airport of Rome, Italy as shown in Figure 6. Although that figure shows how
the great circle route takes you from JFK north over Labrador and Newfoundland, that
route is shorter by more than 1400 miles than a route that simply takes the flight along a
line of latitude.
Figure 6. The Great Circle Route from New York to Rome.
Once we have identified great circles with straight lines, we can examine what would
replace on spheres those five Elements postulates:
1’.
2’.
3’.
4’.
5’.
Construct a segment between two points (not its opposite pole).
Extend a line.
Construct a circle given its center and radius.
All right angles are equal.
That any two lines intersect.
It takes little thought to see that the first four postulates remain essentially as in the
Elements and that they are reasonable on a sphere. A segment is, of course, an arc of
a great circle just as in plane geometry a segment is a piece of a straight line. The final
postulate is, however, very different. It tells us that we have no parallel lines on a sphere.
7
In fact, any two distinct great circles intersect in two points that are poles (that is, 180○ )
apart. This statement may seem strange, but you need only do some experimenting with
a globe to see that it is indeed the case as in Figure 7.
C
A
B
a
b
Figure 7. Postulate 5’ on a Sphere7
In order to continue, we need some vocabulary. Where two great circles meet on a sphere,
they form an angle between those curves. This angle corresponds to and is measured by
the plane angle between the tangents to the great circles at that point, as in Figure 8.
Figure 8. Spherical Angle AP B
A lune is a region bounded by two great circles as in Figure 9.8 The name lune is
etymologically related to lunar, pertaining to the moon, and the lune’s shape clearly justifies
this name.
Figure 9. Lune P AP ′ B
7
Figure 7 was drawn with the Cinderella software as are some other figures that follow. In these views
points and lines on the back of the sphere also appear. These include points at the opposite ends of
diameters through points on the front of the sphere.
8
Some authors refer to a lune as the solid slice taken from the sphere — like a section of an orange.
Here we follow other authors in considering it only the surface region.
8
Note that the angle AOB suggests a way of identifying the area of a lune. That angle,
which is the same as the angle at P , is a fraction of the distance around O. Thus we need
only find the angle of a lune, say x○ , on a sphere with area S to calculate the lune’s area.
x
It is 360
⋅ S. To calculate the area of a lune on a sphere with radius r, you need the formula
for S. It is S = 4πr2 , a formula derived by Archimedes 2200 years ago.9
Finally, a spherical triangle is a 3-sided figure bounded by arcs of great circles. Figure
9 shows a spherical triangle together with the image created on the opposite side of the
sphere by diameters from each corner of the original △ABC.
A
b
c
B
a
C
Figure 10. Spherical triangle ABC.
There is an interesting feature of that shadow triangle on the opposite side of the sphere.
Although its angles are equal to those of the original triangle and it is equal in area to
△ABC, it is not congruent to it. It is a mirror image with orientation reversed and like a
glove, it cannot be made to fit its partner.10
In Figure 9, △P AB is also a spherical triangle and I will use that triangle to suggest the
formula for the area of any spherical triangle. Clearly, the area of that triangle is half the
x
x
area of the lune, whose area we determined is 360
S. Thus the area of △P AB is 720
S.
Now consider what kind of angles are ∠P AB and ∠P BA. They are like meridian lines
meeting the equator, where I hope you will accept that they are both right angles. This
tells us something very interesting about △ABC on Figure 9. The sum of its angles is more
than 180○ . In fact, if ∠P is x○ , the angle sum is 180 + x degrees. That x, the amount over
180○ for a spherical triangle is called the spherical excess. It turns out that every spherical
triangle has such a spherical excess, whether or not they have right angles in them, and
that spherical excess is related to the area of the triangle.
9
By fortunate chance the Danish philologist Johan Ludvig Heiberg discovered parchments in Constantinople (now Istanbul) that included the only known copy of The Method of Mechanical Theorems
of Archimedes. In it Archimedes describes how he found the formulas for the area, S = 4πr2 , and the
volume of a sphere, V = 34 πr3 , as well. His method used calculus-like techniques long before the calculus
was invented.
10
For vocabulary enthusiasts, a mirror image is an enatiomorph and enatiomorphisms play an important
role in crystallography.
9
c
A
C
B
b
a
Figure 10.5 Spherical triangle ABC with sides extended.
On Figure 10.5, the great circle sides of spherical ∆ABC have been extended. Notice
that each of those angles determines a lune. If you add all of those lunes together you will
have the surface of half of the sphere, but the triangle itself will have been counted two
extra times.11 Thus you have the geometric situation regarding areas:
the hemisphere + 2 ∆ABC = Lune A + Lune B + Lune C.
This translates into algebraic terms with S the surface of the sphere:
1
A
B
C
S + 2∆ABC =
S+
S+
S
2
360
360
360
and we have the following algebraic simplification:
A
B
C
S+
S+
S−
360
360
360
B
C
A
S+
S+
S−
∆ABC =
720
720
720
A + B + C − 180
∆ABC =
S
720
2∆ABC =
1
S
2
180
S
720
(4.1)
(4.2)
(4.3)
Recall that we named A + B + C − 180 the triangle’s spherical excess. If you designate
that spherical excess in degrees, x, and the area of any spherical triangle T , you have the
following:
x
x
x
S and T =
4πr2 =
πr2
720
720
180
the latter formula the result of the fact that the area of a sphere is S = 4πr2 .
T=
This finally provides you with the tools necessary to solve the farmer’s problem with
which you began this chapter. You need only find the spherical excess for those surveyor’s
angle measures and use the formula to calculate the area of the property. The actual
calculation is left for the exercises.
11
In making up the area of the hemisphere, it is necessary to take the shadow area of one of the lunes.
10
5
The Other Alternative to Postulate 5
Recall that a third alternative to Euclid’s Postulate 5 was offered. It was:
(3) Through a given point more than one line can be drawn parallel to a given line.
The geometry that this leads to is called hyperbolic geometry and it is more difficult to
demonstrate. Figure 11 gives one way of doing so.
d
b
C
c
B
A
a
Figure 11. Postulate 5 in the Hyperbolic Plane
As on the sphere, hyperbolic lines are curved but, if you examine them closely, you will
see that each of them is perpendicular to the edge of the region containing them. Think
of that region made infinitely large so that the circle represents the “edge” of space. Then
you can think of those lines as being perpendicular to that “edge”, as they are on this
diagram. In order for that to be the case, the lines through most given points will appear
as curved, as they are on the figure. Only lines through the center of the figure would
appear straight.
Under these conditions we do indeed have more than one line through a point outside a
given line that does not intersect it. In particular, two of those lines would meet the line
through A and B at the edge of space. as in Figure 12. Those lines we would designate as
parallel to line AB.
c
C
B
A
a
b
Figure 12. Two lines parallel to AB in the Hyperbolic Plane
11
This is like those parallel railroad tracks that appear to meet at the horizon.
6
Distance on a Sphere
As a kind of bonus, I offer, this time without derivation, a formula for calculating distance
on a sphere. With it you can calculate the mileage between locations for which you have
longitude and latitude. The formula as I will write it will give the distance in nautical
miles. Nautical miles are used — as you might expect — by sailors. Ships’ speeds are
given in knots and a knot is one nautical mile per hour.
A nautical mile is the length of one minute of arc along the equator. You will be asked
in the exercises to calculate how many feet there are in a nautical mile.
One other thing is necessary to translate longitude and latitude into values acceptable
to mathematical computation. Longitude values are given in east or west coordinates
measured from Greenwich and latitude values north and south of the equator. In order
to make these measures acceptable to calculation (and GPS devices) west and south are
negative, east and north positive.
Here then is the formula. Given points with longitude and latitude coordinates (a, b)
and (c, d), the distance in nautical miles along the shorter great circle arc joining them is:
60 cos−1 [sin b ⋅ sin d + cos b ⋅ cos d ⋅ cos(a − c)]
It is this kind of information that is calculated very quickly and very accurately by GPS
devices.
Exercises
(1.1) There are many proofs of the Pythagorean Theorem. In fact, Elisha Scott Loomis offered
367 of them — enough, that is, for a year of daily proofs — in his book The Pythagorean
Proposition, first published in 1927. That edition is now worth a great deal of money to
historians of science. Some of these proofs are very simple and everyone should know at least
one of them. Perhaps the simplest was proved by United States President James Garfield.
Look up this proof and write it out.
(1.2) Look up and follow the proof of the Elements. Wikipedia is a good source for this proof and
provides a thorough explanation. Reduce this explanation to its key steps.
(1.3) Look up the proof of the Pythagorean Theorem in the Heath translation of Euclid’s Elements.
You can access a version of this text through Gutenberg.
(2.1) The words axiom and postulate are synonyms. See if you can determine a technical difference
between the words.
(2.2) Access the 1723 translation of the Elements by John Keil through Google. Where does Keil
list what we have called Proposition 5 of Book I? Why do you suppose this difference arises?
12
(3.1) Of course, the Earth is only approximately a sphere. There are two ways that it differs from
being exactly spherical. Consider the first of those ways. The highest mountain on earth
is Mount Everest at about 29,000 feet and the farthest down you can get, in the Mariana
Trench, is about 36,200 feet deep. What percent of the earth’s radius, about 3960 miles, is
this difference?
(3.2) You can buy raised relief globes that have mountain ranges raised above the surface of the
oceans. If the water were removed, we would have deep trenches in the ocean and these could
be represented as indentations. Toy rubber balloons have a skin thickness of less than .5 mm
≈ .02 inch. Find the percent this is for a 5-inch diameter balloon and compare this with your
answer in 3.1. What does this answer say about the accuracy of raised relief globes?
(3.3) How far can you see on the Earth? The answer is not far when landforms and vegetation
get in the way, but it is a reasonable question at sea or from the seashore. Suppose your eyes
are about 50 feet above sea level. This might be on land along the seashore or from a ship
crow’s nest. Find approximately how far away is your horizon.
(3.4) Devise a general formula to use to estimate the distance you can see in miles, given your
height in feet above sea level.
(3.5) The earth is
any point on
are currently
minutes, and
coordinated so that, as Global Positioning System (GPS) devices make clear,
it may be located with extreme accuracy. Find the coordinates of where you
residing. Give those coordinates in three forms: (a) degrees, (b) degrees and
(c) degrees, minutes and seconds.
(3.6) Most people know that the so-called prime meridian of longitude passes through Greenwich,
England, a London suburb. Where is the longitude-latitude origin, the point that corresponds
to (0,0) on graphs? Where is the point (180○ ,0○ )?
(3.7) In one of the later Oz books by Frank Baum, Dorothy climbs down through the center of the
Earth and comes out in Oz, which is in the middle of the Great Sandy Desert of Australia.
She started, of course, from Kansas. Determine if these two locations are indeed poles apart.
(3.8 Calculate the area of the farmer’s land with which this chapter began.
a
C
A
b
B
c
Figure 13. A spherical triangle with three right angles
(3.9) I did not prove the formula for area of a spherical triangle on a sphere with area S. To further
justify that formula, however, consider the spherical triangle of Figure 13. Each of its angles
is a right angle.
(a) Find its area in terms of S using the area formula.
(b) How many of these triangles could you fit on the sphere without overlapping?
(c) Show that your answer in (b) supports your answer in (a).
13
(4.1) We considered in this chapter the only three possibilities for the number of lines through
a given point parallel to a given line: none, 1, and more than one. There are no other
possibilities. This approach is similar to a method of proof called proof by cases or proof by
exhaustion. You list all the possibilities and eliminate all but one, then that one must provide
your solution. The problem is that we often forget possibilities. (Politicians and lawyers often
encourage you to forget them.) In the following mathematical examples, supply the missing
alternative. Then think up examples of your own:
(a) x > 4 or x < 4.
(b) y is either an integer or a fraction.
(c) Positive integer z is either prime or composite.
(6.1) The length of the equator is 24,901.5 statute miles. (Statute miles are the miles with which
we make land measurements.)
(a) Using this value, what is the radius of the earth at the equator?
(b) We noted that a minute of arc along the equator is equal to a nautical mile. How many
feet are there then in a nautical mile?
(c) How does the nautical mile compare with the statute mile? Is it the same, shorter or
longer?
(d) If there is a difference in (c), what factor can be used to convert nautical miles to statute
miles?
(6.2) Write a calculator program to accept longitude and latitude for two different locations and use
your program to calculate the length of flights between the following cities whose approximate
values are given. Have your calculator provide both nautical miles and statute miles between
five of these locations. (Be especially careful to get longitude and latitude in the correct order
and with the correct sign.)
City
New York
Los Angeles
London
Moscow
Tokyo
Rio de Janeiro
Sydney
Johannesburg
Longitude
74.0○ W
118.3○ W
7.6○ W
37.6○ E
139.7○ E
43.2○ W
151.2○ E
28.1○ W
Latitude
40.7○ N
34.2○ N
51.5○ N
55.8○ N
35.7○ N
22.9○ S
33.9○ S
26.2○ S
(6.3) Use Google Earth or some other program to find the exact location in degrees, minutes and
seconds of where you live. Then find another place of special interest to you and record it in
the same way. Use your program to find the distance between these locations. (You must be
very careful in recording these values this, because coordinates are usually given in the order
latitude, longitude, which does not conform to our usual sense of (x,y) graph coordinates.)
(6.4) Earlier in this chapter I claimed that the great circle flight from New York to Rome saved
over 1400 miles against a flight following closely a line of latitude. Determine the length of
the great circle flight.
14
(6.5) Find a way to determine the length of a New York-Rome trip that approximately follows
the 41○ latitude line and compare your answer with the great circle distance you calculated
in 5.4.
(6.6) We have taken great circle routes as our “straight line” routes between places on the earth
because those are the shortest routes. More generally shortest routes are called geodesics. In
the plane, of course, the geodesic is a straight line. Here is a famous geodesic problem posed
originally by the famous puzzle constructor, Henry Ernest Dudeney, in 1903: In a rectangular
room with dimensions, 30’ x 12’ x 12’, a spider is located in the middle of one 12’ ?12’ wall
one foot away from the ceiling. A fly is in the middle of the opposite wall one foot away from
the floor. If the fly remains stationary, what is the shortest total distance (i.e., the geodesic)
the spider must crawl in order to capture the fly? He can crawl along walls, floor or ceiling.
This problem is best solved by drawing and cutting out the six walls with correct proportions
and rearranging them in various ways that would allow you to calculate the spider’s route.
ceiling
wall
wall
floor
wall
spider
fly
wall
The spider and the fly
15