Geometry

There is no royal road to Geometry.

— Euclid to King Ptolemy

82

His method of “exhaustion” anticipated the development of integral calculus byalmost 2000 years! In his treatise Quadrature of the Parabola, he applied the methodof exhaustion to derive the theorem presented on page 79, expressing the area of asegment of a parabola in terms of the area of a particular inscribed triangle.Archimedes was unable to find the area of a general segment of the other conics,the ellipse and the hyperbola, because these involve transcendental functions notknown in his time.

Archimedes also made significant contributions to applied mathematics andengineering, including his famous law of hydrostatics. He developed a variety ofpractical devices such as the lever which he used to catapult huge quarter-ton rocksagainst an attacking Roman fleet during the siege ofSyracuse in 212 B.C.

Of his countless contributions to many branches ofmathematics, Archimedes was most proud of hisdiscovery that the volume of a sphere is two-thirdsthe volume of the cylinder in which it is inscribed.(His method is described in Archimedes Strikes Againon page 82.) Archimedes requested that this discoveryand the corresponding diagram be carved on his tomb.This request must have been granted, for the greatRoman orator Cicero said that he had restoredArchimedes’ tomb and the engraving when he wasquaestor in Sicily.

The volume of a sphere is two-thirdsthe volume of the right circularcylinder in which it is inscribed.

The Mathematician Who Anticipated Calculus by Almost 2000 Years

HistoricalDigression

Archimedes 287-212 B.C.

Give me a place to stand onand I will move the earth.

When historians of mathematics attempt to identify the greatest mathematician of all time, they inevitably produce a list

including three giants: Archimedes, Newton, andGauss. Although Archimedes lived in ancient Greece,his mathematical techniques were comparable inrigor to those of the greatest mathematicians of the17th and 18th centuries. Among his many treatises wasone titled On Spirals in which he presented twenty-eight propositions about spirals — including onewhich now bears his name. (Archimedes’ spiral canbe used to trisect any angle; the trisesction problemis discussed in Three Impossible Problems of AncientGreece, p. 219.)

Lengths, Areas, and Volumes Arguments both True & False

The circumference C and area A of a circle with radius r can be calculated usingthese familiar formulas:

C = 2 π r and A = πr2

From these formulas, we can readily deduce the formula A = .

Here is another intuitive method to see why thisis true. We divide the circle into n equal “pieslices” (no pun intended).

Each pie slice looks a lot like a triangle with base and height r (especially when

n is very large). Then each triangle has area . As there are n such pie slices, the

total area of the circle is .

Part 2: The Area of a Segment of a Parabola

Some readers may think that the foregoingargument is bogus. However, this is the methodby which the ancient Greeks computed the area.Calculus of some sort is necessary to make thisargument rigorous, but even without calculusthe Greeks were able to prove some remarkableresults. For example, Archimedes wished to find the area of a segment of a parabola.(A parabola is a plane curve similar to the curve y = ax2, for some real number a. Asegment of a parabola is the closed region bounded by that parabola and any straightline.) Archimedes gave a rigorous proof that the area of the segment of the parabolais four thirds the area of a triangle having the same base and equal height.

rC2

A C

B

height

The full circle dividedinto n equal parts

One slice of the fullcircle is one of n equal

parts

r

Cn

nC

2nrC

This entire argument can be betterexpressed in the language of limits.In fact, if you work out what is reallygoing on, you’ll rediscover thefamous result:

limx

limx

xx x

x→∞ →

( ) = =sinsin1 1 1

0or .

This example is a great motivationfor why we might expect the abovelimit to be true.

Part 1: The Area of a Circle

83

Geometry

parabolic segment = colored region

n rCn

rC2 2

=

84

Geometry

2πr n

2πr n

2πr 4

2πr n

2πr 4

π 2 2r

Part 3: The Surface Area of a Sphere

You might be feeling queasy about the lack of rigor in this style of argument —and rightly so. Here is a similar argument that gives a wrong answer. Can you findthe flaw? (This is trickier than the usual “find-what’s-wrong-with-this-proof”puzzles, as this argument doesn’t claim to be rigorous in the first place.)

We divide the sphere of radius r into two hemispheres at its equator, and partition

the equator into n segments, each of length . We draw great circles through

these partition points and the north and south poles of the sphere. Now imaginethe sphere sliced along these great circles, peeled back and flattened out like a

Mercator projection (for readers with a background in geography).

Each of these little triangles has base and height one-quarter the circumference

of the sphere, , and hence has area

1

2

2 2

4 2

2 2π π πr

n

r r

n

=

Since there are 2n triangles, the total areais , which is the correct formula forthe surface area of the sphere — NOT!!!Oh no! Some of you know that the surfacearea of a sphere of radius r is 4πr2, so weare off by a factor of . What has gonewrong? How can the method that workedso well in Part 1 have failed us in Part 3?Philosophical pseudomathematics tells usthat two wrongs do not make a right; butwhat do a right and a wrong make?

TH

E FA

R SID

E C

OPY

RIG

HT

© FA

R W

OR

KS, IN

C.

distributed by Universal Press Syndicate.

Reprinted w

ith permission. A

ll rights reserved.

4 π

85

Geometry

Food for Thought

q. a) Define the “radius” r of a square to be the length of thealtitude from the center of the square to one of its sides. Showthat the formula from Part 1 (A = rC/2) still holds.b) How should the “radius” of a regular n-gon be defined sothe formula still holds?c) What should the “radius” of a cube be for the formula in Part 4 to hold?How about a regular tetrahedron? Why are all of these answers similar?

r. The “hypervolume” of a “four-dimensional sphere” of radius r is Hr

=π 2 4

2.

Can you use a method similar to Parts 1 and 4 to find the “surface volume”?

s. The “hypervolume” of the n-dimensional sphere of radius r has a beautiful

formula: π ( / )

( / )!

n r

n

n2

2. Notice how this works when n =2 and n = 4. Can you see

how it works for n = 0? Can you make sense of it for n = 1 and n =3?

(What is (1/2)!? What is (3/2)!? Is (3/2)! = (3/2) × (1/2)!?)

r

Part 4: The Volume of a Sphere

We can add to this confusing stew by working out the volume of the sphere. Thereader should be able to show (by the same argument as in Part 1, this time bydissecting the sphere into n almost-pyramids withvertices at the center of the sphere) that if a spherehas surface area A and radius r, then the volume

V of the sphere is given by V = .

As the volume of the sphere is V =

and the surface area is A = 4πr2, the formula iscorrect. The method appears to be working again.

Some of this material appeared in Mathematical Mayhem (Vol. 4, Issue 4, (March/April1992)). Part of the erroneous argument is from the Mathematical Digest (No. 86, (Jan.1992)), 15. For more information on either of these publications, see the AnnotatedReferences.

EXPERTSONLY

Ar3

34πr3

86

GeometryArchimedes Strikes Again

You’ve probably memorized the formula for the volume V of a sphere (in terms of

the radius r : V r=4

3

3π ), but you probably haven’t seen why it is true. Here is an

intuitive proof, originally by Archimedes.

First we recall one fact: the volume of a right

circular cone with base of radius r and height h

is V r h=1

32π .

This is a special case of something more

general. If you have a strange plane

shape S with area B, and a point P at a

height h above the plane, then the cone

with base S and vertex P consists of ev-

erything that lies on a line segment with

one endpoint in S and one endpoint at P.

Then the volume of this cone is

This is a generalization of the fact that

the area of a triangle with base b and

height h is

Can you propose the further generalization to four dimensions?

V Bh= 1

3

V Bh= 1

3 .

A bh= 1

2 .

87

GeometryArchimedes Strikes Again (cont’d)

Archimedes computed the volume of a hemisphere follows. Place your hemisphere(of radius r) on a flat table A. Beside it, place a circular cone with base of radius r,and height r, balanced perfectly on its vertex.

On another table (call it B), place a cylinder with base of radius r, and height r.

Now consider a cross-section over table A at height h.

We are now in a position to execute Archimedes’ beautiful heuristic proof.

The cross-section of the cone is a circleof radius h. (The proof uses the factthat the shorter sides of a 90˚-45˚-45˚triangle are equal.)

As shown in the diagram, the cross-section of the sphere is a circle of radius

.r h2 2−

88

GeometryArchimedes Strikes Again (cont’d)

Thus, the total cross-sectional area at height h above table A is

π π πr h h r2 22

2 2−

+ =

↑ ↑ ↑hemisphere inverted cone total cross-section

The cross-section of the cylinder is always a circle of radius r, so the total cross-sectional area above table B is also πr2.

Here’s the key step: for each height, the cross-sectional areas are the same, so thetotal volume must be the same. (You might want to ponder this for a while andconvince yourself that it is reasonable.)

Thus: volume of hemisphere + volume of cone = volume of cylinder

so, volume of hemisphere + (πr2)r = ( πr2 )r

From this, we conclude that the volume of a hemisphere of radius r is πr3.Therefore the volume V of a sphere of radius r is given by:

And the proof is complete! This clever argument surely deserves a shouted“Eureka”, but I’m not sure if I’d run naked through the streets over it.

If you are comfortable with integral calculus, you will recognize how Archimedes’ideas anticipated its fundamental concepts (such as Fubini’s Theorem). And if youhaven’t seen integral calculus before, keep this example in mind for when youdo!

32

13

V r=4

3

3π

89

GeometryFood for Thought

q. Here is a long-famous problem:Suppose I have a large sphere, and I cut a cylindrical hole right through thecenter, and I’m left with a bracelet of height 2r. What is its volume?

There appears to be some information missing in the problem – I haven’ttold you how large the sphere is. But that’s actually a clue: the answershouldn’t be dependent on the radius! If you believe that, then you canalready guess the correct answer: if the radius is actually r, then the cylindricalhole must be “infinitely small”, or basically there should be no hole at all.Then the “bracelet” is a sphere of radius r, which has volume (4/3)πr3. Ifthe answer should always be (4/3)πr3, how would you hope to prove it? Oneway is to set the bracelet on a table beside a ball of radius r, and hope thatArchimedes’ trick with the hemisphere works, and that the cross-section ofthe sphere at any height has the same area as the cross-section of the bracelet.Can you make this work?

EXPERTSONLY

r. There is a similar idea for finding the surface area of a sphere of radius r.Nestle the sphere in a cylinder of radius r and height 2r. Then the surfacearea of the sphere between height h

1 and h

2 (where h

1 and h

2 are between 0

and 2r) is exactly the surface area of the cylinder between height h1 and h

2.

I like to think of it as follows: imagine that the sphere is semi-transparent,and a strong light is shining from the axis of the cylinder. Then this lightprojects the sphere onto the cylinder, and this projection is “area-preserving.”

Can you think of an explanation of why this is true, in thespirit of Archimedes, without using any fancy integrals?

90

Geometry

This is the basis for the Lambert equal-area projection map of theworld, which shows the relative areas of countries correctly, althoughthe distances are distorted, especially near the poles.

The Lambert Cylindrical Equal-Area Projection

91

Geometry

B

q

s

Late at night, a thiefwith a flashlight hasclimbed exactlyhalfway up a ladderin order to breakinto a house.

The ladder slides outfrom under him, leavinghim ignominiously on theground.

r

Unfortunately (or fortunately,depending on your point of view),he hasn’t secured the base of theladder. (Don’t try this at home!)

A

© Taisa Dorney. Reprinted with permission

I am thankful to Ed Barbeau of the University of Toronto for introducing me to this classic.

A police officer running to the scene observes the attempted burglary. Through thedarkness, all she can see is the path of the flashlight. However, she is able todeduce from this the nature of the misadventure and later provide surprisinglyaccurate testimony at the burglar’s trial.

What path did the flashlight take? Did it travel in a straight linebetween point A and point B? Before looking at the answer (inthe Solution to the Falling Ladder Problem, p. 168), you maywant to think about it, and come up with a guess of your own.

What would be the path of a point on the ladder but not at thehalfway point?

The Falling

Ladder Problem

92

GeometryAn Easy Proof that arctan 1/3 + arctan 1/2 = 45˚

Here is a slick non-algebraic proof of the identity arctan 1/3 + arctan 1/2 = 45˚.

Consider the triangle in the Cartesian plane with vertices at A (0,0), B(1,3), andC(2,1), and the point D(1,0).

There are several ways of seeing that ABC isa 45˚-45˚-90˚ triangle. (For example,

AC = BC = 1

2AB.) Hence ∠ABC = 45˚.

But ∠ABC = ∠ABD + ∠DBC.

Can you see why tan(∠ABD) = 1/3 andtan (∠CBD) = 1/2? If you can, then you’vebasically completed the proof!

The especially intrepid reader might want to try to find similar proofs of thefollowing:

a) arctan 1 + arctan 2 + arctan 3 = 180˚.

b) arctan arctan arctan 1

2

1

3

1

7= +

c) More generally, if a and b are positive integers with a > b,

arctan arctan arctan 1 1

a b a

b

a ab–= +

− +2 1

For another example of a proof by diagram, see Fifteen Degrees of SeparationFifteen Degrees of SeparationFifteen Degrees of SeparationFifteen Degrees of SeparationFifteen Degrees of Separationon the following page.1

Many of these “proofs by diagram” have been collected in Nelsen’s Proofs Without Words.See page 39 of Nelsen’s book for a proof (due to Edward M. Harris) of identity a).(For bibliographic information, see the Annotated References.)

A(0,0)

B(1,3)

C(2,1)

D(1,0)

93

Geometry

Everyone who has learned trigonometry has memorized the sines, cosines, andtangents of “easy” angles such as 0˚, 30˚, 45˚, 60˚, and 90˚. Using these, one canuse the subtraction formulas for sine, cosine and tangent

to compute the trigonometric values for 15˚and 75˚. But they are much easier (andmore fun) to work out from scratch, as follows.

Fifteen Degrees of Separation

sin(x – y) = sin x cos y – cos x sin ycos(x – y) = cos x cos y + sin x sin y

tan( )tan tan

tan tanx y

x y

x y− = −

+1

q. What are the lengths of segments x and y?

r. Using them, can you work out the sine, cosine, and tangent of 15˚ and 75˚?

s. Which triangle has larger area, A or C?

This “proof without words” is due to Larry Hoehn of Austin Peay State University(College Math. J., Sept. 2004, p.282). Another beautiful proof of this fact, by theToronto high school teacher Doug Ailles, appeared in the first edition.

For trigonometric values of other angles, see The Pythagorean Pentagram,The Golden Mean, & Strange Trigonometry (p. 158, 18˚) and ThreeImpossible Problems of Ancient Greece (p. 219, 20˚), and the “Food forThought” after that section (1˚).

We glue together two of our favoriteright triangles, 1-1- 2 (labelled A,with angles 45˚-45˚-90˚) and 1- 3 -2(labelled B, with angles 30˚-60˚-90˚).

We then glue on a new triangle C sothat their union is a big 45˚-45˚-90˚triangle. Note that C is a 15˚-75˚-90˚triangle.

31

x

2

45˚

45˚

1

60˚

30˚

15˚

y

zAAAAA BBBBB

CCCCC

One of the Purest Forms of Mental ExerciseOne of the Purest Forms of Mental ExerciseOne of the Purest Forms of Mental ExerciseOne of the Purest Forms of Mental ExerciseOne of the Purest Forms of Mental Exercise

Personal ProfilePersonal ProfilePersonal ProfilePersonal ProfilePersonal Profile

a-Ping Yee was born in Toronto,Canada, but soon moved toWinnipeg, Manitoba, where he

enrolled at St. John’s-Ravenscourt, one of thetop private schools in the country. Not longafter arriving, Ping started to make a namefor himself. Between grades four and six,he took the multiple-choice CanadianNational Math League competition gearedtoward sixth graders; he achieved perfectscores in each of three consecutive years.John Barsby, a dedicated math teacher fromthe Upper School of SJR, noticed Ping andtook him under his wing. He continued towork one-on-one with Ping throughout Ping'sschooling.

In Grade 4, Ping also started doing ScienceFair projects, later entering them in theManitoba Schools’ Science Symposium, the provincial science fair. In Grade 8,his project on chaos, fractals, and nonlinear systems (an area of study just beginningto attract public attention) won him a place at the Canada-Wide Science Fair, aswell as an appearance on the TV program The Nature of Things. The trip to theCanada-Wide Fair in St. John’s, Newfoundland was his first solo trip away fromhome. He considers this experience a significant turning point in his life, openinghis eyes to new people and new experiences.

In Grade 9, Ping again competed at the Canada-Wide Science Fair, this time witha project on evolving computer-simulated life-forms. A prize there earned him atrip to the London International Youth Science Fortnight that summer. Once againhe had a wonderful time, meeting people from over 50 other countries.

In Grade 10, besides winning another trip to the London Fortnight, he won theEuclid competition, a national competition for students two years his senior. Alongwith other successes, this achievement ensured him a place on the Canadian Teamto the International Mathematical Olympiad (IMO) in Sigtuna, Sweden, where heearned an Honorable Mention.

Ka-Ping Yee (Canada)Born February 26, 1976

K

94

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95

The next year, in lieu of competing at the IMO, Ping decided to go to Shad Valley,an intense four-week program for gifted Canadian science students. In his finalyear of high school, he earned the highest score on both senior national mathematicalcompetitions. These achievements won him a place at that year’s IMO in Istanbul,Turkey, where he won a Gold Medal.

Meanwhile, Ping was also pursuing his interest in computers. In 1991 and 1992 hewon the International Computer Problem-Solving Contest run by the Universityof Wisconsin, a contest in which competitors aimed to produce the highest calibreprograms in the fastest possible time. After entering the University of Waterloo’scompetitive Computer Engineering program in the fall of 1993, he joined Waterloo’steam to the Association for Computing Machinery programming contest. Thatteam went on to win the international finals in Phoenix, Arizona, competing againstundergraduate teams from across North America and elsewhere. In his first year atWaterloo, Ping also took part in the North American Putnam MathematicalCompetition and received an Honorable Mention.

Ping has long had a fascination with the idea of applying pattern recognition tobrainwaves, a process that with practice might enable us to communicatetelepathically with computers and, to a certain extent, with other people.Nanotechnology is another fetish; he would love to explore the field of molecularmanufacturing when he has the necessary background in chemistry and biology.

Ping’s other current obsessions include Japanese animation (anime) and other thingsJapanese, graphic stories, devilsticks, and eighties synth-pop. He has also recentlybeen attracted to a set of philosophies called “extropianism” that he discovered onthe Internet. Seeking boundless expansion for the human race through technologiessuch as space travel, artificial intelligence, cryonics, and nanotechnology,extropianism questions traditional limits such as finite lifespans and rejects blindfaith in favor of dynamic optimism. Ping finds that extropianism ties in well withhis personal philosophy of getting involved with as many things and meeting asmany people as possible.

Ping finds that his numerous successes all stem from his attitude toward life. Hislove of mathematics is one facet of an open attitude toward learning and experience:“Math is one of the purest forms of mental exercise — yet both the results and themethods of mathematics have applications everywhere. It is a great way to learn tosolve problems, since you get volumes of essential knowledge at the same time.Math is a boundless universe that requires no more than an eager mind to explore,where you never have to worry about explosive decompression.”

Combinatorics revisited

Strange as it may sound, the power of mathematics restson its evasion of all unnecessary thought and on itswonderful saving of mental operations.

— Ernst Mach