Tài liệu ADVANCED HIGH-SCHOOL MATHEMATICS pdf

1 Advanced Euclidean Geometry 1 1.1 Role of Euclidean Geometry in High-School Mathematics 1 1.2 Triangle Geometry.. In the diagram to the right, △ABCis a right triangle, segments [AB] an

David B SurowskiShanghai American SchoolSingapore American School

January 29, 2011

The present expanded set of notes initially grew out of an attempt toflesh out the International Baccalaureate (IB) mathematics “FurtherMathematics” curriculum, all in preparation for my teaching this dur-ing during the AY 2007–2008 school year Such a course is offered onlyunder special circumstances and is typically reserved for those rare stu-dents who have finished their second year of IB mathematics HL intheir junior year and need a “capstone” mathematics course in theirsenior year During the above school year I had two such IB math-ematics students However, feeling that a few more students wouldmake for a more robust learning environment, I recruited several of my2006–2007 AP Calculus (BC) students to partake of this rare offeringresulting The result was one of the most singular experiences I’ve had

in my nearly 40-year teaching career: the brain power represented inthis class of 11 blue-chip students surely rivaled that of any assemblage

of high-school students anywhere and at any time!

After having already finished the first draft of these notes I becameaware that there was already a book in print which gave adequatecoverage of the IB syllabus, namely the Haese and Harris text1 whichcovered the four IB Mathematics HL “option topics,” together with achapter on the retired option topic on Euclidean geometry This is avery worthy text and had I initially known of its existence, I probablywouldn’t have undertaken the writing of the present notes However, astime passed, and I became more aware of the many differences betweenmine and the HH text’s views on high-school mathematics, I decidedthat there might be some value in trying to codify my own personalexperiences into an advanced mathematics textbook accessible by andinteresting to a relatively advanced high-school student, without beingconstrained by the idiosyncracies of the formal IB Further Mathematicscurriculum This allowed me to freely draw from my experiences first as

a research mathematician and then as an AP/IB teacher to weave some

of my all-time favorite mathematical threads into the general narrative,thereby giving me (and, I hope, the students) better emotional and

1 Peter Blythe, Peter Joseph, Paul Urban, David Martin, Robert Haese, and Michael Haese, Mathematics for the international student; Mathematics HL (Options), Haese and Harris Publications, 2005, Adelaide, ISBN 1 876543 33 7

intellectual rapport with the contents I can only hope that the readers(if any) can find some something of value by the reading of my stream-of-consciousness narrative.

The basic layout of my notes originally was constrained to the fiveoption themes of IB: geometry, discrete mathematics, abstract alge-bra, series and ordinary differential equations, and inferential statistics.However, I have since added a short chapter on inequalities and con-strained extrema as they amplify and extend themes typically visited

in a standard course in Algebra II As for the IB option themes, myorganization differs substantially from that of the HH text Theirs isone in which the chapters are independent of each other, having verylittle articulation among the chapters This makes their text especiallysuitable for the teaching of any given option topic within the context

of IB mathematics HL Mine, on the other hand, tries to bring outthe strong interdependencies among the chapters For example, the

HH text places the chapter on abstract algebra (Sets, Relations, andGroups) before discrete mathematics (Number Theory and Graph The-ory), whereas I feel that the correct sequence is the other way around.Much of the motivation for abstract algebra can be found in a variety

of topics from both number theory and graph theory As a result, thereader will find that my Abstract Algebra chapter draws heavily fromboth of these topics for important examples and motivation

As another important example, HH places Statistics well before ries and Differential Equations This can be done, of course (they didit!), but there’s something missing in inferential statistics (even at theelementary level) if there isn’t a healthy reliance on analysis In my or-ganization, this chapter (the longest one!) is the very last chapter andimmediately follows the chapter on Series and Differential Equations.This made more natural, for example, an insertion of a theoreticalsubsection wherein the density of two independent continuous randomvariables is derived as the convolution of the individual densities Asecond, and perhaps more relevant example involves a short treatment

Se-on the “random harmSe-onic series,” which dovetails very well with thealready-understood discussions on convergence of infinite series Thecute fact, of course, is that the random harmonic series converges withprobability 1

I would like to acknowledge the software used in the preparation ofthese notes First of all, the typesetting itself made use of the indus-try standard, LATEX, written by Donald Knuth Next, I made use of

three different graphics resources: Geometer’s Sketchpad, Autograph, and the statistical workhorse Minitab Not surprisingly, in the chapter

on Advanced Euclidean Geometry, the vast majority of the graphicswas generated through Geometer’s Sketchpad I like Autograph as ageneral-purpose graphics software and have made rather liberal use ofthis throughout these notes, especially in the chapters on series anddifferential equations and inferential statistics Minitab was used pri-marily in the chapter on Inferential Statistics, and the graphical outputsgreatly enhanced the exposition Finally, all of the graphics were con-verted to PDF format via ADOBER

8 PROFESSIONAL(version 8.0.0) I owe a great debt to those involved in the production

of the above-mentioned products

Assuming that I have already posted these notes to the internet, Iwould appreciate comments, corrections, and suggestions for improve-ments from interested colleagues and students alike The present ver-sion still contains many rough edges, and I’m soliciting help from thewider community to help identify improvements

Naturally, my greatest debt of

gratitude is to the eleven students

(shown to the right) I conscripted

for the class They are (back row):

Eric Zhang (Harvey Mudd),

Jong-Bin Lim (University of Illinois),

Tiimothy Sun (Columbia

sity), David Xu (Brown

Univer-sity), Kevin Yeh (UC Berkeley),

Jeremy Liu (University of

Vir-ginia); (front row): Jong-Min Choi (Stanford University), T.J Young(Duke University), Nicole Wong (UC Berkeley), Emily Yeh (University

of Chicago), and Jong Fang (Washington University) Besides ing one of the most stimulating teaching environments I’ve enjoyed over

provid-my 40-year career, these students pointed out countless errors in thisdocument’s original draft To them I owe an un-repayable debt.

My list of acknowledgements would be woefully incomplete withoutspecial mention of my life-long friend and colleague, Professor RobertBurckel, who over the decades has exerted tremendous influence on how

First draft: April 6, 2007

Second draft: June 24, 2007

Third draft: August 2, 2007

Fourth draft: August 13, 2007

Fifth draft: December 25, 2007

Sixth draft: May 25, 2008

Seventh draft: December 27, 2009

Eighth draft: February 5, 2010

Ninth draft: April 4, 2010

1 Advanced Euclidean Geometry 1 1.1 Role of Euclidean Geometry in High-School Mathematics 1

1.2 Triangle Geometry 2

1.2.1 Basic notations 2

1.2.2 The Pythagorean theorem 3

1.2.3 Similarity 4

1.2.4 “Sensed” magnitudes; The Ceva and Menelaus theorems 7

1.2.5 Consequences of the Ceva and Menelaus theorems 13 1.2.6 Brief interlude: laws of sines and cosines 23

1.2.7 Algebraic results; Stewart’s theorem and Apollo-nius’ theorem 26

1.3 Circle Geometry 28

1.3.1 Inscribed angles 28

1.3.2 Steiner’s theorem and the power of a point 32

1.3.3 Cyclic quadrilaterals and Ptolemy’s theorem 35

1.4 Internal and External Divisions; the Harmonic Ratio 40

1.5 The Nine-Point Circle 43

1.6 Mass point geometry 46

2 Discrete Mathematics 55 2.1 Elementary Number Theory 55

2.1.1 The division algorithm 56

2.1.2 The linear Diophantine equation ax + by = c 65

2.1.3 The Chinese remainder theorem 68

2.1.4 Primes and the fundamental theorem of arithmetic 75 2.1.5 The Principle of Mathematical Induction 79

2.1.6 Fermat’s and Euler’s theorems 85

v

2.1.7 Linear congruences 89

2.1.8 Alternative number bases 90

2.1.9 Linear recurrence relations 93

2.2 Elementary Graph Theory 109

2.2.1 Eulerian trails and circuits 110

2.2.2 Hamiltonian cycles and optimization 117

2.2.3 Networks and spanning trees 124

2.2.4 Planar graphs 134

3 Inequalities and Constrained Extrema 145 3.1 A Representative Example 145

3.2 Classical Unconditional Inequalities 147

3.3 Jensen’s Inequality 155

3.4 The H¨older Inequality 157

3.5 The Discriminant of a Quadratic 161

3.6 The Discriminant of a Cubic 167

3.7 The Discriminant (Optional Discussion) 174

3.7.1 The resultant of f (x) and g(x) 176

3.7.2 The discriminant as a resultant 180

3.7.3 A special class of trinomials 182

4 Abstract Algebra 185 4.1 Basics of Set Theory 185

4.1.1 Elementary relationships 187

4.1.2 Elementary operations on subsets of a given set 190 4.1.3 Elementary constructions—new sets from old 195

4.1.4 Mappings between sets 197

4.1.5 Relations and equivalence relations 200

4.2 Basics of Group Theory 206

4.2.1 Motivation—graph automorphisms 206

4.2.2 Abstract algebra—the concept of a binary oper-ation 210

4.2.3 Properties of binary operations 215

4.2.4 The concept of a group 217

4.2.5 Cyclic groups 224

4.2.6 Subgroups 228

4.2.7 Lagrange’s theorem 231

4.2.8 Homomorphisms and isomorphisms 235

4.2.9 Return to the motivation 240

5 Series and Differential Equations 245 5.1 Quick Survey of Limits 245

5.1.1 Basic definitions 245

5.1.2 Improper integrals 254

5.1.3 Indeterminate forms and l’Hˆopital’s rule 257

5.2 Numerical Series 264

5.2.1 Convergence/divergence of non-negative term series265 5.2.2 Tests for convergence of non-negative term series 269 5.2.3 Conditional and absolute convergence; alternat-ing series 277

5.2.4 The Dirichlet test for convergence (optional dis-cussion) 280

5.3 The Concept of a Power Series 282

5.3.1 Radius and interval of convergence 284

5.4 Polynomial Approximations; Maclaurin and Taylor Ex-pansions 288

5.4.1 Computations and tricks 292

5.4.2 Error analysis and Taylor’s theorem 298

5.5 Differential Equations 304

5.5.1 Slope fields 305

5.5.2 Separable and homogeneous first-order ODE 308

5.5.3 Linear first-order ODE; integrating factors 312

5.5.4 Euler’s method 314

6 Inferential Statistics 317 6.1 Discrete Random Variables 318

6.1.1 Mean, variance, and their properties 318

6.1.2 Weak law of large numbers (optional discussion) 322 6.1.3 The random harmonic series (optional discussion) 326 6.1.4 The geometric distribution 327

6.1.5 The binomial distribution 329

6.1.6 Generalizations of the geometric distribution 330

6.1.7 The hypergeometric distribution 334

6.1.8 The Poisson distribution 337

6.2 Continuous Random Variables 348

6.2.1 The normal distribution 350

6.2.2 Densities and simulations 351

6.2.3 The exponential distribution 358

6.3 Parameters and Statistics 365

6.3.1 Some theory 366

6.3.2 Statistics: sample mean and variance 373

6.3.3 The distribution of X and the Central Limit The-orem 377

6.4 Confidence Intervals for the Mean of a Population 380

6.4.1 Confidence intervals for the mean; known popu-lation variance 381

6.4.2 Confidence intervals for the mean; unknown vari-ance 385

6.4.3 Confidence interval for a population proportion 389 6.4.4 Sample size and margin of error 392

6.5 Hypothesis Testing of Means and Proportions 394

6.5.1 Hypothesis testing of the mean; known variance 399 6.5.2 Hypothesis testing of the mean; unknown variance 401 6.5.3 Hypothesis testing of a proportion 401

6.5.4 Matched pairs 402

6.6 χ2 and Goodness of Fit 405

6.6.1 χ2 tests of independence; two-way tables 411

Advanced Euclidean Geometry

1.1 Role of Euclidean Geometry in High-School

Mathematics

If only because in one’s “further” studies of mathematics, the results

(i.e., theorems) of Euclidean geometry appear only infrequently, this

subject has come under frequent scrutiny, especially over the past 50years, and at various stages its very inclusion in a high-school mathe-matics curriculum has even been challenged However, as long as wecontinue to regard as important the development of logical, deductivereasoning in high-school students, then Euclidean geometry provides aseffective a vehicle as any in bringing forth this worthy objective

The lofty position ascribed to deductive reasoning goes back to atleast the Greeks, with Aristotle having laid down the basic foundations

of such reasoning back in the 4th century B.C At about this time Greekgeometry started to flourish, and reached its zenith with the 13 books

of Euclid From this point forward, geometry (and arithmetic) was anobligatory component of one’s education and served as a paradigm fordeductive reasoning

A well-known (but not well enough known!) anecdote describes

for-mer U.S president Abraham Lincoln who, as a member of Congress,had nearly mastered the first six books of Euclid By his own admis-sion this was not a statement of any particular passion for geometry,but that such mastery gave him a decided edge over his counterparts

is dialects and logical discourse

Lincoln was not the only U.S president to have given serious thought

1

to Euclidean geometry President James Garfield published a novel

proof in 1876 of the Pythagorean theorem (see Exercise 3 on page 4)

As for the subject itself, it is my personal feeling that the logical

arguments which connect the various theorems of geometry are every

bit as fascinating as the theorems themselves!

So let’s get on with it !

1.2 Triangle Geometry

1.2.1 Basic notations

We shall gather together a few notational conventions and be reminded

of a few simple results Some of the notation is as follows:

c

△ABC ∼= △A′B′C′ The triangles △ABC and △A′B′C′ are congruent

△ABC ∼ △A′B′C′ The triangles △ABC and △A′B′C′ are similar

1.2.2 The Pythagorean theorem

fundamen-tal results is the well-known

states that a2+ b2 = c2 in a right

triangle with sides a and b and

hypotenuse c The figure to the

right indicates one of the many

known proofs of this fundamental

result Indeed, the area of the

“big” square is (a + b)2 and can be

decomposed into the area of the

smaller square plus the areas of the

four congruent triangles That is,

(a + b)2 = c2 + 2ab,

which immediately reduces to a2 + b2 = c2

Next, we recall the equally

well-known result that the sum of the

interior angles of a triangle is 180◦

The proof is easily inferred from the

diagram to the right

Exercises

Proportional Segments, i.e.,

given the right triangle △ABC as

indicated, then

h2 = pq, a2 = pc, b2 = qc

2 Prove that the sum of the interior angles of a quadrilateral ABCD

is 360◦

3 In the diagram to the right, △ABC

is a right triangle, segments [AB]

and [AF ] are perpendicular and

equal in length, and [EF ] is

BC, b = AB, c = AB, and

de-duce President Garfield’s proof1 of

the Pythagorean theorem by

com-puting the area of the trapezoid

BCEF

1.2.3 Similarity

In what follows, we’ll see that many—if not most—of our results shallrely on the proportionality of sides in similar triangles A convenientstatement is as follows

Similarity Given the similar

tri-angles △ABC ∼ △A′BC′, we have

1 James Abram Garfield (1831–1881) published this proof in 1876 in the Journal of Education (Volume 3 Issue 161) while a member of the House of Representatives He was assasinated in 1881

by Charles Julius Guiteau As an aside, notice that Garfield’s diagram also provides a simple proof

of the fact that perpendicular lines in the planes have slopes which are negative reciprocals.

Proof Note first that △AA′C′

and △CA′C′ clearly have the same

areas, which implies that △ABC′

and △CA′B have the same area

(being the previous common area

plus the area of the common

trian-gle △A′BC′) Therefore

= area△A′BC′area△CA′B

=

1

2h′ · BC′ 1

2h′ · BC

= BC′BC

In an entirely similar fashion one can prove that A′B

A′C′

AC .Conversely, assume that

A′B

BC′

BC .

In the figure to the right, the point

C′′ has been located so that the

seg-ment [A′C′′] is parallel to [AC] But

then triangles △ABC and △A′BC′′

are similar, and so

C'

B

A'

i.e., that BC′′ = BC′ This clearly implies that C′ = C′′, and so [A′C′]

is parallel to [AC] From this it immediately follows that triangles

△ABC and △A′BC′ are similar.

BC Then △ABC ∼ △A′B′C′

2 In the figure to the right,

3 Let △ABC be a given triangle and let Y, Z be the midpoints of[AC], [AB], respectively Show that (XY ) is parallel with (AB).(This simple result is sometimes called the Midpoint Theorem)

4 In △ABC, you are given that

Assuming that the area of △ABC

is 1, compute the area of△XY Z as

a function of x

Z

Y

X C B

A

5 Let ABCD be a quadrilateral and let EF GH be the quadrilateralformed by connecting the midpoints of the sides of ABCD Provethat EF GH is a parallelogram

6 In the figure to the right, ABCD is

a parallelogram, and E is a point

on the segment [AD] The point

F is the intersection of lines (BE)

and (CD) Prove that AB× F B =

7 In the figure to the right, tangents

to the circle at B and C meet at the

point A A point P is located on

the minor arc BC and the tangent˘

to the circle at P meets the lines

(AB) and (AC) at the points D and

E, respectively Prove that DOE =c

C are colinear, then







> 0 if AB and−→ BC are in the same direction−→

< 0 if AB and−→ BC are in opposite directions.−→

2 IB uses the language “sensed” rather than the more customary “signed.”

This implies in particular that for signed magnitudes,

AB

Before proceeding further, the reader should pay special attention

to the ubiquity of “dropping altitudes” as an auxiliary construction

Both of the theorems of this

subsec-tion are concerned with the following

configuration: we are given the

trian-gle △ABC and points X, Y, and Z on

the lines (BC), (AC), and (AB),

respec-tively Ceva’s Theorem is concerned with

the concurrency of the lines (AX), (BY ),

and (CZ) Menelaus’ Theorem is

con-cerned with the colinearity of the points

X, Y, and Z Therefore we may regard these theorems as being “dual”

The proof of Ceva’s theorem will be greatly facilitated by the lowing lemma:

fol-Lemma Given the triangle

△ABC, let X be the intersection of

a line through A and meeting (BC).

Let P be any other point on (AX).

Proof In the diagram to the

right, altitudes BR and CS have

been constructed From this, we see

△BRX ∼ △CSX

Note that the above proof doesn’t depend on where the line (AP ) tersects (BC), nor does it depend on the position of P relative to theline (BC), i.e., it can be on either side

in-Ceva’s Theorem Given the triangle △ABC, lines (usually called Cevians are drawn from the vertices A, B, and C, with X, Y , and Z,

being the points of intersections with the lines (BC), (AC), and (AB), respectively Then (AX), (BY ), and (CZ) are concurrent if and only if

AZ

Y A = +1.

Proof Assume that the lines in question are concurrent, meeting inthe point P We then have, applying the above lemma three times,that

Y A.

To prove the converse we need to

prove that the lines (AX), (BY ),

and (CZ) are concurrent, given

(CQ) ∩ (AB) Then (AX), (BY ),

and (CZ′) are concurrent and so

AZ′

Z′B · BX

Y Z = 1,which forces

AZ′

Z′B =

AZ

ZB.This clearly implies that Z = Z′, proving that the original lines (AX), (BY ),and (CZ) are concurrent

Menelaus’ theorem is a dual version of Ceva’s theorem and concernsnot lines (i.e., Cevians) but rather points on the (extended) edges of

the triangle When these three points are collinear, the line formed

is called a transversal The reader can quickly convince herself thatthere are two configurations related to △ABC:

As with Ceva’s theorem, the relevant quantity is the product of thesensed ratios:

AZ

Y A;

in this case, however, we see that either one or three of the ratios must

be negative, corresponding to the two figures given above

Menelaus’ Theorem Given the triangle △ABC and given points

X, Y, and Z on the lines (BC), (AC), and (AB), respectively, then

X, Y, and Z are collinear if and only if

Case 1 We assume first that

X, Y, and Z are collinear and drop

altitudes h1, h2, and h3 as indicated

in the figure to the right Using

ob-vious similar triangles, we get

AZ

ZB · BX′

X′C · CY

Y A = −1

It follows that BXXC = BXX′ C′, from which we infer easily that X = X′, and

so X, Y, and Z are collinear

Case 2 Again, we drop altitudes from

A, B, and C and use obvious similar

1.2.5 Consequences of the Ceva and Menelaus theorems

As one typically learns in an elementary geometry class, there are eral notions of “center” of a triangle We shall review them here andshow their relationships to Ceva’s Theorem

sev-Centroid In the triangle △ABC

are drawn so that (AX) bisects

[BC], (BY ) bisects [CA], and

(CZ) bisects [AB] That the lines

(AX), (BY ), and (CZ) are

con-current immediately follows from

Ceva’s Theorem as one has that

Next, note that if we apply the Menelaus’ theorem to the triangle

△ACX and the transversal defined by the points B, Y and the centroid

P , then we have that

Therefore, we see that the distance of a triangle’s vertex to the centroid

is exactly 1/3 the length of the corresponding median

Orthocenter In the

trian-gle △ABC lines (AX), (BY ), and

(CZ) are drawn so that (AX) ⊥

(BC), (BY ) ⊥ (CA), and (CZ) ⊥

(AB) Clearly we either have

Incenter In the triangle △ABC lines

(AX), (BY ), and (CZ) are drawn so

that (AX) bisects BAC, (BY ) bisectsc

ABC, and (CZ) bisects Bc CA As wec

show below, that the lines (AX), (BY ),

and (CZ) are concurrent; the point of

concurrency is called the incenter of

△ABC (A very interesting “extremal”

property of the incenter will be given in

Exercise 12 on page 153.) However, we shall proceed below to giveanother proof of this fact, based on Ceva’s Theorem

Proof that the angle bisectors of △ABC are concurrent Inorder to accomplish this, we shall first prove the

Angle Bisector Theorem We

line segment [BP ] (as indicated to

the right) Then

AB

AP

P C ⇔ ABP = Pc BC.c

Proof (⇐) We drop altitudes

from P to (AB) and (BC); call the

points so determined Z and Y ,

re-spectively Drop an altitude from

B to (AC) and call the resulting

BX

P Y .Therefore,

(⇒) Here we’re given that ABBC = APP C Let

P′ be the point determined by the angle

bisector (BP′) of ABC Then by whatc

has already been proved above, we have

inter-of intersection inter-of the bisector inter-of CBA with [AC] Finally, let Z bec

the point of intersection of the exterior angle bisector at C with

the line (AB) Show that X, Y, and Z are colinear.3

3 What happens if the exterior angle bisector at C is parallel with (AB)?

3 Given △ABC and assume that X is on (BC), Y is on (AC) and

Z is on (AB) Assume that the Cevians (AX) (BY ), and (CZ)are concurrent, meeting at the point P Show that

4 Given the triangle △ABC with incenter P , prove that there exists

a circle C (called the incircle of △ABC) with center P which isinscribed in the triangle △ABC The radius r of the incircle isoften called the inradius of △ABC

5 Let △ABC have side lengths a = BC, b = AC, and c = AB,and let r be the inradius Show that the area of △ABC is equal

to r(a+b+c)2 (Hint: the incenter partitions the triangle into threesmaller triangles; compute the areas of each of these.)

6 Given the triangle △ABC Show that the bisector of the internalangle bisector at A and the bisectors of the external angles at Band C are concurrent

7 Given △ABC and points X, Y, and

Z in the plane such that

of concurrency of two of the perpendicular bisectors is equidistant

to all three of the vertices.) If P is the circumcenter, then thecommon value AP = BP = CP is called the circumradius

of the triangle △ABC (This is because the circumscribed circle

containing A, B, and C will have radius AP )

9 △ABC has side lengths AB = 21, AC = 22, and BC = 20

Points D and E are on sides [AB] and [AC], respectively such

that [DE] k [BC] and [DE] passes through the incenter of △ABC

Compute DE

10 Here’s another proof of Ceva’s

the-orem You are given △ABC and

concurrent Cevians [AX], [BY ],

and [CZ], meeting at the point P

Construct the line segments [AN ]

and [CM ], both parallel to the

Ce-vian [BY ] Use similar triangles to

C B

11 Through the vertices of the triangle △P QR lines are drawn lines

which are parallel to the opposite sides of the triangle Call the

new triangle△ABC Prove that these two triangles have the same

centroid

12 Given the triangle △ABC, let C be the inscribed circle, as in

Exercise 4, above Let X, Y, and Z be the points of tangency

of C (on the sides [BC], [AC], [AB], respectively) and show that

the lines (AX), (BY ), and (CZ) are concurrent The point of

concurrency is called the Gergonne point of the circle C (This

is very easy once you note that AZ = Y Z, etc.!)

13 In the figure to the right, the dotted

segments represent angle bisectors

Show that the points P, R, and Q

are colinear

14 In the figure to the right, three

cir-cies of the same radius and centers

X, Y and Z are shown intersecting

at points A, B, C, and D, with D

the common point of intersection of

all three circles

Show that

(a) D is the circumcenter of

△XY Z, and that

(b) D is the orthocenter of△ABC

(Hint: note that Y ZCD is

(i) △A′B′C′ ∼ △ABC and that the ratios of the correspondingsides are 1:2

(ii) △A′B′C′ and △ABC have the same centroid

(iii) The four triangles determined within △ABC by △A′B′C′are all congruent

(iv) The circumcenter of △ABC is the orthocenter of △A′B′C′

The triangle △A′B′C′ of △ABC formed above is called the dial triangle of △ABC

me-17 The figure below depicts a hexagram “inscribed” in two lines ing the prompts given, show that the lines X, Y, and Z are colin-ear This result is usually referred to Pappus’ theorem

Us-Step 1 Locate the point G on the lines (AE) and (F B); we shallanalyze the triangle △GHI as indicated below.4

Step 2 Look at the transversals, applying Menelaus’ theorem toeach:

4 Of course, it may be that (AE) and (F B) are parallel In fact, it may happen that all analogous choices for pairs of lines are parallel, which would render the present theme invalid However, while the present approach uses Menelaus’ theorem, which is based on “metrical” ideas, Pappus’ theorem

is a theorem only about incidence and colinearity, making it really a theorem belonging to “projective geometry.” As such, if the lines (AE) and (BF ) were parallel, then projectively they would meet

“at infinity;” one could then apply a projective transformation to move this point at infinity to the finite plane, preserving the colinearity of X, Y , and Z

[DXB], so GX

XI

IDDH

18 This time, let the hexagram be

in-scribed in a circle, as indicated to

the right By producing edges [AC]

and [F D] to a common point R

and considering the triangle△P QR

prove Pascal’s theorem, namely

that points X, Y , and Z are

co-linear (Proceed as in the proof

of Pappus’ theorem: consider the

transversals [BXF ], [AY D], and

[CZE], multiplying together the

factorizations of−1 which each

pro-duces.)

19 A straight line meets the sides [P Q], [QR], [RS], and [SP ] of thequadrilateral P QRS at the points U, V, W, and X, respectively.Use Menelaus’ theorem to show that

20 The diagram to the right shows

three circles of different radii with

centers A, B, and C The points

X, Y , and Z are defined by

inter-sections of the tangents to the

cir-cles as indicated Prove that X, Y ,

Y X

A

B C

21 (The Euler line.) In this exercise you will be guided through theproof that in the triangle △ABC, the centroid, circumcenter, andorthocenter are all colinear The line so determined is called theEuler line

In the figure to the right, let G be the centroid of △ABC, andlet O be the circumcenter Locate P on the ray OG so that GP :−→

OG = 2 : 1

(a) Let A′ be the intersection of

(AG) with (BC); show that

△OGA′ ∼ △P GA (Hint:

re-call from page 13 that GA :

GA′ = 2 : 1.)

(b) Conclude that (AP ) and (OA′)

are parallel which puts P on the

altitude through vertex A

(c) Similarly, show that P is also

on the altitudes through

ver-tices B and C, and so P is the

orthocenter of △ABC

1.2.6 Brief interlude: laws of sines and cosines

In a right triangle △ABC, where Cc

is a right angle, we have the familiar

trigonometric ratios: setting θ =

We can extend the definitions of

the trigonometric functions to

ar-bitrary angles using coordinates in

the plane Thus, if θ is any given

angle relative to the positive x-axis

(whose measure can be anywhere

between −∞ and ∞ degrees, and if

(x, y) is any point on the terminal

ray, then we set

5 A fancier way of expressing this is to say that by similar triangles, the trigonometric functions are well defined.

Law of Sines Given triangle

△ABC and sides a, b, and c, as

b is also equal to the above.

Law of Cosines Given triangle

△ABC and sides a, b, and c, as

in-dicated, we have

c2 = a2 + b2 − 2ab cos C

Proof Referring to the

dia-gram to the right and using the

Pythagorean Theorem, we infer

3 In the quadrilateral depicted at the

right, the lengths of the diagonals

are a and b, and meet at an angle θ

Show that the area of this

quadri-lateral is 12ab sin θ (Hint: compute

the area of each triangle, using the

Law of Sines.)

abθ

4 In the triangle to the right, show

that c =

√

1 + i +√

1− i4

given the triangle △ABC, together

with the edge BX, as indicated in

the figure to the right Then

cos θ = b

2 − s2 − p2

Equating the two expressions and noting that a = r +s eventually leads

to the desired result

Corollary [Apollonius

Theo-rem] We are given the triangle

△ABC, with sides a, b, and c,

to-gether with the median BX, as

in-dicated in the figure to the right.

Then

b2 + c2 = 2m2 + a2/2

If b = c (the triangle is isosceles),

then the above reduces to

m2 + (a/2)2 = b2

This follows instantly from Stewart’s Theorem

Exercises

1 Assume that the sides of a triangle are 4, 5, and 6

(a) Compute the area of this triangle

(b) Show that one of the angles is twice one of the other angles

2 (The Golden Triangle) You are

given the triangle depicted to the

3 Let △ABC be given with sides a = 11, b = 8, and c = 8 Assumethat D and E are on side [BC] such that [AD], [AE] trisect BAC.cShow that AD = AE = 6.

4 You are given the equilateral

trian-gle with sides of unit length,

de-picted to the right Assume also

that AF = BD = CE = r for

some positive r < 1 Compute the

area of the inner equilateral

gle (Hint: try using similar

trian-gles and Stewart’s theorem to

com-pute AD = BE = CF )

1.3 Circle Geometry

1.3.1 Inscribed angles

Lemma If a triangle △ABC is inscribed in a circle with [AB] being a

diameter, then A CB is a right angle.c

Proof The diagram to the right

makes this obvious; from 2θ + 2φ =

180, we get θ + φ = 90◦

The measure of an angle inscribed

in a circle is one-half that of the

inscribed arc.

Proof We draw a diameter, as

indicated; from the above lemma,

we see that θ1 + ψ = 90 This

quickly leads to φ1 = 2θ1 Similarly

φ2 = 2θ2, and we’re done

Before proceeding, we shall find

the following concept useful We

are given a circle and points A, B,

and P on the circle, as indicated

to the right We shall say that the

angle AP B opens the arcc AB.˘

A degenerate instance of this is

when B and P agree, in which

case a tangent occurs In this case

we shall continue to say that the given angle opens the arc AB.˘

As an immediate corollary to the Inscribed Angle Theorem, we getthe following:

open the same are are equal.

1 In the diagram to the right, the arc

˘

AB has a measure of 110◦ and the

measure of the angle ACB is 70c ◦

Compute the measure of ADB.c 6

2 Let [AB ] be a diameter of the circle C and assume that C is agiven point If ACB is a right angle, then C is on the circlec C

3 Let C be a circle having center

O and diameter d, and let A, B,

and C be points on the circle If

we set α = BAC, then sin α =c

BC/d (Hint: note that by the

inscribed angle theorem, BAC =c

POC What is the sine of Pc OC?)c

4 In the given figure AF = F C and

P E = EC

(a) Prove that triangle △F P A is

isosceles

(b) Prove that AB + BE = EC

5 A circle is given with center O The

points E, O, B, D, and E are

col-inear, as are X, A, F, and C The

lines (XC) and (F D) are tangent

to the circle at the points A and D

respectively Show that

(a) (AD) bisects BAC;c

(b) (AE) bisects BAX.c