Lectura

share no common factor with m other than

^{J .}

Then for each x E S, there is a unique y E m such that xy

^=-

I

^(mod

m).

For example, i f

m =

1 2, then

S = {

1 , 5 , 7 ,

I I}.

^The

"multiplicative inverse" modulo 1 2 of each element x E S turns out to be

x:

^{5 · 5 =-}

I

(mod 1 2) , 7 · 7 =-

I

(mod 1 2 ) , etc.

2.3.21 There are infinitely many primes. Of the many proofs of this important fact, perhaps the oldest was known to the ancient Greeks and written down by Eu­

clid. It is a classic argument by contradiction. We start by assuming that there are only finitely many primes

PI,P2,P3, ... ,PN.

Now (the ingenious crux move ! ) any

three

(not necessarily distinct) elements of

T

^{is in}

T

and that the product of any three elements of U is

2.3.24 Complete the solution started in Example 1 . 1 .2 on page

I.

2.3 M ETHODS O F A R G U M ENT 51 2.3.25 If you haven 't worked on it already, look at Problem 2.2. 1 3 on page 37. The correct answer is

(n2 + n +

2 ) /2. Prove this by induction.

2.3.26 It is easy to prove that the product of any three consecutive integers is always divisible by 6, for at least one of the three integers is even, and at least one is divisible by 3. Done ! This is a very easy proof, but as an

exercise,

prove the assertion using induction. It is less fun, but good practice.

2.3.27 Prove that a set with

n

elements has 2n subsets, including the empty set and the set itself. For example, the set

{a, b, c}

has the eight subsets

0,

{a}, {b}, {c}, {a,b}, {a,e}, {b,c}, {a,b,c}.

2.3.28 Prove the formula for the sum of a geometric series:

an-I +an-2 + .. . + 1 = ( -- ^{an - I} _{a - I} ) ^.

2.3.29 Prove that the absolute value of the sum of sev­

eral real or complex numbers is at most equal to the sum of the absolute values of the numbers. Note: you will need first to verify the truth of the triangle in­

equality , which states that

la + bl

^:S

lal + Ibl

^{for any}

real or complex numbers

a, b.

2.3.30 Prove that the magnitude of the sum of several vectors in the plane is at most equal to the sum of the magnitudes of the vectors.

2.3.33 Prove Bernoulli's Inequality, which states that if

x

^>

-I, x

^{i= 0 and}

n

is a positive integer greater this characterization of

f(n)

using induction.

2.3.35 (Bay Area Mathematical Olympiad, 2006) Suppose that

n

squares of an infinite square grid are colored grey, and the rest are colored white. At each step, a new grid of squares is obtained based on the

previous one, as fol lows. For each location in the grid, examine that square, the square immediately above, and the square immediately to the right. If there are two or three grey squares among these three, then in the next grid, color that location grey; otherwise. color it white. Prove that after at most n steps all the squares in the grid will be white. that each proposition holds for all positive integers 11.

(a) II

+

_{13 +}

15 + .. . +

1211- 1 = 1211' probably discovered that if a triangle drawn on the co­

ordinate plane has vertices at lattice points, then

A = 2B+J - 1. I

2.4 Other Important Strateg ies

where A . B and

J

respectively denote the area, number of boundary lattice points and number of interior lat­

tice points of this triangle. This is a special case of Pick's Theorem, which holds for any polygon with vertices at lattice points (including non-convex poly­

gons). Prove Pick's Theorem with induction. Easy version: assume that it is true for triangles (i.e., as­

sume the base case is true). Harder version: prove that it is true for triangles first !

2.3.39 Here are a few questions about Example 2.3.9 on page 48, where we proved that the vertices of any triangulation can be 3-colored so that no two adjacent vertices are the same color.

(a) Our proof was "nonconstructive," in that it showed that the coloring existed, but did not show how 10 achieve it. Can you come up with a

"constructive" proof; i.e., can you outline a col­

oring

algorithm

that will work on an arbitrary polygon?

(b) On page 49, we raised the question whether any arbitrary triangulation of a polygon has a

"boundary triangle," i.e.. a triangle that cuts the original polygon into two, rather than three, pieces. It sure seems obvious. Prove it.

(c) In the diagram for the triangulation proof, the

"central" triangle split the polygon into two parts, called L and R. What if the central tri­

angle had split the polygon into

three

^parts?

2.3.40 Consider a

2 1 999

_x

2 1 999

square. with a single

I

^x

I

square removed. Show that no matter where the small square is removed. it is possible 10 tile this

"giant square minus tiny square" with ells (see Exam­

ple 1 .3. 1 9 on page 10 for another problem involving tiling by ells).

2.3.41 (IMO 1 997) An n x n matrix (square array) whose entries come from the set

S =

^{{ I .}

2, .

^{. .}

^{, 2n-1 }}

is called a

silver

matrix if, for each

i =

^I

. ^{. . .} .

ⁿ

.

^{the ith}

row and the ith column IOgether contain all the mem­

bers of S. Show that silver matrices exist for infinitely many values of 11 .

Many strategies can be applied at different stages of a problem, not just the beginning.

Here we focus on just a few powerful ideas. Learn to keep them in the back of your

2 . 4 OTH ER I M PO RTANT STRATEG I E S 53 mind during any investigation. We will also discuss more advanced strategies in later chapters.

Draw a Picture!

Central to the open-minded attitude of a "creative" problem solver is an awareness that problems can and should be reformulated in different ways. Often, just translating something into pictorial form does wonders. For example, the monk problem (Exam­

ple

2. 1 .2

on page

1 7)

had a stunningly creative solution. But what if we just interpreted the situation with a simple distance-time graph?

First Day s Path

BAM Noon

It's obvious that no matter how the two paths are drawn, they must intersect some­

where !

Whenever a problem involves several algebraic variables, it is worth pondering whether some of them can be interpreted as coordinates. The next example uses both vectors and lattice points. (See Problem

2.2. 1 7

on page

37

and Problem

2.3.38

on page

52

for practice with lattice points.)

Example 2.4.1 How many ordered pairs of real numbers

(s, t)

with

0

^<

s, t

^<

1

are there such that both

3s + 7t

and

5s + t

are integers?

Solution : One may be tempted to interpret

(s, t)

as a point in the plane, but that doesn't help much. Another approach is to view

(3s + 7t, 5s + t)

as a point. For any

s, t

we have

(3s + 7t, 5s + t)

=

(3, 5)s + (7, 1 )t .

The condition

0

^<

s, t

^<

1

means that

(3s + 7t, 5s +t)

is the endpoint of a vector that lies

inside

the parallelogram with vertices

(0, 0) , (3, 5) , (7, 1 )

and

(3, 5) + (7, 1 )

=

( 1 0, 6) .

The picture below illustrates the situation when

s

=

0.4, t

=

0.7.

... ... · . . . . . . . . . · . . . . . . . . .

^{: : : : : : : : : (10.6)}

... ... ... · . . . . · . . . .

^{: : (3;5) : .}

. .

^:

... ... ... · . . . . · . . . . · . . . . · . . . . ...

^:

....

^:

...

^:

...

^:

...

: . . . (3;S)s+(?,I)t. : . . . . :

...

^:

...

^:

...

..

_{� ' '' '} ^{� �} (1.,1)1 :

•• ··•··· .•...•...•...•...

(0.0)

Since both

3s + 7t

and

5s + t

are to be integers,

(3s + 7t, 5s + t)

is a lattice point. Con­

sequently, counting the ordered pairs

(s, t)

with 0 <

s, t

^< 1 such that both

3s + 7t

and

5s + t

are integers is equivalent to counting the lattice points inside the parallelogram.

This is easy to do by hand; the answer is 3 1 . •

This problem can be generalized nicely using Pick's Theorem (Problem 2.3.38 on page 52). See Problem 2.4.22 below.

Pictures Don't Help? Recast the Problem in Other Ways!

The powerful idea of converting a problem from words to pictures is just one aspect of the fundamental peripheral vision strategy. Open your mind to other ways of reinter­

preting problems. One example that you have already encountered was Example 2. 1 .7 on page 20, where what appeared to be a sequence of numbers was actually a sequence of

descriptions

of numbers. Another example was the locker problem (Example 2.2.3 on page 29) in which a combinatorial question metamorphosed into a number theory lemma. "Combinatorics � Number Theory" is one of the most popular and produc­

tive such "crossovers," but there are many other possibilities. Some of the most spec­

tacular advances in mathematics occur when someone discovers a new reformulation.

For example, Descartes ' idea of recasting geometric questions in a numeric/algebraic form led to the development of analytic geometry, which then led to calculus.

Our first example is a classic problem.

Example 2.4.2 Remove the two diagonally opposite comer squares of a chessboard.

Is it possible to tile this shape with thirty-one 2

x

1 "dominos"? (In other words, every square is covered and no dominos overlap.)

Solution: At first, this seems like a geometric/combinatorial problem with many cases and subcases. But it is really just a question about counting colors. The two comers that were removed were both (without loss of generality) white, so the shape we are interested in contains 32 black and 30 white squares. Yet any domino, once it

2.4 OTH E R I M PO RTANT STRATEG I ES 55 is placed, will occupy exactly one black and one white square. The 3 1 dominos thus require 3 1 black and 3 1 white squares, so tiling is impossible. _

The idea of introducing coloring to reformulate a problem is quite old. Never­

theless, it took several years before anyone thought to use this method on the Gallery Problem (Problem 2.2.26 on page 38). This problem was first proposed by Victor Klee in 1 973, and solved shortly thereafter by Vaclav Chvatal. However, his proof was rather complicated. In 1 978, S. Fisk discovered the elegant coloring argument that we present below.7 If you haven 't thought about this problem yet, please do so before reading the solution below.

Example 2.4.3

Solution to the Gallery Problem:

If we let g(n) denote the minimum number of guards required for an n-sided gallery, we get g(3) = g(4) = g(5) = 1 and g( 6) = 2 by easy experimentation. Trying as hard as we can to draw galleries with

"hidden" rooms, it seems impossible to get a 7-sided or 8-sided gallery needing more than two guards, yet we can use the idea of the 6-sided, two-guard gallery on page 38 to create a 9-sided gallery, which seems to need three guards. Here are examples of an 8-sided and 9-sided gallery, with dots indicating guards.

If this pattern persists, we have the tentative conjecture that g(n) = Ln/3J . A key difficulty with this problem, though, is that even when we draw a gallery, it is hard to be sure how many guards are needed. And as n becomes large, the galleries can become pretty complex.

A coloring reformulation comes to the rescue: Triangulate the gallery polygon.

Recall that we proved, in Example 2.3.9 on page 48, that we can color the vertices of this triangulation with three colors in such a way that no two adjacent vertices are the same color. Now, pick a color, and station guards at all the vertices with that color. These guards will be able to view the entire gallery, since every triangle in the triangulation is guaranteed to have a guard at one of its vertices ! Here is an example of a triangulation of a 1 5-sided gallery.

7 See [22] for a nice discussion of Chvatal's proof and [3 1 ] for an exhaustive treatment of this and related problems.

This procedure works for all three colors. One color must be used at l

n/3

J vertices at most (since otherwise, each color is used on more than l

n/3

J vertices, which would add up to more than

n

vertices). Choose that color, and we have shown that at most

l

n/3

J guards are needed.

Thus,

g( n)

:::; l

n/3

J . To see that

g( n)

_{= l}

n/3

J , we need only produce an example, for each

n,

that requires l

n/3

J guards. That is easy to do; just adapt the construction used for the 9-gon above. If l

n/3

J

=

r, just make r "spikes," etc. _

The next example (used for training the 1 996 USA team for the IMO) is rather contrived. At first glance it appears to be an ugly algebra problem. But it is actually something else, crudely disguised (at least to those who remember trigonometry well).

Example 2.4.4 Find a value of x satisfying the equation

= 6x + 8 �.

Solution : Notice the constants

5 , 6, 8

and the

-

^{x2 and} ± x terms. It all looks like

3-4-5

triangles (maybe

6-8- 10

triangles) and trig. Recall the basic formulas:

sin2 e + cos2 e

= 1 ,

_{SIn 2 -}

. !!.-

^_

2 '

!!.-

^_

^{+ cos e}

cos 2

-2 ·

Make the trig substitution

x =

cos e . We choose cosine rather than sine, because

± cos e is involved in the half-angle formulas, but ± sin e is not. This substitu­

tion looks pretty good, since we immediately get

- x2 =

sin e, and also

1

± cos e

±

x =

± cos e

=

2 . Thus the original equation becomes

5

v'2

(

^sin

� ⁺

^cos

�)

⁼

⁶

^{cos e}

^{+ 8}

^{sin e . (9)}

2.4 OTH E R I M PORTANT STRATEG IES 57 Now we introduce a simple trig tool: Given an expression of the form a cos 0 + b sin O , write

2a 2 cos O + 2b 2 sin o^{+ b + b}

)

^.

This is useful, because a

and b are respectively the sine and cosine

+ b2 + b2

of the angle a := arctan(a/b).

Consequently,

a

acos 0 + b sin O = + b2 (sin a cos 0 + cos a sin 0)

= In particular, we have

sinx + cosx = v'2 sin

(

^{x +}

�) .

Applying this, equation (9) becomes (note that + 82 = 1 0) 5 v'2v'2 sin

( ^�

⁺

�)

^{= 10}

(�

^{cos 0 +}

�

^{sin 0}

)

^.

Hence, if a = arctan(3/4), we have

sin

( ^�

⁺

�)

^{= sin(a + 0).}

Equating angles yields 0 = 1r/2 - 2a. Thus

x = cos O = sin(2a) = 2 sin a cos a = 2

(�) (�)

⁼

^��.

^•

Converting a problem to geometric or pictorial forms usually helps, but in some cases the reverse is true. The classic example, of course, is analytic geometry, which converts pictures into algebra. Here is a more exotic example: a problem that is geo­

metric on the surface, but not at its core.

Example 2.4.5 We are given n planets in space, where n is a positive integer. Each planet is a perfect sphere and all planets have the same radius R. Call a point on the surface of a planet private if it cannot be seen from any other planet. (Ignore things such as the height of people on the planet, clouds, perspective, etc. Also, assume that

the planets are not touching each other.) It is easy to check that if

n

₌ 2, the total private area is

4nR2,

which is just the total area of one planet. What can you say about

n

= 3? Other values of n?

Partial Solution : A bit of experimentation convinces us that if

n

= 3, the total private area is also equal to the total area of one planet. Playing around with larger

n

suggests the same result. We conjecture that the total private area is always equal exactly to the area of one planet, no matter how the planets are situated. This appears to be a nasty problem in solid geometry, but must it be? The notions of "private"

and "public" seem to be linked with a sort of duality; perhaps the problem is really not geometric but

logical.

We need some "notation." Let us assume that there is a universal coordinate system, such as longitude and latitude, so that we can refer to the

"same" location on any planet. For example, if the planets were little balls floating in a room, the location "north pole" would mean the point on a planet that is closest to the ceiling.

Given such a universal coordinate system, what can we say about a planet P that has a private point at location x? Without loss of generality, let x be at the "north pole." Clearly, the centers of all the other planets must lie on the south side of the P's "equatorial" plane. But that renders the north poles of these planets public, for their north poles are visible from a point in the southern hemisphere of P (or from the southern hemisphere of any planet that lies between). In other words, we have shown pretty easily that

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