Colourful Mathematics 10 9789636977726

Table of contents :
Contents
Mathematical reasoning 10
1. What does it imply? 10
2. The pigeonhole principle 21
3. Arrangement (ordering) problems 29
4. Picking problems 32
Computing the root 36
1. Rational numbers, irrational numbers 36
2. The identities (laws) of the square root 40
3. Applying the identities (laws) of the square root 44
4. The nth root of numbers 50
5. The identities (laws) of the nth root . 53
The quadratic equation 60
1. The quadratic equation and function 60
2. The quadratic formula 64
3. The zero product form. The relation between the roots and the coefficients 69
4. Equations of higher degree which can be reduced to quadratic equations 74
5. Quadratic inequalities 80
6. Parametric quadratic equations (higher level courseware) 84
7. Equations involving square roots 90
8. Quadratic simultaneous equations 96
9. Arithmetic and geometric mean 101
10. Extreme value exercises (higher level courseware) 106
11. Problems leading to quadratic equations 110
Geometry 116
Widening the knowledge about circles 116
1. Reminder 116
2. The theorem of the central and inscribed/tangent-chord angles 117
3. The theorem of inscribed angles; the arc of viewing angles 121
4. The theorem of inscribed quadrilaterals (higher level courseware) 125
The similarity transformation and its applications 129
1. Parallel intercepting lines, parallel intercepting line segments (higher level courseware) 129
2. The angle bisector theorem (higher level courseware) 135
3. The transformation of central dilation (or homothety) 137
4. The similarity transformation 141
5. Similarity of figures; the simple cases of similar triangles 143
6. A few applications of similarity 147
7. The ratio of the area of similar planar figures 154
8. The ratio of the volume of similar solids 158
Trigonometric functions of acute angles 161
1. Determining distances with the help of similarity 161
2. Trigonometric functions of acute angles 164
3. Relations between the trigonometric functions of acute angles 168
4. Trigonometric functions of special angles 172
5. Determining several data of a triangle with the help of trigonometric functions 175
6. Calculations in the plane and in space with the help of trigonometric functions 180
Vectors 184
1. The concept of a vector; the sum and the difference of vectors; scalar multiplication of vectors (reminder) 184
2. Expressing vectors as the sum of components in different directions 188
3. Applying vectors in the plane and in space 194
4. Vectors in the coordinate system, the coordinates of a vector, operations with vectors given with coordinates 199
Trigonometric functions 204
1. The definition and the simple properties of the sine and the cosine function 204
2. The graph of the sine function 209
3. The graph of the cosine function, equations, inequalities 214
4. The tangent and the cotangent function 221
5. Compound exercises and applications . 228
6. Geometric applications 232
Calculation of probability 238
1. Events 238
2. Operations with events 243
3. Experiments, frequency, relative frequency, probability 248
4. The classical model of probability .. 251

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Colourful mathematics

10

There is a unique code to be found on the inside of the back cover of this book, which can be activated on the following website: www.mozaweb.com The activation gives access to the electronic version of the published document under the conditions mentioned on the web page.

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Colourful mathematics

József Kosztolányi István Kovács Klára Pintér János Urbán István Vincze

M he ati s textbook

10

Mozaik Education - Szeged, 2015

Authors:

Consultants:

Translated by:

Editor: Typography:

illustrated by:

Photos:

JÓZSef Kosztolányi PhD ■ associate professor István Kovács ■ secondary grammar school teacher Klára Pintér ■ college senior lecturer János Urbán PhD ■ secondary grammar school teacher István Vincze • secondary grammar school teacher József Németh PhD • associate professor József né Szigeti • secondary school teacher

Orsolya Koros-Rónai

Katalin Tóth Tamás Reményfy István Ábrahám Mozaik Archívum - Margit Hilbert Phd; International photo agencies All rights reserved, including copying, the publishing of shortened or lengthened versions of the book. The whole book or any part thereof cannot be copied in any form (microfilm, photocopy or other carrier) withouth the written consent of the publishing company.

isbn Copyright:

978 963 697 772 6 Mozaik Education - Szeged, 2015

Mathematical reasoning 1. 2. 3. 4.

What does it imply? ........................................................... The pigeonhole principle ..................................................... Arrangement (ordering) problems ....................................... Picking problems ...............................................................

10

21 29 32

Computing the root 1. 2. 3. 4. 5.

Rational numbers, irrational numbers .................................. The identities (laws) of the square root ................................ Applying the identities (laws) of the square root .................... The nth root of numbers ..................................................... The identities (laws) of the nth root ......................................

36 40 44 50

53

The quadratic equation 1. The quadratic equation and function .................................... 60 2. The quadratic formula ........................................................ 64 3. The zero product form. The relation between the roots and the coefficients ............................................................ 69 4. Equations of higher degree which can be reduced to quadratic equations ........................................................ 74 5. Quadratic inequalities ......................................................... 80 6. Parametric quadratic equations (higher levelcourseware) ...... 84 7. Equations involving square roots ......................................... 90 8. Quadratic simultaneous equations ....................................... 96 9. Arithmetic and geometric mean .......................................... 101 10. Extreme value exercises (higher level courseware) ............... 106 11. Problems leading to quadratic equations .............................. 110

Geometry Widening the knowledge about circles 1. Reminder........................................................................... 2. The theorem of the central and inscribed/tangent-chord angles .................................... 3. The theorem of inscribed angles; the arc of viewing angles ... 4. The theorem of inscribed quadrilaterals (higher level courseware) ...................................................

The similarity transformation and its applications 1. Parallel intercepting lines, parallel intercepting line segments (higher level courseware) .................................................. 6

116 117

121 125

129

2. 3. 4. 5. 6. 7. 8.

The angle bisector theorem (higher level courseware) ........... 135 The transformation of central dilation (or homothety) ............ 137 The similarity transformation ............................................... 141 Similarity of figures; the simple cases of similar triangles ...... 143 A few applications of similarity ............................................ 147 The ratio of the area of similar planar figures ........................ 154 The ratio of the volume of similar solids .............................. 158

Trigonometric functions of acute angles 1. Determining distances with the help of similarity .................. 161 2. Trigonometric functions of acute angles ............................... 164 3. Relations between the trigonometric functions of acute angles .................................................................. 168 172 4. Trigonometric functions of special angles 5. Determining several data of a triangle with the help of trigonometric functions .................................................. 175 6. Calculations in the plane and in space with the help of trigonometric functions .................................................. 180 Vectors 1. The concept of a vector; the sum and the difference of vectors; scalar multiplication of vectors (reminder) ........................... 2. Expressing vectors as the sum of components in different directions .......................................................... 3. Applying vectors in the plane and in space ........................... 4. Vectors in the coordinate system, the coordinates of a vector, operations with vectors given with coordinates ....................

184

188 194 199

Trigonometric functions 1. The definition and the simple properties of the sine and the cosine function ...................................................... 2. The graph of the sine function ............................................. 3. The graph of the cosine function, equations,inequalities ......... 4. The tangent and the cotangent function ............................... 5. Compound exercises and applications ................................. 6. Geometric applications .......................................................

204

209 214

221

228 232

Calculation of probability 1. 2. 3. 4.

Events............................................................................... Operations with events....................................................... Experiments, frequency, relative frequency,probability ........... The classical model of probability .......................................

238 243 248

251 7

Guide to use the course book The notations and highlights used in the book help with acquiring the courseware.

- The train of thought of the worked examples show samples how to understand the methods and processes and how to solve the subsequent exercises. - The most important definitions and theorems are denoted by colourful highlights. - The parts of the courseware in small print and the worked examples noted in claret colour help with deeper understanding of the courseware. These pieces of knowledge are necessary for the higher level of graduation. - Figures, the key points of the given lesson, review and explanatory parts along with interesting facts of the history of mathematics can be found on the margin. The difficulty level of the examples and the appointed exercises is denoted by three different colours:

Yellow: drilling exercises with basic level difficulty; the solution and drilling of these exercises is essential for the progress. Blue: exercises the difficulty of which corresponds to the intermediate level of graduation. Claret: problems and exercises that help with preparing for the higher level of graduation. These colour codes correspond to the notations used in the Colourful mathematics workbooks of Mozaik Education. The workbook series contains more than 3000 exercises which are suitable for drilling, working on in lessons and which help with preparing for the graduation. The end results of the appointed exercises can be found on the following website: www.mozaik.info.hu. Website www.mozaweb.com offers more help material for processing with the course book. ♦ ♦ ♦

Mathematics, “ratio” and logic way of thinking are probably the most efficient tools of cognition of our world, which sometimes associate with unexplainable phenomena. These are inseparable from Homo sapiens and these make the everyday activities complete. A few thoughts from those who have experienced all this: "So what is ratio the human intellect created logic out of? It is obvious that it is ‘there in ’ the nature, otherwise it would not be possible to understand the nature with the help of rational tools. Ratio unites humans, animals and nature. ” (Imre Kertész, Nobel prlzewinning Hungarian writer) “Through the centuries the collective awareness of mathematicians created its own universe. Where it is I do not know - and I think the word "where" also loses its meaning here but I can assure the reader: this mathematical universe is all too real to those who live in it. The mankind could pierce into the mystery of the surrounding world the most deeply right by means of Mathematics. ” (Ian Stewart) “The strict proof is usually the last step! Before that many conjectures are needed, and for these aesthetic belief is extremely important. ” (Roger Penrose) The Authors wish productive work and learning.

René Descartes (1569-1650) was looking for a general method to solve problems in his work "Regulae ad directionem ingenii’’ (Rules for the Direction of the Mind). His idea was firstly to express every problem as a mathematical problem, then secondly to express every mathematical problem as an algebraic one, and then finally to solve these in the form of equations.

For the development of our way of thinking it is also necessary to observe the structure, the schemes of the reflective action. Henceforth we are going to apply a few methods which give a useful strategy for solving many problems. The more strategies we know, the greater our chance to be able to solve the problems facing us.

MATHEMATICAL REASONING

1. What does it imply? Necessary, sufficient, necessary and sufficient conditions Example 1 The team of the Timbertoes and the Shortlegs played a football match; there were no special happenings (there were no own goals, both teams took part, it did not finish earlier, etc.). Who won if the reporter said the following at the end of the commentary? a) The Timbertoes scored one goal more than the Shortlegs did. b) The Timbertoes scored. c) One of the players of the Shortlegs was sent off. d) The Timbertoes scored more goals than the Shortlegs did. Solution (a) If the Timbertoes scored one goal more than the Shortlegs did, then the Timbertoes won for sure. Thus scoring one goal more than the opponent is a sufficient condition for the Timbertoes to win.

a sufficient but not necessary condition

TlflBERTOES

SHORTLEGS

C . n □ . u

But the Timbertoes can win in other ways too, not only by scoring one goal more, i.e. scoring one goal more than the opponent is not a necessary condition for the Timbertoes to win.

To summarise: scoring one goal more than the opponent is a sufficient but not necessary condition for the Timbertoes to win.

Solution (b) If the Timbertoes scored, it is still possible that they lost the match if the opponent scored more goals. Thus the fact that the Timbertoes scored does not imply that they won. So scoring is not a sufficient condition for the Timbertoes to win.

a necessary but not sufficient condition

TlflBERTOES

I

SHORTLEGS

But on the other hand: if the Timbertoes won, then they must have scored, otherwise they could not have won. So scoring is a necessary condition for the Timbertoes to win.

: 3

To summarise: scoring is a necessary but not sufficient condition for the Timbertoes to win.

10

Solution (c) The fact that one of the players of the Shortlegs was sent off does not

not a necessary and not a sufficient condition

imply that the Timbertoes won, thus one player of the opponent being sent off is not a sufficient condition for the Timbertoes to win. The fact that the Timbertoes won does not imply that one of the players of the Shortlegs was sent off, thus one player of the opponent being sent off is not a necessary condition for the Timbertoes to win.

To summarise: one of the players of the opponent being sent off is not a necessary and not a sufficient condition for the Timbertoes to win.

Solution (d) If the Timbertoes scored more goals than the Shortlegs did, then they must have won; thus the fact that the Timbertoes scored more goals than the opponent implies that they won. So scoring more goals than the opponent is a sufficient condition for the Timbertoes to win.

a necessary and sufficient condition

If the Timbertoes won, it was only possible so that they scored more goals than the Shortlegs, i.e. the fact that the Timbertoes won implies that they scored more goals than the Shortlegs did. So scoring more goals than the opponent is a necessary condition for the Timbertoes to win.

Let us denote a proposition (e.g. the Timbertoes won) by p and another proposition (e.g. the Timbertoes scored more goals than the opponent) by q.

To summarise: scoring more goals than the opponent is a necessary and sufficient condition for the Timbertoes to win. Note: A condition can be necessary or not necessary, sufficient or not sufficient, there are a total of 2 • 2 = 4 options.

STATEMENT: If the Timbertoes won, then the Timbertoes scored

more goals than the opponent. (The Timbertoes won only if the Timbertoes scored more goals than the opponent.) Scoring more goals than the opponent is a necessary condition for the Timbertoes to win. Notation: The Timbertoes won => the Timbertoes scored more goals than the opponent. THE CONVERSE OF THE STATEMENT: If the Timbertoes scored more goals than the opponent, then the Timbertoes won. Scoring more goals than the opponent is a sufficient condition for the Timbertoes to win. Notation: The Timbertoes scored more goals than the opponent => the Timbertoes won.

g is a necessary condition of p p=>g

q is a sufficient condition of p

Q=>P

THE STATEMENT AND ITS CONVERSE: The Timbertoes won if and only if the Timbertoes scored more goals than

q is a necessary and sufficient condition ofp p«g

the opponent. Scoring more goals than the opponent is a necessary and sufficient condition for the Timbertoes to win. Notation: The Timbertoes won the Timbertoes scored more goals than the opponent.

Statement: p=>q The converse of the statement: q=>p

11

MATHEMATICAL REASONING

Example 2 How can we decide whether a natural number n is divisible by 24? Five students gave the following answers: Ann: The natural numbers needs to be divisible by 4. Ben: The natural number n needs to be divisible by 48. Chris: The natural number/? needs to be divisible by 5. Dana: The natural number/? needs to be divisible by 4 and 6. Edward: The natural number/? is divisible by 3 and 8. Let us examine these answers.

i—(

divisible by 4

28

Solution (Ann) Indeed, if a natural number is divisible by 24, then it is also divisible by 4. It means that divisibility by 4 is a necessary condition of the

)

divisible by 24

divisibility by 24.

48

24

16

But there is a number divisible by 4, e.g. 28, which is not divisible by 24. Therefore if a number is divisible by 4, then it is not sure that it is div^ible by 24, i.e. the divisibility by 4 does not imply the divisibility by 24. In other words: the divisibility by 4 is not a sufficient condition of the divisibility by 24.

72 32

Figure 1

24 I n => 4 I n 4In 241 n 4f/7 => 24|r?

H

divisible by 24

To summarise: the divisibility by 4 is a necessary but not sufficient Condition Of the divisibility by 24. (Figure 1)

Solution (Ben) There is a number divisible by 24, e.g. 72, which is not divisible by 48. Therefore if a number is divisible by 24, then it is not sure that it is divisible by 48, i.e. the divisibility by 24 does not imply the divisibility by 48. In other words: the divisibility by 48 is not a necessary condition

|

divisible by 48 -

24 48

96

72

of the divisibility by 24.

However if a natural number is divisible by 48, then it is also divisible by 24. It means that the divisibility by 48 is a sufficient condition of the divisibility by 24.

Figure 2

48 I n => 24 I n 241/7 =£> 48ln

To summarise: the divisibility by 48 is a sufficient but not necessary condition of the divisibility by 24. (Figure 2)

Divisibility by 24 is a necessary but not sufficient condition for divisibility by 48.

12

Solution (Chris) There is a natural number divisible by 24, e.g. 24, which is not divisible by 5. Therefore if a number is divisible by 24, then it is not sure that it is divisible by 5, i.e. the divisibility by 24 does not imply the divisibility by 5. It means that the divisibility by 5 is not a necessary condition of the divisibility by 24.

^-( divisible by 24

Furthermore there is a natural number divisible by 5, e.g. 25, which is not divisible by 24. Therefore if a number is divisible by 5, then it is not sure that it is divisible by 24, i.e. the divisibility by 5 does not imply the divisibility by 24.

{ divisible by5;

Figure 3

In other words: the divisibility by 5 is not a sufficient condition of the divisibility by 24. 5 I n & 24 I n 24 I n 5In

To summarise: the divisibility by 5 is not a necessary and not a sufficient condition of the divisibility by 24. (Figure 3)

Solution (Dana) Indeed, if a natural number is divisible by 24, then it is also divisible by 4 and 6. It means that the divisibility by 4 and 6 is a necessary condition of the divisibility by 24. But there is a number divisible by 4 and 6, e.g. 12, which is not divisible by 24. Therefore if a number is divisible by 4 and 6, then it is not sure that it is divisible by 24, i.e. the divisibility by 4 and 6 does not imply the divisibility by 24.

In other words: the divisibility by 4 and 6 is not a sufficient condition of the divisibility by 24. To summarise: the divisibility by 4 and 6 is a necessary but not a sufficient condition of the divisibility by 24. (Figure 4) Note: In general it is not true that if a and b are natural numbers for which both a and b are divisors of n, then a • b is a divisor of n.

241/7 => 4 In and 6In 41/7 and 61/7 =#> 24|n

Solution (Edward) Indeed, if a natural number is divisible by 24, then it is also divisible by 3 and 8. It means that the divisibility by 3 and 8 is a necessary condition of the divisibility by 24. If a natural number is divisible by 3 and 8, then it is also divisible by 24. It means that the divisibility by 3 and 8 is a sufficient condition of the divisibility by 24.

24 | n => 3 I n and 8 I n 3 I n and 81 n => 24 I n

To summarise: an arbitrary natural number n is divisible by 24 if and only if it is divisible by 3 and 8. The divisibility by 3 and 8 is a necessary and sufficient condition of the divisibility by 24.

241/7 31n and 8In

Note: For all natural numbers a and b, which are relatively primes, it is true that both a and b are divisors of n if and only if a ■ is a divisor of n.

13

MATHEMATICAL REASONING

Example 3 An 8x8 chessboard can be tiled by 2x1 dominoes in several ways. One such possible tiling can be seen in figure 5. a) If one corner (1x1 square), of the chessboard is removed, can the remaining board be tiled by dominoes? b) If two adjacent squares are removed from the top left corner of the chessboard, can the remaining board be tiled by dominoes? c) If two arbitrary but adjacent squares of the chessboard are removed, can the remaining board be tiled by dominoes? d) If the two top corners of the chessboard are removed, can the remaining board be tiled by dominoes? e) If two opposite corners of the chessboard are removed, can the remaining board be tiled by dominoes? f) If two arbitrary and differently coloured squares of the chessboard are removed, can the remaining board be tiled by dominoes? ......

Figure 5 It cannot be tiled: proof by deduction.

Solution (a) It can be seen that with the help of 2 x 1 dominoes only an even number of squares can be tiled. After removing one corner of the chessboard 8-8-1 =63 squares are left, which is an odd number, therefore it cannot be tiled by dominoes. Since originally the number of squares was even, in order to be left with an even number of squares also an even number of squares should be removed. (Figure 6)

The statements below, implied by the above, have the same meaning:

Figure 6

- If we remove an odd number of squares from the chessboard, the remaining board cannot be tiled by 2x 1 dominoes.

It can be tiled: proof by giving an example.

- If the remaining board can be tiled by dominoes, then an even number of squares are left. - An even number of remaining squares is a necessary condition for the remaining board to be able to be tiled by dominoes.

Solution (b) After removing two adjacent squares from the top left comer of the chessboard the remaining board can be tiled by dominoes for example as seen in figure 7.

Solution (c) If two arbitrary but adjacent squares of the chessboard are removed, then let us take a 2 x 8 rectangle in the chessboard according to figure 8 (the longer side of this large rectangle should be perpendicular to the longer side of the rectangle created by the two squares removed). This rectangle can be tiled by dominoes for example so that the dominoes are parallel with the rectangle removed. The remaining whole rows or columns of the chessboard can be tiled by dominoes. So if two adjacent squares of the chessboard are removed, the remaining board can be tiled by dominoes.

Figure 7 It can be tiled: proof by giving an example.

The statements below, implied by the above, have the same meaning: - If two adjacent squares of the chessboard are removed, the remaining board can be tiled by dominoes.

- The fact that two adjacent squares of the chessboard are removed implies that the remaining board can be tiled by dominoes. - Removing two adjacent squares from the chessboard is a sufficient condition for the remaining board to be able to be tiled by dominoes.

14

Solution (d)

It can be tiled: proof by giving an example.

If the two top corners of the chessboard are removed, the remaining board can be tiled by dominoes for example as seen in figure 9.

The statements below, implied by the above, have the same meaning: - It is not true that if the remaining board can be tiled by dominoes, then it is certain that two adjacent squares are removed.

- The fact that the remaining board can be tiled by dominoes does not imply that two adjacent squares are removed. - Removing two adjacent squares from the chessboard is not a necessary condition for the remaining board to be able to be tiled by dominoes.

Therefore removing two adjacent squares from the chessboard is a sufficient but not necessary condition for the remaining board to be able to be tiled by dominoes.

Solution (e) Let us try the tiling. We assume that it is not possible, but it seems difficult to prove, because the dominoes can be placed in many ways. The following idea helps: one domino tile always covers one light and one dark coloured square on the chessboard, i.e. there should be the same number of light and dark coloured squares on the board part that can be tiled. Since the squares in the two opposite corners of the chessboard have the same colour, in figure 10 there are 30 light and 32 dark coloured squares among the remaining squares, thus it cannot be tiled by dominoes. The statements below, implied by the above, have the same meaning: - If two squares with the same colour are removed, the remaining board cannot be tiled by dominoes. - If the remaining board can be tiled by dominoes, then equal number of light and dark coloured squares is left.

Figure 10

- Equal number of light and dark coloured squares remaining is a necessary condition for the remaining board to be able to be tiled by dominoes.

Notes: 1. The idea above, where we solved the problem by colouring the squares, can also be useful when solving other exercises. 2. Another approach, which can also be used often:

Let us place the 8 x 8 chessboard in a coordinate system so that its lower left corner is in the (0; 0) point of the coordinate system. Let the coordinates of a square be the coordinates of its lower left corner. The sum of the coordinates of all the squares of the chessboard is even, since every x- and every y-coordinate appears eight times in the sum. (Figure 11) If we remove the squares with the coordinates (0; 0) and (7; 7), the sum of the coordinates of the remaining squares stays even.

The coordinates of the two squares belonging to one domino are either (x; y) and (x; y + 1), or (x; y) and (x + 1; y). The sum of the four coordinates in both cases is 2x + 2y+ 1, which is odd for sure. The remaining 62 squares could be tiled by 31 dominoes, since their number is odd; the sum of the coordinates of the 31 dominoes is also odd.

Figure 11

The tiling cannot be done, because the sum of the coordinates of the squares to tile is even, while the sum of the coordinates of the dominoes is odd.

15

Solution (f) While removing two differently coloured squares in several different ways let us try to tile the remaining board. We assume that the tiling is always possible, but because of the large number of cases it seems to be difficult to prove. After stringing the squares of the chessboard according to figure 12 the chessboard can be tiled by dominoes along the string. If we remove two squares, the string is cut into two pieces, and since the squares removed are differently coloured, these pieces consist of even number of squares. So both pieces can be tiled by dominoes along the string. The statements below, implied by the above, have the same meaning:

Figure 12

- If one dark and one light coloured square of the chessboard are removed, the remaining board can be tiled by dominoes. - The fact that one dark and one light coloured square of the chessboard are removed implies that the remaining board can be tiled by dominoes. - Removing one dark and one light coloured square of the chessboard is a sufficient condition for the remaining board to be able to be tiled by dominoes. Items e) and f) together have the following meaning:

- By removing two squares of the chessboard the remaining board can be tiled by dominoes if and only if one dark and one light coloured square is removed.

Figure 13

- Removing one dark and one light coloured square of the chessboard is a necessary and sufficient condition for the remaining board (after removing two squares) to be able to be tiled by dominoes.

While giving the reasoning for the exercises let us observe when we used examples, counter-examples or deduction for the proof.

Note: In figure 13 it can be seen that two light and two dark coloured squares of the chessboard can be removed so that the remaining board cannot be tiled by 2x1 dominoes.

16

Proof by giving an example, by giving a counter-example, by deduction. Proof by contradiction Example 4

The following proposition can be read in the Biology course book: Every zebra is striped. “What does it imply?” - asked the teacher. Four students gave the answers below: Adam: If an animal is a zebra, then it is striped. Barbara: If an animal is striped, then it needs to be a zebra. Celia: If an animal is striped, then it is possible that it is a zebra. Dolly: If an animal is not striped, then it cannot be a zebra. Let us decide whose answer is correct and why.

Statement: If an animal is a zebra, then it is striped. The converse of the statement: If an animal is striped, then it is a zebra.

Solution (Adam) Adam’s “If an animal is a zebra, then it is striped.” statement is true, because every zebra is striped. The proposition of the course book implies Adam’s proposition.

Proof by deduction.

Solution (Barbara) Proof by giving a counter-example.

Barbara’s “Ifan animal is striped, then it needs to be a zebra.” statement is false, because there is an animal which is striped but not a zebra, for example a clownfish. We verified that Barbara’s statement was false with the help of a counter­

An animal being striped is a necessary but not sufficient condition for an animal to be a zebra.

example.

Solution (Celia) Celia’s “If an animal is striped, then it is possible that it is a zebra.” statement is true, because for example a zebra is striped and is indeed a zebra. We verified that Celia’s statement was true with the help of an example.

Proof by giving an example.

Solution (Dolly) Dolly’s “Ifan animal is not striped, then it cannot be a zebra.” statement is true; we are going to prove it as follows. Let us assume that the statement is false.

I. there is an animal, which is not striped and is still a zebra, so there e. is a zebra which is not striped. Since every zebra is striped according to the Biology course book, this statement is false. Figure 14

It is a contradiction, thus the original statement is true. ♦ ♦ ♦

We verified the last statement of the example with the help of proof by contradiction. The steps of the proof by contradiction: 1. Let us assume the negative of the statement. 2. Let us draw conclusions. 3. We end up with a contradiction. 4. Thus the original statement is true. 17

MATHEMATICAL REASONING

With the help of figure 16 let us brush up the definition and the properties of the special quadrilaterals.

Example 5 Which statements below are true and which are false? a) There is a rhombus that is a rectangle. b) There is a kite in which all the angles are of different measure. c) If the diagonals of a quadrilateral are perpendicular, then it is a square. d) If the diagonals of a quadrilateral are not of equal length, then it is not a rectangle.

Solution (a) The statement “There is a rhombus that is a rectangle.” is true. (Figure 15)

It can be proven with the help of an example: a square is such a rhombus which is also a rectangle. A square is a rhombus, because all of its sides are of equal length, and it is also a rectangle, because all of its angles are of equal measure.

Figure 15

Solution (b) The statement “There is a kite in which all the angles are of different measure.” is false.

Having two angles of equal measure is a necessary condition for a quadrilateral to be a kite.

It can be verified by deduction. A kite is an axially symmetric quadri­ lateral about one of its diagonals. So it has two opposite angles, which are of equal measure, therefore it is not possible that all its angles are of different measure.

Solution (c) Having perpendicular diagonals is not a sufficient condition for a quadrilateral to be a square.

The statement “If the diagonals of a quadrilateral are perpendicular, then it is a square.” is false. It can be verified by a counter-example. For example the kite, in which not all the sides are of equal length, is such a quadrilateral in which the diagonals are perpendicular to each other, but it still not a square.

Having equally long diagonals is a necessary condition for a quadrilateral to be a rectangle.

Solution (d) The statement “//' the diagonals of a quadrilateral are not of equal length, then it is not a rectangle." is true.

Let us examine whether the necessary conditions are sufficient, and whether the not sufficient conditions are necessary in the example.

It can be verified by proof by contradiction.

Let us assume the negative of the statement. There is a quadrilateral in which the diagonals are not of equal length, and it is still a rectangle. About rectangles we know that the diagonals are of equal length, so this is not possible. It is a contradiction, thus the statement is true.

Figure 16

18

Exercises 1. Write down the converse of the following statements, and decide whether they are true or false. a) If it is raining in the town, then the road is wet. b) If I leave home, then I lock the door. c) If he is Winnie-the-Pooh, then he is a bear. d) If I win the national competition of secondary schools, then I am admitted to the university.

e) If I have a ticket, I can see the play in the theatre.

2. Write down the converse of the following theorems, and decide whether they are true or false. a) If a number is divisible by 4, then it is divisible by 2. b) If a number is a terminating decimal, then it is a rational number. c) If a triangle is a right-angled triangle, then the square of its longest side is equal to the sum of the squares of the other two sides. d) If at least one of two numbers is 0, then the product of the two numbers is 0. 3. What type of condition is B of A? (Necessary but not sufficient; necessary and sufficient; sufficient but not necessary; not necessary and not sufficient.) B: air; a) A: human life, b) A: I can withdraw money from an ATM, B: I have a debit card; B: Big Ben chimes; c) A: it is noon in London, B: I will pass my graduation exams; d) A: I will be admitted to the university, B: Thomas runs world record e) A: Thomas wins the Olympic Games in the Olympic final of 100 m race. 100 m race, 4. What type of condition is B of A? (Necessary but not sufficient; necessary and sufficient; sufficient but not necessary; not necessary and not sufficient.) B: both numbers are even; a) A: the sum of two numbers is even, b) A: the difference of two numbers B: the two numbers leave the same remainder when divided by 3; is divisible by 3, B: all sides of a quadrilateral are c) A: a quadrilateral is a square, of equal length; B: one of the numbers is even, the other d) A: the product of two numbers is even, one is odd; e) A: a quadrilateral is a parallelogram, B: the diagonals of a quadrilateral are perpendicular to each other 5. When rounding x to the nearest thousand we get 4000. What type of conditions of it are the following? a) 3400 < x < 4550; b) 3890 < x < 4390; c) 3500 < x < 4499; d) 3500 < x < 4500; e) 3999 < x < 4999; f) 3501 < x < 4500. 6. 10 exercises should be solved at an exam; each of them is worth 10 points. The exam is

failed if someone gets less than 40% of the total points, otherwise it is passed. Formulate necessary but not sufficient; sufficient but not necessary; necessary and sufficient; not necessary and not sufficient conditions for the exam to be passed. 7. Find necessary but not sufficient; sufficient but not necessary; necessary and sufficient

conditions in everyday life and in Mathematics. 19

MATHEMATICAL REASONING

8. There is a snail resting on every square of a 7x7 chessboard. For an indication every snail crawls to an adjacent square (adjacent squares have a common side). Is it possible that after this there is again exactly one snail on every square? 9. We want to tile a 7x7 square by 3 x 1 triminoes (without overlaps and gaps) so that we

can leave out 2 squares at most. Find a necessary, a sufficient, a necessary and sufficient condition of tiling. 10. A rectangular forest can be seen in the figure; every point represents a tree. A woodpecker is sitting on the tree A, a robin is sitting on the tree B. Both of the birds fly to the

8

nearest tree in the direction north-south or east-west in every minute. Is it possible that both of them are sitting on the same tree at one time? Find a necessary, a sufficient, a necessary and sufficient condition of the meeting with respect to the initial position of the birds. 11. Can the edges of a cube be numbered by numbers 1, 2, 3, 4, 5, ..., 11, 12 so that we get

the same sums for every vertex when adding up the numbers written on the three edges starting from the vertex? Find a necessary condition for the conductibility of the correct numbering with respect to the numbers written on the edges. Is it also sufficient? Find a sufficient condition.

12. We coloured each of the 36 squares of a 6x6 square yellow or blue. After that we have the right to change the colour of every square in any row or column at once (yellow to blue, blue to yellow). We can repeat this step any number of times. Is it possible after any of the initial situations to colour every square yellow? Try the initial situations shown in the figure.

situation so that if it is satisfied, then following the correct steps it is possible to colour every square yellow. -i

P u z z I e

We want to decide whether the following statement is true or false in the case of each of the four envelopes in the figure:

If an envelope is stuck down, then there is a 40 euro cent stamp on it. At least how many envelopes need to be turned round to decide? (If we can see the front of an envelope, we do not know whether it is stuck down; if we can see the back of it and it is turned down, then it is also stuck down.) V_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ /

20

2. The pigeonhole principle Example 1

There are 2 rabbits and 3 pigeons in the magician’s hat. The magician conjures the animals out of the hat at random. So far he has conjured 3 animals. Which of the following statements are true for sure with respect to the conjured animals? A\ There are two rabbits. B: There are two pigeons. C: There are two animals of the same type. D: There is a pigeon.

Solution A and B can be both true and false, therefore it is not true for sure.

The three animals taken out can be: 2 rabbits and 1 pigeon, 1 rabbit and 2 pigeons, 3 pigeons.

C is true for sure, because there are two types of animals, so it is possible

to conjure two animals so that they are of different types, but the type of the third animal will surely be the same as one of the previous two. D is true for sure, because there are 2 rabbits in the hat, so no matter

how we take out three animals, there needs to be a pigeon among them.

Example 2

There are seven registers for coins in a cash register: for 1, 2, 5,10, 20, 50 euro cents and for 1 euro. At the beginning the cash register is empty; the customers keep paying with coins. a) At least how many coins should be there in the cash register so that there is a register with at least two coins? b) At least how many coins should be there in the cash register so that there is a register with at least 11 coins? Solution (a) If there were 7 coins in the cash register, then it would be still possible to have exactly one coin in each of the 7 registers. But if there were 8 coins, then there must be a register in which there are at least two coins.

Therefore there should be at least 8 coins so that there is at least one register for sure in which there are at least two coins. ♦ ♦ ♦

We can also formulate this way of thinking generally: The pigeonhole principle: We place n objects in k pigeonholes. If n > k, then there will be a pigeonhole for sure in which there are at least two objects.

pigeonhole principle

The statement is obviously true, since if in all the k pigeonholes there was at most one object, then there would be a total of k objects at most. 21

MATHEMATICAL REASONING

Solution (b) If there were 70 coins in the cash register, then it would be still possible to have exactly 10 coins in each of the 7 registers. But if there are 71 coins, then there is a register for sure in which there are at least 11 coins. Since if in all 7 registers there were at most 10 coins, then there would be a total of 70 coins at most. Therefore there should be at least 71 coins so that there is at least one register for sure in which there are at least 11 coins. ♦ ♦ ♦

Based on this we can formulate the pigeonhole principle more generally: We place n objects in k pigeonholes. If n >k~p then there will be a pigeonhole for sure in which there are at least p + 1 objects.

The statement is obviously true, since if in all the k pigeonholes there were at most p objects, then there would be a total of k p objects at most. Note: The English “pigeonhole principle” expression represents the problem well.

Example 3 30 shots are on a square shaped target with 5 dm long sides. Let us show that there are two shots for sure the distance of which is at most 1.5 dm.

Solution With the help of line segments parallel with the sides let us divide the square with the 5 dm long sides into 25 pieces of squares with 1 dm long sides. Among the 30 points there are two for sure which are in the same square with 1 dm long sides. (Figure 17)

5 dm

The diagonal of the small square is V2 < 1.5, thus the distance of the two points in the small square must be less than 1.5 dm.

Example 4 Is it true that there are two people in any company of six so that these two have the same number of acquaintances in this company? (People are acquainted with each other mutually; people are not acquainted with themselves.)

Figure 17

Solution Let us draw a graph: let us assign the points of the graph to the people, and let us connect two points with an edge if the two people corresponding to the points are acquainted. Let us write the number of acquaintances, i.e. the edges starting from the points, next to the points. This number is the degree of the point in the graph. Regarding 22

Figure 18

the acquaintanceships among the 6 people we realise that we can always find two people who have the same number of acquaintances. (Figure 18)

We can realise that nobody has more than 5 acquaintances. The number of acquaintances can be 0, 1, 2, 3, 4, 5. But if someone has 5 acquain­ tances, then he/she knows everyone in the company, so there cannot be a person who does not know anyone. Similarly if someone does not have acquaintances, then there cannot be a person who knows everyone. Therefore 0 and 5 cannot appear at once, i.e. there are 5 possibilities regarding the number of acquaintances. Since there are 6 people, there must be two among them who have the same number of acquaintances.

Example 5 We wrote down five positive whole numbers. Is it true that there are two among them for sure the difference of which is divisible by 4?

Solution The possible remainders when dividing the positive whole numbers by 4 are as follows: 0, 1, 2 or 3. Considering five numbers there must be two among them, which leave the same remainder when divided by 4. If two numbers leave the same remainder when divided by 4, then their difference is divisible by 4. Therefore we can surely find two among the five numbers the difference of which is divisible by 4.

a = 4k + m b = 4/ + m a-b = 4(k-l) U 4\a-b

Example 6 Is it true that we can always choose some (one, two, three or four) numbers out of any four positive whole numbers so that the sum of the chosen numbers is divisible by 4?

Solution Let the four numbers be: a, b, c, d. Let us consider the following four numbers: a, a + b,a + b + c,a + b + c + d.

Let us divide them by 4, and let us examine the remainders; these can be 0, 1, 2 or 3. If there is a 0 remainder among them, then we are ready, because this number is divisible by 4, and since it is the sum of some of the original numbers, we found some numbers the sum of which is divisible by 4.

23

MATHEMATICAL REASONING

If there is no 0 remainder among them, then there are 3 different remainders and 4 numbers, so there must be two numbers which leave the same remainder when divided by 4. When subtracting these two numbers the difference is divisible by 4, and it is also true that it is the sum of some of the original numbers.

Therefore there must be some numbers among the original numbers the sum of which is divisible by 4. Note: Two players write down positive whole numbers in turns on a piece of paper. The player who can find some numbers (possibly even one) - among the numbers on the paper after the opponent wrote down his/her number the sum of which is divisible by 4. Let us play! Who has a winning strategy, and after at most how many steps is the game finished, if the players pay attention and count correctly?

Someone has a winning strategy if he/she can find a step so that he/she wins no matter what the step of the other player is.

Solution: Instead of the numbers let us examine the remainders they leave when divided by 4. Let us represent the steps of the game on a directed graph. A number is denoted by a blue circle, if the starting player wins with it, and it is denoted by a green square, if the second player wins. (Figure 19) starting player

second player

starting player

second player

the remainder of the first number when divided by 4

the remainder of the second number when divided by 4

the remainder of the third number when divided by 4

the remainder of the fourth number when divided by 4

Figure 19 If the first player writes down 0, the second player chooses that number and immediately wins. If the first player writes down 1, and the second player writes down 2, then no matter what the first player writes down, the second player wins. If the first player writes down 2, and whether the second player writes down 1 or 3, he/she wins in any case. If the first player writes down 3, the second player writes down 2, thus the second player wins for sure.

If the two players realise for sure that there are some numbers the sum of which is divisible by 4, then the game will definitely finish in four steps.

................................

24

Example 7 We write down the first 100 positive whole numbers on separate pieces of paper and we put these pieces of paper into a hat. Blindfolded and without replacement we draw 55 numbers from the hat. Is it true that there are surely two numbers among the drawn ones the difference of which is exactly 9?

Solution I Let us list the numbers which leave the same remainder when divided by 9. 0 as a remainder: these are 11 numbers, 9, 18, 27, 36, 45, 54, 63, 72, 81,90, 99 1 as a remainder: these are 12 numbers, 1, 10, 19, 28,37,46, 55, 64, 73,82,91, 100

2 as a remainder: these are 11 numbers, 2, 11, 20, 29, 38, 47, 56, 65, 74, 83, 92 3 as a remainder: these are 11 numbers, 3, 12, ...,93 4 as a remainder: 4, 13, ...,94 these are 11 numbers, 5 as a remainder: these are 11 numbers, 5, 14, ...,95 6 as a remainder: these are 11 numbers, 6, 15, ...,96 7 as a remainder: these are 11 numbers, 7, 16, ...,97 8 as a remainder: these are 11 numbers. 8, 17, ...,98 Since we drew 55 numbers, there are 9 different remainders, and 55 > 6 • 9, thus according to the pigeonhole principle there must be a remainder so that at least 7 numbers leave the same remainder when divided by 9. If this remainder is 1, then we drew 7 out of the 12 numbers leaving 1 as a remainder. And this is only possible if there are two numbers among them which are adjacent ones in the list below, i.e. their difference is exactly 9: 1. 10. 19, 28, 37, 46. 55. 64. 73. 82, 91, 100.

If the 7 numbers leave a different remainder, then we drew 7 out of 11 numbers leaving the same remainder; similarly it is true that there are two among them which are adjacent ones in the list above, i.e. their difference is exactly 9. Therefore there are certainly two numbers among the drawn ones the difference of which is exactly 9.

Solution II Let us denote the drawn numbers as follows: x1; x2, ■■■’x55-

The set A contains the drawn numbers, and the set B contains the numbers 9 greater than the elements of the set A: A: I < x, < x2 < x3 < ... < x55 - 100.

B: 10 < x, + 9 < x2 + 9 < ... < x55 + 9 < 109. The elements of A u B are greater than or equal to 1 and less than or equal to 109, therefore there can be at most 109 types of elements. There are 110 elements in the set A and the set B together, there are at most 109 elements in the set A u B. which means that the set A must have an element which is also in the set B: Xj = Xj + 9, which means that there are two numbers among the drawn ones the difference of which is 9.

25

MATHEMATICAL REASONING

Exercises 1. There are 3 rabbits and 5 pigeons in the magician’s hat. The magician conjures the animals

out of the hat at random. So far he has conjured 5 animals. Which of the following statements are true for sure with respect to the conjured animals?

A: There are two rabbits. B: There are two pigeons. C: There are two animals of the same type. D: There is a pigeon. a) What can we say about the 3 animals that remained in the hat? b) At least how many animals should be drawn so that the statement A would be true

for sure? c) At least how many animals should be drawn so that the statement B would be true

for sure? d) At least how many animals should be drawn so that the statement C would be true

for sure? e) At least how many animals should be drawn so that the statement D would be true for sure? f) At least how many animals should be drawn so that the statement A and the statement B would be true for sure? g) At least how many animals should be drawn so that the statement A or the statement D would be true for sure? 2. Out of the 33 students of the class 15 learn Spanish, 13 learn German and 5 learn French

as a foreign language. About which foreign language group can we say that there are at least two students in the group for sure who were born in the same month? 3. 33 students of a class wrote a language test, and they received grades A to E. Show that

there are 7 students who received the same grade. 4. There are three types of balls in a box: 14 red, 17 blue and 21 green balls. At least how many balls should be drawn without replacement at random so that there would be a) at least two of the same colour; b) at least one of each of the three colours; c) at least two green ones; d) at least two red ones among the drawn balls? 5. There are three types of balls in a box: 4 red, 7 blue and

11 green balls. We have drawn 5 balls without replacement and at random. Which of the following statements are true for sure with respect to the balls left in the box? There A: are no red balls left in the box. B: is a ball of each colour left in the box. C: is a blue ball left in the box. D: are balls of at least two different colours left in the box. 26

6. a) In the mess in Gladys’ drawer there are 5 pairs of different gloves (the glove for the

left hand is different from the one for the right hand). At least how many gloves should she pull out without looking there so that there would be a matching pair for sure among the pulled ones? b) In the mess in Sue’s drawer there are 5 pairs of alike gloves (the glove for the left hand is different from the one for the right hand). At least how many gloves should she pull out without looking there so that there would be a matching pair for sure among the pulled ones? 7. There are four types of apples in a box, equal amount of each type, 100 pieces all together. How many apples should be taken out blindfolded so that there would be at least 10 apples of one of the types for sure? How many apples should be taken out blindfolded so that there would be at least 1 apple of each type for sure? 8. In a dark room there are 12 red and 12 blue socks in a drawer. a) At least how many socks shall I take out of the drawer so that there would be at least two socks of the same colour for sure? b) At least how many socks shall I take out so that there would be two red socks among them for sure? 9. Let us assume that there are a number of blue and

a number of red socks in a drawer, and we know that I have to take out at least as many socks to have a pair of socks of the same colour for sure as many at minimum to have two socks of different colours. How many socks are there in the drawer?

10. Let us write “1” on one disc, “2” on two discs, “3” on three discs and so on, “30” on thirty discs. Let us put the discs in a box, and then let us draw discs at random without replacing them. At least how many discs should be drawn so that there would be at least 10 discs for sure which have the same number written on? 11. 100 people attend at a ball; each of them has blond or brown hair. They settle down in the four rooms as they like. We are looking for the largest n for which we can definitely find

a room in which there are n people with the same hair colour. 12. 10 shots are on a square shaped target with 6 dm long sides. Show that there are two shots

for sure the distance of which is at most 3 dm. 13. 7 shots are on a regular hexagon shaped target with 40 cm long sides. Show that there are

at least two among them the distance of which is not greater than 40 cm.

14. Place 30 points in a square with 1-unit-long sides. Is it true that we can always find 4 points which can be covered by a disc with 0.25-unit-long radius? (It is ok if a part of the disc is outside the square.) 15. 18 straight lines are given in the plane so that there are no parallel ones among them. Is it true

that there always are two among them so that the angle included between them is at most 10°? 16. A rectangle is intersected by 13 straight lines so that each straight line divides the rectangle

into two quadrilaterals the ratio of the areas of which is 1:5. Show that there are 4 straight lines which intersect each other at one point. 17. 100 flies are flying around ina5mx6mx3 room. Is it true that there always are two

among them so that the distance between them is less than 2 m?

27

18. We randomly choose 2001 points inside a cube with 4-unit-long sides. Show that we can

choose 32 points out of these so that no matter in which order we connect the 32 points by creating a closed polygon, the perimeter of this polygon cannot be greater than 32 ■ V3. 19. Some people shake hands in a company of 10. Then everyone is asked how many people

he/she shook hands with. Are there always two people who shook hands with the same number of people in the company? 20. 8 teams were playing in a championship; each team played against all the other teams once.

During the championship it was counted how many matches each of the teams had played till then, and it was stated that each of the teams had played different numbers of matches. Was this statement true? 21. At least how many whole numbers should be written down so that there would be two

among them for sure the difference of which is divisible by 8? 22. a) Is it true that if we write down 15 consecutive whole numbers, then there must be two

among them the sum of which is divisible by 15? b) Is it true that if we write down 15 positive whole numbers, then there must be two among them the sum of which is divisible by 15? c) At most how many whole numbers can be written down so that neither the sum nor the difference of any two is divisible by 15? 23. In the expression 1-2-3-4-5-6-7-...- 2001 we can give the order of operations

by appropriately placing pairs of parentheses. Prove that there are such places for the parentheses which give the same result. 24. Is it true that we can always choose some (one, two, three, four or five) numbers out of any

five positive whole numbers so that the sum of the chosen numbers is divisible by 5? Which number can be written instead of 5? 25. Two players write down positive whole numbers in turns on a piece of paper. The player,

who can find some numbers - among the numbers on the paper after the opponent write down his/her number - the sum of which is divisible by 5, wins.

Let us play! Who has a winning strategy, and after at most how many steps is the game finished, if the players pay attention and count correctly? 26. Is there a natural number starting with 2001 which is divisible by 17? 27. Is there a natural number consisting only of digits 0 and 1 which is divisible by 17? 28. We write down the first 100 positive whole numbers on separate pieces of paper and we put

these pieces of paper into a hat. Blindfolded and without replacement we draw 55 numbers from the hat. Is it true that there are surely two numbers among the drawn ones the difference of which is exactly 10? Examine the problem for 11 and then also for 12 instead of 10. Puzzle The seven volumes of an encyclopaedia are on a shelf in the following order: 1, 5, 6, 2, 4, 3, 7. How can these be put into ascending order if we can grasp three volumes at once and these can moved at once to the left or to the right of the others or between any two but without changing the order of the three? (We can obviously perform this action several times.) X___________________________________________________________________________________________________________________________________________________ /

28

3. Arrangement (ordering) problems Arrangement (ordering) of distinct objects Example 1

Andrew, Bruce, Chris, Denis and Evan arrive to a party. They are waiting in front of the door to decide in which order they should enter. In how many different orders can they enter the room if one person can step in at once? Solution We can select the first person to enter in 5 different ways. Since four people entering as second can be selected to each person entering as first, the person entering as second can be selected from only the 4 people still standing in front of the door, i.e. in 4 different ways, thus the people entering as first and as second can be selected in 5 • 4 different ways. Since the person entering as third can be selected from only the 3 people still standing in front of the door, i.e. in 3 different ways, thus the people entering as first, second and third can be selected in 5 • 4 • 3 different ways.

Since the person entering as fourth can be selected from only the 2 people still standing in front of the door, i.e. in 2 different ways, thus the people entering as first, second, third and fourth can be selected in 5 ■ 4 • 3 • 2 different ways. The person entering as fifth can only be the 1 person still standing in front of the door, thus all the possible entering orders of the five people are: 5 • 4 ■ 3 • 2 ■ 1 = 120.

Example 2

Andrew, Bruce, Chris, Denis and Evan sit down at a round table in a party. We only concentrate on the neighbours to the right and to the left. For example the arrangements in figure 20 are considered the same. There is no special place, e.g. for an honoured guest, at the table. In how many different ways can the guests sit down? Solution I We make Andrew sit. This way we have determined his place at the “rotatable” table (after this the table cannot be rotated anymore, as if it were a straight table). Thus we can make the remaining 4 people sit in 4 • 3 ■ 2 • 1 different ways. Therefore 5 people can sit down at a round table in 4 • 3 • 2 • 1 = 24 different ways.

Solution II If we are thinking so that the first person can be seated in 5 different ways, then the others can be seated in 4 • 3 • 2 • 1 different ways; there would be a total of 5 • 4 • 3 • 2 • 1 possibilities.

Figure 20

29

However among those arrangements, where everyone sits to the place to his neighbour to the right, give the same order of seating. They can do this 5 times in a row before they get back to their initial places, i.e. every seating arrangement appears 5 times among the 5 • 4 ■ 3 • 2 • 1 possibilities.

Thus the number of distinct seating arrangements is: 5

The number of arrangements, if there are like objects Example 3 How many five-letter “words” can be made from the set of letters a, a, a, m and m?

Solution If we write the like letters in different forms, for example A, a, a, M, m, then we are looking for all the possible arrangements of 5 distinct signs which, based on the above, is 5 ■ 4 • 3 ■ 2 ■ 1. Since these three types of the letter “a” represent the same letter in the word, the cases, which differ in the different orders of the distinct letters “a”, are not considered as different cases. The letters “a” can be exchanged in 3 • 2 • 1 different ways. Regarding the arrangement of the letters “a” these 3 • 2 ■ 1 = 6

cases count as 1, if the letters “a” are alike. Thus only — of the so far 6 counted cases mean actually distinct cases with respect to the letters “a”. The same way the different cases of the arrangements of the letters “m” are not distinct either, they can be exchanged in 2 • 1 different ways.

Thus only yj of the so far counted cases mean actually distinct cases with respect to the letters “m”.

Let us also count the possibilities with the help of the graph representation!

Therefore the number of five-letter words made from the letters a, a, a, m and m is: (3-2-l)-(21)

Exercises 1. 4 married couples arrive at a party. a) In how many different orders can they enter the room if they arrive at once? b) In how many different orders can they enter the room if every husband steps in right after his wife? c) In how many different ways can they sit down at a round table? d) In how many different ways can they sit down at a round table if every husband sits next to his wife? 30

2. How many different tunes with 7 notes can be produced by the computer program which

plays notes C, C, G, G, G, A, A in all the possible orders? (All the notes are of equal length.) 3. Which one is greater: the number of the seven-letter words that can be created from the

letter set of (a, a, a, a, b, b, b) or the number of the six-letter words that can be created from this same letter set? 4. There are 3 green and 3 blue holders in a set of soft-boiled egg holders.

These can be strung on a rod either with the bottom up or down. In how many different orders can these holders be strung on a rod?

5. A boot can be laced up in many ways. A boot and a lacing can be seen in the figure, but the bootlace can be laced inside the boot in several ways. In how many different ways? 6. Four cars each have a number plate removed.

a) In how many different ways can the number plates be put back? b) In how many different ways can the four number plates be put back so that exactly one will be in the correct place? c) In how many different ways can the four number plates be put back so that exactly three will be in the correct place? 7. A secretary wrote 5 letters and also addressed 5 envelopes for those who the letters were for, but she did not have time to put the letters in the envelopes. The office assistant put the letters in the envelopes at random. a) In how many different ways could he put the letters in the envelopes? b) In how many different ways can the letters be put in the envelopes so that exactly one letter will be in the correct place? c) In how many different ways can the letters be put in the envelopes so that exactly three letters will be in the correct place? 8. How many necklaces can we string using 3 white and 5 red beads if same coloured beads are alike? Andrew’s solution: First we plan the places of the white beads, then we put the red ones

on the three arcs defined by the white ones. There are 5 possibilities:

0 + 0 +5; 0 + 1 +4; 0 + 2 + 3; 1 + 1+3; 1 + 2 +2. Therefore 5 different necklaces can be strung. Bruce’s solution: 8 different beads can be strung on

a round necklace in7-6-5-4-3-21 different ways. Now there are 5 red like beads and 3 white like beads, so it should be divided by 5 • 4 ■ 3 ■ 2 ■ 1 and also by 3 ■ 2 • 1, thus the result will be 7 possibilities. Therefore 7 different necklaces can be strung.

Whose solution is correct? 9. In how many different orders can 5 basketball players with different heights enter the court

if they enter one after the other and none of them can go between two taller players? 31

MATHEMATICAL REASONING

4. Picking problems Now we have to pick some out of several objects where the order is also important.

Picking some out of different objects Example 1

Ann, Barbara, Cissy and Dana take part in a prize draw. One book and one pen will be drawn so that everyone can win at most one prize. They put papers with their names into a hat, and one name will be drawn; she will be the winner of the book. Then another one will be drawn from the papers left in the hat; she will be the winner of the pen. How many results can the draw have? Solution The first name can be drawn in 4 different ways, i.e. 4 people can win the book.

Since the paper with the name of the first winner is not replaced in the hat, the second name can be drawn in only 3 different ways, e. 3 people can win the pen. i. Since 3 possible second winners belong to each of the 4 possible first winners, there are a total of 4 ■ 3 = 12 possibilities.

Picking some out of different types of objects (there are several objects of each type) Example 2 Andrew, Bruce, Chris and Denis take part in a prize draw. One book and one pen will be drawn so that everyone can win more prizes. They put papers with their names into a hat, and one name will be drawn; he will be the winner of the book. Then they replace the paper with the name of the winner into the hat, and one paper will be drawn again; he will be the winner of the pen. How many results can the draw have?

Solution The first name can be drawn in 4 different ways, i.e. 4 people can win the book.

Since the paper with the name of the first winner is replaced in the hat, the second name can be drawn in 4 different ways too, i.e. 4 people can win the pen. Since 4 possible second winners belong to each of the 4 possible first winners, there are a total of 4 • 4 = 42 = 16 possibilities.

32

Exercises 1. How many flags similar to the Hungarian flag (with three stripes) can be created using five

different colours if every colour is used at most once? 2. How many flags similar to the Hungarian flag (with three stripes) can be created using five

different colours if each colour can be used many times but no two adjacent colours can be the same? 3. We wrote down digits 1,2,3,4,5,6,7, 8,9 on separate cards and we put these cards into a hat.

a) In how many different ways can we draw 5 cards in a row if after the draw the cards are not replaced in the hat? b) In how many different ways can we draw 5 cards in a row if after each draw the cards

are replaced in the hat?

4. We roll a die four times in a row. a) In how many cases are there no two equal numbers among the rolled ones? b) How many times shall we roll the die to get two equal numbers for sure?

5. Andrew, Bruce, Chris, Dave, Evan and Frank take part in a draw where 3 prizes will be

handed out: a book, a pen and a ball. In how many different ways can it be done if a) one person can get only one prize; b) one person can get several prizes?

6. Using the digits 1, 2, 3, 4 we create four-digit numbers where every digit can be used

several times. a) How many different numbers can be created? b) In how many of these numbers are the first and the last digits equal?

c) In how many of these numbers is the last digit not greater than the first one? d) How many of these numbers are greater than 2000? e) How many of these numbers are divisible by 4?

7. We roll a die three times in a row; the rolled numbers in order are the digits of a three-digit number. a) How many three-digit numbers can be rolled this way? b) How many even three-digit numbers can be rolled this way?

c) How many three-digit numbers divisible by 25 can be rolled this way?

8. In the one-column football pool blank 1, 2 or X is written in 13 + 1 places. In how many different ways can this football pool blank be filled in? 9. A company called TOPIC would like to colour its name so that each letter is uni-coloured. At least how many colours are needed if they want to have a differently coloured nameplate each day of the year? 10. How many eight-digit numbers can be created using only digits 3 and 8 in which both digits appear?

MATHEMATICAL REASONING

11. How many number plates can there be in Hungary if the only restriction is that it should

consist of three letters (of the English alphabet) and three digits? 12. How many positive whole numbers with at most 6 digits are there containing the digit 1 ?

13. What can the following fractions be equal to? a) 2

b) 2

2 2

2 2 2

14. The letters of the Morse alphabet consist of signs ■ and

Could an alphabet with 44 letters

be written using Morse codes no longer than 3 signs? 15. How many numbers are there between — and — which can be shown on the display 2 8 of an 8-digit calculator? (The leading 0 is not displayed before the decimal point; however the decimal point is displayed.) 16. How many four-digit natural numbers can there be a) in number system with base 2; b) in number system with base 5; c) in number system with base 12? 17. In the 1940s the Bell Telephone Company was planning ten-digit phone numbers so that

the first three digits were the code of the region. The first digit could be between 2 and 9, the second one could be 0 or 1, and the third one could be anything apart from 0. How many different regional codes were possible? P u z zTe There are 8 glasses next to each other on a table. The first four are empty, and there is orange juice in the last four. How could we manage to have an empty and a full glass next to each other alternately if we are allowed to touch only two glasses? k____________________________________________________________________________ __ _________________ ___________________ 7

34

Computing the root From the ancient times till today everyday life has brought up more and more complex problems for the people dealing with calculations. The “mathematicians” made up new methods and concepts while solving the exercises that arose.

1. Rational numbers, irrational numbers Definition: The numbers that can be written as a fraction of two whole numbers are called rational numbers.

rational number

gZ;^ *0}

2|g2 => 2|g.

We get that bothp and q are even. It contradicts our assumption that they are relative primes, therefore the statement is true. Note: With the methods above it can be proven that for every positive whole number a, which is not a square number, Va is an irrational number.

38

Definition: The union of the set of rational numbers and the set of irrational numbers is the set of real numbers.

the set of real numbers

The symbol of the set of real numbers is R, and the set of irrational numbers is usually denoted by Q* (or Qc or I). Using these notations: R = Qu Q*. The place of certain irrational numbers can also be constructed on the number line.

For example v'T can be constructed by using the Pythagorean theorem as sketched in figure 4.

At the same time there are irrational numbers, for example n, which cannot be constructed.

Exercises 1. Give the following numbers as decimals.

2. Rewrite the following numbers as the quotient of two whole numbers.

a) 3.142;

Z?) 3.142;

c) 3.142;

d) 3.142.

3. Prove that the following numbers are irrational.

a) Jl;

b)>/2 + l-,

c)a/3-1;

d)j2+jl.

4. If the unit is known, construct line segments with the following length:

b) ViO;

a)

c) n/17 ;

d) V1956.

5. Give an irrational number which is greater than 0.99, but less than 1.

p u z 11 e

The quotient of which two whole numbers can be the recurring decimal 1.9? \ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ /

39

COMPUTING THE ROOT

2. The identities (laws) of the square root In year 9 in connection with the functions we defined the square root of non-negative numbers.

Definition: If a > 0, then Va (the square root of a) denotes the

square root

non-negative number the square of which is a.

(x/ö)2=o, if a > 0

Similarly to exponentiation, when taking the square root of something there are certain arithmetic rules, identities, which apply. Identity I

Theorem:

The square root of a product is equal to the product of the square roots of the factors.

Proof Since in the set of non-negative numbers squaring and taking the square root are one-to-one, we are allowed to compare the squares of the expressions at the two sides of the equality.

Nicolas Chuquet French physician wrote Vl4 as follows in his work “The Science of Numbers in Three Parts” published in 1484: R214.

definition

(ÆTfVi.;

exponentiation identity III

definition

(Æ.Æ)21(Æ)2(Æ)21Oi>.

Since the squares of the two expressions are equal, the original expressions are equal too.

Example 1 Calculate the value of the product V3 ■ VÏ2.

Solution Let us rewrite the product with the help of one radical sign according to identity I:

40

Theorem:

The square root of a fraction is equal to the quotient of the square root of the numerator and the square root of the denominator.

I~a _ Ja

JTjT

if a > 0 and b > 0.

Proof Let us compare the squares of the non-negative expressions at the two sides exponentiation of the equality. identity IV

definition

l

Hl.

I

a I

fP

definition

2

2

' JeJ I (xtf (x/h)2

a ~b

Since both expressions are non-negative, and their squares are equal, therefore the expressions are equal too.

Example 2

Calculate the value of the quotient

_ /12 V3

.

Solution Based on identity II the expression can be rewritten as follows:

J\2 _ VT = Identity III

Theorem:

The operation of taking the square root and exponentiation are interchangeable.

(Va) = x/a^, where a > 0 and k e Z.

a > 0; ke Z

Proof In this case it is again worth comparing the squares of the positive expressions at the two sides of the equality. exponentiation identity V

exponentiation identity V

definition

definition

[(Æ)‘ V (æH [(Æ)2 ]‘L‘ ; (T7)2V‘. Both sides are positive, and their squares are equal, therefore the expressions are equal too.

Example 3 Which of the following is greater: (V3) or (Vö) ? Solution _ Since (x/3) = x/3^ = V243 and (5/5) = x/i^ = Vl25, and the function Vx (x > 0) is strictly increasing, (V3) is greater.

41

Example 4 Let us perform the following operations.

^(73 + 75)-(273-75); 6)(>/7+4)2;

c) (Til-76)-(TH+ 76). Solution a) Let us multiply each term of the first sum by the terms of the second sum, and let us apply identity I: (V3 + V5) • (2V3 - V5) = >/3 • 2V3 - V3 • V5 + V5 • 2a/3 - (V5 )2= = 2-3-a/15+2V15-5 = 1+V15. b) Let us apply the identity regarding the square of a two-term sum:

(a + b)2 = a2 + 2ab + b2

(V7 + 4)2= (V7 )2+ 2 • V7 • 4 + 42 = 7 + 8 • >/7 +16 = 23 + 877. c) Since (a — b)-(a + b) = a2 - b2. therefore

(VH - V6) • (ViT+V6) = (ViT)2-(V6 )2= 11 - 6 = 5.

(a-b)(a + b) = a2-b2

Example 5 Let us calculate the values of the following expressions. a) 77l3-2• 77l3+2;

b) (^8 + J\5-78-715) .

Solution a) Let us carry out the multiplication under one square root sign

according to identity I: >/713-2 • VV13 + 2 = V(V13-2)(V13 +2) =

= a/(Vb)2 -22

= 5/13-4 =5/9=3.

b) Let us perform the squaring, and then let us write the double of the

product under one square root sign:

(>/8 + 715 -5/8-5/15)’= (0-b)2 = a2-2ab + b2

= 8 + 5/i5-2-A/8 + Vi5-5/8-5/15+8-715 = = 16-2-5/(8+ 5/15)-(8-a/15) = = 16-2 ■ Vs2-(T15)2 = = 16-2-764-15 = 16-14 = 2.

42

Exercises 1. Perform the following operations. d) 7^-75;

b) 750-72;

g)y[l-(732 - 78);

, t >/3-(V6 + V24) /0---------- 7=------->/2

2. Which number is greater: a) V3->/5 or

b) (5/3)

or 7Ï4-72;

^>/85/2?

V12

5/2

(V2)3 735

76

TÎ5

3. Perform the following operations. a) (2-V3 + l)-(3-4-V3);

b) (73 + 2-72-l)-(3-5/2-73);

c) (2-V5-3-V2)2;

d) (3-77+75 )2;

e) (3-V5 + 2-V3)-(3-V5-2->/3);

f) (T2 + T3)3.

4. Calculate the values of the following expressions.

Puzzle Which number is greater: ^3-72-72 or 73-73 ?

43

COMPUTING THE ROOT

3. Applying the identities (laws) of the square root Writing expressions under one radical sign Example 1 Which number is greater: a) 3- V5 or 5-V2;

b) - or—-? 3 4

Solution (a) Let us rewrite each expression.

The number on the left: 3-V5 =79-75=745, the number on the right:

5 • VT = VTJ ■ VT = a/50.

In the second case there is a greater number under the radical sign. Since the function 44 is strictly increasing, this expression is the greater one.

Solution (b) Let us rewrite each expression. The number on the left: 44_^44_ = l6= [2

3 -V9

V9

\3’

the one on the right:

Vio _ Vio _ To _ 4

" V16

V 16

if N8'

e2 164 5 15 K Since — = — and - = —, the number on the left is the greater one. 3 24 8 24 6

Example 2 Write the following expressions in a simpler form by writing factors under the square root sign. a) a• 4a, a>0; b) a,b > 0; c) (l-7io)-,|^ + 1.

\V10-1

a \b

Solution a) a-4a = 4a2 ■ 4a = 4a2 - a - 444

b

14 _ I42~

a yb

44

\ a2

14 _ ib~ - a _ 14

\b

ya2-b

ya

c) Since 1 - VTÔ < 0:

(i-Vw)-

h

_

5/U10

a2

= -G/w-1). IVïô+i _

1)

|(Vki-i)2-(Vio + i) 7__1

= -7(VlÖ-l)-(ViÖ + l) = -Vio-l = -y/9 = -3.

Factoring out from under the square root sign If there is the square of a number under the square root sign, then it is easy to perform the operation, since the following is derived from the definition: yfa2 =\a\. If after factorisation one of the factors of the number under the square root sign is a square number, then it is worth factoring it out from under the square root sign.

Example 3 Let us carry out the following operations: Vl8 + V8 - V50.

Solution Let us factorise the numbers so that one of the factors is a square number, and then let us apply identity I. Vf8 + V8 - a/50 = 79^2 + 74^2 - V25^2 = 3 • a/2 + 2 ■ V2 - 5 • V2 = 0.

Example 4 80 40 20 10 5 1

Let us calculate the following product: (V8Ö + a/50-V20-V8)-(a/Ï25-745-VÏ8).

Solution Let us find square numbers under the radical sign. (V80 + a/50 - V20 - V8) ■ (V125 - V45 - V18) =

2 2 2 2 5

80 = 24 5 = (22)2 ■ 5

= (V16-5+a/25-2-V4^5-V4^2)-(V25-5->/9^5-V9^2). By applying the identity regarding the square root of a product and the definition we get that (4-a/5+5-V2-2-V5-2-V2)-(5-V5-3->/5-3-V2) =

(0 + b) • (0 - b) = a2 - b2

= (2.75+3 V2) (2-75-3.V2) = (2.V5)2-(3.M =

= 4-5-9-2 = 20-18 = 2. 45

COMPUTING THE ROOT

Example 5 Let us perform the following addition:

Solution The expression has a meaning only if a > 0. Let us factorise under the square root signs: + J4a3 + x/9a3 = Xa2 - a + J(2a)~ a + ^(.3a)~ -a =

= | a I • 4a +1 2a I • 4a +13a I • 4a.

Since a > 0, therefore I a I = a, therefore after the addition the expression looks as follows: a • 4a + 2a ■ 4a + 3a • 4a = 6a ■ 4a.

Rationalising the denominator of a fraction Example 6 Let us expand the following fractions so that there will be no square root sign in the denominator. . 3 m 7 i 2 34 a) -¡=-, b) —¡=; c)----- 7=; d) -¡=------- . a/7 3-V5 3 + V2 V3-2-V5 Solution a) It is suitable if we multiply through the fraction by 4Ï : 3

4Ï _3-4ï

4d 4i~

7

b) In this case multiplying through by 4Ï is effective:

7

45 _ 1-45 _ 7-45

3-4Î' 45 ~

3-5

~

15

c) In the case of this fraction multiplying through by 42 is not

effective any more.

As there is a sum in the denominator, therefore by thinking of the identity (a + b) ■ (a - b) = a2 - b2, let us expand by (3- V2): 2

3-42 _ 2X3-42} _

_ 2X3-42) = 2X3-42)

9-2

”

7

d) Similarly to the previous one when multiplying through by

(V3 + 2-V5) the result is: V3 + 2-V5 _ 34-(>/3 + 2->/5) _

34

V3-2-V5'V3 + 2-V5_(^)2_(2.^)2 “

34-(V3 + 2-V5) 3-4-5 After rationalising the denominator, the fraction could be simplified. We should not be surprised that the result is negative, since in the original denominator a/3 < 2 • a/5 = a/20.

Example 7

Which number is greater:

----- 7= or -r=----- 7=?

V7 + V2

VÏÔ + a/2

Solution Let us rationalise the denominator of both fractions: 5___ V7-V2 _ 5-(V7-V2) V7+V2'V7-V2 ’ (V7)2_(V^)2 _ 5-(V7-V2) = 5-(V7-V2) =

7-2 8

5

8-(V10->/2)

V10 -V2 =

Tio+Ti'Tio-Ti " (vn>)2_(v?): " = 8(710-72) = 8 (710-72) = 10-2

K

8

Since the square root function is strictly increasing on the set of non­ negative real numbers, therefore V10 > V7, and thus the second fraction is greater.

47

COMPUTING THE ROOT

Example 8 Let us perform the following addition: 3

1

|

y[a +1

Vô-1 Solution

The expression has a meaning only if a > 0 and a + 1. Let us rationalise the denominators: 3 a/a +1 a/a

3 • a/a + 3

-1 a/a + 1

a—1

and 1 a/a

+1

a/a

-1 _ a/a -1

a/a

—1

A—1

By performing the addition: 3-a/a +3

a/a-1

A-l

A —1

_ 4-a/a + 2

A-l

Example 9

i i . i ... 4-a/o-6 16-a/o Let us calculate the value of the expression ----- =— + —=—, 3 2-VÔ + 1 2-a/o-1 if a = -. . .

Solution Before the substitution let us simplify the expression by rationalising the denominators: 4-a/a-6^ 16-a/a _ 2 • a/a + 1

2 • a/a -1

4-a/a—6 2-a/a—1^ 16-a/a (2a/Ô + 1)-(2-a/Ô-1)=4o-1

2-a/a + 1 2-a/a-1

2-a/a + 1

2-a/a-1 2-a/a + 1

(4-a/a-ô)-(2-a/a-1)

16-a/a-(2-a/a + 1)

4a-1

4a-1

8a-4• a/a -12• a/a+6 +32a +16• a/a _ 40a + 6 ”

4a-1

Then by substituting: 40a + 6

4a-1 ■■■■■■■■■............

48

40 - - + 6 4

30 + 6 3-1

= 18.

4a-1 ’

Exercises 1. Which number is greater: b) 4-V3 or 5-V2;

a) 6-\Î2 or 3-VS;

c) 2-Ji or 3-V3;

3-V5 7-V2n /) —or 6 ’ 2. Write the following expressions into a simpler form by writing factors under the square root sign.

v V2? V15 — or—;

]

I—

¿2“ Ib b)----- . —, where a, b > 0; c) (a + b)b \ cr

a)where a>0; a

, where a + b>0;

3.

4. Rationalise the denominators of the following fractions.

,2 ,

.

V3-V2

J)

,

V2-5-V3 3-

,,

,14

,,5

W3-Æ'

al ST

2-V5-V3 . r~ /— ’ 5-V3 + V5

, 8)

33

.

S-Æ+JT

;Æ-3’ V^-3 I 2-Va+4

5. Compare the following fractions: which one is greater?

6_________ 8 VÏÏ + V5’ s/b + VT

6__________ 8 2-V3-V6’ VÏ4-V6’

4____________ 6 C 2-V2 + 2-V3’ 2-V3 + 3-V2'

6. Perform the following operations.

4-V3-1

2-V3 + 4

a----- 7=------------ t=—;

>/3

V3

a/2+3

s[a + 3 s/a + 12- \[a +15 _ d) —j=---------- 7=------- 1---------------, where a > 0, a s/a+2 ya-2 a-4

,>/x+2 a/x-1

,>/7-1

s/x

dx + \

s/x+l

s/x+3

>/7-2

V7 + 1

, 4;

,

s/x-3

7. Calculate the substitution value of the expressions, if y = y.

,

1

1

s/y+2

2-yJy

,,

3

c)—j=------ + -7=--------- 7=—;

V5

V2

>/7-2

,>/7 + 1

6-V5-1

b)—j=^ +----- 3=—;

\/y + 2

Jÿ-2

3--/y + l

3-sjy-i

49

V7-1

COMPUTING THE ROOT

4. The /7th root of numbers We could define the square root of the non-negative numbers, because every non-negative number can be generated as the square of another non-negative number. (Figure 5)

Let us examine the function f: R—> R, f(x) = x}. This function is a strictly increasing odd function defined on the set of real numbers. Its range is the set of real numbers. (Figure 6)

If we reflect the graph of this function about the straight line y = x, then we get the graph of the inverse of the function f(x) = x3.

To every real number a the function resulting this way assigns the real number the third power (cube) of which is a.

Similarly to the square root this number is called the third root (cube root) of the number a, and it is denoted by y/a.

Definition: The cube root of the real number a is the real number the third power of which is a.

(y/a) = a, a e R. 23 = 8, therefore ^8=2.

Based on the above we can define the root in a general form too, but we have to distinguish the cases of even and odd roots. In the case of even roots the definition is similar to the definition of the square root.

(-3)3 = -27, therefore 3/^27 = -3.

50

Definition: The 2&,h (k e N+l root of the non-negative number a is the non-negative number the 2fc,h power of which is a.

(23/u) =a, if «>0. For example 3/16 = 2 because 24 - 16; ^729 = 3, because 36 = 729.

Definition: The (2k + l)th (k e N+) root of the real number a is the real number the (2k + l)th power of which is a. a>0

(2k^)2k+i=a.

For example: >¡8 = 2, because 23 = 8 and (-3)5 = -243.

v-243 = -3, because

Christoff Rudolff was a Czech mathematician who was working in Vienna. For the first time he used the sign V for the square root, the sign wJ for the cube root, and the sign W for the fourth root in his work published in 1525.

In general in the case of y/a, n is called the index.

In the case of the square root the index is usually not written. The difference between the definitions regarding the even and the odd index results in 2V^r=i«i and 2k+J/a2k+i = a For example: 3/(-3)7 = -3;

3/(-3)8 =3, but or

4l/(_5)41

but

Example 1

Let us calculate the results of the following root computations.

a) 3/i;

b) O;

c) 3/125;

e) 3/-100 000;

f) 3/0;

g) 3H.

d) 3/128;

Solution a) 1 - l9, therefore 3/1=1;

b) -8 = (-2)3, therefore 3^-8 =-2;

c) 125 = 53, therefore 3/125=5; d) 128 = 27. therefore 3/128 = 2; e) -100000 = (-10)5, therefore 3/-100000 =-10;

0 = 07, therefore 3/0=0; g) -1 is a negative number, therefore 3/-1 does not have a meaning.

51

COMPUTING THE ROOT

The mathematicians of China in the 13th century could compute the root of any arbitrary number by solving the equation xn = a.

Example 2

Let us compute the roots.

b) Vô3 ;

a) tfa4;

c)

d) 1Vo3°

Solution a) \la4 has a meaning for every real number, y/a4 = | a I. b) yjld also has a meaning for every real number, y/ci3 - a.

c)

also has a meaning for every real number. Since a12 = (a2)6,

therefore Va12 = I a21 = a2. d) '^1a30 has a meaning for every real number. Since a30 = (a3)10,

therefore 'Va30 = I a31. e)

has a meaning if a > 0. Since ¿zl9 = (W) , therefore 3V>=W.

f) \J-a4 has a meaning if -a4 > 0, and it is only true if a = 0, thus

3/0 = 0.

Exercises 1. Calculate the results of the following root computations.

g) V=TÎ2;

e) V-125;

h) 3/1000;

i) V-216.

2. Perform the following root computations. c) %J(-c)5-,

a)

g) i)

Puzzle Express the following numbers using the number 4 at most four times with operators (exponentiation and roots are allowed).

10; 14; 18; 22; 32. \ >

52

5. The identities (laws) of the nth root While proving the following important theorems (identities) we are using that, similarly to the functions x i-> x2, x H> 4x, x i-> x3 and x x/x, the functions x t-> xn and x i-> 4x are also strictly monotonically

increasing on the set of non-negative numbers (Figure 7). Identity I

Theorem:

The nth root of a product is equal to the product of the nth roots of the factors.

4cTb = 4a • 4b

\Ja • b = 4a • 4b, where a, b > 0; n e IN; n > 2.

a,b>0; nelN; n>2

Proof Because of the strict monotonicity of the previously mentioned functions we can compare the nth powers of the expressions at the two sides of the equality. exponentiation identity III

definition

definition

Since the two powers are equal, therefore the original ones are equal too.

Example 1 Let us calculate the value of the product ^/8-^4. Figure 7

Solution Based on identity I: 4% ■ ^4 = \8 • 4 = x/32 = 2. Theorem:

Identity II

The nth root of a fraction is equal to the quotient of the nth roots of the numerator and the denominator. 4b

, where a > 0; b > 0; n e IN; n > 2.

\b 4b o>0; jb>0; nelN; n>2

Proof Let us raise both sides to the power of n again. exponentiation identity IV

definition

definition

Since the two powers are equal, therefore the statement is true.

Example 2 Let us calculate the value of the fraction

Solution

— 3/48

Based on identity II: -^=-

V 3

53

COMPUTING THE ROOT

Theorem:

The operation of taking the root and exponentiation are interchangeable.

'\Jak =

, where A 6 Z; a > 0; n 6 IN; n > 2.

Proof Let us raise both sides to the power of n again. definition

exponentiation identity V

exponentiation identity V

definition

Since the two powers are equal, the original expressions are equal too.

Example 3 Let us calculate the value of the power (^/3) . Solution Let us apply identity III: (^3) =

= 32 = 9.

Example 4 Let us calculate the exact value of the product \/6-V20 ■ ^6 +V20. Solution Let us write the product under one radical sign; then rewrite the special product that results:

^6-V20 • ^6 + ^20 = ^(6-V20)-(6 + a/20) = = ^62-(a/20)2 = ^/36-20 = \/16 = 2.

54

Writing expressions under one radical sign

writing expressions under one radical sign

The operation of writing expressions under one radical sign can be done similarly to writing expressions under on square root sign.

arfb = y]an -b o,b>0;/?eN;z?>2

Example 5 Which one is greater: 3-3/5 or 2-3/17 ?

Solution After raising the factors in front of the radical to the power of three let us write them under the radical sign:

3• 3/5 = 3/33 ■ 3/5 = ^3^5 = 3/B5, in the case of the other number

2-3/17 = 3/2^ • 3/17 = 3/23 • 17 = 3/136 Therefore the second number is the greater one.

Factoring out from under the radical sign The operation of factoring out from under the radical sign can also be done the known way.

factoring out from under the radical sign

Example 6 In the following expressions let us factor out as many factors as possible from under the radical sign.

Solution a) Let us form fourth powers under the radical signs and then let us take their roots:

16 = 24

3/32+4/162 = 3/16-2 + 3/81-2 =2-4/24-3-4/2 = 5-^/2.

81 = 34

b) Because of the fifth root the expression has a meaning for every real a. Let us form fifth powers under the radical signs:

si32a6 - sja11 = ^¡25a5a - s]al0a =

= 2a-sfa-a2-%/a.

We cannot combine, at best we can factor out: 2a ■ sla -a2 -s[a = \/a ■ (2a - a2) = a • 3/a • (2 - a). c) The expression has a meaning for every real a and b > 0. Let us

form the suitable powers under the radical signs:

3^F_3/^F+3/^ = = sla6b6a2b — 3/a4Z?4a2Z?3 + sja5b15a2b4 =

=IaI•

• s]a2b-1 a I • b ■ sla2b2 + ah3 • sla2b4.

55

Rationalising the denominator

rationalising the denominator

The operation of rationalising the denominator of a fraction can be done similarly to rationalising the denominator in the case of computing the square root. Because of the complexity of the problem we only concentrate on the case of monomial (one-term) denominators.

Example 7

12 23 Let us rationalise the denominators of the fractions: a) -^=; b) —77==. ^3 2-^2 Solution We have to expand the expressions so that we can easily compute the roots in the denominators. . T t u 3/^ 12 12-^9 12 ^/9 . 3/T a) Let us expand by v3. -=••-?= =—=--------- = 4-v9. 3 b) The expression has a meaning if a

by

0. Let us expand the fraction

:

Further identities (laws) of the /7th root Theorem: Identity IV

= n'y/a

The root of a root can also be written as calculating the product of the original root indexes and thus taking this root of the expression under the radical sign.

o>0;m,ne IN;m,n>2

Proof Let us raise both sides to the power of mn. exponentiation identity V

mn

definition

n

definition

definition

-im

Since the results are equal the original expressions are equal too.

Example 8 ___ Let us rewrite the number V3-x/3 with one radical sign.

Solution Let us write 3 under the fifth root, and then let us apply the previous identity: ____ _______

VTW - VVF • ^3 = Vx/3^3 - w.

56

Theorem:

The root index and the exponent can be simplified or expanded respectively. , where a > 0; n, k e IN; n, k >2;

m e Z.

Proof Let us raise both sides to the power oink. exponentiation identity V

definition

exponentiation identity V

definition

(n'^am k )" k=am k.

The powers of the two sides are equal; therefore the original expressions are equal too.

Example 9 Let us perform the operations: a) ^3 •

;

Solution a) Let us expand the root indexes to the least common multiple of 4 = 3 • 1^3. b) Using the previous method:

-3, then it is strictly monotonically increasing. (3) Atx = -3 the function has a minimum, and its value is /(-3) = -2.

Figure 2

THE QUADRATIC EQUATION

According to the above we can in general state the following: The assignment rule of the quadratic function in general is as follows: f: R —> R, f(x) = ax2 + bx + c, where a g R \ {0}; b, c g R.

If a > 0, then the graph is a parabola open upwards. If a < 0, then it is a parabola open downwards.

It is practical to rewrite the assignment rule of the quadratic functions into the following completed square form:

f: R —> R, f(x) = a(x - u)2 + v, where a g R \ {0}; u, v g R. The standard parabola is stretched by a factor of a, and is translated by the vector v(m; v) so that the vertex of the parabola will be at the point C(u; v). A real number x, for which f(x) = 0, is called a zero of the function. A quadratic function can have 0,1 or 2 zeroes, since the parabola, depending on its position, can intersect the x-axis at two points at most. (Figure 3)

Example 2 Let us determine the value of q in the following function so that the substitution value of the function will be positive for every real number x.

f(x) = x2 + 2qx + q2 + 2q - 4. Solution Since the coefficient of x2 is positive, the parabola is open upwards, thus it is possible that it takes a positive value for every real number. At first let us transform the assignment rule of the function by completing the square:

/(x) = x2 + 2qx + q2 + 2q - 4 = (x + q)2 + 2q - 4. Since the first term of this expression is a quadratic one, and it is non­ negative in every case, the substitution value of the function will be positive in every case, if the following inequality is satisfied:

2q - 4 > 0, 9 >2.

So this is how we have to choose the value of q because of the given conditions. Then the graph of the function is translated along the y-axis in the positive direction.

Exercises 1. Complete the square in the following expressions. x 2~4x + 4: a)

b) x2 - 6x + 8;

c)x2 + 8x-2;

d) 2x2 + 8x- 5;

e)-x2 + 8x-2;

f)-3x2 + 6x+l.

2. Plot the graphs of the following functions defined on the set of real numbers in one

coordinate system. a) J\(x) = x2, f2(x) = 2x2, /3(x) = 2(x - 1 )2, f4(x) = 2(x- l)2 + 1; b) /¡(x) = x2, f2(x) = ~ ■ x2, /3(x) = | ■ (x + l)2, /4(x) = | • (x + 1 )2 - 2;

c) f\(x) = x2, /2(x) = -3x2, /3(x) = -3(x - 1 )2, /4(x) = -3(x - 1 )2 - 1. 3. Plot the graphs of and characterise the following functions defined on the set of real

numbers. /(x) = x2-4x + 3; a)

fe)/(x) = —x2-4x +3;

c) f(x) - 2x2- 4x + 3.

Determine the zeroes of the functions, if they exist, and the maximum or minimum places and values. 4. In the following functions defined on the set of real numbers determine the value of q so

that the function will have 0; 1; 2 zeroes. n)/(x) = x2-4x + ^; /?)/(x) = -x2-4x + ^;

c)/(x) = 2x2 - 4x + ¿?.

5. In the function f: R —> R, /(x) -x2 + px + q determine the values of p and q so that the function a) takes its minimum at the place x = 1, and its value is 2; b) takes its minimum at the place x = -1, and its value is -2; c) takes its minimum at the place x = 4, and its value is -3. 6. How shall we choose the value of q so that every value of the following functions is

positive on the set of real numbers? a) f(x) = x2 + 6x + q-, b) f(x) = x2 - 4x + q;

c) f(x) - 2x2 + 8x + q.

7. How shall we choose the value of 0, then the equation has two solutions. If — = 0, then a a c there is one solution, x = 0. There is no solution if — X] = 0 and x2 = 3. Example 4 Let us solve the equation 2x2-8x = 10 on the set of real numbers.

Solution By rearranging the equation so that the right-hand-side is equal to 0, and then by completing the square on the left-hand-side: 2x2 -8x-10 = 0, 2(x2-4x)-10 = 0, 2(x2-4x + 4-4)-10 = 0, 2(x-2)2-8-10 = 0, 2(x-2)2 =18, (x-2)2 =9.

By applying the method used in the previous examples, we get from this: 7(x-2)2 = 3, from which | x - 21 = 3,

which impliesx-2 = 3 orx-2 = -3, i.e.Xj = 5 andx2 = -1.

When checking by substitution we can see that both roots are solutions of the original equation. 65 I ................................

0, then by taking the square root of both sides we

solve the equation:

7(2ax + b)2 = -Jb2 -4ac, 12ax + b I = \Jb2 -4ac, 2ax + b = +\lb2 —4ac,

Formula:

2ax — —b ± y/b2 — 4ac, -b + yjb2 -4ac

It can be seen that in this case the equation has two real solutions.

The result is called the quadratic formula or the square root formula of the quadratic equation. The expression Z?2 - 4ac examined when solving the equation is called the discriminant of the equation, and it is generally denoted by D. It can be seen that the number of solutions of the quadratic equation on the set of real numbers depends on this.

discriminant

Based on this the following theorem can be formulated: Theorem:

The quadratic equation ax2 + bx + c = 0 (a * 0) has

, —b ± \Jb2 — 4ac - two real roots: x( 2 =-------- -—----- ------ , ’ 2a if D = b2 - 4ac > 0; Z> , - one real root: x =------ , if D - b~ - 4ac = 0; 2a

discriminant = Latin word meaning: a distinguishing feature or characteristic D = b2- 4ac

- no real roots, if D = b2 - 4ac < 0.

The necessary and sufficient condition of the existence of a real root is: D>0. If a - 1, then the quadratic equation can also be written as follows:

x2 + px + q = 0, where p and q are arbitrary real numbers. Then the formula has the following form: _-p±ylp2-4q X1’2" 2~ Before applying the formula we should always set one side of the equation equal to 0 and put the terms into order based on the degree of the variable.

Example 5

Let us solve the equation 4x2 = 3 - x. Solution Let us rearrange the equation so that there is only 0 on the right-hand-side, and let us put the terms on the left-hand-side into descending order based on the indices of the unknown: 4x2 + x - 3 = 0. By substituting the values a-4, ¿>=1, c = -3 into the formula:

-l±x/l2 +48 -£i 2 — 1.2

Q

-1±7 —

8

g

3 4’ -1

3 i i.e. x> = —, x9 = -l. 1 4 2 Checking by substitution:

4.2.+ 3_3 = o 16

and

41-1-3 = 0.

4

67

Exercises 1. Solve the following equations. a) x2 = 121; c) x2 - 256 = 0;

b) 3x2 = 27; d) 2x2 + 50 = 0.

2. Solve the following equations. a) (x + l)2 = 4; c) (3x - l)2 - 16 = 9;

b) (2x + l)2 = 25; d) (1-x)2 + 25 = 16.

3. Solve the following equations with a) x2 + 7x = 0; c) 2x2 + 6x = x2 - 8x;

help of factorisation. b) 3x2 - 8x = 0; d) (x - 3)2 - 2(x - 3) = 0.

4. Complete the square in the following equations, and then solve them. a) x2 - 3 = 2x; b) x2 + 4x = 5; c) 2x2 = 4x + 23; d) -2x - 3 = x2.

5. Solve the following equations with the help of the quadratic formula.

a) x2 =-2x + 3; c) -x2 + 10 = 3x;

Z?)x2 = 4x-3; d) 2x2 = -8 + 8x.

6. Rearrange and then solve the following equations. a) (2x-4)-(x-2) = 12x + 8; b) 47-y(3y + 4) = 2(17 - 2y) - 62; c) (v + 2)-(v-3) + (v + 3)-(v-2) = 20; d) (w - 3)2 + (m + 4)2 - (w - 5)2 = 17w + 24.

7. In the equation ox2 + 2x + 3 = 0 determine the value of the parameter a so that the equation has a) two distinct real roots; c) no solution on the set of real numbers.

b) one real root;

8. In the equation x2 + bx + 2 = 0 determine the value of the parameter b so that the equation has a) two distinct real roots; b) one real root; c) no solution on the set of real numbers. 9. In the equation x2 - 4x + c = 0 determine the value of the parameter c so that the equation has

a) two distinct real roots; c) no solution on the set of real numbers.

—

b) one real root;

3. The zero product form. The relation between the roots and the coefficients Recognizing when to use the quadratic formula can significantly speed up the solutions of such exercises. However these equations also have other forms the solutions of which can be found more easily.

Example 1 Let us solve the equation 2(x - 3) • (x +1 ) = 0.

Solution If we expanded the brackets, then we obtain the quadratic equation 2x2 - 4x - 6 = 0, and after applying the formula we get xj = 3 and x2 = -1 as the solutions. These two solutions can also immediately be found from the original form of the equation, since the left side is equal to zero if and only if one of its terms is zero:

x - 3 = 0 or x + 1 = 0. From these equations we can easily find the two roots.

From this line of reasoning the answer to the question below follows:

Example 2 Let us factorise the polynomial 2x2 - 3x - 2.

Solution The solution can be found based on the previous example if at first we try to find the roots of the quadratic equation 2x2 - 3x - 2 = 0. According

to the formula these are: xt = ~ and x2 = 2. With the help of the roots

we can write a product which takes the value zero at the same places. We also have to make sure that after performing the multiplication every coefficient should be equal to the corresponding one in the original example; therefore we have to multiply the equation by 2. Thus the equation is:

2-ix + ij-(x-2) = 0, which can also be written as (2x + l)(x - 2) = 0. The product on the left-hand-side is the expression we were looking for: 2x2 - 3x - 2 - (2x + l)(x - 2).

We can make sure of the correctness of this when performing the multiplication.

THE QUADRATIC EQUATION

In general it is true that if a quadratic polynomial has a real root, then it can be factorised: ax2 + bx + c = a(x — xt) ■ (x - x2).

Definition: The form a(x - xj)-(x - x2) = 0 is called the zero

zero product form

product form of the quadratic equation. ♦ ♦ ♦

When examining the quadratic formula we can observe further simple connec­ tions. If the quadratic equation has real solutions, then these are: —b + ylb2 -4ac X] =---------------------2a

, and

-b - \]b~ -4ac x2 =

2a

The following expressions result for the sum and the product of the two roots: —b + \lb2 -4ac

-b - 'jb1 ~ 4ac

*1 + *2 =-------------- o----------------- -+--------------- n-----------------

2a

2a

-2b b = ^T =----- ’ 2a a

-b + \lb2 - 4ac -b - \lb2 -4ac

Xl-x2 =

la

la

_ (-fe)2-(¿>2-4uc) _ 4ac _ c

4a2

4a2

a

The following relations exist between the roots and the coefficients of the quadratic equation ax2 + bx + c = 0:

b . C %i+x? =— and x-i-Xo = о a

, X]+x2 = — and a

c Xi'X2 = —. a

These relations are also called Vieta’s formulae.

Vieta’s formulae

70

With the help of the Vieta’s formulae we can justify our statement ax2 + bx + c = a(x -x,)■(x - x2) relating to the zero product form. If we factor out a from the general form of the quadratic equation, and we substitute /2 c the expressions appearing in the formulae for — and —, then the following transformations can be done: a a

ax2 + bx + c = a|x2 4----- x + — | = a[x2 - (x, +x2)x + Xi-x2] = I « aJ = a(x2 -X]-x-x2-x + X]-x2) = a(x-X])-(x-x2).

Note: In the case of a = 1 the above equation has the form

X2 + bx + C = X2 - (X| + X2)X + X] • x2,

i.e. -b = X\ + x2 and c = Xj • x2, so the sum of the two roots is equal to the negative of the coefficient of the term of degree one, and the product of the two roots is equal to the constant term. If we are looking for the solutions of the quadratic equation on the set of whole numbers, then using these relations we can easily get the roots with trial and error too. E.g. in the case of the equation x2 - 5x + 6 = 0, X] + x2 = 5, X[ -x2 = 6, from which the following are result: X] = 2 and x2 = 3.

This method can help us when checking or when plotting graphs of functions.

Example 3

The equation x2 + x - 6 = 0 is given on the set of real numbers. Let us determine the sum of squares of its roots. Solution The equation has two distinct real roots, since the discriminant is: £) = 1 - 4 • 1 • (-6) = 25 > 0. The sum of squares of the roots can be expressed in the following way:

François Viète (or Vieta, as he is often known by his Latinised name) (1540-1603), French mathematician. His occupation was a lawyer, and he became the counsel and councillor of Kings Henry III and Henry IV. He was the first to apply letters to denote the coefficients, and thus he could give the relation between the roots and the coefficients in a form similar to the above.

%i2 + x22 = (x1 + x2)2 _ 2xj-x2.

By using the relation between the roots and the coefficients: x,2

4-x22 = (-I)2-2 • (-6)= 13.

We could have answered the question also by solving for the roots and by calculating their sum of squares.

Example 4 Which real number shall we write instead of the parameter c so that the quadratic equation 2x2 - 4x + c = 0 has two positive roots?

Solution First we have to examine what condition is necessary for the existence of a real root of the equation. It is satisfied if: D = 16-8c>0, c0, i.e. c>0. 2 Both of the roots of the equation will be positive if 0 < c < 2.

Example 5 The roots of the equation x2 - 2x-15 = 0 are x-, and x2. Without calculating the roots let us give a quadratic equation the roots of which will be Xj + 3 and x2 + 3.

Solution According to the relation between the roots and the coefficients:

x, + x2 = 2; xrx2 = -15.

We can choose the coefficient of the quadratic term of the equation in question as 1. Thus we can look for the form x2 + px + q = 0 of the equation. Let us examine Vieta’s formulae for this equation too:

x2 + px + q = 0;

Xt + x2 = -p; Xt • x2 = q.

X] + 3 + x2 + 3 = -p; (xj + 3)-(x2 + 3) = 9.

After performing the operations on the left side, and then by substituting the values observed above: X[ + x2 + 6 = -p;

xpx2 + 3xj + 3x2 + 9 = q. 2 + 6 = -p; -15 + 3-2 + 9 = 9.

Then we obtain p = -8 and q = 0 for the two parameters in question. Thus a suitable quadratic equation is:

x2 - 8x = 0. The validity of our train of thought can also be justified by determining the roots. The two roots of the original equation are:

X] = -3 and x2 = 5.

And the roots of the quadratic equation x2 - 8x = 0 are:

X! = 0 and x2 = 8.

Exercises 1. Solve the following equations on the set of real numbers. a) (x - 2) • (x + 2) = 0;

b) (3x - 2) • (4x + 2) = 0;

c) 2(-x-2)-(4-x) = 0;

d) -2(2x + 2) • (x - 4) = 0.

2. Give quadratic equations the roots of which are the number pairs below.

b) -3 and 5;

a) 2 and 4;

. 2 3

.3 4

and V3;

c) — and —;

d)

c) -2x2 — 7x + 15;

d) -2x2 — 7x- 3.

3x2 + 3x - 6 C’ 2x2+3x-2;

d) ----- x----------- .

e) a + b and a - b.

3. Factorise the following expressions,

a) x2 - 8x + 15;

b) 2x2 - 14x + 20;

4. Simplify the following fractions.

,

a)

x2-2x-3 x2-4x

+3

;

,.

x2+2x-3

b) —-------------- ;

-x2+4x-3

,,

6x2+x-2

-2x2 +5x-2

5. Determine the following for the equation -2x2 + 6x - 4 = 0: , >

u

11 10

a) the sum of squares of the roots;

b) the sum — + — ;

x> x2 c) the sum —+ —; x2 x.

d) the difference of squares of the roots.

x,

x2

6. In the equation x2 + px + q = 0 express the following in terms of p and q: a) the sum of the roots of the equation; b) the difference of the roots of the equation;

c) the product of the roots; d) the sum of squares of the roots;

e) the difference of squares of the roots. 7. In the equation -x2 + 2x + c = 0 determine the value of c so that

a) the equation has two equal roots; b) the equation has two positive roots;

c) the equation has two negative roots; d) one of the roots of the equation is 0;

e) one of the roots of the equation is positive and the other one is negative;

f) the equation does not have a real root. 8. In the equation x2 - (4k — 8)x + k2 - 2k + 2 = 0 determine the value of k so that the roots

of the equation differ only in their sign. 9. Give the quadratic equation the roots of which differ from the roots of the equation

x2 - 5x + 6 = 0 only in their sign. 10. Give the quadratic equation the roots of which are 2 greater than the roots of the equation

15x2 - 19x + 6 = 0.

73

THE QUADRATIC EQUATION

4. Equations of higher degree which can be reduced to quadratic equations The search for the solutions of equations of different types proved to be an exciting quest for mathematicians throughout history. The mathematicians of the Antiquity could already cope with solving quadratic equations; in the 16th century formulae were found for cubic equations (equations of degree three) and quartic equations (equations of degree four) (Cardano and Ferrari), but it took quite a long time to make a step forward with solving equations of even higher degree. It has been thought for a long time that such a formula can be found for any equation that determines the roots of the equation with the help of the basic operations and the coefficients. Évariste Galois, who had a tragically short life, managed to disprove this false theory.

Évariste Galois

Let us find stories from the life of Galois.

Évariste Galois (1811-1832), French mathematician is reputed as the one who laid the foundations for modern algebra. He was a revolu­ tionary genius of a period following revolutionary times. However the quick historical changes of the time did not favour those who came up with lateral thinking. Many of his significant works had to face incomprehension, and this lack of understanding often soured him. Because of his political affiliations he was imprisoned. On one occasion someone made an offending comment about his loved one and therefore he invited the person to a duel. The last night before the duel the youngster, who loved Mathematics fanatically, summarised the most important results he had found on a few pages, and sent it to one of his friends. The next day, at the young age of 20, he suffered a fatal injury. Quote from the work of Galois

’

:7¡Cr

■'—

The work was published long after his death, in 1846. In one part of his work he showed which types of equations can be solved algebraically, i.e. with the help of only the four basic operations and by taking the corresponding root.

With this he refined the theorem proven by Niels Henrik Abel (1802-1829), Norwegian mathematician, according to which in general there exist no formulae for equations of degree five or equations of even higher degree. There are however such special equations of higher degree, which can be solved. The principle of the solution is always the same: reducing the degree of the equation while doing the transformations and substitutions, and thus the roots can be determined with the help of the already known formulae.

Example 1

Let us solve the equation x4 - 5x2 + 4 = 0 on the set of real numbers.

Solution Let us realise that (x2 ) = x4, therefore it is practical to introduce a new unknown: y - x2. In this case the equation will have the following form: y2 - 5y + 4 = 0.

By solving with the help of the formula we obtain the solutions below: = 5±7(-5)2-4-l-4

i h = 4>

= 4 = x2: X[ = 2, x2 = -2; y2 - 1 = x2: x3 = 1, x4 = -1.

Therefore an equation of degree four will have four solutions; we can make sure they are valid by checking. ♦ ♦ ♦

It is worth mentioning a statement which is true in general: an equation of degree n can have at most n real solutions.

It is easy to find an example when there exist no real solutions. Such an equation is the following: x2 = -1.

An equation of degree n can have n real solutions at most.

Example 2

Let us solve the equation 8(x-1)6- 215(x - 1 )3 - 27 = 0 on the set of real numbers. Solution Instead of expanding the brackets it is again practical to start with introducing a new unknown.

If we introduce y = (x - l)3, then because ^(x - l)3 J = (x -1)6 we get the quadratic equation 8y2 - 215y - 27 - 0. By solving this: 215±J(-215)2+4-8-27 J>’2=---------------- U---------------

,11 i Therefore (x - 1) - —, x - 1 = —, thus one of the roots is: x, = —. 8 2 1 2 And in the other case (x - l)3 = 27, x - 1 = 3, where we obtain the root x2 = 4. We can make sure of these are valid roots by substituting and

checking.

y=(x-1)3

THE QUADRATIC EQUATION

Example 3 Let us solve the equation (x2 + x + 3) • (x2 + x + 1) -15 = 0.

Solution After expanding the brackets the equation of degree four

x4 + 2x3 + 5x2 + 4x - 12 = 0 would result, but we cannot solve it with any of the methods we have learnt so far. It is practical to introduce a new unknown variable. Let y = x2 + x + 3, thus the other factor will be y - 2. The equation looks as follows:

y(y-2)-15 = 0, y2-2y- 15 = 0. Its solutions are:

2 ± 7(-2)2 +4 • 1 • 15 [ Ji = -3, 3’1.2 =----- ------- Z------------- = . 2 [y2=5By substituting the resulting values, if yj = -3, then: x2 + x + 3 = -3,

x2 + x + 6 = 0. The discriminant of this equation is: D = 1 - 24 = -23, therefore it does not have a real solution.

In the other case, if y2 - 5, then the equation to solve is: x2 + x + 3 = 5, x2 + x - 2 = 0. From this we get the following: -l±x/l2 +4-1-2 fxi=l, *1,2 =---------- z---------- = 0 2 |x2 = -2.

Let us check by substituting into the original equation: (1 + 1 + 3)-(l + 1 + 1)- 15 = 0,

(4-2 + 3)-(4-2+ 1)- 15 = 0. So both roots are solutions of the original equation too. ♦ ♦ ♦

In connection with the example it is worth mentioning that if the equation has a solution x = a on the set of whole numbers, then by substituting it into the equation: a4 + 2a3 + 5a2 + 4a - 12 = 0, a4 + 2a3 + 5a2 + 4a = 12,

a(a3 + 2a2 + 5a + 4) = 12.

The equality can have the above form.

Since there is the product of two whole numbers on the left side of the equation, therefore it is true that if the equation has a solution on the set of whole numbers, then it can only be one of the divisors of 12. It is also satisfied in our case, since the two solutions, 1 and -2, are divisors of 12.

We can use this property of equations with integer coefficients when looking for the roots, there are only finitely many divisors of the constant term and so only finitely many possibilities.

Example 4 Let us find the real solutions of the equation below. x4 + x3-7x2-x + 6 = 0

Solution By using the statements made in the previous example let us examine whether the equation has a solution on the set of whole numbers.

These whole numbers are among the divisors of 6. The divisors of 6 are: ±1; ±2; ±3; ±6. We have to examine these eight possibilities.

It is immediately implied that both -1 and 1 are solutions of the equation, since: (-l)4 + (-l)3-7(-l)2-(-l) + 6 = 1-1-7 + 14-6 = 0;

14+ P-7 ■ l2- 1 +6 = 1 + 1- 7-1+6 = 0. It also means - when thinking of the zero product form of the equations - that when factorising then factor (x-l)(x + 1 ) = x2 — 1

appears. Considering this let us apply the following grouping in the equation: x4 + x3 — 7x2 - x + 6 = x4 - x2 + x3 - x - 6x2 + 6 = = x2(x2 - 1) + x(x2 - 1) - 6(x2 - 1) = = (x2- l) (x2 + x-6),

thus the equation has the following form:

(x2 — l)-(x2+ x-6) = 0.

This product will be 0 if one of its factors is 0. The first factor gives the roots Xj 2 = ±1 found earlier, therefore if there exist further solutions these are given by the roots of the quadratic equation x2 + x-6 = 0. These are as follows:

—1 ± Vl2 +4-1 -6 x3,4 -

-

2

[x3 = 2, = 1

o

[x4 = -3.

Therefore this equation will have four real and at the same time integer solutions: xe {-3; -1; 1;2},

we can make sure these are valid by checking.

”................

{¡g>

THE QUADRATIC EQUATION

Also a group of equations of higher degree exists for which further methods can work too. The following exercise gives such an example.

Example 5 symmetric equation

Let us find the solutions of the equation below. 2x4 - 3/3 -x2 - 3x + 2 = 0.

Solution Let us realise that in this equation the coefficients appearing symmetrically to the right and to the left of the term in the middle are equal. In such a case it is practical to divide both sides of the equation by x2. The value x = 0 obviously does not satisfy the equation; therefore we can divide by it. After this let us group the terms with equal coefficients:

2x2-3x-l-3- —+ 2--y = 0, X xz

1 , , 1 Let us introduce a new unknown: y=x+—. Then y2 = x2+2h—y, thus j x x2 x2 + — = y2 - 2. By substituting these into the equation: x2 2(y2-2)-3y- 1=0, 2y2 - 3y - 5 = 0. By solving the resulting quadratic equation:

3±V(-3)2-4-2-(-5) 2-2

We get the same result by recalling the statement already proven in the topic of functions, according to which:

In view of y, x can also be determined.

If y( = -1, then x + — = -l, i.e. x2 + x + 1 = 0. In this case no real solutions x result, because the discriminant of the equation will be a negative number: D = 1 -4 = -3.

> 2, if x * 0. If y-> = —, then x + — = —, i.e. 2x2 - 5x + 2 = 0. Its roots are: 2 2 x 2 5±7(-5)2-4-2-2 x. , =-------- —------ ----------------------- = 1

X| ~2’ I

By checking we can confirm that these are solutions of the original equa­ tion too. ♦ ♦ ♦

Because of the symmetry of the coefficients the polynomial on the left side of the solved equation is called a symmetric polynomial.

It can be shown that for the solutions of such equations it is true that if a number is a solution, then the reciprocal of this number also satisfies the equation.

Exercises 1. Solve the following equations. a) x4-16 = 0;

Z>)x3 + 27 = 0;

c) x6

64 = 0.

2. Find the roots of the following equations.

a) x4-x2- 1 = 0;

b) x6-2x3-3 = 0;

c) 2x8 - 7x4 - 4 = 0.

3. For which real numbers are the following equations true? a) (x + l)4 - (x + l)2 - 1 = 0; b) 2(x-2)6 - 3(x-2)3 - 2 = 0;

c) 2(2x- 1)8-(2x- l)4- 1 =0.

4. Solve the following equations. a) (x2 + x) • (x2 + x + 1) - 2 = 0; b) (x2 + 2x) • (x2 + 2x - 1) = 6;

c) (x2 - x + l)-(x2 - x— 1) - 3 = 0. 5. Give such equations, and then put the unknowns into descending order based on their

exponents, the solutions of which are the following. a){l;2;3};

{-1;-2;-3; ^1};

c) {-1; 1; -3; 4).

6. Solve the following equations. First find a solution on the set of whole numbers.

a) x3 - 3x2 - x + 3 = 0; b) x4-2x3 - 2x2 - 2x-3 = 0;

c) x4 - x3 - x2 - x - 2 = 0.

7. Solve the following symmetric equations. a) x4 - 2x3 + 2x2 — 2x + 1 =0; b) 2x4 - 5x3 + 4x2 - 5x + 2 = 0;

c) 3x4 - 16x3 + 26x2 - 16x + 3 = 0.

8. Solve the following equations on the set of real numbers. a) y6- 5y3 = -6; b) (z2 -2z)-(2z-z2 + 1) = 2;

c) x4 - 2x3 + 2x2 — 3x + 1 = -x4 + 3x3 — 2x2 + 2x - 1.

THE QUADRATIC EQUATION

5. Quadratic inequalities We can find the relation of any real numbers a and b. There are three possible cases: a b. If we connect expressions with the signs , , then an inequality results. If these expressions are quadratic, then we are talking about quadratic inequality

quadratic inequalities.

When solving quadratic inequalities the fact that the graphs of these expressions define parabolas in the coordinate system, plays an essential role. Plotting the graphs can help a lot in every case with finding and representing the solution set in question.

Example 1

Let us solve the inequality x2 - 9 < 0. Solution I Let us plot the graph of the function determined by the expression on the left-hand-side. (Figure 4) We are looking for those real numbers x, at which the function takes negative values. The zeroes of the function are: V| = -3; x2 - 3, Figure 4

therefore the suitable real numbers are the elements of the open interval ]-3; 3[.

Solution II Solving the algebraic way, by taking the square root: x2 0 and x + 3 < 0,

x>3 and x R, x x2 - 2x - 3 is a parabola open upwards (Figure 6), the solution of the inequality will be the range between the two zeroes. Therefore Figure 6

-1 0.

By finding the roots of the equation y2 + y - 6 = 0 :

-1± 712-4-1 -(-6) [yj=-3, J1.2 =------------2Z------------ = [y2 = O2. Considering that the graph defined by the function on the left side is a parabola open upwards (Figure 7), the solution of the inequality is: y < -3 or y > 2. By substituting the variable x we have to find the solutions of two inequalities. None of the real numbers satisfy the condition x2 < -3, since the square of a real number is always non-negative.

By solving the inequality x2 > 2:

I x I > a/2 .

82

The rules found for linear inequalities in our earlier studies stay true when solving quadratic inequalities, i.e.: (1) We can add/subtract the same real number to/from both sides of the inequality, or multiply/divide both sides by the same positive real number, the relation stays true.

If a > b, then:

a+c>b+c, a-c>b-c, ad>bd, a b d d

(2) Multiplying/dividing by a negative number changes the direction of the inequality to the opposite one. (3) If we multiply both sides of the inequality by 0, then the set of solutions may change.

where c e R and d e R+.

(4) When squaring or taking the reciprocal we have to examine the signs of the original expressions and only after that can we state something about whether the inequality changes direction.

Exercises 1. Solve the following inequalities. a,)x2-160;

c) x2 - 64 < 0.

2. Find the solutions of the following inequalities.

a)x2-x-l>0;

Z?) x2 - 2x - 3 < 0;

c) 2x2 - 7x - 4 < 0.

3. For which whole numbers are the following inequalities true? a) x2 -4x-4 > -3x- 3; b) 2x2 - 4x-3 0;

Z?) x4 - 2x2 - 3 < 0;

c) 2x6- 7x3 -4 < 0.

5. Solve the following inequalities.

b) x-i------ 2 0.

83

THE QUADRATIC EQUATION

6. Parametric quadratic equations (higher level courseware) In view of the coefficients we can state the following about the equation ax2 + bx + c = 0: (1)

If all the coefficients are equal to zero, i.e. a = b = c = 0, then the equality is satisfied by any real number, since the left side is always equal to zero irrespectively of x.

(2)

If a - b = 0 and c 0, then it has no solution, since the left side is always a non-zero real number irrespectively of x.

(3)

If a = 0 and b # 0, then the equation is of degree one, which has one solution, and it is x = ——. b

(4) If a 0, then a quadratic equation results, the solutions of which depend on the value of the discriminant D = b2 — 4ac. If D > 0, then there exist two distinct real roots, if D = 0, then the equation has one real root, if D < 0, then the equation does not have a solution on the set of real numbers.

If we do not know the particular values of the coefficients in the equation to solve, but we denote them by letters, then we are talking about a parametric equation. When solving parametric equations we have to analyse all the possible cases according to the conditions mentioned above.

Example 1 For which values of the parameter a will the roots of the equation x2 + 4ax + 2a2 + 3o -1 = 0 be equal?

Solution We can decide about the condition of the example by examining the discriminant, since we get equal roots if the discriminant is zero. Since the discriminant is: D = (4a)2 - 4(2a2 + 3a - 1) = Sa2 - 12a + 4,

when solving the quadratic equation 8a2 - 12a + 4 = 0 the following two values are obtained for a: and a-> = —. 2 2 1 So in the cases of ax = 1 and a2“ = — We cani 2 the roots will be equal. -■ at = 1

realise that in these cases the left-hand-side of the equation is a perfect square and the parabola touches the x-axis: This is how a computer program of the quadratic formula works.

in the case of aj = 1

x2 + 4x + 4 = (x + 2)2;

in the case of a2 = -^ x2 + 2x + 1 = (x + 1 )2.

Example 2

Let us solve the following equation (a is a real parameter):

ax2 + (2a + 1)x + a + 2 = 0.

Solution Let us first examine the possible cases. If a = 0, then the equation is of degree one, which can be solved easily:

x + 2 - 0, x = -2.

If a * 0, then the equation is a quadratic one, and the number of roots

depends on the discriminant: D - (2a + l)2 - 4a(a + 2) = -4a + 1. ♦

then the two real roots of the equation are:

If-4a + 1 > 0, i.e.

xl,2

la

then the equation has the following form:

♦

So in this case it is possible to complete the square on the left-hand­ side of the equation, and it has only one solution, namely x - -3. ♦

If-4a + 1 < 0, i.e. a > —, then the equation will have no solution 4 on the set of real numbers.

We might realise that Vieta’s formulae do not provide information about the actual existence of real roots, since we can always calculate the values of 2a + 1 X] + x2 =---------- and a

if the condition a a real root.

a+2

xt -x2 =------------ , a

0 is satisfied, even if the equation does not have

It means that when solving parametric equations, with the help of the discriminant we always have to check whether the equation has real root(s) or not. 85

THE QUADRATIC EQUATION

Example 3 For which one-digit positive integer values of p, q will one root of the equation x2 + px + q = 0 be four times its other root? Solution The roots of the equation are: -p-Vp2-4p

-p+W-^

and

2

2

2

According to the conditions relating to the roots we can write the following: _______ _______ 1 -p-y/p2-4q _ —p + y]p2—4q

2

2

After rearranging and simplifying it: ~4p-4■ yjp2 -4q = -p + jp2-4q, -3p = 5-ylp2 -4q.

The left side of the resulting equation is always negative, but the right side is a non-negative number, thus there are no positive parameters p and q which would satisfy this equation. But the conditions relating to the roots can be satisfied in a different way too: _______ _______ ^-p + \lp2-^q _-p~yJp2-4q 2

2

After rearranging and then simplifying it: ~4p + 4• yjp2 ~4q =-p-yip2 -4q,

3p = 5-ylp2 -4q.

By squaring both sides of the equation: 9p2 = 25p2 -lOOq,

!00^ = 16p2,

Since we are looking for the solutions on the set of one-digit positive whole numbers, q can only be a number divisible by 4. I.e. q = 4 or q = 8. We can easily ascertain that an integer solution can be found forp only in the first case, when q = 4, then p = 5.

So the equation that satisfies the conditions of the example is x2 + 5x + 4 = 0. After solving the equation the two roots are: Xj = -1 and x2 = -4.

One of the roots is indeed four times the other root.

Example 4

Let us determine the value of the parameter m so that the value of the following quadratic expression will be negative for any x.

mx2 + (m -1 )x + m -1. Solution We can rephrase the condition of the example as follows. We have to give the value of the parameter m so that (1) the expression written is still quadratic,

(2) the parabola is open downwards, (3) the quadratic equation mx2 + (m—l)x + m-1=0 defined by it will have no solution. These conditions mean that the graph of the function defined by the expression is a parabola open downwards which does not intersect the x-axis of the coordinate system. (Figure 8)

Condition (1) and (2) are ensured by m < 0. Condition (3) is satisfied if the discriminant of the equation is negative,

(m — l)2 - 4m(m - 1 ) < 0, (m - 1) ■ (m - 1 - 4m) < 0,

(m - 1 ) • (-3m - 1 ) < 0.

The product will be negative if the two factors have opposite signs. I.e.: m - 1 < 0 and m - 1 > 0 and

In the first case

1

m < 1 and m

1 3

1

Figure 9

1 i.e. m < —, (Figure 9) 3 o-

in the second case

m> 1

40

o

and m>--, i.e. m > 1. (Figure 10) 3

—*—i1 o 3

By plotting conditions (1), (2) and (3): (1) + (2)

+■ 1

Figure 10

■o

(3)

(3) ■o

+

—i------- F _1 0 3

o+■ 1

+ Figure 11

The set of the parameters m which satisfy all three conditions simultaneously:

I

m I me R and m < — 3

87

...................................

THE QUADRATIC EQUATION

Example 5 Let us give the value of the parameter m so that the following inequality has no solution. (m + 1)x2-2(m- 1)x + 3m-3 < 0. Solution Let us rephrase the condition. If the inequality written does not have a solution, then it means that the inequality (m + 1 )x2 - 2(m - l)x + 3m - 3 > 0

is satisfied by every real x. Let us examine the conditions for which it is true. Since in the case of m + 1 - 0 the function of the left side will be the linear function x i—> 4x - 6, which is not positive for every x, therefore it must be the case m # -1.

In this case the left-hand-side is a quadratic function, which is non­ negative for every x if the image of the function is a parabola open upwards, which can have at most one point in common with the x-axis. e. I.

m + 1 > 0, that is m > -1.

Secondly, it can have at most one real root, which is defined by the value of the discriminant. We must have that D = [2(m — l)]2 - 4(m + 1) • (3m - 3) < 0.

It defines a quadratic inequality in terms of m.

After expanding and rearranging: 4m2 - 8m + 4 - 12m2 — 12m + 12m + 12 < 0, -8m2 - 8m + 16 < 0.

Dividing both sides of the inequality by -8: m2 + m — 2 > 0.

Let us find the zeroes of the equation m2 + m - 2 = 0. -1 ± Jl2-4-l-(-2) i m, = -2, '”1,2 =------------ Z------------ = , 2 [m2 = 1.

ma1 -1

0

Since the graph of the expression on the left-hand-side of the inequality m2 + m - 2 > 0 is a parabola open upwards, the solution of the inequality is: m < -2 or 1 < m.

Also considering the condition m > -1 we obtained earlier, the quadratic inequality of the example will have no solution, if

1

me [1; oo[.

Figure 12

The solution represented on a number line can be seen in figure 12.

88

Exercises 1. Determine the value of the parameter a in the following equations so that the equation

will have two identical roots. a) x2 - ax + 4 = 0;

b) x2 - 3x + a = 0;

c) x2 - ax + a = 0;

d) x2 - ax + a + 1 = 0.

2. Solve and discuss (analyse) the following equations. a) x2-ax+ 4 = 0; b) x2 — bx + a = 0;

c) ax2 - bx + 5 = 0;

d) ax2 - bx + a + I = 0.

3. Solve the following equations. a)

1--- 1

x+a

1

b) -------- 1 x—a x+a

x

1

c) ------ +------x+a x—a

x

4. Determine the value of the parameter a so that the following equations have no real solutions.

a) x2 - ax + 4a = 0; c) ax2 - (a - V)x + 5 = 0.

b) x2 - ax + a - 4 = 0;

5. Determine the value of the parameter m so that the value of the following expression is negative for any x: 7 mx2 + (m — l)x + 1. 6. Determine the value of the parameter m so that the value of the following expression is positive for any x: 9 mx2 + mx — 5.

7. Determine the value of the parameter m so that the value of the expression is negative for any x: 9 -mx2 + (m — l)x + m + 2.

89

THE QUADRATIC EQUATION

7. Equations involving square roots Equations, in which the unknown is under the radical sign are called radical equations. However, we are going to restrict ourselves to equations involving square roots.

(^)2=o,

where a>0 and 4a>0

While solving equations involving square roots we always try to eliminate the expressions involving square roots from the equation. It is usually done with the help of squaring.

When performing this operation problems can often come up which make it especially important to check the roots. Namely after squaring equalities originally not satisfied can become true. For example: -2 2, but if we square both sides, then (-2)2 = 22 = 4.

Therefore this operation is not listed among the equivalent transformations, and care must be taken when applying it.

Example 1

Let us solve the equation 4x-2 -2 = 0.

Solution At first we give the largest subset of the set of real numbers in which we can look for the solution, i.e. we determine the domain of the equation.

domain: x>2

Since there cannot be a negative number under the square root sign, we must have x - 2 > 0, i.e. x > 2.

After rearranging the equation: 4x-2 = 2.

Then both sides are non-negative; in such cases squaring is an equivalent transformation. Performing this: x - 2 = 4,

x = 6. The resulting root is indeed a solution, because on one hand it is an element of the domain, and on the other hand it satisfies the equation:

76-2-2 = 74-2 = 2-2 = 0.

Example 2

For which real numbers is the following true: x - V2x + 1 = 1 ? Solution We can state that the equation has a meaning only in the case of numbers domain: x> — 2

for which 2x + 1 > 0, i.e. x > —. 2

Let us rearrange the equation so that only the expression with the square root stays on one side of the equation. -yj2x +1 = 1 - x; >j2x +1 = x -1.

After the rearrangement we multiply both sides by -1, thus the left-handside will surely take only non-negative values, and if the equation has a solution, then the right-hand-side will also take a non-negative value. So it is also true for the solution that: x - 1 > 0,

x> 1. Under such conditions squaring also counts as an equivalent transfor­ mation. 2x + 1 = x2 - 2x + 1,

x2 - 4x = 0, x(x - 4) = 0. This equation has two roots: x, = 0 and x2 = 4. Only the root x2 = 4 satisfies the given conditions, and this value is indeed a solution of the equation: 4-V2-4 + 1 =4-5/9 =4-3 = 1.

In the case of Xj = 0:

O-a/2-0 + 1 = 0-1 = -1 * 1. It can be seen that in this case the originally false statement would become true as a result of squaring. During the transformation of the equation, the equation a/2x +1 = x - 1 appeared.

Let us plot the graphs of the functions x >—> y/2x +1 and x in a common coordinate system. (Figure 13)

x- 1

The false root which occured during the solution would be a solution of the equation -a/2x +1 = x - 1. The graphs of these plotted in a common coordinate system can be seen in figure 14. 91

................................

THE QUADRATIC EQUATION

Example 3 Let us solve the equation 7x + 9 -7x =1. Solution The domain of the equation: x + 9>0, i.e. x> -9 1 x>0. x>0 J In this example we fall short of having no square root sign in the equation by simply squaring.

If we rearrange the equation to the form 7x + 9 = 7x + 1 then there will be non-negative numbers on both sides. Then by squaring: x + 9 = x + 2a/x +1. Let us rearrange it so that only the expression with the square root stays on one side of the equation:

7x = 4. From here by squaring both sides we get x = 16. It is indeed a solution, since: 716 + 9-716=5-4 = 1.

Example 4 Let us solve the equation 7x-3 + 72-5% = -2. Solution We immediately realise that this equation cannot have a root, since if it had, then - because of the definition of the square root - there would be a non-negative number on the left-hand-side of the equation, and the right-hand-side is negative, therefore the equality cannot be satisfied by any real number.

............

92

Example 5 For which real numbers is the following equation true:

Vx2 + 6x + 9 + 7x2-4x + 4 = 6 ?

Solution Let us realise that there are the squares of two-term expressions under the square root signs. At the same time it also means that the domain of the equation is the set of real numbers. Let us apply the identity y](x + 3)2 + -\J(x — 2)2 — 6,

I x + 31 + I x- 2| = 6. Because of the definition of the absolute value:

. I x + 3, if x>—3, x + 3| = [-x-3, if x < -3.

i x-2, if x > 2, I x-21 = j-x + 2, if x/1=1; and 0 = 02; >/o = 0.

The substitution method Example 2 Let us solve the simultaneous equations of example 1 using the algebraic method.

Solution (a) By applying the substitution method we express y from the first equation; then by substituting it into the second equation we get the following quadratic equation: 2 _ 2 v2 2x2-x-1=0.

By applying the formula:

-(-l)±7(-l)2-4-2-(-l)_l±3 *1,2 =

4

2-2

Thus the solutions of the simultaneous equations are: (x; y)x = (1; 4), (x; y)2 = i-|; Checking: see the graphical solution.

Solution (b) We express y from the first equation; then by substituting it into the second equation we obtain the following equation: X = (xf, 4

X = x\

By rearranging the equation so that there is 0 on one side and factorising the algebraic expression: x — x4 =0, x(l - x3) = 0.

Since a product is 0 if and only if one of its factors is 0, the solutions of the equation are: x, =0 or 1 -x3 =0,

x3 = l,

x2 = 1. Thus the solutions of the simultaneous equations are:

(x;y)| = (1; 1),

(x; y)2 = (0; 0).

Checking: see the graphical solution.

Example 3 Give the two numbers the sum of which is 15, and the product of which is 56. Solution Let us denote the two numbers in question by x and y. According to the conditions we can write down the following quadratic simultaneous equations: x + y = 151 x-y = 56 J

By applying the substitution method: by expressing y from the first equation, and then by substituting it into the second equation we get the following equation: x(15 -x) = 56. 97

Ordered value pairs need to be given as the solution of simultaneous equations. If the simultaneous equations remain the same when swapping the unknowns, then the solutions will also be symmetric.

By rearranging the equation, and then by using the formula: -x2 + 15x - 56 = 0, _ -15±7152-4-(-l)-(-56) _[x1=7, Xl’2 2(-l) U=8.

By substituting the value y the following is resulting: y, = 8 and y2 = 7. So the number pairs in question are the following: (x;y)i = (7; 8), (x; y)2 = (8; 7).

Indeed: 7 + 8 = 15 and 7-8 = 56.

Manipulating simultaneous equations further If it is difficult to isolate an unknown from simultaneous equations, we can manipulate the equations in different ways. Then it may be easier to express one unknown in terms of another.

Example 4

Let us solve the following simultaneous equations: x + 3xy - y = 51 2x-xy + 2y = 4j Solution No matter which unknown we express from any of the equations, a fractional expression results in all cases. But if we add three times the second equation to the first equation, then a simpler equation results, from which it is easier to express the chosen unknown.

x + 3xy - y = 5 6x -3xy + 6y = 12 7x + 5y = 17, 17-7x y =---------5

By substituting it into to the first equation and rearranging we get the following quadratic equation:

5 5 21x2-63x-42 = 0, _ -(-63)±7(-63)2 -4-21-42 f X] = 1, *1’2 ” FT1 “ |x2 = 2. By substituting we also find the solutions of the simultaneous equations:

(x; ^=(1; 2), Checking:

1 + 31-2-2 = 5; 3 2 + 3-2- — 5

Introducing a new unknown There are simultaneous equations which can become significantly simpler when substituting suitably chosen expressions for new unknowns.

Example 5

Let us solve the following simultaneous equations on the set of positive real number pairs: 2(x + y) = 3xyl x2+y2=5 f Solution Let us rewrite the second equation as follows, with the help of the special identities: 2(x + y) = 3xyl (.¥ + y)2 - 2xy = 5

J

After this it is practical to introduce the following new variables: a = x + y, b = xy. Thus the simultaneous equations are: 2a = 3b]

a2-2b = 5 f

2i7 From the first equation: b = —. Based on this the following quadratic

equation results from the second equation, by solving which the values of a and b can be expressed:

a2_2.Z£_5 = o, 3 3a2-4a -15 = 0, -(-4)±7(-4)2-4-3-(-5) _

fll,2 _

2-3

«I =3,

5 a2 = — 3

Thus: (a;b){ = (3; 2),

Accordingly we obtain two pairs of simultaneous equations, which can easily be solved. In the first case:

x + y = 31 xy = 2j’ From the first equation: y = 3 - x, by substituting it into the second one:

x(3-x) = 2, -x2 + 3x-2 = 0.

THE QUADRATIC EQUATION

Its solution is:

_ —3 ±-^32 — 4 • (—1) • (—2) _ J Xj = 2, X1’2 2-(-l) [x2=l. Thus the solutions of the simultaneous equations in this case are:

(x;y), = (2; 1),

(x; y)2 = (1; 2).

In the second case: 5

x + y ——

3 10 '

Since we have to solve the simultaneous equations on the set of positive real number pairs, in this case there are no newer solutions. By checking: 2 • (2 + 1) = 3 • 2 • 1 and l2 + 22 = 5.

Exercises 1. Solve the following pairs of simultaneous equations. b)

a) x + y = 12]

xy = 35j;

x+y =7 1

c)

x2+y2=25j’

xy = 24) x2+y2 = 52 J

2. Solve the following pairs of simultaneous equations graphically. Find the solutions also by the algebraic way. b)

a) x-y-0 ]

x + y = 21

c) y-4x = x2+4l

x2-y=0f

y-2 = x

J

3. Solve the following pairs of simultaneous equations by manipulating the equations. a) x + y + xy - 5 xy +1 = x + y J ’

b) x + 2y + 3xy = ll]

c) 4(x + y) = 5xy-10

x(y-2) = y(x-l)-lj

2x-y + 2xy = 4 J ’

4. Solve the following pairs of simultaneous equations by introducing new unknowns. b) x + y + yjx + y = 12

a) x + y + xy-23

126 -----= x + y xy

;

5. Solve the following pairs a) x2 — y2 = 2(x + y) 1

x2 +y2 = 10(x-y)j’

c) 24(x + y) = 7xyl

x2+y2=100j‘

x2+y2 =53

simultaneous equations. b) x2y + xy2 =20

1 + 1_5 ■; x

y

4

>

c)

7 i v/x-7

5__ ! Vx-7

—i

4 +6

3 \Jy + 6

5 3

=~

_ 13 6

9. Arithmetic and geometric means While examining different amounts we have already come across the expressions average, mean. These are mostly in connection with the concept of the arithmetic mean, where the available numbers are to be added and the sum is to be divided by the number of the numbers.

The concept of the mean can be more general than that.

In the case of two numbers the mean is a third number that according to some rule falls between the two numbers, if the two numbers are distinct. The mean of two equal numbers is equal to the given numbers. The assignment rules can differ; therefore the means can give different values too.

Example 1

Two numbers a and b are given. Find a number which is greater than the first by the same value that it is smaller than the second. Solution Let the number in question be x, therefore (Figure 18): b-a 2

x - a = b - x,

b-a 2

-i--------------- 1--------------- H

2x = a + b,

a

a+b x =------- .

a+b 2

b

Figure 18

2

The number resulting this way is called the arithmetic mean of the numbers a and b, and it is denoted by A(tz, b). In statistics and in our everyday life the expression mean (or average) is used for this concept.

Since later it will be important to be able to compare the given arith­ metic mean to other means, we give the definition for non-negative numbers. arithmetic mean

Definition: The arithmetic mean of two non-negative numbers is half of the sum of the two numbers: A(a, b) =

a+b

2

This notion can also be defined for more than two numbers. The arith­ metic mean of the non-negative numbers af, a2; ■ ■■', atl is: ¿Zi + a^ +... + a„ A = —----- ------------ -. n 101

Example 2 Two positive numbers a and b are given. Find a number which is greater than the first one as many times as it is smaller than the second one. Solution Let the number in question be x, thus according to the conditions: x _ b x

a

x1 = ab, x = Jab.

The resulting number is called the geometric mean of the two numbers. geometric mean

Definition: The geometric mean of two non-negative numbers is the square root of the product of the two numbers: G(a, b) = \[ab.

In general the geometric mean of the non-negative numbers a^', a2', ...;anis: G = n]ai -a2-...-a„.

Example 3 Let a = 3 and b = 12. Let us determine the arithmetic and the geometric means of the two numbers. Let us represent the results on a number line. Solution The arithmetic and the geometric means of the two given numbers are: A(3,12) = ^^ = 7.5;

G(3,12) = VT12 = 6.

Representing the numbers on a number line: G(3,12)

/4(3,12)

4----------- ♦----------- h

4------- 1--------- 1------- *-

2

9

3

4

10

11

12

Figure 19

Based on the representation we can state that the following is satisfied by the given numbers:

3< G(3, 12) < A(3, 12) < 12. ♦ ♦ ♦

The following question might arise: is a similar relation true for arbitrary non-negative real numbers a and b? Our previous conjecture leads to a special inequality, which is called the inequality of arithmetic and geometric means, often abbreviated to AM-GM inequality. Theorem:

For two non-negative real numbers a and b:

inequality of arithmetic and geometric means

Proof Since both sides of the inequality are non-negative, after squaring a statement follows which is equivalent to the original statement.

a

b

, , a2 + 2ab + b2 ab 2. a

The AM-GM inequality might be suitable for determining the minimum or maximum of different amounts.

Example 5 Let us form a rectangle with the help of a 2 m long string. How shall we choose the sides of the rectangle so that its area will be maximized?

Solution Let us denote the length of the adjacent sides of the rectangle by x and 1 — x (Figure 22). The area of the rectangle is: i = x(l — x). Both factors of the two-term product are non-negative, since these represent the length of the sides of the rectangle. Let us write the inequality of arithmetic and geometric means in terms of these two factors:

G = ^77-77
f(q)/f(x) R, f(x) - ax2 + bx + c are parabolas. If a > 0, then the parabola is open upwards (Figure 24); it has a minimum. If a < 0, then the parabola is open downwards (Figure 25); it has a maximum.

The minimum or the maximum of the function is its extreme value.

Example 1

Let us determine the extreme values of the function f(x) = x2 + 2x - 3, if ojxeR;

h)x6[0;2];

c)xg[-3;-2].

Solution Let us complete the square in the assignment rule of the function. This step will prove to be efficient later on too, when we are looking for the extreme values of quadratic expressions.

/(x) = x2 + 2x - 3 = (x + l)2 - 1 - 3 = (x + l)2 - 4. The graph of the function is shown in figure 26. a) The function takes its smallest value at the place where the squared

term is smallest, i.e. when it is equal to 0. It is satisfied in the case of x = -1. Then the minimum value of the function is: /(-1) = -4. b) On the interval [0; 2] the graph is strictly monotonically increasing.

It takes its minimum at x = 0, and its value here is /(0) = -3. It takes its maximum at x = 2, its value is /(2) = 5. c) On the interval [-3; -2] the graph is strictly monotonically decreasing,

thus it takes its maximum at the left end-point of the interval, and it takes its minimum at the right end-point. The maximum place is at x = -3, its value is /(-3) = 0. The minimum place is at x = -2, its value is /(-2) = -3.

Example 2

We divide a 10 cm long line segment into two parts, and we draw squares above the parts. When will the sum of the areas of the two squares be the smallest? Solution Let us denote the length of the side of one of the squares by x, then the length of the side of the other one will be 10 - x, as in figure 27. Then the sum of the areas of the squares is: r(x) = x2 + (10 — x)2,

X2 (10-x)2

which defines a quadratic function in terms of x on the interval [0; 10]. We are looking for its minimum. First let us complete the square: t(x) - x2 + (10-x)2 = x2 + 100 - 20x + x2 = 2(x2-10x + 50) =

x

10-x

Figure 27

= 2[(x-5)2 + 25] = 2(x-5)2 + 50.

This function takes its smallest value at the place x = 5, i.e. we have to exactly bisect the line segment with length 10 cm, and we have to draw two squares of equal measure; the area of each square will be 25 cm2.

Example 3

Two cyclists are riding towards the crossing on two perpendicular roads at uniform speed. They set out at once: one of them at a speed of 30 km/h from a distance of 20 km, the other one at a speed of 40 km/h from a distance of 10 km. When and where will they be the closest to each other?

Solution Let the time in question be x. measured in hours. Then the cyclist riding on one of the roads covers a distance of 30x, while the other cyclist covers a distance of 40x km. (Figure 28)

30x

The current distance of the two cyclists can be calculated with the help of the Pythagorean theorem: J(x) = 7(20-30x)2 +(10-40x)2.

tf(x) 20 - 30x

This function takes its minimum value when the following function does: d2(x) = (20 - 30x)2 + (10 - 40x)2.

Let us denote this function by/'(x), and after squaring let us complete the square. f(x) = 400-1200x + 900x2 + 100-800x + 1600x2 =

= 2500x2 -2000x + 500 =

= 2500

k

= 25Oofx - —+100.

the distance of the two cyclists

10-40%

40x

Figure 28

THE QUADRATIC EQUATION

The function will have its minimum at x = —, i.e. the two cyclists will be 2 5 the closest to each other in — hours = 24 minutes. 5

This minimum distance will be V100 = 10 km. By this time the cyclist with a speed of 40 km/h will have ridden through the crossing.

Example 4 The price of gemstones and the square of their mass are directly proportional. We cut a stone of 1 gram, the price of which is 100 euro, into two parts. To what extent might the value of the gemstone decrease?

Solution Let the mass of the two resulting stones be x and 1 -x measured in grams. Based on the data it can be established that the total value of the two parts is: y(x) = 100x2 + 100(1 — x)2.

We are looking for the minimum of this expression. After expanding the bracket and then by completing the square the following quadratic function results:

y(x) = 100x2 +100(1—x)2 = = 100x2 +100(1 - 2x + x2 ) = 200x2 - 200x + 100 = = 200(x2 - x) + 100 = 200

+ 100 =

= 2Oofx-|l +50.

Since the image of this function is a parabola open upwards, it has a mini­

mum when the quadratic term is equal to 0, i.e. in the case of x = —. It means that the value of the original gemstone decreases to the largest extent when it is cut into exactly half. In this case the total value will be 50 euro, i.e. it will be exactly half of the original price.

To summarise: When determining the extreme values of quadratic expressions at first we complete the square: fix) = ax2 + bx + c = a

This function takes its extreme value at the place x =------ , ¿2 2a and its value will be------ + c. 4a

This extreme value is a minimum if a > 0, and a maximum if a < 0.

Exercises 1. Determine the extreme values of the following functions. a)f(x) = x2-4; b) g(x) - -x2 + 2; c) h(x) = 2(x - l)2 + 2. 2. Find the extreme values of the following functions on the set of real numbers. a) fix) = -x2 — 2x - 4; b) gix) - -2x2 + 6x + 2; c) hix) - 3x2 - 2x + 2. 3. Determine the extreme values of the function x i-> 2x2 + 4x - 6, if aJreR; b) x e [-3; -2]; c)xe[-2;l].

4. Give quadratic functions which have a minimum at the point a)A(l;0); b) Bi-2-2)c)C(3;-5). 5. Give quadratic functions which have a maximum at the point a)A(0;-10); b) BU\-2)-, c)C(4;-3). 6. Break down 30 as the sum of two numbers so that the sum

of squares of the terms will be the smallest possible. 7. We inscribe rectangles into a right-angled triangle with legs

of 10 cm and 15 cm as shown in the figure. Give the sides of the inscribed rectangle with the maximum area.

8. We divide a 20 cm long line segment into two parts, and then we draw semicircles above the parts as diameters. Give the minimum of the sum of the areas of the two semicircles. 9. Two beetles are moving towards the origin on two distinct axes of the coordinate system.

One of them sets off from the point A(40; 0) and it covers 4 units in every second. The other one sets off from the point B(0; 30) and it covers 2 units in every second. In how many seconds will they be the closest to each other, if they set off at the same time?

Puzzle

How does the length of the different coloured lines change, if we always half the line segments and we draw semicircles above them? 109

11. Problems leading to quadratic equations Throughout History there have been time periods in which contem­ porary mathematical results have been particularly significant. This is especially typical of the Ancient Greek society and of its art works.

Pythagoras played an important role in the Greek mathematics. He was bom on the isle of Samos around 580 BC. History does not contain much information about his life, but more information is known about his thoughts, the school founded by him and his followers (the Pythagoreans). While looking for harmony in Nature they proved several significant mathematical theorems which they sometimes believed to have mystical properties. It was fundamental for them that the nature and substance of all things was the number. Among these they especially honoured the value ascribed to ratios, which was introduced among historical ideas under the name golden section (or golden ratio). It is probably not surprising that this value pops up in several figures of Nature, in the pattern of leaves, in the sequence of musical sounds and in works of art.

Leonardo: The Virgin and Child with Saint Anne

It is a ratio, which - according to many people - reflects the most beautiful, most harmonious relationship between the whole and its part.

Ratios in the regular pentagon

An ancient exercise in connection with it is as follows:

Example 1

A1 m long line segment is given. Let us divide it into two parts so that the ratio of the shorter one to the longer one is equal to the ratio of the longer one to the whole segment.

b

c+b

a

Solution Let us denote the length of the longer line segment in question by x. According to the conditions of the example: x _ 1-x

o+b

Figure 29

1

x

One can assume that the Pythagoreans had already known this construction, since this proportion can also be found in the regular pentagon, which was used as a distinctive symbol.

After rearranging the equation we obtain a quadratic equation.

Let us prepare a short presentation about the golden section.

Since we are interested in the length of a line segment, the only solution for this example is:

X2 = 1-x,

x2 + x-l = 0,

110

Example 2 How many sides does the convex polygon have which has 7 more diagonals than sides?

Solution Let us denote the number of sides of the polygon by n. — 3) The number of the diagonals of the polygon is: ---- ----- , since /z - 3 diagonals can be drawn from any of the vertices, and while going round along all n vertices we take every diagonal into account exactly twice.

According to the conditions of the example: n+7=

n(n - 3)

2

In +14 = n2 — 3n,

n2 -5/?-14 = 0, 5±x/25 + 4-14 ”1,2 =--------- Z--------Among the resulting two roots only 7 can satisfy the conditions of the example. Then the number of the sides is 7, and the number of the diagonals will be 14. (Figure 30)

Example 3 The sum of a number plus twice another number is 36, the difference of their squares is 75. Let us determine these numbers.

Solution Let us denote the two numbers by x and y. According to the conditions: x + 2y= 36; x2-y2 = 75. The two equations together define quadratic simultaneous equations in two variables. A possible way of solving it is the substitution method learnt earlier, the idea of which is to isolate one of the variables from one equation and to substitute it into the other.

In this case it is practical to use this method with the first equation. It implies the following: x = 36 - 2y. By substituting it into the second equation a quadratic equation results, which can be solved the usual way:

(36 - 2y)2-y2 = 75,

1296- 144y + 4y2 — y2 = 75, 3y2 — 144y + 1221 = 0.

THE QUADRATIC EQUATION

According to the formula:

_ 144±V1442 -4-3-1221 _i Ji = 11 y’’2” 6 “l.v2=37. After this we can determine the value of x:

x1 = 36-22= 14 and x2 = 36 - 74 =-38. So two number pairs satisfy the conditions of the example:

and x2 ~ _38, y2 - 37.

X] = 14, y] = 11

Example 4

The class collected 720 euros for an excursion to rent a bus. Later on it turned out that two students could not take part in the excursion, so they got their money back, but each of the others had to pay 4 euros more. How many students were there in the class? (The cost of the bus does not decrease if there are fewer students.) Solution Let us denote the number of students in the class by x. So everyone j 720 , u 720 paid ----- euros. Later on the cost was -------- euros per person, and x x-2 it was 4 euros more than the previous cost, therefore: 720 720 _ x -2

x

By dividing both sides of the equation by 4, and then multiplying by x(x- 2): 180x - 180(x - 2) = x(x— 2),

180x - 180x + 360 = x2 - 2x,

x2 - 2x - 360 = 0. According to the formula:

2 ±^4+ 1440 ^1,2 =----------- Ô----------- =

2

fx1=20, i«

[x2 = -18.

As only the positive solution has meaning, there are 20 students in the class, and it indeed satisfies the conditions. ♦ ♦ ♦

To summarise: We can say that we answered the questions raised with the help of quadratic equations. At first we always decided which value to be calculated to choose as the unknown, and how it is related to the other values. Based on these we set up the equation. The correctness of the calculated value should be checked based on the relations in the exercise.

Exercises 1. Give the number for which the following is true: if we multiply the number by the number

1 greater, then the product is 25 greater than the original number. 2. A rectangle shaped plot with an area of 800 m2 at the riverside is enclosed with a 100 m long fence. Give the length of the sides of the plot. (There is no fence along the side which is next to the river.) 3. The area of a right-angled triangle is 55 cm2, its hypotenuse is x/221. Give the length

of the legs.

4. How much time does a diver jumping from a 10 m high platform have to perform an acrobatic exercise before entering the water? For a free-falling object on Earth, the relation between the distance covered and the time elapsed can be calculated based on the relation .s = 5i2, where s is the distance covered in metres, and t is the time elapsed in seconds.

5. A rectangle can be divided into two parts so that one of the parts is a square and the other part is a rectangle similar to the original one. Determine the ratio of the sides of the original rectangle. 6. One of the sides of a rectangular shaped room is 1.5 m longer than its other side. The area

of the room is 22 m2. How many metres of skirting is needed for the parquet, if we do not place skirting at the 90 cm wide door? 7. We plan an esplanade around a rectangular shaped flowerbed with 40 m and 60 m long sides. How wide should the esplanade be, if we wanted the flowerbed to be the 80% of the resulting garden? 8. Give the polygon in which the difference of the number of the sides and the number of the diagonals is equal to the quotient of the sum of the interior angles and 90°. 9. The total number of the diagonals of two convex polygons is 158, the sum of their interior

angles is 4320°. Give the number of sides of the polygons. 10. While preparing for my language exam I filled in the same amount of tests every day from

a book containing 720 tests. If I had solved 20 tests more each day, then I would have finished it 3 days earlier. How many days did I spend with filling in the tests? 11. An event planner company had 3840 euros income on

one occasion. Another time, when it sold the tickets for 30 euros more, although sold 4 tickets less, its income was 4200 euros. How many tickets were sold at each event and what were the unit prices? 12. A car was driven for 540 km. If the same distance were

covered at an average speed 10 km/h less, then it would take one hour longer. Give the average speed of the car. 13. Two trains are departing towards each other from towns A and B 300 km away from each other. The train, that departed from town A, arrived in town B 3 hours after the two trains met, and the other train arrived in town A 1 hour and 20 minutes after the two trains met.

Give the speed of the trains. 113 >......................

THE QUADRATIC EQUATION

14. A 4 m wide and 3 m high trailer truck would like to drive through the tunnel along the centre­

line of the road. The cross-section of the tunnel matches the shape of the graph of the quadratic •

9

function /(x) = -—xz + 4, where the unit is 1 metre on each coordinate axis. Can the trailer truck drive through the tunnel? 15. We can determine the depth of a ravine, if we measure the time we can hear the stone dumping

after letting it fall. How deep is the ravine, if we experience this time difference to be 10 s? (The value of gravitational acceleration is g - 9.81 m/s2, the speed of sound is 340 m/s.) 16. The length of the sides of a rectangular shaped fishpond are 120 m and 160 m. A pile is

driven along one of the diagonals of the rectangle from where the two end-points of the shorter side are seen at a right angle. What distance is the pile from the sides of the fishpond? 17. The area of a rectangular shaped and grass-covered park is 4000 m2. Two concrete roads

parallel with the sides intersect each other in the park. The area of one of the roads is 400 m2, the area of the other one is 250 m2. What percentage of the park’s area is covered by grass? 18. One part of the 3000 euro fund generated 700 euros of interest in a year, its other part

generated 1800 euros of interest. The interest rate of this latter part was 2% more than the interest rate of the first one. At what % interest rate and what amount of money did we hold? 19. Teams of 4 take part in a school cycling competition. Each school could enter the competi­

tion with one team only. At the start of the competition the participants introduced themselves to each other: every student shook hands with everyone else who was not a schoolmate, thus there were 336 handshakes all together. At the farewell party the day after the competi­ tion every student toasted a glass of cola with everyone else. How many clinks were there? 20. The chess players of two schools competed. Everyone played against everyone else once.

At first the games amongst the players of each school were played, and thus there were a total of 36 games. When the students of the two schools were playing against each other, then there were 42 games. How many students of each school took part in the competition? Puzzle

-

-

-

-

-

Is it possible that the ratio of a smaller number to a greater number is the same as the ratio of the greater one to the smaller one?

114

Geometry Based on the clay pots, which were found in 1936 in the course of diggings around Susa, located in today’s Iran, it seems so that the Babylonians used the intercept theorem and the concept of similarity about 4000 years ago. In the 5th century BC Hippocrates of Chios used the relation between the inscribed and the central angles of the circle, and the similarity in his proofs, and could construct the geometric mean of two line segments.

In the 3rd century BC Aristarchus of Samos performed such calculations and approximations in his only surviving work with the title

WIDENING THE KNOWLEDGE ABOUT CIRCLES

WIDENING THE KNOWLEDGE ABOUT CIRCLES 1. Reminder In our earlier studies we got familiar with many important concepts and facts in connection with circles. We summarise these now. Definition: A circle (boundary line of the circle) is a set of points in the plane, which are at a given distance from a given point in the plane. Definition: A tangent line of a circle is a straight line in the plane of the circle, which has exactly one point in common with the circle.

Theorem:

Exactly one tangent line can be drawn at each point of a circle, and the radius drawn to the point of tangency is perpendicular to the tangent line.

Definition: If the vertex of an angle is the centre of a given circle, then the angle is called the central angle of the circle.

(In figure 1 a is the central angle belonging to the arc AB.) Theorem:

If ia and ip are the lengths of the arcs belonging Z cz to the central angles a and P, then .

lp

Theorem:

P

If ta and tp are the areas of the sectors belonging t oc to the central angles a and /J, then — = tp P

The length of the arc belonging to the central angle a (rad) in a circle with radius r is:

Thales’ theorem: If we connect the two end-points of any of the diameters of a circle with any other point of the circle, then we get a right-angled triangle the hypotenuse of which is the diameter of the circle. The converse of the Thales’ theorem: The centre of the circumscribed circle of a right-angled triangle is the midpoint of the hypotenuse.

2. The theorem of the central and inscribed/tangent-chord angles Definition: If the vertex of a convex angle is on the circum­ ference of a given circle, and the two arms of the angle are on two chords of the circle, then the angle is called an inscribed angle of the circle. (Figure 2)

Definition: If the vertex of a convex angle is on the circum­ ference of a given circle, and one of the arms of the angle is on one chord of the circle and the other arm is a tangent line of the circle, then the angle is called a tangent-chord angle of the circle. (Figure 3)

The part of the boundary line of the circle, which is inside the angular domain, is the arc intercepted by or corresponding to the given inscribed angle.

The following are also typical phrasings: the inscribed angle corresponding to a given arc, the inscribed angle subtended by a given arc. In figure 2 the angle a is an inscribed angle corresponding to the arc AB, in figure 3 the angle /5 is a tangent-chord angle corresponding to the arc CD.

We can make an interesting observation when we recall the Thales’ theorem (Figure 4). The central angle corresponding to the arc AB (which in this case is a semicircle) is 180°, and the inscribed angles corresponding to this same arc are 90°, i.e. they are equal to the half of the central angle. Now we are going to prove that it is also true in general, i.e. the following theorem holds: Theorem:

A

f"

The measure of any inscribed angle or any tangent­ chord angle corresponding to a given arc in a given circle is half of the measure of the central angle corresponding to the same arc.

"

.0 __ K. 180°7

B

Figure 4

Proof One central angle and infinitely many inscribed angles and tangent­ chord angles correspond to a given arc in a given circle; therefore we have to consider the different mutual positions of the inscribed angles/tangent-chord angles and the central angle. The proof has more steps to match these positions.

(1) The centre of the circle lies on one of the arms of the inscribed angle. (Figure 5) The angle AOB is an exterior angle of the isosceles triangle COB, therefore /3 = AOB< = OBC< + BCO< = 2a.

117

(2) The centre of the circle is an interior point of the angular domain of the inscribed angle. (Figure 6)

Let us draw the diameter of the circle passing through the point C, and let its other end-point opposite C be D. The position of the inscribed and central angles corresponding to the arcs AD and DB matches the previous cases, thus /3] = AOD< = 2ACD< = 2al, p2 = DOB< = 2 -DCB< = 2a2, which implies P - j3| + P2 - 2at + 2a2 - 2(at + a2) = 2a. Figure 6

(3) The centre of the circle is a point outside the angular domain of the inscribed angle. (Figure 7)

Let us draw the diameter CD again. The position of the inscribed and central angles corresponding to the arcs DA and DB matches the ones in case (1); therefore /3, = DOB< = 2 • DCB< = 2a{, /32 = DOA< = 2 • DCA< = 2a2, thus P = px - p2- 2a} - 2a2 - 2(cq - a2) = 2a.

In the first three steps we have shown the theorem for all possible positions of inscribed angles. In the next step we are going to prove it for the tangent-chord angles.

Figure 7

(4) Let a denote the tangent-chord angle subtended by the arc AB. (Figure S)

X

r A

a= 90°

a< 90°

90° < a

Figure 8

a) If a < 90°, and OT is the altitude belonging to the base of the isosceles triangle AOB, then the tangent-chord angle with vertex A and the angle AOTare angles with mutually perpendicular arms, thus AOT mA. 3. We construct the perpendicular bisector of the line segment AB —>fAB. 4. mA n fAB = O is the centre of the circle in question.

As we can construct the tangent-chord angle with vertex A and with a measure of a on both sides of the straight line AB, therefore two suitable arcs are resulting, which are located symmetric about the straight line AB (Figure 14). The resulting open arcs (not containing the end-points) are the two arcs of viewing angles with a measure of a of the line segment AB. After solving example 1 we have to examine whether there are any planar points other than the points of the arcs of viewing angles from which the line segment AB can be seen at angle a. In figure 15 it can be seen that the viewing angle is greater than a when measured at a point inside the arc, and is less when measured at a point outside the arc. 122

As a consequence of the theorem of inscribed angles we have proven the following theorem: Theorem:

i

i

The set of points in the plane from which a given line segment AB of the plane can be seen at a given angle a (0° < a < 180") is as follows: two open arcs that are symmetric about the straight line AB.

In figure 16 the shapes of the arcs of viewing angles can be seen depending on the measure of angle a.

Figure 76

Example 2

Let us construct a triangle if one side, the angle opposite and the altitude belonging to the given side are given. Solution The side BC = a, the angle a and ma are given. (Figure 17) The side BC can be seen at a given angle a from A, therefore A lies on one of the arcs of viewing angles with a measure of a of the line segment BC. A is at a distance of ma from the straight line BC, therefore it must also lie on one of the two straight lines parallel with the straight line BC and being at a distance of ma from it.

The points corresponding to the third vertex of the triangle are then resulting as the intersection points of the constructible arcs of viewing angles and the also constructible straight lines parallel with the straight line BC and being at a distance of ma from it. (Figure 18)

4, 2 or 0 solutions can result for A, depending on how many points the parallel straight lines and the arcs have in common. The resulting triangle however - to the tune of congruence - is unam­ biguously defined.

123

Exercises 1. Let us take a line segment. Construct the set of points from which the line segment can

be seen at an angle of a) 45°; b) 60°;

c) 90°;

d) 120°;

e) a.

2. The measure of the angles of a triangle are a) 30°, 60°, 90°; b) 40°, 60°, 80°; c) 36°, 58°, 86°;

d) a, /3, y.

What is the included angle between the straight lines of the sides and the tangents of the circumscribed circle of the triangle passing through the vertices? Calculate the angles of the triangle defined by the tangent lines. 3. Place a chord in a circle the length of which is equal to the radius of the circle. Give the

angle this chord can be seen at from the points of the circle. 4. The angle included between two tangent lines passing through a point is 76°. Give the angle

the chord connecting the points of tangency can be seen at from the points of the circle. 5. Take a straight line, two points not lying on the straight line and a convex angle. Construct

a point on the straight line from which the line segment defined by the given points can be seen at angle a. 6. Construct a point inside a right-angled triangle from which each of the two legs can be seen

at 120°. Give the angle the hypotenuse can be seen at from this point. 7. Construct a triangle if the length of one side and the median

belonging to this side, and the measure of the angle opposite the given side are given. 8. The auditorium of a theatre can be seen from above in the

figure. There are side-boxes along the longer side of the auditorium. Find the side-box (the point of the longer side of the auditorium) from which the stage can be seen at the largest

side-boxes

J CD

S C/5

auditorium

1________ side-boxes

Puzzle

A circle with 4 cm radius is given without its centre, and there is an isosceles right-angled triangle shaped ruler with a leg of 20 cm. Find the centre of the circle given with the help of this ruler and a pencil. (Except for the ruler and the pencil nothing else can be used.) X____________________________________________________________________________ /

4. The theorem of inscribed quadrilaterals (higher level courseware) We saw it earlier that there can be a circle drawn around any triangle. It is not so for quadrilaterals, since for example there is no such circle for any concave quadrilateral which would pass through all four vertices. There are also such quadrilaterals among the convex quadrilaterals - for example the rhombi which are not squares which do not have a circumscribed circle. (Figure 19) Henceforth we are going to examine those quadrilaterals around which a circle can be drawn.

Definition: The quadrilaterals, which have circumscribed circles, are called inscribed quadrilaterals.

It is equivalent to the following: The quadrilaterals, the sides of which are chords of a circle, are called inscribed quadrilaterals.

inscribed quadrilateral

Let us consider an inscribed quadrilateral and let us draw its circum­ scribed circle. (Figure 20)

Let us examine two opposite angles of the inscribed quadrilateral ABCD\ a and y. a is the inscribed angle corresponding to the arc BD containing the vertex C, y is the inscribed angle corresponding to the arc BD containing the vertex A, thus resulting from the theorem of the central and inscribed/tangent-chord angles the corresponding central angles are 2a and 2y. However the sum of these is the complete angle, i.e.

2a + 2y = 360°, which implies a+y = 180°.

Since the sum of the interior angles of the quadrilateral is 360°, thus the sum of the other two opposite angles is also 180°.

The

theorem of inscribed quadrilaterals: The sum of two

opposite angles in any inscribed quadrilateral is 180°.

It can be proven that the converse of the theorem is also true, i.e. Theorem:

If the sum of two opposite angles in a quadrilateral is 180", then the quadrilateral is an inscribed quadri­ lateral.

The sum of two opposite angles being 180° is a necessary and sufficient condition for a quadrilateral to be an inscribed quadrilateral. The combination of the theorem and its converse: Theorem:

A quadrilateral is an inscribed quadrilateral if and only if the sum of its opposite angles is 180".

125

WIDENING THE KNOWLEDGE ABOUT CIRCLES

Example 1 Let us prove that a trapezium has an axis of symmetry not passing through any of its vertices if and only if the trapezium is an inscribed quadrilateral. Solution Let us assume that the trapezium has an axis of symmetry not passing through its vertices. From the symmetry it is resulting that the angles on each base of the trapezium are equal, i.e. a = ¡5 and y = 8 (Figure 21). Thus 2a+2y=360°, therefore a+ y- \80°. Figure 21

Half of the statement is resulting from the converse of the theorem of inscribed quadrilaterals. For the sake of the proof in the other direction let us assume that the trapezium is an inscribed quadrilateral. Then

a+y=180° and /3 + 0; b>0)

Geometric mean (altitude) theorem: The length of the altitude belonging to the hypotenuse in a rightangled triangle, is the geometric mean of the length of the two line segments the altitude divides the hypotenuse into.

geometric mean (altitude) theorem

148

Example 1 Two line segments with length a and b are given. Let us construct a line segment with a length of Jab.

Solution Based on Thales’ theorem and the geometric mean (altitude) theorem the process of the construction is as follows (Figure 50):

1. Constructing a circle above the line segment AC with a length of a + b as diameter.

2. Constructing a perpendicular at B to AC; it intersects the circle at D. 3. BD=Jab.

Example 2

Let us prove with the help of geometry that the arithmetic mean of two positive numbers is always greater than or equal to the geometric mean of the same numbers, i.e. if a and b are positive numbers, then ^-^->Jab. What numbers is the equality satisfied by? 2 Solution Based on the geometric mean (altitude) theorem and the first example the proof can be seen in figure 51.

Figure 51

It can also be seen in the figure that the equality is satisfied if and only if a = ¿>. ♦ ♦ ♦

2. Let us now consider the similarity of one of the smaller triangles and the original right-angled triangle.

Since CTB^~ACBh, therefore CB __ TB AB ~ CB

which implies CB2 = AB TB.

Using the notation of figure 49: a2 = cp or a- y[cp.

If we start with the similarity ATC^~ACB^, then the following is resulting: b2 - cq or b = y/cq. The relation that resulted for the leg of the right-angled triangle is the geometric mean (leg) theorem.

Geometric mean (leg) theorem: The length of the leg of a rightangled triangle is the geometric mean of the length of the hypotenuse and the length of the perpendicular projection of the leg on the hypotenuse.

geometric mean (leg) theorem

149

THE SIMILARITY TRANSFORMATION AND ITS APPLICATIONS

With the help of the geometric mean (leg) theorem we can give another proof for the Pythagorean theorem, according to which if the length of the legs of a right-angled triangle are a and b, and the length of the hypotenuse is c, then a2 + ¿2 = c2. Using the notation of figure 49 a2 + b2 = cp + eq = c(p + q) = c2.

Example 3 The ratio of the length of the legs in a right-angled triangle is 5:3. The altitude belonging to the hypotenuse divides the hypotenuse into two line segments; one of these is 4 cm longer than the other one. Let us calculate the length of the hypotenuse and the altitude belonging to the hypotenuse.

c

Figure 52

Solution Let a and b denote the length of the legs, c the length of the hypotenuse, x the length of the shorter part of the hypotenuse measured in centimetres. (Figure 52)

Based on the geometric mean (leg) theorem a2 = ex and b2 = c(x + 4).

Since a: b = 3 :5, therefore ex

a2 _ b2

c(x + 4)

_

x x+4

_ 9 25’

which implies x = — cm = 2.25 cm. 4 17 The length of the hypotenuse is: c = 2x + 4 = — cm = 8.5 cm.

The length of the altitude belonging to the hypotenuse, based on the geometric mean (altitude) theorem, is: m = Jx(x + 4) = .

v

150

U 4

cm = — cm = 3.75 cm. 4

The tangent-secant theorem (higher level courseware) Let us take a circle and a point P outside the circle. Let us draw a tangent line and a secant line through the point P to the circle; let the point of tangency be E, let the intersection points of the secant line and the circle be A and B. (Figure 53)

Figure 53

Our aim is to find the relation between the tangent line segment PE and the secant line segments PA and PB. Based on the theorem of inscribed/tangent-chord angles ABE< = PEA sin 16°14’ = 0.2784 + 0.0011 =0.2795.

Solution (b)

On Triangles.

Based on the table in figure 71:

cos 79°30’= 0.1822 -> cos 79°31’ = 0.1822 - 0.0003 = 0.1819.

Figure 71

The sine and cosine of angles 0’

10’

20’

50’

1’

60’

2’

3’

4’

5’

6’

7’

8’

9’

ein 16°14’

9 9 9 8 8

11 11 11 11 11

14 14 14 14 14

17 17 17 17 17

20 20 20 20 20

23 23 23 23 23

26 26 26 25 25

O.27Ô4 +O.OO11

8 11 14 8 C© 14

20 20 19 19 19

22 22 22 22 22

25 25 25 25 25

7’

8’

9’

0.1794 0.1851 2022 1994 1965 2193 2164 2136 2334 2363 2306 2504 2532 2476

0.1880 2051 2221 2391 2560

0.1908 2079 2250 2419 2588

0.2672 2840 3007 3173 3338

0.2700 2868 3035 3201 3365

0.2728 2896 3062 3228 3393

0.2756 2924 3090 3256 3420

74" 73“ 72“ 71“ 70"

30’

20’

10’

0’

0.1736 1908 2079 2250 2419

0.1765 1937 2108 2278 2447

15“ 16" 17” 18" 19“

0.2588 2756 2924 3090 3256

0.2616 0.2644 2812 2952 2979 3118 3145 3283 3311

50’

40’

79" CD 6 78" 3 6 3 6 77" 3 6 76" 75" 3 6

10» 11“ 12“ 13“ 14“

60’

30’

40’

3 3 3 3 3

6 6 6 6 5

r

2’

8 8 8

11 11 11

14 14 14

17 17 17 17 16

3’

4’

5’

6’

165

0.2795 cos

79’31’

O

1 P>??

-0.0003

0.1Ô19

TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES

Solution (c) Based on the tangent and cotangent table in figure 72:

tg 80°41’ = 6.096.

Figure 72

The tangent and cotangent of angles 0’

1’

2’

3’

4’

5’

6'

7’

8'

9’

10’

79”0’ 79°10' 79»20’ 79»30’ 79°40’ 79°50’

5.145 5.226 5.309 5.396 5.485 5.576

5.153 5.234 5.318 5.404 5.494 5.586

5.161 5.242 5.326 5.413 5.503 S.595

S.169 5.250 5.335 5.422 5.512 5.605

5.177 5.259 5.343 5.431 5.521 5.614

5.185 5.267 5.352 5.440 5.530 5.623

5.193 5.276 5.361 5.449 5.539 5.633

5.201 5.284 5.369 5.458 5.549 5.642

5.209 5.292 5.378 5.466 5.558 5.652

5.217 5.301 5.387 5.475 5.567 5.662

5.226 5.309 5.396 S.485 5.576 5.671

10°50’ 10°40' 10»30’ 10»20’ 10»10’ 10°0’

80°0’ 80»10’ 80“20’ 80»30’ 80“40’ 80“50’

5.671 5.681 5.769 5.779 5.871 5.881 5.976 5.986 6.084 CÉL09ÎD 6.197 6.209

5.691 5.789 5.892 5.997 6.107 6.220

5.700 5.799 5.902 6.008 6.118 6.232

5.710 5.810 5.912 6.019 6.129 6.243

5.720 5.820 5.923 6.030 6.140 6.255

5.730 5.830 5.933 6.041 6.152 6.267

5.740 5.840 5.944 6.051 6.163 6.278

5.749 5.850 5.954 6.062 6.174 6.290

5.759 5.861 5.965 6.073 6.186 6.302

5.769 5.871 5.976 6.084 6.197 6.314

9"50’ 9°40’ 9°30’ 9»20’ 9“10’ 9»0’

81°0' 81"10’ 81°20’ 81°30’ 81 ”40’ 81“50'

6.314 6.435 6.561 6.691 6.827 6.968

6.326 6.447 6.573 6.704 6.841 6.983

6.338 6.460 6.586 6.718 6.855 6.997

6.350 6.472 6.599 6.731 6.869 7.012

6.362 6.485 6.612 6.745 6.883 7.026

6.374 6.497 6.625 6.758 6.897 7.041

6.386 b.510 6.638 6.772 6.911 7.056

6.398 6.522 6.651 6.786 6.925 7.071

6.410 6.535 6.665 6.799 6.940 7.085

6.423 6.548 6.678 6.813 6.954 7.100

6.435 6.561 6.691 6.827 6.968 7.115

8»50’ 8”40’ 8°30’ 8°20’ 8°10’ 8»0’

10’

9’

8’

7’

6’

5’

4’

3’

2’

1’

0’

Solution (d) In figure 73 another part of the previous table can be seen. Based on this: ctg 73° = 0.3057 -> ctg 73°3’= 0.3057 - 0.0009 = 0.3048. The tangent and cotangent of angles

ctg 73°3’

O.3O57 -O.OOO9 0.304Ô

•

O’

10’

20’

30’

40’

50’

60’

0° 1" 2» 3° 4"

0.0000 0175 0349 0524 0699

0.0029 0204 0378 0553 0729

0.0058 0233 0407 0582 0758

0.0087 0262 0437 0612 0787

0.0116 0291 0466 0641 0816

0.0145 0320 0495 0670 0846

0.0175 0349 0524 0699 0875

5” 6° 7» 8° 9°

0.0875 1051 1228 1405 1584

0.0904 1080 1257 1435 1614

0.0934 1110 1287 1465 1644

0.0963 1139 1317 1495 1673

0.0992 1169 1346 1524 1703

0.1022 1198 1376 1554 1733

10° 11" 12° 13° 14°

0.1763 1944 2126 2309 2493

0.1793 1974 2156 2339 2524

0.1823 2004 2186 2370 2555

0.1853 2035 2217 2401 2586

0.1883 2065 2247 2432 2617

0.1914 2095 2278 2462 2648

15» 16“ 17" 18» 19»

0.2679 2867 3057 3249 3443

0.2711 2899 3089 3281 3476

0.2742 2931 3121 3314 3508

0.2773 2962 3153 3346 3541

0.2805 2994 3185 3378 3574

60’

50’

40’

30’

20’

166

1’

2’

3’

4’

5’

6’

7’

8’

9’

89» 88“ 87" 86» 85»

3 3 3 3 3

6 6 6 6 6

9 9 9 9 9

12 12 12 12 12

15 15 15 15 15

17 17 18 18 18

20 20 20 20 21

23 23 23 23 23

26 26 26 26 26

0.1051 1228 1405 1584 1763

84» 83» 82» 81» 80»

3 3 3 3 3

6 6 6 6 6

9 9 9 9 9

12 12 12 12 12

15 15 15 15 15

18 18 18 18 18

21 21 21 21 21

24 24 24 24 24

26 27 27 27 27

0.1944 2126 2309 2493 2679

79° 78° 77° 76° 75°

3 3 3 3 3

6 6 6 6 6

9 9 9 9 9

12 12 12 12 12

15 15 15 15 16

18 18 18 18 19

21 21 21 21 22

24 24 24 25 25

27 27 27 28 28

0.2836 0.2867 74» 3026 73° 3217 3249 72» 3411 3443 71» 3607 3640 70»

3 3 3 3 3

6 9 13 6 C€>13 6 10 13 6 10 13 7 10 13

16 16 16 16 16

19 19 19 19 20

22 22 22 23 23

25 25 26 26 26

28 29 29 29 30

4’

5’

6’

7’

8’

9’

10’

0'

1’

2’

3’

Example 2

Give the length of the straight slope that elevates 18 m from its starting point till its end point and has an angle of inclination (to the horizontal) of 10°. Give the distance of the starting point to the end point of the slope measured horizontally.

X

Figure 74

Solution Based on figure 74

— = sinlO°, I

from which the length of the slope is / = —— m = 103.69 m. sin 10° 0.1736

If x denotes the horizontal distance, then — = cos 10°, I

which implies X = I • cos 10° = 103.69 • 0.9848 m = 102.12 m.

Exercises 1. In a right-angled triangle a and b denote the length of the legs, c denotes the length of the hypotenuse, and a is the angle opposite the leg with the length of a. Calculate the sine, cosine, tangent and cotangent of a, if a) a = 3, b = 4;

c) a = 1, c = >/2;

b) a = 5, c = \3;

d) b = —, c = 2.

2. Determine the sine, cosine, tangent and cotangent of the following angles to four decimal

places. a) 36°;

b) 52°40’;

c) 11°32’;

0 it is unidirectional with a, in the case of a < 0 it is in the opposite direction as a.

2. If a = 0, then a-a = 0 for any real number a.

It can be proven that the identities below are satisfied for the multiplica-

Exercises 1. Six vectors defined by the four vertices of a square can be seen

in the figure. Choose the ones that are a) equal; b) parallel;

c) negatives.

2. Express the following vectors by taking the vectors of the

previous exercise as a basis. a) a + b;

b) c-d;

c) b + d;

d) a + b + d;

e) b + e;

f) d-e;

g) c + d;

h) e +f;

i) (e-c) + a;

J) (e-f) + (3-b).

3. Take a different from 0, and construct the following vectors. _ 3 a) 2 a;

b) -3 a;

e)~(3-a);

f)-j2-Us-a\

c)------ a;

4

4. Take the non-parallel vectors a and b different from the zero vector. Construct the following

vectors. a)a + b;

b)a-b;

e)-(a + b);

f) 2-

d) a——;

c)2 a + b;

5. Give a simpler form of the following vectors.

a) 2-(a+ b} + (3-a-b);

b,

2

c) -4-(b + 3-d) — \/2-(—d — 4-b). 6. Prove that if it is satisfied for the vectors a, b, c, d that b - a = c - d.

=----- b H------- d,J then

v3

3

7. The edge vectors starting from the vertex O of the cube shown in the figure are a, b, c.

Express the following vectors with the help of these three vectors. a) OF;

b) OD;

c) OG;

d) GB;

e) DE;

f) FD;

g) DA;

h) CF.

187

3

2. Expressing vectors as the sum of components in different directions Relation between parallel vectors Let a and b be parallel vectors not equal to the zero vector. The vectors ;—; • a and —L • b are parallel vectors with unit length.

lai

aP = ^-o lai

|Z>|

Definition: If a * 0, then a = —— • a is the unit vector uni\a | directional with a.

(¿7*0)

|ae| =1

a = I a I • oe

If ae and be are unidirectional, then according to the definition of the equality of vectors they are equal, i.e.

which implies

If ae and be are in the opposite directions, then they are each other’s negatives, which implies that

With the simple train of thought above we have proven the following theorem: Theorem:

If a and b are parallel vectors not equal to the zero vector, then there exists exactly one real number a different from 0, so that b - a • a. IKI If a and b are unidirectional: a = -—- > 0.

lai

b If they are in the opposite directions: a = --— < 0. Ia|

The theorem directly implies that all the vectors parallel with a vector a (a * 0) can be deduced as the scalar multiplication of a.

................................

188

Example 1

a and b are parallel vectors in the opposite directions, lol = V3, Id I = 5. Let us express a in terms of b. Solution

-

=

1

A 1 A According to the condition ' ' =ï ■

ae and be are

negative vectors, i.e. After the substitution

which implies

Expressing planar vectors as the sum of components in different directions Example 2 The midpoint of the side AB of the triangle ABC is F. Let us express CF as the sum of vectors parallel with CA and CB.

Solution For the sake of simpler notation:

G4 = a, CB = b and CF = f.

So that we can use the parallelogram rule of the addition of vectors let us draw the two midlines of the triangle passing through F. Let the midpoint of the side AC be D, and the midpoint of the side BC be E. (Figure 101)

About the midlines of the triangle we know that they are parallel with the corresponding side of the triangle, and their length is half the length of the side considered. So the quadrilateral CDFE is a parallelogram. Therefore f = CD+ CE.

However since D and E are midpoints, therefore CD = —a

2

and

CE = —b.

2

So

189

Example 3

Let the trisection point closer to the side B in the triangle AB in example 2 be W, and the vector pointing from C to H be h. Let us express h as the sum of vectors parallel with the vectors a and b. Solution Let us draw again a straight line through H parallel with the sides AC and BC of the triangle, let the resulting intersection points be K and L. (Figure 102)

Based on the intercept theorem: CK = --a

3

and

CL=-b.

3

Thus

1 -, 3

2 3

h = — -a + — ■b.

The vector pointing to the other trisection point of the line segment AB: Figure 102 r

2-a + b

3

a + 2-b

Example 4 The midpoint of the hypotenuse AB of the right-angled triangle ABC is F. Let CA = a, CF = f and CB = b. Let us break ¿into components parallel

with a and f.

(Figure 103)

Solution Example 2 implies that

7

1 2

1 Ü 2

f = — ■« -l— b,

b = 2-f — a.

This algebraic type of solution can be represented graphically too, if we reflect the triangle about the point F. (Figure 104)

190

Example 5

We hold a 2 kg ball hanging on a non-tensile rope with the horizontal force F at equilibrium, as shown in figure 105 (g = 10 m/s2). Give the force F and the tension force acting on the rope. Solution The ball is at equilibrium, therefore the resultant force of the force K applied in the rope and the weight G of the ball (vector sum) is equal to the negative of F, i.e. is equal to the force -F. Since | G| - m • g = 20 N and |-F| = |f|, therefore based on figure 106: |F| = wg-tan30° = 20~ = 11.55 N. x/3 Based on the above and the figure:

\k\=

sin 30°

=2-|f| = 23.1

N.

♦ ♦ ♦

The previous examples were the special cases of a very important theorem. Theorem:

If the non-parallel vectors a and b different from the zero vector are given, then every vector v of the plane defined by them can be expressed in the form v - a- a + ft-b,

where a and P are unambiguously determined real numbers depending only on v in the case of any given v. Figure 107

Proof Case I: v is parallel with one of the vectors. If v is parallel with a, then, according to the theorem previously proven, there exists one and only one real number for which v = a-a. Then v = a-a + Ob. The statement is also similarly resulting if v is parallel with b: _ If v = 0, then a = fl = 0.

v = 0 a + ft b.

Case II: v is not parallel with either a or b. Let us take the vectors a, b and v from a common starting point. In this case v is in one of the four angular domains defined by the straight lines of a and b, i.e. exactly one of the four cases shown in figure 107 is fulfilled. It can also be seen in the figure that if we draw straight lines through the end-point of v parallel with the straight lines of the two vectors given, then a parallelogram is resulting in each case the diagonal vector of which is v. And it means that vcan be expressed as the sum of vectors parallel with the given vectors a and b, i.e. v = a-a + p- b. Based on the theorem proven for parallel vectors it is obvious that in the case of a given v a and p are unambiguously defined. With this we have finished the proof.

191

VECTORS

Definition: If for the vectors a, b and v of the plane v = a-a + p-b, where a and P are real numbers, then v is the linear combination of the vectors a and b.

linear combination of vectors

The theorem can also be phrased in a different way with this new concept: Theorem:

In the plane: 2 base vectors.

If the non-parallel vectors a and b different from the zero vector are given, then every vector v of the plane defined by them can unambiguously be expressed as the linear combination of the vectors a and b.

In this case the vectors a and b are usually called base vectors. It can be shown with a train of thought similar to the train of thought of the proof above that the spatial equivalent of this planar theorem also holds.

Exercises 1. e and a are parallel vectors, and I e I = 1. Give I a I, if a) a = 2-e;

b) a = -3-e;

c)e=y/3-a',

2 _

e)----- a = 4-e.

3

2. The vector pointing from the vertex A of the regular triangle ABC to the orthocentre M is m, the vector pointing to the midpoint F of the side BC is/ Prove that m and/ are parallel vectors. If in = a f, then give a. 3. Let A, B, C, D, E be arbitrary points of the plane. Determine the value of the real number a, if AB + BC+ CD = a XdE + EA).

4. In the parallelogram ABCD let AB = b and AC = c. With the help of b and c express the vector (so as the linear combination of b and c) pointing from the vertex A to the midpoint of the side BC.

5. Divide the sides of the triangle ABC into four equal parts. The point of division on the side AB closer to A is N, the one closer to B is M. Express CM and CM with the help of CA

and CB (so as the linear combination of CA and CB). 6. In the triangle ABC AB = 4, BC = 5, CA = 6. The interior angle bisector starting from the

vertex A intersects the side BC at P. Give the value of a and [3, if AP = a ■ AB + / ■ AC. 7. We took the point P on the extension of the side BC beyond C of the triangle ABC so that

BP: CP = 9:5. Break AP into components parallel with AB and AC.

8. A body with a mass of 2 kg is anti-frictionally sliding down a slope with an angle of inclination of 30°. Break the gravity force acting upon the body into components perpendicular to and parallel with the slope. a) Give the magnitude of these components. b) Give the acceleration of the body.

193

VECTORS

3. Applying vectors in the plane and in space Many geometric exercises and problems can be solved quickly and graphically, “in an elegant way” with the help of vectors. The application of vectors makes it also possible to treat the planar and spatial problems in a uniform way. Henceforth we are going to see such examples. However first we introduce an important and a quite useful concept from the point of view of the solution of the example.

Let the point O be the origin of a reference system. Let us assign the OP to every single point P of the plane (space). (Figure 109)

Thus we gave a one-to-one correspondence between the points and the vectors of the plane (space) since (1) there exists exactly one directed line segment pointing from Oto P

for any point P, and it unambiguously defines the OP',

Figure 109

(2) for any vector in the plane (space) there is exactly one representative with O as the starting point, and the end-point of this is unambigu­ ously defined.

The origin of the reference system is called the reference point O,

reference point, position vector

OP is called the position vector of the point P. The position vector of the point O is the zero vector.

The usual notation of the position vector of the point P: OP = p.

Example 1 Let us prove by applying vectors that the medians of a triangle intersect each other at one point.

Solution

C

We have already proven this statement in two different ways. We are going to show a third proof with the help of vectors.

Let the position vectors pointing from an arbitrary but fixed reference point O to the vertices of the triangle ABC be a, b, c, let the position vector of the midpoint F of the side AB bcf. and the position vector of the trisection point S of the median CF closer to F be s. (Figure 110)

Our aim is to express s with the help of a, b and c. Based on example 2 on page 189 Figure 110

— ■a + —b.

2

2

(1)

Based on example 3 on page 190

(2) 194

Let us substitute (1) into (2), then by applying the identities relating to the vector operations let us transform the resulting expression.

_

2 i1 3 1^2

_

1 /A 1 _ 2 J 3

1 _ 3

1 / 3

1 _ 3

a+b+c

3

The following results: _ a+b+c s =------------ .

3

We can get the position vector of the trisection point closer to the side on another median so that we switch the role of the vectors a,_b, c accordingly. However this switch does not change the sum a + b + c, thus the result stays the same. And it means that no matter which median’s trisection point closer to the side we consider, the position vector of each of them is s, i.e. the same point trisects the medians.

So with the help of the vectors we simultaneously proved that the medians intersect each other at one point and also that this point is the trisection point of the medians closer to the side. ♦ ♦ ♦

Definition: The line segment connecting a vertex of the tetra­ hedron (triangular based pyramid) with the centroid of the face opposite is the median of the tetrahedron.

The tetrahedron has four medians.

median of the tetrahedron

Example 2 Let us prove that the medians of the tetrahedron intersect each other at one point, at the point of division dividing the median into four equal parts and the one that is closest to the face.

Solution The position vectors of the vertices of the tetrahedron ABCD, in point of a fixed reference point O are a, b, c, d respectively, the position vector of the centroid SD of the triangle ABC is sD, and the position vector of the point of division 5 (dividing the median into four equal parts) of the median DSd closer to is s. (Figure 111) Based on example 1 :

(3) If the position vector of the midpoint F of the median

7=

2

(4)

DSd is/, then Figure 111

VECTORS

Since .S’ bisects the line segment FSD, therefore

(5)

2 Let us first substitute (4) into (5).

3?n + d

s = --------

’

2

(6)

(6) actually gives the position vector of the corresponding point of division (dividing the median into four equal parts) of the median considered in terms of the position vectors of the end-points. Let us substitute (3) into (6). a + b +c

-t

+

a + b +c +d

4

The position vector of the point of division (dividing the median into four equal parts) of the median DSd and closer to the face is _ a + b +c + d s —------------------ .

4

The above expression of ? remains unchanged in point of switching the roles of the position vectors of the vertices, therefore the position vector of the point of division (dividing the median into four equal parts) of any median closer to the face is ?. With this we have proven that the medians of the tetrahedron intersect each other at one point. Definition: The intersection point of the medians of the tetra­ hedron is the centroid of the tetrahedron.

Example 3 Let us prove that the midlines of any quadrilateral bisect each other.

Solution Reminder: the line segments connecting the midpoints of the opposite sides of a quadrilateral are called the midlines of the quadrilateral.

Let the position vectors of the vertices of the quadrilateral ABCD relating to a fixed reference point O be 5, b, c, d respectively. (Figure 112)

The position vector of the midpoint of the side AD is ° + b +c

~

, the position

vector of the midpoint of the side BC is —-—. The position vector of the midpoint F of the midline connecting these is ■7 _

} ~

196

a+d 2

b+c

2

2

_ a + b +c + d ~ 4 ’

Figure 112

We can give the midpoint of the midline connecting the midpoints of the sides AB and CD the same way, except for swapping the position vectors of the vertices in the corresponding way. However the expression off remains unchanged when switching the roles of the position vectors of the vertices, thus the midpoints of the two midlines have the same position vector, namely/. With this we have shown the statement. It is an important consequence of the statement just proven that the midpoints of the sides of any quadrilateral define a parallelogram (Figure 113). It can easily be seen if we take into account that the diagonals of the quadrilateral defined by the midpoints are the midlines of the original quadrilateral about which we have just proven that they bisect each other.

Example 4

Let us consider a line segment AB, and let us fix a reference point 0. Let R be the point of the line segment AB for which AR:RB = p:q. Let us give the position vector of the point R with the help of the position vectors of the points A and B.

Solution Using the notation of figure 114: r -a + AR.

The AR is unidirectional with AB, and the condition implies that AR = —AB. p+q

Since AB = b - a, therefore AR = — p+q

VECTORS

Let us substitute it into the expression of r above. P+q

p+q

(p + q)-a+ p-b - p-a

q-a + p-b

p+q

P+Q

Midpoint: P = q = 1.

So the following is resulted:

Trisection points: P = 1, a+b .. -3-a+2-b a) a + b', b)a-b', c) b - a; a) 2 a + 3 b; e)------- ; j) --------------- . 3. The vertices of the triangle ABC are 4(-l; 1), 8(2; 3), C(0; -3). The triangle ABC is taken to the triangle AB'C' by the transformation of central dilation with the origin as the centre

and A as the scale factor. Give the coordinates of the vertices of the image triangle, if the value of A is a) -1;

b) 2;

C>3'

d) -3;

A. The vertices of the parallelogram ABCD are denoted by letters according to the positive

orientation, and the coordinates of three vertices are given: 4(1; 5), 8(-2; 3), C(-3; -5). Calculate the coordinates of the vertex D. 5. Give the coordinates of the a) midpoint;

b) trisection point

of the line segment defined by the points 4(2; -4) and 8(1; 7).

.....................

202

Trigonometrie functions Astronomy, sailing and other practical needs evoked the first “tables of chords”: the ancestors of the trigonometric functions. The great scientist of the Hellenistic period, Ptolemy (2nd century AD) created such a table ranging from Oto 180 degrees by increments of 30 minutes. The sine and cosine tables that we know today were first created by the Indian mathematicians. The tangent and cotangent function were introduced by the Arabic scientists in the middle Ages. Even the book of Ptolemy survived in Arabic translation only with the title “Almagest”. This book was an important tool for the astronomers for almost a thousand years.

François Viète (1540-1603), the French mathematician of the Renaissance period contributed greatly to the exploration of the relations between the trigonometric functions and their identities.

sinx

TRIGONOMETRIC FUNCTIONS

1. The definition and the simple properties of the sine and the cosine function In geometry we have already got familiar with the trigonometric functions of the acute angles. Now we are going to extend these definitions to any arbitrary rotation angle, naturally by paying attention that the old definition holds for the acute angles.

Our aim is to allow talking about the sine and the cosine of an arbitrary real number, thus we choose the following way. The rotation angles can be measure with real numbers, it is called the radian measure. The sine and the cosine of the angle should also mean the sine and the cosine of the real number serving as the radian measure of the angle.

Let us recall what we learnt about the radian measure. Let us now place the rotation angle in the coordinate system so that its vertex coincides the origin, one of its arms coincides the positive half of the x-axis (it is sometimes called the “initial arm”, and the other arm is called the “terminal arm”).

2n (rad) = 360° y (rad) = 90°

| (rad) = 60°

The unit vector e pointing to the terminal arm of the angle can be derived from the unit vector i pointing to the positive direction of the x-axis so that the i is rotated about the origin through the given angle in the corresponding (positive or negative) direction. (Figure 1) y

V x ,

y

2n

471

3

3

\ 0

* 0 \

/ i

X \

/ V

-it

X\> / 7tj

X

\ 3

Figure 2

We choose the signed length of the arc covered by the end-point of the unit vector during the rotation as the angular measure. For example 271 4-77

71

angles with radian measures -7t,—,—,— can be seen in figure 2. 7t

Now let a be an acute angle, i.e. 0 < a < —, and let e denote the vector 2 resulting when rotating the base vector i about the origin through the angle a. (Figure3) Let us draw a perpendicular at the end-point P of the unit vector e to the x-axis; let its base point be T. The length of the hypotenuse OP of the right-angled triangle OPT is 1, therefore PT = sin a and OT - cos a. It can also be phrased so that the first coordinate of the unit vector e is cos a, its second coordinate is sin a.

204

We also settle the general case according to this, thus we give the following definition:

y

Definition: If a is an arbitrary real number, then cos a is equal / a

to the first coordinate of the vector e, and sin a is equal to its second coordinate, where e is resulting by rotating the unit vector i about the origin through the angle with a radian measure of a. (Figure 4)

Ve

i

* X

(cos a; sin a)

Figure 4

Example 1 7T

3/r

2/1

TC

Let us give the sine and the cosine of the numbers —, —, —, — 2 2 3 3

Solution The vector

through

tc

is resulting by rotating the i into the positive direction

2>tc

and e2 is resulting by rotating it through — (Figures). Thus cos—= 0, sin —=-1. 2 2

cos—= 0, sin—= 1, 2 2

271

The angle included between the vector e3 and i is —, the angle n

and i is---- . (Figure 6) 3 By using the known values of the sine and the cosine of the acute angles the following results: V5 2n 1 . 2n a/3 cos— = —, sin — = —, cos 2 3 2 3 2 2 included between

The properties of the sine and the cosine function In view of the definition it is worth collecting the most important properties of the sine and the cosine function.

(1)We have seen before that if 0 < a < —, then sin1 2 a + cos2 a = 1. 2 Based on the definition of the sine and the cosine function and the Pythagorean theorem it can easily be proven that for an arbitrary real number a the Pythagorean identity holds, i.e. sin2 a + cos2 a = 1.

sin2 a + cos2 a = 1

(2) If a vector e includes an angle of a with i, then after rotating it about O into the positive direction through a full rotation the same vector e is resulting, so sin (a + 2n) = sin a, cos (a + 2tt) = cos a.

sin (a + 27r) = sin a

cos (a + 2zr) = cos a

This property is usually phrased as the sine and the cosine function are periodic with period 2n. In general if we add or subtract an integer multiple of 2n to or from a, the value of the sine and the cosine do not change. 205

y e/ /

/ a\

X

(3) If the vector e is resulting by rotating the vector i through the angle a, and then we rotate e through 7t, i.e. we rotate it further by a half rotation, then the resulting vector e’ is the mirror image of e about the origin, i.e. it is its negative (Figure 7). Therefore sin (a + 7t) - -sin a,

7 /e

cos (a + 7v) = -cos a. (4) If we rotate the unit vector e(ef, e2) with an angle of rotation of a 7T through — further, then the coordinates of the resulting vector e’

Figure 7

sin (a + tf) = -sin a

are (~e2; et) (Figures). Therefore

cos (a + 7t) = -cosa

sin

= cosa,

sin a.

cos

(5) There is an important and interesting relation between the coordinates of the vector e with an angle of rotation of a and the vector e’ with

an angle of rotation of

- a). It can easily be realised that the

two vectors are each other’s mirror images when reflected about the straight line with the equation y = x. According to this the coordinates are switched (Figure 9). so

Figure 8

Figure 9

(6) If the coordinates of the vector e with an angle of rotation of a are (q; e2), then the coordinates of the vector e’ with an angle of rotation of -a are (gp -e2) (Figure 1O). Thus sin (-a) = -sin a

sin (-a) = -sin a,

cos (-a) = cosa

cos (-a) = cos a. The vector e’ and e are each other’s mirror images when reflected about the x-axis. 206

(7) Let us reflect the vector efq; e2) with an angle of rotation of a about the y-axis, then its coordinates are (-ef, e2) (Figure 11). Thus the following identities can be seen: sin (77- a) = sin a

sin (77 - a) = sin a,

cos (77- a) = -cos a

cos (77 - a) = - cos a.

Example 2 Let us calculate the values of the following expressions. , . 77 77 .> 77 3?7 > 77 . 377 . 577 a) sin—-cos—; b) sin—+ cos—; c) sin—+ sin—+sin—. 3 3 2 2 4 4 4

Solution It is worth calculating based on the definition (Figure 12): , . n n 73 i 73 a) sin —-cos—=-------- = —;

3

3

2

2

4

,

_

,

, . . 77 2

377 2

, . 77

. 377

. 577

4

4

4

b) sin—r-cos— = 1 + 0 = 1;

1 72

1 72

1 72

1 72

c) sin----1- sin----- F Sin---- = —7= + —¡=----- f= = —r= . ♦ ♦ ♦

Naturally we can also talk about the sine and the cosine of 200°, 300° 156°. These mean the coordinates of the unit vector resulting when rotating the vector i through the corresponding angle measured in degrees.

Example 3

Let us calculate the following values. a) sin 120°; b) cos 120°; c) sin (-60°); Solution ra) sin 120° = sin 60° = —; 2

d) cos 150°.

b) cos 120° = -cos 60° =

73 c) sin (-60°) = -sin 60° = ——;

2

;

A d) cos 150° =-cos 30° = -—.

2

207

TRIGONOMETRIC FUNCTIONS

So far we calculated such sine and cosine values, which can directly be calculated from the definition or based on the definition and by using the known values of the sine and the cosine of the acute angles 30°, 45°, 60°. In other cases we need a pocket calculator or a table to calculate the trigonometric functional values.

The tables contain the trigonometric functional values of the acute angles. With the help of the known properties, identities the calculation of the sine and the cosine of any angle originates in the acute angles.

Example 4 By using a table let us calculate the following trigonometric functional values. a) sin 190°; b) cos 200°; c) sin 313°; d) cos(-71°). Solution a) sin 190° = sin(10° + 180°) =-sin 10° =-0.1736; b) cos 200° = cos(20° + 180°) = -cos 20° = -0.9397; c) sin 313° = sin (313°- 360°) = sin (-47°) = -sin 47° = -0.7314; d) cos (-71°) = cos 71° = 0.3256. ♦ ♦ ♦

It is worth tabulating the possible signs of the sine and the cosine function on the interval [0; 27r[: 77

r

1

377

0

]0:|[

sin

0

+

1

+

0

-

-1

-

cos

1

+

0

-

-1

-

0

+

77

2

T

Exercises 1. Calculate the following values. 77

277

377

77

77

77

a) cos—Feos— + cos—; b) sin- -sin -sin —;

4

4

4

6

3

2

c) cos 11° + cos22° + cos33°.

2. Verify the following identities. a) cos (2k- a) - cos a; b) sin (277 - a) = -sin a. 3. Which one is greater: a) cos 1 or sin 1;

b) cos 3 or sin 3?

4. Determine the sign of the following differences without using data tables or a calculator. a) sin 123°-sin 132°; b) sin 58° - cos 58°; c) cos 35° - cos 40°.

208

2. The graph of the sine function We became familiar with the definition of the sine function for an arbitrary real number, in other words we defined the function f: R —> R, /(x) = sin x,

and we examined its basic properties. Now we are going to plot the graph of the function. It is worth performing it gradually, step by step, because we get a more precise picture, and meanwhile we can use the basic properties of the sine function. I. Let us first plot the graph of the function for 0 < x < —.

Let us tabulate the known functional values.

By using a data table of function or a pocket calculator and calculating further 4-6 values the graph is forming. (Figure 13) Based on the definition we accept that the value of the sine function

is strictly increasing on the interval [0; —], i.e. a greater functional

value belongs to a greater number. As a unit vector e is turning from the vector i to the unit vector j pointing to the direction of the y-axis (x is increasing from 0 to —), so is the second coordinate of the vector

gradually increasing from 0 to l. It is worth fixing two more important properties.

y

y=xzzz

(1) If 0 < x < —, then sinx < x, i.e. the curve is below the straight line with 1

the equation y = x.

i'

(2) The function sinx is concave on the interval [0; — ] i.e.

if 0 < xx < x2

. x,+x9 sinx, + sinx, then sin —----- - >------ !---------- 2 2

z

✓

/y = / / 'ST'-' ’

X2 Xi + X2

in X

*1

X

2

Figure 14

The concavity graphically means that on the interval [0; — ] the curve of the

sine function is above the chord connecting the two points of the graph. (Figure 14) The two properties can also be proven.

Property (1), i.e. that in the case of 0 < x < — sinx < x, can be seen in figure 15. 2 The end-point of the rotating unit vector e covers a quarter circle between

Oand-. 2 7T Let us reflect it together with e standing in the position 0 < x < — about the x-axis. 2 Since the length of the arc PQ in the circle with a radius of OQ = 1 and with the centre O is x, therefore the length of the arc PQP' is 2x. Based on PT = sinx, PP’ = 2• sinx and it is obviously true that PP’ is less than the length of the arc PQP\ so 2 • sin x < 2x, which implies sin x < x.

209

TRIGONOMETRIC FUNCTIONS

It is again worth sketching a diagram to prove the property (2). (Figure 16)

In the circle with the centre O and with a radius of OQ = 1 let the length of the arc PQ be x,. the length of the arc RQ be x2 (xj < x2), and let S be the midpoint

of the arc PR. then the length of the arc SQ is

+ X~ .

Based on the definition of the sine function PT = sinxb RC = sinx2, and . Xj +x2 SV = sin —----- - . 2 The length of the midline FE of the trapezium UTPR is sln'V| + sinA2 ant| since F is closer to O than S, therefore FE < SV, and this is exactly what we wanted to justify.

II. In the next step let us draw the graph of the sine function for Fl ----- < x < 0. Let us use the identity (6) of the previous lesson. 2 71 71 If now 0 < x < —, then---- < -x < 0 and sin (-x) = -sin x, which means

2 2 that the graph is symmetric about the origin.

The reflection implies that the sine function is convex on the interval [-yl 0], i.e. between two points the graph it is below the chord

connecting the points. (Figure 17) ♦ ♦ ♦

Let us now continue with plotting the graph of the sine function

III. .

, F

271

on the interval — < x < — . 2 2 For this let us use the identity (7) of the previous lesson. TT

IT

71

If — ----- and sinf/r-x) = sinx. 2 2 2 2 The identity geometrically means that the graph of the sine function is symmetric about the straight line that can be described with the equation 71 x = — and is parallel with the y-axis. It can be seen from the fact that 71 x and 77-x are equidistant from the — on its two sides, and according

to the identity the functional values are equal at these places. (Figure 18) It also implies that the sine function is concave on the interval [0; 7i]. ♦ ♦ ♦

According to the identity (2) of the previous lesson the sine function is periodic with the period 2n, so sin(x + 2zr) = sinx. It means that by translating the graph of the function plotted on the 71

371

interval - — < x < — by 2tc or by integer multiples of 2/r, the curve

of the sine function is resulting on the whole number line. (Figure is) 210

Example 1 Let us solve the following equations on the set of real numbers. a) sin x = 0; b) sin x = 1; c) sin x - -1. Solution a) When plotting the sine function we could see that it takes the value of 0 at two places in one period, for example on the interval

sinx: (1) it is defined for every real number; its range is the closed interval [—1; 1] (here we graphically accept it, higher level mathematical tools are necessary for the proof);

(2) it is periodic with a period of 2n; (3) it is strictly increasing on the interval (4) it is strictly decreasing on the interval [ y;

; y];

b

(5) it is an odd function, i.e. sin(-x) = -sinx for any real x, and thus its graph is symmetric about the origin; 71

(6) it has a maximum at the places —i- 2kn (k g Z), and its maximum value is 1; (7) it has a minimum at the places------ 12nn (n g Z), and its minimum value is-1; (8) it has zeros at the places wtt (m g Z). 211

TRIGONOMETRIC FUNCTIONS

Example 2 Let us solve the equation sinx =

y 1-

—^-1

y = sin X z \

7t

n

6

2

y

1 2

X 6

on the set of real numbers.

Solution It is again worth looking for the solution in one period first. Graphically the equation can be rephrased so that we are looking for that place or those places where the curve of the sine function intersects the straight line parallel with the x-axis at the height y = -. (Figure 20)

Figure 20

We know that the sine function is strictly increasing on the interval 2

2

and sin — = —, so — is the only solution here. On the 6 2 6

interval — R, g(x) = sin 2x.

Solution (a) According to the method learnt in year 9 it is worth thinking that the 71

function/ takes every value at a place — greater than the sine function does. For example: 2

x = y, /iyj=sinO = O;

x = 2n, /(2ff) = sin^-=-1.

It means that the graph of sin x should be translated by — to the positive

positive direction along the x-axis. Thus the graph of the function /results. (Figure21)

212

Solution (b) In the case of the exercise b) a similar solution helps. Here the function g takes the same value at a place one half less than the sine function. For example:

Graphically it means that we shrink the graph of the sine function perpendicularly to the y-axis along the x-axis to its half, thus the graph of the function g results. (Figure 22)

Exercises 1. Solve the following equations on the set of real numbers.

a) sinx =—;

1 2

1

b) sinx =—;

c) I sin x I = —.

2

2

2. Plot the graphs of the following functions defined on the set of real numbers.

0, if cosx cosLx + — I might help. Let us

translate the graph of the cosine function by — to the negative direction along the x-axis. Let us now choose the interval

; —] to be one period,

and by taking it into account after the translation to the negative direction

the graph of the corresponding period of the function cos x + —

results

on the interval [------ ; — J. Let us plot the straight line parallel with the x-axis

4

4

V2

and with the equation y = — . (Figure 29)

Figure 29 JI The straight line intersects the graph of the function at the places---- and 0 on this interval, thus all the solutions are as follows:

- — + 2kn < x < 2kn (k g Z). 2

Here we have taken the periodicity with a period of 2w into account.

Solution (c) At first it is practical to realise that for any real number x: -----< -1 < cosx < 1 < —. 2 2 So we are only interested in the sign of the values of the sine function taken on the interval [-1; 1], more precisely we ask: when is it negative on this interval?

It is easy to answer: if cos x < 0. This inequality is easy to solve, all the solutions are as follows:

- + 2kn < x < — + 2kn (k g Z). 2

2

The examples above show well that being familiar with the properties of the corresponding trigonometric functions is very important for the solution of the equations, inequalities containing trigonometric functions. 219

TRIGONOMETRIC FUNCTIONS

Example 6 Let us assume that 0 < x < y. Which one is greater: cos (sin x) or sin (cos x)?

Solution We know that if 0 < x < —, then 2

sin X < X.

Since 0 < cosx < 1 < —, therefore 2 sin (cosx) < cosx. The cosine function is strictly decreasing on the interval [0; y], therefore cos (sin x) > cosx.

These imply that sin (cosx) < cos (sin x).

Exercises 1. Solve the following equations on the set of real numbers. a) sinx = cosx; b) sinx = sin(x + zr); c) sin2x—1 = 0.

2. Solve the following inequalities on the set of real numbers. a) sin2x > —;

2

b) cos2x 0.

3. Plot the graphs of the following functions defined on the set of real numbers. a) J\x) - cosx + 1; b) g(x) = 3cosx; c) h(x) - 1 - 2cosx; d) k(x) = Icosxl.

4. Plot the graphs of the following functions defined on the set of real numbers. a) /(x) = cos(x + zr);

b) g(x) -

c) h(x) = cos

*d) k(x) = cos(x - it) ■ sgn(sin x).

*5. Solve the following equations, inequalities on the set of real numbers.

a) x2 + 1 = sinx;

220

b) x + — 0 and tan(-x) = -tanx, and it is the 2 2 y-coordinate of the point P in figure 33. If we reflect P about the x-axis, and we connect the mirror image of Q with O, then the ray OQ includes an angle with a radian measure of exactly x with the positive half of the x-axis, and due to the reflection the y-coordinate of the point Q is tanx. It means that a representation is good in this case too.

We can show an important property of the tangent function based on the representation we have just got familiar with.

If---- tan—!----- Z. 2 2

Figure 35

The idea applied when representing the tangent might help with showing it. (Figure 36) In the unit circle with the centre O the length of the arc TQX is xb the length of the arc TQ2 is x2, and if g3 is the midpoint of the arc 2j22, then the length of the arc TQ2 is: x, +x2

2

It implies that tanx, = P1r, tanx2 = P2T, tan

X] + x2

2 tanx j + tgx2 If F denotes the midpoint of the line segment P\P2, then FT = 2 Since in the triangle OP\P2 OP2 is an angle bisector, it divides the side PjP2 in the ratio of the adjacent sides. Since PtO < P2O, PtP2 < P2P2, and P2T< FT, this is what we wanted to verify.

Figure 36

♦ ♦ ♦

7T

The tangent function is odd, thus if---- < x < 0, then its graph comes 2 ji about by reflecting the curve previously plotted for 0 < x < — about the origin. By using that the tangent is periodic with the period n we can plot the graph of the whole function. (Figure 37)

It is worth emphasizing that the tangent function is strictly increasing

on the open interval Figure 37

................................

and it takes every real number as value

(exactly once), i.e. its range is the set R. 224

Plotting the cotangent function Example 2

Let us represent the values of the cotangent function with the help of a tangent line parallel with the x-axis at the point (0; 1) of the unit circle with the origin as the centre, if 0 < x < n. Solution Let us draw the tangent line, and let us take a point R on the circle so that the radian measure of the angle ROQ is x fo < x < (Figure38)

The radian measure of the angle OPT is also x, since it is an alternate angle for the angle ROQ.

In the right-angled triangle OTP the length of the leg OT is 1, so PT = cotx. If x = —, P and T coincide, thus PT = 0 and cot — = 0. 2 2

If — cotx, x ± kit (ke Z).

For this we only have to think over that according to the property (4)

tan

-cotx,

thus if x * kit (k g Z):

cot x = - tan

So the graph of the cotangent function is resulting from the graph

of the of the tangent by translating it by — to the negative direction along the x-axis and by reflecting the resulting curve about the x-axis. (Figure 39)

The cotangent function is strictly decreasing on the interval 0 cotx 1.

Solution (a) The inequality a) can be written as follows based on the definition of the absolute value: -1 < tanx < 1.

It is satisfied on ]-—; — [if and only if-— — > 1, then cosx < 1, there is no solution here. If — < x < 0, 2 . 2 then cos x > 0, so there is no solution here either, and there is no solution 71

in the case ofx< — -1. 2 So according to the results the equation has one root only on the

interval [0; y]. We cannot give the exact value of the root with an elementary method, its approximate value is: x = 0.74.

Solution (c) It is again worth considering the equation c) first on the interval n 71 ---- < x < —. 2 2

7T The only solution here is x = 0, since tan 0 = 0, and if 0 < x < —, then 71 2 tan x > x, and if — < x < 0, then tan x < x. 2

Since the tangent function is strictly increasing on every interval with 71

71

the form of- — + kn < x < — + kTt (keZ) and it indeed takes every real value on this interval, here the function x i—> x is also increasing,

228

the equation has a root (at least one) on every such interval (with the help of higher level methods it can be verified that it has exactly one -i i root on every such interval.) The root from the interval J — ; —[ is approximately: x ~ 4.49. 2

So the equation has two roots on the interval

2

0. Then we can divide both sides of the inequality by x: I x + — < 2-sin(x + y). X Since x is positive about the left side we know that x + — > 2 and the x equality is satisfied only in the case of x = 1. And on the right side the sine value is at most 1, thus 2 ■ sin(x + y) < 2.

229

TRIGONOMETRIC FUNCTIONS

So the inequality is satisfied only if both sides are equal to 2, i.e. x = 1 and

sin (1 + y) = 1,

h-r,+2;] •I

which is true if and only if l+y=|+2£/r (teZ), i.e

i •I

y = - - 1 + 2ta Qt 6 Z). 2

J

X

PA-’)*

In the case of positive x the coordinates of the points satisfying the inequality are:

fl;y-l + 2m iteZ. It similarly results that in the case of x < 0 the following points are the solutions (Figure 46):

•i

f-1; -| + l + 2tol keZ.

Figure 46

Example 4 Let us assume that a point P is moving with uniform angular velocity along the circumference of a circle with the radius A. Let us place the circle in the coordinate system so that its centre is the origin, and let us denote the angular velocity by co. It means that the point rotates by an angle with a radian measure co over a time unit (for example in 1 second). We also assume that the point is moving in the positive direction along the circumference of the circle. Let us examine the movement of the point Q which is the orthogonal projection Of the point P tO the y-axiS. (Figure 47)

Solution Let cp0 denote the angle included between the radius OP and the positive half of the x-axis in the starting position of the point P. Then after t seconds the included angle between the ray OP and the positive half of the x-axis is co t + cp0. The position of the projected point Q is given by the y-coordinate of the point P. The y-coordinate of the vector OP is A sin (ft)-i + 0;

4. Give the largest subset of the set R of real numbers on which the following expressions

have a meaning.

a) tan ; tan 3x

b)

- cot2 3x;

c) Vtan3x;

1 - tan x 1 + tan x

5. Verify that if a, ft, y are the angles of a triangle, then

231

TRIGONOMETRIC FUNCTIONS

6. Geometric applications We have already proven some relations relating to triangles with the help of the trigonometric functions of the acute angles. Back then we had to examine the cases of the acute triangles, the right-angled triangles and the obtuse triangles separately. The extension of each of the trigonometric functions to the set of real numbers makes it possible to give the relations and the formulas in a uniform form for all types of triangles.

The area of a triangle We have proven that if a and b are the length of two sides of a triangle, y is the measure of the angle included between the two sides, and A is the area of the triangle, then ab-siny .. nno --------- -, if y 90°.

Based on the above it is again true that the relation between the side of the triangle, the interior angle opposite it and the radius of the circumscribed circle holds in the case of any arbitrary 0° < a< 180°: a = 2Rsina.

(2)

Naturally a similar relation holds for the other two sides of the triangle. ♦ ♦ ♦

Other formulae are also resulting for the area of the triangle as a direct consequence of the relations valid for the above arbitrary triangle. Out of these it is always worth using the formula with the help of which we can calculate the area the easiest and the fastest by substituting the particular data. We are now going to prove three formulae.

232

Let A denote the area of a triangle, let a, b, c denote the length of its sides, let a, P, y, denote the measure of the opposite interior angles respectively, and let the length of the radius of the circumscribed circle be R. With this notation:

the area of a triangle

Proof I. Since a = 2R ■ sin a and b = 2R • sinp, therefore aZ?-siny _ (27?-sinct)-(27?-sin/3)-siny

2

2 = 27?2 ■ sin a ■ sin p • sin y.

(We have already proven it for an acute triangle in connection with an example.)

11. Since c = 27? • siny: siny = —. By substituting it into (1): 27? ab ■ sin y

2

abc

2

AR

III. Let us start with I, and let us use (2):

2-sin ct (27?-sin ct)2 • sin/3-sin y _ a2 ■ sin /3 ■ sin y

2 ■ sin ct

2 • sin ct

Example 1 Let us calculate the area of the triangle if the length of one of its sides is 4 cm, and two of its angles are 30° and 45°.

Solution The third angle of the triangle is 105°. We know the length of one of the sides of the triangle and the measure of its angles, thus we can apply the area formula III just proven. The example did not fix the angle opposite the side given, so we have three options.

(1) The 30° angle is opposite the side given. Then the area of the triangle is (it can be used for the calculations that sin 105° = sin 75°):

’ bill Jv

x

2 = 8->/2-0.9659 cm2 = 10.93 cm2. 233

TRIGONOMETRIC FUNCTIONS

(2) The 45° angle is opposite the side given. Then

16 ■--0.9659 2

42 • sin30°-sin 105° 2-sin 45°

2 x/2 (3) The 105° angle is opposite the side given. Then

2-sin 105° 4 • V2

2-0.9659 2

n m

2

= --------- cmz ~ 2.93 cm . 1.9318

'I

Example 2

The length of the diagonals of a parallelogram are 8 cm and 6 cm, the smaller angle included between the diagonals is 60°. Let us calculate the area of the parallelogram.

Solution Since the diagonals of the parallelogram bisect each other, and sin 60° = sin 120°, therefore, according to the formula I relating to the area of the triangle, the diagonals divide the parallelogram into four triangles with equal areas (Figure 49). Thus the area is:

Figure 49

Notes: 1. It can be shown the above way that if e and /are the length of the two diagonals of the parallelogram, and the angle included between the diagonals is ®, then its area is . A e/sin

A = D. ♦ ♦ ♦

Definition: The complement of the event A is the event which occurs exactly when the event A does not occur. (Figure 3)

Jele: A.

The complement of the event A consists of the elementary events of the sample space which are not in A.

The complement of the complement of any event is the original event: A = A.

The complement of the certain event is the impossible event. The complement of the impossible event is the certain event.

equal events

Example 2 We roll a die and we are looking for the complement of the following events: 4: the result of the roll is at most 3; B: the result of the roll is at least 3; C: the number rolled is less than 3.

Solution A: the result of the roll is at most 3, A = {1, 2, 3}, then A: the number rolled is greater than 3, A = {4, 5, 6}. B: the result of the roll is at least 3, B - {3, 4, 5, 6}, then

B: the number rolled is less than 3, B = {1, 2} = C.

C: the number rolled is less than 3, C = {1, 2}, then C: the number rolled is at least 3, C = {3, 4, 5, 6} = B.

Definition: The union of the arbitrary events A and B is the event, which occurs exactly when A or B occurs. (Figure 4)

Notation: Au B.

Theorem:

Every event can be expressed as the union of ele­ mentary events.

For example when rolling a die:

{2}u{4, 6} = {2, 4,6} =

= {the result of the roll is an even number} = {2} u {4} o {6}.

Definition: The intersection of the arbitrary events A and B is the event, which occurs exactly when A and B occur. (Figure 5) Notation: AnB.

For example when rolling a die:

{the result of the roll is at most 3 and the result of the roll is at least 3} = = {1, 2, 3} n {3, 4, 5,6} = {3}.

Definition: The arbitrary events A and B are mutually exclusive if they do not occur at once, i.e. AnB = 0. (Figure 6) For example when rolling a die:

A: the result of the roll is at most 2, B: the result of the roll is at least 4; AnB= {1, 2}n {4, 5, 6} = 0.

244

Identities of the operations in connection with the events, which are true due to the identities known for sets:

Commutativity Associativity

Union

Intersection

AuB = BuA

AnB = BnA

(AuB)uC=Au(BuC)

(A^B)nC = An{BnC)

An(8uC) = (A^B)u(AnC)

Distributivity

A u (BnC) = (A u B) n (A u C)

Idempotency

AuA =4

AnA =A

Impossible event

A u0 = A

An0 = 0

Certain event

AoH = H

AnH = A

AoA = H

AnA= 0

AuB=AnB

AnB=AoB

Identities in relation with the complement event

De Morgan’s laws Example 3

We roll a die. Let A, B, C, D be the following events: A: the result of the roll is an even number; 8: the result of the roll is at most 3; C: the result of the roll is at least 3; D: the result of the roll is an odd number. Determine the following events. ojAuB; b)BuC', C)AuD; d)AnB\ e)BnC', f)AnD.

Solution a) Au B = {the result of the roll is an even number or at most 3} = {the result of the roll can be 2,4, 6,1, 3} = {the result of the roll is not 5}. Ao B= {2,4,6} u {1,2, 3} = {1,2, 3,4,6}. b) Bu C = {the result of the roll is at most 3 or at least 3} = {the

result of the roll could be any of the numbers}, this is the certain event. BuC= {1,2, 3} o {3,4,5, 6} = {1,2, 3,4, 5,6}= R c) Au D = {the result of the roll is an even number or an odd number},

this is the certain event. Au D = {2, 4, 6} o {1, 3, 5} = {1, 2, 3, 4, 5, 6} = H. d) A n B = {the result of the roll is an even number and at most 3} =

= {the result of the roll is 2}. AnB={2, 4,6} n {1.2,3} = {2}. e) B n C = {the result of the roll is at most 3 and at least 3} = {the result

of the roll is 3}. BnC= {1,2,3} n {3,4,5,6} = {3}. f) AoD = {the result of the roll is an even number and an odd number},

this is the impossible event. An£>= {2,4,6} n {1,3,5} = 0. 245

CALCULATION OF PROBABILITY

Example 4

A shot reaches a square shaped target at random. Let A, B, C mean the events that the shot is in the red coloured parts of the square respectively.

Let us draw the following events: AuB; AkjC', Ar\B', Br^C. Solution The solution can be seen in figure 7.

Figure 7

Example 5 There are three different lamps in a room. Let A mean the event that the ceiling lamp blows out, let B mean the event that the floor lamp blows out, and let C mean the event that the reading lamp blows out. Express the following with the help of the events A, S, C and the operations. a) All lamps blow out. b) None ofthe lamps blows out. c) A lamp blows out. d) Exactly one lamp blows out. e) There is a lamp which gives out light.

Solution a) {all lamps blow out} = An Bn C. b) {none of the lamps blow out} = Ac\BciC.

c) {a lamp blows out} = {the ceiling lamp blows out or the floor lamp

blows out or the reading lamp blows out} = {it is not true that none

of the lamps blows out} = AuBuC = AnBnC. d) {exactly one lamp blows out} = {the ceiling lamp blows out but not

the rest, or the floor lamp blows out but not the rest, or the reading lamp blows out but not the rest} = .4 n B n C u ,4 n /1 n C u .4 n B n C. e) {there is a lamp which gives out light} = {either the ceiling lamp does

not blow out, or the floor lamp does not blow out, or the reading lamp does not blow out} = {it is not true that all of them blow out} =

= AuBuC = AnBnC.

Exercises 1. Give the complement of the following events. Give the set of elementary events belonging to the event and to its complement. A: After rolling a die the result of the roll is at least 4. B : After rolling a die the number rolled is not greater than 2. C: After rolling two dice there is a 6 among the numbers rolled. D: When flipping three coins none of them land on tails. E: When flipping three coins all of them land on the same side. 2. When rolling a die let A, B. C. D denote the following events: A: the number rolled is a prime; B: the number rolled is even; C: the number rolled is less than 4; D: the number rolled is at least 4. Describe the following events.

AnB; AuB; Cui); Ct)D;

B n C;

Cui):

Cull

3. A die is rolled four times in succession; the results of the rolls are written down next to each other in order and thus a four-digit number results. Let A, B, C, D denote the following events: A: the four-digit number is even; B: the four-digit number is divisible by 3; C: the result of all four rolls is 6; D: the result of the fourth roll is 6. Describe the following events.

AnB; AnC; AnD; CnD; Cu£); AuC:

BnC; BnC.

4. We select a student from a class at random. Let A, B, C denote the following events: A: the selected student is a girl; B: the selected student learns German; C: the selected student sings in the choir. a) Describe the following events: A n (B o C); A n B n C. b) Under what conditions is it true that AnBnC = A, or A = C?

5. We shoot at a circular shaped target at random. Let A, B, C, D denote the following events: A: the shot is in the left part of the target; B: the shot is in the lower part of the

target; C: the shot is in the lower left quarter of the target; D: the shot is in the circle concentric with the target but its radius is half of it. Colour the parts corresponding to the following events in the figure. AnB; AuB; AuC: AnC; AuD; BoD; BuC;

AuC; BriD;

AuD.

6. A roundabout can be entered from Titmouse street, Finch street and Throstle street. Let A, B, C denote the following events: A: a car is arriving from Titmouse street; B: a car is arriving from Finch street; C; a car is arriving from Throstle street. Express the following with the help of the events A, B, C and the operations. a) A car is arriving from exactly one street to the roundabout. b) One car is arriving from each street to the roundabout. c) One car is arriving from Titmouse street to the roundabout. d) A car is arriving from exactly two streets to the roundabout. e) There is a street from which no car is arriving to the roundabout. f) A car is arriving from each street to the roundabout. g) No car is arriving from Throstle street to the roundabout.

247

3. Experiments, frequency, relative frequency, probability Suppose we have an experiment where we roll a die and we observe the number rolled. Let the event A be that the result of the roll is 6. We repeated the experiment 100 times and we counted that the event A occurred 15 times, i.e. the frequency of the event A is 15. The event 15 3 A occurred in -----= — of the number of the experiments, i.e. the 100 20 15 relative frequency of the event A is----- . 100

Definition: If out of n experiments the event A occurred k times frequency, relative frequency

(k < n), then k is called the frequency of the event A,

and — is called the relative frequency of the event A. n

Example 1 Let us roll a die 100 times. Let us give the frequency and the relative frequency of the following events. A. The result of the roll is an even prime. B: The result of the roll is an odd prime. C: The number rolled is a prime. 0: The number rolled is at most 6. Solution result of the roll is: 1:

15

2: flHW

17

3:

wHhii

5:

WHMI » 0, therefore 0 < — < 1. n ♦ The relative frequency of the certain event is 1, because the certain event occurs in every experiment, thus k = n.

Galileo Galilei (1564-1642): the work with the title Considerazione sopra il Giuoco dei Dadi

(Thoughts about Dice Games) was found in his legacy, in which he dealt with such problems.

♦ The relative frequency of the union of mutually exclusive events is the sum of the relative frequency of the terms.

By conducting many similar sequences of experiments we can observe that the relative frequency of a given event deviates around a number. The more experiments we conduct, the less the deviation is in general. The probability of the event A is considered to be the number around which the relative frequency is deviating.

probability

Andrej Nyikolajevics Kolmogorov (1903-1987), Russian mathematician was the modern mathematical founder of the calculation of probability. His work with the title Basics

The notation for the probability of the event A is: P(A). Note: We do not define the probability but we determine it with axioms, which are resulting corresponding to the experience based on the properties of the relative frequency.

Axiom I: If A is an arbitrary event, then P(A) > 0.

of Probability Theory

Axiom II: P(H) = 1.

was first published in 1933 in German, then in 1936 in Russian. In this work he wrote down the axioms relating to the probability.

Axiom III: If A and B are arbitrary events for which An B = 0, then P(A —, the probability of drawing a red ball is indeed greater for him when drawing from the black hat compared to drawing from the green hat. Figure 9

ANDREW

BRUCE

253

If they put the balls together, then there will be a total of 11 + 9 = 20 balls in the black hat, out of these the number of red balls is: 5 + 6 = 11,

thus the probability of drawing a red ball is

There will be a total of 7 + 14 = 21 balls in the green hat, out of these 12 3 + 9 = 12 are red, thus the probability of drawing a red ball is —. 11 12 21 Since — < —, thus after putting the balls together the probability of drawing a red ball is greater when drawing from the green hat compared to drawing from the black hat.

It is possible that: a c x — > — and b d y

v

This weirdness, that despite the probability of drawing a red ball was greater when drawing from the black hats separately, after putting the balls together the probability became greater when drawing from the green hat, is called Simpson’s paradox.

Exercises 1. A die is rolled once. Give the probability of the following events. A: The result of the roll is an even prime. B: The result of the roll is an odd prime. C: The number rolled is a prime. D: The number rolled is at most 6.

2. Flip two coins and find the probability of the following events: the number of heads on the

two coins is 0, 1, 2. 3. We put 4 pieces of paper in a hat with the numbers 1, 2, 3,4 written on them respectively.

The experiment is that we draw a piece of paper from the hat three times in succession at random, and then we write down the number and replace the piece of paper. Thus a threedigit number results. Give the probability of the following events. A: The three-digit number is even. B: The three-digit number is divisible by 3. 4. Which one is more probable: there is at least one 1 when rolling two dice or there is at least

two 2s when rolling four dice? 5. We draw one card from the Hungarian pack of 32 cards at random. (In a Hungarian pack

of cards there are 4 suits: Hearts, Bells, Leaves and Acorns, the numbering includes VII, VIII, IX, X, Under, Over, King and Ace.) Let A, B, C, D denote the following events: A: the card drawn is Hearts; B: the card drawn is VII; C: the card drawn is King or Ace; D: the card drawn is Bells or Leaves or Acorns. Give the probability of each of the events. Write down the following events and find their probability. Anfi; AuB; AnC; AnD; CnO; Cu/i: Au C;

BnC; BnC.

6. There are 8 adjacent parking places in a car park. Every day 4 cars are parking in one

parking place each for 8 hours. A bus is arriving with holiday-makers and the bus can only park if there are four free parking places next to each other. Give the probability that the bus can park, if the car drivers chose parking places with equal probability. 7. There are 15 gossiping boys in a class. One of them shared gossip with another one, and

this latter one peddled the gossip to a third one, who does not know who the second person heard it from, and peddles it further. Everyone selects a gossiping boy from the class with equal probability whom he peddles the gossip to. Give the probability that they peddle the gossip 10 times without one of them hearing it more than once. 254

8. In a game the aim is to roll a 6. We can decide whether we roll a regular die and the number rolled is the result or we roll two dice and the sum of the numbers rolled is the result. Which one shall we choose for a higher chance to get a 6? 9. We paint the face with 1 dot on one regular die and we paint the face with 4 dots on another

regular die. Now give the probability that the sum of the numbers rolled will be 7 when rolling the two dice at once.

10. What is the probability that the two dice need to be rolled at least 4 times until the number of dots on the two dice is 7? 11. What is the most probable sum when rolling two regular octahedra (numbered from 1 to 8)? (The probability of rolling each number is the same in the case of both octahedra.) 12. There are 7 balls in a box, numbered from 1 to 7. We drew the

7 balls in succession without replacement. What is the probability that we first drew the odd numbers? 13. There are 10 balls in a mug, numbered from 1 to 10. Hansel goes to the mug and takes out

one ball at random. Then comes Gretel and takes out one ball, different from the previous one. What is the probability that the product of the numbers on the two balls is even? 14. Tommy chose two numbers from the set {8, 9, 10} at random, and he added them up.

Charlie chose two numbers from the set {3, 5, 6} at random, and he multiplied them. Give the probability that Tommy’s result is greater than Charlie’s. 15. The net of four cubes can be seen in the figure; there is a number on each face of the cubes.

We build the cubes, and then first you choose a cube, then me. Then we roll our dice and who rolls the greater number is the winner. Which cube is it worth choosing?

16. Let us start with the oldest magic square that can also be seen in the figure. Let us assume that 9 chess players are ranked according to knowledge: 1, 2, 3, 4, 5, 6, 7, 8, 9 and are grouped into three teams according to the rows of the magic square. Which one is the best team? Two teams are playing with each other so that every player of one of the teams plays against every player of the other team.

17. Three dice are rolled. The player A wins if the numbers rolled with the three dice are all different, and the player B wins if the numbers rolled are all the same. The role of which player shall we choose if we want to win with a higher chance? 18. Chris recommends the following game to you: You roll two regular dice. If the product of the two numbers rolled is odd or divisible by 5, then you win, otherwise Chris wins. For whom is the game more advantageous, and what is your chance of winning?

255

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