Lecture Notes On Mathematical Olympiad Courses
Lecture Notes onMathematicalOlympiad CoursesFor Junior Section Vol. 17600 tp.indd 111/4/09 1:57:55 PM
Mathematical Olympiad SeriesISSN: 1793-8570Series Editors: Lee Peng Yee (Nanyang Technological University, Singapore)Xiong Bin (East China Normal University, China)PublishedVol. 1A First Step to Mathematical Olympiad Problemsby Derek Holton (University of Otago, New Zealand)Vol. 2Problems of Number Theory in Mathematical Competitionsby Yu Hong-Bing (Suzhou University, China)translated by Lin Lei (East China Normal University, China)ZhangJi - Lec Notes on Math's Olymp Courses.pmd211/2/2009, 3:35 PM
Xu JiaguVol. 6MathematicalOlympiadSeriesLecture Notes onMathematicalOlympiad CoursesFor Junior Section Vol. 1World Scientific7600 tp.indd 211/4/09 1:57:55 PM
Published byWorld Scientific Publishing Co. Pte. Ltd.5 Toh Tuck Link, Singapore 596224USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601UK office: 57 Shelton Street, Covent Garden, London WC2H 9HEBritish Library Cataloguing-in-Publication DataA catalogue record for this book is available from the British Library.Mathematical Olympiad Series — Vol. 6LECTURE NOTES ON MATHEMATICAL OLYMPIAD COURSESFor Junior SectionCopyright 2010 by World Scientific Publishing Co. Pte. Ltd.All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means,electronic or mechanical, including photocopying, recording or any information storage and retrievalsystem now known or to be invented, without written permission from the Publisher.For photocopying of material in this volume, please pay a copying fee through the CopyrightClearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission tophotocopy is not required from the publisher.ISBN-13 978-981-4293-53-2 (pbk) (Set)ISBN-10 981-4293-53-9(pbk) (Set)ISBN-13 978-981-4293-54-9 (pbk) (Vol. 1)ISBN-10 981-4293-54-7(pbk) (Vol. 1)ISBN-13 978-981-4293-55-6 (pbk) (Vol. 2)ISBN-10 981-4293-55-5(pbk) (Vol. 2)Printed in Singapore.ZhangJi - Lec Notes on Math's Olymp Courses.pmd111/2/2009, 3:35 PM
PrefaceAlthough mathematical olympiad competitions are carried out by solving problems, the system of Mathematical Olympiads and the related training courses cannot involve only the techniques of solving mathematical problems. Strictly speaking, it is a system of mathematical advancing education. To guide students who areinterested in mathematics and have the potential to enter the world of Olympiadmathematics, so that their mathematical ability can be promoted efficiently andcomprehensively, it is important to improve their mathematical thinking and technical ability in solving mathematical problems.An excellent student should be able to think flexibly and rigorously. Here theability to do formal logic reasoning is an important basic component. However, itis not the main one. Mathematical thinking also includes other key aspects, likestarting from intuition and entering the essence of the subject, through prediction,induction, imagination, construction, design and their creative abilities. Moreover,the ability to convert concrete to the abstract and vice versa is necessary.Technical ability in solving mathematical problems does not only involve producing accurate and skilled computations and proofs, the standard methods available, but also the more unconventional, creative techniques.It is clear that the usual syllabus in mathematical educations cannot satisfythe above requirements, hence the mathematical olympiad training books must beself-contained basically.The book is based on the lecture notes used by the editor in the last 15 years forOlympiad training courses in several schools in Singapore, like Victoria JuniorCollege, Hwa Chong Institution, Nanyang Girls High School and Dunman HighSchool. Its scope and depth significantly exceeds that of the usual syllabus, andintroduces many concepts and methods of modern mathematics.The core of each lecture are the concepts, theories and methods of solvingmathematical problems. Examples are then used to explain and enrich the lectures,and indicate their applications. And from that, a number of questions are includedfor the reader to try. Detailed solutions are provided in the book.The examples given are not very complicated so that the readers can understand them more easily. However, the practice questions include many from actualv
viPrefacecompetitions which students can use to test themselves. These are taken from arange of countries, e.g. China, Russia, the USA and Singapore. In particular, thereare many questions from China for those who wish to better understand mathematical Olympiads there. The questions are divided into two parts. Those in PartA are for students to practise, while those in Part B test students’ ability to applytheir knowledge in solving real competition questions.Each volume can be used for training courses of several weeks with a fewhours per week. The test questions are not considered part of the lectures, sincestudents can complete them on their own.K. K. Phua
AcknowledgmentsMy thanks to Professor Lee Peng Yee for suggesting the publication of this thebook and to Professor Phua Kok Khoo for his strong support. I would also like tothank my friends, Ang Lai Chiang, Rong Yifei and Gwee Hwee Ngee, lecturers atHwaChong, Tan Chik Leng at NYGH, and Zhang Ji, the editor at WSPC for hercareful reading of my manuscript, and their helpful suggestions. This book wouldbe not published today without their efficient assistance.vii
This page intentionally left blank
Abbreviations and H EUROPERUSMOSSSMOSMOSSSMO(J)UKJMOUSAMOAmerican High School Mathematics ExaminationAmerican Invitational Mathematics ExaminationAsia Pacific Mathematics OlympiadOlympics Mathematical Competitions of Allthe Soviet UnionAustralia Mathematical CompetitionsBritish Mathematical OlympiadChina Mathematical OlympiadChina Mathematical Competition for SecondarySchoolsChina Mathematical Competitions for SecondarySchools except for CHNMOLCanada Mathematical OlympiadHungary Mathematical CompetitionInternational Mathematical OlympiadJapan Mathematical OlympiadKiev Mathematical OlympiadMoscow Mathematical OlympiadNorth Europe Mathematical OlympiadAll-Russia Olympics Mathematical CompetitionsSingapore Secondary Schools Mathematical OlympiadsSingapore Mathematical OlympiadsSingapore Secondary Schools Mathematical Olympiadsfor Junior SectionUnited Kingdom Junior Mathematical OlympiadUnited States of American Mathematical Olympiadix
Abbreviations and NotationsxNotations for Numbers, Sets and Logic RelationsNN0ZZ QQ Q 0R[a, b](a, b) A BA BA BA Ba Athe set of positive integers (natural numbers)the set of non-negative integersthe set of integersthe set of positive integersthe set of rational numbersthe set of positive rational numbersthe set of non-negative rational numbersthe set of real numbersthe closed interval, i.e. all x such that a x bthe open interval, i.e. all x such that a x biff, if and only ifimpliesA is a subset of Bthe set formed by all the elements in A but not in Bthe union of the sets A and Bthe intersection of the sets A and Bthe element a belongs to the set A
ContentsvPrefaceAcknowledgmentsviiAbbreviations and Notationsix1Operations on Rational Numbers12Monomials and Polynomials73Linear Equations of Single Variable134System of Simultaneous Linear Equations195Multiplication Formulae276Some Methods of Factorization357Absolute Value and Its Applications418Linear Equations with Absolute Values479Sides and Angles of a Triangle5310Pythagoras’ Theorem and Its Applications5911Congruence of Triangles6512Applications of Midpoint Theorems7113Similarity of Triangles77xi
xiiContents14Areas of Triangles and Applications of Area8515Divisions of Polynomials93Solutions to Testing Questions101Index169
Lecture 1Operations on Rational Numbers1. Basic Rules on Addition, Subtraction, Multiplication, DivisionCommutative Law:a b b a ab baAssociative Law:a b c a (b c) (ab)c a(bc)Distributive Law:ac bc (a b)c c(a b)2. Rule for Removing BracketsFor any rational numbers x, y,(i)x (y) x y, x ( y) x y;(ii)x (y) x y, x ( y) x y.(iii) x ( y) xy; ( x) y xy; ( x) ( y) xy;( 1)n 1 for odd n, ( 1)n 1 for even n.(iv) If the denominators of the following expressions are all not zeros,thenxx xx xx ; ; . yyyy yy3. Ingenious Ways for Calculating Make a telescopic sum by using the following expressions:111 ,k(k 1)k µk 1¶11 11 ,k(k m)m k ·k m 1111 .k(k 1)(k 2)2 k(k 1) (k 1)(k 2) By use of the following formulae:(a b)2 a2 2ab b2 ;a2 b2 (a b)(a b);a3 b3 (a b)(a2 ab b2 );a3 b3 (a b)(a2 ab b2 ), etc.1
Lecture 12Operations on Rational NumbersExamplesµ¶3¶2µ113Example 1. Evaluate ( 5) 2 ( 1)1999 .52µ¶3¶2µ11Solution ( 5)2 23 ( 1)199952µ¶112 5 8 ( 1)1254111 8 4 1 31 31 .5552below:Example 2. There are five¶µ operational expressions1 1 1 1 ;(i)(2 3 5 7)2 3 5 7(ii)( 0.125)7 · 88 ;(iii) ( 11) ( 33) ( 55) ( 66) ( 77) ( 88);µ¶2 µ ¶27537(iv) ;13"µ 13 ¶# µµ¶µ¶¶764416246(v) 9 0.666 .75981247Then the expression with maximal value is(A) (i),(B) (iii),(C) (iv),(D) (v).Solutionµ¶1 1 1 1(i) (2 3 5 7) 2 3 5 7 105 70 42 30 247;(ii) ( 0.125)7 · 88 (0.125 8)7 8 8;(iii) ( 11) ( 33) ( 55) ( 66) ( 77) ( 88) 55¶ 66 77 88 11 22 242;µ 11¶ 2 33 µ27537(iv) 62 32 45;13"µ 13 ¶# µµ¶ µ¶¶764416246(v) 9 0.66675981247 1 10 10;Thus, the answer is (A).
Lecture Notes on Mathematical Olympiad3.Example 3. 123456789 999999999 Solution123456789 999999999 123456789 (1000000000 1) 123456789000000000 123456789 123456788876543211.Example 4. The value of13579is ( 13578)(13580)(C) 1,(D) 13578.( 13579)2(A) 1,(B) 13579,SolutionBy use of (a b)(a b) a2 b2 , we have13579 ( 13578)(13580)13579 13579.(13579)2 (135792 1)( 13579)2The answer is (B).Example 5.Solution832 173 83 66 172.By use of the formula a3 b3 (a b)(a2 ab b2 ),832 17383 66 172 (83 17)(832 83 17 172 )83 66 172100 (83 66 172 ) 100.83 66 172Example 6. Evaluate(4 7 2)(6 9 2)(8 11 2) · · · · · (100 103 2).(5 8 2)(7 10 2)(9 12 2) · · · · · (99 102 2)Solutionn, we haveFrom n(n 3) 2 n2 3n 2 (n 1)(n 2) for any integer(4 7 2)(6 9 2)(8 11 2) · · · · · (100 103 2)(5 8 2)(7 10 2)(9 12 2) · · · · · (99 102 2)(5 6)(7 8)(9 10) · · · · · (101 102) (6 7)(8 9)(10 11) · · · · · (100 101) 5 102 510.Example 7.200920082 200920072 200920092 2.
Lecture 14SolutionOperations on Rational Numbers200920082200920072 200920092 2200920082(200920072 1) (200920092 1)200920082 (20092006)(20092008) (20092008)(20092010)1200920082200920082 . (20092008)(20092006 20092010)2(200920082 )2 Example 8. 3 1 111111 2 6 12 20 30 42 56.Solutionµ¶1 1111113 2 6 12 20 30 42 56µ¶1111 ··· 3 1 2 2 3 3 47 8·µµ¶ µ¶¶ 11 11 1 3 1 ··· 22 37 8µ¶11 3 1 2 .88111111 .3 15 35 63 99 143¶µ1 111 for any positive integer k, soSince k(k 2)2 k k 2Example 9. EvaluateSolution111111 3 15 35 63 99 143111111 1 3 3 5 5 7 7 9 µ9 11 11 13·µ¶ µ¶¶ 111 11 11 ··· 21 33 511 13· 116 1 .21313Example 10. If ab 0, then the relation in sizes of (a b)2 and (a b)2 is(A) (a b)2 (a b)2 ; (B) (a b)2 (a b)2 ;(C) (a b)2 (a b)2 ; (D) not determined.
Lecture Notes on Mathematical Olympiad5Solution From (a b)2 a2 2ab b2 a2 2ab b2 4ab (a b)2 4ab (a b)2 , the answer is (C).1Example 11. If 1 a 0, then the relation in sizes of a3 , a3 , a4 , a4 , ,a1 isa11(A) a4 a3 a3 a4 ;aa11(B) a a4 a4 a3 ;aa11(C) a3 a4 a4 a3 ;aa11(D) a3 a4 a4 a3 .aa1From 1 a 0 we have 0 a4 a3 1 , soa1 a4 a3 and a3 and a4 a4 , the answer is (C).aSolutionTesting Questions(A)1.Evaluate 1 ( 1)1 ( 1)2 ( 1)3 · · · ( 1)99 ( 1)100 .2.Evaluate 2008 20092009 2009 20082008.3.1From 2009 subtract half of it at first, then subtract of the remaining num311ber, next subtract of the remaining number, and so on, untilof the42009remaining number is subtracted. What is the final remaining number?4.Find the sum11111 .5 7 7 9 9 11 11 13 13 155.Find the sum11111 .10 40 88 154 2386.Evaluateµ¶µ¶1 11111 ··· ··· 3 4200922008µ¶µ¶1 1111 1 1 ··· ··· .3 420092 32008
Lecture 16Operations on Rational Numbers111 ··· .1 2 1 2 31 2 · · · 517.Find the sum8.Let n be a positive integer, find the value of1 2n n 111 2 1 1 2 3 2 11 · · · · · · · · · .2 2 2 3 3 3 3 3n nnnn9.10.Evaluate 12 22 32 42 · · · 20082 20092 .Find the sum 11 192 1993 19994 199995 1999996 19999997 199999998 1999999999.Testing Questions(B)32 1 52 1 72 1992 1 2 2 ··· 2.23 1 5 1 7 199 11.Calculate2.After simplification, the value of342 1 1 · (1 2) (1 2)(1 2 3) (1 2 3)(1 2 3 4)100(1 2 · · · 99)(1 2 · · · 100)is a proper fraction in its lowest form. Find the difference of its denominatorand numerator. ··· 111 ··· .1 2 3 2 3 4100 101 1023.Evaluate4.Find the sum12350 ··· .1 12 141 22 241 32 341 502 5045.Evaluate the expression2280212 ··· .12 10 50 22 20 50802 80 50
Lecture 2Monomials and PolynomialsDefinitionsMonomial: A product of numerical numbers and letters is said to be a monomial. In particular, a number or a letter alone is also a monomial, for example,16, 32x, and 2ax2 y, etc.Coefficient: In each monomial, the part consisting of numerical numbers andthe letters denoting constants is said to be the coefficient of the monomial, like 32in 32x, 2a in 2ax2 y, etc.Degree of a Monomial: In a monomial, the sum of all indices of the lettersdenoting variables is called the degree of the monomial. For example, the degreeof 3abx2 is 2, and the degree of 7a4 xy 2 is 3.Polynomial: The sum of several monomials is said to be a polynomial, its eachmonomial is called a term, the term not containing letters is said to be the constant term of the polynomial. The maximum value of the degree of terms in thepolynomial is called degree of the polynomial, for example, the degree is 2 for3x2 4x 1, and 5 for 2x2 y 3 2y. A polynomial is called homogeneous whenall its terms have the same degree, like 3x2 xy 4y 2 .Arrangement of Terms: When arranging the terms in a polynomial, the termscan be arranged such that their degrees are in either ascending or descending order,and the sign before a term should remain attached to when moving it. For example,the polynomial x3 y 3 1 2xy 2 x3 y should be arranged as x3 y 3 x3 y 2xy 2 1or 1 2xy 2 x3 y x3 y 3 .Like Terms: Two terms are called like terms if they have the same constructionexcept for their coefficients, like in 4ax2 y and 5bx2 y.Combining Like Terms: When doing addition, subtraction to two like terms,it means doing the corresponding operation on their coefficients. For example,4ax2 y 5bx2 y (4a 5b)x2 y and 4ax2 y 5bx2 y (4a 5b)x2 y.7
Lecture 28Monomials and PolynomialsOperations on PolynomialsAddition: Adding two polynomials means:take all terms in the two polynomials as the terms of the sum;(i)combine all the like terms if any;(ii)(iii) arrange all the combined terms according to the order of ascending or descending degree.Subtraction: Let P and Q be two polynomials. Then P Q meanschange the signs of all terms in Q to get Q at first;(i)take all terms in the two polynomials P and Q as the terms of P Q;(ii)(iii) combine all the like terms if any;(iv) arrange all the combined terms according to the rule mentioned above.Rule for Removing or Adding Brackets:The rule for removing or adding brackets is the distributive law. For example, toremove the brackets in the expression 2x(x3 y 4x2 y 2 4), then 2x(x3 y 4x2 y 2 4) 2x4 y 8x3 y 2 8x,and to add a pair of bracket for containing the terms of the expression 4x5 y 2 6x4 y 8x2 y 2 and pick out their common factor with negative coefficient, then 4x5 y 2 6x4 y 8x2 y 2 2x2 y(2x3 y 3x2 4y).Multiplication:(i)For natural numbers m and n,am · an am n ; (am )n amn ; (ab)n an bn ;(ii)When two monomials are multiplied, the coefficient of the product is theproduct of the coefficients, the letters are multiplied according to the rulesin (i);(iii) When two polynomials are multiplied, by using the distributive law, get asum of products of a monomial and a polynomial first, and then use thedistributive law again, get a sum of products of two monomials;(iv) Three basic formulae in multiplication:(i)(a b)(a b) a2 b2 ;(ii)(a b)2 a2 2ab b2 ;(iii) (a b)2 a2 2ab b2 .ExamplesExample 1. Simplify 3a { 4b [4a 7b ( 4a b)] 5a}.Solution3a { 4b [4a 7b ( 4a b)] 5a} 3a { 4b [8a 6b] 5a} 3a { 3a 2b} 2b.
Lecture Notes on Mathematical Olympiad9or3a { 4b [4a 7b ( 4a b)] 5a} 8a 4b [4a 7b ( 4a b)] 4a 3b ( 4a b) 2b.Note: We can remove the brackets from the innermost to outermost layer, orvice versa.Example 2. Simplify the expression 4{(3x 2) [3(3x 2) 3]} (4 6x).Solution Taking 3x 2 as whole as one number y in the process of thesimplification first, we have4{(3x 2) [3(3x 2) 3]} (4 6x) 4{y [3y 3]} 2y 4{ 2y 3} 2y 6y 12 6(3x 2) 12 18x.Example 3. Evaluate 9xn 2 8xn 1 ( 9xn 2 ) 8(xn 2 2xn 1 ), wherex 9, n 3.Solution 9xn 2 8xn 1 ( 9xn 2 ) 8(xn 2 2xn 1 ) 8xn 1 8x. By substituting x 9, n 3, it follows thatn 2the expression 8xn 1 8xn 2 8 (81 9) 576.Example 4. Given x3 4x2 y axy 2 3xy bxc y 7xy 2 dxy y 2 x3 y 2for any real numbers x and y, find the value of a, b, c, d.Solution4x2 y and bxc y must be like terms and their sum is 0, sob 4, c 2.axy 2 7xy 2 0 and 2xy dxy 0 for every x and y yields a 7 0 and3 d 0, soa 7, d 3.Thus, a 7, b 4, c 2, d 3.Example 5. Given that m, x, y satisfy (i) 32 (x 5)2 5m2 0; (ii) 2a2 by 1and 3a2 b3 are like terms, find the value of the expression½ ¾·7 23 21 23 2222x y 5m x y xy x y 3.475xy 6.275xy .816416
Lecture 210Monomials and PolynomialsSolution The condition (i) implies (x 5)2 0, 5m2 0, so x 5, m 0.The condition (ii) implies y 1 3, i.e. y 2. Therefore· ¾½1 23 23 27 2222x y 5m x y xy x y 3.475xy 6.275xy816416½¾3 27 21 23 222 x y x y xy x y 3.475xy 6.275xy8164167 21 23 23 2 x y x y xy x y 3.475xy 2 6.275xy 2816416¶µ¶µ73131911 x2 y 3 6xy 2 8 16 1644040 x2 y 10xy 2 (52 )(2) 10(5)(22 ) 250.Example 6. Given that P (x) nxn 4 3x4 n 2x3 4x 5, Q(x) 3xn 4 x4 x3 2nx2 x 2 are two polynomials. Determine if there exists an integern such that the difference P Q is a polynomial with degree 5 and six terms.Solution P (x) Q(x) (n 3)xn 4 3x4 n x4 3x3 2nx2 3x 3.When n 4 5, then n 1, so that 3x4 n 3x3 0, the difference hasonly 5 terms.When 4 n 5, then n 1, so that P (x) Q(x) 3x5 x4 7x3 22x 3x 3 which satisfies the requirement. Thus, n 1.Example 7. Expand (x 1)(x 2)(x 3)(x 4).Solution(x 1)(x 2)(x 3)(x 4) [(x 1)(x 4)] · [(x 2)(x 3)] (x2 4x x 4)(x2 3x 2x 6) [(x2 5x 5) 1][(x2 5x 5) 1] (x2 5x 5)2 1 x4 25x2 25 10x3 10x2 50x 1 x4 10x3 35x2 50x 24.Example 8. Expand (5xy 3x2 Solution1 21y )(5xy 3x2 y 2 )22Considering the f
the above requirements, hence the mathematical olympiad training books must be self-contained basically. The book is based on the lecture notes used by the editor in the last 15 years for Olympiad training courses in several schools in Singapore, like Victoria Junior College, Hwa Chong Institution, Nanyang Girls High School and Dunman High School.
Introduction of Chemical Reaction Engineering Introduction about Chemical Engineering 0:31:15 0:31:09. Lecture 14 Lecture 15 Lecture 16 Lecture 17 Lecture 18 Lecture 19 Lecture 20 Lecture 21 Lecture 22 Lecture 23 Lecture 24 Lecture 25 Lecture 26 Lecture 27 Lecture 28 Lecture
In a district, a school provides the venue of the regional olympiad. Partic-ipants who are awarded gets to participate in the national olympiad. The olympiads take place in a festive manner and the national level olympiad is known as BdMO(Bangladesh Mathematical Olympiad). Around 40 partici-
English International Olympiad (EIO) 4. General Knowledge International Olympiad (GKIO) 5. International Drawing Olympiad (IDO) 6. National Essay Olympiad (NESO) Overall question paper pattern includes academic syllabus questions,
DAVCAE Olympiad Registration for ICT, General Science and Math's Olympiads. We are pleased to inform you that we have decided to conduct ICT, Math's and Science Olympiad for the Academic session 2021-22. 1. ICT Olympiad ICT Olympiad will be organized/conducted at two levels. (age group) Age Group I: Students from grade V to VIII.
The official program of the Olympiad 0.3. The 28th Pan African Mathematical Olympiad PAMO (incorporating the Pan African Mathematical Olympiad for Girls PAMOG) will take place in Monastir, Tunisia from 11 to 20 March 2021. 0.4. The official languag
GEOMETRY NOTES Lecture 1 Notes GEO001-01 GEO001-02 . 2 Lecture 2 Notes GEO002-01 GEO002-02 GEO002-03 GEO002-04 . 3 Lecture 3 Notes GEO003-01 GEO003-02 GEO003-03 GEO003-04 . 4 Lecture 4 Notes GEO004-01 GEO004-02 GEO004-03 GEO004-04 . 5 Lecture 4 Notes, Continued GEO004-05 . 6
Moscow Math Olympiad runs since 1935). Still, for all these years the “most main” olympiad in the country was traditionally and actually the Moscow Math Olympiad. Visits of students from other towns started the expansion of the range of the Moscow Math Olympiad to the whole country, an
Joanne Freeman – The American Revolution Page 3 of 265 The American Revolution: Lecture 1 Transcript January 12, 2010 back Professor Joanne Freeman: Now, I'm looking out at all of these faces and I'm assuming that many of you have
