Preview Pathfinder for Olympiad Mathematics by Vikash Tiwari, V. Seshan (2017) Preview Pathfinder for Olympiad Mathematics by Vikash Tiwari, V. Seshan (2017) Preview Pathfinder for Olympiad Mathematics by Vikash Tiwari, V. Seshan (2017) Preview Pathfinder for Olympiad Mathematics by Vikash Tiwari, V. Seshan (2017) Preview Pathfinder for Olympiad Mathematics by Vikash Tiwari, V. Seshan (2017)
Pathfinder for OLYMPIAD MATHEMATICS Pathfinder for Vikash Tiwari | V Seshan EXTENSIVE PEDAGOGY Solved Problems Build-up Your Understanding Check Your Understanding Challenge Your Understanding Vikash Tiwari has been teaching students for Mathematical Olympiads (Pre RMO, RMO, INMO, and IMOTC) and other examinations like KVPY and JEE for the last 20 years He is a renowned resource person in the field of Mathematics in India He conducts Olympiad training camps for several organizations such as Kendriya Vidyalaya Sangathan, Delhi Public School (DPS), etc He has devoted himself to the service of mankind via the medium of mathematics and has come up with this “first of its kind” book, which is an inventory of all essential concepts required to ace the Mathematics Olympiad at various levels The students have found his methods of problem-solving and teaching to be both insightful and intriguing He has been instrumental in the success of several medal-winning students who have made our country proud in various International Mathematics Olympiads Fulbright Teacher Awardee (USA-1970) Presidential Awardee (1987) Advisor to National Science Olympiad Foundation (Since 1989) Rotary (Int.) Awardee (1992) PEE VEE National Awardee (2000) Ramanujan Awardee (2008) in.pearson.com Spine: 22mm OLYMPIAD MATHEMATICS Vikash Tiwari V Seshan Tiwari Seshan Additional resources available at http://www.pearsoned.co.in/Olympiad/Mathematics Size: 203x254mm Cover Image: vlastas/Click Bestsellers Shutterstock V Seshan is one of the key resource persons nominated by CBSE to provide Olympiad training across India He is a popular teacher and retd Principal and Director of Bhartiya Vidya Bhavan, Baroda Centre He is well known for his unique ability in teaching Mathematics with utmost conceptual clarity With teaching experience spanning over 40 years, he has been instrumental in setting-up the Olympiad Centre in Tata Institute of Fundamental Research, Mumbai He has also been awarded with various medals, honors and recognitions by prestigious universities and institutions from across the globe These bear testament to his immense contribution to the field of mathematics Many of his students have taken part in both national and international Mathematics Olympiad and have also won gold and silver medals to their credits A few of his key recognitions are listed below Pathfinder for O LYMPIAD MATHEMATICS This book has been prepared in line with the requirements of national and international Olympiad examinations The questions are carefully chosen to suit the needs of Olympiad aspirants and to provide highest level of clarity for mathematical concepts This book also provides deep insights about the origin of important formulae and equations by eminent Mathematicians The exercises are designed and graded from simple to difficult level to enable the students’ to build, check and challenge their understanding ISBN: 9789332568723 Title Sub Title Edition Authors / Editors Name With CD Red Band Territory line URL Price mQuest About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world We believe learning opens up opportunities, creates fulfilling careers and hence better lives We hence collaborate with the best of minds to deliver you class-leading products, spread across the Higher Education and K12 spectrum Superior learning experience and improved outcomes are at the heart of everything we This product is the result of one such effort Your feedback plays a critical role in the evolution of our products and you can contact us – reachus@pearson.com We look forward to it A01_Olympiad Mathematics_FM.indd 8/11/2017 5:21:37 PM This page is intentionally left blank A01_Olympiad Mathematics_FM.indd 8/11/2017 5:21:38 PM Pathfinder for Olympiad MATHEMATICS Vikash Tiwari V Seshan A01_Olympiad Mathematics_FM.indd 8/11/2017 5:21:38 PM Copyright © 2017 Pearson India Education Services Pvt Ltd Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128, formerly known as TutorVista Global Pvt Ltd, licensee of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time ISBN: 9789332568723 eISBN: 9789352862757 Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No 1, Block-C, Sector-16, Noida 201 301,Uttar Pradesh, India Registered Office: 4th Floor, Software Block, Elnet Software City, TS-140, Block & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India Fax: 080-30461003, Phone: 080-30461060 Website: in.pearson.com, Email: companysecretary.india@pearson.com A01_Olympiad Mathematics_FM.indd 8/11/2017 5:21:38 PM Brief Contents Preface ������������������������������������������������������������������������������������������������������������������������������������������������ xi Acknowledgements ����������������������������������������������������������������������������������������������������������������������������� xii About the Authors������������������������������������������������������������������������������������������������������������������������������� xii 1. Polynomials ���������������������������������������������������������������������������������������������������������� 1.1 2. Inequalities ���������������������������������������������������������������������������������������������������������� 2.1 3. Mathematical Induction ������������������������������������������������������������������������������������� 3.1 4. Recurrence Relation ������������������������������������������������������������������������������������������� 4.1 5. Functional Equations������������������������������������������������������������������������������������������� 5.1 6. Number Theory���������������������������������������������������������������������������������������������������� 6.1 7. Combinatorics ����������������������������������������������������������������������������������������������������� 7.1 8. Geometry ������������������������������������������������������������������������������������������������������������� 8.1 Answer Keys ���������������������������������������������������������������������������������������������������������������������������������� AK.1 Appendix ���������������������������������������������������������������������������������������������������������������������������������������� AP.1 Logarithms Table ���������������������������������������������������������������������������������������������������������������������������� LT.1 Photo Credits ��������������������������������������������������������������������������������������������������������������������������������� PC.1 A01_Olympiad Mathematics_FM.indd 8/11/2017 5:21:38 PM This page is intentionally left blank A01_Olympiad Mathematics_FM.indd 8/11/2017 5:21:38 PM Contents Prefacexi Acknowledgementsxii About the Authors Chapter Polynomials Polynomial Functions Division in Polynomials Remainder Theorem and Factor Theorem Fundamental Theorem of Algebra Polynomial Equations Vieta’s Relations Symmetric Functions Common Roots of Polynomial Equations Irreducibility of Polynomials Chapter Inequalities Basic Rules Weirstras’s Inequality Modulus Inequalities Sum of Squares (SOS) Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean Weighted Means Power Mean Inequality Rearrangement Inequality Chebyshev’s Inequality Cauchy–Schwarz Inequality Hölders Inequality Some Geometrical Inequalities Jensen’s Inequality A01_Olympiad Mathematics_FM.indd xii 1.1 1.1 1.2 1.3 1.3 1.7 1.9 1.16 1.22 1.24 2.1 2.1 2.3 2.4 2.6 2.11 2.22 2.24 2.26 2.27 2.29 2.33 2.35 2.36 8/11/2017 5:21:38 PM viii Contents Chapter Mathematical Induction 3.1 Introduction3.1 First (or Weak) Principle of Mathematical Induction 3.2 Second (or Strong) Principle of Mathematical Induction 3.13 Chapter Recurrence Relation 4.1 Introduction4.1 Classification4.1 First Order Linear Recurrence Relation 4.3 First Order Non-linear Linear Homogeneous Recurrence Relation with Constant Coefficient of Order ‘2’ General Form of Linear Homogeneous Recurrence Relation with Constant Coefficients General Method for Non-Homogeneous Linear Equation Chapter Functional Equations 4.7 4.12 4.14 4.15 5.1 Function5.1 Functional Equation 5.3 Chapter Number Theory 6.1 Divisibility of Integers 6.1 Euclids Division Lemma 6.4 Greatest Common Divisor (GCD) 6.4 Primes6.8 Fundamental Theorem of Arithmetic 6.13 Number of Positive Divisors of a Composite Number 6.13 6.21 Modular Arithematic Complete Residue System (Modulo n)6.27 Some Important Function/Theorem 6.28 A01_Olympiad Mathematics_FM.indd 8/11/2017 5:21:39 PM Contents ix Scales of Notation Greatest Integer Function Diophantine Equations 6.35 6.39 6.45 Chapter Combinatorics 7.1 Definition of Factorial 7.1 Basic Counting Principles 7.2 Combinations7.13 The Bijection Principle 7.33 7.34 Combinations with Repetitions Allowed Definition of Permutation (Arrangements) 7.39 Introduction to Circular Permutation 7.57 Division and Distribution of Non-identical Items in Fixed Size 7.64 Number of Integral Solutions 7.69 Binomial, Multinomial and Generating Function 7.72 Application of Recurrence Relations 7.78 Principle of Inclusion and Exclusion (PIE) 7.81 Derangement7.93 Classical Occupancy Problems 7.98 Dirichlet’s (Or Pigeon Hole) Principle (PHP) 7.104 Chapter Geometry 8.1 Angle8.1 Congruent Triangles 8.7 Triangle Inequality 8.16 Ratio and Proportion Theorem (or Area Lemma) 8.22 Mid-point Theorem 8.26 Basic Proportionality Theorem (Thales’ Theorem) 8.29 Similar Triangles 8.38 Baudhayana (Pythagoras) Theorem 8.44 Quadrilaterals8.55 A01_Olympiad Mathematics_FM.indd 8/11/2017 5:21:39 PM 1.38 Chapter And similarly, we can find by the second equation 16 25 y = 1 − 2 x +y (2) By adding Eqs (1) and (2), we get 25( x + y ) = Let, Now x2 144 1 + 2 x +y + 16 1 − 2 x +y (3) 1 = t so that x + y = t +y 25 144 16 = + t (1 + t ) (1 − t ) ⇒ 144t(1 - t)2 + 16t(1 + t)2 = 25(1 - t2)2 ⇒ 32t(5t2 - 8t + 5) = 25(t4 - 2t2 + 1) Dividing both sides by t2, we get 1 32 5 t + − 8 = 25 t + − t t Putting t + = α in the above equation, we get t 26 25a2 - 160a + 156 = 0 ⇒ α = , 5 ⇒ t + =α = t or 26 ⇒ 5t2 - 6t + = 0 or 5t2 - 26t + = Since the discriminant of 5t2 - 6t + = is 36 - 100 < 0, there is no real root 5t2 - 26t + = 0, the roots are and Thus, x2 + y2 = or x2 + y2 = 5 1 1 If x + y = 5, then 5x 1 + = 12 and 5y 1 − = 5 5 Thus, by solving, we get x = 2 and y = 1 If x + y = then 5x(1 + 5) = 12 and 5y(1 − 5) = Thus, by solving, we get x= M01_Polynomials_C01.indd 38 and y= −1 8/11/2017 1:37:00 PM Polynomials 1.39 The two solution are x = 2, y = and x = Aliter: Let z = x + iy ⇒ −1 , y= 5 x + y = | z|2 = z ⋅ z x − iy Now Eq (1) + i Eq (2) ⇒ x + iy + = 12 + 4i x + y2 z ⇒ 5 z + = 12 + 4i ⇒ z − (12 + 4i ) z + = z ⋅ z ⇒ z = 12 + 4i ± (12 + 4i ) − 100 12 + 4i ± 28 + × 48i = 2(5) 10 12 + 4i ± 64 − 36 + × × 6i 12 + 4i ± (8 + 6i ) = 10 10 = + i, − i 5 −1 ⇒ ( x, y ) ≡ ( 2,1), , 5 = Problem 18 Solve the system ( x + y )( x + y + z ) = 18 ( y + z )( x + y + z ) = 30 ( z + x )( x + y + z ) = L in terms of L Where x, y, z, L ∈ + Solution: Adding the three equations, we get 2(x + y + z)2 = 48 + 2L x + y + z = 24 + L or Dividing the three equation by x + y + z = 24 + L , we get x+ y = 18 24 + L ,y+z = 30 24 + L ,z+ x = 24 24 + L Also, by solving, we get x= y= and z= ( 24 + L ) − 30 24 + L ( 24 + L) − L 24 + L 24 + L − 18 24 + L = = = L−6 24 + L 24 − L 24 + L L+6 24 + L , , where < L < 24 Problem 19 Solve: x + y - z = 4(1) Where x, y, z ∈ M01_Polynomials_C01.indd 39 x2 - y2 + z2 = -4(2) xyz = (3) 8/11/2017 1:37:02 PM 1.40 Chapter Solution: From Eq (1), (x - z) = (4 - y) ⇒ x2 - 2xz + z2 = 16 - 8y + y2 ⇒ (x2 + z2 - y2) - 2xz + 8y - 16 = ⇒ xz = 2(2y - 5) ( x2 + z2 - y2 = -4)(4) ∴ From Eqs (3) and (4), we get y × 2(2y - 5) = ⇒ 2y2 - 5y - = ⇒ (2y + l)(y - 3) = ⇒ Putting the value of y = − x−z =4 and xz = -12 y=− or y = in Eqs (1) and (3), we get 2 1 ( x + z ) = ( x − z ) + 4xz = − 48 < 2 So, y = is the only valid solution for y x - z = 1, xz = 2(5) ⇒ (x + z)2 = (x - z)2 + 4xz = ⇒ x + z = ±3(6) Solving Eqs (5) and (6), we get x = 2 and z = 1 or x = -1 and z = -2 So, the solution is x = 2, y = and z = or, x = -1, y = 3, z = -2 Problem 20 Solve: 3x(x + y - 2) = 2y y(x + y - 1) = 9x Solution: 3x(x + y - 2) = 2y(1) y(x + y - 1) = 9x (2) Multiplying Eqs (1) and (2), we get 3xy(x + y - 2)(x + y - 1) = 18xy ⇒ 3xy[(x + y - 2)(x + y - 1) - 6] = ⇒ 3xy[(x + y)2 - 3(x + y) - 4] = ⇒ 3xy(x + y - 4)(x + y + 1) = 0(3) So, x = or y = or x + y = or x + y = -1 Putting x + y = 4 in Eq (1), we get 6x = 2y ⇒ y = 3x ⇒ x = 1, y = M01_Polynomials_C01.indd 40 8/11/2017 1:37:02 PM Polynomials 1.41 Putting x + y = -1 in Eq (1), we get y= −9 x −7 x = −1 (As x + y = -1) ⇒ −9 ⇒x= ,y= 7 Also, x = 0 ⇔ y = −9 Thus, the solutions are (0, 0), (1, 3), , 7 Problem 21 Solve: xy + x + y = 23 yz + y + z = 31 zx + z + x = 47 Solution: We know xy + x + y = 23 (1) yz + y + z = 31 (2) zx + z + x = 47(3) Adding in both sides of Eq (1), we get xy + x + y + = 24 ⇒ (x + l)(y + 1) = 24(4) Similarly, we get (y + 1)(z + 1) = 32 (5) and (z + 1)(x + 1) = 48(6) By multiplying Eqs (4), (5) and (6), we get (x + l)2(y + 1)2(z + 1)2 = 24 × 32 × 48 ⇒ (x + l)(y + l)(z + 1) = ±(24 × 8) Since none of (x + 1), (y + 1) and (z + 1) is zero, we get z + = ±8 x + = ±6 y + = ±4 Thus, we have two solutions x = 5, y = 3, z = and x = -7, y = -5, z = -9 3 Problem 22 Find all the solutions of the system of equations y = 4x − 3x, z = 4y − 3y and x = 4z3 − 3z 3 Solution: If x > 1, then y = x + 3x(x − 1) > x > x > 1, z = 4y3 − 3y = y3 + 3y(y2 − 1) > y3 > y > and x = 4z3 − 3z = z3 + 3z(z2 − 1) > z3 > z > Thus, z > y > x > z, which is impossible, ⇒ x ≤ and, again, x < −1, and lead to x > y > z > x, so x ≥ −1 So, |x| ≤ 1, |y| ≤ 1, |z| ≤ And hence, we can write x = cosθ, where ≤ θ ≤ π M01_Polynomials_C01.indd 41 8/11/2017 1:37:02 PM 1.42 Chapter Now, y = cos3θ − cos θ = cos 3θ, z = 4y3 − 3y = cos3 3θ − cos 3θ = cos × 3θ = cos 9θ and x = 4z3 − 3z = cos3 9θ − cos 9θ = cos × 9θ = cos 27θ Since trigonometric functions are periodic, it is possible Thus, cos θ = cos 27θ ⇒ cos θ − cos 27θ = ⇒ sin 14θ sin 13θ = ⇒ sin 14θ = or sin 13θ = so θ = kπ 13 where k = 0, 1, 2, ,12, 13 kπ where k = 1, 2, ,13 14 and the solution is (x, y, z) = (cosθ, cos 3θ, cos 9θ) where θ takes all the above values or θ = Problem 23 Let, x = p, y = q, z = r and w = s be the unique solutions of the system of linear equations x + aiy + ai2 z + ai3 w = ai4 , i = 1, 2, 3, Express the solution of the following system in terms of p, q, r and s x + ai2 y + ai4 z + ai6 w = ai8 , i = 1, 2, 3, Assume the uniqueness of the solution Solution: Consider: the quadratic equation p + qt + rt2 + st3 = t4 t4 − st3 − rt2 − qt − p = or Now, by our assumption of the problem, a1 a2, a3 and a4 are the solution of this equation and hence, σ1 = a1 + a2 + a3 + a4 = s σ2 = (a1 + a2) (a3 + a4) + a1a2 + a3a4 = −r σ3 = a1a2 (a3 + a4) + a3a4 (a1 + a2) = q σ4 = a1a2a3a4 = −p The second system of equation is (t ) − w (t )3 − z (t ) − y (t ) − x = Putting t = u, we have u4 − wu3 − zu2 − yu − x = and the roots can be seen to be a12 , a22 , a32 and a42 σ = a12 + a22 + a32 + a42 = w and ⇒ w = ( ∑ ) − 2∑ a j = s + 2r i< j σ2 = ∑ i< j ai2 a 2j = −z or z = −∑ i< j M01_Polynomials_C01.indd 42 ai2 a 2j = − ∑ a j + ( ∑ ) ∑ a j ak − 2a1a2 a3 a4 i< j i< j c 3 c But, by Eq (3), \ S = 27 × M01_Polynomials_C01.indd 44 a = 24 c 8/11/2017 1:37:06 PM Polynomials 1.45 Aliter: Let b = B k then system of equations becomes a2 + ak + k2 = 25 or a2 + k2 - 2ak cos150° = 52 k2 + c2 = or k2 + c2 - 2kc cos90° = 32 c2 + ac + a2 = 16 ⇒ a2 + c2 - 2ac cos120° = 42 Now consider a ΔABC of sides 3, 4, and a point P in it such that AP = a, BP = k, CP = c Now consider ab + 2bc + 3ca = a k + kc + 3ca(1) C Area of ΔABC = k P c 150° 120° a A 1 1 kc + ac sin120° + ak sin150° = × × 2 2 3ac kc + + ak = 4 ⇒ 3kc + 3ca + ak = 24 ⇒ ⇒ ab + 2bc + 3ca = 24 (as 3k = b) Problem 26 Solve: log3(log2 x) + log1/3 ⋅ (log1/2 y) = xy2 = Solution: We have, log3(log2x) + (log1/3 ⋅ log1/2y) = ⇒ log3(log2x) - log3(log1/2y) = log x ⇒ log3 =1 ⇒ log1/ y log x = 31 log1/ y ⇒ log2x = 3log1/2y ⇒ log2 x = -3log2y = -log2 y3 ⇒ log2 xy3 = ⇒ xy3 = But, we have xy2 = So, by using the above equation, we get y = satisfy the parent equations and x = 64 Which Problem 27 Solve: log2 x + log4 y + log4 z = log3 y + log9z + log9 x = log4 z + log16 x + log16 y = Solution: We know that, loga x = log(an)(xn) So, log2x = log22 x2 = log4 x2, log3 y = log32 y2 = log9 y2 log4 z = log42 z2 = log16 z2 So, log2x + log3y + log4z = M01_Polynomials_C01.indd 45 8/11/2017 1:37:07 PM 1.46 Chapter ⇒ log4 x2yz = ⇒ x2yz = 42 = 16 (1) (2) (3) y xz = 81 Similarly, z xy = 256 and 2 x yz × y xz × z xy = 16 × 81 × 256 Hence, ⇒ (xyz)4 = 24 × 34 × 44 xyz = 24 as x, y, z > Dividing Eqs (1), (2), and (3) by xyz = 24, we get 16 = ,y 24 x= ,y= = x ⇒ 81 256 and z = 24 24 27 32 ,z = Problem 28 Find all real numbers x and y satisfying log3x + log2 y = 2; 3x − 2y = 23 Solution: By observation one solution is x = 3, y = As log33 + log22 = and 33 - 22 = 23 If x < 3, then log3x < Since, log3x + log2y = 2, log2y > and y > Hence, 3x < 33 = 27 and 2y > 22 = ⇒ 3x − 2y < 27 − = 23 So, x cannot be less than If x > 3, then log3x > and log2y < and so y < 2, 3x > 33 = 27 and 2y < 22 = So 3x − 2y > 27 − = 23 So, x cannot be greater than Hence, x = ⇒ y = Here, the only solution for the given equation is x = and y = Check Your Understanding Find the value of Find the value of Solve: 2+ + 2+ + 2− − 2− 444445 × 888885 × 444442 + 444438 using algebra 444444 x − x + ≥ − x 4 Let a, b, g be the roots of x3 - x2 - = Then find the value of 1+ α 1+ β 1+ γ + + 1−α 1− β 1− γ Show that (x − 1)2 is a factor of xm+1 − xm − x + Find all real solution x of the equation x10 − x8 + 8x6 − 24x4 + 32x2 − 48 = M01_Polynomials_C01.indd 46 8/11/2017 1:37:08 PM Polynomials 1.47 Solve x 99 + x 98 + x 97 + 3x 96 + + x + = in 8 Prove that + x111 + x222 + x333 + x444 divides + x111 + x222 + x333 +…+ x999 If x, y, z are rational and strictly positive and if 1 + = show that x y z x2 + y2 + z2 1 1 + + = , find x y z p a+ b+ c is rational 10 If a2x3 + b2y3 + c2z3 = p5, ax2 = by2 = cz2 and only in terms of p 11 If ax3 = by3 = cz3 and 1 + + = 1; prove that ax + by + cz = a + b + c x y z 12 Prove that, if (x, y, z) is a solution of the system of equations, x + y + z = a, 1 1 + + = Then, at least one of the numbers x, y, z is ‘a’ x y z a 13 If one root of the equation 2x2 – 6x + k = is (a + 5i) where i2 = –1; k, a ∈ , find the values of ‘a’ and ‘k’ 14 If x3 + px2 + q = 0, where q ≠ has a root of multiplicity 2, prove that 4p3 + 27q = 15 If f(x) is a quadratic polynomial with f(0) = 6, f(1) = and f(2) = 0, find f(3) 16 Show that, if a, b, c are real numbers and ac = 2(b + d), then, at least one of the equations x2 + ax + b = and x2 + cx + d = has real roots 17 Given any four positive, distinct, real numbers, show that one can choose three numbers A, B, C among them, such that all the quadratic equations have only real roots or all of them have only imaginary roots Bx2 + x + C = 0; Cx2 + x + A = 0; Ax2 + x + B = 18 Show that the equation x4 – x3– 6x2 – 2x + = cannot have negative roots 19 If a, b, c, d ∈ such that a < b < c < d, then show that, the roots of the equation (x – a)(x – c) +2(x – b)(x – d) = are real and distinct 20 Find the maximum number of positive and negative real roots of the equation x4 + x3 + x2 – x – = 21 If P(x) = ax2 + bx + c and Q(x) = –ax2 + bx + c, where ac ≠ 0, show that the equation P(x) ⋅ Q(x) = has at least two real roots 22 Let f(x) be the cubic polynomial x3 + x + 1; suppose g(x) is a cubic polynomial, such that g(0) = –1 and the roots of g(x) = are squares of the roots of f(x) = Determine g(9) 23 If p, q, r, s ∈ , show that the equation ( x + px + 3q)( x + rx + q)( − x + sx + 2q) = has at least two real roots = 24 If tn denotes the nth term of an AP and 1 , tq , then show that tpq is a root = q p of the equation (p + 2q – 3r)x2 + (q + 2r – 3p)x + (r + 2p – 3q) = 25 If p and q are odd integers, show that the equation x2 + 2px + 2q = has no rational roots 26 Show that there cannot exist an integer n, such that n3 − n + divides n3 + n2 + n + M01_Polynomials_C01.indd 47 8/11/2017 1:37:10 PM 1.48 Chapter 27 If sn = + q + q2 + … + qn and Sn = + n + 1 n + 1 n + 1 prove that s2 + s1 + + 28 Solve for x, y, z, the equations a= 1+ q 1+ q + + 1+ q + n n + 1 n + sn = Sn n + xy yz xz ,b = , and c = ( a, b, c ≠ 0) x+ y y+z x+z 29 Solve and find the non-trivial solutions x2 + xy + xz = y2 + yz + yx = z2 + zx + zy = 30 Solve: x2 + xy + y2 = y2 + yz + z2 = 19 z2 + zx + x2 = 31 Determine all solutions of the equation in , (x2 + 3x – 4)3 + (2x2 – 5x + 3)3 = (3x2 – 2x – 1)3 32 Show that there is no positive integer, satisfying the condition that (n4 + 2n3 +2n2 + 2n + 1) is a perfect square 33 Find the possible solutions of the system of equations: ax = (x + y + z)y; ay = (x + y + z)z; az = (x + y + z)x 34 If a and b are given integers, prove that the systems of equations, x + y + 2z + 2t = a and 2x - 2y + z - t = b has a solution in integers x, y, z, t 35 Show that 2x3 − 4x2 + x − cannot be factored into polynomials with integer coefficients 36 The product of two of the four roots of the equation x4 + 7x3 - 240x2 + kx + 2000 = is −200, determine k 37 The product of two of the four roots of x4 − 20x3 + kx2 + 590x − 1992 = is 24, find k 38 Let a, b, c, d be any four real numbers not all equal to zero Prove that the roots of the polynomial f (x) = x6 + ax3 + bx2 + cx + d can not all be real 39 If a, b, c and p, q, r are real numbers, such that for every real number x, ax2 + 2bx + c ≥ and px2 + 2qx + r ≥ 0, then prove that apx2 + bqx + cr ≥ for all real number x 40 Find a necessary and sufficient condition on the natural number n, for the equation xn + (2 + x)n + (2 − x)n = to have an integral root 41 Given that α, b, and γ are the angles of a right angled triangle Prove that sin α sin β sin (α − β) + sin b sin γ sin (β − γ) + sin γ sin a sin (γ − α) + sin (α − β) sin (β − γ) sin (γ − α) = 42 For a given pair of values x and y satisfy x = sin α, y = sin β, there can be four different values of z = sin (α + β): (i) Set up a relation between x, y, and z not involving trigonometric functions or radicals (ii) Find those pairs of values (x, y) for which z = sin (α + β) takes on fewer than four distinct values M01_Polynomials_C01.indd 48 8/11/2017 1:37:10 PM Polynomials 1.49 43 Suppose, a, b, and c are three real numbers, such that the quadratic equation x2 - (a + b + c)x + (ab + bc + ca) = has roots of the form a ± iβ, where a > and b ≠ are real numbers [here, i = −1 ] Show that (i) the numbers a, b, and c are all positive (ii) the numbers a , b, and c , form the sides of a triangle 44 Find the number of quadratic polynomials ax2 + bx + c, which satisfy the following conditions: (i) a, b, c, are distinct (ii) a, b, c ∈ {1, 2, 3, …, 999} (iii) (x + 1) divides (ax2 + bx + c) 45 Show that there are infinitely many pairs (a, b) of relatively prime integers (not necessarily positive) such that both quadratic equations x2 + ax + b = and x2 + 2ax + b = have integer roots. [INMO, 1995] 46 If the magnitude of the quadratic function f (x) = ax2 + bx + c never exceeds for ≤ x ≤ 1, prove that the sum of the magnitudes of the coefficients cannot exceed 17 47 Suppose that -1 ≤ ax2 + bx + c ≤ for -1 ≤ x ≤ 1, where a, b, c are real numbers, prove that -4 ≤ 2ax + b ≤ for -1 ≤ x ≤ 48 Find the polynomial p(x) = x2 + px + q for which max | P ( x ) | is minimal x∈[ −1,1] 49 Find real numbers a, b, c for which |ax2 + bx + c| ≤ ∀ |x| < and a + 2b is maximal 50 Let a, b, c, ∈ and a < and all roots of x3 + ax2 + bx + c = are negative real numbers Prove that b + c < Challenge Your Understanding xp( x − 1) = ( x − 30) p( x ) ∀x ∈ », find all such polynomial p(x) Find a polynomial p( x ) if it exist such that xp( x − 1) = ( x + 1) p( x ) Let f ( x ) be a quadratic function suppose f ( x ) = x has no real roots Prove that f ( f ( x )) = x has also no real roots If ( ax + bx + cx + dx + e) ∀x ∈ » where a, b, c, d , e ∈ » Prove that | a, | b, | c, | d , | e Prove that a + ab + b ≥ 3( a + b − 1) ∀a, b ∈ » Let p( x ) = x + x + x + x + Find the remainder on dividing p( x ) by p( x ) Find the remainder when x 2025 is divided by ( x + 1)( x + x + 1) If A, B, C, …, a, b, c, …, K are all constants, show that all the roots of the equation A2 B2 C2 H2 + + + + = x + K are real x−a x−b x−c x−h Prove that there does not exist a polynomial, p( x ) = a0 + a1 x + a2 x + + an x n , such that p(0), p(1), p( 2),… are all prime numbers M01_Polynomials_C01.indd 49 8/11/2017 1:37:14 PM 1.50 Chapter 10 Solve the following equations for real ‘x’ depending upon real parameter ‘a’: (a) x + a + x = a (b) x − a − x = a (c) a − a + x = x 11 The polynomial ax + bx + cx + d has integral coefficients a, b, c, d with ad odd and bc even Prove that all roots cannot be rational 12 If roots of x + ax + bx + ax + = has real roots then find the minimum value of a2 + b2 13 If the coefficient of x k upon the expansion and collecting of terms in the expres- ( sion … ( ( x − )2 − 2) − ) 2 − − is ak , then find a0 , a1 , a2 , a3 and a2 k n times 14 Prove that the equations x2 − 3xy + 2y2 + x − y = and x2 − 2xy + y2 − 5x + 7y = imply the equation xy − 12x + 15y = 15 If a and b are integers and the solutions of the equation y − 2x − a = and y2 − xy + x2 − b = are rational, then prove that the solutions are integers 16 Solve the following system of equations for real numbers a, b, c, d, e: 3a = (b + c + d)3, 3b = (c + d + e)3, 3c = (d + e + a)3, 3d = (e + a + b)3, 3e = (a + b + c)3. [INMO, 1996] 17 Solve for real numbers x and y, simultaneously the equations given by xy2 = 15x2 + 17xy + 15y2 and x2y = 20x2 + 3y2 18 Solve the system of equations in integers: 3x2 – 3xy + y2 = 7, 2x2 – 3xy + 2y2 = 14 19 In the sequence a1, a2, a3, …, an, the sum of any three consecutive terms is 40; if the third term is 10 and the eighth term is 8; find the 2013th term 20 A sequence has first term 2007, after which every term is the sum of the squares of the digits of the preceding term Find the sum of this sequence upto 2013 terms 21 Find a finite sequence of 16 numbers, such that (a) it reads same from left to right as from right to left (b) the sum of any consecutive terms is –1 (c) the sum of any 11 consecutive terms is +1 22 A two-pan balance is inaccurate since its balance arms are of different lengths and its pans are of different weights Three objects of different weights A, B and C are each weighed separately When they are placed on the left pan, they are balanced by weights A1, B1, and C1 respectively When A and B are placed on the right pan, they are balanced by A2 and B2, respectively Determine the true weight of C in terms of A1, B1, C1, A2 and B2 [USA MO, 1980] 23 If a and b are two of the roots of x + x − = 0, prove that ab is a root of x6 + x4 + x3 - x2 − = [USA MO, 1977] M01_Polynomials_C01.indd 50 8/11/2017 1:37:15 PM Polynomials 1.51 24 If P(x), Q(x), R(x), and S(x) are all polynomials, such that P(x5) + xQ(x5) + x2R(x5) = (x4 + x3 + x2 + x + 1) S(x), prove that (x − 1) is a factor of P(x) [USA MO, 1976] The generalization of the above problem is: if P0(x), P1(x), …, P(n-3) (x), n ≥ and S(x) are polynomials, such that P0(xn) + xP1(xn) + … + xn−3 P(n−3)(xn) = (xn-1 + xn-2 + … + x + 1)S(x) then (x − 1) is a factor of Pi, (x) for all i 25 If x5- x3 + x = a, prove that x6 ≥ 2a - 1. [INMO, 1994] 26 The solutions x1, x2, and x3 of the equation x3 + ax + a = 0, where a is real and a ≠ 0, satisfy x12 x22 x32 + + = −8, find x1, x2, and x3. x2 x3 x1 [AMTI, 1994] 27 Let p(x) be a polynomial with degree 2008 and leading coefficient such that p(0) = 2007, p(1) = 2006, p(2) = 2005, …, p(2007) = 0; determine p(2008) 28 If P(x) denotes a polynomial of degree n, such that P ( k ) = for k = 1, 2, 3, …, n + 1, determine P(n + 2) k k for k = 0, 1, 2,…, k +1 n, determine P(n + 1). [USA MO, 1975] 30 Let a, b and c denote three distinct integers and let P denote a polynomial having all integral coefficients Show that it is impossible that P(a) = b, P(b) = c and P(c) = a. [USA MO, 1974] 29 If P(x) denotes a polynomial of degree n, such that P (k )= 31 Let, ai, i = 1, 2, …, n be distinct real numbers b1 b2, …, bn be real numbers, n such that the product ∏ ( + b j ) is the same for each i Prove that the product j =1 n ∏ ( + b j ) is also constant for all j i =1 32 In the polynomial P(x) = xn + a1xn−1 + … + an-1 x + 1, the coefficients a1, a2, …, an–1 are non-negative and it has n real roots Prove that P(2) ≥ 3n 33 Determine all polynomials of degree n with each of its (n + 1) coefficients equal to ±1, which have only real roots 34 Let p(x) be polynomial over and at three distinct integers it takes ±1 value, prove that it has no integral root 35 Let a, b be the roots of x2 - 6x + = Prove that α n + β n ∈ » ∀ n∈ » , also prove that + (α n + β n ) ∀n ∈ » 36 Let P(x) be a polynomial with real coefficients such that P(x) ≥ for every real x Prove that P ( x ) = f1 ( x ) + f ( x ) + + f n ( x)2 [Putnam, 1999] 37 Is it possible to find three quadratic polynomials f (x), g(x), h(x) such that the equation f (g(h(x))) = has eight roots 1, 2, 3, 4, 5, 6, 7, 8? [Russian MO, 1995] M01_Polynomials_C01.indd 51 8/11/2017 1:37:16 PM 1.52 Chapter 38 Let P(z) = az3 + bz2 + cz + d, where a, b, c, d are complex numbers with |a| = |b| = |c| = |d| = Show that |P(z)| ≥ for at least one complex number z satisfying |z| = 39 Consider two monic polynomials f (x) and g(x) of degree and respectively over real numbers Let there be an interval (a, b) of length more than such that both f (x) and g(x) are negative for x ∈(a, b) and both are positive for x < a or x > b Prove that there is a real number ‘a’ such that f (a) < g(a) 40 Let P1(x) = x2 - and Pj(x) = P1(Pj-1(x)) ∀ j = 2, 3, … Show that for any positive integer n, the roots of the equation Pn(x) = x are real and distinct. [IMO, 1976] 41 Find all polynomials f satisfying f (x2) + f (x) f (x + 1) = ∀x ∈ » 42 Find all polynomials P(x), for which P ( x ) ⋅ P ( x ) = P ( x + x )∀x ∈ » 43 Find all polynomials f (x) such that f ( x ) ⋅ f ( x + 1) − f ( x + x + 1) = ∀x ∈ » 44 Find all polynomials f (x) such that f ( x ) ⋅ f ( − x ) − f ( x ) = ∀x ∈ » 45 Prove that if a polynomial of degree over is equal to +1 or -1 for different integers then it is irreducible over 46 Prove that ( x − a1 ) ( x − a2 ) ( x − an ) + is irreducible over 47 Prove that ( x + 12 )( x + 22 )… ( x + n2 ) + is irreducible over 48 Let a1 , a2 ,… , an ∈ » are distinct, find them for which ( x − a1 )( x − a2 )… ( x − an ) + can be expressible as product of two polynomials with integral coefficients ( a)| | p= (b)| for a, b ∈ », a < b; If 49 Let p( x ) be a polynomial over such that | p= a+b p( x ) = has rational root a, then prove that a − b = or and α = 50 Let a1 , a2 , … , an and b1 , b2 , …, bn be two distinct collections of n positive integers, where each collection may contain repetitions If the two collections of integers + a j (1 ≤ i < j ≤ n) and bi + b j (1 ≤ i < j ≤ n) are the same, then prove that n is a power of ⋅ M01_Polynomials_C01.indd 52 8/11/2017 1:37:18 PM ... page is intentionally left blank A01 _Olympiad Mathematics_ FM.indd 8/11/2017 5:21:38 PM Pathfinder for Olympiad MATHEMATICS Vikash Tiwari V Seshan A01 _Olympiad Mathematics_ FM.indd 8/11/2017 5:21:38... all those using this book Vikash Tiwari V Seshan A01 _Olympiad Mathematics_ FM.indd 11 8/11/2017 5:21:39 PM Acknowledgements First and foremost, we thank the Pearson group for motivating us and rendering... for my (our) sake Vikash Tiwari About the Authors Vikash Tiwari has been teaching students for Mathematical Olympiads (Pre RMO, RMO, INMO and IMOTC) and other examinations like KVPY and JEE for