- Page 3 and 4: About the Authors Titu Andreescu re
- Page 5 and 6: Titu Andreescu University of Texas
- Page 7 and 8: About the Authors Titu Andreescu re
- Page 9 and 10: Titu Andreescu University of Texas
- Page 11 and 12: vi Contents 2 Complex Numbers in Tr
- Page 13 and 14: viii Contents 5.7 Problems Involvin
- Page 15 and 16: x Preface plex numbers and the oper
- Page 17 and 18: Notation Z the set of integers N th
- Page 19 and 20: 2 1. Complex Numbers in Algebraic F
- Page 21 and 22: 4 1. Complex Numbers in Algebraic F
- Page 23 and 24: 6 1. Complex Numbers in Algebraic F
- Page 25 and 26: 8 1. Complex Numbers in Algebraic F
- Page 27 and 28: 10 1. Complex Numbers in Algebraic
- Page 29 and 30: 12 1. Complex Numbers in Algebraic
- Page 31 and 32: 14 1. Complex Numbers in Algebraic
- Page 33 and 34: 16 1. Complex Numbers in Algebraic
- Page 35 and 36: 18 1. Complex Numbers in Algebraic
- Page 37 and 38: 20 1. Complex Numbers in Algebraic
- Page 39 and 40: 22 1. Complex Numbers in Algebraic
- Page 41: 24 1. Complex Numbers in Algebraic
- Page 45 and 46: 2 Complex Numbers in Trigonometric
- Page 47 and 48: 2.1. Polar Representation of Comple
- Page 49 and 50: and and Arg z3 = 2.1. Polar Represe
- Page 51 and 52: 2.1. Polar Representation of Comple
- Page 53 and 54: and Example. Let z1 = 1 − i and z
- Page 55 and 56: Problem. Compute 2.1. Polar Represe
- Page 57 and 58: Verify your results using the algeb
- Page 59 and 60: The cube roots of the number z are
- Page 61 and 62: iii) For n = 4, the fourth roots of
- Page 63 and 64: 2.2. The n th Roots of Unity 47 Pro
- Page 65 and 66: 2.2. The n th Roots of Unity 49 Now
- Page 67 and 68: 2.2. The n th Roots of Unity 51 Bec
- Page 69 and 70: 3 Complex Numbers and Geometry 3.1
- Page 71 and 72: 3) arg(z − a) = arg(b − a); z
- Page 73 and 74: 3.1. Some Simple Geometric Notions
- Page 75 and 76: since the triangle M2OM1 is negativ
- Page 77 and 78: 3.1.5 Angle between two lines 3.1.
- Page 79 and 80: 3.1. Some Simple Geometric Notions
- Page 81 and 82: hence Likewise, 3.2. Conditions for
- Page 83 and 84: 3.3. Similar Triangles 67 2) The po
- Page 85 and 86: 3.3. Similar Triangles 69 Triangles
- Page 87 and 88: 1 e) + z − z1 1 + z − z2 1 = 0,
- Page 89 and 90: 3.4. Equilateral Triangles 73 Proof
- Page 91 and 92: Using Proposition 2, we have 3.4. E
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or equivalently Let α = A − Bi 2
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Indeed, the equation is � � �
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Applying formula (2) we find that 3
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3.6. The Circle 83 Proof. Using car
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the equation of the circle with cen
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Figure 3.15. 3.6. The Circle 87 Pro
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4 More on Complex Numbers and Geome
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4.1. The Real Product of Two Comple
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since n� εk = 0. k=1 4.1. The Re
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4.1. The Real Product of Two Comple
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4.2. The Complex Product of Two Com
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1) Since 4.2. The Complex Product o
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4.3. The Area of a Convex Polygon 1
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4.4. Intersecting Cevians and Some
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the proposition we obtain It follow
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4.5 The Nine-Point Circle of Euler
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Figure 4.6. If Gs is the midpoint o
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Proof. Let us prove that α satisfi
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4.6. Some Important Distances in a
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Using the real product we can write
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4.7. Distance between Two Points in
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4.8. The Area of a Triangle in Bary
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4.8. The Area of a Triangle in Bary
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BA2 A2C BA3 A3C 4.8. The Area of a
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= = αβγ 8(s − α)(s − β)(s
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Figure 4.11. where R is the circumr
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This is equivalent to |x| 2 = R 2
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4.9. Orthopolar Triangles 131 sides
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or 4.9. Orthopolar Triangles 133 Ca
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It follows that A ′ 1 , B′ 1 ,
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4.10. Area of the Antipedal Triangl
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Relations (2) and (3) imply that 4.
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The point G with coordinate 4.11. L
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1) In the relation (4) we choose M
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Using the formula (1) is equivalent
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4.11. Lagrange’s Theorem and Appl
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4.12. Euler’s Center of an Inscri
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4.13. Some Geometric Transformation
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4.13.4 Rotation 4.13. Some Geometri
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4.13. Some Geometric Transformation
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4.13. Some Geometric Transformation
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4.13. Some Geometric Transformation
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5 Olympiad-Caliber Problems The use
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or Solution. The given equality can
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5.1. Problems Involving Moduli and
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Solution. We have if and only if i.
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Problem 10. Prove that � � �
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and 1 z1 On the other hand, we have
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5.1. Problems Involving Moduli and
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Prove that for any z ∈ A there is
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5.2. Algebraic Equations and Polyno
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and 5.2. Algebraic Equations and Po
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5.3. From Algebraic Identities to G
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5.3. From Algebraic Identities to G
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Hence, the inequality (1) is equiva
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5.3. From Algebraic Identities to G
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5.3. From Algebraic Identities to G
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Using the rotation formulas, we obt
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Figure 5.1. The angle between the l
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Figure 5.3. 5.4. Solving Geometric
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Then 5.4. Solving Geometric Problem
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Using the rotation formula we obtai
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5.4. Solving Geometric Problems 201
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if and only if That is, 5.4. Solvin
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the coordinates of points M and N a
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We find Likewise, we have 5.4. Solv
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5.4. Solving Geometric Problems 209
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5.4. Solving Geometric Problems 211
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5.4. Solving Geometric Problems 213
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5.5. Solving Trigonometric Problems
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hence and and 5.5. Solving Trigonom
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we derive the formulas: 5.5. Solvin
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5.6. More on the n th Roots of Unit
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5.6. More on the n th Roots of Unit
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5.6. More on the n th Roots of Unit
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) From the relation z + ω n � z
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5.7. Problems Involving Polygons 22
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5.7. Problems Involving Polygons 23
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We obtain 5.7. Problems Involving P
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5.7. Problems Involving Polygons 23
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5.8. Complex Numbers and Combinator
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Passing to the absolute value it fo
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Problem 5. The following identity h
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5.8. Complex Numbers and Combinator
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5.9. Miscellaneous Problems 245 Her
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5.9. Miscellaneous Problems 247 whe
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5.9. Miscellaneous Problems 249 Add
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and the values of z for which the m
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6 Answers, Hints and Solutions to P
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where Since we have a ≤ 2, as des
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hence E ∈ R. 33. Notice that and
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6.1. Answers, Hints and Solutions t
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6.1. Answers, Hints and Solutions t
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and 6.2. Solutions to the Olympiad-
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6.2. Solutions to the Olympiad-Cali
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and hence = as desired. Problem 30.
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6.2. Solutions to the Olympiad-Cali
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6.2. Solutions to the Olympiad-Cali
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with roots zk, k = 1, 5. Then 6.2.
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It follows that 6.2. Solutions to t
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6.2. Solutions to the Olympiad-Cali
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The last relation can be written as
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In our case, for n = 662 we obtain
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6.2. Solutions to the Olympiad-Cali
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or 6.2. Solutions to the Olympiad-C
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6.2. Solutions to the Olympiad-Cali
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6.2. Solutions to the Olympiad-Cali
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with respect to x,weget n� k=1 6.
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6.2. Solutions to the Olympiad-Cali
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6.2. Solutions to the Olympiad-Cali
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We have MB =|zM − ε| = MC =|zM
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since t ∈[0,π]. From (2) it foll
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and find where y j = n m = (ε −
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If z a −1 ± i = , then 2 6.2. So
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or 6.2. Solutions to the Olympiad-C
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308 Glossary Binomial equation: An
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310 Glossary Isometric transformati
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312 Glossary half-angle formulas: t
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314 References [9] Andreescu, T., K
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316 References [46] Retali, V., Big
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318 Index of Authors Jinga, D.: 5.1
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320 Subject Index extended argument