Complex numbers and polynomials - University College Cork

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Moreover, since t 2 ≥ 4 is a requirement,(x, y) ∈ ∪ −2≤t≤2 {(x, y) : tx + y + 2 ≤ 0}= {(x, y) : −2x + y + 2 ≤ 0} ∪ {(x, y) : 2x + y + 2 ≤ 0}.It’s easily seen that the lines −2x + y + 2 = 0, 2x + y + 2 = 0 are tangentsto the parabola x 2 = 4(y − 2) at the points (4, 6), (−4, 6), respectively, andintersect at (0, −2). Hence{(x, y) : −2x + y + 2 ≤ 0} ∪ {(x, y) : 2x + y + 2 ≤ 0} ⊂ {(x, y) : 4(y − 2) ≤ x 2 },and soΓ ⊂ {(x, y) : −2x + y + 2 ≤ 0} ∪ {(x, y) : 2x + y + 2 ≤ 0}.Plainly, equality holds here. What we seek is the square of the distance fromΓ to the origin. It’s clear from geometric considerations that the distance isgiven by the distance from the origin to one of the tangent lines, i.e., it is|0 + 0 + 2|√(±2)2 + 1 = √ 2 . 5Hence the required minimum value is 4/5.2. USAMO75. Let p be a poly of degree n and suppose thatDetermine p(n + 1).p(k) =k , k = 0, 1, 2, . . . , n.k + 1Consider the poly q(x) = (x + 1)p(x) − x. Then (?)q(x) = an∏(x − k)for some a. But q(−1) = 1. Hence a can be determined, whencek=0p(n + 1) = (−1)n+1 (n + 1).n + 23. IRMO89. Let a be a positive real number, and letb = 3 √a + √ a 2 + 1 + 3 √a − √ a 2 + 1.30
Prove that b is a positive integer if, and only if, a is a positive integer of theform 1 2 n(n2 + 3), for some positive integer n.By direct computation it can be seen that b is a root of the cubic x 3 + 3x − 2a.Hence, if b is a positive integer, then so is 2a = b 3 + 3b. But the latter is even.Hence, a is a positive integer. Conversely, if a is of the form 1 2 n(n2 + 3), forsome positive integer n, then n is also a root of the same cubic. But this cubichas only one real root. Thus b = n.4. IRMO95. Suppose that a, b and c are complex numbers, and that all threeroots z of the equationx 3 + ax 2 + bx + c = 0satisfy |z| = 1 (where | | denotes absolute value). Prove that all three roots wof the equationx 3 + |a|x 2 + |b|x + |c| = 0also satisfy |w| = 1.Denote the roots of x 3 + ax 2 + bx + c = 0 by p, q, r. ThenHence |c| = 1, andThus−a = p + q + r, b = pq + qr + rp, −c = pqr.|b| = |(pqr)( 1 r + 1 p + 1 )| = |¯r + ¯p + ¯p| = |a|.qx 3 + |a|x 2 + |b|x + |c| = x 3 + |a|(x 2 + x) + 1 = (x + 1)(x 2 + (|a| − 1)x + 1);and (?) since ||a| − 1| ≤ 2, the roots of the quadratic x 2 + (|a| − 1)x + 1 arecomplex numbers of unit modulus, the result follows.5. Prove that x 4 + x 3 + x 2 + x + 1 is a factor of x 44 + x 33 + x 22 + x 11 + 1.Since x 5 − 1 = (x − 1)(x 4 + x 3 + x 2 + x + 1) = 0 we see that x 5n = 1 for everyinteger n. So x 44 + x 33 + x 22 + x 11 + 1 = x 8.5 x 4 + x 6.5 x 3 + x 4.5 x 2 + x 2.5 x + 1 =x 4 + x 3 + x 2 + x + 1 = 0. Thus every root of x 4 + x 3 + x 2 + x + 1 is a root ofx 44 + x 33 + x 22 + x 11 + 1, and the Remainder Theorem does the rest.6. The coeffs of the cubic ax 3 + bx 2 + cx + d are integers, ad is odd and bc is even.Prove that at least one root is irrational.Suppose all roots are rational, and apply the rational roots theorem. Accordingto this, if p/q is a rational root with p, q in their lowest form, then p divides31
Page 1 and 2: Enrichment Lectures 2010Some facts
Page 3 and 4: For,[ a b−b a] [ c d+−d c] [ a
Page 5 and 6: 3 Linear and quadratic polynomialsH
Page 7 and 8: 3.3 Heron’s problemGiven two dist
Page 9 and 10: We can also conclude that the graph
Page 11 and 12: 7. Let P, Q be two points not neces
Page 13 and 14: 2. Let p be a arbitrary quartic. Us
Page 15 and 16: with equality iff α = β = γ. Thu
Page 17 and 18: Plainly, the pair a = 1, b = 2 work
Page 19 and 20: Note especially that if a is ration
Page 21 and 22: and, for k ≥ 2,c k =Again,( mk)b
Page 23 and 24: 2. For which n ∈ N is x 2 +x+1 a
Page 25 and 26: 13. IRMO95. Suppose that a, b and c
Page 27 and 28: q = −3uv, r = u 3 + v 3 for u, v.
Page 29: To utilize this, replace x by ρ co

Complex numbers and polynomials - University College Cork

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