# 102 Combinatorial Problems from the Training of the USA IMO by Titu Andreescu

By Titu Andreescu

"102 Combinatorial difficulties" includes rigorously chosen difficulties which have been utilized in the learning and trying out of the us foreign Mathematical Olympiad (IMO) crew. Key positive aspects: * offers in-depth enrichment within the very important parts of combinatorics via reorganizing and embellishing problem-solving strategies and methods * issues comprise: combinatorial arguments and identities, producing features, graph concept, recursive relatives, sums and items, chance, quantity concept, polynomials, idea of equations, advanced numbers in geometry, algorithmic proofs, combinatorial and complex geometry, sensible equations and classical inequalities The e-book is systematically prepared, steadily construction combinatorial abilities and strategies and broadening the student's view of arithmetic. apart from its sensible use in education academics and scholars engaged in mathematical competitions, it's a resource of enrichment that's sure to stimulate curiosity in numerous mathematical parts which are tangential to combinatorics.

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B ( ) powss 18th April 2013 14:32 World Scientific Book - 9in x 6in Chapter 4 Identities, Inequalities and Expressions (34) Problem. Without calculating the value, prove that 1 3 5 99 1 · · · ··· · < . 2 4 6 100 10 Solution. Note that 1 2 3 4 2 3 4 < 5 ..

2 2 In order to obtain real solutions we need the discriminant of the quadratic, namely 1 + 6x − 3x2 to be non-negative. That is, x+1± 0 ≤ 1+6x−3x2 = 1−3(x2 −2x) = 4−3(x2 −2x+1) = 4−3(x−1)2 , which means that (x − 1)2 ≤ 34 . Since we need x to be an integer, we conclude that (x − 1)2 = 0 or 1, which means x can only be one of 0, 1, or 2. Solving this equation with x = 0 yields y = 0 or 1; with x = 1 yields y = 0 or 2; and with x = 2 yields y = 1 or 2. Thus, the solution set is (x, y) = (0, 0), (1, 0), (0, 1), (1, 2), (2, 1), and (2, 2).

Multiplying by A2 + AB + B 2 yields A3 − B 3 = A2 + AB + B 2 , which means that A2 + AB + B 2 = x + 20 − (x + 1) = 19. Also, we see that (A − B)2 = A2 + AB + B 2 = 1. Subtracting, we get 3AB = 19 − 1 = 18, or AB = 6. This implies that (x + 20)(x + 1) = A3 B 3 = 63 = 216 which yields the quadratic equation x2 + 21x − 196 = 0. This is equivalent to (x + 28)(x − 7) = 0. Therefore, the solution is x = −28 or x = 7. (38) Problem. Find the product of x, y, and z if {x, y, z} is the solution set for the given equations: x + y + z = 1, x2 + y 2 + z 2 = 2, x3 + y 3 + z 3 = 3.