Problems!

  1. Let \(H\) be the orthocenter of triangle \(ABC\). Prove that, if \(\dfrac{AH}{BC}=\dfrac{BH}{CA}=\dfrac{CH}{AB}\), the triangle is equilateral.

  2. Let a,b,ca, b, c be the roots of x3x2x1x^3-x^2-x-1. Prove that a2014b2014ab+b2014c2014bc+c2014a2014ca\dfrac{a^{2014}-b^{2014}}{a-b}+\dfrac{b^{2014}-c^{2014}}{b-c}+\dfrac{c^{2014}-a^{2014}}{c-a} is an integer.

  3. Let AA and BB be two subsets of S={1,...,2000}S=\{1,...,2000\} such that AB3999|A|\cdot|B|\ge3999. For a set XX, let XXX-X denote the set {xyx,yX}\{x-y|x,y\in{X}\}. Prove that (AA)(BB)(A-A)\cap(B-B) is not an empty set.

#Algebra #Geometry #Combinatorics #OlympiadMath #ProblemSolving

Note by José Marín Guzmán
6 years, 11 months ago

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Comments

Solution to Problem 1

2RcosAa=2RcosBb=2RcosCc\frac{2RcosA}{a}=\frac{2RcosB}{b}=\frac{2RcosC}{c}

By cosine rule,

b2+c2a22abc=c2+a2b22abc=a2+b2c22abc\frac{b^2+c^2-a^2}{2abc}=\frac{c^2+a^2-b^2}{2abc}=\frac{a^2+b^2-c^2}{2abc}

Hence a=b=ca=b=c

Souryajit Roy - 6 years, 11 months ago
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