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2006 AIME I Problems/Problem 8

Revision as of 21:32, 28 November 2006 by 1337h4x (talk | contribs) (Solution)

Problem

Hexagon $ABCDEF$ is divided into four rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}.$ Given that $K$ is a positive integer, find the number of possible values for $K.$



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Solution

Let $x$ denote the common side length of the rhombi. Let $y$ denote one of the smaller interior angles of rhombus $\mathcal{P}$. Then $x^2sin(y)=\sqrt{2006}.$$K=x^2\sin(2y) \Rightarrow K=2x^2siny\cdot cosy \Rightarrow K= 2\sqrt{2006}\cdot cosy.$ Thus $K$ is any positive integer on ($0, 2\sqrt{2006}$). $2\sqrt{2006}\approx 89.58$. Hence, the number of positive values for $K$ is ${089}$.

Solution provided by 1337h4x

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