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Difference between revisions of "2006 AIME I Problems/Problem 8"

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== Solution ==
 
== Solution ==
{{solution}}
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Let <math>x</math> denote the common side length of the rhombi.
 
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Let <math>y</math> denote one of the smaller interior angles of rhombus <math> \mathcal{P} </math>. Then <math>x^2sin(y)=\sqrt{2006}.</math><math>
 
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\mathcal{T}=x^2\sin(2y) \Rightarrow \mathcal{T}=2x^2siny\cdot cosy \Rightarrow \mathcal{T}= 2\sqrt{2006}\cdot cosy.</math> Thus <math>K</math> is any positive integer on (<math>0, 2\sqrt{2006}</math>). <math>2\sqrt{2006}\approx 89.58</math>. Hence, the number of positive values for <math>K</math> is <math>{089}</math>.
 
== See also ==
 
== See also ==
 
* [[2006 AIME I Problems]]
 
* [[2006 AIME I Problems]]
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 17:14, 28 November 2006

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}.$$\mathcal{T}=x^2\sin(2y) \Rightarrow \mathcal{T}=2x^2siny\cdot cosy \Rightarrow \mathcal{T}= 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 ${089}$.

See also

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