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

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Let <math>x</math> denote the common side length of the rhombi.
 
Let <math>x</math> denote the common side length of the rhombi.
 
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>  
 
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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  K=x^2\sin(2y) \Rightarrow K=2x^2siny\cdot cosy \Rightarrow K= 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>.
+
  K=x^2\sin(2y) \Rightarrow K=2x^2siny\cdot cosy \Rightarrow K= 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>\boxed{089}</math>.
  
 
Solution provided by 1337h4x
 
Solution provided by 1337h4x

Revision as of 23:31, 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}.$$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 $\boxed{089}$.

Solution provided by 1337h4x

See also

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