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

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== Problem ==
 
== Problem ==
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Hexagon <math> ABCDEF </math> is divided into four rhombuses, <math> \mathcal{P, Q, R, S,} </math> and <math> \mathcal{T,} </math> as shown. Rhombuses <math> \mathcal{P, Q, R,} </math> and <math> \mathcal{S} </math> are congruent, and each has area <math> \sqrt{2006}. </math> Let <math> K </math> be the area of rhombus <math> \mathcal{T}. </math> Given that <math> K </math> is a positive integer, find the number of possible values for <math> K. </math>  
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[[Hexagon]] <math> ABCDEF </math> is divided into four [[rhombus]]es, <math> \mathcal{P, Q, R, S,} </math> and <math> \mathcal{T,} </math> as shown. Rhombuses <math> \mathcal{P, Q, R,} </math> and <math> \mathcal{S} </math> are [[congruent (geometry) | congruent]], and each has [[area]] <math> \sqrt{2006}. </math> Let <math> K </math> be the area of rhombus <math> \mathcal{T}</math>.  Given that <math> K </math> is a [[positive integer]], find the number of possible values for <math> K</math>.
  
  
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== Solution ==
 
== Solution ==
 
Let <math>x</math> denote the common side length of the rhombi.
 
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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Let <math>y</math> denote one of the smaller interior [[angle]]s of rhombus <math> \mathcal{P} </math>. Then <math>x^2\sin(y)=\sqrt{2006}</math>.  We also see that <math>\displaystyle K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y</math>.  Thus <math>K</math> can be any positive integer in the [[interval]] <math>(0, 2\sqrt{2006})</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 089.
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<math>2\sqrt{2006} = \sqrt{8024}</math> and <math>89^2 = 7921 < 8024 < 8100 = 90^2</math>, so <math>K</math> can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for <math>K</math> is 089.
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Solution provided by 1337h4x
 
  
 
== See also ==
 
== See also ==
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* [[2006 AIME I Problems/Problem 7 | Previous problem]]
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* [[2006 AIME I Problems/Problem 9 | Next problem]]
 
* [[2006 AIME I Problems]]
 
* [[2006 AIME I Problems]]
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Trigonometry Problems]]

Revision as of 12:34, 29 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 $x^2\sin(y)=\sqrt{2006}$. We also see that $\displaystyle K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y$. Thus $K$ can be any positive integer in the interval $(0, 2\sqrt{2006})$. $2\sqrt{2006} = \sqrt{8024}$ and $89^2 = 7921 < 8024 < 8100 = 90^2$, so $K$ can be any integer between 1 and 89, inclusive. Thus the number of positive values for $K$ is 089.

See also

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