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

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[[Image:2006AimeA8.PNG]]
 
[[Image:2006AimeA8.PNG]]
  
== Solution ==
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== Solution 1==
 
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 [[angle]]s of rhombus <math> \mathcal{P} </math>. Then <math>x^2\sin(y)=\sqrt{2006}</math>.  We also see that <math>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>.  
 
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>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>.  
 
<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 <math>\boxed{089}</math>.
 
<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 <math>\boxed{089}</math>.
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==Solution 2==
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Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where <math>w*h=\sqrt{2006}</math>. The height of rhombus T would be 2h, and the width would be <math>\sqrt{w^2-h^2}</math>. Substitute the first equation to get <math>\sqrt{\frac{2006}{h^2}-h^2}</math>. Then the area of the rhombus would be <math>2h * \sqrt{\frac{2006}{h^2}-h^2}</math>. Combine like terms to get <math>2 * \sqrt{2006-h^4}</math>
  
 
== See also ==
 
== See also ==

Revision as of 22:29, 21 October 2020

Problem

Hexagon $ABCDEF$ is divided into five 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 $K$.

2006AimeA8.PNG

Solution 1

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 $K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y$. 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 $\boxed{089}$.


Solution 2

Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where $w*h=\sqrt{2006}$. The height of rhombus T would be 2h, and the width would be $\sqrt{w^2-h^2}$. Substitute the first equation to get $\sqrt{\frac{2006}{h^2}-h^2}$. Then the area of the rhombus would be $2h * \sqrt{\frac{2006}{h^2}-h^2}$. Combine like terms to get $2 * \sqrt{2006-h^4}$

See also

2006 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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