# 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> | ||

− | 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 | + | 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. |

Solution provided by 1337h4x | Solution provided by 1337h4x |

## Revision as of 23:33, 28 November 2006

## Problem

Hexagon is divided into four rhombuses, and as shown. Rhombuses and are congruent, and each has area Let be the area of rhombus Given that is a positive integer, find the number of possible values for

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## Solution

Let denote the common side length of the rhombi. Let denote one of the smaller interior angles of rhombus . Then Thus is any positive integer on (). . Hence, the number of positive values for is 089.

Solution provided by 1337h4x