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 00:33, 29 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