|
|
| Line 1: |
Line 1: |
| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 8]] |
| − | [[Hexagon]] <math> ABCDEF </math> is divided into five [[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>.
| |
| − | | |
| − | | |
| − | [[Image:2006AimeA8.PNG]]
| |
| − | | |
| − | == Solution ==
| |
| − | 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>.
| |
| − | <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.
| |
| − | | |
| − | == See also ==
| |
| − | {{AIME box|year=2006|n=II|num-b=7|num-a=9}}
| |
| − | | |
| − | [[Category:Intermediate Combinatorics Problems]]
| |
