Difference between revisions of "2006 AIME I Problems/Problem 8"
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[[Image:2006AimeA8.PNG]] | [[Image:2006AimeA8.PNG]] | ||
| − | == Solution == | + | == 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== | ||
| + | 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 23:29, 21 October 2020
Contents
Problem
Hexagon 
is divided into five 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 .
Solution 1
Let 
denote the common side length of the rhombi.
Let 
denote one of the smaller interior angles of rhombus 
. Then 
. We also see that 
. Thus can be any positive integer in the interval 
.
and 
, so can be any integer between 1 and 89, inclusive. Thus the number of positive values for is 
.
Solution 2
Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where 
. The height of rhombus T would be 2h, and the width would be 
. Substitute the first equation to get 
. Then the area of the rhombus would be 
. Combine like terms to get 
See also
| 2006 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
