Difference between revisions of "2006 AIME I Problems/Problem 8"
I_like_pie (talk | contribs) |
Jackshi2006 (talk | contribs) (→Solution 2) |
||
(20 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Hexagon <math> ABCDEF </math> is divided into | + | [[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 == | + | == Solution 1== |
− | {{ | + | 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 <math>\boxed{089}</math>. | ||
+ | |||
+ | |||
+ | ==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>. This expression equals an integer K. <math>2006-h^4</math> specifically must be in the form <math>n^2/2</math>. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of <math>n^2</math> for <math>2006-h^4</math>. Now, quick testing shows that <math>44^2 < 2006</math> and <math>45^2>2006</math>, but we must also test <math>44.5^2</math>, because the product of two will make it an integer. <math>44.5^2</math> is also less than <math>2006</math>, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us <math>44*2+1=</math> <math>\boxed{089}</math> | ||
+ | |||
+ | |||
+ | -jackshi2006 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2006|n=I|num-b=7|num-a=9}} | |
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 23:37, 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 . This expression equals an integer K. specifically must be in the form . There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of for . Now, quick testing shows that and , but we must also test , because the product of two will make it an integer. is also less than , so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us
-jackshi2006
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.