Difference between revisions of "2006 AIME I Problems/Problem 8"
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==Solution 2== | ==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{89}</math> | 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{89}</math> | ||
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-jackshi2006 | -jackshi2006 |
Revision as of 22:36, 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.