2006 AIME I Problems/Problem 8
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 within . 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
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.