2006 AIME I Problems/Problem 8
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 .
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 .
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
To determine the possible values of we must determine the maximum and minimum possible areas.
In the case where the rhombi are squares, we have implying the minimum possible positive-integer-valued area is
Denote the length and We have by the Pythagorean Theorem, which implies and The first equation yields Plugging into the second, we have The maximal value of occurs when the height of is minimized, which means Plugging back up, we have We have thus our answer is
|2006 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|