2007 AIME II Problems/Problem 8
Problem
A rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if
- (i) all four sides of the rectangle are segments of drawn line segments, and
- (ii) no segments of drawn lines lie inside the rectangle.
Given that the total length of all lines drawn is exactly 2007 units, let be the maximum possible number of basic rectangles determined. Find the remainder when is divided by 1000.
Solution
Solution 1
Denote the number of horizontal lines as , and the number of vertical lines as . The number of basic rectangles is . . Substituting, we find that .
FOIL this to get a quadratic, . Use to find the maximum possible value of the quadratic: . However, this gives a non-integral answer for . The closest two values that work are and .
We see that . The solution is .
Solution 2
We realize that drawing vertical lines and horizontal lines, the number of basic rectangles we have is . The easiest possible case to see is vertical and horizontal lines, as . Now, for every 4 vertical lines you take away, you can add 5 horizontal lines, so you basically have the equation maximize.
Expanded, this gives . From you get that the vertex is at . This is not an integer though, so you see that when , you have and that when x=6, you have . , so the maximum integral value for x occurs when . Now you just evaluate which is .
See also
2007 AIME II (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.