Difference between revisions of "2007 AIME II Problems/Problem 8"
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− | Denote the number of horizontal lines as <math>x</math>, and the number of vertical lines as <math>y</math>. The number of basic rectangles is <math>(x - 1)(y - 1)</math>. <math>5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}</math>. Substituting, we find that <math>(x - 1)\left(-\frac 54x + \frac{2003}4\right)</math>. | + | Denote the number of horizontal lines drawn as <math>x</math>, and the number of vertical lines drawn as <math>y</math>. The number of basic rectangles is <math>(x - 1)(y - 1)</math>. <math>5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}</math>. Substituting, we find that <math>(x - 1)\left(-\frac 54x + \frac{2003}4\right)</math>. |
[[FOIL]] this to get a quadratic, <math>-\frac 54x^2 + 502x - \frac{2003}4</math>. Use <math>\frac{-b}{2a}</math> to find the maximum possible value of the quadratic: <math>x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201</math>. However, this gives a non-integral answer for <math>y</math>. The closest two values that work are <math>(199,253)</math> and <math>(203,248)</math>. | [[FOIL]] this to get a quadratic, <math>-\frac 54x^2 + 502x - \frac{2003}4</math>. Use <math>\frac{-b}{2a}</math> to find the maximum possible value of the quadratic: <math>x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201</math>. However, this gives a non-integral answer for <math>y</math>. The closest two values that work are <math>(199,253)</math> and <math>(203,248)</math>. |
Latest revision as of 21:17, 24 February 2021
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 drawn as , and the number of vertical lines drawn 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.