Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 AIME II Problems/Problem 8"

(→‎Solution 2: tex)
m
 
(3 intermediate revisions by 3 users not shown)
Line 10: Line 10:
 
== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
−
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>.  
  
−
We see that <math>252 \cdot 198 = 49896 > 202 \cdot 247 = 49894</math>. The solution is <math>896</math>.
+
We see that <math>252 \cdot 198 = 49896 > 202 \cdot 247 = 49894</math>. The solution is <math>\boxed{896}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
Line 20: Line 20:
  
 
Expanded, this gives <math>-20x^{2}+222x+222^{2}</math>. From <math>-\frac{b}{2a}</math> you get that the vertex is at <math>x=\frac{111}{20}</math>. This is not an integer though, so you see that when <math>x=5</math>, you have <math>-20*25+222*5+222^{2}</math> and that when x=6, you have <math>-20*36+222*6+222^{2}</math>. <math>222 > 20*11</math>, so the maximum integral value for x occurs when <math>x=6</math>. Now you just evaluate <math>-20*36+222*6+222^{2}\mod 1000</math> which is <math>{896}</math>.
 
Expanded, this gives <math>-20x^{2}+222x+222^{2}</math>. From <math>-\frac{b}{2a}</math> you get that the vertex is at <math>x=\frac{111}{20}</math>. This is not an integer though, so you see that when <math>x=5</math>, you have <math>-20*25+222*5+222^{2}</math> and that when x=6, you have <math>-20*36+222*6+222^{2}</math>. <math>222 > 20*11</math>, so the maximum integral value for x occurs when <math>x=6</math>. Now you just evaluate <math>-20*36+222*6+222^{2}\mod 1000</math> which is <math>{896}</math>.
 +
 +
 +
== Video Solution ==
 +
[https://youtu.be/wofmtnOkpro?si=_gcBKGGlB5LKUPCd 2007 AIME II #8]
 +
 +
[https://mathproblemsolvingskills.wordpress.com/ MathProblemSolvingSkills.com]
 +
  
 
== See also ==
 
== See also ==
Line 25: Line 32:
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 14:45, 23 September 2025

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 $N$ be the maximum possible number of basic rectangles determined. Find the remainder when $N$ is divided by 1000.

Solution

Solution 1

Denote the number of horizontal lines drawn as $x$, and the number of vertical lines drawn as $y$. The number of basic rectangles is $(x - 1)(y - 1)$. $5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}$. Substituting, we find that $(x - 1)\left(-\frac 54x + \frac{2003}4\right)$.

FOIL this to get a quadratic, $-\frac 54x^2 + 502x - \frac{2003}4$. Use $\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201$. However, this gives a non-integral answer for $y$. The closest two values that work are $(199,253)$ and $(203,248)$.

We see that $252 \cdot 198 = 49896 > 202 \cdot 247 = 49894$. The solution is $\boxed{896}$.

Solution 2

We realize that drawing $x$ vertical lines and $y$ horizontal lines, the number of basic rectangles we have is $(x-1)(y-1)$. The easiest possible case to see is $223$ vertical and $223$ horizontal lines, as $(4+5)223 = 2007$. Now, for every 4 vertical lines you take away, you can add 5 horizontal lines, so you basically have the equation $(222-4x)(222+5x)$ maximize.

Expanded, this gives $-20x^{2}+222x+222^{2}$. From $-\frac{b}{2a}$ you get that the vertex is at $x=\frac{111}{20}$. This is not an integer though, so you see that when $x=5$, you have $-20*25+222*5+222^{2}$ and that when x=6, you have $-20*36+222*6+222^{2}$. $222 > 20*11$, so the maximum integral value for x occurs when $x=6$. Now you just evaluate $-20*36+222*6+222^{2}\mod 1000$ which is ${896}$.


Video Solution

2007 AIME II #8

MathProblemSolvingSkills.com


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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_8&oldid=261183"