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 "2000 AIME I Problems/Problem 4"

(→‎Solution 2 Length-chasing (Angle-chasing but for side lengths))
(→‎Solution 2 Length-chasing (Angle-chasing but for side lengths))
Line 29: Line 29:
 
==Solution 2 Length-chasing (Angle-chasing but for side lengths) ==
 
==Solution 2 Length-chasing (Angle-chasing but for side lengths) ==
  
−
We set the side length of the smallest square to 1, and set the side length of square <math>A4</math> in the previous question to a. We do some "side length chasing" and get <math>4a - 4 = 2a + 5</math>. Solving, we get <math>a = 4.5</math> and the side lengths are <math>61</math> and <math>69</math>. <math>2(61 + 69) = 260</math>
+
We set the side length of the smallest square to 1, and set the side length of square <math>a_4</math> in the previous question to a. We do some "side length chasing" and get <math>4a - 4 = 2a + 5</math>. Solving, we get <math>a = 4.5</math> and the side lengths are <math>61</math> and <math>69</math>. <math>2(61 + 69) = \boxed{260}</math>
  
 
== See also ==
 
== See also ==

Revision as of 18:01, 12 January 2023

Problem

The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.

[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]

Solution 1

Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$, and let $l,w$ represent the dimensions of the rectangle.

The picture shows that \begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ a_6 + a_9 &= a_7 + a_8.\end{align*}

Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$.

We can guess that $a_1 = 2$. (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\boxed{260}$.

Solution 2 Length-chasing (Angle-chasing but for side lengths)

We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some "side length chasing" and get $4a - 4 = 2a + 5$. Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$. $2(61 + 69) = \boxed{260}$

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_4&oldid=186462"