Difference between revisions of "2000 AIME I Problems/Problem 4"

Latest revision as of 23:31, 18 January 2024

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 1.2 (more detail)

We can just list the equations: $\begin{align*} s_3 &= s_1 + s_2 \\ s_4 &= s_3 + s_1 \\ s_5 &= s_4 + s_3 \\ s_6 &= s_5 + s_4 \\ s_7 &= s_5 + s_3 + s_2 \\ s_8 &= s_7 + s_2 \\ s_9 &= s_8 + s_2 - s_1 \\ s_9 + s_8 &= s_7 + s_6 + s_5 \end{align*}$ We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows $\begin{align*} s_4 &= 2s_1 + s_2 \\ s_5 &= 3s_1 +2s_2 \\ s_6 &= 5s_1 + 3s_2 \\ s_7 &= 4s_1 + 4s_2 \\ s_8 &= 4s_1 + 5s_2 \\ s_9 &= 3s_1 + 6s_2 \\ \end{align*}$ Since $s_9 + s_8 = s_7 + s_6 + s_5 \implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),$ $2s_2 = 5s_1 \implies \frac{2}{5}s_2 = s_1.$ Since the side lengths of the rectangle are relatively prime, we can see that $s_1 = 2$ and $s_2 = 5.$ Therefore, $2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = \boxed{260}.$ ~peelybonehead

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 $69$ . Thus, the perimeter of the rectangle is $2(61 + 69) = \boxed{260}.$

@@ Line 2: / Line 2: @@
 The diagram shows a [[rectangle]] that has been dissected into nine non-overlapping [[square]]s. Given that the width and the height of the rectangle are relatively prime positive integers, find the [[perimeter]] of the rectangle.
-<center><asy>draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36));
+<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></center>
+draw((34,45)--(41,45));</asy>
-== Solution ==
+== Solution 1 ==
 Call the squares' side lengths from smallest to largest <math>a_1,\ldots,a_9</math>, and let <math>l,w</math> represent the dimensions of the rectangle.
-The picture shows that:
+The picture shows that
-<cmath>\begin{eqnarray*}
+<cmath>\begin{align*}
-a_1+a_2 &=& a_3\\
+a_1+a_2 &= a_3\\
-a_1 + a_3 &=& a_4\\
+a_1 + a_3 &= a_4\\
-a_3 + a_4 &=& a_5\\
+a_3 + a_4 &= a_5\\
-a_4 + a_5 &=& a_6\\
+a_4 + a_5 &= a_6\\
-a_2 + a_3 + a_5 &=& a_7\\
+a_2 + a_3 + a_5 &= a_7\\
-a_2 + a_7 &=& a_8\\
+a_2 + a_7 &= a_8\\
-a_1 + a_4 + a_6 &=& a_9\\
+a_1 + a_4 + a_6 &= a_9\\
-a_6 + a_9 &=& a_7 + a_8\end{eqnarray*}</cmath>
+a_6 + a_9 &= a_7 + a_8.\end{align*}</cmath>
-With a bit of trial and error and some arithmetic, we can use the last equation to find that <math>5a_1 = 2a_2</math>; [[without loss of generality]], let <math>a_1 = 2</math>. Then solving gives <math>a_9 = 36</math>, <math>a_6=25</math>, <math>a_8 = 33</math>, which gives us <math>l=61,w=69</math> (relatively prime), and the perimeter is <math>2(61)+2(69)=\boxed{260}</math>.
+Expressing all terms 3 to 9 in terms of <math>a_1</math> and <math>a_2</math> and substituting their expanded forms into the previous equation will give the expression <math>5a_1 = 2a_2</math>.
+We can guess that <math>a_1 = 2</math>. (If we started with <math>a_1</math> odd, the resulting sides would not be integers and we would need to scale up by a factor of <math>2</math> to make them integers; if we started with <math>a_1 > 2</math> even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives <math>a_9 = 36</math>, <math>a_6=25</math>, <math>a_8 = 33</math>, which gives us <math>l=61,w=69</math>.  These numbers are relatively prime, as desired. The perimeter is <math>2(61)+2(69)=\boxed{260}</math>.
+== Solution 1.2 (more detail) ==
+We can just list the equations:
+<cmath>\begin{align*}
+s_3 &= s_1 + s_2 \\
+s_4 &= s_3 + s_1 \\
+s_5 &= s_4 + s_3 \\
+s_6 &= s_5 + s_4 \\
+s_7 &= s_5 + s_3 + s_2 \\
+s_8 &= s_7 + s_2 \\
+s_9 &= s_8 + s_2 - s_1 \\
+s_9 + s_8 &= s_7 + s_6 + s_5 \end{align*}</cmath>We can then write each <math>s_i</math> in terms of <math>s_1</math> and <math>s_2</math> as follows
+<cmath>\begin{align*}
+s_4 &= 2s_1 + s_2 \\
+s_5 &= 3s_1 +2s_2 \\
+s_6 &= 5s_1 + 3s_2 \\
+s_7 &= 4s_1 + 4s_2 \\
+s_8 &= 4s_1 + 5s_2 \\
+s_9 &= 3s_1 + 6s_2 \\
+\end{align*}</cmath>
+Since <math>s_9 + s_8 = s_7 + s_6 + s_5 \implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),</math> <cmath>2s_2 = 5s_1 \implies \frac{2}{5}s_2 = s_1.</cmath>Since the side lengths of the rectangle are relatively prime, we can see that <math>s_1 = 2</math> and <math>s_2 = 5.</math> Therefore, <math>2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = \boxed{260}.</math>
+~peelybonehead
+==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>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>. Thus, the perimeter of the rectangle is <math>2(61 + 69) = \boxed{260}.</math>
 == See also ==
@@ Line 29: / Line 57: @@
 [[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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