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Difference between revisions of "2022 AMC 12A Problems/Problem 3"
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<math>\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E</math>
 
<math>\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E</math>
  
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==Solution 1 (Perimeter of Square)==
+
==Solution 2 (Area and Perimeter of Square)==
  
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Note that the perimeter of the square is the sum of 4 pairs of dimensions (8 values are added). Adding up all of the dimensions (<math>2+7+5+6+2+3+1+6+2+4</math>) gives us 38. We know that as the square's side length is an integer, the perimeter must be divisible by 4. Testing out by subtracting all 5 pairs of dimensions from 38, only <math>2\times4</math> works (<math>38-2-4=32=8\cdot4</math>), which corresponds with <math>\boxed{\textbf{(B) }B}</math>.
+
The area of this square is equal to <math>6 + 8 + 30 + 14 + 6 = 64</math>, and thus its side lengths are <math>8</math>. The sum of the dimensions of the rectangles are <math>2 + 7 + 5 + 6 + 2 + 3 + 1 + 6 + 2 + 4 = 38</math>. Thus, because the perimeter of the rectangle is <math>32</math>, the rectangle on the inside must have a perimeter of <math>6 \cdot 2 = 12</math>. The only rectangle that works is <math>\boxed{\textbf{(B) }B}</math>.
 +
 
 +
~mathboy100
 +
 
 +
==Solution 2 (Perimeter of Square)==
 +
 
 +
Note that the perimeter of the square is the sum of four pairs of dimensions (eight values are added). Adding up all dimensions gives us <math>2+7+5+6+2+3+1+6+2+4=38</math>. We know that as the square's side length is an integer, the perimeter must be divisible by <math>4</math>. Testing out by subtracting all five pairs of dimensions from <math>38</math>, only <math>2\times4</math> works since <math>38-2-4=32=8\cdot4</math>, which corresponds with <math>\boxed{\textbf{(B) }B}</math>.
  
 
~iluvme
 
~iluvme
  
−
==Solution 2 (Area, Perimeter of Square)==
+
==Solution 3 (Observations)==
 +
Note that rectangle <math>D</math> must be on the edge. Without loss of generality, let the top-left rectangle be <math>D,</math> as shown below:
  
−
The area of this square is equal to <math>6 + 8 + 30 + 14 + 6 = 64</math>, and thus its side lengths are <math>8</math>. The sum of the dimensions of the rectangles are <math>2 + 7 + 5 + 6 + 2 + 3 + 1 + 6 + 2 + 4 = 38</math>. Thus, because the perimeter of the rectangle is <math>32</math>, the rectangle on the inside must have a perimeter of <math>6 \cdot 2 = 12</math>. The only rectangle that works is <math>\boxed{\textbf{(B) }B}</math>.
+
<b>DIAGRAM READY SOON</b>
 +
 
 +
We conclude that <math>x=1,</math> so we can determine Rectangle <math>A.</math>
 +
 
 +
Continuing with a similar process, we can determine Rectangles <math>C,E,</math> and <math>B,</math> in this order. The answer is <math>\boxed{\textbf{(B) }B}</math> as shown below.
  
−
~mathboy100
+
<b>DIAGRAM READY SOON</b>
  
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==Solution 3 (Observations)==
+
~MRENTHUSIASM
  
 
==Solution 4 (Observations)==
 
==Solution 4 (Observations)==

Revision as of 09:22, 14 December 2022

Problem

Five rectangles, $A$, $B$, $C$, $D$, and $E$, are arranged in a square as shown below. These rectangles have dimensions $1\times6$, $2\times4$, $5\times6$, $2\times7$, and $2\times3$, respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle? [asy] fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,mediumgray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); [/asy] $\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E$

Solution 2 (Area and Perimeter of Square)

The area of this square is equal to $6 + 8 + 30 + 14 + 6 = 64$, and thus its side lengths are $8$. The sum of the dimensions of the rectangles are $2 + 7 + 5 + 6 + 2 + 3 + 1 + 6 + 2 + 4 = 38$. Thus, because the perimeter of the rectangle is $32$, the rectangle on the inside must have a perimeter of $6 \cdot 2 = 12$. The only rectangle that works is $\boxed{\textbf{(B) }B}$.

~mathboy100

Solution 2 (Perimeter of Square)

Note that the perimeter of the square is the sum of four pairs of dimensions (eight values are added). Adding up all dimensions gives us $2+7+5+6+2+3+1+6+2+4=38$. We know that as the square's side length is an integer, the perimeter must be divisible by $4$. Testing out by subtracting all five pairs of dimensions from $38$, only $2\times4$ works since $38-2-4=32=8\cdot4$, which corresponds with $\boxed{\textbf{(B) }B}$.

~iluvme

Solution 3 (Observations)

Note that rectangle $D$ must be on the edge. Without loss of generality, let the top-left rectangle be $D,$ as shown below:

DIAGRAM READY SOON

We conclude that $x=1,$ so we can determine Rectangle $A.$

Continuing with a similar process, we can determine Rectangles $C,E,$ and $B,$ in this order. The answer is $\boxed{\textbf{(B) }B}$ as shown below.

DIAGRAM READY SOON

~MRENTHUSIASM

Solution 4 (Observations)

Let's label some points. [asy] fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,mediumgray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("A",(0,0),S); label("B",(3,0),S); label("C",(7,0),S); label("D",(7.5,3),S); label("E",(7.5,7.8),S); label("F",(5.5,7.8),S); label("G",(-.5,7.8),S); label("H",(-.5,5),S); [/asy] By finding the dimensions of the middle rectangle, we need to find the dimensions of the other 4 rectangles. By doing this, there is going to be a rule.

Rule: $AB + BC = CD + DE = EF + FG = GH + AH$

Let's make a list of all the dimensions of the rectangles from the diagram. We have to fill in the dimensions from up above, but still apply to the rule.

$AB\times AH$

$CD\times BC$

$EF\times DE$

$GH\times FG$

By applying the rule, we get $AB=2, BC=6, CD=5, DE=3, EF=2, FG=6, GH=1$, and $AH=7$.

By substitution, we get this list

$2\times 7$

$5\times 6$

$2\times 3$

$1\times 6$

This also tells us that the diagram is not drawn to scale, lol.

Notice how the only dimension not used in the list was $2\times 4$ and that corresponds with B so the answer is, $\textbf{(B) }B.$

~ghfhgvghj10 & Education, the study of everything.

See Also

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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