# 2022 AMC 12A Problems/Problem 3

## 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? $[asy] size(150); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); [/asy]$ $\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E$

## Solution 1 (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: $[asy] size(175); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("7",midpoint((0,7)--(5.3,7)),N,blue); label("x",midpoint((5.3,7)--(7,7)),N,blue); label("2",midpoint((0,4.5)--(0,7)),W,blue); label("x+5",midpoint((0,0)--(0,4.5)),W,blue); label("D",midpoint((0,7)--(5.3,4.5)),red); [/asy]$ It is clear 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. $[asy] size(175); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("7",midpoint((0,7)--(5.3,7)),N,blue); label("1",midpoint((5.3,7)--(7,7)),N,blue); label("2",midpoint((0,4.5)--(0,7)),W,blue); label("6",midpoint((0,0)--(0,4.5)),W,blue); label("6",midpoint((7,7)--(7,2.5)),E,blue); label("2",midpoint((7,2.5)--(7,0)),E,blue); label("2",midpoint((5.3,4.5)--(5.3,7)),E,blue); label("4",midpoint((5.3,4.5)--(5.3,2.5)),E,blue); label("1",midpoint((5.3,2.5)--(7,2.5)),S,blue); label("5",midpoint((0,4.5)--(3,4.5)),N,blue); label("2",midpoint((3,4.5)--(5.3,4.5)),N,blue); label("5",midpoint((0,0)--(3,0)),S,blue); label("3",midpoint((3,0)--(7,0)),S,blue); label("2",midpoint((3,0)--(3,2.5)),W,blue); label("4",midpoint((3,2.5)--(3,4.5)),W,blue); label("2",midpoint((3,2.5)--(5.3,2.5)),S,blue); label("D",midpoint((0,7)--(5.3,4.5)),red); label("A",midpoint((5.3,7)--(7,2.5)),red); label("C",midpoint((0,4.5)--(3,0)),red); label("E",midpoint((3,2.5)--(7,0)),red); label("B",midpoint((3,4.5)--(5.3,2.5)),red); [/asy]$ ~MRENTHUSIASM

## Solution 4 (Observations)

Let's label some points: $[asy] size(175); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("A",(0,0),SW); label("B",(3,0),S); label("C",(7,0),SE); label("D",(7,2.5),(1,0)); label("E",(7,7),NE); label("F",(5.5,7),N); label("G",(0,7),NW); label("H",(0,4.5),W); [/asy]$ By finding the dimensions of the middle rectangle, we need to find the dimensions of the other four rectangles. We have a rule: $$AB + BC = CD + DE = EF + FG = GH + AH.$$ Let's make a list of all dimensions of the rectangles from the diagram. We have to fill in the dimensions from up above, but still apply to the rule: \begin{align*} AB&\times AH \\ CD&\times BC \\ EF&\times DE \\ GH&\times FG \end{align*} 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 \begin{align*} 2&\times 7 \\ 5&\times 6 \\ 2&\times 3 \\ 1&\times 6 \\ \end{align*} (This also tells us that the diagram is not drawn to scale.)

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

~ghfhgvghj10 & Education, the study of everything.

## Video Solution (CREATIVE THINKING)

~Education, the Study of Everything

## Video Solution (Smart and Simple)

~Math-X

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