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2022 AMC 12A Problems/Problem 3

Revision as of 13:09, 26 November 2022 by Mathboy100 (talk | contribs) (→‎Solution 3 (Area, Perimiter of Square))

Problem 3

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 1 (List)

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.

Solution 2 (Perimeter of Square)

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 ($2+7+5+6+2+3+1+6+2+4$) 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 $2\times4$ works ($38-2-4=32=8\cdot4$), which corresponds with $\boxed{\textbf{(B) }B}$.

~iluvme

Solution 3 (Area, 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 $6$. The only rectangle that works is $\boxed{\textbf{(B) }B}$.

~mathboy100

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