Difference between revisions of "2022 AMC 12A Problems/Problem 3"

Revision as of 09:36, 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? $[asy] size(150); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); [/asy]$ $\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E$

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:

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: $[asy] size(175); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("$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 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.

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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