Difference between revisions of "2022 AMC 12A Problems/Problem 3"
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− | = | + | ==Problem== |
− | |||
+ | Five rectangles, <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, and <math>E</math>, are arranged in a square as shown below. These rectangles have dimensions <math>1\times6</math>, <math>2\times4</math>, <math>5\times6</math>, <math>2\times7</math>, and <math>2\times3</math>, respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle? | ||
<asy> | <asy> | ||
− | fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle, | + | 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((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); | ||
draw((3,0)--(3,4.5)); | draw((3,0)--(3,4.5)); | ||
Line 9: | Line 11: | ||
draw((5.3,7)--(5.3,2.5)); | draw((5.3,7)--(5.3,2.5)); | ||
draw((7,2.5)--(3,2.5)); | draw((7,2.5)--(3,2.5)); | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
</asy> | </asy> | ||
+ | <math>\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E</math> | ||
+ | ==Solution 1 (Area and Perimeter of Square)== | ||
− | + | 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)== | |
− | <math> | + | 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 | |
− | <math> | + | ==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: | ||
+ | <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 <math>x=1,</math> so we can determine Rectangle <math>A.</math> | ||
− | <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. |
+ | <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: <cmath>AB + BC = CD + DE = EF + FG = GH + AH.</cmath> | ||
+ | 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: | ||
+ | <cmath>\begin{align*} | ||
+ | AB&\times AH \\ | ||
+ | CD&\times BC \\ | ||
+ | EF&\times DE \\ | ||
+ | GH&\times FG | ||
+ | \end{align*}</cmath> | ||
By applying the rule, we get <math>AB=2, BC=6, CD=5, DE=3, EF=2, FG=6, GH=1</math>, and <math>AH=7</math>. | By applying the rule, we get <math>AB=2, BC=6, CD=5, DE=3, EF=2, FG=6, GH=1</math>, and <math>AH=7</math>. | ||
By substitution, we get this list | By substitution, we get this list | ||
+ | <cmath>\begin{align*} | ||
+ | 2&\times 7 \\ | ||
+ | 5&\times 6 \\ | ||
+ | 2&\times 3 \\ | ||
+ | 1&\times 6 \\ | ||
+ | \end{align*}</cmath> | ||
+ | (This also tells us that the diagram is not drawn to scale.) | ||
− | <math>2\times | + | Notice how the only dimension not used in the list was <math>2\times 4</math> and that corresponds with B so the answer is, <math>\boxed{\textbf{(B) }B}.</math> |
− | + | ~ghfhgvghj10 & Education, the study of everything. | |
− | + | ==Video Solution (CREATIVE THINKING)== | |
+ | https://youtu.be/YNJ_0dq7gU4 | ||
− | + | ~Education, the Study of Everything | |
− | + | ==Video Solution (Smart and Simple)== | |
+ | https://youtu.be/7yAh4MtJ8a8?si=mFXNAtcL16AYj139&t=445 | ||
− | + | ~Math-X | |
− | + | ==See Also== | |
+ | {{AMC12 box|year=2022|ab=A|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:40, 25 October 2023
Contents
Problem
Five rectangles, ,
,
,
, and
, are arranged in a square as shown below. These rectangles have dimensions
,
,
,
, and
, respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle?
Solution 1 (Area and Perimeter of Square)
The area of this square is equal to , and thus its side lengths are
. The sum of the dimensions of the rectangles are
. Thus, because the perimeter of the rectangle is
, the rectangle on the inside must have a perimeter of
. The only rectangle that works is
.
~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 . We know that as the square's side length is an integer, the perimeter must be divisible by
. Testing out by subtracting all five pairs of dimensions from
, only
works since
, which corresponds with
.
~iluvme
Solution 3 (Observations)
Note that rectangle must be on the edge. Without loss of generality, let the top-left rectangle be
as shown below:
It is clear that
so we can determine Rectangle
Continuing with a similar process, we can determine Rectangles and
in this order. The answer is
as shown below.
~MRENTHUSIASM
Solution 4 (Observations)
Let's label some points:
By finding the dimensions of the middle rectangle, we need to find the dimensions of the other four rectangles. We have a rule:
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:
By applying the rule, we get
, and
.
By substitution, we get this list
(This also tells us that the diagram is not drawn to scale.)
Notice how the only dimension not used in the list was and that corresponds with B so the answer is,
~ghfhgvghj10 & Education, the study of everything.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution (Smart and Simple)
https://youtu.be/7yAh4MtJ8a8?si=mFXNAtcL16AYj139&t=445
~Math-X
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.