Difference between revisions of "2022 AMC 12A Problems/Problem 3"
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==Solution 3 (Observations)== | ==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: | 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> | ||
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. | 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 | ~MRENTHUSIASM | ||
Line 60: | Line 100: | ||
label("$H$",(0,4.5),W); | label("$H$",(0,4.5),W); | ||
</asy> | </asy> | ||
− | By finding the dimensions of the middle rectangle, we need to find the dimensions of the other | + | 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 \\ | |
− | Let's make a list of all | + | 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== | ==See Also== | ||
{{AMC12 box|year=2022|ab=A|num-b=2|num-a=4}} | {{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.