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Difference between revisions of "2022 AMC 12A Problems/Problem 3"
m (→‎Video Solution (CREATIVE THINKING))
 
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=Solution 1 (List)=
+
==Problem==
−
Let's label some points.
 
  
 +
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,mediumgray);
+
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));
−
 
−
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>
 
</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)==
  
−
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.  
+
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>.
  
−
Rule: <math>AB + BC = CD + DE = EF + FG = GH + AH</math>
+
~mathboy100
  
−
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.
+
==Solution 2 (Perimeter of Square)==
  
−
<math>AB\times AH</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>.
  
−
<math>CD\times BC</math>
+
~iluvme
  
−
<math>EF\times DE</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>GH\times FG</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 7</math>
+
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>
  
−
<math>5\times 6</math>
+
~ghfhgvghj10 & Education, the study of everything.
−
 
 
−
<math>2\times 3</math>
 
−
 
 
−
<math>1\times 6</math>
 
  
−
This also tells us that the diagram is not drawn to scale, lol.  
+
==Video Solution (Creativity)==
 +
https://youtu.be/YNJ_0dq7gU4
  
−
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>\textbf{(B) }B.</math>
+
~Education, the Study of Everything
  
−
~ghfhgvghj10 & Education, the study of everything.
+
==Video Solution (Smart and Simple)==
 +
https://youtu.be/7yAh4MtJ8a8?si=mFXNAtcL16AYj139&t=445
  
 +
~Math-X
  
−
 
+
==See Also==
−
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 (<math>2+7+5+6+2+3+1+6+2+4</math>) 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 <math>2\times4</math> works (<math>38-2-4=32=8\cdot4</math>), which corresponds with B so the answer is, <math>\boxed{\textbf{(B) }B}</math>.
+
{{AMC12 box|year=2022|ab=A|num-b=2|num-a=4}}
−
 
+
{{MAA Notice}}
−
~iluvme
 

Latest revision as of 18:26, 22 September 2024

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 (Creativity)

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

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. AMC logo.png

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