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Difference between revisions of "2021 AMC 10B Problems/Problem 11"

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<math>\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64</math>
 
<math>\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64</math>
  
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==Solution 1:==
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==Solution 1==
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Let the side lengths of this rectangular pan be <math>m</math> and <math>n</math>. It follows that <math>(m-2)(n-2) = \frac{mn}{2}</math>. This gives <math>(m-4)(n-4) = 8</math> after some manipulation, so <math>(m, n) = (5, 12), (6, 8)</math>. By inspection, <math>5 \cdot 12 = \boxed{\textbf{(D) }60}</math> maximizes the number of brownies.  
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Let the side lengths of the rectangular pan be <math>m</math> and <math>n</math>. It follows that <math>(m-2)(n-2) = \frac{mn}{2}</math>, since half of the brownie pieces are in the interior. This gives <math>2(m-2)(n-2) = mn \iff mn - 4m - 4n + 8 = 0</math>. Adding 8 to both sides and applying [[Simon's Favorite Factoring Trick]], we obtain <math>(m-4)(n-4) = 8</math>. Since <math>m</math> and <math>n</math> are both positive, we obtain <math>(m, n) = (5, 12), (6, 8)</math> (up to ordering). By inspection, <math>5 \cdot 12 = \boxed{\textbf{(D) }60}</math> maximizes the number of brownies.
  
 
~ ike.chen
 
~ ike.chen
  
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==Solution 2:==
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==Solution 2==
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Let the dimensions of the rectangular pan be <math>x</math> and <math>y</math>. The number of interior pieces is <math>(x-2)(y-2)</math> because you cannot include the border, and the number of pieces along the perimeter is <math>\frac{xy}{2}</math> (THIS PART IS FLAWED ~ anonymous user).
 
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Setting these two expressions equal, we have <math>\frac{xy}{2}=(x-2)(y-2) \Rightarrow xy=2(xy-2y-2x+4) \Rightarrow xy-4x-4y+8=0</math>
 
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Applying SFFT (Simon's Favorite Factoring Trick), we get <math>(x-4)(y-4)=8</math>. Doing a bit of trial-and-error, we see that <math>xy</math> is maximum when <math>x=5</math> and <math>y=12</math>, which gives us a maximum of <math>60</math> brownies. <math>\Rightarrow \boxed{\textbf{(D) }60}</math>.
 
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~Bryguy
 
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==Solution 3==
 
 
Obviously, no side of the rectangular pan can have less than <math>5</math> brownies beside it. We let one side of the pan have <math>5</math> brownies, and let the number of brownies on its adjacent side be <math>x</math>. Therefore, <math>5x=2\cdot3(x-2)</math>, and solving yields <math>x=12</math> and there are <math>5\cdot12=60</math> brownies in the pan. <math>64</math> is the only choice larger than <math>60</math>, but it cannot be the answer since the only way to fit <math>64</math> brownies in a pan without letting a side of it have less than <math>5</math> brownies beside it is by forming a square of <math>8</math> brownies on each side, which does not meet the requirement. Thus the answer is <math>\boxed{\textbf{(D) }60}</math>.
 
Obviously, no side of the rectangular pan can have less than <math>5</math> brownies beside it. We let one side of the pan have <math>5</math> brownies, and let the number of brownies on its adjacent side be <math>x</math>. Therefore, <math>5x=2\cdot3(x-2)</math>, and solving yields <math>x=12</math> and there are <math>5\cdot12=60</math> brownies in the pan. <math>64</math> is the only choice larger than <math>60</math>, but it cannot be the answer since the only way to fit <math>64</math> brownies in a pan without letting a side of it have less than <math>5</math> brownies beside it is by forming a square of <math>8</math> brownies on each side, which does not meet the requirement. Thus the answer is <math>\boxed{\textbf{(D) }60}</math>.
  
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~ pi_is_3.14
 
~ pi_is_3.14
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==Video Solution by TheBeautyofMath==
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https://youtu.be/L1iW94Ue3eI
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~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/wltQ9m07kzg
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 +
~Interstigation
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==Video Solution by Challenge 25==
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https://youtu.be/Gf5YNjxsaoA?t=531
 +
 +
==Solution 3 (Simon's favorite factoring trick)==
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Let <math>m</math>be the number of brownies on the top horizontal side of the pan and let <math>n</math> be the number of brownies on the leftmost vertical side of the pan. The total number of brownies in the interior is then <math>(m-2)(n-2).</math> Then, the number of brownies on the perimeter would be <math>2m+2n-4.</math> Expanding the left side we get: <math>mn-2m-2n+4.</math> Bringing the right side of the equation to the left side and adding 8 to both the sides we get <math>(n-4)(m-4)=8.</math> Making a table, we get that the highest number possible would be 60 brownies.
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-nambha
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2021|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:02, 11 September 2024

Problem

Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?

$\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64$

Solution 1

Let the side lengths of the rectangular pan be $m$ and $n$. It follows that $(m-2)(n-2) = \frac{mn}{2}$, since half of the brownie pieces are in the interior. This gives $2(m-2)(n-2) = mn \iff mn - 4m - 4n + 8 = 0$. Adding 8 to both sides and applying Simon's Favorite Factoring Trick, we obtain $(m-4)(n-4) = 8$. Since $m$ and $n$ are both positive, we obtain $(m, n) = (5, 12), (6, 8)$ (up to ordering). By inspection, $5 \cdot 12 = \boxed{\textbf{(D) }60}$ maximizes the number of brownies.

~ ike.chen

Solution 2

Obviously, no side of the rectangular pan can have less than $5$ brownies beside it. We let one side of the pan have $5$ brownies, and let the number of brownies on its adjacent side be $x$. Therefore, $5x=2\cdot3(x-2)$, and solving yields $x=12$ and there are $5\cdot12=60$ brownies in the pan. $64$ is the only choice larger than $60$, but it cannot be the answer since the only way to fit $64$ brownies in a pan without letting a side of it have less than $5$ brownies beside it is by forming a square of $8$ brownies on each side, which does not meet the requirement. Thus the answer is $\boxed{\textbf{(D) }60}$.

-SmileKat32

Video Solution by OmegaLearn (Simon's Favorite Factoring Trick)

https://youtu.be/vWlRQiyt0c8

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/L1iW94Ue3eI

~IceMatrix

Video Solution by Interstigation

https://youtu.be/wltQ9m07kzg

~Interstigation

Video Solution by Challenge 25

https://youtu.be/Gf5YNjxsaoA?t=531

Solution 3 (Simon's favorite factoring trick)

Let $m$be the number of brownies on the top horizontal side of the pan and let $n$ be the number of brownies on the leftmost vertical side of the pan. The total number of brownies in the interior is then $(m-2)(n-2).$ Then, the number of brownies on the perimeter would be $2m+2n-4.$ Expanding the left side we get: $mn-2m-2n+4.$ Bringing the right side of the equation to the left side and adding 8 to both the sides we get $(n-4)(m-4)=8.$ Making a table, we get that the highest number possible would be 60 brownies.

-nambha

See Also

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 10 Problems and Solutions

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