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

(Solution 2:)
(solution 2 is exactly the same as solution 1 except x,y are used instead of m,n. Added more detail to solution 1.)
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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>
  
==Solution 1:==
+
==Solution 1==
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.  
+
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 - 2m - 2n - 4 = 0</math>. Adding 8 to both sides and applying [[Simon's Favorite Factoring Trick]], we obtain <math>(m-2)(n-2) = 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
  
==Solution 2:==
+
==Solution 2==
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> <math>\textbf{(THIS PART IS FLAWED - ANONYMOUS USER)}</math>.
 
 
 
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>
 
 
 
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>.
 
 
 
~Bryguy
 
 
 
==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>.
  

Revision as of 23:13, 13 February 2021

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 $2(m-2)(n-2) = mn \iff mn - 2m - 2n - 4 = 0$. Adding 8 to both sides and applying Simon's Favorite Factoring Trick, we obtain $(m-2)(n-2) = 8$. 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

See Also

2021 AMC 10B (ProblemsAnswer KeyResources)
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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