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> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | Let the | + | 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 | |
| − | + | ==Solution 2== | |
| + | 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>. | ||
| − | Solution by | + | -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 <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. | ||
| + | |||
| + | -nambha | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2021|ab=B|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:02, 11 September 2024
Contents
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?
Solution 1
Let the side lengths of the rectangular pan be 
and 
. It follows that 
, since half of the brownie pieces are in the interior. This gives 
. Adding 8 to both sides and applying Simon's Favorite Factoring Trick, we obtain 
. Since and are both positive, we obtain 
(up to ordering). By inspection, 
maximizes the number of brownies.
~ ike.chen
Solution 2
Obviously, no side of the rectangular pan can have less than 
brownies beside it. We let one side of the pan have brownies, and let the number of brownies on its adjacent side be 
. Therefore, 
, and solving yields 
and there are 
brownies in the pan. 
is the only choice larger than 
, but it cannot be the answer since the only way to fit brownies in a pan without letting a side of it have less than brownies beside it is by forming a square of 
brownies on each side, which does not meet the requirement. Thus the answer is 
.
-SmileKat32
Video Solution by OmegaLearn (Simon's Favorite Factoring Trick)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by Interstigation
~Interstigation
Video Solution by Challenge 25
https://youtu.be/Gf5YNjxsaoA?t=531
Solution 3 (Simon's favorite factoring trick)
Let be the number of brownies on the top horizontal side of the pan and let be the number of brownies on the leftmost vertical side of the pan. The total number of brownies in the interior is then 
Then, the number of brownies on the perimeter would be 
Expanding the left side we get: 
Bringing the right side of the equation to the left side and adding 8 to both the sides we get 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
