Difference between revisions of "2021 AMC 10B Problems/Problem 11"
Scrabbler94 (talk | contribs) (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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==Solution 1== | ==Solution 1== | ||
− | 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 - | + | 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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~ pi_is_3.14 | ~ 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 |
Latest revision as of 20:23, 10 July 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