Difference between revisions of "2021 AMC 10B Problems/Problem 11"
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| + | ==Video Solution by Challenge 25== | ||
| + | https://youtu.be/Gf5YNjxsaoA?t=531 | ||
==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}} | ||
Revision as of 20:19, 6 November 2022
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
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. 
