Difference between revisions of "2001 AMC 10 Problems/Problem 11"
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Thus, the <math> 100^\text{th} </math> ring has <math> 8 \times 100 = \boxed{\textbf{(C)}\ 800} </math> unit squares. | Thus, the <math> 100^\text{th} </math> ring has <math> 8 \times 100 = \boxed{\textbf{(C)}\ 800} </math> unit squares. | ||
+ | |||
+ | ===Solution 3 (Less Rigorous)=== | ||
+ | |||
+ | Notice that the first ring around the center square contains <math> 8 </math> unit squares, the second ring contains <math> 16 </math> unit squares, the third contains <math> 24 </math> unit squares, and so on. The number of squares in the <math> n^\text{th} </math> ring is determined by the expression <math> 8 \times n </math>. Thus, the number of unit squares in the <math>100^\text{th}</math> ring is equal to <math> 8 \times 100 </math>, which equals <math> \boxed{\textbf{(C) }800} </math> unit squares. | ||
+ | |||
+ | -Darth_Cadet | ||
+ | |||
+ | ==Video Solution by Daily Dose of Math== | ||
+ | |||
+ | https://youtu.be/y52knpoCVYo?si=dYATo3Zxoj4obeMV | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
== See Also == | == See Also == |
Latest revision as of 21:41, 15 July 2024
Contents
Problem
Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains unit squares. The second ring contains
unit squares. If we continue this process, the number of unit squares in the
ring is
Solution
Solution 1
We can partition the ring into
rectangles: two containing
unit squares and two containing
unit squares.
There are unit squares in the
ring.
Thus, the ring has
unit squares.
Solution 2
We can make the ring by removing a square of side length
from a square of side length
.
This ring contains unit squares.
Thus, the ring has
unit squares.
Solution 3 (Less Rigorous)
Notice that the first ring around the center square contains unit squares, the second ring contains
unit squares, the third contains
unit squares, and so on. The number of squares in the
ring is determined by the expression
. Thus, the number of unit squares in the
ring is equal to
, which equals
unit squares.
-Darth_Cadet
Video Solution by Daily Dose of Math
https://youtu.be/y52knpoCVYo?si=dYATo3Zxoj4obeMV
~Thesmartgreekmathdude
See Also
2001 AMC 10 (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.