Difference between revisions of "2001 AMC 10 Problems/Problem 11"
m (small fixes) |
m (minor spacing fix) |
||
Line 2: | Line 2: | ||
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 <math> 8 </math> unit squares. The second ring contains <math>16</math> unit squares. If we continue this process, the number of unit squares in the <math>100^\text{th}</math> ring is | 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 <math> 8 </math> unit squares. The second ring contains <math>16</math> unit squares. If we continue this process, the number of unit squares in the <math>100^\text{th}</math> ring is | ||
− | |||
<math>\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404</math> | <math>\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404</math> | ||
Revision as of 21:20, 16 March 2020
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.
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.