Difference between revisions of "2012 AMC 10A Problems/Problem 20"
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| − | First, there is only one way for the middle square to be black because it is not affected by the rotation. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is <math>1</math> case with all black squares. There are four cases with one white square and all <math>4</math> work. There are six cases with two white squares, but only the <math>2</math> with the white squares | + | First, there is only one way for the middle square to be black because it is not affected by the rotation. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is <math>1</math> case with all black squares. There are four cases with one white square and all <math>4</math> work. There are six cases with two white squares, but only the <math>2</math> with the white squares diagonal from each other work. There are no cases with three white squares or four white squares. Then the total number of ways to color the corners is <math>1+4+2=7</math>. In essence, the edges work the same way, so there are also <math>7</math> ways to color them. The number of ways to fit the conditions over the number of ways to color the squares is |
<cmath>\frac{7\times7}{2^9}=\boxed{\textbf{(A)}\ \frac{49}{512}}</cmath> | <cmath>\frac{7\times7}{2^9}=\boxed{\textbf{(A)}\ \frac{49}{512}}</cmath> | ||
Revision as of 10:53, 4 January 2013
- The following problem is from both the 2012 AMC 12A #15 and 2012 AMC 10A #20, so both problems redirect to this page.
Problem
A 
x square is partitioned into 
unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated 
clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?
Solution
First, there is only one way for the middle square to be black because it is not affected by the rotation. Then we can consider the corners and edges separately. Let's first just consider the number of ways we can color the corners. There is 
case with all black squares. There are four cases with one white square and all 
work. There are six cases with two white squares, but only the 
with the white squares diagonal from each other work. There are no cases with three white squares or four white squares. Then the total number of ways to color the corners is 
. In essence, the edges work the same way, so there are also 
ways to color them. The number of ways to fit the conditions over the number of ways to color the squares is
![\[\frac{7\times7}{2^9}=\boxed{\textbf{(A)}\ \frac{49}{512}}\]](http://latex.artofproblemsolving.com/d/1/2/d12be61744c1adb4e336be847dcbf4e9a9ea7d06.png)
See Also
| 2012 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| 2012 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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 12 Problems and Solutions | |
