2001 AIME II Problems/Problem 9
Problem
Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is 
, where 
and 
are relatively prime positive integers. Find 
.
Solution 1
We can use complementary counting, counting all of the colorings that have at least one red 
square.
- For at least one red
square:
- There are four squares to choose which one will be red. Then there are
ways to color the rest of the squares. 
ways to color the rest of the squares. 
- For at least two squares:
- There are two cases: those with two red squares on one side and those without red squares on one side.
- The first case is easy: 4 ways to choose which the side the squares will be on, and
ways to color the rest of the squares, so 32 ways to do that. For the second case, there will by only two ways to pick two squares, and 
ways to color the other squares. 
ways to color the rest of the squares, so 32 ways to do that. For the second case, there will by only two ways to pick two squares, and 
ways to color the other squares. 
- For at least three squares:
- Choosing three such squares leaves only one square left, with four places to place it. This is
ways.
ways.- For at least four squares, we clearly only have one way.
By the Principle of Inclusion-Exclusion, there are (alternatively subtracting and adding) 
ways to have at least one red 
square.
There are 
ways to paint the 
square with no restrictions, so there are 
ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a red square is 
, and 
.
Solution 2
We consider how many ways we can have 2*2 grid
: All the girds are red--
case
: One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid 
cases
: Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, 
cases.
Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. 
cases
Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are 
Four unit squares are blue, no overcomunted case will be considered. there are 
Five unit squares are blue, 
cases in all
Sum up those cases, there are 
cases that a 2*2 grid can be formed.
In all, there are possible ways to paint the big square, so the answer is 
leads to 
~bluesoul
Solution 3 (Case Work)
![\[\begin{array}{|c|c|c|} \hline C_{11} & C_{12} & C_{13}\\ \hline C_{21} & C_{22} & C_{23}\\ \hline C_{31} & C_{32} & C_{33}\\ \hline \end{array}\]](http://latex.artofproblemsolving.com/7/1/d/71d4337d26b978be0aa7017a136a716f0c2f9fbc.png)
Case 1: If the 3-by-3 unit-square grid has exactly 2-by-2 red square
Assume the 2-by-2 red square is at 
. To make sure there are no more 2-by-2 red squares, 
can't both be red and 
can't both be red. Meaning that there are 
coloring methods for and . 
can be colored with either colors. However, the coloring method where 
are all red needs to be removed. For exactly one red square at , there are 
coloring methods. As there are locations for the 2-by-2 red square on the 3-by-3 unit-square grid, if the 3-by-3 unit-square grid has exactly 2-by-2 red square, there are 
coloring methods.
See also
| 2001 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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