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Difference between revisions of "2015 AIME II Problems/Problem 5"

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Each corner square is adjacent to <math>2</math> other squares, edge squares to <math>3</math> other squares, and center squares to <math>4</math> other squares. In the problem, the second square can be any square that is not the first, which means there are <math>n^2-1</math> to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is <math>\frac{2}{n^2-1}(\frac{4}{n^2}) +\frac{3}{n^2-1}(\frac{4(n-2)}{n^2}) +\frac{4}{n^2-1}(\frac{n^2-4n+4)}{n^2})</math>.
 
Each corner square is adjacent to <math>2</math> other squares, edge squares to <math>3</math> other squares, and center squares to <math>4</math> other squares. In the problem, the second square can be any square that is not the first, which means there are <math>n^2-1</math> to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is <math>\frac{2}{n^2-1}(\frac{4}{n^2}) +\frac{3}{n^2-1}(\frac{4(n-2)}{n^2}) +\frac{4}{n^2-1}(\frac{n^2-4n+4)}{n^2})</math>.
  
−
Simplifying, we get <math>\frac{4}{n(n+1)}</math> which we can set to be less than <math>\frac{1}{2015}</math>. By introspection, we find that the first such <math>n</math> is <math>\boxed{090}</math>.
+
Simplifying, we get <math>\frac{4}{n(n+1)}</math> which we can set to be less than <math>\frac{1}{2015}</math>. By inspection, we find that the first such <math>n</math> is <math>\boxed{090}</math>.
  
 
==See also==
 
==See also==
 
{{AIME box|year=2015|n=II|num-b=4|num-a=6}}
 
{{AIME box|year=2015|n=II|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:43, 4 February 2016

Problem

Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$.

Solution 1

Call the given grid "Grid A". Consider Grid B, where the vertices of Grid B fall in the centers of the squares of Grid A; thus, Grid B has dimensions $(n-1) \times (n-1)$. There is a one-to-one correspondence between the edges of Grid B and the number of adjacent pairs of unit squares in Grid A. The number of edges in Grid B is $2n(n-1)$, and the number of ways to pick two squares out of Grid A is $\dbinom{n^2}{2}$. So, the probability that the two chosen squares are adjacent is $\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1)}{\frac{n^2(n^2-1)}{2}} = \frac{4}{n(n+1)}$. We wish to find the smallest positive integer $n$ such that $\frac{4}{n(n+1)} < \frac{1}{2015}$, and by inspection the first such $n$ is $\boxed{090}$.


Solution 2

Consider a $3x3$ grid, where there are $4$ corner squares, $4$ edge squares, and $1$ center square. A 4x4 grid has $4$ corner squares, $8$ edge squares, and $4$ center squares. By examining simple cases, we can conclude that for a grid that is $n \times n$, there are always $4$ corners squares, $4(n-2)$ edge squares, and $n^2-4n+4$ center squares.

Each corner square is adjacent to $2$ other squares, edge squares to $3$ other squares, and center squares to $4$ other squares. In the problem, the second square can be any square that is not the first, which means there are $n^2-1$ to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is $\frac{2}{n^2-1}(\frac{4}{n^2}) +\frac{3}{n^2-1}(\frac{4(n-2)}{n^2}) +\frac{4}{n^2-1}(\frac{n^2-4n+4)}{n^2})$.

Simplifying, we get $\frac{4}{n(n+1)}$ which we can set to be less than $\frac{1}{2015}$. By inspection, we find that the first such $n$ is $\boxed{090}$.

See also

2015 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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