2015 AIME II Problems/Problem 5

Revision as of 21:21, 18 March 2018 by Maheshbabuchapati (talk | contribs) (Solution 2)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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 $3 \times 3$ grid, where there are $4$ corner squares, $4$ edge squares, and $1$ center square. A $4 \times 4$ 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 $\boxed{090}$.

Solution 3

There are 3 cases in this problem. Number one, the center squares. Two, the edges, and three, the corners. The probability of getting one center square and an adjacent square is $\frac{(n-2)(n-2)}{n^2}$ multiplied by $\frac{4}{n^2 -1}$. Add that to the probability of an edge and an adjacent square( $\frac{4n-8}{n^2}$ multiplied by $\frac{3}{n^2-1}$) and the probability of a corner and an adjacent square( $\frac{4}{n^2}$ multiplied by $\frac{2}{n^2-1}$) to get $\frac{4n^2-4n}{n^4-n^2}$. Simplify to get $\frac{4}{n^2+n}$. With some experimentation, we realize that the smallest value of n is $\boxed{090}$.

See also

2015 AIME II (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS