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

Latest revision as of 16:40, 1 July 2023

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 $3 \times 3$ 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 $\boxed{090}$ .

Solution 4 (cheese)

Notice how a chosen unit square on the grid has 4 vertically & horizontally adjacent squares around it (not counting corners or sides.) That's $\frac{4}{n^2}$ . Using this, we rewrite $\frac{1}{2015}$ as $\frac{4}{8060}$ . Notice that the denominator $8060$ is really close to $90^2$ , and the problem is asking for the least positive integer less than $\frac{1}{2015}$ . Therefore, the closest possible estimation is $\boxed{090}$ . We can check this by adding in our corners and sides. Easy multiplication and simplification finds us with $\boxed{090}$ as the correct answer.

~orenbad

Video Solution

https://www.youtube.com/watch?v=9re2qLzOKWk&t=304s

~MathProblemSolvingSkills.com

@@ Line 3: / Line 3: @@
 Two unit squares are selected at random without replacement from an <math>n \times n</math> grid of unit squares. Find the least positive integer <math>n</math> such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than <math>\frac{1}{2015}</math>.
-==Solution==
+==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 <math>(n-1) \times (n-1)</math>. 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 <math>2n(n-1)</math>, and the number of ways to pick two squares out of grid A is <math>\dbinom{n^2}{2}</math>. So, the probability that the two chosen squares are adjacent is <math>\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1)}{\frac{n^2(n^2-1)}{2}} = \frac{4}{n(n+1)}</math>. We wish to find the smallest positive integer <math>n</math> such that <math>\frac{4}{n(n+1)} < \frac{1}{2015}</math>, and by inspection the first such <math>n</math> is <math>\boxed{090}</math>.
+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 <math>(n-1) \times (n-1)</math>. 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 <math>2n(n-1)</math>, and the number of ways to pick two squares out of Grid A is <math>\dbinom{n^2}{2}</math>. So, the probability that the two chosen squares are adjacent is <math>\frac{2n(n-1)}{\binom{n^2}{2}} = \frac{2n(n-1)}{\frac{n^2(n^2-1)}{2}} = \frac{4}{n(n+1)}</math>. We wish to find the smallest positive integer <math>n</math> such that <math>\frac{4}{n(n+1)} < \frac{1}{2015}</math>, and by inspection the first such <math>n</math> is <math>\boxed{090}</math>.
+==Solution 2==
+Consider a <math>3 \times 3</math> grid, where there are <math>4</math> corner squares, <math>4</math> edge squares, and <math>1</math> center square. A <math>4 \times 4</math> grid has <math>4</math> corner squares, <math>8</math> edge squares, and <math>4</math> center squares. By examining simple cases, we can conclude that for a grid that is <math>n \times n</math>, there are always <math>4</math> corners squares, <math>4(n-2)</math> edge squares, and <math>n^2-4n+4</math> center squares.
+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 inspection, we find that the first such <math>n</math> is <math>\boxed{090}</math>.
+==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 <math>\frac{(n-2)(n-2)}{n^2}</math> multiplied by <math>\frac{4}{n^2 -1}</math>. Add that to the probability of an edge and an adjacent square( <math>\frac{4n-8}{n^2}</math> multiplied by  <math>\frac{3}{n^2-1} </math>) and the probability of a corner and an adjacent square( <math>\frac{4}{n^2} </math> multiplied by  <math>\frac{2}{n^2-1} </math>) to get  <math>\frac{4n^2-4n}{n^4-n^2} </math>. Simplify to get  <math>\frac{4}{n^2+n} </math>. With some experimentation, we realize that the smallest value of n is <math>\boxed{090}</math>.
+==Solution 4 (cheese)==
+Notice how a chosen unit square on the grid has 4 vertically & horizontally adjacent squares around it (not counting corners or sides.) That's <math>\frac{4}{n^2}</math>. Using this, we rewrite <math>\frac{1}{2015}</math> as <math>\frac{4}{8060}</math>. Notice that the denominator <math>8060</math> is really close to <math>90^2</math>, and the problem is asking for the least positive integer less than <math>\frac{1}{2015}</math>. Therefore, the closest possible estimation is <math>\boxed{090}</math>. We can check this by adding in our corners and sides. Easy multiplication and simplification finds us with <math>\boxed{090}</math> as the correct answer.
+~[[OrenSH|orenbad]]
+==Video Solution==
+https://www.youtube.com/watch?v=9re2qLzOKWk&t=304s
+~MathProblemSolvingSkills.com
 ==See also==
 {{AIME box|year=2015|n=II|num-b=4|num-a=6}}
 {{MAA Notice}}

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

Page

Toolbox

Search