Difference between revisions of "2015 AIME II Problems/Problem 5"
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− | Consider a <math>3x3</math> grid, where there are <math>4</math> corner squares, <math>4</math> edge squares, and <math>1</math> center square. A 4x4 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. | + | Consider a <math>3x3</math> grid, where there are <math>4</math> corner squares, <math>4</math> edge squares, and <math>1</math> center square. A <math>4x4</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 | + | 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>. | 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>. |
Revision as of 17:25, 8 February 2016
Contents
Problem
Two unit squares are selected at random without replacement from an grid of unit squares. Find the least positive integer
such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than
.
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 . 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
, and the number of ways to pick two squares out of Grid A is
. So, the probability that the two chosen squares are adjacent is
. We wish to find the smallest positive integer
such that
, and by inspection the first such
is
.
Solution 2
Consider a grid, where there are
corner squares,
edge squares, and
center square. A
grid has
corner squares,
edge squares, and
center squares. By examining simple cases, we can conclude that for a grid that is
, there are always
corners squares,
edge squares, and
center squares.
Each corner square is adjacent to other squares, edge squares to
other squares, and center squares to
other squares. In the problem, the second square can be any square that is not the first, which means there are
to choose from. With this information, we can conclude that the probability that second unit square is adjacent to the first is
.
Simplifying, we get which we can set to be less than
. By inspection, we find that the first such
is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.