Difference between revisions of "2012 AMC 12A Problems/Problem 23"
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==Video Solution by Richard Rusczyk== | ==Video Solution by Richard Rusczyk== | ||
− | https://www.youtube.com/watch?v= | + | https://www.youtube.com/watch?v=FiUZTlzA7lg |
== See Also == | == See Also == | ||
{{AMC12 box|year=2012|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2012|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:25, 27 April 2020
Problem
Let be the square one of whose diagonals has endpoints and . A point is chosen uniformly at random over all pairs of real numbers and such that and . Let be a translated copy of centered at . What is the probability that the square region determined by contains exactly two points with integer coefficients in its interior?
Solution
The unit square's diagonal has a length of . Because square is not parallel to the axis, the two points must be adjacent.
Now consider the unit square with vertices and . Let us first consider only two vertices, and . We want to find the area of the region within that the point will create the translation of , such that it covers both and . By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.
For to contain the point , must be inside square . Similarly, for to contain the point , must be inside a translated square with center at , which we will call . Therefore, the area we seek is Area.
To calculate the area, we notice that Area Area by symmetry. Let . Let be the midpoint of , and along the line . Let be the intersection of and within , and be the intersection of and outside . Therefore, the area we seek is Area. Because all have coordinate , they are collinear. Noting that the side length of and is (as shown above), we also see that , so . If follows that and . Therefore, the area is Area.
Because there are three other regions in the unit square that we need to count, the total area of within such that contains two adjacent lattice points is .
By periodicity, this probability is the same for all and . Therefore, the answer is .
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=FiUZTlzA7lg
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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