Difference between revisions of "2015 AMC 10A Problems/Problem 25"
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Divide the boundary of the square into halves, thereby forming <math>8</math> segments. Without loss of generality, let the first point <math>A</math> be in the bottom-left segment. Then, it is easy to see that any point in the <math>5</math> segments not bordering the bottom-left segment will be distance at least <math>\dfrac{1}{2}</math> apart from <math>A</math>. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least <math>0.5</math> apart from <math>A</math> is <math>\dfrac{0 + 1}{2} = \dfrac{1}{2}</math> because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.) | Divide the boundary of the square into halves, thereby forming <math>8</math> segments. Without loss of generality, let the first point <math>A</math> be in the bottom-left segment. Then, it is easy to see that any point in the <math>5</math> segments not bordering the bottom-left segment will be distance at least <math>\dfrac{1}{2}</math> apart from <math>A</math>. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least <math>0.5</math> apart from <math>A</math> is <math>\dfrac{0 + 1}{2} = \dfrac{1}{2}</math> because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.) | ||
− | If the second point <math>B</math> is on the left-bottom segment, then if <math>A</math> is distance <math>x</math> away from the left-bottom vertex, then <math>B</math> must be up to <math>\dfrac{1}{2} - \sqrt{0.25 - x^2}</math> away from | + | If the second point <math>B</math> is on the left-bottom segment, then if <math>A</math> is distance <math>x</math> away from the left-bottom vertex, then <math>B</math> must be up to <math>\dfrac{1}{2} - \sqrt{0.25 - x^2}</math> away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is |
<cmath>\frac{1}{\left( \frac{1}{2} \right)^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left( \frac{1}{4} - \frac{\pi}{16} \right) = 1 - \frac{\pi}{4}.</cmath> | <cmath>\frac{1}{\left( \frac{1}{2} \right)^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4\left( \frac{1}{4} - \frac{\pi}{16} \right) = 1 - \frac{\pi}{4}.</cmath> | ||
Revision as of 22:36, 14 February 2018
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[hide]Problem 25
Let be a square of side length . Two points are chosen independently at random on the sides of . The probability that the straight-line distance between the points is at least is , where , , and are positive integers with . What is ?
Solution 1
Divide the boundary of the square into halves, thereby forming segments. Without loss of generality, let the first point be in the bottom-left segment. Then, it is easy to see that any point in the segments not bordering the bottom-left segment will be distance at least apart from . Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least apart from is because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
If the second point is on the left-bottom segment, then if is distance away from the left-bottom vertex, then must be up to away from the left-middle point. Thus, using an averaging argument we find that the probability in this case is
(Alternatively, one can equate the problem to finding all valid with such that , i.e. is outside the unit circle with radius )
Thus, averaging the probabilities gives
Our answer is .
Solution 2
Let one point be chosen on a fixed side. Then the probability that the second point is chosen on the same side is , on an adjacent side is , and on the opposite side is . We discuss these three cases.
Case 1: Two points are on the same side. Let the first point be and the second point be in the -axis with . Consider a point on the unit square on the Cartesian plane. The region has the area of . Therefore, the probability that is .
Case 2: Two points are on two adjacent sides. Let the two sides be on the x-axis and on the y-axis and let one point be and the other point be . Then and the distance between the two points is . As in Case 1, is a point on the unit square . The area of the region is and the area of its complementary set inside the square (i.e. ) is . Therefore, the probability that the distance between and is at least is .
Case 3: Two points are on two opposite sides. In this case, the probability that the distance between the two points is at least is obviously .
Thus the probability that the probability that the distance between the two points is at least is given by Therefore , , and . Thus, and the answer is
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 10 Problems and Solutions |
2015 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.