Difference between revisions of "2015 AMC 12A Problems/Problem 23"
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==Solution== | ==Solution== | ||
− | + | Divide the boundary of the square into halves, thereby forming 8 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 5 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 0.5 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 at least <math>\dfrac{1}{2} - \sqrt{0.25 - x^2}</math> away from that same vertex. Thus, using an averaging argument we find that the probability in this case is | |
+ | <cmath>\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4(\frac{1}{4} - \frac{\pi}{16}) = 1 - \frac{\pi}{4}.</cmath> | ||
+ | |||
+ | (Alternatively, one can equate the problem to finding all valid <math>(x, y)</math> with <math>0 < x, y < \dfrac{1}{2}</math> such that <math>x^2 + y^2 \ge \dfrac{1}{4}</math>, i.e. (x, y) is outside the unit circle with radius 0.5.) | ||
+ | |||
+ | Thus, averaging the probabilities gives | ||
+ | <cmath>P = \frac{1}{8} (5 + \frac{1}{2} + 1 - \frac{\pi}{4}) = \frac{1}{32} (26 - \pi).</cmath> | ||
+ | |||
+ | Our answer is <math>\textbf{(A)}</math>. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2015|ab=A|num-b=22|num-a=24}} |
Revision as of 13:53, 5 February 2015
Problem
Let be a square of side length 1. 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 and . What is ?
$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}}\ 62 \qquad\textbf{(E)}\ 63$ (Error compiling LaTeX. ! Extra }, or forgotten $.)
Solution
Divide the boundary of the square into halves, thereby forming 8 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 5 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 0.5 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 at least away from that same vertex. 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. (x, y) is outside the unit circle with radius 0.5.)
Thus, averaging the probabilities gives
Our answer is .
See Also
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 |