Difference between revisions of "2015 AMC 10A Problems/Problem 25"
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Our answer is <math>\boxed{\textbf{(A) } 59}</math>. | Our answer is <math>\boxed{\textbf{(A) } 59}</math>. | ||
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+ | == Solution 2 == | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}} | {{AMC10 box|year=2015|ab=A|num-b=24|after=Last Problem}} | ||
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Revision as of 14:10, 21 March 2015
Contents
[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
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. is outside the unit circle with radius )
Thus, averaging the probabilities gives
Our answer is .
Solution 2
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 |
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