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| − | Problem 25
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| − | Let S be a square of side length 1. Two points are chosen independently at random on the sides of S. The probability that the straight-line distance between the points is at least \tfrac12 is \tfrac{a-b\pi}c, where a, b, and c are positive integers with \gcd(a,b,c)=1. What is a+b+c?
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| − | \textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63
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| − | [edit]Solution
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| − | Divide the boundary of the square into halves, thereby forming 8 segments. Without loss of generality, let the first point A 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 \dfrac{1}{2} apart from A. Now, consider choosing the second point on the bottom-right segment. The probability for it to be distance at least 0.5 apart from A is \dfrac{0 + 1}{2} = \dfrac{1}{2} because of linearity of the given probability. (Alternatively, one can set up a coordinate system and use geometric probability.)
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| − | If the second point B is on the left-bottom segment, then if A is distance x away from the left-bottom vertex, then B must be at least \dfrac{1}{2} - \sqrt{0.25 - x^2} away from that same vertex. Thus, using an averaging argument we find that the probability in this case is \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 - \fra...
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| − | (Alternatively, one can equate the problem to finding all valid (x, y) with 0 < x, y < \dfrac{1}{2} such that x^2 + y^2 \ge \dfrac{1}{4}, i.e. (x, y) is outside the unit circle with radius 0.5.)
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| − | Thus, averaging the probabilities gives P = \frac{1}{8} (5 + \frac{1}{2} + 1 - \frac{\pi}{4}) = \frac{1}{32} (26 - \pi).
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| − | Our answer is \textbf{(A)}.
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