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−  Problem 25
 
   
−  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 straightline distance between the points is at least \tfrac12 is \tfrac{ab\pi}c, where a, b, and c are positive integers with \gcd(a,b,c)=1. What is a+b+c?
 
−  \textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63
 
−  [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 bottomleft segment. Then, it is easy to see that any point in the 5 segments not bordering the bottomleft segment will be distance at least \dfrac{1}{2} apart from A. Now, consider choosing the second point on the bottomright 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.)
 
−  If the second point B is on the leftbottom segment, then if A is distance x away from the leftbottom 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.)
 
−  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)}.
 