2015 AMC 12A Problem 23

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23.

Let SS be a square of side length 1.1. Two points are chosen independently at random on the sides of S.S. The probability that the straight-line distance between the points is at least 12\dfrac12 is abπc,\dfrac{a - b\pi}{c}, where a,a, b,b, and cc are positive integers and gcd(a,b,c)=1.\gcd(a, b, c) = 1. What is a+b+c?a + b + c?

5959

6060

6161

6262

6363

Answer: A
Concepts:geometric probabilitycasework
Difficulty rating: 2380
Solution:

The second point is on the same side as the first with probability 14,\dfrac14, on the opposite side with probability 14,\dfrac14, and on an adjacent side with probability 12.\dfrac12.

Opposite sides: the distance is at least 1121 \ge \dfrac12 always, probability 1.1.

Same side: for points (a,0)(a, 0) and (b,0),(b, 0), the condition ab12|a - b| \ge \dfrac12 has probability 14.\dfrac14.

Adjacent sides: for points (a,0)(a, 0) and (0,b),(0, b), the condition a2+b212\sqrt{a^2 + b^2} \ge \dfrac12 is the region outside a quarter-circle of radius 12,\dfrac12, with probability 114π(12)2=1π16.1 - \dfrac14\pi\left(\dfrac12\right)^2 = 1 - \dfrac{\pi}{16}.

The total probability is 141+1414+12(1π16)=26π32. \begin{aligned} &\dfrac14\cdot 1 + \dfrac14\cdot\dfrac14 \\ &\quad {}+ \dfrac12\left(1 - \dfrac{\pi}{16}\right) \\ &\quad = \dfrac{26 - \pi}{32}. \end{aligned} Thus a+b+c=26+1+32=59.a + b + c = 26 + 1 + 32 = 59.

Thus, the correct answer is A.

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