2015 AMC 10A Problems/Problem 25

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 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 $a+b+c$ ?

$\textbf{(A) }59\qquad\textbf{(B) }60\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$

Solution

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.)

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 $\frac{1}{\frac{1}{2}^2} \int_0^{\frac{1}{2}} \dfrac{1}{2} - \sqrt{0.25 - x^2} dx = 4(\frac{1}{4} - \frac{\pi}{16}) = 1 - \frac{\pi}{4}.$

(Alternatively, one can equate the problem to find 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 $P = \frac{1}{8} (5 + \frac{1}{2} + 1 - \frac{\pi}{4}) = \frac{1}{32} (26 - \pi).$

Our answer is $\textbf{(A)}$ .

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2015 AMC 10A Problems/Problem 25

Problem 25

Solution