2015 AMC 12A Problems/Problem 23

Revision as of 13:52, 5 February 2015 by Michelangelo (talk | contribs) (Solution)

Problem

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 $\frac12$ is $\frac{a-b\pi}{c}$, where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$. What is $a+b+c$?

$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}}\ 62 \qquad\textbf{(E)}\ 63$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

Solution

Each segment of half of the length of a side of the square is identical, so arbitrarily choose one.

The portion of the square within $0.5$ units of a point on that segment is $0.5+d+\sqrt{.25-d^2}$ where $d$ is the distance from the corner. The integral from $0$ to $0.5$ of this formula resolves to $\frac{6+\pi}{8}$, so the probability of choosing a point within $0.5$ of the first point is $\frac{6+\pi}{32}$. The inverse of this is $\frac{26-\pi}{32}$, so $a+b+c= \boxed{\textbf{(A)}\ 59}$.

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 12 Problems and Solutions
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