2014 AMC 10B Problems/Problem 19

Revision as of 19:02, 20 February 2014 by DivideBy0 (talk | contribs) (Solution: fixed 120 degree sentence)

Problem

Two concentric circles have radii $1$ and $2$. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?

$\textbf{(A) }\frac{1}{6}\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }\frac{2-\sqrt{2}}{2}\qquad\textbf{(D) }\frac{1}{3}\qquad\textbf{(E) }\frac{1}{2}\qquad$

Solution

Pick any arbitrary point on the circle, and it is clear that the second point must lie on the arc of 120 degrees, on the opposite end of the circle from the point. Therefore the probability is $120/360=1/3$

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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