Question

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?

$\ \dfrac{1}{6}$

$\ \dfrac{1}{4}$

$\ \dfrac{2-\sqrt{2}}{2}$

$\ \dfrac{1}{3}$

$\ \dfrac{1}{2}$

Solution:

First, without loss of generality, we could choose some point on the outer circle. Then, the second point can be chosen in a region on the other circle.

This region is such that it has a line that intersects the circle, so the edge of the region is such that the chord is perpendicular with the inner circle.

If we look at the angle at the center, we can see that it has 2 right triangles where the adjacent side is $1$ and the hypotenuse is $2,$ making $\cos \left(\dfrac \theta 2\right) = \dfrac 12.$

Thus, $\dfrac \theta 2 = 60^\circ,$ making $\theta = 120 ^\circ .$

Therefore, the probability is $\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$

Thus, the correct answer is D .

2014 AMC 10B Problem 19

Solution:

Problem 19 in Other Years