2008 AMC 12B Problem 21

Question

Two circles of radius

1

are to be constructed as follows. The center of circle

A

is chosen uniformly and at random from the line segment joining

(0, 0)

to

(2, 0).

The center of circle

B

is chosen uniformly and at random, and independently of the first choice, from the line segment joining

(0, 1)

to

(2, 1).

What is the probability that circles

A

and

B

intersect?

\dfrac{2 + \sqrt{2}}{4}

\dfrac{3\sqrt{3} + 2}{8}

\dfrac{2\sqrt{2} - 1}{2}

\dfrac{2 + \sqrt{3}}{4}

\dfrac{4\sqrt{3} - 3}{4}

With centers

(a, 0)

and

(b, 1),

the circles intersect iff

\sqrt{(a - b)^2 + 1} \le 2,

i.e.

|a - b| \le \sqrt{3}

In the square

[0, 2]^2

of area

4,

find the area with

|a - b| \le \sqrt{3}

by removing the two corner triangles where

|a - b| \gt \sqrt{3}

Let the centers be

(a, 0)

and

(b, 1)

with

a, b \in [0, 2].

The circles (radius

1

each) intersect iff the distance between centers is at most

2:

\begin{aligned} &\sqrt{(a - b)^2 + 1} \le 2 \\ &\iff |a - b| \le \sqrt{3}. \end{aligned}

The pairs

(a, b)

fill the square

[0, 2]^2

of area

4.

The failing region

|a - b| \gt \sqrt3

is two right triangles, each with legs

2 - \sqrt3,

of total area

(2 - \sqrt3)^2 = 7 - 4\sqrt3.

So the favorable area is

4 - (7 - 4\sqrt3) = 4\sqrt3 - 3,

and the probability is

\frac{4\sqrt3 - 3}{4}.

Thus, the correct answer is E.

Problem 21 in other years