If two points, A and B, are randomly placed on the circumfer

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If points A and B are randomly placed on the circumference of a circle with circumference 12pi inches, what is the probability that the length of chord AB will be at least 6 inches?

(A) 1/(2pi)
(B) 1/pi
(C) 1/3
(D) 2/pi
(E) 2/3

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Answer: E
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by regor60 » Thu Feb 16, 2017 11:33 am
Radius of circle of circumference of 12pi = 6 by 2(pi)r formula

Drawing the circle and plotting points A and B on circle and connecting center and two points, and given minimum chord of 6, we have an equilateral triangle

Therefore, the angles of the triangle are 60 degrees and the sum of the outside angles are 120 to complete the 180 diameter.

A&B have to be plotted from the minimum 6 inch chord position sweeping out to diameter to satisfy problem statement.

This means that 120 degrees of 180, or [spoiler]2/3[/spoiler] satisfy the question. Same logic applies for the other half of the circle.

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by Brent@GMATPrepNow » Thu Feb 16, 2017 3:22 pm
Brent@GMATPrepNow wrote:If points A and B are randomly placed on the circumference of a circle with circumference 12pi inches, what is the probability that the length of chord AB will be at least 6 inches?

(A) 1/(2pi)
(B) 1/pi
(C) 1/3
(D) 2/pi
(E) 2/3
regor60's solution is perfect.
Below is a very similar solution (with a few diagrams to help students visualize the solution)

Let's first determine the details of this circle.
For any circle, circumference = (diameter)(pi)

The circumference of the given circle is 12pi inches, so we can write: 12pi inches= (diameter)(pi)
This tells us that the diameter of the circle = 12 inches
It also tells us that the radius of the circle = 6 inches


Okay, now let's solve the question.
We'll begin by arbitrarily placing point A somewhere on the circumference.
Image



So, we want to know the probability that a randomly-placed point B will yield a chord AB that is at least 6 inches long.
So, let's first find a location for point B that creates a chord that is EXACTLY 6 inches.
Image



There's also another location for point B that creates another chord that is EXACTLY 6 inches.
Image



IMPORTANT: For chord AB to be greater than or equal to 6 inches, point B must be placed somewhere along the red portion of the circle's circumference.
Image


So, the question really boils down to, "What is the probability that point B is randomly placed somewhere on the red line?"
To determine this probability, notice that the 6-inch chords are the same length as the circle's radius (6 inches)
Image


Since these 2 triangles have sides of equal length, they are equilateral triangles, which means each interior angle is 60 degrees.
Image


The 2 central angles (from the equilateral triangles) add to 120 degrees.
This means the remaining central angle must be 240 degrees.
Image

This tells us that the red portion of the circle represents 240/360 of the entire circle.
So, P(point B is randomly placed somewhere on the red line) = 240/360 = 2/3

Answer: E
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by Matt@VeritasPrep » Fri Feb 17, 2017 1:20 am
Worth noting that randomness on a circle is a little more treacherous than you'd think. There's a good discussion in Mosteller (in the solution to problem #25) about the definition of "at random" with respect to "chords chosen at random" on a circle and how this can change the answer to a problem.

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by Brent@GMATPrepNow » Fri Feb 17, 2017 6:48 am
Hi Matt,

I checked the Mosteller link, and it seems to discuss the issues related to randomly selecting a chord, whereas my question involves choosing 2 random points that will create a chord.

The Mosteller link (in part c) seems to suggest that, if we create a chord by choosing 2 random points (as my question suggests), then the answer to my original question is [spoiler]2/3[/spoiler]

Am I missing something?

Cheers,
Brebt
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