Probability Question?

Question

I have this last problem for homework and I can't figure it out.

Consider a unit circle with an inscribed equilateral triangle. A chord is selected at random. What is the probability that the chord is longer than a side of the triangle?

Wouldn't it be very hard to measure how each chord that could be drawn to make sure it is longer than a side but less than the diameter.

Answer 1

If you let the radius be 1, then the side of the triangle has length sqrt(3).

Draw a chord AB with angle AOB = θ
AB = sqrt(2 - 2cosθ)
AB > 3
2 - 2 cosθ > 3
cosθ < -1/2
120 < θ < 240

In other words, for a given starting point A, there is a 120 degrees range in which B can be positioned for AB to be longer than sqrt(3).
That's a probability of 1/3

Answer 2

It all hinges on what you mean by selecting a chord "at random". A few answers above have assumed this means independently selecting two points uniformly from the circle, which is certainly a possible interpretation, and in that case the probability is 1/3 as they said.

However, there are other ways in which we might select a chord at random. For instance, we could select an angle uniformly from 0° to 180° and an offset from the centre of the circle uniformly from -r to +r. If this offset is x, the length of the chord will be 2√(r^2-x^2). This is greater than r√3 when |x| < r/√2, so the probability in this case is (2r/√2) / (2r) = 1/√2.

So the answer depends on how precisely you make your random choice of chord. Though I agree the answer they're most likely looking for is 1/3.

Answer 3

A chord can have a central angle from 0 to 180 degrees. A chord randomly selected is larger than the side of the inscribed triangle when the central angle is larger than 120 degrees. Since the largest central angle for a chord is 180 degrees, the probability = (180-120)/180 = 1/3.

Answer 4

It's not so bad. This is a completely geometric problem, and need not involve trig or algebra. That's a bit nutty.

Draw your circle, triangle, and the first point of the chord. Since it doesn't matter where we put the triangle, we can assume the point we just picked is actually vertex. This is EXCEEDINGLY helpful.

Refer to this picture:
http://math.colgate.edu/~kellen/interspace/circ.gif

Call X the first point. The question is where can we put the other point (call it Y)?

We need XY to be longer than a side length of the triangle, meaning the segment must go through the triangle (meaning y is between the other two vertices). The angle of the triangle is 60 degrees, meaning its arc on the circle (between the other two vertices) is 120 degrees.

Since Y is randomly distributed on the circle, we know that the chances of it landing there are 120 / 360 = 1/3.

Answer 5

Yes.