What is the probability that a randomly drawn chord will be longer than the radius of the circle?

Prove it.

(In reply to

re(4): Different Approach (Continued) by SilverKnight)

I'm repeating alot here, but anyways...

We pick a random point, and draw a radius to that point. Now we pick another random point, draw a chord and another radius to the second point. We now have a triangle, with the chord, two radii, and angle A between the two radii.

Now we solve for angle A at which the chord is equal to the radii. This would be an equilateral triangle, with all three sides equal (chord=radius=radius) and all three angles equalling 60 degrees. For all angles A greater than 60 degrees, the chord is longer than the radii, for all A less than 60 degrees, the chord is shorter than the radii.

Now to calculate the probablity that the chord is greater than the radii, we calculate the probality that A is greater than 60 degrees (pi/3 radians). Because an equivalent chord could be drawn from the same starting point to the other side of the circle, we will divide the circle in two halves, each with 180 degrees. The probablity that A is greater than 60 is (180-60)/180=120/180=2/3.

I saw someone saying we dont know how the chords are distributed according to angle. I didnt really follow that but this might help. We can assume there are an equal number of infinite points over any two equal proportions of the circumfrence. If you think of angle A in radians rather than degrees, it might help. Radians measure the amount of the cicumfrence encompassed by the angle, so any two equal angles A will have an equal number of infinite points and an equal number of infinite possible chords.

I dont know if I explained this very clearly, I usually reach peak lucidity after 4 beers, and am presently dead sober, however I think this makes sense and it all adds up to 2/3.