You choose two points on a circle, at random. What is the probability that the chord joining the two chosen points is longer than the radius?
Hint so that you can still maybe get it: Connect the two random points with the center, then think about when the chord is equal/larger/smaller than the radius.
answer is here \/
If the two points make a 60-degree angle with the center, then the chord will be equal to the radius (because the three points form an equilateral triangle).
It doesn't matter where we put the first point, but it does matter where we put the second point. If the second point is within 60 degrees of the first angle, then the chord will be shorter than the radius. So the arc of the circle where we can put the second point has measure 120 degrees (the 60 degrees goes on either side of the point), with a 1/3 chance of putting the point there. To find the probability that the chord is longer, just subtract 1/3(chance of the chord being shorter) from 1 to get \(\boxed{2/3}\) .
A diagram would be kinda useful but this is the best explanation I can give.
"this is the best explanation I can give" Not just your best, I think it's the best anybody could give, simple and elegant. Or is simplicity already an element of elegance? Anyway, good work! I was attempting a solution based on the ratio of the circumference of the circle being 2π radii. I quickly bogged down, and then realized I was completely on the wrong tack.
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