ANSWERS: 3
  • From any given point we have two regions of chords that are laess than rt3 in length (in black). The area of this region is 2*(segment - 2*green triangle) Area of a segment = (theta*r^2)/2 Theta/2 = arcsin ((rt3)/2) = Pi/3 Theta = 2*Pi/3 Segment area = ((2*Pi/3)*1^2)/2 = Pi/3 Area of green triangle = (base*height)/2 Base = (rt3)/2 Height = a a^2 + b^2 = c^2 a = rt(1^2 - ((rt3)/2)^2) = rt(1/4) = 1/2 Area of green triangle = (((rt3)/2)*(1/2))/2 = (rt3)/8 Black region = 2*(segment - 2*green triangle) = 2*(Pi/3 - 2*(rt3)/8) ~ 1.23 As a percentage: (100*black region)/(Pi*r^2) ~ 39% Thus those greater than rt3 = 100% - 39% = 61%
  • Here's one way to think of it: Consider the circle such that it is divided into three vertical stripes such that any chord passing through there will have a length greater than rt3. It turns out that the area of this region is 1/2 and an answer of 1/2. Here's another way to think of it: Express it in polar coordinates. Draw in every single chord of exact length rt3. This will form a circle around an area in the middle of the circle that does not have any chords through it, since they'd have to be of length greater than rt3. Now consider the distribution of the midpoints of all possible chords. Using polar coordinates for uniform randomness, the area of that circle is 1/4 that of the big circle (radius is half that of the big circle). This leads to 1/4 as an answer. See the diagram! Consider the midpoints of all the chords. 1/4 of them would fall into the red circle. Thus the answer is 1/4. Also, see Farino's answer. I'm not sure if it is right....
  • There is no right answer because a percentage is calculated as a proportion: i.e. how many chords have a length greater than the square root of three. But how do you count an infinite number of chords? You have to specify how your chords are distributed. Here's three ways: One way is to join two points on the circumference, one way is to pick arbitrary lines that cross the circle, one way is to pick a chord a certain distance from the centre: Each of these give different results. That's three, but there are uncountable numbers of results. I can pick an arbitrary point twice the radius from the centre and choose a chord to pass through that point. That's four ... and so on. EDIT: Here's the proof that you get different answers: Working for method 1: Pick two points on the circumference. Assume that the points are picked separately and at two arbitrary angles each uniformly distributed in the range [0,360) degrees anticlockwise from the x-axis. The length of the chord is related to the angle subtended at the center. The angle subtended at the centre for a length of root three is 120 degrees. An angle greater than 120 will result in a length greater than root three. What is the probability that the two points subtend a angle at the center greater than 120 degrees? Rotate the circle til A is on the x-axis. B is still equally distributed about the circle. If B's new angle is between 120 and 240 the chord matches the test. That's a probability of one third. So the answer, method 1, is 1/3. Working for method 3: Pick a distance d from the center in the range [0,1] Pick a point on the circle radius d, making an angle in the range [0,360) degrees anticlockwise from the x-axis. Draw the chord tangent to the circle radius d, through the chosen point. The angle won't make a difference to the length of the chord. To make the length greater than root three, need d to exceed one half. If d is chosen uniformly distributed, exactly half of the choices will have this property. So the answer, method 3, is 1/2. As I said it just depends what distribution you choose and without more information, you cannot choose a correct answer.

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