Posted 3 Jul 2013 · Report post There is a circle with a radius of 1 on the ground. A straight line intersects the circle. What is the probability that the length of the chord created by the intersection of the circle and the line is greater than or equal to sqrt(3)? 0 Share this post Link to post Share on other sites

0 Posted 7 Jul 2013 · Report post Select a point at random, uniformly within the circle. That is, select x and y uniformly on [-1, 1]. Repeat if the point (x, y) lies outside the circle. Draw the chord through the selected point that is perpendicular to the line from the point to the origin. If the point lies a distance 0.5 from the origin, its chord has length sqrt(3). If the point lies inside a concentric circle with radius 0.5 its chord will have greater length. That probability is 1/4. And this method has the best rationale. I have three possible and perfectly acceptable solutions to this question each depending on what assumptions in your model you choose to make. Only two of the three possibilities have been correctly (at least according to my logic) found. Joyandwarmfuzzies- 1/3 K-Man - 1/2 But there is still another possibility that I would like examined. these are the solutions. Well done everyone! 0 Share this post Link to post Share on other sites

0 Posted 3 Jul 2013 · Report post 1/3 + sqrt(3) / (2*pi) 0 Share this post Link to post Share on other sites

0 Posted 3 Jul 2013 · Report post The probability distribution depends on the method of drawing the line through the circle. My answer of 1/2 is based upon the assumption that every distance from the chord to the center of the circle is equally likely. Depending on the method of drawing the line this assumption may or may not hold true. For example, the following method of drawing the line through the circle would satisfy the assumption: 1) Uniformly pick the orientation (for example by picking a point on the circle and calling it the "top of the circle") 2) Uniformly pick a number x between 0 and 1 and draw the "vertical" line x units away from the center (it's vertical in relation to the "top" of the circle - the line is drawn parallel to line connecting the top and the center of the circle) Other methods of drawing the line through the circle will produce different probability distributions, so maybe the OP should define the method of drawing the line. 0 Share this post Link to post Share on other sites

0 Posted 3 Jul 2013 · Report post 1/16 or 6.25% The length of the sqrt(3) chord can form a right triangle with the midpoint of the circle as one angle, the end of the chord as another angle, and the midpoint of the chord as the right angle. Since the radius of the circle is one, the length of the distance from the midpoint of the chord to the midpoint of the circle should be equal to the probability that the length of the chord created by the intersection of the circle and a random line is greater than or equal to sqrt(3). This should be true irrespective of orientation. Good ole Pythagoras said that a^{2}+b^{2}=c^{2}. Since b equals sqrt(3)/2 and c is 1, a should be equal to 1/16. 0 Share this post Link to post Share on other sites

0 Posted 3 Jul 2013 (edited) · Report post Yeah, I agree with k-man that different methods of drawing give different answers, but overall I think I agree with JAWF on the answer... The way I draw a chord is usually to pick a point on the circumference and then pick another point and connect them, so using that method and assuming each point on the circumference is equally likely, the second point must be >120 deg away from the first point for the chord to have length >sqrt(3), so the probability would be 1/3. Edited 3 Jul 2013 by Yoruichi-san 0 Share this post Link to post Share on other sites

0 Posted 6 Jul 2013 (edited) · Report post The probability can reasonably vary by a factor of two, depending on the description of "randomness" for the line. There is a strong argument for "best" implementation of randomness. It leads to one of the answers given above. an answer not given above. The reason it's best is that it gives the same result with circles of any radius, and arbitrary center locations. It also has a physical implementation that is satisfyingly random in nature. What is it? Edited 7 Jul 2013 by bonanova Correction. 0 Share this post Link to post Share on other sites

0 Posted 6 Jul 2013 · Report post I have three possible and perfectly acceptable solutions to this question each depending on what assumptions in your model you choose to make. Only two of the three possibilities have been correctly (at least according to my logic) found. Joyandwarmfuzzies- 1/3 K-Man - 1/2 But there is still another possibility that I would like examined. 0 Share this post Link to post Share on other sites

0 Posted 7 Jul 2013 · Report post Select a point at random, uniformly within the circle. That is, select x and y uniformly on [-1, 1]. Repeat if the point (x, y) lies outside the circle. Draw the chord through the selected point that is perpendicular to the line from the point to the origin. If the point lies a distance 0.5 from the origin, its chord has length sqrt(3). If the point lies inside a concentric circle with radius 0.5 its chord will have greater length. That probability is 1/4. And this method has the best rationale. 0 Share this post Link to post Share on other sites

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There is a circle with a radius of 1 on the ground. A straight line intersects the circle. What is the probability that the length of the chord created by the intersection of the circle and the line is greater than or equal to sqrt(3)?

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