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Thread: How to generate this?

  1. #1
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    How to generate this?

    Given Euclidean geometry, how to generate for any pair of distinct points x and y, for any factor 0 \leq \lambda \leq 1, two halfplanes X and Y, where X consists of all points v with \lambda d(v,x) < (1 - \lambda) d(v,y)? In the simple case where \lambda = 1 - \lambda = .5 , it's easy. You just draw a line through x and y, half it and draw the perpendicular line through the center of the line. But how do you do it otherwise?
    Last edited by Richard; Feb 11th 2010 at 04:44 PM.
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  2. #2
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    Quote Originally Posted by Richard View Post
    Given Euclidean geometry, how to generate for any pair of distinct points x and y, for any factor 0 \leq \lambda \leq 1, two halfplanes X and Y, where X consists of all points v with \lambda d(v,x) < (1 - \lambda) d(v,y)? In the simple case where \lambda = 1 - \lambda = .5 , it's easy. You just draw a line through x and y, half it and draw the perpendicular line through the center of the line. But how do you do it otherwise?
    If \lambda\ne1/2, the regions X and Y will not be half-planes, but the interior and exterior of a circle.

    First, you have to locate two points s and t on the line joining x and y, so that \lambda d(s,x) = (1 - \lambda) d(s,y) and \lambda d(t,x) = (1 - \lambda) d(t,y). One of these points, s, will be on the line segment between x and y, dividing it in the ratio 1 - \lambda to \lambda. The other point, t, will be on the far side of y (if \lambda<1/2), and on the far side of x (if \lambda>1/2).

    Next, construct a circle for which the line segment from s to t is a diameter. If \lambda<1/2 then X will be the region outside this circle. If \lambda>1/2 then it will be the region inside the circle.
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  3. #3
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    How to generate this---question of understanding

    Thank you for this reply to my threat. I am quite puzzled now. For, if I'm not wrong, on Euclidean metric with n dimensions, \lambda d(v,x) = (1 - \lambda) d(v,y) just in case \Sigma_{1 \leq i \leq n} (\lambda v)^2 - (\lambda x)^2 = \Sigma_{1 \leq i \leq n} ((1 - \lambda)v)^2 - ((1 - \lambda)y)^2. But the latter equation rewrites to a quadratic equation of the form: \Sigma_{1 \leq i \leq n} (2\lambda - 1) x^2\ +\ ax\ +\ k = 0 , which does not define a circle, if I am not wrong. It would be great if you could help me out of my state of puzzlement.
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    Quote Originally Posted by Richard View Post
    Thank you for this reply to my threat. I am quite puzzled now. For, if I'm not wrong, on Euclidean metric with n dimensions, \lambda d(v,x) = (1 - \lambda) d(v,y) just in case \Sigma_{1 \leq i \leq n} (\lambda v)^2 - (\lambda x)^2 = \Sigma_{1 \leq i \leq n} ((1 - \lambda)v)^2 - ((1 - \lambda)y)^2. But the latter equation rewrites to a quadratic equation of the form: \Sigma_{1 \leq i \leq n} (2\lambda - 1) x^2\ +\ ax\ +\ k = 0 , which does not define a circle, if I am not wrong. It would be great if you could help me out of my state of puzzlement.
    I was thinking in terms of two-dimensional space when I wrote that reply. In n dimensions it will be an n-sphere. The equation you give is essentially correct. If x = (x_1,x_2,\ldots,x_n) then the equation is \textstyle\sum_{1 \leqslant i \leqslant n} (2\lambda - 1) x_i^2 + a_ix_i + k = 0 . Write this as \textstyle\sum_{1 \leqslant i \leqslant n} \bigl (x_i + \tfrac{a_i}{2(2\lambda - 1)}\bigr)^2  = \text{const.} to see that it is indeed the equation of a sphere.
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