# How to generate this?

• Feb 11th 2010, 04:34 PM
Richard
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?
• Feb 12th 2010, 07:53 AM
Opalg
Quote:

Originally Posted by Richard
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.
• Feb 12th 2010, 10:36 AM
Richard
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.
• Feb 12th 2010, 10:47 AM
Opalg
Quote:

Originally Posted by Richard
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.