# Thread: How to generate this?

1. ## How to generate this?

Given Euclidean geometry, how to generate for any pair of distinct points $\displaystyle x$ and $\displaystyle y$, for any factor $\displaystyle 0 \leq \lambda \leq 1$, two halfplanes $\displaystyle X$ and $\displaystyle Y$, where $\displaystyle X$ consists of all points $\displaystyle v$ with $\displaystyle \lambda d(v,x) < (1 - \lambda) d(v,y)$? In the simple case where $\displaystyle \lambda = 1 - \lambda = .5$, it's easy. You just draw a line through $\displaystyle x$ and $\displaystyle y$, half it and draw the perpendicular line through the center of the line. But how do you do it otherwise?

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

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

Next, construct a circle for which the line segment from $\displaystyle s$ to $\displaystyle t$ is a diameter. If $\displaystyle \lambda<1/2$ then $\displaystyle X$ will be the region outside this circle. If $\displaystyle \lambda>1/2$ then it will be the region inside the circle.

3. ## 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, $\displaystyle \lambda d(v,x) = (1 - \lambda) d(v,y)$ just in case $\displaystyle \Sigma_{1 \leq i \leq n} (\lambda v)^2 - (\lambda x)^2 = \Sigma_{1 \leq i \leq n} ((1 - \lambda)v)^2 - ((1 - \lambda)y)^2$. But the latter equation rewrites to a quadratic equation of the form: $\displaystyle \Sigma_{1 \leq i \leq n} (2\lambda - 1) x^2\ +\ ax\ +\ k = 0$, 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.

4. 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, $\displaystyle \lambda d(v,x) = (1 - \lambda) d(v,y)$ just in case $\displaystyle \Sigma_{1 \leq i \leq n} (\lambda v)^2 - (\lambda x)^2 = \Sigma_{1 \leq i \leq n} ((1 - \lambda)v)^2 - ((1 - \lambda)y)^2$. But the latter equation rewrites to a quadratic equation of the form: $\displaystyle \Sigma_{1 \leq i \leq n} (2\lambda - 1) x^2\ +\ ax\ +\ k = 0$, 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 $\displaystyle x = (x_1,x_2,\ldots,x_n)$ then the equation is $\displaystyle \textstyle\sum_{1 \leqslant i \leqslant n} (2\lambda - 1) x_i^2 + a_ix_i + k = 0$. Write this as $\displaystyle \textstyle\sum_{1 \leqslant i \leqslant n} \bigl (x_i + \tfrac{a_i}{2(2\lambda - 1)}\bigr)^2 = \text{const.}$ to see that it is indeed the equation of a sphere.