0
$\begingroup$

I'm given that the distance between two points in the Beltrami-Klein model is $$d(XY)=\frac{1}{2}ln\Big(\frac{\overline{XQ}\cdot\overline{YP}}{\overline{XP}\cdot \overline{YQ}}\Big)$$ where $P$ and $Q$ are ideal points lying on the boundary of the unit disc, and $\overline{XQ}$ denotes the standard Euclidean distance between a point $X$ inside the unit disc and an ideal point $Q$.

Given the ideal points $P=(0,1),Q=(0,-1)$, and points $A=(0,0)$ and $B=(0,\frac{1}{2})$. I am asked to find the midpoint $M=(0,m)$ between $A$ and $B$.

I was able to successfully compute $d(AB)=\frac{1}{2}ln(3)$. However, I'm not sure how to actually compute the midpoint of the given points.

I was given a solution which states if we could calculate $d((0,0),M)$, then set that solution equal to $d(AB)$, we would be able to obtain the midpoint. But, if you do this you end up just computing $d((0,0),M)=B$ which makes no sense in determining the midpoint.

Is there a general formula to compute the midpoint of a given line in the Klein model? Also, why is it that it's sufficient to measure the distance between the ideal points and ordinary points in the model with the standard Euclidean distance?

$\endgroup$
0
$\begingroup$

Assume that $M$ is the midpoint of $XY$. Then $d(M,X)=d(M,Y)$ has to hold, so: $$ \frac{XQ\cdot MP}{XP\cdot MQ} = \frac{MQ\cdot YP}{MP\cdot YQ} $$ has to hold, and we must have: $$\left(\frac{MP}{MQ}\right)^2 = \frac{XP\cdot YP}{XQ\cdot YQ} $$ so it is quite easy to construct the midpoint of $XY$ with straightedge and compass, but in general it is not the "euclidean" midpoint $\frac{X+Y}{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy