Prove that the median level of the dual of a set of non-degenerate points is formed by the duals of all points in the set.

Question

Given a non-degenerate set of points, if we take the dual of each point and look at the median level of this line arrangement, how can one prove that the median level includes the duals of all the points at some point. Note that in this case, point (a, b) has the dual y=ax-b and the median level is a piece-wise linear function in the dual plane that has an equal number of lines above and below it at all times.

Answer 1

First of all, an example of a MLL (Median Level Line) for a family of $n=5$ lines :

The MLL is the red dotted broken line. Please note that on the left, the vertical order of the 5 lines : blue, brown, orange, purple, green, is inverted on the right. The MLL coincides with the median line (orange) at the beginning and at the end.

Let us write the equations of lines $(L_k)$ under the form :

$$(L_k) : \ \ \ \ y=a_kt+b_k, \ \ k=1,2,\cdots n, \ \ \text{where} \ n=2m+1.$$

We have taken variable $t$ instead of $x$ because a kinematics interpretation will help. Consider $n=2m+1$ runners on a road. Let us interprete abscissa as time and ordinates as position. In this way, assuming that person n° $k$ runs at the uniform speed $a_k$, starting at position $b_k$ on the road, the set of lines can be interpreted as the time-space trajectories of these runners. We consider that the lines are in general position, therefore that all speeds $a_k$ are different ; WLOG, we can index the lines in such a way that, up to a time-shift, we have

$$a_1 > a_2 > \cdots > a_n \tag{1a}$$

and

$$b_1 < b_2 < \cdots < b_n \tag{1b}$$

(consider that the runners are given a "handicap" by letting them start from a position $b_k$ with a "remoteness" taking into account their speed $a_k$ as can be seen on the graphical representation).

Let $t_{p,q}$ the (unique) time instant where runners n° $p$ and $q$ ($p \ne q$) are situated simultaneously at the same point of the road. Let $t_{min}=\min{t_{p,q}}$ and $t_{max}=\max{t_{p,q}}$.

With our assumptions (1a) and (1b), we have $t_{min}>0$. We will take, in a symmetrical manner, an ending time $t_{end}>t_{max}$. Let us consider the median runner $R_m$ (which is, at starting time the one with median speed). For $t<t_{min}$ (and even some time later), it is this medial line $(L_m)$ which defines the MLL. At times $t>t_{end}$ (but hopefully before), there is again complete coincidence between $(L_m)$ and the MLL (as can be seen on the graphic).

Meanwhile, what happens ? It is like in a relay race : runner n° $m$ gives its "baton" (stick) to the first runner he meets (either "hit" by him or being "hitted" from the rear) ; on its turn, the new holder of the baton will give it to the runner he will meet next, and so on... until the baton comes finally back into the hands of the initial holder n° $m$...

This establishes the existence of this red line in this context (an existence that can be obtained of course by different means).

Now, the final touch. Why are all other lines $(L_k), \ k \ne m$ sharing at least a common segment with median line MLL ? Because this line divides the plane into an upper $U$ region and a lower $L$ region and that all these lines have one of their endpoints (for $t=t_{min}$ and $t=t_{max}$) in $L$ and the other in $U$. This is a topological explanation.

But we can say it with our kinematic framework.

Any runner n° $p$ that runs faster than the runner n° $m$ (which means that $p>m$), although having started earlier will progressively reach a certain number of the runners. Assuming that it never reaches the baton-holder, it would mean that, at the end, he stays behind runner n° $m$ : contradiction. Same reasoning for a runner who is running slower than runner n° $m$.

Remark : we can assume all the slopes are $\ge 0$. If this is not the case, we can replace all of them by $y=(a_k-a_{min})x+b_k$ where $a_{min}$ is the smallest slope. This linear operation preserves the median line in the sense that one retrieves the original median line by doing the inverse operation.

Remark : Interesting things can be found into this document..