Segments on a plane, what curve do the intersections tend to?

Question

In a Cartesian diagram, given a size $s$, suppose I create $m$ segments as such:

I connect $(0,s/m)$ with $(s,0)$; $(0,2s/m)$ with $(s-s/m,0)$; ... ; $(0,s)$ with $(s/m,0)$.

For example, if $s=4$ and $m=4$, we connect:

$\begin{align*} (0,1)&\rightarrow(4,0)\\ (0,2)&\rightarrow(3,0)\\ (0,3)&\rightarrow(2,0)\\ (0,4)&\rightarrow(1,0)\\ \end{align*}$

Now, let's consider the locus of the intersections between adjacent segments. As $m\to\infty$, the locus should define a curve.

What type of curve is it? Does it have an analytical expression and a name?

Answer 1

A Mathematica picture, first of all:

With[{m = 30, s = 2}, Graphics[Table[
     Line[{{0, k s/m}, {s - (k - 1) s/m, 0}}], {k, m}]]]

What you're asking for here is the envelope of your line segments. Without loss of generality, let's consider $s=1$ and replace the discrete parameter $k/m$ with a continuous parameter $t$, and let $m\to \infty$; thus, we want the envelope of the line segment joining the points $(0,t)$ and $\left(1-t, 0\right)$. The equation for the generator lines is

$$y=\frac{t}{t-1}x+t$$

Generating parametric equations for the envelope is easy. First, we differentiate the generators of the envelope with respect to $t$ and solve for $x$, to yield the $x$-component of the parametric equation. We obtain

$$x=(t-1)^2$$

Obtaining the $y$-component is as simple as substituting the expression for the $x$-component into the equation for the generators, and solving for $y$; this yields

$$y=t^2$$

From the parametric equations $((t-1)^2\quad t^2)^\top$, we can eliminate the parameter and obtain the Cartesian equation

$$(x-y)^2-2 x-2 y+1=0$$

which through usual means can be recognized as a parabola.