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