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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?

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As a side note, if anyone can suggest an online resource where I could plot this problem, I will gladly include a picture. – Sklivvz Nov 15 '11 at 21:07
You should look into the Astroid: – A Walker Nov 15 '11 at 21:11
@AWalker thanks, this is it: – Sklivvz Nov 15 '11 at 21:21
Note that the astroid (as noted on that MathWorld page) is the result in the case where the segments between the $x$ and $y$ axes have constant length; that's not the case you were asking about. – joriki Nov 16 '11 at 4:44
up vote 2 down vote accepted

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


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


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


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.

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The final equation can also be written as $(x-1)^2+(y-1)^2-2xy=1$ which is manifestly invariant under $x\Leftrightarrow y$ but not as obviously a parabola. Take your pick. – Ross Millikan Nov 16 '11 at 3:50

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