**Preamble:** Consider the curve $\mathcal{C}$ formed by the segments $CD$, $DE$, $EB$ and the arc $BC$ presented below:

Next, let $F$ be a "perspective" point and $l$ a line segment whose other end point is $A$. Suppose that we draw similar curves for each point $T$on the line segment $l$, such that the bottom segment $DE$ in the picture is orthogonal to the ray originating from $F$ and passing through $T$. Below is my best attempt to graph that this looks like at the end points of a segment from $(0, 0)$ to $(2, 0)$. (N.B., the distances are not exact in the picture, as I don't know how to use GeoGebra well).

**Question:** I would like to determine the bounding curve for the shape which is obtained by the moving/painting area. That is, what is the minimal curve $\mathcal{M}$ which contains all of the "moving" curves $\mathcal{C}$?

It is quite clear that portion of the bounding curve are given by the segments $CD$, $DA$ of the leftmost area, and by the segments $AE$, $EB$ of the righmost area. What I'm struggling with is how to determine the bounding shape of the moving arc given that the line $l$ (the line segment which curve $\mathcal{C}$ moves along) and the perspective point $F$ can be be arbitary?

## Best Answer

A point of vocabulary: The generic name of this kind of "bounding curve" is "envelope".

This is not a full answer: it gives, under the parametric representation (1) the biggest part of the equation of the upper envelope.

Take a look at the following picture (obtained with Desmos: you can see an animated version here.

The locus of $C$ is the red curve, the locus of $D$ is the purple curve.

The parameters are

the radius $r$ of the circular arc,

the aperture angle $2a$ of this arc,

the ordinate $b$ of point $F$.

The locus of point $C$ is given under the form of parametric equations that can be seen on the upper left of the figure:

$$\begin{cases}x&=&r\cdot\cos\left(\arctan\left(\frac{-b}{t}\right)+a\right)+t\\y&=&r\cdot\sin\left(\arctan\left(\frac{-b}{t}\right)+a\right)\end{cases} \ \text{where} \ t \ \text{is the abscissa or the arc's center}\tag{1}$$

For point $D$ change $a$ into $-a$.

Explanationabout the angles found in (1): let $N$ be the midpoint of arc $LM$. As points $F,G,N$ are aligned, if we denote by $t$ the abscissa of $G$:$$\text{angle of GN w.r.t.} \ x \ \text{axis} \ = \ \angle AGF = \arctan \frac{AF}{AG}=\arctan \frac{-b}{t}$$

To this angle, we add (resp. subtract) angle $a$ in order to obtain point $M$ (resp. point $L$).

The animation shows that, in the first quadrant, it is essentially this point $C$ that governs the "envelope", with an exception around the origin (in the animation here it is the case for $x<0.5$).

This disruption, calling a specific analysis, slightly increases when $a$ becomes important (say $>\pi/4$). The analysis of this exceptional zone looks possible but more complicated...

Waiting for your reactions.