# Determine the bounding curve of a moving circular sector

analytic geometrycurveseuclidean-geometrygeometryproblem solving

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?

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 $$-a$$.

Explanation about 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...