For a while now, I've been looking for a proof of the following fact, so I can include it in my educative Master's thesis:
"The caustic of a deltoid, where the light source source is at infinity, i.e. the incoming light rays are parallel, is an astroid (no matter what the direction of the incoming light rays is)."
I have been trying to use established formulas to determine a caustic, but the parametrisations I know for the deltoid seem to produce enormous equations and I am looking for a more elegant argument, e.g. using the (classic) geometry of the deltoid and the astroid, which I understand perfectly (see e.g. Lockwood, "A Book of Curves"), or the fact that they are hypocycloids.
Interestingly, this is only the case for the astroid and deltoid 'couple', the caustic of an astroid has little to do with a hypocycloid with five cusps for example.
I'm not asking for a full proof or anything, but if people happen to be able to point me in the right direction I would be very grateful. Thanks in advance!
The deltoid is the locus of a point $P$ belonging to a circle, which rolls without slipping inside another circle with triple radius. If we take the larger circle of radius $3$ and centered at the origin $O$ of a coordinate system, the smaller circle has unit radius and its center $C$ lies on a circle of radius $2$.
If we set $\theta=\angle AOC$ (where $A=(3,0)$), then $\angle PCH=3\theta$. From that one can easily found the coordinates of $P$ as a function of parameter $\theta$: $$ P_x=2\cos\theta+\cos2\theta,\quad P_y=2\sin\theta-\sin2\theta. $$
Tangency point $H$ between the circles is the instantaneous center of rotation: it follows that line $PH$ is the normal to the deltoid at $P$. A light ray $SP$, parallel to the $x$-axis, is then reflected to a ray $PN$ such that $\angle HPN=\angle SPH$. A simple calculation shows that line $PN$ forms an angle $\angle PMO=\theta$ with the $x$-axis.
If the reflected line $PN$ intersects the $x$-axis at $M$ and line $x=-1$ at $N$, a straightforward computation gives: $$ O'M=P_x+P_y\cot\theta+1={4}\cos\theta, \quad O'N=P_y+(P_x+1)\tan\theta={4}\sin\theta, $$ where $O'=(-1,0)$. It follows that segment $MN$ has a fixed length of $4$.
But it is a well known result that the envelope of a segment of fixed length, with its endpoints sliding on the sides of a right angle, is an astroid (dotted in the diagram), with radius equal to the length of the segment.