12
$\begingroup$

Let $P$ be a point in the plane outside the unit circle. There is a unique point $Q$ on the circle such that a light ray from $P$ is reflected in the circle at $Q$ and emerges parallel to the $x$-axis. Is it possible to construct $Q$ using a ruler and compass?

If $P$ is a point on the piece of paper in this picture, the image of $P$ will appear to an observer at infinity to be at a point easily constructed from $Q$.

cylindrical mirror

I would also be interested in a proof that such a construction is impossible.

$\endgroup$
  • 1
    $\begingroup$ cool picture (+1) $\endgroup$ – tired Dec 20 '16 at 14:26
  • $\begingroup$ I'm shocked ! WOW!!! $\endgroup$ – Arman Malekzadeh Dec 20 '16 at 14:58
  • $\begingroup$ Related: en.wikipedia.org/wiki/Anamorphosis $\endgroup$ – Michael Hoppe Dec 20 '16 at 15:04
  • $\begingroup$ If $P$ corresponds to the complex number $z=x+iy$ with $x>0$ or $y>1$, and $Q$ corresponds to $w$ with $|w|=1$, then you require $$w^2 = \frac{z-w}{|z-w|}.$$ I've no idea if $w$ is constructible. Nice question! $\endgroup$ – Bob Pego Dec 20 '16 at 18:07
  • $\begingroup$ Take $w=s+it$ in my previous comment, multiply by $|z-w|\bar w^2$ and take imaginary part. Then because $w\bar w=1$ we require $$0=\Im (z\bar w^2-\bar w)=y(s^2-t^2)-2xst+t, \quad s=\sqrt{1-t^2}.$$ Moving $2xst$ over and squaring gives a quartic polynomial in $t$. Typically solving such a polynomial requires taking a cube root, which may not always be constructible. This is not a proof, but it seems probable there is no general construction. $\endgroup$ – Bob Pego Dec 21 '16 at 17:14
0
$\begingroup$

I don't have an answer to your question (which I like a lot!), but I have two thoughts, the first related to @Bob Pego's comment: at a purely algebraic level, the requirement that the normal vector split the ray and its reflection is satisfied by four points on the circle in general (corresponding to Bob's 4th degree equation, and showing that this equation didn't introduce extraneous roots). Here's a (very rough) picture of the situation, where each solution is colored differently: enter image description here The red dot is P; the four circle-intersections are (I hope) clear.

The second comment is that whether perhaps the question you asked is the wrong one: if we actually want to construct an anamorphic image, we already KNOW what we want the picture in the cylinder to look like. From an eye-point E, we can trace to the arropriate cylinder point and reflect to determine where on the canvas to put the appropriate mark.

If, on the other hand, you want to make an anamorphic-image-viewer, where you start with the colored canvas and compute the anamorphic image that will be displayed, then your question is the right one. To do that task in practice is relatively simple: you do the thing I described in the first case for a few hundred points in a grid. Each little rectangle in your grid is mapped almost linearly, so you extend the (inverse) mapping by bilinear interpolation on a cell-by-cell basis. I suspect it can be done easily in real time on a modern graphics card...but that takes us far from the domain of constructability with ruler and compass.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.