# Construct point on a circle such that the reflection in that point is horiztonal

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 $$Q$$.

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

• cool picture (+1) Commented Dec 20, 2016 at 14:26
• I'm shocked ! WOW!!! Commented Dec 20, 2016 at 14:58
• Commented Dec 20, 2016 at 15:04
• 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! Commented Dec 20, 2016 at 18:07
• 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. Commented Dec 21, 2016 at 17:14

Starting from @Narasimham's nice solution, his/her last equation can be solved analytically.

Using the tangent half-angle substitution $$\theta_Q=2 \tan ^{-1}(t)$$, the equation write $$y_P \,t^4+ 2( a+2 x_P)\,t^3-6y_P\, t^2+2 ( a-2 x_P)\,t+y_P=0$$ which can be solved with radicals.

"More than likely", this equation has two distinct real roots and two complex conjugate non-real roots since $$\Delta=-256\left(a^6+3\left(5 y^2-4 x^2\right)a^4+48 \left(x^2+y^2\right)^2a^2-64 \left(x^2+y^2\right)^3 \right)$$

• Fine. Can impossibility of construction by R&C be established similar to what is done for the angle trisection ? Commented Feb 25, 2020 at 19:21

Yes, we can. It seems to me a locus finding problem in reflection geometry. Incident ray direction and incident ray segment length can be taken $$2 \theta$$ anti-clockwise and constant $$b$$ respectively when reflected ray QR is horizontal.

Directly from the figure at left parametric equation of locus of P of two components from $$(a=1,b)$$ is the spiral

$$(x_P,y_P)=(a\cos\theta_Q+b\cos 2\theta,\,a\sin \theta_Q + b\sin 2\theta_Q)\tag 1$$

which ( appearance similar to Limacon ) is an algebraic curve of fourth degree (not converted here to cartesian coordinates ) sketched below:

When $$b$$ segment length is perturbed the yellow band is obtained with rulings of all possible incident rays.

For Ruler and Compass choose a point R, draw a line parallel to x-axis to cut mirror M at Q which is the unique point. Drawing a perpendicular to OQN, transferring angle $$\theta$$ to left of normal OQN by cutting off arc length QR with compass to find P, is simple construction.

EDIT1:

If the above is an indirect method we can directly find coordinates of Q:

Using above parametrization and from the condition that the angular bisector of QP, QO is QR we arrive at the required unique $$\theta_Q$$ implicit condition to be satisfied for given P$$(x_P,y_P).$$ Calculation detail is omitted.

$$y_P \cos 2 \theta_Q- x_P \sin 2 \theta_Q + a \sin \theta_Q =0 \tag2$$

Numerical solution by Newton-Raphson can determine $$\theta_Q$$.

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: 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.