# 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) – tired Dec 20 '16 at 14:26
• I'm shocked ! WOW!!! – Arman Malekzadeh Dec 20 '16 at 14:58
• – Michael Hoppe Dec 20 '16 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! – Bob Pego Dec 20 '16 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. – Bob Pego Dec 21 '16 at 17:14