construction of a chord that is trisected by a point 
Find a construction of a chord through a point P such that
  P divides the chord in the ratio 1:2 in any given circle.

Cleary not all points P work, so I'm trying to find the construction when it is possible. What is invariant about all such chords that would be useful in finding a construction?
 A: 
Assume $P$ is a point where such a chord (labeled here as $\overline{QR}$ could be constructed. Draw a circle with center $O$ that passes through $P$.  By symmetry, $A$ is the other chord trisector.  Take $x=QP=PA=AR$.
Let $r$ be the radius of the larger circle and $r'$ the radius of the smaller circle.  Also draw $\overline{QBO}$ and extend it to $C$.  By the intersecting secant theorem: $$QB\cdot QC=QP\cdot QA\\(r-r')(r+r')=x(2x)\\r^2-r'^2=2x^2\\x=\sqrt{\frac{r^2-r'^2}{2}}$$
So $x$ is constructable from $r$ and $r'$ (I remember doing it for a level in Euclidea and it was gross but not impossible), and drawing a circle with radius $x$ centered at $P$ will identify $Q$ on the outer circle.
To find this point $Q$ in the plane we have to have the calculated distance $x$ match or exceed the gap from $P$ to the given circle.  To wit,
$x=\sqrt{\frac{r^2-r'^2}{2}}\ge (r-r')$
Square both sides of the inequality and factor the difference of squares:
$\frac{(r+r')(r-r')}{2}\ge (r-r')^2$
$r+r'\ge2(r-r')$
Thence
$3r'\ge r$
This says that $P$ must be on or outside the circle concentric with the given one and having radius one-third as large, a constraint we expect on intuitive grounds.
ETA: https://www.youtube.com/watch?v=KxnrR_Dg8Tg is a walkthrough of the Euclidea level I mentioned.  The goal of that is "backwards" in that they are trying to find $P$ given $Q$, but they do the same job of constructing $x$.
A: 
Define $r := |OR|$ and $p := |OP|$.
Note that, by the Power of a Point theorem,
$$-|PX||PY|= \operatorname{pow}P := p^2-r^2 \tag{1}$$
We also have $|OP'|=r^2/p$ (why?) and $$|PQ|=\frac14(|OP'|-|OP|)=\frac1{4p}(r^2-p^2) \tag{2}$$
Then,
$$\begin{align}
|PX|^2 &= |PQ|^2 + |QX|^2 \tag{3}\\
&= |PQ|^2 + r^2 - ( p + |PQ| )^2 \tag{4}\\
&= r^2 - p^2 - 2 p |PQ| \tag{5}\\
&= \frac12(r^2 - p^2) \tag{6}\\
&= \frac12 |PX||PY| \tag{7}
\end{align}$$
So that $|PY| = 2|PX|$, as desired. $\square$
