construction of a chord that is trisected by a point

Question

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?

Answer 1

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 $x$.

Answer 2

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. $\square$