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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?

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  • The power is constant. So, if for example, you construct the chord that is perpendicular and it has length $2c$. Then you need to solve for $a$ in $2a^2=2a\cdot a= c^2$ to get the length of one of the intervals of the sought chord. Once you have it draw the circle with radius $a$ and center $P$ and the intersections with the circle gives you the possible locations of one of the end-points of the chord. Do you know how to solve quadratic equations geometrically or at least this particular one? Think of a square and its diagonal. – conditionalMethod Nov 08 '19 at 20:38
  • Do all chords pass through a single point on the circle? – Narasimham Nov 09 '19 at 21:15

2 Answers2

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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 $x$.

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    Computing $2x$ such that $(2x)^2=(r+r')(r-r')$ can be done in this way: Draw the segment $r+r'$ and extend it further by $r-r'$. Call $M$ the point where one interval ends and the other begins. Draw a circle with the long segment of length $2r$ as diameter. Lift the perpendicular to this segment at $M$. The segment from the intersection of this line with the circle to $M$ has length $2x$. – conditionalMethod Nov 08 '19 at 20:54
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    What this is using is that the square of the altitude of a right-angle triangle to its hypotenuse is the product of the two segments in which the hypotenuse is divided by the base of the altitude. – conditionalMethod Nov 08 '19 at 20:57
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    @conditionalMethod Thanks. I just added a link to the video solution that shows how $x$ can be constructed just by choosing some point on the boundary of the circle. Take back everything I said about it being gross -- I am evidently a newb at compass-and-straightedge constructions! –  Nov 08 '19 at 21:04
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    The step that can fail in the construction is when intersecting the circle of radius $x$ with the given circle. They could not intersect. For them to intersect one must have $r-r'\leq x\leq r+r'$. For example, when $P=O$ we have $r'=0$ and then it requires $r\leq r/\sqrt{2}\leq r$. – conditionalMethod Nov 08 '19 at 21:04
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    Feel free to roll hack my edit where I prove the condition for the construction to hold. It's not very surprising, really. – Oscar Lanzi Nov 08 '19 at 23:51
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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. $\square$

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