Euclidea 3 9.8 Chord Trisection

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Construct a chord of the larger circle through the given point (on circumference of larger circle) that is divided into three equal segments by the smaller circle (circles are concentric)

I'm having trouble finding a method to solve this geometric problem (construction). Google translation of a Russian discussion suggested using Thales theorem and parallels. Any thoughts?

enter image description here

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AC is the chord to be trisected.

AB is just any supporting line.

Circles 1, 2 and 3 are used to make 3 equal sections on AB such that $AB_1 = B_1B_2 = B_2B_3$.

$B_2Q$ and $B_1P$ are lines parallel to $B_3C$. They will create 3 equal intercepts on AC such that $AP = PQ = QC$.

OP is then the radius of the required circle.

enter image description here

I mis-interpreted the question but I will leave that on for future reference. The following is another attempt.


WLOG, we can let one end point of the required chord (of circle O) be at A, as shown. enter image description here $AP’$ is just any chord on the green circle with midpoint $X’$.

As $P’$ moves along the green circle, it can be shown that the locus of $X’$ is the dotted circle (touching the green circle internally and with radius = R/2).

Suppose that $AP$ is the required chord such that it cuts the red circle at U and V where AU = UV = VP. Then AU : UM = 2k : 1k.

Let AT be the tangent to the red circle. Then, by the power of A, we have $AT^2 = (2k)(4k)$. That is $AT =2\sqrt 2 k$.

This means $AT : AU = \sqrt 2 : 1$. U is therefore located because AT is fixed.

In short, the way to create the required chord through A (on the outer circle) is:-

(1) From A, draw a tangent to the inner circle touching it at T.

(2) Create a right isosceles triangle AZT using AT as hypotenuse.

(3) Let the circle (centered at A, radius = AZ) cut the inner circle at U.

(4) Draw the line AUVP cutting the inner circle at U, V and the outer circle at P.

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Solution

Here is the solution to the problem for those interested. Thank you for the help people.