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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

Andriy K
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Jeremy Corrin
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  • You have to say something about the ratio $R_1/R_2$ of the circles' radii. Because, if the radius of the inner circle is to small, there is no solution to the problem... – Jean Marie Jul 03 '16 at 08:43
  • Thus you must assume that $R/r \leq 3$. One more question : are you looking for a straight edge and compass only solution ? – Jean Marie Jul 03 '16 at 08:50

2 Answers2

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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.

Mick
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  • I believe you misunderstood the problem. You are given two circles, and the goal is to find a chord which is trisected by them, – Wojowu Jul 03 '16 at 09:28
  • @Wojowu Have uploaded another version. – Mick Jul 03 '16 at 16:36
  • @JeremyCorrin Have uploaded another version. – Mick Jul 03 '16 at 16:37
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    Looks good, though it might help readers if the construction for the question asked is given first, and the "misinterpreted but kept for future reference" part appears at the end of the answer. That is, rather than, "Here's an answer ... several inches of text/figure ... oops, it should be ... ," let the answer say, "Here's the answer ... figure and text ... and by the way, the old answer (before I corrected my misunderstanding) was ... ." – David K Jun 01 '17 at 17:39
  • @DavidK Thanks for the advice. I am going to divide the two versions clearly. – Mick Jun 01 '17 at 18:11
  • Why does $AT^2=(2k)(4k)$ hold? – T. Verron May 24 '18 at 08:03
  • @T.Verron The "power of a point" says the product of the two secants from a point A (i.e. $\ AU \times AV$) is equal to the product of the tangents from A (i.e. $AT \times AT$). – Mick May 24 '18 at 13:04
  • Wow... that's got to be dark magic. Thanks! – T. Verron May 24 '18 at 18:22
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Solution

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

Jeremy Corrin
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