# Circle Chord Sequence

This is my first post, so be nice!

When I was in my first Geometry class in high school, I asked the teacher the following:

Given a circle of radius 2a, find the length of the chord running parallel to the diameter of the circle such that the semicircle cut by the chord is divided into two regions of equivalent area.

The teacher looked at me as if he knew the answer but then paused, puzzled, and told me that we should work on it after class. So we did, and it was at this moment that I was introduced to trigonometry. Yet we did not find a solution and the problem was filed into the back of my memory for some time. A year and a half later, at the end of Algebra 2/Trig, the problem re-entered my awareness and I took another crack at it. After maybe an hour of work I found the solution! I was so excited that I asked my teacher if I could present it to the class on the second to last day of school, and she consented. So I presented it, but crap I forgot my notes and I screwed up the work leading to the solution and embarrassed myself in front of everybody. I cleaned it up after class when I had more time, but by then only the real math enthusiasts were left. Anyway, that was completely tangential, as I would now like to pose the question that I came here with:

Given a circle of radius 2a, find the sequence of real numbers given by the length of successive iterations of slicing the segment resulting from a chord which runs parallel to the diameter of the circle such that the area of the segment is halved with each successive slice.

Oh, and the answer to the first question is something like 0.71... or 0.79..., I have forgotten by now. Bonus points to the most elegant solution of the first problem.

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Wlog. the radius is $1$. The area of the cap subtending angle $\alpha$ from the center is $A(\alpha)=\frac12(\alpha-\sin\alpha)$. The corresponding segment length is $2\sin\frac\alpha2$. You are asking for the sequence $(x_n)_{n=1}^\infty$ such that $x_n=2\sin\frac{\alpha_n}2$ and $A(\alpha_n)=\frac\pi2\cdot2^{-n}$. This can only be solved numerically. The first few approximations (after $x_0=2$) are: $$1.8295420351460716385409955906837076169, 1.5455097261234120202206332289590795207, 1.2678613848545969240544214118382796268, 1.0255171578787055775354919807723448528, 0.82317890720583251879913780639731025539, 0.65785871279582689026079790969165393609, 0.52435985537171345597016857982005882625, 0.41728294853497515917670605958970406336, 0.33174364554487851240387578246924654023, 0.26357709084095859727510715888869346114, 0.20933695838483619922846766000771974415, 0.16621859822603799414074748448500084143, 0.13196163045501028275182369205869205768, 0.10475492671579429585769886806519067206, 0.083152493022244344061029851545052090070, 0.066002402792776475648886352412199553896, 0.052388254185767766597240438200662186974, 0.041581640994898317827871218807935473777, 0.033003898345400309144536032806404103535, 0.026195475475160737053195360050566833997.$$ For big $n$ (that is small $\alpha$) one can employ the approximation $A(\alpha)\approx \frac1{12}\alpha^3\approx\frac1{12}x^3$ (from $\sin x=x-\frac16x^3+\frac1{120}x^5\mp\ldots$), hence $$x_n\approx \sqrt[3]{12\cdot\frac\pi2\cdot 2^{-n}} =\frac{\sqrt[3]{6\pi}}{(\sqrt[3]2)^n}.$$ For $n=20$, this approximation gives us $0.02619592\ldots$ instead of the last value in the list above.

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The area of a segment of a circle of radius $r$ subtending an angle $\theta$ at he center is

$$\frac12 r^2 (\theta - \sin{\theta})$$

For the initial segment (one-half a semicircle), the angle $\theta_1$ satisfies

$$\theta_1-\sin{\theta_1} = \frac{\pi}{2}$$

Subsequent angles satisfy

$$\theta_{n+1}-\sin{\theta_{n+1}} = \frac12 (\theta_{n}-\sin{\theta_{n}})$$

All of these equations require some numerical solution. The first equations yields $\theta_1 \approx 2.30988$. The length of the chord is then $2 r \sin{\theta_1/2}$, which, for $r=1$ is $\approx 1.82954$.

Note, however, that you can deduce the behavior of the chord length for large $n$ by observing that

$$\theta_n-\sin{\theta_n} \sim \frac{\theta_n^3}{6}$$

Then

$$\theta_{n+1} \approx 2^{-1/3} \theta_n$$

which shows that, with each successive iteration, the chord length decreases by a factor of $2^{1/3}$, or about $26\%$.

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