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The circle below has radius 1 , and the longer circular are joining $A$ and $B$ is twice as long as the chord $A B$. Find the length of the chord $A B$, correct to 18 decimal places.
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This is somewhat of a trick question. Sorry! It seemed like a good idea at the time.

Draw a perpendicular from $O$ to $A B$, meeting $A B$ at $M$. Let $\theta=$ $\angle A O M$. Standard trigonometry shows that the length of $A B$ is $2 \sin \theta$. The shorter arc joining $A$ and $B$ has length $2 \theta$, so the longer arc has length $2 \pi-2 \theta$. The longer arc is twice the chord, and therefore
$2 \pi-2 \theta=4 \sin \theta .$ We can use the Newton Method to solve this equation as it stands. Let $f(\theta)=2 \sin \theta+\theta-\pi$. Then $f^{\prime}(\theta)=2 \cos \theta+1$, and the Newton Method recurrence is
$\theta_{n+1}=\theta_{n}-\frac{2 \sin \theta_{n}+\theta_{n}-\pi}{2 \cos \theta_{n}+1} .$
An ordinary calculator will only handle this to 8 or 9 places. The scientific calculator that comes bundled with Microsoft Windows can handle about 30 .

But we can find the answer without doing any work by looking back at a calculation done in the notes. We are solving $\pi-\theta=2 \sin \theta$. Note that by the symmetry of the sine function, we have $\sin \theta=\sin (\pi-\theta)$. Let $x=\pi-\theta$. Then our equation is equivalent to $x=2 \sin x$.

It so happens that in the notes this equation is solved to high accuracy. The positive root of this equation, to 19 places, is given there as $x=$ $1.8954942670339809471$. But the length of the chord is $2 \sin \theta$, that is, $2 \sin x$, and that is equal to $x$.

by Platinum (102k points)

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