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The problem is pretty simple to state. Draw a circumference of radius r, and fix a point on the perimeter. The question is: What is the average length of all the chords defined by that fixed point and any other random point on the circumference?

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  • $\begingroup$ Agreed with answers. "Random" is anything but. Why would anyone close this?! $\endgroup$
    – humn
    Mar 15, 2023 at 11:46

2 Answers 2

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The question depends on how you pick a random point. If we pick uniformly so that the angle between the fixed point, the centre and the random point is uniform then the answer is

$\dfrac{4r}\pi$

To see this let the angle above be $\theta$ then

The length of the chord is $2r\sin(\theta/2)$ and the average is $$\frac1\pi\int_0^\pi 2r\sin(\theta/2)\ d\theta=\frac1\pi[-4r\cos(\theta/2)]_0^\pi=\frac{4r}\pi$$

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    $\begingroup$ Right now I'm struggling to understand why it is $2r\sin\theta$ and not $2r\sin(\theta/2)$ $\endgroup$
    – Evargalo
    Mar 15, 2023 at 8:55
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    $\begingroup$ @evargalo Apologies, you are correct. I shall amend. $\endgroup$
    – Daniel S
    Mar 15, 2023 at 8:58
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We can use the law of cosines to arrive at the length of the chord that subtends an angle of $\theta$ at the center. We would just have to average it over the whole circle.

The average is

$$M=\frac{\int\limits_{0\le \theta\le 2\pi}\sqrt{2r^2-2r^2\cos(\theta)}d\theta}{2\pi}$$ $$\implies M=\frac{\int\limits_{0\le \theta\le 2\pi}\sqrt 2 r\sqrt{1-\cos(\theta)}d\theta}{2\pi}=\frac{8r}{2\pi}=\frac{4r}{\pi}$$

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