Average length of chords with a fixed end [closed]

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?

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

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 $$\frac1\pi\int_0^\pi 2r\sin(\theta/2)\ d\theta=\frac1\pi[-4r\cos(\theta/2)]_0^\pi=\frac{4r}\pi$$

• Right now I'm struggling to understand why it is $2r\sin\theta$ and not $2r\sin(\theta/2)$ Mar 15, 2023 at 8:55
• @evargalo Apologies, you are correct. I shall amend. Mar 15, 2023 at 8:58

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

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