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From a specific point A on a circle's circumference, am drawing various chords. a) what is the sum of length of such chords and b) what is the average length of such chords?

Assuming a radius r, I'm trying to solve a) in the below manner:

One specific chord length = 2rCos(Theta)

To get the sum, I'm integrating over -pi/2 to pi/2 ( 2rCos(Theta) dTheta )

But i don't think this is right. Please point me in the correct direction.

Thanks.

LostAgain
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  • The sum of the lengths of the chords is heavily dependent on how many chords you draw, and how. For the interesting question of average length, we need to decide how the second point is chosen. Is the angle uniformly distributed? Is the second point uniformly distributed? In geometric probability in particular, one has to be very careful to specify precisely. – André Nicolas Nov 15 '13 at 05:52
  • there are infinite such chords. – LostAgain Nov 15 '13 at 23:25
  • For your second question, it was clear that you intended that all the chords through $A$ be drawn. However, my question has to do with how they are generated. Do we pick the second point with uniform distribution over the circle? Do we pick its $x$-coordinates uniformly? Or perhaps the $y$-coordinates? Or do we pick say $A=(r,0)$, and choose the angle that $PA$ makes with the $x$-axis uniformly from $-\pi/2$ to $\pi/2$? The choice we make will strongly influence the answer. – André Nicolas Nov 15 '13 at 23:33
  • infinite chords can be drawn within the circle, with each chord starting from point A (on the circle's circumference) and ending on another point on the circumference. am not sure how to upload images from my pc, else I could have better illustrated it. – LostAgain Nov 16 '13 at 01:42
  • I have written out a solution of the problem, making what I consider a reasonable assumption about the way the second endpoint is generated. – André Nicolas Nov 16 '13 at 02:09

2 Answers2

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There are many possible answers, depending on the probability distribution of the parameter that defines the other endpoint of the chord. That is a standard issue in problems of geometric probability.

We give one interpretation, and derive the answer under that interpretation. Other interpretations will yield a different answer.

Imagine that we select the other endpoint $B$ by choosing $\theta$ with uniform distribution in the interval $[0,2\pi]$, and letting $B$ be the point obtained by rotating the point $A$ counterclockwise through the angle $\theta$.

Then the length $X$ of the chord $AB$ is, by basic trigonometry, $2r\sin(\theta/2)$. It follows that $$E(X)=\int_0^{2\pi} 2r\sin(\theta/2)\frac{1}{2\pi}\,d\theta.$$ Integrate. We get $$E(X)=\frac{4r}{\pi}.$$

André Nicolas
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If you imagine many chords of a circle arranged in a stack, with appropriate lengths, it can be made into a circle. so, the total of all the chords will give us the area of the circle. This can be easily proved and imagined. For example, take a rectangle. Have you ever wondered why the area is l x b?? It is because we are adding the length breadth number of times. Hence applying the same logic to a circle, we can understand that if we add the lengths of all the chords in a circle we would get the circle's area..pi*R^2. The total number of chords would be equal to 2*R as the length of the circle is 2*R. Hence the average chord length = (pi*R^2)/(2*R). => pi*r/2

Aditya Ravindran
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