What is the probability distribution function of the distnace between the centre of a regular hexagon of radius R and any point within it?

I have done the following and would appreciate if you could criticise my ideas:

Assuming a hexagon with radius $R$, and the distance between the centre and any point in the hexagon to be denoted by $d$ we can write the pdf in two parts:

$P(d)=$ (circumference of a circle with radius $d$)/ Hexagon area, $d<\frac{\sqrt(3)}{2}*R^2$

$P(d)=$ (6 * arch length)/ Hexagon area, $\frac{\sqrt(3)}{2}*R^2<d<R$

Green area describes the first part, while the red area describes the second

  • $\begingroup$ What attempts have you made at solving this problem? Are the difficulties conceptual (setting up the calculation) or are they mechanical (computing the relevant areas)? $\endgroup$ – whuber Dec 2 '14 at 21:27
  • $\begingroup$ See the edit :) $\endgroup$ – Pioneer83 Dec 2 '14 at 21:47
  • $\begingroup$ the distribution of the distance would depend on the distribution of points within the region. You don't specify one. One might assume uniformity, but if that's what you mean, you should be explicit. $\endgroup$ – Glen_b -Reinstate Monica Dec 2 '14 at 23:40

In closed form this will be : $P(d)=\left\{\begin{array}{cc} \frac{4\pi{d}}{3\sqrt{3}R^2} & 0\leq{d}\leq\frac{\sqrt{3}{R}}{2} \\ \frac{4\pi{d}-24d\cos^{-1}{\frac{\sqrt{3}R}{2d}}}{3\sqrt{3}R^2} & \frac{\sqrt{3}{R}}{2}\leq{d}\leq{R} \\ \end{array}\right. $

Integrating the function will give 1.0

  • 1
    $\begingroup$ I worked through it and I agree with your answer. $\endgroup$ – Flounderer Dec 3 '14 at 4:34

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