related question: probablity of random pick up three points inside a regular triangle which form a triangle and contain the center

What is the probability that a (possibly degenerate) triangle made by three randonly chosen points on the perimeter of an n-gon contains the centre of an n-gon?

For a square, there is a $\frac{1}{16}$ chance that the points are in configuration a, $\frac{3}{16} $ for configuraion b, and $\frac{3}8$ for c and d. The probility that the points contain the center is $0$ for a and c, $\frac{1}3$ for b (since the center is contained iff one point is on each side of the line TF1 and arbitrarily taking the square to have unit sides yields $2\int_0^1 a-a^2 \mathrm{d} a=\frac{1}3$) and $\frac{1}2$ for d (center contained iff B1 is the opposite side of the line through D1H1 to C1, $\int_0^1 a \mathrm{d}a=\frac{1}2$).Therefore, if I have somehow not made an error, the probability is $\frac{1}4$. enter image description here

[edited] The limiting case of a circle is $\frac{2}{\tau}\int_0^{\frac{\tau}2}\frac{a}{\tau}\mathrm{d}a=\frac{1}4$ (using $\tau=2\pi$ just to be controversial)

  • $\begingroup$ I presume by "$n$-gon" you mean a regular $n$-gon? $\endgroup$ – joriki Jul 18 '12 at 11:23
  • $\begingroup$ Please mark edits as such visibly if they make part of an existing answer look wrong. Thank you. $\endgroup$ – joriki Jul 19 '12 at 5:48

The answer is in fact always $\frac14$, independent of $n$; you forgot to multiply by $2$ for symmetry and divide by $\tau$ for normalization in the circle result.

Wherever the first point is chosen, the "diameter" on which it lies divides the $n$-gon into two symmetric halves, and the second and third points must be in opposite halves. The line connecting them must also lie above the centre (as seen from the first point), and if the second point is at a distance $x$ from the first point along the perimeter (in units of the length of the perimeter), there's an admissible range of length $\frac12-x$ for the third point. Thus the probability is

$$2\int_0^{1/2}\left(\frac12-x\right)\mathrm dx=2\int_0^{1/2}x\,\mathrm dx=\frac14\;,$$

where the factor of $2$ is for symmetry because the second and third point can be interchanged.

  • $\begingroup$ "Wherever the first point is chosen, the "diameter" on which it lies divides the n-gon into two symmetric halves". If $n$ is odd and the point is neither a vertex nor the middle point of a side I doubt this is true. I think your answer works perfectly for $n$ even and this is a special case of a theorem of Wendel: see e.g. mathoverflow.net/questions/33112/… $\endgroup$ – Gilles Bonnet May 10 '14 at 12:14
  • $\begingroup$ In the case $n$ is odd, Wendel's result does not apply because we loose the central symmetry. This is the same problem as in math.stackexchange.com/questions/72977/… where the probability turned out to be $<4$. $\endgroup$ – Gilles Bonnet May 10 '14 at 12:16

The answer is 1/4. I did a slight variation in the above answer to make things look easier, if x is length at which the second point lies. The third point can only lie in the equal sector on opposite side with the same angle of the sector and length of arc being x ( Two sector are made by two diameters. Thus 2∫x dx limits from 0 to 1/2 gives 1/4.


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