3
$\begingroup$

What is probability that $3$ random point in $2n$ reguler polygon contains center of gravity

enter image description here

I contain the diagram to make up my explanation. Left one is contain the center of gravity,while right one doesn't.

$$\binom{2n}{n}$$ is the whole case and I found that when $n=1$,Probablity $P_{1} =$ Not defined.And $n=2$,$P_{2}=1$. I also get $P_{3}=\frac{7}{10}$ and $P_{4}=\frac{4}{7}$, but I can't certain about this.

Also I found $$\lim_{n\rightarrow\infty}P_{n} =\frac{1}{4}$$

Plz help me to find $P_n$

|cite|improve this question
$\endgroup$
  • $\begingroup$ made me think about this : youtube.com/watch?v=OkmNXy7er84 probably not that related but good video though $\endgroup$ – P. Quinton Sep 4 '18 at 14:28
  • $\begingroup$ @P.Quinton Yes I think about this from that vidio~!!!!. But I can't do my own to do this...... I tried about to get all cases using symmetry, It's value is very awkward.... $\endgroup$ – user366725 Sep 4 '18 at 14:30
  • $\begingroup$ What if the center of gravity lies on the border of the triangle? Does that count as contained? $\endgroup$ – Mike Earnest Sep 4 '18 at 14:41
  • $\begingroup$ @MikeEarnest Yes, that's why I wrote that $P_2=1$ $\endgroup$ – user366725 Sep 4 '18 at 14:42
5
$\begingroup$

In order for the three points to not contain the center in their convex hull, two of the points must be at most $2\pi\cdot\frac{(n-1)}{2n}$ radians anti-clockwise from the third point. There are $2n$ possibilities for the location of the third point, and for each of these are $\binom{n-1}2$ ways to choose two points from the $n-1$ points in that arc. Therefore, $$ 1-P_n = \frac{2n\binom{n-1}{2}}{\binom{2n}3}=\frac{3(n-1)(n-2)}{(2n-1)(2n-2)}=\frac{3(n-2)}{2(2n-1)}, $$ $$ P_n = \frac{n+4}{4n-2}. $$

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.