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Three distinct vertices are choose at random from the vertices of a given regular polygon of $2n+1$ sides. Each of those three vertices will determine a triangle. What is the probability that the center of the polygon lies inside the triangle determined by three vertices of the polygon?

Note: All vertices of a regular polygon lie on a common circle (the circumscribed circle), the center of this circle is the center of the polygon.

P.S: why is this question asked specially for a odd sided regular polygon, will the answer differ for a even sided regular polygon?

I have just found a difficult proof of a Much more general question (Probability of a fixed point in a convex region being inside a triangle formed by any three point form the convex region). I am not interested in such generality, can one give a much simpler proof this special case (the convex region being the odd-sided polygon and the fixed point the center of the polygon).

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    $\begingroup$ With an even sided regular polygon, the centre lies on several diagonals - would that count as inside or outside? $\endgroup$
    – Henry
    Commented Jul 2, 2016 at 22:48
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    $\begingroup$ For an odd regular polygon, the centre would always be exterior or interior to the triangle. For an even regular polygon, it could be on the boundary. $\endgroup$
    – Joffan
    Commented Jul 2, 2016 at 22:48

1 Answer 1

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Call the $(2n+1)$-gon $P$, and label an arbitrary vertex $A$. We assume without loss of generality that $A$ is one of the chosen vertices. Let the other two chosen vertices in clockwise order be $B$ and $C$, respectively, and suppose there are $k$ edges of $P$ contained within minor arc $BAC$. Because $\triangle ABC$ contains the center of $P$, we have $\angle BAC < 90^\circ$. This implies $k\cdot \frac{180}{2n+1} < 90$; hence $1\le k \le n$.

For a fixed $k$ between $1$ and $n$ inclusive, it is easy to verify that there are $k$ choices of $B$ and $C$ such that exactly $k$ edges of $P$ are contained within minor arc $BAC$ . This gives us a total of $\sum_{k=1}^n k = n(n+1)/2$ valid choices for $B$ and $C$. Because we have $\binom{2n}{2}$ ways to choose $B$ and $C$ from all the remaining vertices of $P$, the probability that a randomly chosen triangle contains the center of $P$ is $$ \frac{\frac{n(n+1)}{2}}{\binom{2n}{2}} = \frac{n+1}{2(2n-1)}. $$

Note: This probability implies there are a total of $\frac{n(n+1)(2n+1)}{6} = \sum_{k=1}^n k^2$ triangles that contain the center of $P$. Is there also a bijective proof that counts this directly?

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  • $\begingroup$ I did not understand how you wrote $k \times \frac{180}{2n+1} \lt 90$ can you please explain $\endgroup$ Commented Aug 30, 2017 at 2:52
  • $\begingroup$ @EkaveeraKumarSharma This is a necessary condition, but not sufficient. $\endgroup$
    – x100c
    Commented Nov 7, 2019 at 12:53
  • $\begingroup$ how can u say for each fixed value of k no of choices are fixed? $\endgroup$
    – maveric
    Commented Apr 30, 2021 at 18:14

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