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The statement is:

n points on a circle divide it in $2^{n-1}$ regions.

I have used geometry to figure out that indeed, this statement is false, but I was wondering how to dis-prove this statement using simple and logical mathematics.

What I did for this, is if $n$ points, then there will be $^nC_2$ lines. Intersections in the lines can be $0$ or $n-1$ or $n$ only.

Am I going in the right direction and how should I go on with this to find something about the regions the lines divide the circle into.

Edit: I don't want to dis-prove this with a counter example, I am thinking of doing this in the way I have started, with logic and simple mathematics.

PradyumanDixit
  • 897

2 Answers2

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Well, to disprove such a statement, just give a counterexample.

Alternatively, you can calculate the explicit expression for the number of regions. We do this by induction.

Suppose there are already $(n-1)$ points, and $N_{n-1}$ regions. now we add a point, and gradually draw all the additional $(n-1)$ lines, while tracking the number of regions. Connecting this new point with another one that is $k$ points away (i.e. $k$ points lie between them) creates $k(n-k-2)$ intersections, as the $k$ points on one side and $(n-k-2)$ points on the other side are all connected. These intersections, including the starting/ending point of the segment, divides the new line into $k(n-k-2)+1$ parts. Each part of the segment separates a region. So, $$N_n = N_{n-1}+ \sum_{k=0}^{n-2}\left(1+ k(n-k-2)\right)=N_{n-1}+n-1+\sum_{k=1}^{n-3}k(n-k-2).$$ We know that $N_2$ is $2$, $N_3=4$, $N_4=8$. Applying this formula gives $N_5=16$, which agrees with my counting. Continue to compute $N_6$ and the result deviates from powers of $2$.

As a side note, the summation above can be shown to be $$ \frac{n^3}6 - n^2 + \frac{11n}6 - 1, $$ by realizing it must be a cubic polynomial, and computing the coefficients. And so the expression for $N_n$ reduces to $$\begin{align} N_n &= 1+\sum_{m=1}^{n} \frac{m^3}6 - m^2 + \frac{17m}6 - 2\\ &= \frac{n^4}{24} - \frac{n^3}{4} + \frac{23n^2}{24} - \frac{3n}4 \end{align}$$

Trebor
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Hint
There are $n$ points on the circle.
How many lines do the points form?
How many points of intersection do the lines form?
How many little lines do the lines divided into by those points?
And finally, use $V-E+F=2$.

For a counterexample, use $n=6$.

Kemono Chen
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