Double Trouble

Logic Level 3

Suppose there is/are $n$ points on the circumference of the circle, define $f(n)$ as the number of pieces for which a circle is divided with these points are joined by chords such that no more than 2 chords intersect at the same point.

The diagram above shows that $f(1) = 1, f(2) =2, f(3) =4$ , $f(4) = 8, f(5) = 16$ . Find the value of $f(6)$ .

32 31 64 33

1 solution

Isaac Buckley
Jul 10, 2015

Let's find a general formula for $f(n)$ by using Eulers theorem.

Eulers theorem is $V-E+F=2$ Where $V$ , $E$ and $F$ denote the number of vertices, edges and faces respectively.

$f(n)=F-1$ since the outside of the circle is also counted as a face.

After finding $V$ and $E$ in terms of $n$ we deduce:

$f(n)=1+\frac{n(n-1)(n^2-5n+18)}{24}.$

$f(6)=\boxed{31}$

I would like to note a very similar problem has been posted before and can be found here .

The amazing Otto Bretscher has a solution in which he explains briefly how you can go about finding $V$ and $E$ .

(Where the hell is Otto? He's been inactive for a month and I miss his incredible problems.)

Double Trouble

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