0
$\begingroup$

Find the no. of parts in which the diagonals of a convex decagon divide the interior if no three diagonals are concurrent inside the decagon.

I have no idea as to how I can solve this.
All I know is that the no. of diagonals is given by $\frac{n(n-3)}{2}$ and the no. of intersections by $\binom{n}{4}$ but I don't know how to use this data to find the answer.

It would be great if I could get a hint to proceed with this question.

  • 0
    What exactly are those _ways to divide_? Say, I have a square with two diagonals; in how many ways do they divide it, and what are those ways? – 2017-01-20
  • 0
    @IvanNeretin more precisely no. of parts in which the diagonals divide the figure, ie 4 in your example. (Made the edit, thanks!) – 2017-01-20

1 Answers 1

1

Think of your decagon as a planar graph and apply the Euler's formula for planar graphs (some may refer to it as a formula for polyhedra; it's all the same): $$V-E+F=2$$ $V$ is the numbers of vertices of our graph. Those are the vertices of decagon and intersections of diagonals, so you seem to already know how to count them.

$E$ is the number of graph's edges. Think of it this way: we have our decagon's edges and diagonals (we know exactly how many), but then we also have intersections, and each intersection seems to break two segments in halves, so the number of edges increases by 2 for each intersection.

$F$ is the number of faces, i.e. the parts into which this graph splits the plane. This is what you want to find. Don't forget to subtract 1 to account for the exterior part (yes, it also counts).

So it goes.