Number of Regions in Plane Defined by Given Number of Lines

Theorem
The maximum number $L_n$ of regions in the plane that can be defined by $n$ straight lines in the plane is:
 * $L_n = \dfrac {n \left({n+1}\right)} 2 + 1$

Setting up a Recurrence Rule
First we consider the plane with no lines at all. This has one region, so $L_0 = 1$.

Now when we have one line, we divide the plane into two regions, so $L_1 = 2$.

Now consider the $n$th line.

This increase the number of regions by $k$ iff it splits $k$ of the old regions.

It can split $k$ of the old regions iff it hits the existing lines on the plane in $k-1$ places.

Two straight lines can intersect in at most one point.

So the new line can intersect the $n-1$ old ones in at most $n-1$ different points.

Therefore $k \le n$.

So we see that $L_n \le L_{n-1} + n$.

Now, it is always possible to place the $n$th line so that:
 * It is not parallel to any of the others, and therefore intersects all the other $n-1$ lines;
 * It does not go through any of the existing intersection points (so intersects them all in different places).

Thus we see that $L_n \ge L_{n-1} + n$.

Hence the recurrence:
 * $L_n = L_{n-1} + n$

Solution of Recurrence
Using induction, we show that $L_n = \dfrac {n \left({n+1}\right)} 2 + 1$.

The base case is straightforward:
 * $L_0 = 1 = \dfrac {0 \left({0+1}\right)} 2 + 1$
 * $L_1 = 2 = \dfrac {1 \left({1+1}\right)} 2 + 1$

Now assume the induction hypothesis:
 * $L_k = \dfrac {k \left({k+1}\right)} 2 + 1$

and try to show:
 * $L_{k+1} = \dfrac {\left({k+1}\right) \left({k+2}\right)} 2 + 1$

Hence the induction step:

Hence the result by induction.

History
This was published in.