Show that $n$ lines separate the plane into $\frac{n^2+n+2}{2}$ regions

Question

Show that $n$ lines separate the plane into $\frac{n^2+n+2}{2}$ regions if no two of these lines are parallel and no three pass through a common point.

I know we start with the base case, where, if we call the above equation P(n), P(0), for 0 lines would be 0. But I really have no idea how to begin the inductive step. How do we know what k+1 we're supposed to arrive at?

Thanks!

Answer 1

Here is the way I usually think about this (it sort of uses induction).

With $0$ lines, there is $1$ region and no intersections of lines.

Each time a line is added and it crosses $k$ other lines it adds $k+1$ regions and $k$ intersections. Another way of looking at this is that for each line and $k$ intersections added, $k+1$ regions are added (the number of added lines and intersections).

Therefore, the number of regions is $1+\text{the number of lines}+\text{the number of intersections}$. With $n$ lines, there are $\binom{n}{2}$ intersections (if no two lines are parallel and no three lines are coincident).

Thus, the number of regions is $\binom{n}{2}+n+1=\frac{n(n-1)}{2}+n+1=\frac{n^2+n+2}{2}$.

Answer 2

Suppose we draw $n$ straight lines on the plane so that every pair of lines intersects (but no $3$ lines intersect at a common point). Into how many regions do these $n$ lines divide the plane?

With $n = 1$ we divide the plane into $2$ regions. With $n = 2$ we have $4$ regions; with $n = 3$ we get $7$ regions. A fourth line will meet the other $3$ lines in $3$ points and so traverse $4$ regions, dividing them into $2$ parts and adding $4$ new regions. In general the $n^{th}$ line will add $n$ new regions:

$$u(1) = 2$$ $$u(2) = 4$$ $$u(3) = 7$$ $$u(4) = 11$$

And so on, where $u(n) =$ number of regions with $n$ lines.

We get the recurrence relationship:

$$u(n+1) = u(n) + (n+1)$$

We get the following chain of equations:

$$u(n) - u(n-1) = n$$ $$u(n-1) - u(n-2) = n-1$$ $$u(n-2) - u(n-3) = n-2$$ $$\vdots$$ $$u(4) - u(3) = 4$$ $$u(3) - u(2) = 3$$ $$u(2) - u(1) = 2$$

Adding these equations, we get: $$u(n) - u(1) = 2 + 3 + 4 + ..... + (n-1) + n$$

All other terms on the left cancel between rows, and we are left with:

$$u(n) = u(1) + 2 + 3 + 4 + \ldots + n$$

We know, $u(1) = 2$

Thus:

$$u(n) = 1 + (1+2+3+4+ \ldots+n)$$ $$\implies u(n) = 1 + \dfrac{n(n+1)}{2}$$ $$\implies u(n) = \dfrac{n^2 + n + 2}{2}$$

So:

$$u(n) = \dfrac{n^2 + n + 2}{2}$$

Remark $\,$ If you allow parallel lines and more than $2$ lines to intersect at a point, the above relation doesn't hold.

The answer then depends on the number of lines intersecting at a point or the number of lines which are parallel to one another.

Answer 3

There are two ways to arrive at this formula. I will first give the explanation for the proof of the formula by induction followed by a derivation of the same formula by combinatorial methods.

Proof by Induction:

Let $f(n)$ be the maximum number of regions into which $n$ lines divide a plane.

Now, let us add the $(n+1)$th line. Provided none of the lines are collinear and no 3 lines intersect at the same point upon adding the new line, this new $(n+1)$th line would make an intersection with each of the previous $n$ lines. Hence the new line passes thorugh $(n+1)$ regions dividing each of these regions into two. Hence $(n+1)$ new regions are created when the $(n+1)$th line is added. This means that:

$f(n+1)=f(n)+(n+1)$.

Let me demonstrate this with an example:

Assume that we have 2 lines dividing a plane into 4 regions. When we add a 3rd line, it will intersect the previous 2 lines and will be passing through 3 regions: the region on one side of the 1st line, the region between the 1st line and 2nd line, and the region on the other side of the 2nd line. Hence 3 new regions have been created.

So, now we have the conditions: $f(n)=f(n-1)+n$ and $f(0)=1$

$f(n)=f(n-1)+n$

$f(n-1)=f(n-2)+n-1$

$f(n-2)=f(n-3)+n-2$

.....

$f(2)=f(1)+2$

$f(1)=f(0)+1$

$f(0)=1$

Adding the above equations gives:

$f(n)=1+(1+2+3+...+n)$

So, $f(n)=1+\frac{n(n+1)}{2}=\frac{n^2 + n + 2}{2}$

This can also be expressed as:

$f(n)=\binom{n}{2}+\binom{n}{1}+\binom{n}{0}$

Combinatorial Proof:

Each region in the plane (except for the regions which are not bounded below) has exacly one "deepest point" (i.e. a point at the deepest "corner" in that region). This deepest point is the deepest point of only that particular region and not of any of the other 3 regions adjacent to it. We can also notice that an intersection between any two lines will create exactly one deepest point belonging to one particular region. The maximum number of points of intersections created by $n$ lines is $\binom{n}{2}$. Hence, there are $\binom{n}{2}$ deepest points.

Now let us consider the regions which do not have a deepest point. These are the regions which are not bounded below. So we cannot pick out a deepest point for these regions. There are exactly $n+1$ such regions.

So the total number of regions is $\binom{n}{2}+n+1 = \binom{n}{2}+\binom{n}{1}+\binom{n}{0}$.

There is also a 3 dimensional version of this problem which asks for the maximum number of regions into which n planes divide a space. The answer turns out to be $\binom{n}{3}+\binom{n}{2}+\binom{n}{1}+\binom{n}{0}$. This can also be proved by the induction method (considering the number of regions divided by the $(n+1)$th plane) as well as the combinatorial method (using the concept of deepest points).