Dividing a circle into areas

The problem

Suppose you have a circle with n points lying on the perimeter of the circle and lines connecting all of the points. There are a total of

For small values we have:

• n = 0 or 1 gives f(0) = f(1) = 1;
• f(2) = 2 because there is a single diameter, making two areas;
• f(3) = 4;
• f(4) = 8,

f(5) = 16.

It is tempting to conjecture that

.

Lemma

If we already have n points on the circle and add one more point, we draw n lines from the new point to previously existing points. Two cases are possible. In the first case (a), the new line passes through a point where two or more of lines between previously existing points cross. In the second case (b), the new line crosses each of the old lines in a different point. It will be useful to know the following fact.

Lemma. We can choose the new point A so that case b occurs for every of the new lines.

Proof. Notice that, for the case a, three points must be on one line: the new point A, the old point O to which we draw the line and the point I where two of the old lines intersect. Notice that there are finitely many (n) old points O, finitely many points I where two of the old lines intersect and, for each O and I, the line OI crosses the circle in one point other than O. Since the circle has infinitely many points, there is a point A which will be on none of lines OI. Then, for this point A and every of old points O, case b will be true.

This lemma means that, if there are k lines crossing AO, then each of them crosses AO at a different point and k+1 new areas are created by the line AO.

Proof

And here we go. This will be an inductive proof, providing a formula for f(n) in terms of f(n − 1).

In the figure you can see the dark lines connecting
points 1 through 4 dividing the circle into 8 total
regions (i.e., f(4) = 8). This figure illustrates the inductive
step from n = 4 to n = 5 with the dashed lines. When the fifth point is added (i.e., when
computing f(5) using f(4)), this results
in four (n − 1) new lines (the dashed lines in the diagram) being added, numbered 1 through 4, one for each point that they connect
to. The number of new regions introduced by the fifth point can
therefore be determined by considering the number of regions added by
each of the 4 lines. Set i to count the lines we are adding.
Each new line can cross a number of existing lines, depending on which
point it is to (the value of i). The new lines will never cross
each other, except at the new point.

The number of lines that each new line intersects can be determined by
considering the number of points on the "left" of the line and the
number of points on the "right" of the line. Since all existing
points already have lines between them, the number of points on the
left multiplied by the number of points on the right is the number of
lines that will be crossing the new line. For the line to point i,
there are

n − i − 1

i − 1 points

(n − i − 1)(i − 1)

In this example, the lines to i = 1 and i = 4 each cross zero lines,
while the lines to i = 2 and i = 3' each cross two lines (there are two
points on one side and one on the other).

So the recurrence can be expressed as

:

It follows that the solution will be a quartic polynomial in n. Its actual coefficients can be found, by using the five values of f(n) given above.

Discussion of an alternate method

One part of the theorem asserts that the number of regions is maximal if all "inner" intersections of two different chords are simple (exactly two chords pass through a point of intersection of chords Q in the interior). This will be the case if the points on the circle are chosen "in general position". Under this assumption of "generic intersection", the number of regions can also be determined in a non-inductive way, using the formula for the Euler characteristic of a connected planar graph (viewed here as a graph embedded in the 2-sphere S ²).

A planar graph determines a cell decomposition of the plane with f faces (2-dimensional cells), e edges (1-dimensional cells) and v vertices (0-dimensional cells). If the graph is connected then the Euler relation for the 2-dimensional sphere S ²
:
holds. View the diagram (the circle together with all the chords) above as a planar graph G whose edges are the chordal line segments (connecting two "neighbouring" points of intersection of chords) created inside the circle by the collection of

Its vertices are the n points together with the points given as the intersection of two (or more) chords in the interior of the circle. Under the "generic intersection" assumption made above exactly two chords pass through a point .

Call the number of interior points . Then the graph G' has

In the given graph the vertices have degree

The number of regions r_G inside the circle is then

f − 1

It remains to determine the total number of interior intersection points . In the following discussion we will assume that the points have been numbered in counterclockwise increasing order (after choosing some of the points as ). A pair of chords may be viewed as a "doubled" pair
::

where

;

References

External links

• Circle Division by Chords