In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides, in such a way as to maximise the number of areas created by the edges and diagonals, has a solution by an inductive method.

The problem

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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

$\backslash ell(n)=\backslash frac\{n(n-1)\}\{2\}$

lines connecting the points. The mission is to find the maximum number f(n) of areas that the circle can be divided into, as a function of the number n of points.

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

$f(n)\; =\; 2^\{n-1\}.$

But f(6) = 31.

Lemma

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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.

Solution

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Inductive method

The lemma establish an important property for solving the problem. By employing inductive proof, one can provide 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

points on the left and

i − 1 points

on the right, so a total of

(n − i − 1)(i − 1)

lines must be crossed.

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

$f(n)=f(n-1)+\backslash sum^\{n-1\}\_\{i=1\}\backslash left(1+\backslash left(n-i-1\backslash right)\backslash left(i-1\backslash right)\backslash right).$

Which can be easily reduced somewhat to

$f(n)=f(n-1)+\backslash sum^\{n-1\}\_\{i=1\}\backslash left(2-n+ni-i^2\backslash right)$

$=f(n-1)-n^2+3n-2+\backslash sum^\{n-1\}\_\{i=1\}\backslash left(ni-i^2\backslash right)$

This can be further reduced, using the formula for Σ i².

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.

Combinatorics and topology method

The lemma 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). As the graph is connected then the Euler relation for the 2-dimensional sphere S ²

$\backslash ,\; V-E+F=2$

holds. View the diagram (the circle together with all the chords) above as a planar graph. If the general formula for V and E can both be found, the formula for F can also be derived, which will solve the problem.

Its vertices are the n points on circle, referred to as the exterior vertex, together with the points which are the intersection of two chords in the interior of the circle due to the "generic intersection" assumption made above, though in general case there may be more than two chords intersecting at the same point. These are called the interior vertex.

Thus the main task is to find the number of interior vertex. As a consequence of the lemma, any two intersecting chords will uniquely determine an interior vertex. These chords are in turn uniquely determined by the four corresponding endpoints of the chords, which are all exteriror vertex. The fact that four exterior points determine a cyclic quadrilateral, and that all cyclic quadrilateral is a special case of convex quadrilateral confirm it has exactly one intersection point formed by the diagonals(chords). Further, by definition all interior vertex are formed by intersecting chords.

Therefore all interior vertex can be uniquely determined by any combination of four exterior points, and the number of interior points are given by

$V\_\{interior\}\; =\; \{\; n\; \backslash choose\; 4\; \}$

And so

$V\; =\; V\_\{exterior\}\; +\; V\_\{interior\}\; =\; n\; +\; \{\; n\; \backslash choose\; 4\; \}$

The edges are the chordal line segments(which will be explained later) created inside the circle by the collection of chords together with the n circular arcs connecting two adjacent points. Since there is two groups of vertex, exterior, and interior, the edges that are chordal line segments can be further catergorized into three groups:

1. Edges directly(not cut by other chords) connecting two exterior vertex. These are edges between adjacant exterior vertex and form the perimeter of the polygon. Obviously, there are n such edges.
2. Edges connecting between two interior vertex.
3. Edges connecting between a interior and exterior vertex.

To find the number of edges for the remaining groups, consider each interior vertex, which is connected to exactly four edges. This yield

$4\{\; n\; \backslash choose\; 4\; \}$

edges. But as each edge are defined by two endpoint vertex, edges for group 2 are counted twice while group 3 are counted once only(because we only enumerated the interior vertex).

Notice how every chords that are cut by other chords(those that are not group 1 edges) must contain two group 3 edges, the beginning and ending chordal segment. As chords are uniquely determined by two exterior vertex, there are althogether

$2\backslash left(\; \{\; n\; \backslash choose\; 2\; \}\; -\; n\; \backslash right)$

group 3 edges(excluded selection of two adjacant exterior vertex, which are actually group 1 edge).

Sum them and divide by two to get the total number of group 2 and 3 edges. Hence the total number of edges are

$E\; =\; n\; +\; \backslash frac\{4\{\; n\; \backslash choose\; 4\; \}\; +\; 2\backslash left(\; \{\; n\; \backslash choose\; 2\; \}\; -\; n\; \backslash right)\}\{2\}\; +\; n\; =\; 2\{\; n\; \backslash choose\; 4\; \}\; +\; \{\; n\; \backslash choose\; 2\; \}\; +\; n.$

(The first n is group 1 edges, the last n is the n circular arc edges)

Substituting V and E into the Euler relation solved for F, $\backslash ,\; F\; =\; E\; -\; V\; +\; 2$, one then obtains

$F\; =\; \{\; n\; \backslash choose\; 4\; \}\; +\; \{\; n\; \backslash choose\; 2\; \}\; +\; 2.$

Since one of these faces is the exterior of the circle, the number of regions r_G inside the circle is F − 1, or

$r\_G\; =\{n\; \backslash choose\; 4\}+\{n\; \backslash choose\; 2\}+1$

which is the quartic polynomial determined in the inductive proof.

References

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• Conway, J. H. and Guy, R. K. "How Many Regions." In The Book of Numbers. New York: Springer-Verlag, pp. 76-79, 1996.

External links

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• Circle Division by Chords

Geometry | Combinatorics