Posted by nancy on .
find a formula that gives the maximum number of enclosed regions formed by n lines

math 
BeagleBoy,
n2 works for triangles inside of a polygon.

math 
drwls,
Refer to question #4 at the end of this Stanford test:
http://math.stanford.edu/sumac/CompleteApplication.pdf
A formula that seems to work for n = 3 (one area), n = 4 (three areas) and n= 5 (five areas) is :
A = 1 + 2(n3) = 2n 5,
for n an integer greater than or equal to 3.
n is the number of lines, and A is the number of enclosed areas. 
math 
tchrwill,
the number of regions developed within a triangle resulting from the creation of "n" lines being drawn from each vertex to "n" points on the opposite sides derives from N = 3n^2 + 3n + 1.
n lines will divide a circle up into Rn = 1 + n(n + 1)/2 = [n^2 + n + 2]/2 regions.
Regions in ngons
The derivation of the number of regions created within different shaped figures comes in several forms. Consider the following variations.
1Irregular convex polygons divided up by connecting each vertex with every other vertex by straight lines.
2Circles divided up by connecting n randomly located points on the circumference with each other by straight lines. This is essentially joining all n points in sequence and creating all the diagonals between the points.
3Circles divided up by a series of lines that cross one another within the circle without connecting the points on the circle and without any three lines passing through the same point. This is the same as the #2 variation with the exception that the points on the circumference are not joined to one another.
It becomes immediately obvious that the connection of n randomly located points by straight lines within the circle is simply the polygonal ngon inscribed in a circle. Therefore, the number of regions created within the circle is really the number of regions defined within the ngon plus the n arc/chord regions created when inscribing the ngon within the circle. Lets tackle the polygonal ngons, first.
What can we learn from the first set of ngons. One thing that must be recognized is that all our ngons are irregular where all the sides and internal angles are not equal to one another. The triangle is the first ngon, ((n+2) = 3 sides) to be considered. Clearly, it has but one region within its three sides. Our second ngon is obviously the quadrilateral, or 4gon. Joining the opposite vertices clearly divides it up into 4 regions. Thirdly, comes the 5gon which is clearly divided into 11 regions. Our patience is now tested a bit when we attempt to determine the number of regions within the 6gon, 7gon, 8gon, and so on. After much enumeration and erasures, the 6gon has 25 regions, the 7gon, 50 regions, and the 8gon, 91 regions. What can we learn from this information that would enable us to define the number of regions within any ngon? Lets summarize the data derive so far.
n.....................1......2......3......4.......5......6 the nth ngon in the sequence.
Sides..............3......4......5......6.......7......8 (n+2) sides on the ngon
Regions..........1......4......11....25.....50....91
1stDifference......3......7......14....25.....41
2nd Difference.........4......7......11....16
3rd Difference..............3......4........5
4th Difference..................1.......1
An expression can be derived enabling the definition the nth term (or the (n+2)gon) of any finite difference series. The expression is a function of the number of successive differences required to reach the constant difference. If the first differences are constant, the expression is of the first order, i.e., N = an + b. If the second differences are constant, the expression is of the second order, i.e., N(n) = an^2 + bn + c. Similarly, constant third differences derive from N = an^3 + bn^2 + cn + d. Using the data points (n1, N1), (n2,N2), (n3,N3), etc., from our data, we substitute them into N = an^4 + bn^3 + cn^2 + dn + e. as follows:
(n1,N1) = (1,1) produces a(1^4) + b(1^3) + c(1^2) + d(1) + e = 1 or a + b + c + d + e = 1
(n2,N2) = (2,4) produces a(2^4) + b(2^3) + c(2^2) + d(2) + e = 4 or 16a + 8b + 4c + 2d + e = 4
(n3,N3) = (3,11) produces a(3^4) + b(3^3) + c(3^2) + d(3) + e = 11 or 81a + 27b + 9c + 3d + e = 11
(n4,N4) = (4,25) produces a(4^4) + b(4^3) + c(4^2) + d(4) + e = 25 or 256a + 64b + 16c + 4d +e = 25
(n5,N5) = (5,50) produces a(5^4) + b(5^3) + c(5^2) + d(5) + e = 50 or 625a + 125b + 25c + 5d + e = 50
Subtracting successive pairs ultimately leads to the solution of N(n) = [n^4 + 22n^3  59n^2 + 110n  48]/24 where N(n) = the nth term in the series with (n+2) sides to the ngon.
As we observed earlier, the number of regions created by inscribing these ngons within a circle, is simply n more than the number of regions within the ngon itself. This translates into a slightly modified expression for the number of regions created by joining all the vertices of an inscribed ngon within a circle to N(n) = [n^4 + 22n^3  59n^2 + 134n  48]/24 where N(n) = the nth term in the series with (n+2) sides to the inscribed ngon.
It might be of some interest (without derivation) that the number of regions within an ngon is also defined by
N = n/4 + (n1)/2
which is the number of combinations on n things taken 4 at a time plus the number of combinations of (n1) things taken 2 at a time.