Quandaries
and Queries 

David Say you have a cirlcle. Then you draw 2 dots on the circle. Then you connect the dots with lines. The circle is divided into 2 parts. If you do the same with 3 dots and connect each dot to each dot with a line then you get a circle with 4 parts. 4 dots with lines connecting all (6 lines) = 8 parts 5 dots with lines connecting all (10 lines) = 16 parts Ok so far, there seems to be a good pattern forming here 2=2, 3=4, 4=8, 5=16 Now it gets tricky:: 6 dots all connected by lines (15 lines) = NOT 32, Correct answer is 31 spaces 7 dots all connected by lines (21 lines) = 57 spaces >> STRANGE!! I can't figure out the pattern. I don't know higher math well enough to do calculus or anything like that. Is there a formula you can think of to predict how many spaces would be formed by 8 dots/(28 lines) ?? Thank you for your time. David 

Hi David, When you say "6 dots all connected by lines (15 lines) = NOT 32, Correct answer is 31 spaces", it could even be lower. If you place your 6 dots as the vertices of a regular hexagon, then the little triangle in the middle vanishes into a single point, and you are left with 30 parts. So actually, you want a formula for the maximum number of parts into which the circle can be subdivided by joining n points on its circumference. For this you use Euler's formula: E is the number of "edges", that is, the lines joining two points. F is the number of "faces", that is, the parts of the circle and the outside face. What you really want is F  1 (because you don't count the outside face), and the above formula can be rewritten as Of this, V is the easiest to count: you start with n points, and every time you have 4 points on the circle, then you get a point of intersection of the two lines joining opposite pairs of points. There are = n(n1)(n2)(n3)/4*3*2*1 ways to select 4 points amongst your n points on the circle, so you get Counting the edges is a bit more tricky. It is best to fix a vertex and count its "degree", that is, the number of edges incident to it. There are two cases:


