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You are given a family of n pairwise intersecting circles in the plane. No three intersect(share a common point). Find a simple formula for counting the number of regions determined by these circles.

Given n=4 we will have 14 regions

How can I find this pattern while expanding upwards.

OLE
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2 Answers2

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HINT: Suppose that $n$ circles divide the plane into $r_n$ regions. Now you add another circle, $C$, that non-trivially intersects each of the first $n$ circles. Start at any point on $C$ and travel around $C$; you will cross one of the other $n$ circles $2n$ times. If $p_0$ and $p_1$ are two consecutive crossings, the arc of $C$ from $p_0$ to $p_1$ divides one of the original $r_n$ regions into two smaller regions. You do this $2n$ times; how many new regions does that add? That gives you a recurrence $$r_{n+1}=r_n+\text{some simple function of }n\;,$$

and you can fairly easily turn it into an expression for $r_n$ that is just a summation. If you do it right, it’s a summation that you can evaluate pretty easily in closed form to get a formula for $r_n$. If you get stuck, click on the spoiler-protected block below for a further hint.

You add $2n$ new regions, and it follows that $r_{n+1}=r_n+2n$. You also know that $r_1=2$, so $r_2=2+2\cdot1$, $r_3=2+2\cdot1+2\cdot2$, $r_4=2+2\cdot1+2\cdot2+2\cdot3$, and so on. Write down a general expression for $r_n$ involving a summation, and then find a closed form; in this case that’s not hard.

Brian M. Scott
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Assuming that "intersecting circles" meet in two distinct points:

$F=E-V+2=n(2n-2)-n(n-1)+2=n(n-1)+2=n^2-n+2$.

bof
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    can you elaborate $E = n(2n-2)$ ? – null Sep 24 '18 at 00:50
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    Let $C$ be one of the circles. $C$ meets each of the other $n-1$ circles in $2$ points. The $2n-2$ points where $C$ crosses one of the other circles divide the circle $C$ into $2n-2$ arcs, which we consider as edges of a plane graph. There are $2n-2$ such arcs on each of the $n$ circles, making a total of $E=n(2n-2)$ edges in the graph. – bof Sep 24 '18 at 06:55