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What is the greatest number of parts a plane can be divided into using $n$ infinite straight lines? What about $n$ circles?

Can you generalise this into 3-dimensional space, planes and spheres?

For lines, I got $U_{n+1}=U_n+n,$ with $U_0=1.$ And for circles, I got $U_{n+1}=U_n+2n,$ with $U_0=1$ and $U_1=2.$

Am I right so far?

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  • $\begingroup$ If it is $n$, how can it be infinite? $\endgroup$ – hjpotter92 Mar 24 '13 at 16:46
  • $\begingroup$ @Back in a Flash - The line itself is infinite, it is not saying that there is an infinite number of lines $\endgroup$ – user61067 Mar 24 '13 at 16:48
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Lines

For a plane with $n$ lines, consider what happens when you add a line. It divides all the regions through which it passes into two, thus adding one region for each region it passes through. The number of regions it passes through is the number of lines it crosses plus one, that is the number of points it creates plus one for itself. Thus, the number of regions added to the original one, $\binom{n}{0}$, is the number of points, $\binom{n}{2}$, plus the number of lines, $\binom{n}{1}$. Thus, the maximum number of regions is $$ \binom{n}{0}+\binom{n}{1}+\binom{n}{2}=\frac{n^2+n+2}{2} $$


Circles

For a plane with $n$ circles, consider what happens when you add a circle. It divides all the regions through which it passes into two, thus adding one region for each region it passes through. The first circle divides the plane into two. The number of regions it passes through is at least one, if it doesn't intersect any other circles, and up to the number of crossings with other circles. For $n$ circles there can be up to $2\binom{n}{2}$ crossings. Thus, the first circle divides the plane into two, $2\binom{n}{0}$, then the rest of the circles add the number of crossings. Thus, the maximum number of regions is $$ 2\binom{n}{0}+2\binom{n}{2}=n^2-n+2 $$

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  • $\begingroup$ can you please tell me your reasoning behind this? How did you know to use binomials? $\endgroup$ – user61067 Apr 1 '13 at 17:45
  • $\begingroup$ I understand what you are doing, but don't feel as though I could explain it to someone else. $\endgroup$ – user61067 Apr 1 '13 at 17:47
  • $\begingroup$ @user61067: In the case of the lines, $\binom{n}{2}$ is the number of pairs of lines, and each pair of lines forms at most one point of intersection. In the case of circles, $2\binom{n}{2}$, is twice the number of pairs of circles, and each pair of circles forms at most two points of intersection. I used $\binom{n}{1}=n$ and $\binom{n}{0}=1$ not only because they make the formulas look nice, but also they explain the fact that the formulas match powers of $2$ for a few terms. $\endgroup$ – robjohn Apr 2 '13 at 0:28
  • $\begingroup$ thanks again. Is it correct that for planes in space the solution is the same as this but with $\binom{n}{3}$ added? And if so... What does $\binom{n}{3}$ represent? $\endgroup$ – user61067 Apr 2 '13 at 13:09
  • $\begingroup$ @user61067: in $\mathbb{R}^3$, $\binom{n}{3}$ represents the number of points since the intersection of three planes determines a point. However, a simple way to compute the maximum number of regions into which $n$ planes divides $\mathbb{R}^3$ is to notice that the number of regions added by plane $n$ is the number of regions split by plane $n$ which is the number of regions into which the previous $n-1$ planes divides plane $n$. That is, $R_3(n)-R_3(n-1)=R_2(n-1)=\binom{n-1}{2}+\binom{n-1}{1}+\binom{n-1}{0}$. $R_3(0)=1$ implies $$R_3(n)=\binom{n}{3}+\binom{n}{2}+\binom{n}{1}+\binom{n}{0}$$ $\endgroup$ – robjohn Nov 2 '16 at 4:37
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if plane divided by lines, recursion function is $R(n+1)=R(n)+n+1$, where $R(0)=1$, $n\geq0$ Solving recurrence $R(n)=\frac{(n^2+n+2)}{2}$
if plane divided by circles, recursion function is $R(n+1)=R(n)+2n$, where $R(1)=2$, $n>0$ Solving recurrence $R(n)=(n^2-n+2)$

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For a plane divided by lines: $$l_{n+1}=l_{n}+n+1$$

For space divided by planes: $$s_{n+1}=s_{n}+l_{n}$$

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    $\begingroup$ Can you tell me how you got this please? $\endgroup$ – user61067 Mar 24 '13 at 17:30

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