Choice, Chance and Sensible Logic

Win_odd Dhamnekar

Win_odd Dhamnekar

Junior Member
Joined
Aug 14, 2018
Messages
207
a) In how many parts, we can divide an infinite plane by n straight lines of which no two lines are parallel and no three lines are concurrent?

b) In how many parts, we can divide an infinite space by n planes of which no four planes meets in a point and no two planes are parallel?


How to answer both these questions? What are the answers to these questions?


I am working on these questions.

Any math help, math hint, or even correct answer will be accepted.
 
Last edited:
Dhamnekar Winod said:
a) In how many parts, we can divide an infinite plane by n straight lines of which no two lines are parallel and no three lines are concurrent?

b) In how many parts, we can divide an infinite space by n planes of which no four planes meets in a point and no two planes are parallel?

How to answer both these questions? What are the answers to these questions?

I am working on these questions.

Any math help, math hint, or even correct answer will be accepted.
Click to expand...
I would start by finding the answers for n = 1, 2, 3, 4 and looking for a pattern (not necessarily in the numbers themselves, but a way to obtain one from the previous one(s)).

Please show what you have tried, as usual. And if you just want the answer without any effort, why? What is your reason for asking, if not to learn?
 
Dhamnekar Winod said:
a) In how many parts, we can divide an infinite plane by n straight lines of which no two lines are parallel and no three lines are concurrent?
b) In how many parts, we can divide an infinite space by n planes of which no four planes meets in a point and no two planes are parallel?
Any math help, math hint, or even correct answer will be accepted.
Click to expand...
a) In a plane two non-parallel lines determine four distinct areas. Can you explain why?
Now if there is a third such line that does not contain the common point of those two, then how many areas are determined? Explain!
 
I was hoping that Winod might answer our requests. But because this is such a well known problem here goes.
The solution takes a bit of back ground in axiomatic geometry. Lets assume a fixed plane [imath]\Pi[/imath].
If line [imath]\ell_1\subset\Pi[/imath], then that line divides the plane into two disjoint convex sets.
Those sets are called sides of the line. Now if the line [imath]\ell_2\subset\Pi[/imath] such that [imath]\ell_1\cap \ell_2=\{Q\}[/imath].
The line [imath]\ell_2[/imath] also separates [imath]\Pi[/imath] into two half-planes.
The intersections of the four half-planes along with [imath]\ell_1\cup \ell_2[/imath] create four areas of [imath]\Pi[/imath].
Allow me to urge anyone to use a good mathematics library at a local university. A complete discussion of this question
can be found in MATHEMATICS OF CHOICE by Ivan Niven.
The basic idea is that [imath]\Gamma(\Pi,\ell_n)[/imath] is the set of all lines in [imath]\Pi[/imath] no two of which are parallel, no three are concurrent
If [imath]f(n)[/imath] is the number of areas that lines from [imath]\Gamma(\Pi,\ell_n)[/imath] determined by [imath]\ell_1,~\ell_2,~\cdots~\ell_n[/imath]
We have seen that [imath]f(1)=2,~f(2)=4~\&~f(3)=7[/imath]. It turns out that [imath]f(n)=n+f(n-1) [/imath].............(n>1)
That is a recurrent function. see here.


[imath][/imath][imath][/imath]
 
Last edited:
You must log in or register to reply here.
Top