Dissection of Plane with Lines

Problem 1. Let the maximum number of sections of the plane you can divide with 2n2n lines be the function f(n)f(n). Let the maximum number of sections of the plane you can divide with nn pairs of parallel lines be g(n)g(n). Prove that f(n)>g(n)f(n) > g(n) for n>0n > 0.

Problem 2. Let the maximum number of sections of the plane you can divide with 2n2n lines be the function f(n)f(n). Let the maximum number of sections of the plane you can divide with nn pairs of perpendicular lines be g(n)g(n). Prove that f(n)=g(n)f(n)=g(n) for n>0n > 0.


If any of these two problems are actually wrong, feel free to disprove it in the comments.

#ParallelLines #PerpendicularLines #Proofs #Plane #DividingThePlane

Note by Daniel Liu
7 years, 2 months ago

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Comments

Problem 1.\textbf{Problem 1.}

Let f(n)f(n) denote the number of regions a plane is divided by n lines....

Clearly f(1)=2f(1) = 2 , f(2)=4f(2) = 4 and f(3)=7f(3) = 7.

Now let us think about the recursive definition of the function f(n)f(n),when the nth line enters the [plane it intersect each of the previous n1n-1 lines at one point only and in process creates nn new regions.Hence f(n)=f(n1)+nf(n) = f(n-1) + n

Solving this and putting f(1)=2f(1) = 2, we get, f(n) =1+n(n+1)2f(n)\ = 1+\frac{n(n+1)}{2}

So clearly the number of regions a plane is divided by 2n2n lines is f(2n)=1+n(2n+1)=1+2n2+nf(2n) = 1+n(2n+1) = 1+ 2n^{2} + n.

Now let us the consider n pairs of parallel lines...

By inspection we can say that g(1)=3g(1) = 3 and g(2)=9g(2) = 9,

Now let us consider the nth pair of parallel lines being drawn...each line intersects the previous 2n22n - 2 lines to form 2n12n-1 regions!!

So the recursive definition is g(n)=g(n1)+2(2n1)g(n) = g(n-1) + 2(2n-1)

Solving this recurrence relation we get g(n)=1+2n2g(n) = 1 + 2n^{2}. This relation's validity can easily be proved by induction.

Hence f(n)=1+2n2+n>1+2n2>g(n)f(n) = 1+2n^{2} + n > 1+2n^{2} > g(n).

Eddie The Head - 7 years, 2 months ago

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I think perpendicular lines will also follow the same recursion as f(n)f(n) and hence the 2nd identity also correct....

Eddie The Head - 7 years, 2 months ago

Here you have no where mentioned whether the lines can coincide wholly with each other.......

Krishna Viswanathan - 7 years, 2 months ago

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If I understand what you are saying, then yes, more than three lines can intersect at one place. However, that case will not give the maximum number of sections.

Daniel Liu - 7 years, 2 months ago
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