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# Dissection of Plane with Lines

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

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

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

Note by Daniel Liu
3 years, 3 months ago

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$$\textbf{Problem 1.}$$

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

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

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

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

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

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

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

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

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

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

Hence $f(n) = 1+2n^{2} + n > 1+2n^{2} > g(n)$. · 3 years, 3 months ago

I think perpendicular lines will also follow the same recursion as $$f(n)$$ and hence the 2nd identity also correct.... · 3 years, 3 months ago

Here you have no where mentioned whether the lines can coincide wholly with each other....... · 3 years, 3 months ago