**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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest$\textbf{Problem 1.}$

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

Log in to reply

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

Log in to reply

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

Log in to reply

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.

Log in to reply