Cutting a Plane with Nine Lines

Consider the case for 3 lines.In the first place,we only have 2 lines with 4 regions.If we add another line,the most number of intersections of this line with the other lines is 2.Then this line must pass through 3 regions(The region touching 1st intersection,both,and 2nd intersection).Therefore,these 3 regions are divided into 2,which makes an additional 3 regions.Continue this process,we have a total of:$$4+3+4+\ldots+9=46$$ regions

By observation, if k lines cut the plane, there is an increment of k regions from the previous, which is if (k-1) lines cut the plane. This is because there are (k-1) lines to divide the kth line, producing additional [(k-1)+1] regions, or k regions. Now, since the initial number of region is 1, the most number of regions that k lines can cut the plane into is $1+\displaystyle \sum_{i=1}^k n$ , which simplifies to $\frac {k^2}{2} +\frac {k}{2} +1$ . If k = 9, there will be at most 46 regions.

By listing the number of lines and the maximum possible regions in existence as x and y values respectively for some function, the problem can be analyzed as determining a correlation between the number of lines present and the most number of regions. A pattern arises such that the maximum number of regions added by the nth line is n. This is verified by graphing 3 lines so as to maximize the regions of the plane. By maximizing the intersection points, there are 7 planes total, 3 having been added by the introduction of the third line. Once the pattern has been realized, it is simple to derive the formula. As each nth line is added, it adds n more regions. This is just the sum of 1,2,...,9. Using the formula n(n+1)/2, we find that the ninth line being added would result in 45 total regions, but this assumes a value of 0 for the initial term, which is not the case. This means that the original term, 1, must be added back into the answer, giving us 46.

Each successive nth cut will add exactly n regions to the plane. Counting up to 9, there will be exactly 46 regions in the plane after 9 cuts.

We note that for n=0 there is 1 region, n=1 there are two regions, n=2 there are 4 regions, and for n=3 there are 7 regions

from the pattern 1,2,4,7,... we can see that it is an arithmetic sequence with two degrees

so we can write the term into quadratic function $f(x) = ax^{2}+bx+c$ input n=0,1,2 so for n=0 we know that c=1 for n=1, a+b+c=2 then a+b =1 for n=2, 4a+2b+c=4 then 4a+2b=3

with elimination we can find that $a=\frac{1}{2}$ and $b=\frac{1}{2}$ so $f(x)= \frac{1}{2}x^{2} + \frac{1}{2}x +1$ and $f(9)=46$

let the nth line cuts the plane into A {n} parts. this means that when we draw the nth line we add some parts to A {n-1} parts previously formed by the n-1 lines. so we count the no. of parts we add in terms of n then apply reccurence relation and get that n lines divide a plane into n(n+1)/2 parts.

Two lines can divide a plane in maximum 4 regions when they are coplaner.

Three lines can divide a plane in maximum number of regions when it is coplaner and non-coincident with the other two, thus forming a triangle and creating 7 regions. The fourth line, for maximum regions, must also be non coincident with any two of the previous lines.
so, for maximum regions, each new line has to be coplaner as well as non coincident with any two other lines. The sequence goes thus, 1 line _ 2 regions 2 lines_4 regions 3 _ 7 4 _ 11 . . Notice the A.P in the differences of the number of regions. Hence, for nine lines, 46 regions

There must be a sequence between the number of lines and the number of regions.

First, there is a blank plane.

1 line cuts the plane into 2 regions

another line cuts the plane into another 2 regions and become 4 regions

another line intersects with the two previous line and add another 3 regions and become 7 regions

Hence, the sequence between the lines and the number of regions are as the followings.

1 $\rightarrow$ 2

2 $\rightarrow$ 2 + 2 = 4

3 $\rightarrow$ 4 + 3 = 7

4 $\rightarrow$ 7 + 4 = 11

5 $\rightarrow$ 11 + 5 = 16

6 $\rightarrow$ 16 + 6 = 22

7 $\rightarrow$ 22 + 7 = 29

8 $\rightarrow$ 29 + 8 = 37

9 $\rightarrow$ 37 + 9 = 46

Hence, the answer is 46

When we have n lines and add a new line then this new line can cut all the previous n lines and hence contribute (n+1) to the total. This implies something like the triangular numbers.

Let us say that the total is n(n+1)/2 + 1 which is true for n=0,1,2. Assume it is true for n. For the next line we add (n+1) to get n(n+1)/2+1+(n+1)=(n+1)(n+2)/2+1 which follows the induction rule.

Now take n=9 and we get 9*10/2+1 = 46

We can find a general formula for the number of regions when a plane is divided by n lines.

Suppose the number of regions is given by f(n) when there are n lines drawn on the plane. Now draw one more line cutting all the other n lines. There are n points on the additional line and so this line must traverse n+1 of the available f(n) regions, dividing each into 2 parts. It therefore adds n+1 more regions to those present. Therefore:

f(n+1) = f(n) + n + 1

We can write f(n+1) - f(n) = n + 1, and now form a succession of equations as follows:

f(1) - f(0) =  1
f(2) - f(1) =  2
f(3) - f(2) =  3
...................
...................
f(n-1) - f(n-2) =  n-1
f(n)   - f(n-1) =  n

---

f(n) - f(0)  = SUM(1 to n)      [adding all the equations, we get cancellations between lines]

f(n) - 1  = n(n+1)/2

f(n) = n(n+1)/2  + 1

Now there are 9 lines.

f(9) = 9(9+1)/2 + 1

f(9) = 9*10/2 + 1

f(9) = 45 + 1

f(9) = 46

from question, we know that f(0)=1,f1)=2,f(2)=4 so , 0 - 1 1 1 - 2 1/2 2 2 - 4

so,we got the equation f(x)=1+x+(1/2)(x)(x-1) f(9)=1+9 +4(9) =46

answer is 46. first you have 0 lines and 1 region next 1 line and 2 regions, next 2 lines and cut into 4 regions.. likewise 3 lines can cut into 7 regions and 4 lines to 11 regions, 5 lines to 16 regions.. the sequence goes like this.. 6 lines to 22 regions, 7 lines to 29 regions, 8 lines to 37 regions, 9 lines to 46 regions.....

total number of regions that the 9 line will cut the plane will be 1+sigma (n) where n is the number of the lines

so it will be 1+ sigma (9)= 1+45= 46

so here n=9 , answer=[9[9+1]/2]+1

0 lines intersects plane in 1 region

1 line intersects plane in 2 region

similarly,

2 lines \rightarrow 4 regions

3 lines \rightarrow 7 regions

4 lines \rightarrow 11 regions

5 lines \rightarrow 16 regions

6 lines \rightarrow 22 regions

7 lines \rightarrow 29 regions

8 lines \rightarrow 37 regions

9 lines \rightarrow 46 regions

$For\,one\,line,\,the\,plane\,is\,divided\,into\,2\,parts.\\ So,\,let\,us\,de\!fine\:f(1)\,=\,2\\ Similarly\:two\,lines\,divide\,the\,plane\,into\,4\, parts.\: f(2)\,=\,4\\ For\,three\,lines ,\:f(3)\, =\, 7 \\ Hence\, we\,can\,see\,that\,consecutive\,terms\,have\,their\,dif\!ferences\,in\\Arithmetic\, Progression\: ( 2, 3, ... )\\ Hence\,we\, can\,conclude\: f(4)\, =\, 11 \\ The\, Recurrence\,relation\, is\: f(n)\, = \,f(n-1)\, +\, n \\ On\, generalizing,\, f(n)\, is\, found\, to\, be\: \frac{(n-1)(n+2) + 4}{2}\\ Putting\, n=9,\, f(9)\, =\, 46\\ Answer: 46$

To solve this problem, you have to try the problem first, manually, for the first 3 lines on a piece of paper and start dividing the paper into regions using lines.

Let the number of plane regions partitioned be a function of the number of lines, n. Then, if

n = 0 1 2 3 ... f(n)= 1 2 4 7 ...

The trick here is that you have to be able to identify that this kind of progression is a quadratic one. To do so, you have to find the common difference of the terms by subtracting two consecutive terms (like what you do to check if a progression is linear). However, you have to do it twice since it's a second degree progression.

                    f(n)=   1      2     4     7 ...
     First difference of f(n): 1      2      3...
         Second difference of f(n):1      1      1      1....

If f(n) is quadratic, then f(n)= An^2 + Bn + C, where A,B, and C are constants.

If n=0, then f(0)= A(0^2) + B(0) + C = 1

                        --> C = 1

If n=1, then f(1)= A(1^2) + B(1) + C = 2

                        --> A + B + C = 2

                        --> A + B + 1 = 2

                        --> A + B = 1      [EQ. 1]

If n=2, then f(2)= A(2^2) + B(2) + C =2

                        --> 4A + 2B + C = 4

                        --> 4A + 2B + 1 = 4

                        --> 4A + 2B = 3  [EQ. 2]

Adding [EQ. 2] from two times the negative of [EQ. 1] we get:

                       --> 2A = 1

                       --> A = 1/2

Substituting A = 1/2 to [EQ. 1] we get:

                       --> 1/2 + B =1

                       --> B = 1/2

f(n)= (n^2)/2 + n/2 + 1 since A = 1/2 B = 1/2 C = 1.

If n = 9, then f(9) = (9^2)/2 + 9/2 + 1

                        --> f(9) = 81/2 + 9/2 + 1

                        --> f(9) = 46

The maximum number of regions partitioned when a plane is divided by 9 lines is 46.