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You have a very large sheet of paper and are going to draw 15 straight lines across it.

Of course they may intersect each other in lots of places.

What's the most number of regions you can divide the paper into this way? For example with only three lines you can get seven regions as in the picture below.

 

Guest Jul 6, 2018
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Problems like these can be daunting, but it is important to perservere and be observant. I was unsure of the solution, so I decided to attempt to find the solution by drawing 0 straight lines to 5 straight lines. I have created an image of my final solutions to all of these. The red line on each indicates where I added a line from the previous image. 
 

 

I will the notation \(f(n)\) to notate the maximum number of regions generated with straight lines. I could not go much further than this, so I decided to stop here. 

 

\(f(0)=1\\ f(1)=2\\ f(2)=4\\ f(3)=7\\ f(4)=11\\ f(5)=16\\ \)

 

I immediately noticed a pattern with this sequence of numbers. This is what I noticed

 

\(f(0)+1=f(1)\\ f(1)+2=f(2)\\ f(2)+3=f(3)\\ f(3)+4=f(4)\\ f(4)+5=f(5)\)

 

This is a quadratic relationship since the numbers 1,2,3,4, and 5 have a second common difference of 1. This means that there is a quadratic equation out there that represents this exact situation. Let's try and find it, shall we? Let's start this by plugging in that \(f(0)=1\):

 

\(f(n)=an^2+bn+c\)Since we already determined that the above relationship can be observed quadratically, let's plug in some values to develop that quadratic. 
\(f(0)=1;\\ f(0)=a*0^2+b*0+c\)This is the easiest one to simplify.
\(\boxed{1}\hspace{3mm}1=c\)We now know that the constant term of the quadratic is 1.
  

 

Let's plug in the next point, \(f(1)=2\):

 

\(f(1)=2;\\ f(1)=a*1^2+b*1+c\)Now, simplify both sides completely. 
\(2=a+b+c\)We established in \(\boxed{1}\) that \(c=1\), so let's substitute that in here.
\(2=a+b+1\)Substract 1 from both sides. 
\(\boxed{2}\hspace{3mm}1=a+b\) 

 

Let's plug in the next point, \(f(2)=4\):

 

\(f(2)=4;\\ f(2)=a*2^2+b*2+c\)Simplify from here. 
\(4=4a+2b+c\)As aforementioned, \(c=1\), so substitute this information in and simplify completely. 
\(4=4a+2b+1\)Subtract 1 from both sides again.
\(\boxed{3}\hspace{3mm}3=4a+2b\) 
  

 

Our next task is to solve this system of equations. Let's do that. I will solve by substitution here because of the presence of a variable with a coefficient of one. 

 

\(\boxed{2}\hspace{3mm}1=a+b\Rightarrow a=1-b\)

 

Since I have a isolated, it is possible to substitute it in for \(\boxed{3}\):

\(\boxed{2}\hspace{3mm}a=1-b;\\ \boxed{3}\hspace{3mm}3=4(1-b)+2b\)Solve for in \(\boxed{3}\).
\(3=4-4b+2b\) 
\(-1=-2b\) 
\(b=\frac{1}{2}\)Let's substitute this into \(\boxed{2}\).
\(b=\frac{1}{2};\\ \boxed{2}\hspace{3mm}a=1-\frac{1}{2}\) 
\(a=\frac{1}{2}\)All the variables have been solved for now!
  


After all this work, we have determined that \(f(n)=\frac{1}{2}n^2+\frac{1}{2}n+1\) is the quadratic that represents the maximum number of regions for straight lines. In this case, n=15, so let's substitute this in and solve. 

 

\(n=15;\\ f(15)=\frac{1}{2}*15^2+\frac{1}{2}*15+1\)Simplify the right hand side of the equation. 
\(f(15)=\frac{225}{2}+\frac{15}{2}+\frac{2}{2}\)Now, do the addition. 
\(f(15)=\frac{242}{2}=121\text{ regions}\)You're done!
  
TheXSquaredFactor  Jul 6, 2018

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