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For how many positive integers $p$ does there exist a triangle with sides of length $3p-1,$ $3p,$ and $p^2 + 1?$

Guest Apr 22, 2018

Best Answer 

 #1
avatar+1961 
+3

We are going to be using this theorem: the Triangle Inequality Theorem. I think a picture can tell a thousand words, and I think this picture does just that. It may help you understand more than the words I can produce on the page!
 

Source: http://www.mathwarehouse.com/geometry/triangles/images/triangle-inequality-theorem/imagestriangle-inequality-theorempicture-of-triangle-inequality-theorem-and-formula.png

 

If the sum of the shorter side must be greater than the final side, then we must consider three cases:

 

  • The side length 3p-1 is the longest side
  • The side length 3p is the lonest side
  • The side p^2+1 is the longest side

Let's deal with the simplest case. Let's assume that 3p-1 is the longest side. Wait! It cannot be the longest side. Why? Well, 3p-1 will always be less than 3p, so this side length cannot possibly be the shortest side length. The subtraction of one causes me two know this for certain. This leaves us with the next viable possibility: 3p is the longest side length.

 

\(3p-1+p^2+1>3p\)If 3p is the longest side, then I am assuming that the other 2 are shorter sides. We can use the relationship aforementioned via the Triangle Inequality Theorem, and we can solve for one possible range for p. Notice how there is a lot that will cancel out in this inequality.
\(p^2>0\)Take the square root of both sides.
\(|p|>0\)An absolute-value inequality, when a greater-than symbol is involved, results in an or-type solution set. Let's write those out.
\(p<0\text{ or }p>0\)This did not really restrict the options that much. Let's consider the next possibility.
  

 

The next possibility is that p^2+1 is the longest side length.

 

\(3p-1+3p>p^2+1\)This inequality is going to be much harder. Let's just everything to one side of the inequality.
\(6p-1>p^2+1\) 
\(p^2-6p+2=0\)I think it is easiest to pretend, for now, that this inequality is actually a regular equation. In this case, it is not factorable, so we will have to use another method to solve this quadratic.
\(a=1;b=-6;c=2\\ p_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(2)}}{2(1)}\)The rest is a matter of simplification. 
\(p_{1,2}=\frac{6\pm\sqrt{28}}{2}\)The radical can use some simplfiying here.
\(p_{1,2}=\frac{6\pm\sqrt{28}}{2}=\frac{6\pm2\sqrt{7}}{2}\)A common factor between every term is two. Factor that out.
\(p_1=3+\sqrt{7}\\ p_2=3-\sqrt{7}\)Use test points to find the range that satisfies the original inequality. I found that the following range for p satisfied the original equation.
\(3-\sqrt{7} Let's convert these to decimal approximations so that we can determine the number of integer solutions easier.
\(\approx0.35  
  

 

Therefore, every positive integer in between these two numbers is the answer. The integers are 1,2,3,4, and 5. This is a total of 5 integers for p.

TheXSquaredFactor  Apr 22, 2018
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1+0 Answers

 #1
avatar+1961 
+3
Best Answer

We are going to be using this theorem: the Triangle Inequality Theorem. I think a picture can tell a thousand words, and I think this picture does just that. It may help you understand more than the words I can produce on the page!
 

Source: http://www.mathwarehouse.com/geometry/triangles/images/triangle-inequality-theorem/imagestriangle-inequality-theorempicture-of-triangle-inequality-theorem-and-formula.png

 

If the sum of the shorter side must be greater than the final side, then we must consider three cases:

 

  • The side length 3p-1 is the longest side
  • The side length 3p is the lonest side
  • The side p^2+1 is the longest side

Let's deal with the simplest case. Let's assume that 3p-1 is the longest side. Wait! It cannot be the longest side. Why? Well, 3p-1 will always be less than 3p, so this side length cannot possibly be the shortest side length. The subtraction of one causes me two know this for certain. This leaves us with the next viable possibility: 3p is the longest side length.

 

\(3p-1+p^2+1>3p\)If 3p is the longest side, then I am assuming that the other 2 are shorter sides. We can use the relationship aforementioned via the Triangle Inequality Theorem, and we can solve for one possible range for p. Notice how there is a lot that will cancel out in this inequality.
\(p^2>0\)Take the square root of both sides.
\(|p|>0\)An absolute-value inequality, when a greater-than symbol is involved, results in an or-type solution set. Let's write those out.
\(p<0\text{ or }p>0\)This did not really restrict the options that much. Let's consider the next possibility.
  

 

The next possibility is that p^2+1 is the longest side length.

 

\(3p-1+3p>p^2+1\)This inequality is going to be much harder. Let's just everything to one side of the inequality.
\(6p-1>p^2+1\) 
\(p^2-6p+2=0\)I think it is easiest to pretend, for now, that this inequality is actually a regular equation. In this case, it is not factorable, so we will have to use another method to solve this quadratic.
\(a=1;b=-6;c=2\\ p_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(2)}}{2(1)}\)The rest is a matter of simplification. 
\(p_{1,2}=\frac{6\pm\sqrt{28}}{2}\)The radical can use some simplfiying here.
\(p_{1,2}=\frac{6\pm\sqrt{28}}{2}=\frac{6\pm2\sqrt{7}}{2}\)A common factor between every term is two. Factor that out.
\(p_1=3+\sqrt{7}\\ p_2=3-\sqrt{7}\)Use test points to find the range that satisfies the original inequality. I found that the following range for p satisfied the original equation.
\(3-\sqrt{7} Let's convert these to decimal approximations so that we can determine the number of integer solutions easier.
\(\approx0.35  
  

 

Therefore, every positive integer in between these two numbers is the answer. The integers are 1,2,3,4, and 5. This is a total of 5 integers for p.

TheXSquaredFactor  Apr 22, 2018

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