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

#1**+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!

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

#1**+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!

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