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Triangle length and largest angle

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A triangle has sides of length (n2+n+1), (2n+1) and (n2-1), where n > 1.

    (a) Explain why the side (n2+n+1) must be the longest side of the triangle

    (b) Show that the largest angle, θ , of the triangle is 120º.


    2. Relevant equations

    In a triangle of sides a, b and c: a - b < c < a + b.

    3. The attempt at a solution

    (a)

    n2+n+1 > 2n+1
    n2-n > 0
    n(n-1) > 0
    Thus, n < 0, n > 1.

    n2+n+1 > n2-1
    n+1 > -1
    n > -2

    I thought they would both give n > 1. I don't know what other way to show this is true.


    (b) I have no idea what to do here. I thought about vectors, but don't know where to go from there.
     
  2. jcsd
  3. May 1, 2012 #2
    I got (b)! I can just public the sides into the cos rule and rearrange to find theta. The n should cancle out.

    Does anyone know how to do (a)? Please?
     
  4. May 1, 2012 #3

    Curious3141

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    First of all, it should be immediately obvious that for n>1,

    (n2+n+1) > (n2-1)

    Now all you need to prove is that for n>1,

    (n2+n+1) > (2n+1)

    Consider the opposite proposition: (n2+n+1) [itex]\leq[/itex] (2n+1)

    and solve the inequality. What's the possible range for n you get?
     
  5. May 1, 2012 #4

    tiny-tim

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    Hi 314Jason! :smile:
    but you have proved it …

    you've proved it's true if {n < 0 or n > 1} and {n > -2} …

    so, in particular, it's true for {n > 1} ! :wink:
     
  6. May 1, 2012 #5

    Curious3141

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    This is what comes of not reading the OP thoroughly. :redface:
     
  7. May 1, 2012 #6
    So because n > 1 is in both equations, this is true?
     
  8. May 1, 2012 #7

    tiny-tim

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    yes, the question specifies that n > 1, so you only have to prove it for n > 1 :wink:
     
  9. May 1, 2012 #8
    Ok then, thanks!
     
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