Triangle length and largest angle

AI Thread Summary
The discussion focuses on proving that the side length (n^2+n+1) is the longest side of a triangle with sides (n^2+n+1), (2n+1), and (n^2-1) for n > 1. Participants confirm that the inequalities n^2+n+1 > 2n+1 and n^2+n+1 > n^2-1 hold true under the given condition. The largest angle, θ, is determined to be 120º using the cosine rule, with confirmation that the variable n cancels out in the calculations. The conversation emphasizes understanding the inequalities and the implications of the condition n > 1. The thread concludes with a consensus on the validity of the proofs provided for both parts of the homework.
314Jason
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Homework Statement



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.


Homework Equations



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

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.
 
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314Jason said:
(b) I have no idea what to do here. I thought about vectors, but don't know where to go from there.

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?
 
314Jason said:
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?

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) \leq (2n+1)

and solve the inequality. What's the possible range for n you get?
 
Hi 314Jason! :smile:
314Jason said:
(a) Explain why the side (n2+n+1) must be the longest side of the triangle
…(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

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:
 
tiny-tim said:
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:

This is what comes of not reading the OP thoroughly. :redface:
 
tiny-tim said:
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:

So because n > 1 is in both equations, this is true?
 
yes, the question specifies that n > 1, so you only have to prove it for n > 1 :wink:
 
Ok then, thanks!
 

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