If $a^n, b^n$ and $c^n$ form a triangle for all $n \in \Bbb N$, prove that the triangles are isosceles.

Question

If $a^n, b^n$ and $c^n$ form a triangle for all $n \in \Bbb N$, prove that the triangles are isosceles given that $a\geq b\geq c> 0$.

I started on the lines of $a< b+c$. But no idea how to proceed further. Tried using cosine law, but no help.

Hint: if $a>b>c>0$, then $c^n$ is growing slower than $a^n-b^n$ — Jean-Claude Arbaut, May 18 '18 at 18:36
Also recently asked and answered: https://math.stackexchange.com/q/2780470/42969. — Martin R, May 18 '18 at 18:55

score 1 · Answer 1 · answered May 18 '18 at 18:39

1

Hint Look at the possible values of $$\lim_{n \to \infty} \frac{b^n+c^n}{a^n}$$

answered May 18 '18 at 18:39

N. S.

score 0 · Answer 2 · answered May 18 '18 at 18:50

0

let's assume, wolog $a\ge b \ge c$.

To be a triangle we need $c^n > a^n - b^n =(a-b)(a^n + a^{n-1}b + .... + ab^{n-1} + b^n)$

If $a = b$ this is always true but if $a \ne b$. Then:

we need $c^n > a^n - b^n =$

$(a-b)(a^n + a^{n-1}b + .... + ab^{n-1} + b^n) > $

$(a-b)(a^n)$ or in other words we need

$a-b > \frac {c^n}{a^n} = (\frac ca)^n$.

but $0< \frac ca < 1$ so $\lim_{n\to \infty } (\frac ca)^n = 0$.

Which means there is an $N$ so that $(\frac ca)^N < a-b$.

So it is not possible for all $n$ for $a-b > (\frac ca)^n$.

So $a \ne b$ is impossible and $a = b$.

answered May 18 '18 at 18:50

fleablood

1

Whoops.. you were given the inequalities in the exact opposite order. Let me reedit. – fleablood May 18 '18 at 18:52
Understood. All this time I was trying to find a way to prove b=c. As they were the smaller numbers. – Red2504 May 18 '18 at 19:06
Actually I wasn't sure where the proof would go and hearing the question I thought it'd have to do with the smaller sides too. But the proof went where it wanted to go. And in hindsight its obvious, but there is something that intuitively look for the smaller sides. not sure what it is. – fleablood May 18 '18 at 20:26

2 Answers2