MHB Can a Triangle with Sides from This Number Set Avoid Being Isosceles?

[anemone]

Sorry for being late in coming to do the POTW, I just didn't feel well last week due to the prolonged fever, flu, sore throat and cough. To help make up for being late, I will present to you an intriguing problem which I truly hope you are going to have lots of fun solving the problem!

Without further ado, here is this week's POTW:

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A triangle is built with the lengths of its sides chosen from the set $\{2,\,3,\,5,\,8,\,13,\,21,\,34,\,55,\,89,\,144\}$.

Prove that this triangle must be isosceles.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. greg1313
2. Ackbach
3. lfdahl
4. kaliprasad

Solution from Ackbach:

Spoiler

The numbers given are all consecutive Fibonacci numbers. Suppose the triangle is not isosceles. Then it must use three distinct numbers from the set. However, by the triangle inequality, if $a,b,c$ are the three lengths of a triangle, with no side longer than $a$, then $a\le b+c$. That is, according to our assumption, $a$ must be the largest number (since distinct). However, because the numbers are Fibonacci, the Triangle Inequality forces $b$ and $c$ to be the two numbers immediately preceding $a$ in the Fibonacci sequence. But then we get equality in the Triangle Inequality by definition of the Fibonacci sequence. The Triangle Inequality then says that equality happens only in the degenerate case of zero triangle area. That is, you basically don't have a triangle anymore. Therefore, our starting assumption that the triangle is not isosceles is nonsense.