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alexb
Charter Member
2482 posts 
Mar0910, 11:43 AM (EST) 

1. "RE: Acute angled triangles"
In response to message #0

First of all, by the theorem of sines, x<=y<=z, implies X<=Y<=Z. With this in mind you only have to prove that Z <= X + Y. But this is obvious for an acute triangle; for then Z <= 90 <= X + Y. Note that for obtuse triangles the statement is not true. 

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johnm_12
guest

Mar1110, 06:21 PM (EST) 

2. "RE: Acute angled triangles"
In response to message #1

Nearly. You have to prove that Z<X+Y etc. Since the triangle is acute angled, Z < π/2 (or 90  makes no difference) etc. Also, X + Y + Z = π > 2Z (from above). Hence, X + Y > Z or Z < X + Y. Also, not true for Rightangled triangles as well as Obtuse. Interestingly (or not) I have never seen this in print before. I wonder if for all the years that mathematicians have been looking at triangles, they have just never noticed this. There are some interesting consequences here!!! E.g. 0 < π/2  Z <= 3XY/2π Also, if x, y, z is a non trivial solution to Fermat ... z^n = x^n + y^n, n>=2, then (x, y, z) forms an acute angled triangle only if n>=3. 

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alexb
Charter Member
2482 posts 
Mar1110, 06:26 PM (EST) 

3. "RE: Acute angled triangles"
In response to message #2

>Nearly. Yes, got it in a rush. >z^n = x^n + y^n, n>=2, then (x, y, z) forms an acute angled >triangle only if n>=3. This I have seen, although a reference escapes me.


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