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 Subject: "Acute angled triangles" Previous Topic | Next Topic
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johnm_12
Member since Mar-9-10
Mar-09-10, 11:14 AM (EST)    "Acute angled triangles"

 If an acute angled triangle has sides of length x, y and z, where x<=y<=z, show that the corresponding angles, X, Y, and Z, can also form the sides of a triangle.

alexb Charter Member
2482 posts
Mar-09-10, 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. johnm_12 guest
Mar-11-10, 06:21 PM (EST)

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

 Nearly. You have to prove that Z 2Z (from above). Hence, X + Y > Z or Z < X + Y.Also, not true for Right-angled 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/2πAlso, 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. alexb Charter Member
2482 posts
Mar-11-10, 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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