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 November 19th, 2009, 02:09 PM #1 Newbie   Joined: Nov 2009 Posts: 12 Thanks: 0 law of cosines Hello everybody, my question is about pythagorean and "similar"triples. pythagorean triples $(ab,c)\in N$ are triples that confirm the equation $a^2+b^2=c^2$. they can be represented in parametres by this form: $a=2mn, b=m^2-n^2, c=m^2+n^2$ with m>n, and $m,n \in N$ now its my task to find rules for triples in a triangle with integer sidelengths and a angle of 60° oder 120°. with the law of cosines follows: $c^2=a^2+b^2-ab$ or$c^2=a^2+b^2+ab$. do you have some ideas how to find a presentation of parametres? perhaps with an square addition?(is this the right word in english?) one solution for a triangle with a 120°angle is c=7, a=5 and b=3 and i am looking for a formula to generate all these kind of triples. thanks marten
 November 20th, 2009, 08:36 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: law of cosines Assume that one of the shorter sides is a rational multiple of the sum of the other two. Substitute that into c˛=a˛+b˛+ab and see where it takes you.
 November 23rd, 2009, 12:56 PM #3 Newbie   Joined: Nov 2009 Posts: 12 Thanks: 0 Re: law of cosines hi aswoods, thank your for your advice, i dont know if i understood you in the right way, is this what you mean? $a^2+b^2+2ab=c^2\;\;\; |:c^2$ $(\frac{a}{c})^2+(\frac{b}{c})^2+\frac{a}{c}\cdot \frac{b}{c}=1$ substitution: $x^2+y^2+x \cdot y=1$ with $x,y \in Q$ when you use this method for triangle with 90°, you get the formula for the unique circle ($x^2+y^2=1$). but what is $x^2+y^2+x \cdot y=1$? an ellipse?
November 24th, 2009, 09:15 PM   #4
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Re: law of cosines

Quote:
 Originally Posted by MG85 but what is $x^2+y^2+x \cdot y=1$? an ellipse?
Written in the form
$Ax^2 + Bxy +Cy^2 +...=0$
We look at $B^2-4AC$, which is less than zero in this case. So it's an ellipse.
You could rotate the axis, if desired, but I'm not sure how that would help.

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