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   Author  Topic: Triangles  (Read 348 times)
ThudnBlunder
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Triangles  
« on: Jan 1st, 2005, 9:46am »
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Prove that if a triangle has sides of length a,b,c then sides of length 1/(a+b), 1/(b+c), 1/(a+c) can form another.
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towr
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Re: Triangles  
« Reply #1 on: Jan 2nd, 2005, 12:02pm »
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::
Would it be sufficient to note that the sides become more average?
If you keep repeating the mapping the triangle approaches an equilateral one.
 
Of course you could also prove it based on
a < b+c and b < a+c and c < a+b,  
which is a sufficient and necessary condition for non-negative a,b,c forming a triangle.
 
The same thing has to hold for the new three numbers so
1/(a+b) < 1/(a+c) + 1/(b+c)
1/(a+b) < [(a+c) + (b+c)]/[(a+c)(b+c)]  
(a+c)(b+c)< (a+b+2c) * (a+b)  
ab+(a+b)c+c2 < a2+2ab+2ac+ b2+2bc
c2 < (a+b)c + a2+ab+b2
which is easily seen to be true since  
c < (a+b)  
and thus  
c2 < (a+b)c (+something>0)
 
and the same goes for the other permutation of a,b,c in the above relations
::
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ThudnBlunder
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Re: Triangles  
« Reply #2 on: Jan 3rd, 2005, 12:30pm »
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Quote:
Would it be sufficient to note that the sides become more average?  
If you keep repeating the mapping the triangle approaches an equilateral one.

A typically incisive observation from towr.  Wink  However, no such mapping was defined.
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towr
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Re: Triangles  
« Reply #3 on: Jan 3rd, 2005, 1:12pm »
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It may not be defined, but it's intuitively there.. Smiley
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