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        riddles >> easy >> Triangles
        
(Message started by: THUDandBLUNDER on Jan 1^st, 2005, 9:46am)

Title: Triangles
Post by THUDandBLUNDER on Jan 1^st, 2005, 9:46am

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.

Title: Re: Triangles
Post by towr on Jan 2^nd, 2005, 12:02pm

::[hide]
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
[/hide]::

Title: Re: Triangles
Post by THUDandBLUNDER on Jan 3^rd, 2005, 12:30pm

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. ;) However, no such mapping was defined.

Title: Re: Triangles
Post by towr on Jan 3^rd, 2005, 1:12pm

It may not be defined, but it's intuitively there.. :)