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Topic: Triangles (Read 348 times) |
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ThudnBlunder
wu::riddles Moderator Uberpuzzler
The dewdrop slides into the shining Sea
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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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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
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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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Wikipedia, Google, Mathworld, Integer sequence DB
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ThudnBlunder
wu::riddles Moderator Uberpuzzler
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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. However, no such mapping was defined.
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