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Topic: Triangles (Read 503 times) |
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ThudnBlunder
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For i = 1,2 let Ti be a triangle with side lengths ai, bi, ci, and area Ai.
Suppose that
a1 [smiley=eqslantless.gif] a2
b1 [smiley=eqslantless.gif] b2
c1 [smiley=eqslantless.gif] c2
T2 is an acute triangle.
Does it follow that A1 [smiley=eqslantless.gif] A2 ?
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| « Last Edit: Dec 6th, 2004, 3:00am by ThudnBlunder » |
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towr
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Re: Triangles
« Reply #1 on: Dec 6th, 2004, 12:39am » |
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a2+b2=c2+e
for e -> 0, A2 -> 0
So for any positive A1, A2 may be smaller
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| « Last Edit: Dec 6th, 2004, 12:41am by towr » |
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Icarus
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Re: Triangles
« Reply #2 on: Dec 6th, 2004, 2:00am » |
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But for small e, T2 is no longer acute, so the problem is not quite that simple.
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towr
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Re: Triangles
« Reply #3 on: Dec 6th, 2004, 2:28am » |
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Sorry, I read 'has an acute triangle' as 'has an acute angle', rather than as 'is an acute triangle'.
(The former is of course trivially true for any triangle, so I suppose the latter was intended)
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ThudnBlunder
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Re: Triangles
« Reply #4 on: Dec 6th, 2004, 2:58am » |
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on Dec 6th, 2004, 2:28am, towr wrote:Sorry, I read 'has an acute triangle' as 'has an acute angle', rather than as 'is an acute triangle'.
(The former is of course trivially true for any triangle, so I suppose the latter was intended) |
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Sorry for the typo. I have amended it.
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| « Last Edit: Dec 6th, 2004, 7:33am by ThudnBlunder » |
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Icarus
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Re: Triangles
« Reply #5 on: Dec 6th, 2004, 5:47pm » |
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I guess it is that simple after all:
At least one of the three angle conditions
[alpha]1 [le] [alpha]2
[beta]1 [le] [beta]2
[gamma]1 [le] [gamma]2
must hold, since both sets of angles sum up to [pi]. Assume wlog that [gamma]1 [le] [gamma]2.
Since [gamma]2 < [pi]/2, sin [gamma]1 [le] sin [gamma]2.
So A1 = (1/2)a1b1sin [gamma]1 [le] (1/2)a2b2sin [gamma]2 = A2.
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| « Last Edit: Dec 6th, 2004, 5:48pm by Icarus » |
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Barukh
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Re: Triangles
« Reply #6 on: Dec 7th, 2004, 6:47am » |
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Very elegant, Icarus!
A friend of mine gave the following nice argument: consider 2 triangles with the same base and same altitude to the base (= same area). Look at the 4 sides remained. Because the angles at the base are acute, it is easy to see that the biggest and the smallest side belong to the same triangle.
BUT: does it really belong to the Putnam exam section? IMHO, putting it here, drives away the attention of many potential puzzlers.
It seems like THUD&BLUNDER discovered recently an unknown Putnam exam treasure 
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| « Last Edit: Dec 7th, 2004, 6:55am by Barukh » |
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ThudnBlunder
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Re: Triangles
« Reply #7 on: Dec 7th, 2004, 8:15am » |
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Quote:BUT: does it really belong to the Putnam exam section?
It seems like THUD&BLUNDER discovered recently an unknown Putnam exam treasure |
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Aren't those two points of view inconsistent? 
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| « Last Edit: Dec 8th, 2004, 3:14am by ThudnBlunder » |
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