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August 21st, 2013, 10:18 PM   #1
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compare s and t

Triangle ABC (area =1/4) ,with side length a,b,c and the radius of its circumscribed circle =1

given :





compare s and t
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August 22nd, 2013, 05:10 AM   #2
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Re: compare s and t

Let A be area of triangle and R radius of circumscribed circle.
[1]





RMS-AM inequality



Lemma: for a,b,c that abc=1 [2]

Proof:




adding we get [2]


/add 2a+2b+2c







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August 22nd, 2013, 05:28 AM   #3
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Re: compare s and t

it should be s < t,
or t > s
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August 22nd, 2013, 06:01 AM   #4
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Re: compare s and t

RMS-AM equality is only equal in the case that a=b=c.
Using the sine rule and multiplying it with each other we see that
so
If then we have a equiliteral triangle with angles of 60

Which is a contradiction. So this means the sides of the triangle can not be equal and thus there is no equality possible. That's why we should use t>s
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August 22nd, 2013, 06:16 AM   #5
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How large can s/t be?
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August 24th, 2013, 04:08 AM   #6
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Re: compare s and t







we get :







(1)+(2)+(3) and we obtain t > s
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August 24th, 2013, 05:43 AM   #7
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Re: compare s and t

[quote="Albert.Teng"]





I'm confused. If A = the area of triangle ABC, then which are you claiming is true ... A = 1/4 or A > 1/4?
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August 24th, 2013, 08:03 AM   #8
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The intention was presumably to show that a = b = c = 1 is not possible, as the area would not be 1/4.
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August 24th, 2013, 10:21 AM   #9
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Re:

Quote:
Originally Posted by skipjack
The intention was presumably to show that a = b = c = 1 is not possible, as the area would not be 1/4.
Ahh... got it! Once that is out then the AM-GM inequality is true but with equality eliminated.
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