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August 21st, 2013, 10:18 PM  #1 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  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 
August 22nd, 2013, 05:10 AM  #2 
Senior Member Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11  Re: compare s and t
Let A be area of triangle and R radius of circumscribed circle. [1] RMSAM inequality Lemma: for a,b,c that abc=1 [2] Proof: adding we get [2] /add 2a+2b+2c 
August 22nd, 2013, 05:28 AM  #3 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: compare s and t
it should be s < t, or t > s 
August 22nd, 2013, 06:01 AM  #4 
Senior Member Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11  Re: compare s and t
RMSAM 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 
August 22nd, 2013, 06:16 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 21,034 Thanks: 2269 
How large can s/t be?

August 24th, 2013, 04:08 AM  #6 
Senior Member Joined: Apr 2012 Posts: 799 Thanks: 1  Re: compare s and t we get : (1)+(2)+(3) and we obtain t > s 
August 24th, 2013, 05:43 AM  #7 
Senior Member Joined: Feb 2010 Posts: 714 Thanks: 151  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? 
August 24th, 2013, 08:03 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 21,034 Thanks: 2269 
The intention was presumably to show that a = b = c = 1 is not possible, as the area would not be 1/4.

August 24th, 2013, 10:21 AM  #9  
Senior Member Joined: Feb 2010 Posts: 714 Thanks: 151  Re: Quote:
 

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