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February 19th, 2013, 06:13 AM   #1
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Area of triangle AEF

[attachment=0:2yqq22o7]area of triangle AEF.jpg[/attachment:2yqq22o7]
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 February 19th, 2013, 01:38 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,627 Thanks: 2077 S = BC·CD - a - b - c    = (BE + EC)(CF + FD) - BE(CF + FD)/2 - EC·CF/2 - (BE + EC)FD/2    = BE·CF/2 + EC·CF/2 + EC·FD/2 S² = (BE²CF² + CF²EC² + FD²EC²)/4 + (BE·CF²EC + BE·CF·FD·EC + CF·FD·EC²)/2 (a + b + c)² - 4ac = (BE(CF + FD)/2 + EC·CF/2 + (BE + EC)FD/2)² - BE·FD(CF + FD)(BE + EC)                           = (BE(CF + 2FD)/2 + EC·CF/2 + EC·FD/2)² - BE·FD(CF + FD)(BE + EC)                           = (BE²CF² + CF²EC² + FD²EC²)/4 + (BE·CF²EC + BE·CF·FD·EC + CF·FD·EC²)/2 The required result follows.
 February 19th, 2013, 03:44 PM #3 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: Area of triangle AEF I'm consistently amazed and impressed by these problems by Albert Teng and the answers from those who explain them. Have you considered NOT posting the answers right away, Albert?
 February 19th, 2013, 03:50 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,627 Thanks: 2077 I was doing this kind of answer when I was aged twelve. I wonder whether Albert.Teng has a neater method.
 February 19th, 2013, 04:34 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,413 Thanks: 1024 Re: Area of triangle AEF Let k = length(AB) * length(AD). S = k - a - b - c
February 19th, 2013, 10:52 PM   #6
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Re: Area of triangle AEF

Quote:
 Originally Posted by johnr I'm consistently amazed and impressed by these problems by Albert Teng and the answers from those who explain them. Have you considered NOT posting the answers right away, Albert?.
johnr :thank you very much ! OK ! I will not post the answers right away
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Quote:
 Originally Posted by skipjack I was doing this kind of answer when I was aged twelve. I wonder whether Albert.Teng has a neater method.
I am thinking -----

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