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February 19th, 2013, 07:13 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, 02:38 PM   #2
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S = BCCD - a - b - c
= (BE + EC)(CF + FD) - BE(CF + FD)/2 - ECCF/2 - (BE + EC)FD/2
= BECF/2 + ECCF/2 + ECFD/2
S = (BECF + CFEC + FDEC)/4 + (BECFEC + BECFFDEC + CFFDEC)/2

(a + b + c) - 4ac = (BE(CF + FD)/2 + ECCF/2 + (BE + EC)FD/2) - BEFD(CF + FD)(BE + EC)
= (BE(CF + 2FD)/2 + ECCF/2 + ECFD/2) - BEFD(CF + FD)(BE + EC)
= (BECF + CFEC + FDEC)/4 + (BECFEC + BECFFDEC + CFFDEC)/2

The required result follows.
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February 19th, 2013, 04:44 PM   #3
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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?
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February 19th, 2013, 04:50 PM   #4
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I was doing this kind of answer when I was aged twelve. I wonder whether Albert.Teng has a neater method.
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February 19th, 2013, 05:34 PM   #5
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Re: Area of triangle AEF

Let k = length(AB) * length(AD).

S = k - a - b - c
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February 19th, 2013, 11:52 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
--------------------------------------------------------------------------------
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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