I am completely at a loss as to how to approach this problem :(

it's due in about 18 hours, so I'll be working on it only a little bit more, as I also have to sleep sometime tonight

http://i54.tinypic.com/2wh4ln4.png

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- Nov 9th 2010, 09:25 PMjbpellerinarea proof, maybe menelaus
I am completely at a loss as to how to approach this problem :(

it's due in about 18 hours, so I'll be working on it only a little bit more, as I also have to sleep sometime tonight

http://i54.tinypic.com/2wh4ln4.png - Nov 9th 2010, 10:12 PMProve It
What do you mean by $\displaystyle \displaystyle \textrm{[D\,E\,F]}$ and $\displaystyle \displaystyle \textrm{[D\,F\,C]}$?

- Nov 9th 2010, 10:14 PMjbpellerin
Area of triangle DEF and triangle DFC, respectively

- Nov 9th 2010, 10:17 PMProve It
And what work have you done yourself so far? It's not in MHF's policy to knowingly complete work that will be graded - only to give a nudge in the right direction...

- Nov 9th 2010, 10:26 PMjbpellerin
all I've been able to deduce is that [ABF]=[AFC], [BDF]=[FDC]

[AFC]=[AED]+[DEF]+[FDC]

[AFC]=[AED]+1+[BDF]

[ABE]=[ABF]-1

and from there it's just substitution of what I already know and no elimination

and I know we don't provide full solutions, I had some thanks to my name in the past but they seem to be gone since the forum software migration (I spent a long time away) - Nov 10th 2010, 10:23 AMWilmer
Mais tabarnak Monsieur Pellerin, c'est quoi cette hostie de torture ?!

Well, a bit too much for my humble capabilities (plus I hate anything involving Heron's formula!);

so I set up a program looking at AB=AC in 1/4 increments, similarly with BC, and got this:

AB=AC=65 and BC=1/4 (yes, 1/4):

results in areaDEF = .99999815.... (darn close to 1)

and areaCDF = .06249988... (suspiciously close to 5/8)

and AD being 65-1 = 64

So seems you'll get an awfully long skinny isosceles ABC.

Probably won't help you much...but what the heck: any reply is better than none :) - Nov 10th 2010, 12:39 PMjbpellerin
using menalaus

I got AD = 3/2

then menalaus again BE/ED = 5/3

then the base*height ratio thing (forget the name)

I get 5/3 = BE/ED = [BEF]/[DEF] = [BEF]

and then [DFC]=[BDF]=[DEF]+[BEF] = 1+5/3 = 8/3

your solution didn't make sense since it would make [BFD] less than 1, but [BFD] contains [DEF]=1

hurray for sleeping on it then trying again with fresh insight - Nov 10th 2010, 03:39 PMWilmer