# area proof, maybe menelaus

• Nov 9th 2010, 09:25 PM
jbpellerin
area 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 PM
Prove It
What do you mean by $\displaystyle \textrm{[D\,E\,F]}$ and $\displaystyle \textrm{[D\,F\,C]}$?
• Nov 9th 2010, 10:14 PM
jbpellerin
Area of triangle DEF and triangle DFC, respectively
• Nov 9th 2010, 10:17 PM
Prove 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 PM
jbpellerin
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 AM
Wilmer
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.

• Nov 10th 2010, 12:39 PM
jbpellerin
using menalaus
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 PM
Wilmer
Quote:

Originally Posted by jbpellerin
using menalaus I got AD = 3/2 .....
your solution didn't make sense since it would make [BFD] less than 1, but [BFD] contains [DEF]=1

Yep; my bad; at least I got to meet Mr. Menalaus!