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Topic: Geometry Problem
Replies: 17   Last Post: Feb 20, 2008 10:04 AM

 Messages: [ Previous | Next ]
 Carl Barron Posts: 152 Registered: 12/13/04
Re: Geometry Problem
Posted: Feb 20, 2008 12:30 AM

In article <mn.9c387d829325ccd5.77903@nomail.com>, rb
<wnowmailw@nomail.com> wrote:

> > rb wrote :
> >> Kaba :
> >>> rb :
> >>>>
> >>>> http://img180.imageshack.us/img180/8286/20080218181526nq7.gif
> >>>>

> >
> > Seems the drawing is distorted (y scale factor != x scale factor)
> >

> >>>> The angle to find is ECD.
> >>>> The matter is solving it without trigonometric functions.

> >>>
> >>> Interesting problem, couldn't solve it either. I get 4 unknowns but
> >>> only 3 linearly independent equations.

> >>
> >> I'm not sure if an equation system is needed.

> >
> > you may consider 180° = x + y + 20° as an equation ;-)
> >

> >> I guess drawing a few parallels and looking for properties like
> >> angle bisector theorem, Thales' theorem, etc... should do the trick
> >> by showing two similar triangles.

> >
> > Or maybe isosceles triangles, or even equilateral ones...
> >
> > Knowing the answer (ECD=20°) shows ABC is isosceles (AB=BC).
> > But how to proove it directly ???
> > We already have another isosceles triangle if we extend line ED and
> > BC to their intersect point F : FEB is isosceles (angles FEB = FBE)
> > How to use this FEB triangle to get other properties ?

>
> Yes, I was working mainly on this triangle and on a parallel to BC
> passing through D. This gives two 20° angles and also shows the
> bisector theorem in the triangle EBD but I don't know if it's
> necessary...
>

> > The problem here is we have no hint for which point/line would give
> > a specific (hence right, isosceles, equilateral) triangle which
> > would give the searched angles.
> >
> > This is similar to another problem, being discussed two weeks ago
> > on a french newsgroup.
> > I've put it on my Web site, as well as this (unsolved) one.
> >
> > <http://chephip.free.fr/pbm_en/pb226.html>
> >
> > The hint there was to add a point at a specific 20° angle, which
> > results then into isosceles triangles.

>
> Nice one!
> It's really very similar. I wonder if mine is a particular case too.
>

> > Without the hint, we need hard trigonometric calculations.
> > (not very hard but still harder than "angles sum up to 180° in a
> > triangle")
> > These trig equations is the ones that lack to Kaba to solve his system.
> >
> > Unfortunately, I didn't either find a suitable point for this new
> > problem. (the 20° trick works only in the other problem's angle values)
> >
> > Regards.

>
> Thanks for your help ;)
>
>

I lost my notes :) but see
http://chephip.free.fr/pbm_en/sol226.html the last figure looks the
same as your problem in disguise.

Date Subject Author
2/18/08 rb
2/18/08 mensanator
2/18/08 none@here.com
2/18/08 rb
2/19/08 Philippe 92
2/19/08 rb
2/20/08 Carl Barron
2/20/08 Philippe 92
2/20/08 rb
2/20/08 Philippe 92
2/20/08 rb
2/18/08 Carl Barron
2/19/08 Nat Silver
2/19/08 rb
2/19/08 I.M. Soloveichik
2/19/08 rb
2/19/08 briggs@encompasserve.org
2/20/08 Philippe 92