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

 Messages: [ Previous | Next ]
 rb Posts: 5 Registered: 2/19/08
Re: Geometry Problem
Posted: Feb 19, 2008 12:00 PM

> 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.

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