Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

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

 Messages: [ Previous | Next ]
 Philippe 92 Posts: 614 Registered: 1/25/05
Re: Geometry Problem
Posted: Feb 19, 2008 11:16 AM

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 ?

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.
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.

--
Philippe C., mail : chephip+news@free.fr
site : http://chephip.free.fr/ (recreational mathematics)

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