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        riddles >> medium >> Embarassing geometry problem?
        
(Message started by: ecoist on Feb 17^th, 2008, 3:18pm)

Title: Embarassing geometry problem?
Post by ecoist on Feb 17^th, 2008, 3:18pm

Let A, B, and C be collinear points in the plane with B between A and C. On one side of AC erect equilateral triangles ABD and BCE. On the other side of AC erect equilateral triangle ACF. Let P be the intersection of DC and EA. 1) Show that P, B, and F are collinear. 2) Show that angles DPE, EPF, and FPD are each 120 degrees.

Title: Re: Embarassing geometry problem?
Post by Icarus on Feb 17^th, 2008, 8:11pm

A picture might help.

Title: Re: Embarassing geometry problem?
Post by Grimbal on Feb 18^th, 2008, 8:26am

[hide] EA and DC are BF rotated 60° around C resp. A.[/hide]

Title: Re: Embarassing geometry problem?
Post by ecoist on Feb 18^th, 2008, 11:11am

Wonderful, Grimbal, now I'm doubly embarassed! I had no idea how to solve this puzzle until I remembered the solution of a very different problem. Now I see that this puzzle does not require knowledge of this "other" problem; indeed, it has a self-contained, simple solution! Took me awhile, though, to see that your solution also works for 1) as well as 2).

Title: Re: Embarassing geometry problem?
Post by SMQ on Feb 18^th, 2008, 11:40am

There's an even simpler solution for 1: label the point where AE intersects BD as O and the point where CD intersects BE as Q, then [hide]OPQB is similar to APCF and so PB lies along PF[/hide].

--SMQ

Title: Re: Embarassing geometry problem?
Post by ecoist on Feb 18^th, 2008, 2:27pm

Wow, SMQ! I was wrong to think that Grimbal's solution included collinearity! Your observation, SMQ, fills the gap beautifully. Nice work, guys!

(I was stuck on this problem until I remembered the solution of the familiar problem:

If P is a point inside a triangle ABC such that the sum of the lengths of PA, PB, and PC is minimal, then the angles APB, BPC, and CPA are each 120 degrees.)

Title: Re: Embarassing geometry problem?
Post by Grimbal on Feb 19^th, 2008, 12:55am

on 02/18/08 at 14:27:25, ecoist wrote:

Wow, SMQ! I was wrong to think that Grimbal's solution included collinearity!

Er... oops.

Title: Re: Embarassing geometry problem?
Post by ecoist on Feb 20^th, 2008, 5:59am

Now I am triply embarassed! Grimbal's trick does it all! Continuing Grimbal's argument, [hide]let X be the point on DC such that triangle APX is equilateral. Then, rotating DC clockwise about A, X goes to P, B to C, and C to F; whence P, B, and F are collinear![/hide]

Moreover, Grimbal's argument works unchanged if B does not lie on AC! SMQ's argument requires that B lie on AC.

Title: Re: Embarassing geometry problem?
Post by Grimbal on Feb 24^th, 2008, 8:45am

Thanks for revealing all the cleverness of my argument that even I didn't know... ::)

Title: Re: Embarassing geometry problem?
Post by ecoist on Feb 24^th, 2008, 9:29am

Somebody's watching you, Grimbal! Your proof for Fermat's point has appeared at

http://www.cut-the-knot.org/Generalization/fermat_point.shtml

as the 7-th proof, with your nick attached!

Title: Re: Embarassing geometry problem?
Post by Grimbal on Jan 28^th, 2009, 6:26am

They should have said "by Grimbal & ecoist". I gave the solution for the angles and you extended it to the colinearity.