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Topic: Embarassing geometry problem? (Read 2362 times) |
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ecoist
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Embarassing geometry problem?
« on: Feb 17th, 2008, 3:18pm » |
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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.
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Icarus
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A picture might help.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Grimbal
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Re: Embarassing geometry problem?
« Reply #2 on: Feb 18th, 2008, 8:26am » |
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EA and DC are BF rotated 60° around C resp. A.
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ecoist
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Re: Embarassing geometry problem?
« Reply #3 on: Feb 18th, 2008, 11:11am » |
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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).
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SMQ
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Re: Embarassing geometry problem?
« Reply #4 on: Feb 18th, 2008, 11:40am » |
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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 OPQB is similar to APCF and so PB lies along PF.
--SMQ
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--SMQ
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ecoist
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Re: Embarassing geometry problem?
« Reply #5 on: Feb 18th, 2008, 2:27pm » |
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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.)
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Grimbal
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Re: Embarassing geometry problem?
« Reply #6 on: Feb 19th, 2008, 12:55am » |
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on Feb 18th, 2008, 2:27pm, ecoist wrote:| Wow, SMQ! I was wrong to think that Grimbal's solution included collinearity! |
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Er... oops.
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ecoist
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Re: Embarassing geometry problem?
« Reply #7 on: Feb 20th, 2008, 5:59am » |
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Now I am triply embarassed! Grimbal's trick does it all! Continuing Grimbal's argument, 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!
Moreover, Grimbal's argument works unchanged if B does not lie on AC! SMQ's argument requires that B lie on AC.
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Grimbal
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Re: Embarassing geometry problem?
« Reply #8 on: Feb 24th, 2008, 8:45am » |
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Thanks for revealing all the cleverness of my argument that even I didn't know... 
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Grimbal
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Re: Embarassing geometry problem?
« Reply #10 on: Jan 28th, 2009, 6:26am » |
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They should have said "by Grimbal & ecoist". I gave the solution for the angles and you extended it to the colinearity.
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