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        riddles >> medium >> Bizaare way to bisect 120 degrees
        
(Message started by: ecoist on Feb 19^th, 2008, 5:40pm)

Title: Bizaare way to bisect 120 degrees
Post by ecoist on Feb 19^th, 2008, 5:40pm

In the plane, let l be a line and A a point not on l. For each point B on l and form the equilateral triangle ABC, labelled clockwise. Show that the points C all lie on a line making an angle of 60 degrees with l.

Title: Re: Bizaare way to bisect 120 degrees
Post by rmsgrey on Feb 21^st, 2008, 3:14pm

The following proves something slightly different, but it's fairly clear that the two are interchangeable...

[hideb]Take two lines, l,m crossing at a 60 degree angle at point D. Pick a point, A, on the line bisecting the 120 degree angle, and find points E and F where the perpendiculars from A come down to l and m respectively. Obviously, the angle EAF is 60 degrees.

Pick a point B on l and reflect it in AE to get B'. Reflect B' and B in AD to get C and C' respectively.

The angles C'AF, FAC, B'AE and EAB are all equal and in the same sense so BAC=EAF + FAC - EAB=EAF=60 degrees (and similarly for B'AC'). Since AC is a reflection of AB, they are the same length, so triangle ABC is an equilateral triangle (as is AB'C') for any choice of B on l.

Since the images of l under reflection in AE and AD are l and m respectively, C and C' both lie on m.[/hideb]

Title: Re: Bizaare way to bisect 120 degrees
Post by Eigenray on Feb 21^st, 2008, 5:20pm

Here's a solution using vectors:
[hide]C-A = R(B-A), where R is rotation by 60 degrees clockwise. Each point B has the form B₀ + tV, where V is parallel to l, so C = [A-R(A)+R(B₀)] + tR(V) lies on a line in the direction R(V)[/hide].

Title: Re: Bizaare way to bisect 120 degrees
Post by ecoist on Feb 21^st, 2008, 8:32pm

I was just about to remove this problem as redundant! I had posted this problem because I had a solution akin to Eigenray's solution. Later, I noticed that Grimbal's wonderfully simple trick used in "An embarassing geometry problem?" also works here.

[hide]Let B and B' be on l and let C and C' be the corresponding points such that ABC and AB'C' are equilateral triangles labelled clockwise. Rotate the line l about A clockwise 60 degrees to the line m. Then B goes to C and B' goes to C', both C and C' lying on the line m.[/hide]

(I also noticed that Grimbal's trick provides, in my humble opinion, the best proof (as well as the shortest) of the existence of the Fermat Point for triangles.)