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Topic: Bizaare way to bisect 120 degrees (Read 3489 times) |
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ecoist
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Bizaare way to bisect 120 degrees
« on: Feb 19th, 2008, 5:40pm » |
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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.
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| « Last Edit: Feb 19th, 2008, 5:47pm by ecoist » |
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rmsgrey
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Re: Bizaare way to bisect 120 degrees
« Reply #1 on: Feb 21st, 2008, 3:14pm » |
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The following proves something slightly different, but it's fairly clear that the two are interchangeable...
| hidden: | 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. |
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Eigenray
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Re: Bizaare way to bisect 120 degrees
« Reply #2 on: Feb 21st, 2008, 5:20pm » |
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Here's a solution using vectors:
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).
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ecoist
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Re: Bizaare way to bisect 120 degrees
« Reply #3 on: Feb 21st, 2008, 8:32pm » |
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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.
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.
(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.)
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