

Geometric proof that angle trisection by straightedge and compass is impossible - CarolineW
http://terrytao.wordpress.com/2011/08/10/a-geometric-proof-of-the-impossibility-of-angle-trisection-by-straightedge-and-compass/

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btilly
Any time I see this claim I want to bring up the fact that you can do it if
the straightedge is marked in 2 places. See
<http://www.geom.uiuc.edu/docs/forum/angtri/> for a random link that explains
the construction.

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monochromatic
This is exactly the type of thing that has always made me think compass and
straightedge constructions were even sillier than a lot of very obviously
contrived mathematics.

/math major

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psykotic
That's an odd attitude for a math major to have. When geometers in ancient
times thought mechanical curves were lesser second-rank objects, that was
obviously a counterproductive prejudice that for a long time retarded the
development of mathematics. But if you ask me, there is nothing "obviously
contrived" about studying a configuration of lines and circles, their points
of intersection, the lines and circles directly constructible from those
intersections, and so on, ad infinitum. It strikes me as marvelously natural
and beautiful and anything but contrived. That this corresponds to studying
quadratic field extensions is a confirmation of that, if there were ever any
serious doubt.

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monochromatic
What I was saying is that I've always thought it was a silly restriction to
say that you can't put a couple of arbitrary marks on your straightedge. It's
just a matter of taste, I guess. The _connection between_ these constructions
and field theory is interesting, but I derive no pleasure from compass and
straightedge proofs.

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Someone
But you _can_ put _arbitrary_ marks on your straightedge. I haven't seen
proofs needing or even using them, but you are free to place a new point at an
arbitrary distance from a point A. You can then use the compass to transfer
that distance to any line segment.

What you cannot do is put specific points on it, say three equidistant ones,
or two at a distance of PI. If you allowed such constructs, the 'game' would
become extraordinarily dull.

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btilly
No, the rules specify "unmarked straightedge". And marking the straightedge is
not allowed.

If you were allowed to mark the straightedge then you could put 2 random marks
on it. With 2 random marks, you can trisect any angle. The location of the
marks doesn't matter, just their existence.

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Someone
Thanks for the pointer. It led me easily to
<http://planetmath.org/encyclopedia/TrisectionOfAngle.html>. Step 3 in that
construction is not one I had thought of as allowed.

It also, to me, isn't evident to me how one can proof that that Neusis
construction is possible to place the ruler in that way without resorting to
analytical means. I guess somebody must have thought about that. Are you aware
of more rigid geometric proofs?

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strmpnk
At least with a finite number of arcs and lines... true of anything I guess.

Very interesting as it might be for lovers of flat paper, I wonder if this is
true of all topologies. I'd guess not but I'm not mathematician.

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sp332
_At least with a finite number of arcs and lines... true of anything I guess._

Actually, there are a lot of geometrical constructions that are possible with
a finite number of operations with just a straightedge (not a ruler) and
compass. Exactly bisecting an angle is easy. Trisecting is impossible.

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phoobahr
See here I am a little confused. I never really understood why one couldn't do
this:

 _construct an equilateral triangle atop the angle to be trisected_ bisect any
side, construct a new triangle with said 30deg angle giving a line segment
1/sqrt(3) of the original eq tri _bisect the new 60deg angle, constructing a
new 30/60/90 tri

voila, 1/[sqrt(3)_sqrt(3)] == 1/3. You have trisected a line with a compass
and unmarked straightedge

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sp332
If I'm starting with a 40 degree angle, how do I construct an equilateral
triangle on it?

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phoobahr
Given angle(PQR) begin as you would when bisecting the angle:
<http://www.mathopenref.com/constbisectangle.html>.

Your arc intersections with |PQ| and |RQ| (call them p & q) give you an
isosceles triangle pQr.

Construct an equilateral triangle on |pr| similar to:
<http://www.mathopenref.com/constequilateral.html>

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samstave
This should be a vid on Khan Academy.

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Estragon
I don't know about Khan Academy, but a video of the rotations would definitely
be cool.

