The Angle Trisected ???
Andrew Burt
reported the following babble (attributed to AP)...
"Arian's proof apparently establishes a proportion at point zero and
sends it out in all directions. His equation also includes pi on both
sides, which has never been done before."
My first thought (before I read the above) was that he didn't understand all
the criteria for the problem, and was using, e.g., some form of
non-Euclidean geometry, infinite # steps (hey, I found a way to do it
as a limit when I was 14 or something, but I didn't go bug my senator about
it), etc. In this case the babble leads me to believe (without knowing
more than this stuff) that perhaps he thinks a compass can actually
create structure within one point (like, ah, multiple points). Or maybe
you have to move the compass _just_ _so_ ...
And I'm sure we'll all be pleased to know that equations like, say, pi = pi,
have never been "done" before. The kid's a genius! The reporters are
amazing!
Dibs on offering him a scholarship! (Oh, and did I mention, I have a
half page proof of FLT? I'd type it in right now but I don't have the time.)
--
Andrew Burt ab...@du.edu
"But if he was dying he wouldn't bother to carve "Aaaaargh", he'd just say it."
Frank Crary
Andrew Burt <ab...@mnemosyne.cs.du.edu> wrote:
>And I'm sure we'll all be pleased to know that equations like, say, pi = pi,
>have never been "done" before. The kid's a genius! The reporters are
>amazing!
Don't forget the stories about a hydroelectric power plant
proposed for the Boulder mountians last year. According to the
papers, it would use to man-made lakes, one higher than the
other, generate power by the flow from the upper to the lower
lake and then (because there aren't any near-by rivers to speak
of) refill the upper lake by pumping the water back from the
lower one. It took half a dozen stories before one article
finally mentioned that it wasn't supposed to produce power, just
store it to average out the demand. It might be nice if newspapers
required science reporters to have some idea what they were reporting.
Frank Crary
CU Boulder
Dick Dunn
...[Andrew Burt's pooh-pooh-ing of rediscovery of reflexive law:-]
>Don't forget the stories about a hydroelectric power plant
>proposed for the Boulder mountians last year. According to the
...[send water from upper to lower to generate, later pump it back up]
>...It took half a dozen stories before one article
>finally mentioned that it wasn't supposed to produce power, just
>store it to average out the demand. It might be nice if newspapers
>required science reporters to have some idea what they were reporting.
This one is even worse than Frank suggests, because the exact same pumped-
storage technique has been in use for...gee, I'd guess close to 30 years,
pretty near-by at the Cabin Creek facility, up above Georgetown on the way
to Guanella Pass. It's neither novel nor new to the area!
Too bad that "science reporter" is more often than not an oxymoron.
--
Dick Dunn r...@eklektix.com -or- raven!rcd Boulder, Colorado USA
...Mr Natural says, "Get the right tool for the job!"
Dale Gerdemann
> The trisection of any given angle using only a compass and unmarked
>straightedge (in the sense of the ancient Greeks) had been proved impossible;
>the problem belongs to the same class as squaring the circle (i.e. finding
>a length such that the square of that length is equal to the area of a circle
>of given radius) and doubling the cube (finding the length of a cube with
>exactly twice the volume of a given cube). All of these constructions
>involve finding either cube roots or roots of transcendental polynomials,
>which cannot be done by straightedge and compass. However, as many people
>have already pointed out, there are numerous approximations/tricks to solving
>the problem, although none are, in the strict sense, a solution.
>Peter Wang
Does this mean that a solution to one of the problems is also a
solution to the others. In other words, if I give you a compass, a
straightedge and an angle-trisector, can you then square the circle?
Or do I need to give you a senator in addition? :-)
Dale Gerdemann
Hiroshi
<>straightedge (in the sense of the ancient Greeks) had been proved impossible;
Mark Hopkins
>Dibs on offering him a scholarship! (Oh, and did I mention, I have a
>half page proof of FLT? I'd type it in right now but I don't have the time.)
in the margin of my notes here. Unfortunately, I don't have access to any
photo scanners or laser printer software.
###
Mark Hopkins
Alexander Pruss
> Does this mean that a solution to one of the problems is also a
> solution to the others. In other words, if I give you a compass, a
> straightedge and an angle-trisector, can you then square the circle?
No. In technical terms, a compass and a straightedge can only construct
a subset of the algebraic numbers, and adding an angle-trisector will
only increase the set of the algebraic numbers that we can construct.
(This can be easily proved. Just look at the constructions involved
and note that with a straightedge, compass and trisector, every distance
constructed (from the initial) satisfies a polynomial with rational
coefficients.)
However, to square the circle one needs to construct a segment of
length pi, and pi is transcendental.
Alex.
Michael Weiss
> Does this mean that a solution to one of the problems is also a
> solution to the others. In other words, if I give you a compass, a
> straightedge and an angle-trisector, can you then square the circle?
Alexander Pruss writes:
No. In technical terms, a compass and a straightedge can only construct
a subset of the algebraic numbers, and adding an angle-trisector will
only increase the set of the algebraic numbers that we can construct.
...
However, to square the circle one needs to construct a segment of
length pi, and pi is transcendental.
More interesting are two other facts: (1) a circle-squarer does not enable
you to trisect the angle or duplicate the cube (by arguments about
transcendental extension fields); (2) a generalized cube duplicator (i.e.,
the geometrical equivalent of a cube-root extractor for real numbers) does
not enable you to trisect the angle (the famous irreducible case of the
cubic).
I doubt that an angle-trisector will enable you to duplicate the cube, but
I don't see a proof offhand.
Michael Weiss
(2) a generalized cube duplicator (i.e., the geometrical equivalent of
a cube-root extractor for real numbers) does not enable you to trisect
the angle (the famous irreducible case of the cubic).
Someone asked via email for elaboration, since the so-called Cardan
formula for a cubic does express a root as a sum of two cube roots of
quantities in a quadratic extension of the base field.
I specified a cube-root extractor for REAL numbers. Trisecting the angle
leads to a cubic with three real roots. The "Cardan formula" (really
discovered by del Ferro and later independently by Tartaglia) leads to cube
roots of imaginary numbers in this case.
Bombelli was the first to note that the "Cardan formula" still makes sense in
this case, since the imaginary parts of two cube roots will cancel, leaving
a real result.
For a long time, it was an unsolved problem whether one could solve
cubics with three distinct real roots by radicals without a detour through
the complex numbers. This came to be known as the "irreducible case of
the cubic" (nothing to do with irreducibility in the usual sense).
Lipman Bers once told me this was how he thought imaginary numbers were
discovered in the first place. Quadratic equations provide no motivation.
So some quadratics have no roots. OK, they don't have roots! So what? (I
like this pseudo-history better than the actual history, especially since
the same work, Cardan's "Great Art", introduced both imaginary numbers and
"Cardan's formula" to the literature.)
It makes a cute exercise in Galois theory to show the following:
Let p(x) be an cubic with coefficients in Q and irreducible over Q,
with three distinct real roots. (Here I'm using "irreducible" in the
usual sense.) Let F_0 < F_1 < ... < F_n be a tower of extension
fields such that (1) all F_i are subfields of R; (2) F_0 = Q; (3)
F_{i+1} is obtained from F_i by the adjunction of a k_i-th root of an
element of F_i, for some k_i. Then p(x) is irreducible over F_n.
The remaining "reduction question", does an angle-trisector enable you to
duplicate the cube, asks the question:
Is there a tower of extension fields, Q = F_0 < F_1 < ... < F_n, all
subfields of R, with cube_root(2) in F_n, such that F_{i+1} is either a
quadratic extension of F_i or is obtained by adjoining a root of a
cubic with coefficients in F_i and with three real roots?
Answer: also no. (An even easier exercise in Galois theory.)
So the three classical geometric construction problems are all
"independent".