Trisection
Pythicarus
When I took geometry 35 years ago, it was alleged that neither an angle nor a
line could be trisected, using only a compass & straight-edge, using standard
construction techniques. But, like the 4-color map problem, there was no proof
at
that time. The 4-color problem has been solved. But I've always wondered whe-
ther the trisection problem has been solved, or been proven to be impossible.
I
would appreciate any feedback.
John Conway
Your letter has things the wrong way around, because in fact the
trisection problem was proved impossible before 1840!
John Conway
Sarah Seastone
Hi -
There's an entry in the Dr. Math FAQ that takes up this question about
"impossible constructions":
http://forum.swarthmore.edu/dr.math/faq/faq.impossible.construct.html
You'll find a discussion of the 'rules of antiquity' and links to sites on
the Web that go more deeply into the constructions.
- Sarah
At 8:01 AM -0400 6/24/98, Pythicarus wrote:
>When I took geometry 35 years ago, it was alleged that neither an angle nor a
>line could be trisected, using only a compass & straight-edge, using standard
>construction techniques. But, like the 4-color map problem, there was no
>proof
>at
>that time. The 4-color problem has been solved. But I've always wondered
>whe-
>ther the trisection problem has been solved, or been proven to be impossible.
>I
>would appreciate any feedback.
Sarah Seastone
Editor, Archivist, Web Page Designer
The Math Forum
http://forum.swarthmore.edu/
sa...@forum.swarthmore.edu
Joshua Zucker
You write:
When I took geometry 35 years ago, it was alleged that neither an
angle nor a line could be trisected, using only a compass &
straight-edge, using standard construction techniques.
A line (segment) can be trisected (or divided into any number of
pieces) quite easily. One recently discovered method uses a rectangle
(find the center by intersecting the diagonals; drop a perpendicular
from there down to the side; then draw from the midpoint of that side
up to a corner; from the intersection of that segment with the
diagonal, drop another perpendicular; you have now trisected the
side). The classical method uses similar triangles. To trisect (or
n-sect) a segment AB, draw a ray AC and mark off n equal (arbitrary)
lengths along it, AX1, X1X2, X2X3, and so on. Connect Xn to B, and
then draw parallels through each of the other X's to divide the
segment into equal pieces.
The angle trisection, as others have already pointed out, was proved
impossible about 150 years ago. My understanding of the proof is that
it can be shown that solutions to cubic equations (except for certain
special ones with rational or quadratically expressible roots) are not
constructible, and general angle trisectors are solutions to cubic
equations.
--Joshua Zucker
Guy F. Brandenburg
Pythicarus wrote:
> When I took geometry 35 years ago, it was alleged that neither an angle nor a
> line could be trisected, using only a compass & straight-edge, using standard
> construction techniques. But, like the 4-color map problem, there was no proof
> at
> that time. The 4-color problem has been solved. But I've always wondered whe-
> ther the trisection problem has been solved, or been proven to be impossible.
> I
> would appreciate any feedback.
Yes, this has been PROVED to be impossible. The proof was done last century. It can
be done by cheating, however.
Compaq
You cannot trisect a line because it has no determinate length, but, I
think you meant to say line segment, which you can trisect easily. Just
look in any high school geometry book to find out how to divide a segment
into n equal segments.
As for the trisected angle; I, too am curious. I remember reading either
in Marilyn Vos Savant's column or a scientific journal that it has been
proven to be impossible, but until I see the proof I will remain skeptical
and I may give it a try myself.
Jim Wallace
Pythicarus <pythi...@aol.com> wrote in article
<199806240656...@ladder01.news.aol.com>...
John Conway
On 29 Jun 1998, Compaq wrote:
> You cannot trisect a line because it has no determinate length, but, I
> think you meant to say line segment, which you can trisect easily. Just
> look in any high school geometry book to find out how to divide a segment
> into n equal segments.
>
> As for the trisected angle; I, too am curious. I remember reading either
> in Marilyn Vos Savant's column or a scientific journal that it has been
> proven to be impossible, but until I see the proof I will remain skeptical
> and I may give it a try myself.
>
> Jim Wallace
An attitude I've always sympathized with, since I myself dislike
accepting things on authority. But of course the best way to avoid
wasting a tremendous amount of time is to try to understand the
proof, rather than try to find constructions for the trisection.
After all, the proof's been around for one-and-a-half centuries and
been studied by many sceptical people in that time; and by now it's
an undergraduate exercise in Galois Theory.
Here's a method that almost always shows that somebody's construction
doesn't work. Consider what happens as the angle being trisected
continually increases, even past 180 and 360 degrees. The angle
given by the construction should also increase, but at one-third the
rate. Let me illustrate the idea by first considering the problem
of bisection of a varying angle AOB:
C2 B1=B4
C3 /
/ C1
/
B2-------O-------A
/
C4 /
/ C5
B3=B5
I've drawn several positions B1,B2,B3,B4,B5 for the point B,
and the corresponding positions C1,C2,C3,C4,C5 for the "trisecting point" C
given by AOC = AOB/3. Now for four of these positions (namely B1,B3,B4,B5)
the lone OB is the same. However, they yield two distinct corresponding
lines OC. So if we just specified two lines L and M, we'd see two
lines (C1 C4 and C3 C5) that deserve to be called angle-bisectors:
this is why one speaks of internal and external angle-bisectors.
C3 /
/ C1
/
-------O------- <-- L
/
C4 /
/ C5
__
/|
/
M
Any construction for bisecting the angle has to produce BOTH
these lines, if interpreted correctly. For instance, I describe
the standard one in words:
"Choose points A on L, B on M, both at distance R from O.
Then draw intersecting circles of the same radius r centered at A
and B, and then the line joining A and B is the required
bisecting line"
This actually can be done in four ways, since it allows us to
take either A+ or A- for A and B1 or B4 for B:
B1
/
/
/
A+------O-------A-
/
/
/
B4
and you can check that indeed it therefore constructs both the
internal and external angle-bisectors.
In a similar way, one can see that two lines AOA and BOB
should have THREE "trisectors" COC, ie., lines with the property
that AOC = AOB/3:
C B
C /
/ C
A-----O-----A
C /
/ C
B C
and ideally a trisecting construction should be capable of giving
all three.
This, though not yet a proof, is the germ of the proof, the
idea of which is to show that the number of "solutions" of a
geometrical construction is always a power of 2, whereas for
a trisection it must be 3. But the rigorous proof involves
saying this algebraically rather than geometrically, because
this allows us to avoid all sorts of problems, like the fact
that the roots of some of the underlying equations might
be complex.
John Conway