Newsgroups: sci.math
From: ramsay@unixg.ubc.ca (Keith Ramsay)
Subject: Re: Trisect an angle???
Date: Thu, 17 Sep 1992 17:26:11 GMT
In article <19a4g0INNah4@bigboote.WPI.EDU>
cubanski@tensor4.WPI.EDU (David J Cubanski) writes:
>I have heard that such a thing is possible, but have never seen it
>done. Is the technique straightforward enough to be posted?? If not,
>could you supply a reference?
Suppose that we have an angle. Draw a circle around the vertex, and
extend one of the rays of the angle, so that we have a circle centered
at O with points A, B, and C on the circumference, an angle AOB which
we want to trisect, and a diameter AC.
Mark the ruler (or use some equivalent trick, like holding the compass
next to it) with marks at an interval congruent to the radius OA.
Slide the ruler along, with one of the marks on the ray OC beyond C,
and the other mark on the circle, until the ruler passes through the
point B. Draw a line there. So now we have a point D on the circle,
and a point E on the ray OC, such that the length of DE is equal to
the radius of the circle, and E, D, and B are collinear in that order.
The triangles DOB and EDO are isoceles, since two of the sides of each
are congruent to the radius of the circle. Using the fact that the
corresponding angles are equal, and that the sum of the angles of a
triangle is 180 degrees, you can show that the angle AOB is three
times that of OEB.
Keith Ramsay
ramsay@unixg.ubc.ca
==============================================================================
Newsgroups: sci.math
From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz)
Subject: Re: NEW sci.math FAQ: The Trisection of an Angle
Date: Thu, 10 Nov 1994 01:56:19 GMT
In article <39r5t8$hb1@dartvax.dartmouth.edu>, Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly) writes:
> ...
> I think that the solution using a compass and a ruler marked in 2
> places should be put in the FAQ. Unfortunately I do not remember the
> construction off of the top of my head. Does anyone else remember it?
I do. It is due to Archimedes amd consists in drawing a semicircle of
arbitrary radius with base on one of the (extended) lines that form
the angle. Then you draw a segment joining the point of intersection of
the other line with the semicircle (point marked Z in the drawing)
and the extended base of the line in such a way that the distance from
the base to the first intersection point (point Y in diagram below) is
the radius of the circle.
__
- Z
/ / \
Y / j
/ / \
| / |
-X------------------------------
I'm writing this from memory. Any one smart enough to check the angles using
trigonometry?
Alex
==============================================================================
From: jurjus@kub.nl (H. Jurjus)
Newsgroups: sci.math
Subject: Re: NEW sci.math FAQ: The Trisection of an Angle
Date: Thu, 10 Nov 1994 11:48 MET
In Article <39r5t8$hb1@dartvax.dartmouth.edu> "Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)" says:
(...)
> I think that the solution using a compass and a ruler marked in 2
> places should be put in the FAQ. Unfortunately I do not remember the
> construction off of the top of my head. Does anyone else remember it?
>
> Ben Tilly
I'll have a go.
(The originator of this proof is Archimedes, BTW.)
_ ------ _ 'X"
. " / " .
'B'. / .
.o / .
. / .
--o---------------(-1,0)---------(0,0)-----------(1,0)----
'A' 'O' 'E'
(A, B, X on one line)
Suppose X is a point on the unit circle such that /_ XOE is the angle we
would like to 'trisect'. Draw a line AX through a point A on the x-axis
such that AB = 1 (= radius of the circle = BO ), where B is the
intersection-point of the line AX with the circle.
Let a be /_ BAO
Then /_ BOA = a, and /_ XBO = 2a (= /_ BXO )
Therefore:
/_ AOB + /_ BOX + /_ XOE = pi/2 = /_ XBO + /_ BXO + /_ BOX
a + /_ BOX + /_ XOE = 2a + 2a + /_ BOX
Therefore:
/_ XOE = 3a
QED.
(I'm not experienced enough in drawing lines in ASCII:
I think one should draw the line AX.)
H.Jurjus
==============================================================================
From: Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)
Newsgroups: sci.math
Subject: Re: Need angle trisection disproof help.
Date: 24 Jan 1995 02:15:09 GMT
In article <19950123143411.jurjus@pi0220.kub.nl>
jurjus@kub.nl (H. Jurjus) writes:
> In Article <3fp2k1$a8n@dartvax.dartmouth.edu> "Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)" says:
>
> > ... In fact the impossible
> > problems are called that for a reason, namely they have been proven to
> > be impossible with compass and unmarked ruler. If you check the FAQ I
> > believe that they have the method for trisections that uses a compass
> > and a ruler marked at 2 points...
>
> I didn't find it; maybe I looked at the wrong place.
It used to be there, they may have changed it. Here is the
construction...
Suppose that the angle to cut is ABC. Continue the line BC. Now draw a
circle around B with radius equal to the distance between the two
points on the marked straight-edge. Let D be the point where the ray
from B to A intersects this circle.
Now take the straightedge, and slide it along to the point where it is
touching D, one of the marks is touching another point of the circle,
call it E, and the other mark is touching the line BC outside of the
circle at a point that we will call F. Then EBF is exactly 1/3 of the
original angle. (If you draw the diagram, note which edges are the same
length as the distance between the two points so you have isocoles
triangles, and then you just work out angles then it is easy to show
that this works.)
Of course the act of laying your marked straight-edge down in that
fashion cannot be done unless the straight-edge has marks on it...
...
Ben Tilly
==============================================================================
From: Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)
Newsgroups: sci.math
Subject: Re: trisect arbitrary angles HELP!
Date: 26 Apr 1995 05:00:13 GMT
In article <3nk17q$so@falcon.abare.gov.au>
Greg Griffiths writes:
> As I recall, the proof that it is impossible to trisect an angle using
> ruler and compass assumes that the ruler is used merely form the line
> connecting two previously constructed points. {I think that there's a
> proof that all ruler and compass constructions are possible with compass
> alone (assuming that you don't actually want to draw lines just find
> their points of intersection etc.) At the risk of inviting flames (please
> don't, I'll freely admit that my memory is poor) I believe that there is
> a method (know in pre-medieval times) for trisecting angles using ruler
> and compass that involves using the ruler to record a length and then to
> find the points AB such that A and B lie on preconstructed lines or rays
> and AB has the length desired. However I may merely be another victing of
> CRAFT disease (Can't Remember A Flamin' Thing). Sorry I can't provide a
> reference.
In fact this is true, and I have run across at least 2 ways to trisect
an arbitrary angle using compass and a ruler marked in two places. The
one that turns up on sci.math more often goes as follows.
Suppose that a, b, c are three points and your angle is abc. Draw a
circle centered at b with radius equal to the distance between your
markings on the ruler. Extend the edges in the angle so they both
intersect the circle. Extend the line from a to b into a line that
intersects the circle at another point a'. Now slide the ruler along
keeping the ruler touching where the ray from b to c touches the circle
(call that poin c'), and keeping the far mark on the ruler touching the
line from a to b, until the intermediate mark touches the circle at a
point d (and the line at point d'). Then angle a'bd is exactly 1/3 of
angle abc.
Here is why. Draw the diagram and call angle a'bd (= angle d'bd) alpha.
(None of this will make sense without a diagram present in front of
you.) Note that edges d'd and db are the same length so angle dd'b is
also alpha. Therefore angle d'db must be 180-2alpha. Therefore angle
bdc' must be 2alpha. But bd and bc' are both the same length so angle
dc'b is 2alpha. Therefore angle dbc' is 180-4alpha. However we have
180 = angle d'bd + angle dbc' + angle c'ba
so angle c'ba is 3 alpha. However angle c'ba is angle cba, so that
completes the demonstration.
The other construction is in _A History Of Mathematics_ by Boyer, as
well as a number of other solutions by the Greeks which arise if you
allow yourself to use some special curve or other. (With those curves
you can also double the cube and square the circle, which are other
famous problems.)
Ben Tilly