Trisecting a line
Rajesh G
I don't know the full problem definition but it goes
like this.
Can you trisect a line with a ruler and a compass ?
Don't use the compass like a divider and give a solution by
trial and error.
-Rajesh.G
--
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| Rajesh.G (gr...@cse.iitb.ernet.in) | Hostel #6, Room 231 |
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Ken Kelley
In article <E5w7F...@bhishma.cse.iitb.ernet.in>,
SPOILER
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How many times to do I need to post to be a "regular", and get in
on that silly geography puzzle?
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Yes, trisecting a line-segment is very simple (trisecting an infinite
line is a bit harder).
Take your line segment AB.
Choose a point C not on that line.
Draw line AC.
Measure AC, and mark off point D that same distance from C,
on the line AC. Mark off E, the same distance from D.
Draw line EB. Draw lines through C and D, both parallel
to EB. These lines intersect AB at points F and G, which
trisect AB.
This (obviously) works for any positive integer, not just 3.
Since doubling the length of a segment is even simpleer,
it follows that one can construct a segment of any positive
rational length.
I wonder if I just did someone's homework for them.
Oh well, I've never been fond of grading people by their
homework anyway...
Ken
--
Disclaimer: of my several posts to r.p, almost all have contained
an egregious flaw. You have been warned.
Douglas Limmer
gr...@bhishma.cse.iitb.ernet.in (Rajesh G) wrote:
>I don't know the full problem definition but it goes
>like this.
>Can you trisect a line with a ruler and a compass ?
Certainly.
Well, assuming you mean a "line segment".
Draw a ray starting at one end of the segment (which doesn't coincide
with the segment itself).
Pick a distance for your compass -- it doesn't matter what.
Mark off, from the start of your ray, three points on the ray with
each point one compass-distance away from the last.
Draw a line from the third point to the end of the line segment.
Draw lines parallel to that line through the other two points.
The points where those lines intersect the original line segment
trisect the original line segment.
Now, if you wanted to trisect an _angle_, it would be harder (that is,
it's been proven to be impossible).
Douglas Limmer -- lim...@math.orst.edu
"No wonder these mathematical wizards were nuts - went off the beam -
he'd be pure squirrel-food if he had half that stuff in _his_ skull!"
E. E. "Doc" Smith, _Second Stage Lensmen_
Matthew Daly
In article <E5w7F...@bhishma.cse.iitb.ernet.in> gr...@bhishma.cse.iitb.ernet.in (Rajesh G) writes:
>I don't know the full problem definition but it goes
>like this.
>
>Can you trisect a line with a ruler and a compass ?
Nitpick: You probably mean "straightedge" instead of "ruler". The
problem would be trivial with a ruler -- you measure the line and
mark off one-third of that with the ruler.
Yes. Let's say that AB is the line segment that you want to trisect.
Draw another line AC that is suitably long. Using the compass, mark
off points D, E, and F such that AD = DE = EF. (The distance that you
choose is irrelevant.) Draw a line connecting B and F. Now, you want
to draw a line parallel to BF that passes through E. (See below) Call
the point where this line intersects AB as G. The length of BG is one
third of AB. You can measure off that distance on the compass and
find the third trisection point if you like.
For a sidebar: you probably know how to construct a line parallel to a
given line that passes through a point P, but just in case you don't:
Draw a circle around P with radius large enough to intersect the line
at at least one point. Call one of these points S. Using the same
radius and center S, mark off another point on the line and call it T.
Draw another circle with the same radius and center T. The two circles
will intersect in two places, S and another point that you should call
U. Draw PU -- it is parallel to ST. (Indeed, PSTU is a rhombus.)
-Matthew
--
Matthew Daly I feel that if a person has problems communicating
mwd...@kodak.com the very least he can do is to shut up - Tom Lehrer
My opinions are not necessarily those of my employer, of course.
Noam Elkies
In article <E5w7F...@bhishma.cse.iitb.ernet.in>
gr...@bhishma.cse.iitb.ernet.in (Rajesh G) writes:
>Can you trisect a line with a ruler and a compass ?
Yes; one way of doing this (or more generally dividing a segment
into k equal parts) goes all the way back to Euclid.
>Don't use the compass like a divider and give a solution by
>trial and error.
It is, however, possible to give a legitimate construction
that unlike Euclid's uses the compass only as a divider, i.e.
only to measure out equal lenths. Namely:
To trisect line AM, extend an arbitrary ray from A, and mark
off equal lengths AN, NB; then extend BM through M to a point C
such that BM=MC. Then CN trisects AM because CN and AM are
medians of triangle ABC.
B
. .
. .
N .
. . .
. . .
A...........*.....M
. .
. .
. .
. .
..
C
This is surely well-known; I'm particularly fond of this
construction since I discovered it independently in 1978.
Obpuzzle: find a bisection method along the same lines.
Somewhat more challenging: generalize to a method for
dividing a segment into k equal parts.
--Noam D. Elkies (elk...@math.harvard:edu)
Dept. of Mathematics, Harvard University
David Bradley
Rajesh G wrote:
> Can you trisect a line with a ruler and a compass ?
I assume you mean a line segment, but sure, this is easy. Just draw
a line at an angle to the line you want to trisect and mark off
three equidistant segments (shown by the stars below)
-----------------------------------------------
\
\
\
\
*
\
\
\
\
*
\
\
\
\
*
Next, connect the bottom star with the end of the line you are trying
to trisect, and then construct lines parallel to tis line through
each of the "stars". By triangle similarity, you get a trisected
line segemnt.
You may have been thinking about trisecting an angle, which has been
shown not to be constructable with an unmarked straightedge and
compass.
-Dave
-----------------------------------------------------------
David Bradley | Hewlett-Packard Company
email: bra...@fc.hp.com | Graphics Hardware Lab, MS73
phone: (970)229-6992 | 3404 East Harmony Road
FAX: (970)229-6858 | Fort Collins, CO 80525
-----------------------------------------------------------
>
> Don't use the compass like a divider and give a solution by
> trial and error.
>
Amit Chakrabarti
Rajesh G (gr...@bhishma.cse.iitb.ernet.in) writes:
> I don't know the full problem definition but it goes
> like this.
>
There is a very simple Euclidean construction for n-secting a line segment.
This oviously subsumes trisection.
What u cannot trisect is an angle.
AmitC.
________________________________________________________________________
_ The smallest real number________________________ Amit Chakrabarti ____
___ in the interval (0,1] is 1,___________________ IV year B.Tech. _____
_ since any other candidate_______________________ # 253, Hostel 2 _____
___ may be replaced by its square!______________________________________
Utpat Shriram Arvind
Hello,
Trisecting an angle was a problem for some time,
but a schoolboy in Solapur , has devised a method to do
that . I remember to have worked out the procedure.
-shriram
--
___ _____
.'/,-Y" "~-.
l.Y ^.
/\ _\_
Wei-Hwa Huang
lim...@math.orst.edu (Douglas Limmer) writes:
>gr...@bhishma.cse.iitb.ernet.in (Rajesh G) wrote:
>>Can you trisect a line with a ruler and a compass ?
>Certainly.
>Well, assuming you mean a "line segment".
Hmm.
Can you trisect a LINE with *anything*??
--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Now surfing the Internet at 24 hours a week.
Amit Chakrabarti
Utpat Shriram Arvind (shr...@bhishma.cse.iitb.ernet.in) writes:
> Hello,
> Trisecting an angle was a problem for some time,
> but a schoolboy in Solapur , has devised a method to do
> that . I remember to have worked out the procedure.
It must be an approximate trisection method. You can prove algebraically
that trisecting an angle by Euclidean means is impossible.
Douglas Limmer
whu...@ugcs.caltech.edu (Wei-Hwa Huang) wrote:
>lim...@math.orst.edu (Douglas Limmer) writes:
>>gr...@bhishma.cse.iitb.ernet.in (Rajesh G) wrote:
>>>Can you trisect a line with a ruler and a compass ?
>>Certainly.
>>Well, assuming you mean a "line segment".
>Can you trisect a LINE with *anything*??
It depends what you mean by "line". (And, I suppose, "trisect".)
On a circle a "line" is usually a great circle. You could probably
trisect one of those with a compass and straightedge (curvededge?).
You can't split a line into three (continuous) pieces without one of
them being finite (and the other two infinite). (No fair splitting it
lengthwise. :-) You probably shouldn't take "every third point"
either.)
Noam Elkies
In article <330f5f6d...@news.orst.edu>
lim...@math.orst.edu (Douglas Limmer) writes:
>On a circle a "line" is usually a great circle. You could probably
>trisect one of those with a compass and straightedge (curvededge?).
Presumably you mean `on a sphere a "line" is usually a great circle'.
I wonder whether a "straightedge" -- i.e. a device for finding the
great circle through any two non-antipodal points on the sphere --
would suffice to trisect a great circle, or for that matter do any
of the standard constructions (perpendiculars, triangle given three
sides, etc.) in the context of spherical geometry.
Matthew Daly
In article <5en26p$m...@gap.cco.caltech.edu> whu...@ugcs.caltech.edu (Wei-Hwa Huang) writes:
>
>Can you trisect a LINE with *anything*??
If the line were on a sphere, you could.
If it were on the plane, I think that one could draw one mark on the line
and say "Here is one of the trisection points, the other is too far
away to see in any scaling of the plane. :-)
Utpat Shriram Arvind
Hello,
Which is the reference to such a proof. Also you are saying
that we can trisect an angle approximately. Then what is the amount of
error that will be there.
Alexander Bogomolny
You may be amused to know that you can
trisect a line with a compass alone. Check
http://www.cut-the-knot.com/do_you_know/compass.html
Alexander Bogomolny
http://www.cut-the-knot.com
Rajesh G <gr...@bhishma.cse.iitb.ernet.in> wrote in article
<E5w7F...@bhishma.cse.iitb.ernet.in>...
> I don't know the full problem definition but it goes
> like this.
>
> Can you trisect a line with a ruler and a compass ?
>
Right...@nasa.gov
am...@bhishma.cse.iitb.ernet.in (Amit Chakrabarti) wrote:
>Rajesh G (gr...@bhishma.cse.iitb.ernet.in) writes:
>> I don't know the full problem definition but it goes
>> like this.
>>
>> Can you trisect a line with a ruler and a compass ?
>There is a very simple Euclidean construction for n-secting a line segment.
>This oviously subsumes trisection.
>What u cannot trisect is an angle.
Why not???
Take an angle A, and mark off equall distances B & C
Then trisect line segment BC, and draw lines from A to each point
Douglas Limmer
elk...@brauer.harvard.edu (Noam Elkies) wrote:
>lim...@math.orst.edu (Douglas Limmer) writes:
>>On a circle a "line" is usually a great circle. You could probably
>>trisect one of those with a compass and straightedge (curvededge?).
>Presumably you mean `on a sphere a "line" is usually a great circle'.
Oops. (Wait! I meant a "2 dimensional circle"! Yeah, that's it! :-)
>I wonder whether a "straightedge" -- i.e. a device for finding the
>great circle through any two non-antipodal points on the sphere --
>would suffice to trisect a great circle, or for that matter do any
>of the standard constructions (perpendiculars, triangle given three
>sides, etc.) in the context of spherical geometry.
Well, I didn't have a method in mind -- just an idea that since it was
"finite" you might be able to do it.
A spherical "compass" is a rather strange concept, too. (It gives
what we in 3-space would call _two_ circles -- a compass with the
stationary arm at the poles would give matching latitute circles.)
I _think_ the standard "bisect a line segment" procedure works. But I
don't have the tools to try it out. :-)
Amit Chakrabarti
In article <E61u4...@bhishma.cse.iitb.ernet.in>, Utpat Shriram Arvind wrote:
> Which is the reference to such a proof. Also you are saying
>that we can trisect an angle approximately. Then what is the amount of
>error that will be there.
The proof (that trisection of an angle is impossible) can be found in
any good textbook on Algebra or Field Theory. For example, I.N.Herstein's
book "Topics In Algebra" gives a proof in the chapter on Fields.
I do not know any details about any approximate trisection methods.
Someone had made a posting saying that someone else had found a method
for "trisecting" an angle using ruler-and-compass. In the light of the
proof we're talking about it follows that such a method *must* be
approximate.
Dave Seaman
In article <E63GL...@bhishma.cse.iitb.ernet.in>,
Amit Chakrabarti <am...@bhishma.cse.iitb.ernet.in> wrote:
>I do not know any details about any approximate trisection methods.
>Someone had made a posting saying that someone else had found a method
>for "trisecting" an angle using ruler-and-compass. In the light of the
>proof we're talking about it follows that such a method *must* be
>approximate.
No, it doesn't follow. What is impossible is to trisect an arbitrary
angle with compass and *straightedge* (not ruler), using only the
operations that are permitted in classical geometry. In particular,
you are not allowed to make or use markings on the straightedge (which
would turn it into a ruler), and you are not permitted to hold the
compass against the straightedge and use both in the same operation.
--
Dave Seaman dse...@purdue.edu
++++ stop the execution of Mumia Abu-Jamal ++++
++++ if you agree copy these lines to your sig ++++
++++ see http://www.xs4all.nl/~tank/spg-l/sigaction.htm ++++
Seth Breidbart
In article <E61u4...@bhishma.cse.iitb.ernet.in>,
Utpat Shriram Arvind <shr...@bhishma.cse.iitb.ernet.in> wrote:
> Hello,
> Which is the reference to such a proof. Also you are saying
>that we can trisect an angle approximately. Then what is the amount of
>error that will be there.
Given that you can bisect the angle, you can approximate a trisection
precisely as well as you can approximate 1/3 in binary (that is,
exactly in the limit).
ObPuzzle: What's the best you can do, as a function of the number of
operations?
Seth
Wei-Hwa Huang
a...@seaman.cc.purdue.edu (Dave Seaman) writes:
>Amit Chakrabarti <am...@bhishma.cse.iitb.ernet.in> wrote:
>>I do not know any details about any approximate trisection methods.
>>Someone had made a posting saying that someone else had found a method
>>for "trisecting" an angle using ruler-and-compass. In the light of the
>>proof we're talking about it follows that such a method *must* be
>>approximate.
>No, it doesn't follow. What is impossible is to trisect an arbitrary
>angle with compass and *straightedge* (not ruler), using only the
>operations that are permitted in classical geometry. In particular,
>you are not allowed to make or use markings on the straightedge (which
>would turn it into a ruler), and you are not permitted to hold the
>compass against the straightedge and use both in the same operation.
For that matter, it's worth emphasizing that the angle be *arbitrary*.
ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
angle to trisect?
--
Wei-Hwa Huang, whu...@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Metaphors be with you.
Matthew T. Russotto
In article <5esm73$s...@gap.cco.caltech.edu>,
Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
}ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
}angle to trisect?
270.
--
Matthew T. Russotto russ...@pond.com
"Extremism in defense of liberty is no vice, and moderation in pursuit
of justice is no virtue."
Dave Dodson
In article <5esm73$s...@gap.cco.caltech.edu>,
Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
>ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
>angle to trisect?
1080 degrees.
Dave
Noam Elkies
270 degrees, or arguably 540 and multiples thereof :-)
NDE
Matthew Daly
In article <E63GL...@bhishma.cse.iitb.ernet.in> am...@bhishma.cse.iitb.ernet.in (Amit Chakrabarti) writes:
>
>I do not know any details about any approximate trisection methods.
>Someone had made a posting saying that someone else had found a method
>for "trisecting" an angle using ruler-and-compass. In the light of the
>proof we're talking about it follows that such a method *must* be
>approximate.
There IS an exact trisection available with ruler and compass. The
difference between a ruler and a straightedge is like the difference
between a one-pan scale and a balance scale.
Given angle AOB, draw a circle with radius O, and the diameter that is
perpendicular to BO. Now, find a point C on the circle such that AC
is bisected by the diameter. (That is the step that you can do with
a ruler but not a straightedge or compass.) Now, draw a diameter
parallel to AC -- it will trisect the angle as desired.
Matthew Daly
In article <5esmri$l...@wanda.vf.pond.com> russ...@wanda.vf.pond.com (Matthew T. Russotto) writes:
>In article <5esm73$s...@gap.cco.caltech.edu>,
>Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
>}ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
>}angle to trisect?
>
I think that 90 or 180 has that beat for auxiliary marks and lines. Just
to get the ball rolling, I'll count.
A) With the compass center at the apex of the angle, draw an arc across
the angle.
B) Choose a point where the arc meets the angle. With the compass
center at that point (and the same radius), mark a point on the
arc.
C) Draw the line connecting the apex of the angle with that point.
This trisects the angle.
Of course, I'm not completely familiar with the metric for "operations"
of a construction.
Carl Cerecke
Douglas Limmer wrote:
>
> gr...@bhishma.cse.iitb.ernet.in (Rajesh G) wrote:
>
> >I don't know the full problem definition but it goes
> >like this.
>
> >Can you trisect a line with a ruler and a compass ?
>
[explanation snipped]
> Now, if you wanted to trisect an _angle_, it would be harder (that is,
> it's been proven to be impossible).
For the general angle, yes. But there are specific angles for
which trisection *is* possible? What are they?
Cheers,
cdc
Orjan Johansen
Yet another trisection puzzle, solvable this time:
Find an angle which you cannot construct (with the usual tools), but
which you can trisect.
Greetings,
Ørjan.
Ed Murphy
>For the general angle, yes. But there are specific angles for
>which trisection *is* possible? What are they?
Any multiple of 90 degrees, for one.
--
Ed Murphy <mailto:fo...@bayside.net>
http://www.bayside.net/users/ford/
"I see," said Arthur, who didn't.
Carl Witthoft
In article <5esm73$s...@gap.cco.caltech.edu> whu...@ugcs.caltech.edu (Wei-Hwa Huang) writes:
>
>ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
>angle to trisect?
>Metaphors be with you.
I never metamorphosis I didn't like.
--
Carl Witthoft @ Adaptive Optics Associates
ca...@aoainc.com c...@world.std.com 617-864-0201
" Axis-navigo, ergo sum."
Abigail
On 25 Feb 1997 18:32:37 GMT, Carl Witthoft wrote in rec.puzzles:
++ In article <5esm73$s...@gap.cco.caltech.edu> whu...@ugcs.caltech.edu
++ (Wei-Hwa Huang) writes:
++ >
++ >ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
++ >angle to trisect?
++ I vote for 270 degrees.
I prefer 1080 degrees.
Abigail
Seth Breidbart
In article <5etb9m$h...@due.unit.no>,
Orjan Johansen <orj...@lie.matstat.unit.no> wrote:
>Yet another trisection puzzle, solvable this time:
>
>Find an angle which you cannot construct (with the usual tools), but
>which you can trisect.
0.9 degrees comes to mind. (I don't _think_ it's constructible.)
Seth
Noam Elkies
In article <5etb9m$h...@due.unit.no>
orj...@lie.matstat.unit.no (Orjan Johansen) writes:
>Yet another trisection puzzle, solvable this time:
>Find an angle which you cannot construct (with the usual tools), but
>which you can trisect.
Lots of such examples, probably the simplest being multiples of
540/7 degrees.
--Noam D. Elkies (elk...@math.harvard.edu)
Bill Shatzer
In a previous article, Right...@nasa.gov () says:
>am...@bhishma.cse.iitb.ernet.in (Amit Chakrabarti) wrote:
>
>>Rajesh G (gr...@bhishma.cse.iitb.ernet.in) writes:
>>> I don't know the full problem definition but it goes
>>> like this.
>>>
>>> Can you trisect a line with a ruler and a compass ?
>
>>There is a very simple Euclidean construction for n-secting a line segment.
>>This oviously subsumes trisection.
>
>>What u cannot trisect is an angle.
>
>Why not???
>
>
>Take an angle A, and mark off equall distances B & C
>
>Then trisect line segment BC, and draw lines from A to each point
And, thus what you have done is trisect line BC and NOT angle A.
If you doubt this, try your approach with a _really_ large angle A,
say 160 degrees or so. The differences in the three angles will
be visiably apparent.
Cheers,
--
- Bill Shatzer bsha...@orednet.org -
"Oh, what sad times are these when passing ruffians can say 'ni' at will to
old ladies."
Fred the Wonder Worm
In article <5esm73$s...@gap.cco.caltech.edu>,
Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
>
>ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
>angle to trisect?
270 degrees is the obvious one - claim it as is, or if you insist on the
lines in the right spots extend one of the sides of the angle. If we
restrict the question to non-reflex angles (in an attempt to make it
marginally more interesting), then 180 degrees seems the most likely
candidate - two arcs and a line (or two lines, if you want both trisection
lines) do the job. (Likewise for 90 degrees, except another arc is needed
for the second trisection line as well.)
If we insist that the angle not be a multiple of 45 degrees (those stupid
60 degree angles make it too easy), then my suggestion would be 540/7
degrees, requiring three arcs and one line (add another arc and line for
the other trisection point). [45 degrees takes the same amount,
interestingly enough.] If you only want to create an angle equal to 1/3
the given angle (without insisting that it be in the right position),
then you can drop the costs by one arc.
Cheers,
Geoff.
-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@cs.su.oz.au | Gameplayer by vocation.
-------------------------------------------------------------------------------
Dave Seaman
In article <5f0bgh$3...@staff.cs.su.oz.au>,
Fred the Wonder Worm <ft...@cs.su.oz.au> wrote:
>In article <5esm73$s...@gap.cco.caltech.edu>,
>Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
>>ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
>>angle to trisect?
>270 degrees is the obvious one - claim it as is, or if you insist on the
Many people have mentioned 270 degrees, but I think 180 degrees is easier to
trisect (not as easy as 540 degrees or 1080 degrees, though).
Matthew T. Russotto
In article <5f1jro$1...@seaman.cc.purdue.edu>,
Dave Seaman <a...@seaman.cc.purdue.edu> wrote:
}In article <5f0bgh$3...@staff.cs.su.oz.au>,
}Fred the Wonder Worm <ft...@cs.su.oz.au> wrote:
}
}>In article <5esm73$s...@gap.cco.caltech.edu>,
}>Wei-Hwa Huang <whu...@ugcs.caltech.edu> wrote:
}
}>>ObPuzzle: Not counting 0, what is the "easiest" (fewest operations needed)
}>>angle to trisect?
}
}>270 degrees is the obvious one - claim it as is, or if you insist on the
}
}Many people have mentioned 270 degrees, but I think 180 degrees is easier to
}trisect (not as easy as 540 degrees or 1080 degrees, though).
270 can be trisected with two straight-edge operations only.
Joe Keane
In article <5f0bgh$3...@staff.cs.su.oz.au>
In article <5f1v6k$k...@wanda.vf.pond.com>
Matthew T. Russotto <russ...@wanda.vf.pond.com> writes:
>270 can be trisected with two straight-edge operations only.
What if we exclude any rational multiple of a circle?
--
Joe Keane, amateur mathematician
John Rickard
Noam Elkies (elk...@brauer.harvard.edu) wrote:
: I wonder whether a "straightedge" -- i.e. a device for finding the
: great circle through any two non-antipodal points on the sphere --
: would suffice to trisect a great circle, or for that matter do any
: of the standard constructions (perpendiculars, triangle given three
: sides, etc.) in the context of spherical geometry.
No, because there are transformations of the sphere that map great
circles to great circles but do not preserve angles and distances.
(Identify a point on the sphere with the ray through it from the
origin; then great circles correspond to planes through the origin, so
any linear transformation of 3-space gives a transformation of the
sphere that maps great circles to great circles, but it does not
preserve angles etc unless it is orthogonal.)
--
John Rickard
Ivon Kok
da...@PPD.Kodak.COM (Matthew Daly) wrote:
>There IS an exact trisection available with ruler and compass. The
>difference between a ruler and a straightedge is like the difference
>between a one-pan scale and a balance scale.
>Given angle AOB, draw a circle with radius O,
I don't get this. What should be the center of the circle? What should
the radius be? (Radius can't be O, 'cos O is a point).
>and the diameter that is
>perpendicular to BO. Now, find a point C on the circle such that AC
>is bisected by the diameter. (That is the step that you can do with
>a ruler but not a straightedge or compass.) Now, draw a diameter
>parallel to AC -- it will trisect the angle as desired.
>-Matthew
If I take O as the center of the circle, AC is a diameter, and the
last step makes no sense.
PLease clarify.
-Ivon
Fred the Wonder Worm
In that case, my current best is (3/2)*Arccos(1/4), which takes 4 arcs
and 2 lines for the two trisection lines (3 and 1 if only one is desired).
Dr D F Holt
In article <5etb9m$h...@due.unit.no>,
orj...@lie.matstat.unit.no (Orjan Johansen) writes:
>Yet another trisection puzzle, solvable this time:
>
>Find an angle which you cannot construct (with the usual tools), but
>which you can trisect.
>
pi/7.
Given an angle of pi/7, I can construct pi/3 - pi/7 = 4pi/21, and then
bisecting that twice gives pi/21.
Derek Holt.