(which I firmly recommend, BTW, along with his other books),
when I came across a remark that, although a regular heptagon
is not constructible with ruler and compass only, it IS
constructible if you add in an angle-trisecting device!
Well, that sounded fun, as trisections seem to have little to do
with heptasections, on the face of it. No further details or
references were given, but I soon managed to convince myself,
using basic Galois ideas and complex numbers, that it would be
possible, in principle. (That's the 1st exercise for the reader!)
However, it would be a pretty hopeless mess to try to convert
that algebra into a neat geometric construction, so there is
my main query, (and thus 2nd exercise for the reader...)
** Find a simple geometric construction of a regular heptagon
** using ruler, compass and angle-trisector.
-- Testing Taylor
## The math is done right, but is the right math done?
> (which I firmly recommend, BTW, along with his other books),
> when I came across a remark that, although a regular heptagon
> is not constructible with ruler and compass only, it IS
> constructible if you add in an angle-trisecting device!
> Well, that sounded fun, as trisections seem to have little to do
> with heptasections, on the face of it. No further details or
> references were given, but I soon managed to convince myself,
> using basic Galois ideas and complex numbers, that it would be
> possible, in principle. (That's the 1st exercise for the reader!)
> However, it would be a pretty hopeless mess to try to convert
> that algebra into a neat geometric construction, so there is
> my main query, (and thus 2nd exercise for the reader...)
> ** Find a simple geometric construction of a regular heptagon
> ** using ruler, compass and angle-trisector.
See Conway and Guy's "The Book of Numbers". They also
do the regular 13-gon.
> I've been bobbling through Ian Stewart's
> ** "A Cabinet of Mathematical Curiosities",
> (which I firmly recommend, BTW, along with his other books),
> when I came across a remark that, although a regular heptagon
> is not constructible with ruler and compass only, it IS
> constructible if you add in an angle-trisecting device!
> Well, that sounded fun, as trisections seem to have little to do
> with heptasections, on the face of it. No further details or
> references were given, but I soon managed to convince myself,
> using basic Galois ideas and complex numbers, that it would be
> possible, in principle. (That's the 1st exercise for the reader!)
> However, it would be a pretty hopeless mess to try to convert
> that algebra into a neat geometric construction, so there is
> my main query, (and thus 2nd exercise for the reader...)
> ** Find a simple geometric construction of a regular heptagon
> ** using ruler, compass and angle-trisector.
Start with a 30-60-90 triangle OPQ. Then sides OP:PQ:OQ = 1:sqrt(3):2.
Extend side PQ to point R, PQ:PR = 1:3. Then in the triangle OPR,
sides OP:PR:RO = 1:sqrt(27):sqrt(28). Angle POR is then
acos(1/sqrt(28)). Trisect it to form angle POS where S lies on
segment PQ. The radius r of the circumscribed circle is
6*(length of OS), and (length of OR)-(length of OS) = r*cos(2*pi/7).
-- write(*,*) transfer((/17.392111325966148d0,6.5794487871554595D-85, &
6.0134700243160014d-154/),(/'x'/)); end
On Wednesday, October 10, 2012 4:20:56 AM UTC-4, Robin wrote:
> On 10/10/2012 06:44, Bill Taylor wrote: > I've been bobbling through Ian Stewart's > > ** "A Cabinet of Mathematical Curiosities", > > (which I firmly recommend, BTW, along with his other books), > when I came across a remark that, although a regular heptagon > is not constructible with ruler and compass only, it IS > constructible if you add in an angle-trisecting device! > > Well, that sounded fun, as trisections seem to have little to do > with heptasections, on the face of it. No further details or > references were given, but I soon managed to convince myself, > using basic Galois ideas and complex numbers, that it would be > possible, in principle. (That's the 1st exercise for the reader!) > > However, it would be a pretty hopeless mess to try to convert > that algebra into a neat geometric construction, so there is > my main query, (and thus 2nd exercise for the reader...) > > ** Find a simple geometric construction of a regular heptagon > ** using ruler, compass and angle-trisector. See Conway and Guy's "The Book of Numbers". They also do the regular 13-gon.
It seems that any regular polygon with p = 3*2^n + 1 sides should also be constructible. (p prime). e.g. p = 97, 193, etc.
> It seems that any regular polygon with p = 3*2^n + 1 sides should also be constructible. (p prime). e.g. p = 97, 193, etc.
Gauss (at the age of nineteen!) found that the regular n-gon can be
constructed if the odd prime factors of n are distinct Fermat primes
2^{2^k} + 1. The fourth Fermat prime is 65537 and a fellow called
Hermes spent 10 years constructing a regular 65537-gon.
Wantzel (1837) found that Gauss's condition is necessary as well as
sufficient.
This we may learn (without proofs) from the first two sections of
Chapter two of Coxeter's Introduction to geometry.
-- Where are the songs of Summer?--With the sun,
Oping the dusky eyelids of the south,
Till shade and silence waken up as one,
And morning sings with a warm odorous mouth.
> > It seems that any regular polygon with p = 3*2^n + 1 sides should also be constructible. (p prime). e.g. p = 97, 193, etc.
> Gauss (at the age of nineteen!) found that the regular n-gon can be
> constructed if the odd prime factors of n are distinct Fermat primes
> 2^{2^k} + 1.
We know all of this. If p is prime and equals 2^2^k + 1, then
Q(zeta(p)) consists only of a tower of quadratic extensions of Q.
The solutions to the equations that arise from the intersections of
line and circles (straightedge and compass) give rise to linear
or quadratic (tower of) equations. The roots of said equations
when appended to Q yield a quadratic tower of extension fields.
If p = 3*2^n + 1, and we can solve cubics, we can construct
algebraic numebers arising from a cubic field or a tower of
quadratic extensions of a cubic field....
> On Oct 12, 11:21 am, Frederick Williams
> <freddywilli...@btinternet.com> wrote:
> > Pubkeybreaker wrote:
> > > It seems that any regular polygon with p = 3*2^n + 1 sides
> > > should also be constructible. (p prime). e.g. p = 97, 193, etc.
> > Gauss (at the age of nineteen!) found that the regular n-gon can be
> > constructed if the odd prime factors of n are distinct Fermat primes
> > 2^{2^k} + 1.
> We know all of this. If p is prime and equals 2^2^k + 1, then
> Q(zeta(p)) consists only of a tower of quadratic extensions of Q.
> The solutions to the equations that arise from the intersections of
> line and circles (straightedge and compass) give rise to linear
> or quadratic (tower of) equations. The roots of said equations
> when appended to Q yield a quadratic tower of extension fields.
> If p = 3*2^n + 1, and we can solve cubics, we can construct
> algebraic numebers arising from a cubic field or a tower of
> quadratic extensions of a cubic field....
C:\gfortran\clf\nonadecagon>nonadecagon
Error in first cubic:
-1.7763568394002505E-015
-2.4868995751603507E-014
-1.7763568394002505E-015
Error in first root:
-2.2204460492503131E-016
-1.7763568394002505E-015
-8.8817841970012523E-016
Error in coefficients of second cubic:
-2.2204460492503131E-016
-4.4408920985006262E-016
-2.6645352591003757E-015
Error in second cubic:
-6.2172489379008766E-015
-1.7763568394002505E-015
-1.7763568394002505E-015
Error in 2*cos(theta):
0.0000000000000000
-- write(*,*) transfer((/17.392111325966148d0,6.5794487871554595D-85, &
6.0134700243160014d-154/),(/'x'/)); end
>If p = 3*2^n + 1, and we can solve cubics, we can construct
>algebraic numebers arising from a cubic field or a tower of
>quadratic extensions of a cubic field....
The ability to trisect an arbitrary constructible angle allows one to solve _some_ cubics which are irreducible over Q or over a field which is a tower of quadratic extensions of Q,
but it's not clear to me that it gives the means to solve _all_ such cubics.
As a simple example, does the ability to trisect an arbitrary
constructible angle yield a means to solve the cubic equation
x^3 - 2 = 0 -- that is, does it allow construction of the length 2^(1/3)? I suspect not.
Similarly, I don't believe that for all primes p of the form 3*(2^n) + 1, one can construct a regular p-gon using only a
straight-edge, a compass, and an angle-trisector device.
(which I firmly recommend, BTW, along with his other books),
when I came across a remark that, although a regular heptagon
is not constructible with ruler and compass only, it IS
constructible if you add in an angle-trisecting device!
_____________________________________
In my bookshelf, and heartily agree.
Well, that sounded fun, as trisections seem to have little to do
with heptasections, on the face of it. No further details or
references were given, but I soon managed to convince myself,
using basic Galois ideas and complex numbers, that it would be
possible, in principle. (That's the 1st exercise for the reader!)
However, it would be a pretty hopeless mess to try to convert
that algebra into a neat geometric construction, so there is
my main query, (and thus 2nd exercise for the reader...)
** Find a simple geometric construction of a regular heptagon
** using ruler, compass and angle-trisector.
________________________________________
I applaud you for raising this, but http://math.fau.edu/yiu/regularheptagontrisection070424.pdf does exactly this. It forms the cubic equation that has to be solved for the heptagon, and then shows how the internal angle that is need can be formed by trisecting another angle. The equations are the geometry are quite well linked. It looks to be a very good read, a good overview. http://www.tweedledum.com/rwg/heptagon.htm provides a simpler construction alone; as it was devised by Conway its fair to say its not obvious.
Polygons with a prime number of sides p cannot be constructed with straight edge and compass unless p is a Fermat prime. I suspect you would need a device which could take the 1/p part of angle to construct them, which makes it trivial. I can quite believe that a 9gon and 12gon are constructible using an angle trisector. Does an 11gon require solving a quintic (ie a device which divides an angle into 5 equal parts) ? The approach given in the paper should allow the corresponding eqn to be formed for the 11gon, and eventually to characterise which angle constructors are needed for which regular polygons.
-- Testing Taylor
## The math is done right, but is the right math done?
> >If p = 3*2^n + 1, and we can solve cubics, we can construct
> >algebraic numebers arising from a cubic field or a tower of
> >quadratic extensions of a cubic field....
> The ability to trisect an arbitrary constructible angle
> allows one to solve _some_ cubics which are irreducible over Q
> or over a field which is a tower of quadratic extensions of Q,
> but it's not clear to me that it gives the means to solve
> _all_ such cubics.
Yep. I spoke too quickly. Let me think about this.
It would be a nice problem to characterize all p = 3*2^n + 1
that can be constructed with a trisector, ruler, and compass.
> On Oct 13, 3:27 am, quasi <qu...@null.set> wrote:
> > Pubkeybreaker wrote:
> > >If p = 3*2^n + 1, and we can solve cubics, we can construct
> > >algebraic numebers arising from a cubic field or a tower of
> > >quadratic extensions of a cubic field....
> > The ability to trisect an arbitrary constructible angle
> > allows one to solve _some_ cubics which are irreducible over Q
> > or over a field which is a tower of quadratic extensions of Q,
> > but it's not clear to me that it gives the means to solve
> > _all_ such cubics.
> Yep. I spoke too quickly. Let me think about this.
> It would be a nice problem to characterize all p = 3*2^n + 1
> that can be constructed with a trisector, ruler, and compass.
Why not all p = 3**m*2**n? The example of x**3-2 = 0 is invalid
because it doesn't have three real roots. By construction, the
cubics involved in constructing p = 3**m*2**n always have three
real roots, so the trisection method does find all three roots.
-- write(*,*) transfer((/17.392111325966148d0,6.5794487871554595D-85, &
6.0134700243160014d-154/),(/'x'/)); end
>"Pubkeybreaker" <pubkeybrea...@aol.com> wrote in message
>> On Oct 13, 3:27 am, quasi <qu...@null.set> wrote:
>> > Pubkeybreaker wrote:
>> > >If p = 3*2^n + 1, and we can solve cubics, we can construct
>> > >algebraic numebers arising from a cubic field or a tower of
>> > >quadratic extensions of a cubic field....
>> > The ability to trisect an arbitrary constructible angle
>> > allows one to solve _some_ cubics which are irreducible over Q
>> > or over a field which is a tower of quadratic extensions of Q,
>> > but it's not clear to me that it gives the means to solve
>> > _all_ such cubics.
>> >Yep. I spoke too quickly. Let me think about this.
>> It would be a nice problem to characterize all p = 3*2^n + 1
>> that can be constructed with a trisector, ruler, and compass.
>Why not all p = 3**m*2**n?
There aren't many primes of that form!
Did you perhaps mean p = (3^m)*(2^n) + 1 ?
>The example of x**3-2 = 0 is invalid because it doesn't have >three real roots. By construction, the cubics involved in >constructing p = 3**m*2**n
Presumably you mean "constructing a regular p-gon", not
"constructing p".
>always have three real roots, so the trisection method does >find all three roots.
Let's call a regular polygon "tri-constructible" if it's
constructible with the use of a straight-edge, compass, and angle trisector.
Now it's trivial to show that if a regular k-gon is tri-constructible, then so is a regular 3k-gon.
Hence it's,immediate that one can tri-construct a regular k-gon for all k of the form (3^m)*(2^n) where m,n are nonnegative integers with m >= 1 or n >= 2.
But the question wasn't about those kinds of regular polygons.
Rather it was about regular p-gons where p is a prime of the
form 3*(2^n) + 1. Known examples of primes p of that form for which a regular p-gon is tri-constructible include p = 7 and p = 13, however as far as I know, those are the
_only_ primes p of that form which are known to work.
> "James Van Buskirk" <not_va...@comcast.net> wrote:
>>Why not all p = 3**m*2**n?
> There aren't many primes of that form!
> Did you perhaps mean p = (3^m)*(2^n) + 1 ?
Yes, I forgot to append the "+1" there. Just a typo.
> But the question wasn't about those kinds of regular polygons.
> Rather it was about regular p-gons where p is a prime of the
> form 3*(2^n) + 1. Known examples of primes p of that form
> for which a regular p-gon is tri-constructible include
> p = 7 and p = 13, however as far as I know, those are the
> _only_ primes p of that form which are known to work.
If you just go over the technique I believe you will come to the
conclusion that all regular p-gons with prime p = 3**m*2**n+1
are tri-constructible. Think about how the theory of cyclotomic
equations applies here. Try p = 37 (hence tri-construct the
beastagon) yourself, and you'll see how it works.
-- write(*,*) transfer((/17.392111325966148d0,6.5794487871554595D-85, &
6.0134700243160014d-154/),(/'x'/)); end
>> But the question wasn't about those kinds of regular polygons.
>> Rather it was about regular p-gons where p is a prime of the
>> form 3*(2^n) + 1. Known examples of primes p of that form
>> for which a regular p-gon is tri-constructible include
>> p = 7 and p = 13, however as far as I know, those are the
>> _only_ primes p of that form which are known to work.
>If you just go over the technique I believe you will come to the
>conclusion that all regular p-gons with prime p = 3**m*2**n+1
>are tri-constructible. Think about how the theory of cyclotomic
>equations applies here. Try p = 37 (hence tri-construct the
>beastagon) yourself, and you'll see how it works.
Ok, I was skeptical, but when I get a chance, I'll try to work it out.
> James Van Buskirk wrote:
>>I posted p = 19 yesterday:
>>https://groups.google.com/d/msg/sci.math/M81uSlWJjMU/oRZqclg9DewJ > Thanks -- I'll take a closer look.
>>If you just go over the technique I believe you will come to the
>>conclusion that all regular p-gons with prime p = 3**m*2**n+1
>>are tri-constructible. Think about how the theory of cyclotomic
>>equations applies here. Try p = 37 (hence tri-construct the
>>beastagon) yourself, and you'll see how it works.
> Ok, I was skeptical, but when I get a chance, I'll try to
> work it out.
Here is p = 37:
C:\gfortran\clf\nonadecagon>type p37.f90
program p37
implicit none
integer, parameter :: dp = selected_real_kind(15,30)
real(dp), parameter :: pi = 4*atan(1.0_dp)
real(dp), parameter :: theta = 2*pi/37
real(dp), parameter :: x0 = &
sum(2*cos([1,8,27,31,26,23]*theta))
real(dp), parameter :: x1 = &
sum(2*cos([2,16,17,25,15,9]*theta))
real(dp), parameter :: x2 = &
sum(2*cos([4,32,34,13,30,18]*theta))
real(dp), parameter :: x00 = sum(2*cos([1,31]*theta))
real(dp), parameter :: x01 = sum(2*cos([8,26]*theta))
real(dp), parameter :: x02 = sum(2*cos([27,23]*theta))
real(dp), parameter :: x10 = sum(2*cos([2,25]*theta))
real(dp), parameter :: x11 = sum(2*cos([16,15]*theta))
real(dp), parameter :: x12 = sum(2*cos([17,9]*theta))
real(dp), parameter :: x20 = sum(2*cos([4,13]*theta))
real(dp), parameter :: x21 = sum(2*cos([32,30]*theta))
real(dp), parameter :: x22 = sum(2*cos([34,18]*theta))
real(dp), parameter :: x000 = 2*cos(theta)
real(dp), parameter :: x001 = 2*cos(31*theta)
real(dp), parameter :: y2 = (sqrt(148.0_dp)*cos( &
acos(-11/sqrt(148.0_dp))/3)-1)/3
real(dp), parameter :: y1 = (sqrt(148.0_dp)*cos( &
acos(-11/sqrt(148.0_dp))/3+2*pi/3)-1)/3
real(dp), parameter :: y0 = (sqrt(148.0_dp)*cos( &
acos(-11/sqrt(148.0_dp))/3-2*pi/3)-1)/3
real(dp), parameter :: u0 = 2*sqrt(-13*y0-13*y1-11*y2)/3
real(dp), parameter :: c3phi0 = -(115*y0+120*y1+111*y2)/(93*u0)
real(dp), parameter :: y00 = u0*cos(acos(c3phi0)/3)+y0/3
real(dp), parameter :: y02 = u0*cos(acos(c3phi0)/3+2*pi/3)+y0/3
real(dp), parameter :: y01 = u0*cos(acos(c3phi0)/3-2*pi/3)+y0/3
real(dp), parameter :: u1 = 2*sqrt(-11*y0-13*y1-13*y2)/3
real(dp), parameter :: c3phi1 = -(111*y0+115*y1+120*y2)/(93*u1)
real(dp), parameter :: y10 = u1*cos(acos(c3phi1)/3)+y1/3
real(dp), parameter :: y11 = u1*cos(acos(c3phi1)/3+2*pi/3)+y1/3
real(dp), parameter :: y12 = u1*cos(acos(c3phi1)/3-2*pi/3)+y1/3
real(dp), parameter :: u2 = 2*sqrt(-13*y0-11*y1-13*y2)/3
real(dp), parameter :: c3phi2 = -(120*y0+111*y1+115*y2)/(93*u2)
real(dp), parameter :: y21 = u2*cos(acos(c3phi2)/3)+y2/3
real(dp), parameter :: y22 = u2*cos(acos(c3phi2)/3+2*pi/3)+y2/3
real(dp), parameter :: y20 = u2*cos(acos(c3phi2)/3-2*pi/3)+y2/3
real(dp), parameter :: y000 = (y00+sqrt(y00**2-4*y21))/2
real(dp), parameter :: y001 = (y00-sqrt(y00**2-4*y21))/2
write(*,*) 'Error in coefficients of first cubic:'
write(*,*) 1+(x0+x1+x2)
write(*,*) -12-(x0*x1+x1*x2+x2*x0)
write(*,*) 11+x0*x1*x2
write(*,*) 'Error in first cubic:'
write(*,*) x0**3+x0**2-12*x0+11
write(*,*) x1**3+x1**2-12*x1+11
write(*,*) x2**3+x2**2-12*x2+11
write(*,*) 'Error in roots of first cubic:'
write(*,*) y0-x0
write(*,*) y1-x1
write(*,*) y2-x2
write(*,*) 'Error in coefficients of second cubic[0]:'
write(*,*) -y0+(x00+x01+x02)
write(*,*) (y1-1)-(x00*x01+x01*x02+x02*x00)
write(*,*) (y0-2)+x00*x01*x02
write(*,*) 'Error in second cubic[0]:'
write(*,*) x00**3-y0*x00**2+(y1-1)*x00+(y0-2)
write(*,*) x01**3-y0*x01**2+(y1-1)*x01+(y0-2)
write(*,*) x02**3-y0*x02**2+(y1-1)*x02+(y0-2)
write(*,*) 'Error in roots of second cubic[0]:'
write(*,*) y00-x00
write(*,*) y01-x01
write(*,*) y02-x02
write(*,*) 'Error in coefficients of second cubic[1]:'
write(*,*) -y1+(x10+x11+x12)
write(*,*) (y2-1)-(x10*x11+x11*x12+x12*x10)
write(*,*) (y1-2)+x10*x11*x12
write(*,*) 'Error in second cubic[1]:'
write(*,*) x10**3-y1*x10**2+(y2-1)*x10+(y1-2)
write(*,*) x11**3-y1*x11**2+(y2-1)*x11+(y1-2)
write(*,*) x12**3-y1*x12**2+(y2-1)*x12+(y1-2)
write(*,*) 'Error in roots of second cubic[1]:'
write(*,*) y10-x10
write(*,*) y11-x11
write(*,*) y12-x12
write(*,*) 'Error in coefficients of second cubic[2]:'
write(*,*) -y2+(x20+x21+x22)
write(*,*) (y0-1)-(x20*x21+x21*x22+x22*x20)
write(*,*) (y2-2)+x20*x21*x22
write(*,*) 'Error in second cubic[2]:'
write(*,*) x20**3-y2*x20**2+(y0-1)*x20+(y2-2)
write(*,*) x21**3-y2*x21**2+(y0-1)*x21+(y2-2)
write(*,*) x22**3-y2*x22**2+(y0-1)*x22+(y2-2)
write(*,*) 'Error in roots of second cubic[2]:'
write(*,*) y20-x20
write(*,*) y21-x21
write(*,*) y22-x22
write(*,*) 'Error in coefficients of quadratic[00]:'
write(*,*) -y00+(x000+x001)
write(*,*) y21-x000*x001
write(*,*) 'Error in quadratic[00]:'
write(*,*) x000**2-y00*x000+y21
write(*,*) x001**2-y00*x001+y21
write(*,*) 'Error in roots of quadratic:'
write(*,*) y000-x000
write(*,*) y001-x001
end program p37
C:\gfortran\clf\nonadecagon>p37
Error in coefficients of first cubic:
-3.1086244689504383E-015
-1.0658141036401503E-014
2.8421709430404007E-014
Error in first cubic:
1.0658141036401503E-014
2.8421709430404007E-014
-1.0658141036401503E-014
Error in roots of first cubic:
2.4424906541753444E-015
0.0000000000000000
1.3322676295501878E-015
Error in coefficients of second cubic[0]:
-2.4424906541753444E-015
1.7763568394002505E-015
-8.8817841970012523E-016
Error in second cubic[0]:
-2.2870594307278225E-014
-1.2212453270876722E-015
-1.0436096431476471E-014
Error in roots of second cubic[0]:
8.8817841970012523E-016
1.2212453270876722E-015
2.2204460492503131E-015
Error in coefficients of second cubic[1]:
0.0000000000000000
-1.7763568394002505E-015
-5.3290705182007514E-015
Error in second cubic[1]:
-7.9936057773011271E-015
1.7763568394002505E-015
-2.6645352591003757E-015
Error in roots of second cubic[1]:
5.5511151231257827E-016
0.0000000000000000
6.6613381477509392E-016
Error in coefficients of second cubic[2]:
-4.4408920985006262E-016
1.6375789613221059E-015
-6.3837823915946501E-016
Error in second cubic[2]:
-1.6653345369377348E-016
-1.6653345369377348E-016
-1.0824674490095276E-015
Error in roots of second cubic[2]:
3.3306690738754696E-016
-4.4408920985006262E-016
1.5543122344752192E-015
Error in coefficients of quadratic[00]:
-8.8817841970012523E-016
8.8817841970012523E-016
Error in quadratic[00]:
-4.4408920985006262E-016
0.0000000000000000
Error in roots of quadratic:
6.6613381477509392E-016
2.2204460492503131E-016
From the above program you should be able to see the arithmetic
used to get to 2*cos(2*pi/37) via tri-construction. The program
outputs are just to guard against gross errors in the derivation,
of which there were many before I got it right.
-- write(*,*) transfer((/17.392111325966148d0,6.5794487871554595D-85, &
6.0134700243160014d-154/),(/'x'/)); end
>>>If you just go over the technique I believe you will come >>>to the conclusion that all regular p-gons with prime >>>p = 3**m*2**n+1 are tri-constructible.
Seems right.
Thus, Pubkeybreaker was also right, although your claim is
stronger.
>>>Think about how the theory of cyclotomic
>>>equations applies here. Try p = 37 (hence tri-construct the
>>>beastagon) yourself, and you'll see how it works.
beastagon -- haha.
>> Ok, I was skeptical, but when I get a chance, I'll try to
>> work it out.
>Here is p = 37:
> ...
>From the above program you should be able to see the arithmetic
>used to get to 2*cos(2*pi/37) via tri-construction. The program
>outputs are just to guard against gross errors in the derivation
>of which there were many before I got it right.
I'm still thinking about the details, but your conclusion seems right.
> "Bill Taylor" wrote in message > news:fbf050e3-efc6-4c48-82fa-f2a0edb7031e@q5g2000pbk.googlegroups.com...
> > I've been bobbling through Ian Stewart's
> > ** "A Cabinet of Mathematical Curiosities",
> > (which I firmly recommend, BTW, along with his other books),
> > when I came across a remark that, although a regular heptagon
> > is not constructible with ruler and compass only, it IS
> > constructible if you add in an angle-trisecting device!
> I applaud you for raising this, but > http://math.fau.edu/yiu/regularheptagontrisection070424.pdf does exactly > this. It forms the cubic equation that has to be solved for the heptagon, > and then shows how the internal angle that is need can be formed by > trisecting another angle. The equations are the geometry are quite well > linked. It looks to be a very good read, a good overview. > http://www.tweedledum.com/rwg/heptagon.htm provides a simpler > construction alone; as it was devised by Conway its fair to say its not > obvious.
> Polygons with a prime number of sides p cannot be constructed with > straight edge and compass unless p is a Fermat prime. I suspect you would > need a device which could take the 1/p part of angle to construct them, > which makes it trivial. I can quite believe that a 9gon and 12gon are > constructible using an angle trisector. Does an 11gon require solving a > quintic (ie a device which divides an angle into 5 equal parts) ? The > approach given in the paper should allow the corresponding eqn to be > formed for the 11gon, and eventually to characterise which angle > constructors are needed for which regular polygons.
I would have said that the 11-gon, which does require solving a
quintic, could not be solved by 5-section. However, my guess was
dead wrong! Here is the solution to the 11-gon, in both complex
fifth roots and real 5-sections:
Does that mean you can do a 31-gon with angle 5-section and
trisection? How about a 101-gon with angle 5-section, or a 29-gon
with angle 7-section and trisection?
-- write(*,*) transfer((/17.392111325966148d0,6.5794487871554595D-85, &
6.0134700243160014d-154/),(/'x'/)); end
of course, a la Masceroni, one doesn't need a straightedge
... have to use a linkage of compasses & trisectors!
thus:
adding iron to sea only produces a relative deficit
of the other stable elements of the periodic table,
a-hem.
thus:
I haven't seen any defensible explanation of any errors
by Miskolczi, either, nor of the INQUA guy on sealevel;
that is to say, theoretically nor statistically.
thus:
AnIS and GrIS are known to have *only* risen in heighth,
since the IGY (the first international bipolar year,
'57-9; we are still in the second intl. bip.,
as far as I am concerned (-.
thus:
there is another primary product
of combustion of your trademark "fossilized fuels,"
which is by far the greater glass house gas;
is it not?
thus:
the Quaternary Period has seen mostly periods
of glaciation, followed by brief respites,
such as we are (or were) in, now. a big problem
with conceptualizing these things, though, is the assumption
that "glaciation means global cooling," and
the similar nonsequiter consequent to Ahrrenius's 1896 coinage
of "glass house effects," to which has been apended that
strange, geodesic gnomish nomenclature of "global" warming.
