the eggs problem
master1729
master1729
>
> http://sites.google.com/site/tommy1729/home/eggs-probl
> em
>
> have fun
>
> tommy1729
btw i will post about tetration on that site too in the near future.
regards
tommy1729
spudnik
don't tell me what it is, but did you solve it?
> >http://sites.google.com/site/tommy1729/home/eggs-probl
thuis quoth:
Simultaneously, EU Commission advisor Alberto Giovannini, who led the
group that set up the technical transition from national currencies to
the euro, is quoted in today's Italian daily Il Sole 24 Ore stating
unabashedly: "History teaches us that empires are more efficient and
achieve great prosperity, because the imperial model is successful
with an extended geography."
Although much attention has been mis-focused on Greece, LaRouche has
emphasized that the epicenter of the European crisis is not Greece but
Spain, and its Banco Santander. For example, of total German bank
exposure in the eurozone of some 540 billion euros, Greek debt
accounts for only 43 billion, or 8% of the total. Spain, by contrast,
amounts to 240 billion euros, or 44% of the total.
http://larouchepub.com/pr_lar/2010/lar_pac/100210lar_no_banco_bailout.html
thus:
yeah, massless rocks o'light,
built a hugely impenetrable bosonic wall around EinsteinoNewtonianism!
thus:
the photographic record that I saw,
in some rather eclectic compendium of Einsteinmania,
seemed to show quite an effect, I must say;
not that the usual interpretation is correct, though.
Nude Scientist said:
> > "Enter another piece of luck for Einstein. We now know that the light-
> > bending effect was actually too small for Eddington to have discerned
--Another Flower for Einstein:
http://www.21stcenturysciencetech.com/articles/spring01/Electrodynamics.html
--les OEuvres!
http://wlym.com
--Stop Cheeny, Ricw & the ICC in Sudan;
no more Anglo-american quagmires!
http://larouchepub.com/pr/2010/100204rice
William Elliot
> the eggs problem
>
> http://sites.google.com/site/tommy1729/home/eggs-problem
>
> have fun
How long have you been suffering from dyscaplexia?
master1729
spudnik
http://sites.google.com/site/tommy1729/home/eggs-problem
--Another Flower for Einstein:
http://www.21stcenturysciencetech.com/articles/spring01/Electrodynamiics.html
William Elliot
> so, guess nobody can solve it huh.
>
No, unfortunately there's no cure for dyscaplexia,
nor the illiterate reputation it gives you.
Dan Cass
The central ellipse is not a circle.
So the other five ellipses will not have their
centers (midpoints of segment joining vertices)
all located on one circle.
My guess is that in fact there is only one arrangement
since if you start with a circle and five other circles
arranged around it, there will be a gap.
As one continuously changes each original circle to
an ellipse there would seem to be only one size so
that all five ellipses going around will touch.
So it seems to me this is not a "min" problem,
only one of finding the ratio (major axis)/(minor axis)
for the one situation that "works".
master1729
learn to spell !
master1729
and btw , the first non-retarded/insulting reply in this thread ; congr Dan :)
though far from a solution as you would expect from self-declared mathematicians on a math forum ...
> > are the five eggs arrayed like a regular pentagon?
NO
> >
> http://sites.google.com/site/tommy1729/home/eggs-probl
>
> > em
> >
> > --Another Flower for Einstein:
> >
> http://www.21stcenturysciencetech.com/articles/spring0
>
> > 1/Electrodynamiics.html
> >
> > --les OEuvres!
> > http://wlym.com
> >
> > --Stop Cheeny, Ricw & the ICC in Sudan;
> > no more Anglo-american quagmires!
> > http://larouchepub.com/pr/2010/100204rice
>
> The central ellipse is not a circle.
> So the other five ellipses will not have their
> centers (midpoints of segment joining vertices)
> all located on one circle.
Correct.
>
> My guess is that in fact there is only one
> arrangement
> since if you start with a circle and five other
> circles
> arranged around it, there will be a gap.
> As one continuously changes each original circle to
> an ellipse there would seem to be only one size so
> that all five ellipses going around will touch.
>
> So it seems to me this is not a "min" problem,
> only one of finding the ratio (major axis)/(minor
> axis)
> for the one situation that "works".
Wrong.
And no solution.
Oh well , we went from retards insulting to people trying.
Its sad though , such a problem unsolved on a MATHforum.
regards Dan ( why should i mention anyone else hmm )
tommy1729
William Elliot
> learn to spell !
>
The master isn't a dyslexic, he's a dyscaplexic like you.
Dan Cass
So by "no solution" you mean one cannot arrange five ellipses around a sixth, each of the five touching
its two neighbors and the central ellipse?
Or perhaps that this connot be done in a
really "symmetric" way as shown in the diagram?
jbriggs444
It's a taunt. He is pointing out that no one has proffered a solution
yet.
I don't have a solution either. Though it looks like one could
approach a solution with numerical techniques. e.g. find one solution
and metaphorically wiggle each egg to see if a better solution is
possible. Once that zeroes in on an "optimal" solution,
metaphorically rotate the ring of outer eggs around the inner egg to
see if there's anything to be gained in that direction.
Nor do I have a proof that such an approach will yield the optimal
answer. The best I have is a handwave that the situation looks
"stable" to me so that this kind of hill-climbing algorithm should
lead to a global minimum rather than a mere local minimum.
It is clear that a solution exists. By inspection of the drawing (if
you trust it). Or by realizing that a ring of five toothpicks can
enclose a central toothpick but that a ring of five circles cannot
enclose a central circle. Somewhere in between must be a minimal
eccentricity that will allow a ring of 5 identical ellipses to just
barely enclose a 6th.
Mensanator
Hasn't anybody noticed that eggs aren't symetrical?
A
I would like to see the notion of "symmetric" made precise, here.
Given sufficiently oblong ellipses one can always make them into a
ring surrounding an ellipse in the center, if one is allowed to rotate
them in whatever way one wants, but this isn't guaranteed to have any
particular kind of symmetry. It's not at all clear, in the original
statement of the problem, exactly in what ways the ellipses are
allowed to be arranged.
master1729
nonsense , its very clear.
so lame.
you cant solve it => its not clear
hahaha
tommy1729
spudnik
I asked about the regularness, because
you can't really tell by looking at it.
http://sites.google.com/site/tommy1729/home/eggs-prob
thus:
the best odometer for relativity is quaternions,
where the imaginary part is the "vector," and
the real part is the "scalar" in Hamilton's lingua;
see the work-up by Lanczos, _Variational Mechanics_.
Death to the lightcone!
thus:
don't top-post!...
what exactly is not analogous about doppler shifts
of frequency of light waves & sound waves?
presumably, there are no "photonic booms," because
nothing emits light that is going at over "warp factor one."
thus:
sea-level is not rising, globally --
http://www.21stcenturysciencetech.com/Articles%202007/MornerInterview...
-- and warming is mostly equatorial. however,
there is loss of soil,
and that might change *relative* sea-level
in some locations, as well as dysplace some sea!
thus quoth:
Let’s take a look at the complexity of polar bear life. First, the
polar bear has been around for about 250,000 years, having survived
both an Ice Age, and the last Interglacial period (130,000 years ago),
when there was virtually no ice at the North Pole.
Clearly, polar bears have adapted to the changing environment,
as evidenced by their presence today.
(This fact alone makes the polar bear smarter than Al Gore and
the other global warming alarmists. Perhaps the polar
bear survived the last Interglacial because it did not
have computer climate models that said,
polar bears should not have survived!)
http://www.21stcenturysciencetech.com/Global_Warming.html
thus:
the photographic record that I saw,
in some rather eclectic compendium of Einsteinmania,
seemed to show quite a "bending" effect, I must say;
not that the usual interpretation is correct, though.
Nude Scientist said:
Enter another piece of luck for Einstein. We now know that the light-
bending effect was actually too small for Eddington to have discerned
--Another Flower for Einstein:
http://www.21stcenturysciencetech.com/articles/spring01/Electrodynamics.html
--les OEuvres!
http://wlym.com
--Stop Cheeny, Rice & the ICC in Sudan;
no more Anglo-american quagmires!
http://www.larouchepub.com/pr/2010/100204rice-ists_sudan.html
Dan Cass
The question at the original site said nothing
about any required symmetry in the diagram.
It seems all that is required is that
* all six ellipses are congruent (same maj/min axes)
* The five arranged going around the central sixth
ellipse each are tangent to the central ellipse
and are tangent to their two immediate neighbors
going around.
[At least, this formulation seems to give a
genuine problem.]
So if your idea of using longish ellipses can be positioned,
I guess the question is "what's the shortest possible
ellipse for which it can be done?"
[where by "shortest" one means minimal ratio
of major axis to minor axis...]
Tonico
Somebody already said this is a taunt.I agree: this a tease: the
problem is not sufficiently well described.
Now just watch the kid Tommy "laugh" at others...but will he pose a
solution to the problem, whatever it is? I bet he won't, as usual, and
if he does then it will probably be a nonsense or using some data not
specifically given in the OP...as usual.
Second bet: if he poses a "solution", it will be a nonsense.
Tonio
David Rusin
master1729 <tomm...@gmail.com> wrote:
>the eggs problem
>
>http://sites.google.com/site/tommy1729/home/eggs-problem
Let's see if I can summarize the problem and comment on the solution.
We know that six congruent circles fit perfectly around another such circle.
The question is, can we surround an ellipse with 5 congruent ellipses in a
similar way? It's not hard to show that the answer is "yes" when the
ellipse is long and thin, e.g. an arrangement like this:
AAA
E X B
E X B
E X B
D C
D C
D C
So the question becomes, how close to circular can the ellipses be and
still permit a five-around-one configuration; specifically, what is the
minimum permissible ratio r>1 between the lengths of the semi-major
and semi-minor axes of the ellipses?
I don't know the answer but I can show how to calculate it. It will be
an algebraic number.
Here is the basic idea: if X is the central ellipse, then each of the
other ellipses may be obtained from X by a Euclidean motion: there is
a rotation R and a translation by a vector v so that A = R(X) + v .
If we pick a parameterization (x,y) = f(t) of ellipse X, and an
implicitization F(x,y)=0 of it, then a parameterization of A is
(x,y) = R(f(t)) + v , and the points of intersection of A and X
correspond to the values of t for which F( R(f(t))+v ) = 0. We may
use a rational function f for this parameterization, so this last
equation is a polynomial one, and the points of intersection correspond
to its roots. These points of intersection are points of tangency
when the roots are double roots. In particular, we have obtained a
characterization to determine when the ellipses X and A are tangent:
the discriminant D( R, v ) = discrim_t( F( R(f(t))+v ) ) vanishes.
I actually computed this using a rational parameterization of the rotation
matrices (e.g. the upper-left entry is (1-u^2)/(1+u^2) ). Then D(R,v) =
D(u; a,b) is a polynomial of degree 16 in u and degree 8 in a,b.
As a reality check I looked to see what happens for circular ellipses (r=1);
in that case D(u, a,b) = (1+u^2)^8 (a^2+b^2)^3 (a^2+b^2 - 4) . I also
tried a nontrivial example: when r=2, I tried moving the ellipse from
the origin to (2,2), and found that it would be tangent to the central
ellipse precisely when it is rotated by 17.2033 or 70.4949 degrees
counterclockwise.
Note that the ellipses congruent to X form a 3-parameter family, as
we vary v and R; the condition D=0 imposes a single constraint on
the parameters, so the set of tangent ellipses forms a 2-parameter family.
Our configuration of six ellipses (the central one of which we may assume
is the ellipse x^2 + (y/r)^2 = 1 ) thus corresponds to one point in a
15-dimensional space, and the fact that each of the five outer ellipses
is tangent to the central one is equivalent to having this point lie in
a 10-dimensional subvariety.
But we have the additional constraint that adjacent ellipses must also
be tangent. Applying a few Euclidean motions we see that R(X) + v is
tangent to R'(X) + v' iff X is tangent to (R^-1R')(X) + (R^-1(v'-v)).
From the preceding discussion we see that this may be characterized by
the equation D( R^-1R', R^-1(v'-v) ) = 0 .
So the configurations we seek are the points in R^15 which satisfy the
ten equations
D( R_i , v_i ) = 0 and
D( (R_i)^(-1)R_(i+1), (R_i)^(-1)(v_(i+1)-v_i) ) = 0 (i=1,...5)
Note that the rotations R may be parameterized with rational functions
too, so these equations may all be expressed algebraically (i.e. without
trigonometric functions); so these equations describe an algebraic variety
in R^15. It seems to be 5-dimensional, which is reflected in the five
degrees of freedom in the solution set: once a configuration is found, each
of the five outer ellipses still has one degree of freedom in its position.
Now, in this discussion, r was treated as fixed. If we treat r also as
a variable, then the set of all solutions is described by these 10 equations
in now 16 variables. The solution is a 6-dimensional variety (which may be
viewed as a stack of the 5-dimensional varieties we had above, with r=const.)
As r drops towards 1, we see the variety shrink, as the configurations
get tighter -- there are fewer options for the v_i and R_i. The minimal
value of r is one where the 5-dimensional variety is discrete; that is,
this value of r is the r-coordinate of a point on the boundary of the
6-dimensional variety. If we think of the variety as being the zero-locus
of a function F : R^15 -> R^10, then the optimal value of r should
be a point where F' has minimal rank (namely 5). This is characterized
by the vanishing of a set of minors (determinants) in F' and in particular
is described by polynomials in the 15 variables. Then the optimal value of
r should be the r-coordinate of one of the points in this singular variety,
and hence r satisfies a polynomial equation.
I would not in a million years advocate attempting to find the minimal
polynomial for r in this way...
As a practical matter it would not be too hard to write down the 10 equations
in 15+1 variables and find at least one point that satisfies all the equations;
then use a homotopy method to track a path of solutions as r decreases.
For some low value of r the path would end (or more precisely the solutions
would become complex). I think this can be done, but not without actual effort
and I have other things to do right now :-)
dave
Remark: my parameterization of the rotation group is a bit of overkill since
the ellipses have a 180-degree rotational symmetry: if a rotation of theta
gives tangency, so does a rotation by theta+pi. These two angles correspond
to values of u which are negative reciprocals of each other, that is, if
D(u,a,b) = 0 then also D(-1/u, a,b) = 0. So the condition of tangency
can be written in terms of v = u-1/u rather than u. This cuts the degree
of D in half, i.e. it is now of degree 8 in v (as well as of degree 8 in a,b).
I have this polynomial D(v,a,b) and can supply it to anyone who is curious.
spudnik
in space, using ellipsoids (and either a)
eleven around one, or b)
13 around one, a la Newton's go-around of Harriot/Kepler's problem).
> Remark: my parameterization of the rotation group is a bit of overkill since
> the ellipses have a 180-degree rotational symmetry: if a rotation of theta
> gives tangency, so does a rotation by theta+pi. These two angles correspond
> to values of u which are negative reciprocals of each other, that is, if
> D(u,a,b) = 0 then also D(-1/u, a,b) = 0. So the condition of tangency
> can be written in terms of v = u-1/u rather than u. This cuts the degree
> of D in half, i.e. it is now of degree 8 in v (as well as of degree 8 in a,b).
> I have this polynomial D(v,a,b) and can supply it to anyone who is curious.
thus:
ah, the old exploded planet hypothesis;
even Kepler could have been wrong about *some* thing (although,
he was not, about most things: http://wlym.com .-)
> http://www.theregister.co.uk/2010/02/18/lhc_fireup_2010_forecast/
--Another Flower for Einstein:
http://www.21stcenturysciencetech.com/articles/spring01/Electrodynamics.html
spudnik
that the difficulty -- though I suppose it was solved --
with the "thiteen balls around one" problem was that
it was an odd number; so, maybe trying ten or 14 is better,
given the symmetry of the ellipsoids. qua ellipses,
how about 4 aroound one?
> in space, using ellipsoids (and either a)
> eleven around one, or b)
> 13 around one, a la Newton's go-around of Harriot/Kepler's problem).
thus:
a hundred is not enough for such sophistry to be taken-up, dood --
rotating your yardstick to be "a lightyear's duration?"
quaternions already takes care of this "problem," because
time is a "0d" -- as you say -- scalar;
*all* of "4d" vector mechanics is in Hamilton's quaternions.
anyway, orthogonality is indeed generalized to Nd in various ways, but
it just does not mean the same thing as in "(x,y,z;t)."
> http://bandtechnology.com/PolySigned
thus:
death to the lightcone!
> I rather like this term, "funky functional."
thus:
so, Fermatttt could not have done it, either?... try working it
in space, using ellipsoids (and either a)
eleven around one, or b)
13 around one, a la Newton's go-around of Harriot/Kepler's problem).
thus:
ah, the old exploded planet hypothesis;
even Kepler could have been wrong about *some* thing (although,
he was not, about most things: http://wlym.com .-)
--Another Flower for Einstein:
Robert H. Lewis
Sorry to be coming in on this so late. I think this is a cool problem.
My approach is quite similar to Dave Rusin's.
Rather than do the case of five ellipses tangent to a central one, I started with four around the outside of a central one. Let's look for a solution that is symmetrical. Then we can focus on just the central ellipse and one that is tangent to it and lies in the first quadrant.
Let E1 be the ellipse at the origin oriented vertically with major axis a and minor axis b (a is vertical; b is horizontal). Place a second ellipse E2 with the same axes in the first quadrant, tilted so as to be tangent to the first, the y-axis, and the x-axis. Let theta be the angle made by the radius vector to the center of E2. Imagine rotating downward until the second ellipse becomes vertical; call this E2'. Let (h,k) be the center of E2'. Thus a point (x, y) on E2 has been rotated into (x', y') with the usual coordinate transform. (x', y') satisfies
a^2 (x' - h)^2 + b^2 (y' - k)^2 - a^2 b^2 = 0.
By replacing x' and y' with their expressions in terms of sin(theta), cos(theta), x, y, we get the equation for E2. For convenience, let's write this equation as
eq1 = a1*x^2 + b1*y^2 + c1*x + d1*y + e1*x*y + f1. The coefficients are easily computed:
a1 = ct1^2*a^2 + st1^2*b^2 ,
b1 = st1^2*a^2 + ct1^2*b^2 ,
..
f1 = a^2*h^2 + b^2*k^2 - b^2*a^2
where st1 = sine(theta), ct1 = cos(theta).
Let eq2 be the equation of E1:
eq2 = a^2*x^2 + b^2*y^2 - a^2*b^2.
Let (x, y) be the point of tangency of E1 and E2. Tangency is defined by partial derivatives.
Let d1x = Deriv(eq1,x,1), d1y = Deriv(eq1,y,1), d2x = Deriv(eq2,x,1), d2y = Deriv(eq2,y,1).
Let (xt, 0) be the point of tangency of E2 with the x-axis, and (0, yt) the point of tangency of E2 with the y-axis.
We now use a standard trick in this subject, that of expressing st1 and ct1 in terms of tt1 = the tangent of half of theta.
We end up with seven equations in terms of the nine variables x, y, xt, yt, a, b, h, k, tt1.
[d] := [[ b1*yt^2 + d1*yt + f1,
a1*xt^2 + c1*xt + f1,
ct1^2*a^2*xt + st1^2*b^2*xt - ct1*a^2*h + st1*b^2*k,
st1^2*a^2*yt + ct1^2*b^2*yt - st1*a^2*h - ct1*b^2*k,
eq1,
eq2,
d1x*d2y - d1y*d2x ]];
With a resultant computation we eliminate all variables but a, b, and tt1. This is easily done with the Dixon-EDF method (39 seconds). The resultant (irreducible) is
ans = 6912*tt1^24*a^24 - 69120*tt1^22*a^24 + 251136*tt1^20*a^24 - 394240*tt1^18*a^24 + 251136*tt1^16*a^24 - 69120*tt1^14*a^24 + 6912*tt1^12*a^24
+ 10080*tt1^26*b^2*a^22 - 162816*tt1^24*b^2*a^22 + 1000576*tt1^22*b^2*a^22 - 2956288*tt1^20*b^2*a^22 + 4323392*tt1^18*b^2*a^22
- 2956288*tt1^16*b^2*a^22 + 1000576*tt1^14*b^2*a^22 - 162816*tt1^12*b^2*a^22 + 10080*tt1^10*b^2*a^22 + 2367*tt1^28*b^4*a^20 - 118506*tt1^26*b^4*
a^20 + 1263635*tt1^24*b^4*a^20 - 6174456*tt1^22*b^4*a^20 + 15975854*tt1^20*b^4*a^20 - 22136380*tt1^18*b^4*a^20 + 15975854*tt1^16*b^4*a^20
- 6174456*tt1^14*b^4*a^20 + 1263635*tt1^12*b^4*a^20 - 118506*tt1^10*b^4*a^20 + 2367*tt1^8*b^4*a^20 - 82*tt1^30*b^6*a^18 - 24376*tt1^28*b^6*a^18
+ 588156*tt1^26*b^6*a^18 - 5154392*tt1^24*b^6*a^18 + 22177010*tt1^22*b^6*a^18 - 52476528*tt1^20*b^6*a^18 + 69845960*tt1^18*b^6*a^18
- 52476528*tt1^16*b^6*a^18 + 22177010*tt1^14*b^6*a^18 - 5154392*tt1^12*b^6*a^18 + 588156*tt1^10*b^6*a^18 - 24376*tt1^8*b^6*a^18 - 82*tt1^6*b^6*
a^18 - 143*tt1^32*b^8*a^16 - 718*tt1^30*b^8*a^16 + 94111*tt1^28*b^8*a^16 - 1675556*tt1^26*b^8*a^16 + 13041469*tt1^24*b^8*a^16 - 52393906*tt1^22*
b^8*a^16 + 116712883*tt1^20*b^8*a^16 - 151208120*tt1^18*b^8*a^16 + 116712883*tt1^16*b^8*a^16 - 52393906*tt1^14*b^8*a^16 + 13041469*tt1^12*
b^8*a^16 - 1675556*tt1^10*b^8*a^16 + 94111*tt1^8*b^8*a^16 - 718*tt1^6*b^8*a^16 - 143*tt1^4*b^8*a^16 - 22*tt1^34*b^10*a^14 + 324*tt1^32*b^10*a^14
+ 2426*tt1^30*b^10*a^14 - 197788*tt1^28*b^10*a^14 + 3050732*tt1^26*b^10*a^14 - 22164932*tt1^24*b^10*a^14 + 86279782*tt1^22*b^10*a^14 - 185249508*tt1^20*
b^10*a^14 + 236582548*tt1^18*b^10*a^14 - 185249508*tt1^16*b^10*a^14 + 86279782*tt1^14*b^10*a^14 - 22164932*tt1^12*b^10*a^14 + 3050732*tt1^10*b^10*a^14
- 197788*tt1^8*b^10*a^14 + 2426*tt1^6*b^10*a^14 + 324*tt1^4*b^10*a^14 - 22*tt1^2*b^10*a^14 - tt1^36*b^12*a^12 + 26*tt1^34*b^12*a^12 - 515*tt1^32*b^12*a^12
- 4068*tt1^30*b^12*a^12 + 248312*tt1^28*b^12*a^12 - 3718380*tt1^26*b^12*a^12 + 26321684*tt1^24*b^12*a^12 - 101671596*tt1^22*b^12*a^12 + 215441144*tt1^20*
b^12*a^12 - 274074940*tt1^18*b^12*a^12 + 215441144*tt1^16*b^12*a^12 - 101671596*tt1^14*b^12*a^12 + 26321684*tt1^12*b^12*a^12 - 3718380*tt1^10*b^12*a^12
+ 248312*tt1^8*b^12*a^12 - 4068*tt1^6*b^12*a^12 - 515*tt1^4*b^12*a^12 + 26*tt1^2*b^12*a^12 - b^12*a^12 - 22*tt1^34*b^14*a^10 + 324*tt1^32*b^14*a^10
+ 2426*tt1^30*b^14*a^10 - 197788*tt1^28*b^14*a^10 + 3050732*tt1^26*b^14*a^10 - 22164932*tt1^24*b^14*a^10 + 86279782*tt1^22*b^14*a^10 - 185249508*tt1^20*
b^14*a^10 + 236582548*tt1^18*b^14*a^10 - 185249508*tt1^16*b^14*a^10 + 86279782*tt1^14*b^14*a^10 - 22164932*tt1^12*b^14*a^10 + 3050732*tt1^10*b^14*a^10
- 197788*tt1^8*b^14*a^10 + 2426*tt1^6*b^14*a^10 + 324*tt1^4*b^14*a^10 - 22*tt1^2*b^14*a^10 - 143*tt1^32*b^16*a^8 - 718*tt1^30*b^16*a^8 + 94111*tt1^28*
b^16*a^8 - 1675556*tt1^26*b^16*a^8 + 13041469*tt1^24*b^16*a^8 - 52393906*tt1^22*b^16*a^8 + 116712883*tt1^20*b^16*a^8 - 151208120*tt1^18*b^16*
a^8 + 116712883*tt1^16*b^16*a^8 - 52393906*tt1^14*b^16*a^8 + 13041469*tt1^12*b^16*a^8 - 1675556*tt1^10*b^16*a^8 + 94111*tt1^8*b^16*a^8
- 718*tt1^6*b^16*a^8 - 143*tt1^4*b^16*a^8 - 82*tt1^30*b^18*a^6 - 24376*tt1^28*b^18*a^6 + 588156*tt1^26*b^18*a^6 - 5154392*tt1^24*b^18*a^6
+ 22177010*tt1^22*b^18*a^6 - 52476528*tt1^20*b^18*a^6 + 69845960*tt1^18*b^18*a^6 - 52476528*tt1^16*b^18*a^6 + 22177010*tt1^14*b^18*a^6
- 5154392*tt1^12*b^18*a^6 + 588156*tt1^10*b^18*a^6 - 24376*tt1^8*b^18*a^6 - 82*tt1^6*b^18*a^6 + 2367*tt1^28*b^20*a^4 - 118506*tt1^26*b^20*a^4
+ 1263635*tt1^24*b^20*a^4 - 6174456*tt1^22*b^20*a^4 + 15975854*tt1^20*b^20*a^4 - 22136380*tt1^18*b^20*a^4 + 15975854*tt1^16*b^20*a^4
- 6174456*tt1^14*b^20*a^4 + 1263635*tt1^12*b^20*a^4 - 118506*tt1^10*b^20*a^4 + 2367*tt1^8*b^20*a^4 + 10080*tt1^26*b^22*a^2 - 162816*tt1^24*b^22*
a^2 + 1000576*tt1^22*b^22*a^2 - 2956288*tt1^20*b^22*a^2 + 4323392*tt1^18*b^22*a^2 - 2956288*tt1^16*b^22*a^2 + 1000576*tt1^14*b^22*a^2
- 162816*tt1^12*b^22*a^2 + 10080*tt1^10*b^22*a^2 + 6912*tt1^24*b^24 - 69120*tt1^22*b^24 + 251136*tt1^20*b^24 - 394240*tt1^18*b^24
+ 251136*tt1^16*b^24 - 69120*tt1^14*b^24 + 6912*tt1^12*b^24
We may as well let b=1. So the resultant is a function of two variables. Therefore, there is no unique solution. This is easily verified by a small amount of experimentation. Clearly, there are two solutions for any tall and skinny ellipse, and clearly also, there is no solution for a circle. How do we find the critical value when there is exactly one solution?
That is easily answered with resultants. We just compute the discriminant of ans with respect to tt1. That is a polynomial in a alone. Set it equal to 0 and solve numerically. The answer is a = 2.601665311. The correspnding angle tt1 is just about 38 degrees. I made a model by cutting ellipses out of paper and taping them together. The picture is a little off, but see http://home.bway.net/lewis/tangellip.PDF
It would be fun to extend this to the five case.
Robert H. Lewis
Mathematics
Fordham University
Jim Ferry
> That is easily answered with resultants. We just compute the discriminant of ans with respect to tt1. That is a polynomial in a alone. Set it equal to 0 and solve numerically. The answer is a = 2.601665311. The correspnding angle tt1 is just about 38 degrees. I made a model by cutting ellipses out of paper and taping them together. The picture is a little off, but seehttp://home.bway.net/lewis/tangellip.PDF
I copied your expression for "ans" to look for a simple expression for
the critical value of a (a ~= 2.60166531130747548542). I found that
a = sqrt(r + sqrt(r^2 - 1)), where r is the largest real zero of
606528 x^6 - 1113264 x^5 - 4938975 x^4 + 4246864 x^3 + 4984430 x^2 -
3566112 x - 3334223 (r ~= 3.45820103486087299935).
Robert H. Lewis
> <rle...@fordham.edu> wrote:
>
> > That is easily answered with resultants. We just
> > compute the discriminant of ans with respect to tt1.
> > That is a polynomial in a alone. Set it equal to 0
> > and solve numerically. The answer is a =
> > 2.601665311. The correspnding angle tt1 is just
> > about 38 degrees. I made a model by cutting ellipses
> > out of paper and taping them together. The picture
>
> I copied your expression for "ans" to look for a
> simple expression for the critical value of a (a ~=
> 2.60166531130747548542). I found that
> a = sqrt(r + sqrt(r^2 - 1)), where r is the largest
> real zero of 606528 x^6 - 1113264 x^5 - 4938975 x^4 + 4246864 x^3
> + 4984430 x^2 - 3566112 x - 3334223 (r ~= 3.45820103486087299935).
That's interesting. How did you get that?
Robert H. Lewis
Fordham University
Jim Ferry
> > On Feb 25, 3:17 pm, "Robert H. Lewis"
> > <rle...@fordham.edu> wrote:
>
> > > That is easily answered with resultants. We just
> > > compute the discriminant of ans with respect to tt1.
> > > That is a polynomial in a alone. Set it equal to 0
> > > and solve numerically. The answer is a =
> > > 2.601665311. The correspnding angle tt1 is just
> > > about 38 degrees. I made a model by cutting ellipses
> > > out of paper and taping them together. The picture
>
> > I copied your expression for "ans" to look for a
> > simple expression for the critical value of a (a ~=
> > 2.60166531130747548542). I found that
> > a = sqrt(r + sqrt(r^2 - 1)), where r is the largest
> > real zero of 606528 x^6 - 1113264 x^5 - 4938975 x^4 + 4246864 x^3
> > + 4984430 x^2 - 3566112 x - 3334223 (r ~= 3.45820103486087299935).
>
> That's interesting. How did you get that?
>
> Robert H. Lewis
> Fordham University
Using Mathematica, I substituted b = 1 in your expression for "ans",
then took the square root of the discriminant (w.r.t. tt1) and
factored it to get
203460101828121226907016879885732297716927261405191329994549979207940652714549051461582765162496*
a^156*(-1 + a)^168*(1 + a)^168*(1 + a^2)^12*
(4 - 13*a^2 + 8*a^4 + 18*a^6 + 8*a^8 - 13*a^10 + 4*a^12)^2*
(3509 + 15246*a^2 - 77797*a^4 - 170818*a^6 + 783719*a^8 -
460364*a^10 - 682606*a^12 - 460364*a^14 + 783719*a^16 -
170818*a^18 - 77797*a^20 + 15246*a^22 + 3509*a^24)^4*
(151632 - 556632*a^2 - 4029183*a^4 + 5710568*a^6 + 2456300*a^8 -
8614032*a^10 - 40073338*a^12 - 8614032*a^14 + 2456300*a^16 +
5710568*a^18 - 4029183*a^20 - 556632*a^22 + 151632*a^24)^2
This yields the trivial roots 0, 1, -1, i, -i, and three non-trivial
polynomials. The first polynomial has no real roots. The second has
the real root a_2 = 1.69558 (and no others except for -a_2, 1/a_2, and
-1/a_2: all three polynomials have these symmetries). This would be
a pudgier ellipse than your solution, but solving for tt1 yields no
real solutions (perhaps the fact that the polynomial occurs to the
fourth power in the discriminant was a clue?). The third polynomial
has the real root a_3 = 2.60167 (and no others except the trivial
variants). It reduces to a sixth-degree polynomial because (1) it's
even, and (2) it's a "palindrome". Therefore we let a^2 + a^-2 = 2x,
which yields the equation 606528 x^6 - 1113264 x^5 - 4938975 x^4 +
4246864 x^3 + 4984430 x^2 - 3566112 x - 3334223 = 0 for x. Letting
x_3 be the largest real root of this, we have a_3 = sqrt(x_3 +
sqrt((x_3)^2 - 1)).
We may now use a_3 to solve for tt1 (which we'll abbreviate "t"
henceforth). Substituting the exact expression for a_3 into the
equation for t (with b = 1), Mathematica finds that t satisfies 351 -
11124*t^2 + 143262*t^4 - 954020*t^6 + 3535665*t^8 - 7685608*t^10 +
10008484*t^12 - 7685608*t^14 + 3535665*t^16 - 954020*t^18 +
143262*t^20 - 11124*t^22 + 351*t^24 = 0. Using the same trick to
reduce this as above, we find the solution t_3 = sqrt(u_3 -
sqrt((u_3)^2 - 1)), where u_3 is the largest real root of 50359 -
152474*u + 185361*u^2 - 112300*u^3 + 35289*u^4 - 5562*u^5 + 351*u^6
(u_3 ~= 4.29118791995814294122556), which yields t_3 ~=
0.34372112953919005867485 (and no other real values except the trivial
variants), so the critical angle is 2*Arctan(t_3) =
37.93785689151653274240857 degrees.