T.R | Title | User | Personal Name | Date | Lines |
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1230.1 | One demo is worth... | ULTRA::ELLIS | David Ellis | Tue May 01 1990 19:30 | 10 |
| My intuition is that the inradius of a cyclic pentagon is _not_ uniquely
determined by the lengths of its sides and the order in which they occur.
It seems to me that there might be at least one extra degree of freedom.
Informal approach: cut a straw into five segments, not all equal. Draw a
circle of the "right" size and fit the segments into a pentagon inscribed in
the circle. Now move them around and see if you can shift the angles
while keeping the segments inscribed and adjacent. If I were a betting man,
my bet would be there's room to wiggle.
|
1230.2 | Addressing no one in particular, | VMSDEV::HALLYB | Twin Peaks Municipal Software Works | Tue May 01 1990 23:19 | 4 |
| Perhaps you would get better results if you lit candles and chanted out
"Fire, Walk with me" :-)
John
|
1230.3 | working on the easy part | CSSE::NEILSEN | I used to be PULSAR::WALLY | Wed May 02 1990 16:57 | 41 |
| This looked too hard to me, but since .1 raises doubts about uniqueness, I'll
take a shot at it.
Pick a side and call the length s1. Now consider the right triangle formed by
half the side, the radius from the end point and the perpendicular bisector of
the side (a theorem in elementary geometry shows that this will pass through
the center). Call the interior angle of this triangle at the center a1. Then
s1 and a1 are related by
a1 = arcsin ( s1 / ( 2 * r ) )
There are five such equations in six unknowns: a1 ... a5 and r.
There is also a sixth equation relating the ai
2 * ( a1 + a2 + a3 + a4 + a5 ) = 2 * pi
This makes six equations in six unknowns. There is no room for wiggle here, but
see a few comments on non-uniqueness below.
If I actually needed to find r, I would substitute the first five into the
sixth, giving one equation with the one unknown r. Then I could solve
for r numerically.
I presume that .0 is looking for a closed solution. I suspect there is one,
using either the formulas in .0 or the side-side-side trig relations, but I'll
leave that to the more algebraically inclined.
Now for the non-uniqueness:
In principle, the sixth equation should really equate to n * 2 * pi. The case
n=1 is a pentagon. Reply .2 may have been referring to the fact that n=2 is
a pentagram. But .0 explicitly asked about a pentagon, and we may take that
to exclude n > 1.
It is also true that arcsin is not uniquely defined. Most of the cases I
have found lead to five sided figures which are not strictly pentagons. I can
draw one case where r is close to s/2. There are two values of r which will
solve the six equations, corresponding to two pentagons, one of which encloses
the center of the circle and one of which does not. Still no wiggle, but not
a uniquely defined radius.
|
1230.4 | Geometric argument for uniqueness. | CADSYS::COOPER | Topher Cooper | Thu May 03 1990 15:34 | 42 |
| If we make the further assumption that the pentagon in question is
*simple* (i.e., no edges intersect except at a vertex, and all vertexes
[it is *too* a proper plural] have exactly two edges meeting at them),
then it is pretty easy to see that the solution is unique. Multiple
solutions for "r" are spurious.
Imagine that an oracle has presented you with a set of sticks of the
specified length, numbered in clockwise order, and (more oracularly)
a circle of the correct radius. Your job is to place the sticks on
the circle to form the pentagon.
You take the first stick (which, of course, could be any one of them)
and place the "counterclockwise" end of it on some point on the
circle's circumference. This choice of starting point is the last
free choice you have. There are only two points that the other end
can be placed on -- only one of which is clockwise of the first point
(the two points may coincide if the the stick is twice the radius).
You now take the next stick and place it one end on the just placed
end of the first. Once again you have only one choice of where to
place the second end.
This continues through the five sticks until you get to the last stick
whose end-point will just line up with the starting point -- assuming
that your oracle was right.
If she was wrong and the circle is too large, then your actions would
be no less constrained but the final point will come short of the
initial point. If the circle is too small, then the final point will
overshoot. Only one radius will work.
There is *at most* a single solution.
Any additional solutions will be spurious -- the same as other
solutions, negative radii or whatever.
The most interesting property to the analytic solution in .3 is that
the radius *does not depend on the order of the sides*. This is not
obvious from my geometric argument. Anyone with a geometric argument
for this property.
Topher
|
1230.5 | Geometric argument: order doesn't matter. | CADSYS::COOPER | Topher Cooper | Thu May 03 1990 16:24 | 27 |
| RE: .4 (me)
Never mind -- I thought of one myself in the lunch line.
Imagine that we have the sticks in one order within the proper size
circle as in .4. We want to move the sticks around into another order
in such a way so that the size of the circle doesn't need to change.
The question is -- can we always do this?
Let the segement of the circle formed the two radii from the center of
the circle to an edges endpoints be called a 1-wedge. A 2-wedge
consists of two adjacent 1-wedges. We can take any 2-wedge and reverse
the order of its two edgeds by "flipping it over" -- the circle will
be unchanged by this. By using a "bubble sort" we can use a series of
such 2-wedge flips to put the edges in any desired order. QED.
Of course, we can switch any two edges by using either a 2-wedge flip
or a 3-wedge flip, allowing us to use a the more efficient "exchange
sort" rather than "bubble sort" but mathematics doesn't care about
efficiency.
Note that the arguments and formula in .3, .4 and here do not depend
on the figure being a pentagon -- they apply to any simple n-gon.
(with the exception that exchanging two edgess without affecting the
order of the rest may need two i-wedge flips instead of just 1 if n>5).
Topher
|
1230.6 | hmm... I don't think that that's a valid argument | CSSE::NEILSEN | I used to be PULSAR::WALLY | Thu May 03 1990 16:44 | 38 |
| re: .4
Thanks for a good definition of a 'simple' pentagon.
> You take the first stick (which, of course, could be any one of them)
> and place the "counterclockwise" end of it on some point on the
> circle's circumference. This choice of starting point is the last
> free choice you have. There are only two points that the other end
> can be placed on -- only one of which is clockwise of the first point
> (the two points may coincide if the the stick is twice the radius).
OK, if a stick is sufficiently far from twice the radius, then choosing the
counterclockwise point will lead to a non-simple pentagon.
But suppose that the stick length is very nearly twice the radius (and the
other stick lengths are large compared to this difference). Now the clockwise
and counterclockwise choices will be very close, and neither will lead to
a non-simple pentagon. So we have another free choice.
I have not got a good geometric argument for why you can finish the pentagon
in both cases, but the algebraic argument is still good: there is still a
solution radius for the equation.
> The most interesting property to the analytic solution in .3 is that
> the radius *does not depend on the order of the sides*. This is not
> obvious from my geometric argument. Anyone with a geometric argument
> for this property.
Imagine that you have created a solution using your method. You now have 5
isosceles triangles with all the double sides equal. If you put them
together with their unique angles touching, the unique angles will add up to
2 * pi and you will have your inscribed pentagon. Now put them together in
some other order. You will still have an inscribed pentagon, with the same
radius and the unique angles will still add up to 2 * pi. The pentagon will,
in general, be different, but the radius of the circle is determined.
This works for any inscribed n-gon.
|
1230.7 | | TRACE::GILBERT | Ownership Obligates | Thu May 03 1990 20:02 | 10 |
| RE: rearranging the 5 sticks.
The `interior angle' argument (angles add up to 2*pi) applies if the simple
pentagon contains the center of the circle. If it doesn't, a slightly
different argument is needed.
P.S. I'm not convinced by the arguments that the radius is unique. For a
counterexample to exist, I think it's necessary that for one radius, the
circle's center is inside the pentagon, and for another radius, the radius
is outside the pentagon.
|
1230.8 | is the center inside or outside the pentagon | CSSE::NEILSEN | I used to be PULSAR::WALLY | Fri May 04 1990 16:14 | 18 |
|
>The `interior angle' argument (angles add up to 2*pi) applies if the simple
>pentagon contains the center of the circle. If it doesn't, a slightly
>different argument is needed.
Good point. The different argument is the same in mathematical form, but
one cannot visualize it as I did, just pushing rigid wedges together.
>P.S. I'm not convinced by the arguments that the radius is unique. For a
>counterexample to exist, I think it's necessary that for one radius, the
>circle's center is inside the pentagon, and for another radius, the radius
>is outside the pentagon.
Not clear to me which arguments you find unconvincing. In any case, you
are correct that for the one counterexample I have found, the center of the
circle is inside and outside the pentagon for the two different radii. Both
pentagons are still simple, by the earlier definition. So it still looks like
a valid counterexample to me.
|
1230.9 | my thoughts so far | HERON::BUCHANAN | combinatorial bomb squad | Mon May 07 1990 17:13 | 92 |
1230.10 | | HERON::BUCHANAN | combinatorial bomb squad | Wed May 09 1990 21:18 | 40 |
1230.11 | | GUESS::DERAMO | Dan D'Eramo | Thu May 10 1990 13:06 | 14 |
| re .10
>> However, it's worth pointing out that r is not likely to be
>>be particularly clean, for general polygons. For instance, if n=7, and
>>all the edge lengths are the same, then the polygon is known (Gauss) not to be
>>constructible, and r can be a root of an irreducible sixtic equation.
The sixtic equation (is it really called sixtic?) for the
side of a regular 7-gon is not irreducible. x^6 + x^5 +
x^4 + x^3 + x^2 + x + 1 factors into a product of two
cubics, which can then be solved. The roots as you say are
not constructible though.
Dan
|
1230.12 | Credit where credit isn't due. | CADSYS::COOPER | Topher Cooper | Fri May 11 1990 14:36 | 16 |
| RE: .9,.10 (Andrew)
I don't want to take credit where credit is not due. The concept of a
"simple polygon" is a standard one in geometry. I tailored the
definition a little for the application, but informally a simple
polygon is one which:
1) Is non-self interesecting.
2) No adjacent triple of vertexes are co-linear.
I ignored the second requirement since it was implicit in the polygon
being inscribed in a circle. If anyone finds it usefull, keep in mind
that being inscribed in a circle means that "simplicity" is equivalent
to "convexity".
Topher
|
1230.13 | ex | TRACE::GILBERT | Ownership Obligates | Fri May 11 1990 17:49 | 12 |
| FWIW, I considered the five side lengths of a,a,a,a,b.
The `big equation' simplifies to:
2 6 4 2
(B - 16) R + 20 R - 8 R + 1
where B = b/a, and R = r/a. This is a cubic in R^2.
For what real values of B does it have more than one real root?
Does this happen when B < 4 ? (i.e., the longest side can't be
larger than the sum of the other 4 sides).
|
1230.14 | surprise! | HERON::BUCHANAN | combinatorial bomb squad | Fri May 11 1990 19:39 | 28 |
1230.15 | or not a surprise! | HERON::BUCHANAN | combinatorial bomb squad | Fri May 11 1990 19:54 | 9 |
| Of course, a moment's thought tells us that the result I quoted
for b=a in the previous reply is *not* surprising. The three roots
correspond to equilateral triangle (double back), regular pentagon &
regular pentagram respectively.
I'm off for fish & chips
Regards,
Andrew.
|
1230.16 | oops | GUESS::DERAMO | Dan D'Eramo | Sun May 13 1990 15:37 | 33 |
| re .11
>> re .10
>>
>> >> However, it's worth pointing out that r is not likely to be
>> >>be particularly clean, for general polygons. For instance, if n=7, and
>> >>all the edge lengths are the same, then the polygon is known (Gauss) not to be
>> >>constructible, and r can be a root of an irreducible sixtic equation.
>>
>> The sixtic equation (is it really called sixtic?) for the
>> side of a regular 7-gon is not irreducible. x^6 + x^5 +
>> x^4 + x^3 + x^2 + x + 1 factors into a product of two
>> cubics, which can then be solved. The roots as you say are
>> not constructible though.
Oops. I retract what I said in .10. I was quoting a
result from memory and quoted the wrong result. :-(
It's not that x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 is
reducible over the rationals, it may or may not be,
though I doubt it. The result I wanted to quote is that
its zeroes, the seventh roots of unity, are solveable in
radicals. To see this note that if w is a zero of the
polynomial, and a = w + 1/w, then a is a zero of the
cubic polynomial y^3 + y^2 - 2y + 1. This is solveable
in radicals and then w, 1/w = (a +/- sqrt(a^2 - 4)) / 2
is solveable, too.
Hmmmm. I wonder what the least n is such that e^(2 pi i / n)
is not expressiblein radicals, if any. But that's
another topic. :-)
Dan
|
1230.17 | 3 points... | HERON::BUCHANAN | combinatorial bomb squad | Sun May 13 1990 18:36 | 55 |
| 1.) f(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1
is irreducible over the rationals.
Let x = y+1. f(x) is irreducible iff f(y+1) is irreducible.
f(y+1) = y^6 + 6y^5 + 15y^4 + 20y^3 + 15y^2 + 6y + 1
+ y^5 + 5y^4 + 10y^3 + 10y^2 + 5y + 1
+ y^4 + 4y^3 + 6y^2 + 4y + 1
+ y^3 + 3y^2 + 3y + 1
+ y^2 + 2y + 1
+ y + 1
+ 1
f(y+1) = y^6 + 7y^5 + 21y^4 + 35y^3 + 35y^2 + 21y + 7
Now, by Eisenstein's criterion, with p=7, this is irreducible.
(The requirements are:
p must divide every coefficient *except* the leading coeff.
p^2 must not divide the trailing coefficient.)
2.) Yes, I have no problems with the solubility *route*, except
you made a slight slip in your cubic, might I humbly suggest. But I
question also the * motivation* in remark 3 below. The cubic should be:
y^3 + y^2 - 2y - 1.
^
That this works, we can see by expanding the cubic:
w^3 * [(w + 1/w)^3 + (w + 1/w)^2 - 2(w + 1/w) + 1]
= w^6 + 3w^4 + 3w^2 + 1
+ w^5 + 2w^3 + w
- 2w^4 - 2w^2
- w^3
= f(w) = 0. Since w <> 0, a is a root of the cubic.
3.) You look for the smallest n such that e^(2*pi*i/n) is not expressible
by radicals. I don't understand your question.
An extension L:K is radical if L = K(a_1,...,a_m), where for each
a_j, there is an integer n(j) such that (a_j)^n(j) lies in K(a_1,...,a_(j-1)).
So a radical is some such a_j.
Now here, z = e^(2*pi*n) is either going to lie in Q anyway, or it
can be immediately derived by extending Q by directly adding the radical z
itself!
Please tell me if I've misunderstood you.
Regards,
Andrew.
|
1230.18 | | GUESS::DERAMO | Dan D'Eramo | Mon May 14 1990 04:51 | 32 |
| >> 1.) f(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1
>> is irreducible over the rationals.
>>
>> Let x = y+1. f(x) is irreducible iff f(y+1) is irreducible.
Clever proof!
>> The cubic should be:
>>
>> y^3 + y^2 - 2y - 1.
>> ^
Well, the next to last step in my derivation was
y^3 + y^2 + y = 3y + 1
so I suppose you're right. :-)
>> 3.) You look for the smallest n such that e^(2*pi*i/n) is not expressible
>> by radicals. I don't understand your question.
You mean just saying it is 1^(1/n) is enough to make it
"official"? That doesn't seem fair to me. :-) It also
uses an n-th root although the polynomial is of degree
n-1. I.e. I would expect a solveable zero of a degree
six polynomial to require only square, cube, and fifth
roots (most likely nested).
You understood the question, I just didn't think that
that was the answer.
Dan
|
1230.19 | Deramo & Fermat | HERON::BUCHANAN | combinatorial bomb squad | Mon May 14 1990 09:27 | 55 |
| > You mean just saying it is 1^(1/n) is enough to make it
> "official"? That doesn't seem fair to me. :-) It also
> uses an n-th root although the polynomial is of degree
> n-1. I.e. I would expect a solveable zero of a degree
> six polynomial to require only square, cube, and fifth
> roots (most likely nested).
The point is this:
w = 1^(1/n) is a radical
=> Q(w):Q has an abelian Galois group, G.
This is always true, even if x^n-1 is reducible. This is because
every root of the minimum polynomial, what ever it is, will be a zero of x^n-1,
and so will be a power of w. Thus each Q-automorphism of Q(w) is of the form
w -> w^k. Two such Q-automorphisms obviously commute.
Then from this, a fortiori, G is *ALWAYS* solvable. S5 doesn't get
a look in. We're working entirely with abelian groups.
=> we can construct a chain of normal subgroups from I to G, each quotient
being prime cyclic.
=> there exists a corresponding chain of extensions from Q(w) -> Q,
the degree of each of which is prime, p_i
Dan asks, I think: "When is *every* p_i < n?"
Well, how big is G?
say: n = prod(j) (q_j)^(a_j) where q_j are prime.
then E(n) = prod(j) (q_j - 1)*(q_j)^(a_j - 1) is the number of primitive
nth roots of 1. (This is just the number of integers in {1,...,n} which are
coprime to n.)
So from this, it's obvious that p_i < n.
Dan's case with a 7th root, yields E(n) = 6 = 2.3, so it is clear that
a cubic and then a quadratic extension will build w. Or a quadratic followed
by a cubic, equivalently.
Also, Gauss' constructibility argument falls out of this. If you
accept that constructibility with ruler and compasses is equivalent to only
permitting quadratic extensions, then our requirement becomes that:
E(n) is a power of 2.
ie: q_j > 2 => a_j = 1, & q_j-1 a power of 2.
q_j = 2^x + 1 cannot be prime unless x is itself a power of 2, so
Bob's one's uncle, one has the Fermat numbers popping out.
Regards,
Andrew.
|
1230.20 | yup | HERON::BUCHANAN | combinatorial bomb squad | Mon May 14 1990 09:44 | 13 |
| I overlooked a more general allegation that Dan made.
> I.e. I would expect a solveable zero of a degree
> six polynomial to require only square, cube, and fifth
> roots (most likely nested).
Yes, this is correct. A poly of degree n will have a splitting
field of degree dividing n!. If it is solvable, then a chain of extension
fields each of prime degree exists, and such a prime will divide n!, so
is less than n.
Regards,
Andrew.
|
1230.21 | | GUESS::DERAMO | Dan D'Eramo | Mon May 14 1990 14:36 | 3 |
| Thanks Andrew. You answered my questions.
Dan
|
1230.22 | Can't get there from here | CIVAGE::LYNN | Lynn Yarbrough WNP DTN 383-5663 | Fri May 18 1990 14:49 | 39 |
| I think the problem of calculating the circumradius of a (convex) pentagon
is intractable, in that its solution involves solving a polynomial equation
of degree >=5.
The following approach works for triangles:
$maple
# (File 3sides.maple)
# Given the triangle {a,b,c} find the radius of the circumscribed circle.
# ============================================================
# Let A=angle at the center of the circle of the isoceles triangle (r,r,a),
# and B,C, resp. be angles similarly defined. Then sin(A) = a/(2r), etc.
# Since the 3 angles add to 2Pi, sin(A+B+C) = 0.
# (In MAPLE, '"' means the immediately previous result.)
expand(sin(A+B+C)=0);
# Convert cosines to sines:
subs( cos(A)=(1-sin(A)^2)^(1/2),
cos(B)=(1-sin(B)^2)^(1/2),
cos(C)=(1-sin(C)^2)^(1/2), ");
# Convert the sines to functions of the sides and radius:
subs( sin(A)=a/(2*r),
sin(B)=b/(2*r),
sin(C)=c/(2*r), ");
solve(",r);
# A pair of solutions, one negative, are produced by the last operation.
quit;
Extending the approach to 4 sides exhausts the capacity of the 8Meg 3100 I
am using. Extending the approach to 5 sides causes an error message to be
generated which gives a clue to the impossibility of machine symbolic
analysis. So it appears that the only method of solution is numerical
approximation, which can be obtained by iterating values of r in the
equation produced by the substitution steps outlined above, or by other
methods.
If anyone else comes up with a brilliant insight, I would be willing to
try it.
Lynn Yarbrough
|
1230.23 | try this... | HERON::BUCHANAN | combinatorial bomb squad | Fri May 18 1990 15:34 | 13 |
| The solution for n=4 splits into two solutions for s. One of
them is the one given by Peter in the base note, and the second is the same
with A -> -A.
So, if you take one of these, and make the substitution
D -> D*sqrt(4*r^2-E^2) + E*sqrt(4*r^2-D^2) (or something like that)
then square the new equation twice after rearranging, to get rid
of the nasty surds, you should have something which after getting rid of
any r=0 cases, will factor() into some smaller polys, which will then
crack.
Must dash,
Andrew.
|
1230.24 | poly degree 15 found | HERON::BUCHANAN | combinatorial bomb squad | Fri May 25 1990 21:11 | 19 |
| I tried the strategy suggested in -.1, and it gives me a
polynomial of degree 15 in r^2, A^2, B^2, C^2, D^2, E^2. (ie, it would
be of degree 30, but every term is of entirely even degree in each of
the six indeterminates. degree 15 is actually pretty good. The
polynomial is only 239 blocks long as text!
A basic sanity check suggests that it is correct, but I could do
more (eg try E=0, or A+B+C+D). I gave the beast to MAPLE to factorize,
and MAPLE came back after an hour or so, but the polynomial was unreduced.
I still haven't got any MAPLE documentation. Does my experience
mean that the poly is *sure* to be irreducible, or is MAPLE only limited
in its guesses.
Alternatively, can we come up with a plausible argument for why
the thing should be degree 15?
Regards,
Andrew.
|