Re: [math-fun] equilateral elliptical n-gons
Andy is right. The recursive subdivision doesn't work correctly -- it produces pairs of equal sides, but those pairs don't necessarily equal pairs futher away. Back to the drawing board... At 02:37 PM 11/18/2010, Andy Latto wrote:
On Thu, Nov 18, 2010 at 5:10 PM, Henry Baker <hbaker1@pipeline.com> wrote:
[My statement below about "notchy" is *wrong*. Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse.
But as someone pointed out, the recursive perpendicular bisector subdivsion doesn't work. You replace side AB with sides AC and CB, where |AC| = |CB|, and you replace side XY with sides XZ and ZY, with |XZ| = |ZY|, but there's no reason to expect that just because |AB| = |XY|, that |AC| should equal |XZ|.
Andy
I read your notes and did some imagining, and here are my unjustified beliefs: For every n >= 3, and for any ellipse, there is a single continuous family of inscribed equilateral n-gons in the ellipse. For even n = 2k, an inscribed n-gon will have 180-degree symmetry about the center of the ellipse. For any point P on the ellipse, there will be a unique n-gon with vertex P. For odd n = 2k+1, the behavior will be more complex. In low eccentricity ellipses, for any point P on the ellipse, I think there will be a unique n-gon with vertex P, as in the even n case (clearly true for the circle). However, in high eccentricity ellipses, we will see something interesting: Envision the ellipse elongated horizontally. The n-gon will be small in area compared to the ellipse. Near the center of the ellipse, the n-gon will look like a slightly fattened trapezoid. There will be k sides on one arc of the ellipse (say the bottom) and k-1 sides on the opposite arc (top). The remaining 2 sides will cut across the ellipse to complete the closed polygon. The polygon will slide toward one end (say right end) of the ellipse. The nearer vertex (on the bottom arc) will eventually slide to the end of the ellipse. At this point, the n-gon will be crammed into the right end of the ellipse, with its left side vertical. I will guess that this is when the polygon sides are shortest. Then the end point continues to slide upward and the process continues in reverse and upside down. The n-gon moves to the left, and as it reaches the middle of the ellipse, it is upside down from its start position, with k sides on the top arc and k-1 on the bottom. We then slide the n-gon to the left end, and run the closest vertex down across the left end of the ellipse, and the trapezoid flips back to right side up. We then slide the polygon back to its original position, with vertices rotated counterclockwise. Performed quickly, it will look like a small trapezoid bouncing back and forth between the ends of the ellipse, flipping upside down on each bounce. I'm not sure if there is a cutoff point or a smooth transition between the simple rotational behavior of the circle and the bouncing behavior of the highly eccentric ellipse. It is interesting to note that for a point P on ellipse close to the center, this process generates n distinct polygons with vertex P, but if P is an end of the ellipse, only 1 polygon has vertex P. ----- Original Message ----- From: "Henry Baker" <hbaker1@pipeline.com> To: "Andy Latto" <andy.latto@pobox.com> Cc: "math-fun" <math-fun@mailman.xmission.com> Sent: Thursday, November 18, 2010 11:01 PM Subject: Re: [math-fun] equilateral elliptical n-gons Andy is right. The recursive subdivision doesn't work correctly -- it produces pairs of equal sides, but those pairs don't necessarily equal pairs futher away. Back to the drawing board... At 02:37 PM 11/18/2010, Andy Latto wrote:
On Thu, Nov 18, 2010 at 5:10 PM, Henry Baker <hbaker1@pipeline.com> wrote:
[My statement below about "notchy" is *wrong*. Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse.
But as someone pointed out, the recursive perpendicular bisector subdivsion doesn't work. You replace side AB with sides AC and CB, where |AC| = |CB|, and you replace side XY with sides XZ and ZY, with |XZ| = |ZY|, but there's no reason to expect that just because |AB| = |XY|, that |AC| should equal |XZ|.
Andy
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ----- No virus found in this message. Checked by AVG - www.avg.com Version: 10.0.1153 / Virus Database: 424/3266 - Release Date: 11/19/10 ----- No virus found in this message. Checked by AVG - www.avg.com Version: 10.0.1153 / Virus Database: 424/3266 - Release Date: 11/19/10
On Sat, Nov 20, 2010 at 12:44 AM, David Wilson <davidwwilson@comcast.net>wrote:
I read your notes and did some imagining, and here are my unjustified beliefs:
For every n >= 3, and for any ellipse, there is a single continuous family of inscribed equilateral n-gons in the ellipse.
I disagree -- I think there are "less obvious" inscribed equilateral n-gons in ellipsis with high eccentricity, in which one or two edges take a short-cut, circumventing a large piece of the ellipse. (Given a point P on the ellipse and a distance d, there may be *four* points on the ellipse at distance d from P, not just two.) This objection carries through to all the following statements also, I think...
For even n = 2k, an inscribed n-gon will have 180-degree symmetry about the center of the ellipse. For any point P on the ellipse, there will be a unique n-gon with vertex P.
For odd n = 2k+1, the behavior will be more complex. In low eccentricity ellipses, for any point P on the ellipse, I think there will be a unique n-gon with vertex P, as in the even n case (clearly true for the circle). However, in high eccentricity ellipses, we will see something interesting:
Envision the ellipse elongated horizontally. The n-gon will be small in area compared to the ellipse. Near the center of the ellipse, the n-gon will look like a slightly fattened trapezoid. There will be k sides on one arc of the ellipse (say the bottom) and k-1 sides on the opposite arc (top). The remaining 2 sides will cut across the ellipse to complete the closed polygon. The polygon will slide toward one end (say right end) of the ellipse. The nearer vertex (on the bottom arc) will eventually slide to the end of the ellipse. At this point, the n-gon will be crammed into the right end of the ellipse, with its left side vertical. I will guess that this is when the polygon sides are shortest. Then the end point continues to slide upward and the process continues in reverse and upside down. The n-gon moves to the left, and as it reaches the middle of the ellipse, it is upside down from its start position, with k sides on the top arc and k-1 on the bottom. We then slide the n-gon to the left end, and run the closest vertex down across the left end of the ellipse, and the trapezoid flips back to right side up. We then slide the polygon back to its original position, with vertices rotated counterclockwise. Performed quickly, it will look like a small trapezoid bouncing back and forth between the ends of the ellipse, flipping upside down on each bounce. I'm not sure if there is a cutoff point or a smooth transition between the simple rotational behavior of the circle and the bouncing behavior of the highly eccentric ellipse.
It is interesting to note that for a point P on ellipse close to the center, this process generates n distinct polygons with vertex P, but if P is an end of the ellipse, only 1 polygon has vertex P.
----- Original Message ----- From: "Henry Baker" <hbaker1@pipeline.com> To: "Andy Latto" <andy.latto@pobox.com> Cc: "math-fun" <math-fun@mailman.xmission.com> Sent: Thursday, November 18, 2010 11:01 PM Subject: Re: [math-fun] equilateral elliptical n-gons
Andy is right. The recursive subdivision doesn't work correctly -- it produces pairs of equal sides, but those pairs don't necessarily equal pairs futher away.
Back to the drawing board...
At 02:37 PM 11/18/2010, Andy Latto wrote:
On Thu, Nov 18, 2010 at 5:10 PM, Henry Baker <hbaker1@pipeline.com> wrote:
[My statement below about "notchy" is *wrong*. Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse.
But as someone pointed out, the recursive perpendicular bisector subdivsion doesn't work. You replace side AB with sides AC and CB, where |AC| = |CB|, and you replace side XY with sides XZ and ZY, with |XZ| = |ZY|, but there's no reason to expect that just because |AB| = |XY|, that |AC| should equal |XZ|.
Andy
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
----- No virus found in this message. Checked by AVG - www.avg.com Version: 10.0.1153 / Virus Database: 424/3266 - Release Date: 11/19/10
----- No virus found in this message. Checked by AVG - www.avg.com Version: 10.0.1153 / Virus Database: 424/3266 - Release Date: 11/19/10
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush.
Michael Kleber is correct. When I scanned some of this thread, at first I thought that maybe Poncelet's porism would generalize (as Victor Miller suggested:)
Is what you want related to Poncelet's Porism? http://mathworld.wolfram.com/PonceletsPorism.html
There's an argument that would have imply that if there were exactly two solutions to arcs of length L inscribed in an ellipse, for every L, the same phenomenon as in Poncelet's porism would hold: the map of the ellipse to itself gotten by stepping off with steps of length L would be diffeomorphically conjugate to a rotation because the algebraic curve consisting of secants of length L would be an elliptic curve. But feeding the algebra to Mathematica showed this is not true: for each point on the ellipse, there's a degree 4 equation for the other endpoints of the secant, and the curve parametrizing secants of length L has high genus. (approximately genus 16, but I'm not clear enough on algebraic geometry to do it exactly without more work than I'm up for). The complexification of an ellipse is topologically an annulus, and given a length, L, and a starting point, x, you get an image of a 4-way-branching tree mapped to the annulus. As you vary L and x, the branches sometimes intersect and form loops. Sometimes these are on the real ellipse, sometimes just on the complexification. There are typically 32 critical points for the process, where two of the 4 branches coincide. When L is just a little longer than the minor axis of the ellipse so that there are "less obvious" secants of length L, there are 4 real critical points where the secant is "jammed" --- one end is perpendicular to the ellipse, the other makes an angle, and the perpendicular end can be perturbed in either direction while moving the non-jammed end slightly forward. There are lots of other non-obvious complex critical points Anyway, for each n, there is an exponential-in-n number of complex values of L with complex inscribed polygons. Even for the case of real polygons where each secant is chosen to advance the obvious way in the smallest possible step, the polygon can be any of the (n-1) n-pointed stars (where retracing is allowed, e.g. going around a triangle 3 times would count as a 9-pointed star). If Kleber's non-obvious secants are allowed, I think the number of n-gons is exponential-in-n even in the real case. That's because there's always an open set of {(starting point p, length L)} where there are four secants, when p is near an end of the minor axis and L is just longer than the minor axis. For any L < major axis, there's a rotation number associated with the process of advancing counterclockwise to the nearest point at distance L. Choose L so that the rotation number is irrational, and L is just greater than the minor axis. Every orbit for L' near L enters the open set with 4 solutions at regular intervals. By making branching choices at those times, then proceeding "obviously" for a while, and eventually adjusting L slightly to make the polygon close, I think you get exponentially many solutions. Bill Thurston On Nov 20, 2010, at 9:21 AM, Michael Kleber wrote:
On Sat, Nov 20, 2010 at 12:44 AM, David Wilson <davidwwilson@comcast.net>wrote:
I read your notes and did some imagining, and here are my unjustified beliefs:
For every n >= 3, and for any ellipse, there is a single continuous family of inscribed equilateral n-gons in the ellipse.
I disagree -- I think there are "less obvious" inscribed equilateral n-gons in ellipsis with high eccentricity, in which one or two edges take a short-cut, circumventing a large piece of the ellipse. (Given a point P on the ellipse and a distance d, there may be *four* points on the ellipse at distance d from P, not just two.)
This objection carries through to all the following statements also, I think...
I seem to be missing something: isn't it pretty obvious that for large enough n (the number of chords) there will be a continuum of n-gons? Scale the ellipse so that its arc length is 1 and consider only n-gons with chord length d<D so that a circle of radius d about any point of the ellipse cuts the ellipse in only two points. Call an arbitrary point of the ellipse the origin and let x be the signed arc length around the ellipse with respect to this origin, and clockwise corresponding to increasing x. Now consider a point at arc length x and a circle of radius d about x. The circle cuts the ellipse at two points, at arc lengths x' and x", where x'>x>x". Define f(x,d) = x'. For 0<d<D, the function f(x,d) is smooth and monotonic increasing in d. Next define the sequence of functions f_1(x,d) = f(x,d) f_2(x,d) = f(f(x,d),d) f_3(x,d) = f(f(f(x,d),d),d) etc. These are also smooth and monotonic increasing in d. Finally, for any d<D define n by the property |1+x - f_(n+1)(x,d)| >= |1+x - f_n(x,d)| and |1+x - f_(n-1)(x,d)| > |1+x - f_n(x,d)| So n is the best approximation to closure of the n-gon. Well now we have a smooth function f_n(x,d) monotonic in d and it has 1+x in its range. There is a function like this for any x and we can always find a d that gives closure, i.e. f_n(x,d) = 1+x. The arc lengths x thus parameterize a continuous family of n-gons. Veit On Nov 20, 2010, at 1:06 PM, Bill Thurston wrote:
Michael Kleber is correct. When I scanned some of this thread, at first I thought that maybe Poncelet's porism would generalize (as Victor Miller suggested:)
Is what you want related to Poncelet's Porism? http://mathworld.wolfram.com/PonceletsPorism.html
There's an argument that would have imply that if there were exactly two solutions to arcs of length L inscribed in an ellipse, for every L, the same phenomenon as in Poncelet's porism would hold: the map of the ellipse to itself gotten by stepping off with steps of length L would be diffeomorphically conjugate to a rotation because the algebraic curve consisting of secants of length L would be an elliptic curve. But feeding the algebra to Mathematica showed this is not true: for each point on the ellipse, there's a degree 4 equation for the other endpoints of the secant, and the curve parametrizing secants of length L has high genus. (approximately genus 16, but I'm not clear enough on algebraic geometry to do it exactly without more work than I'm up for).
The complexification of an ellipse is topologically an annulus, and given a length, L, and a starting point, x, you get an image of a 4-way-branching tree mapped to the annulus. As you vary L and x, the branches sometimes intersect and form loops. Sometimes these are on the real ellipse, sometimes just on the complexification. There are typically 32 critical points for the process, where two of the 4 branches coincide. When L is just a little longer than the minor axis of the ellipse so that there are "less obvious" secants of length L, there are 4 real critical points where the secant is "jammed" --- one end is perpendicular to the ellipse, the other makes an angle, and the perpendicular end can be perturbed in either direction while moving the non-jammed end slightly forward. There are lots of other non-obvious complex critical points
Anyway, for each n, there is an exponential-in-n number of complex values of L with complex inscribed polygons. Even for the case of real polygons where each secant is chosen to advance the obvious way in the smallest possible step, the polygon can be any of the (n-1) n-pointed stars (where retracing is allowed, e.g. going around a triangle 3 times would count as a 9-pointed star). If Kleber's non-obvious secants are allowed, I think the number of n-gons is exponential-in-n even in the real case. That's because there's always an open set of {(starting point p, length L)} where there are four secants, when p is near an end of the minor axis and L is just longer than the minor axis. For any L < major axis, there's a rotation number associated with the process of advancing counterclockwise to the nearest point at distance L. Choose L so that the rotation number is irrational, and L is just greater than the minor axis. Every orbit for L' near L enters the open set with! 4 solutions at regular intervals. By making branching choices at those times, then proceeding "obviously" for a while, and eventually adjusting L slightly to make the polygon close, I think you get exponentially many solutions.
Bill Thurston On Nov 20, 2010, at 9:21 AM, Michael Kleber wrote:
On Sat, Nov 20, 2010 at 12:44 AM, David Wilson <davidwwilson@comcast.net>wrote:
I read your notes and did some imagining, and here are my unjustified beliefs:
For every n >= 3, and for any ellipse, there is a single continuous family of inscribed equilateral n-gons in the ellipse.
I disagree -- I think there are "less obvious" inscribed equilateral n-gons in ellipsis with high eccentricity, in which one or two edges take a short-cut, circumventing a large piece of the ellipse. (Given a point P on the ellipse and a distance d, there may be *four* points on the ellipse at distance d from P, not just two.)
This objection carries through to all the following statements also, I think...
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I've been playing with inscribing an equilateral triangle inside the standard ellipse (x/a)^2+(y/b)^2=1. In order to keep Macsyma from going nuts, I first pinned down the point (x1,y1)=(a,0). I then eliminated y3,x3,y2, leaving only an equation in x2. Alternatively, we can eliminate y3,x3,x2, leaving an equation in y2. I then eliminated the complex solutions & matched up the real x2's and y2's. Due to symmetry, the x3's would be the same as the x2's. Curiously, the equation involves factors like (a-b) and (a+b), but also sqrt(3a^2+b^2), and most curiously sqrt(3b^2-7a^2). There appears to be some sort of interesting behavior that happens when 3b^2=7a^2. Let a=sqrt(3), b=sqrt(7). Then the ellipse equation is 7xx+3yy=21. The equilateral triangle in this case is (sqrt(3),0), (-3sqrt(3)/4,7/4), (-3sqrt(3)/4,-7/4). This particular ratio of a/b=sqrt(3/7) might be interesting, because (**speculation**) it might produce the equilateral triangle of largest area (relative to the area of the ellipse).
There is a continuum if you vary the starting point, but I was trying to say something about how many there are with a fixed starting point. The number varies according to the starting point, but I think my argument is a good sketch of a proof that for any fixed ellipse that is not a circle and any fixed starting point p on the ellipse, the number of n-gons through p grows exponentially with n. Bill Thurston On Nov 20, 2010, at 3:23 PM, Veit Elser wrote:
I seem to be missing something: isn't it pretty obvious that for large enough n (the number of chords) there will be a continuum of n-gons?
Scale the ellipse so that its arc length is 1 and consider only n-gons with chord length d<D so that a circle of radius d about any point of the ellipse cuts the ellipse in only two points. Call an arbitrary point of the ellipse the origin and let x be the signed arc length around the ellipse with respect to this origin, and clockwise corresponding to increasing x.
Now consider a point at arc length x and a circle of radius d about x. The circle cuts the ellipse at two points, at arc lengths x' and x", where x'>x>x". Define f(x,d) = x'. For 0<d<D, the function f(x,d) is smooth and monotonic increasing in d.
Next define the sequence of functions
f_1(x,d) = f(x,d) f_2(x,d) = f(f(x,d),d) f_3(x,d) = f(f(f(x,d),d),d) etc.
These are also smooth and monotonic increasing in d.
Finally, for any d<D define n by the property
|1+x - f_(n+1)(x,d)| >= |1+x - f_n(x,d)| and |1+x - f_(n-1)(x,d)| > |1+x - f_n(x,d)|
So n is the best approximation to closure of the n-gon.
Well now we have a smooth function f_n(x,d) monotonic in d and it has 1+x in its range. There is a function like this for any x and we can always find a d that gives closure, i.e. f_n(x,d) = 1+x. The arc lengths x thus parameterize a continuous family of n-gons.
Veit
On Nov 20, 2010, at 1:06 PM, Bill Thurston wrote:
Michael Kleber is correct. When I scanned some of this thread, at first I thought that maybe Poncelet's porism would generalize (as Victor Miller suggested:)
Is what you want related to Poncelet's Porism? http://mathworld.wolfram.com/PonceletsPorism.html
There's an argument that would have imply that if there were exactly two solutions to arcs of length L inscribed in an ellipse, for every L, the same phenomenon as in Poncelet's porism would hold: the map of the ellipse to itself gotten by stepping off with steps of length L would be diffeomorphically conjugate to a rotation because the algebraic curve consisting of secants of length L would be an elliptic curve. But feeding the algebra to Mathematica showed this is not true: for each point on the ellipse, there's a degree 4 equation for the other endpoints of the secant, and the curve parametrizing secants of length L has high genus. (approximately genus 16, but I'm not clear enough on algebraic geometry to do it exactly without more work than I'm up for).
The complexification of an ellipse is topologically an annulus, and given a length, L, and a starting point, x, you get an image of a 4-way-branching tree mapped to the annulus. As you vary L and x, the branches sometimes intersect and form loops. Sometimes these are on the real ellipse, sometimes just on the complexification. There are typically 32 critical points for the process, where two of the 4 branches coincide. When L is just a little longer than the minor axis of the ellipse so that there are "less obvious" secants of length L, there are 4 real critical points where the secant is "jammed" --- one end is perpendicular to the ellipse, the other makes an angle, and the perpendicular end can be perturbed in either direction while moving the non-jammed end slightly forward. There are lots of other non-obvious complex critical points
Anyway, for each n, there is an exponential-in-n number of complex values of L with complex inscribed polygons. Even for the case of real polygons where each secant is chosen to advance the obvious way in the smallest possible step, the polygon can be any of the (n-1) n-pointed stars (where retracing is allowed, e.g. going around a triangle 3 times would count as a 9-pointed star). If Kleber's non-obvious secants are allowed, I think the number of n-gons is exponential-in-n even in the real case. That's because there's always an open set of {(starting point p, length L)} where there are four secants, when p is near an end of the minor axis and L is just longer than the minor axis. For any L < major axis, there's a rotation number associated with the process of advancing counterclockwise to the nearest point at distance L. Choose L so that the rotation number is irrational, and L is just greater than the minor axis. Every orbit for L' near L enters the open set wi! th! 4 solutions at regular intervals. By making branching choices at those times, then proceeding "obviously" for a while, and eventually adjusting L slightly to make the polygon close, I think you get exponentially many solutions.
Bill Thurston On Nov 20, 2010, at 9:21 AM, Michael Kleber wrote:
On Sat, Nov 20, 2010 at 12:44 AM, David Wilson <davidwwilson@comcast.net>wrote:
I read your notes and did some imagining, and here are my unjustified beliefs:
For every n >= 3, and for any ellipse, there is a single continuous family of inscribed equilateral n-gons in the ellipse.
I disagree -- I think there are "less obvious" inscribed equilateral n-gons in ellipsis with high eccentricity, in which one or two edges take a short-cut, circumventing a large piece of the ellipse. (Given a point P on the ellipse and a distance d, there may be *four* points on the ellipse at distance d from P, not just two.)
This objection carries through to all the following statements also, I think...
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Thanks for the clarification. Isn't it necessary for the evolute (locus of centers of curvature) to cut the ellipse for there to be 4-point intersections of the ellipse with circles centered on it? That limits you to eccentricity < sqrt(1/2). Even in that case I would be worried about starting points such that the orbit, initially determined without branching by 2-point intersections, would be quasiperiodic and avoid the regions near the ends of the minor axis where the 4-point intersections occur. Veit On Nov 20, 2010, at 7:04 PM, Bill Thurston wrote:
There is a continuum if you vary the starting point, but I was trying to say something about how many there are with a fixed starting point. The number varies according to the starting point, but I think my argument is a good sketch of a proof that for any fixed ellipse that is not a circle and any fixed starting point p on the ellipse, the number of n-gons through p grows exponentially with n. Bill Thurston
I did some more work on embedding an equilateral pentagon (cyclicly, P1,P2,P3,P4,P5; P1=(x1,y1), etc.) into a standard ellipse (x/a)^2+(y/b)^2=1. Once again, I assumed that (x1,y1)=(a,0). The problem with the polynomial equations is that they incorporate all sorts of degenerate cases. For example, P4=P2, where the links P2P3 and P3P4 are the same line segment. We then have an equilateral triangle with an extra link hanging off of one corner. Another degenerate case has P1-P5 all distinct, but the links P2P3 and P4P5 cross each other to the other side. I don't know how to easily eliminate these degenerate cases from the set of equations. For this pentagonal case, there are three interesting factors that enter in: (3*b^2-a^2), (15*b^2-a^2), and (27*b^2-a^2). I assume that these again define critical ratios of b/a. For example, as b gets larger relative to a, the degenerate case where P2P3 crosses P4P5 collapses to where P3P4,P2P3,P4P5 all become the same link, and we have just an equilateral triangle. So this behavior can only happen when b/a is small enough. In order to eliminate these degenerate solutions, we may have to put in additional constraints. For example, in the case of the pentagon with P1=(a,0), the (squared) distances P1P3 and P1P4 are equal. I don't know how many degenerate solutions we can eliminate, but we can add equality constraints for many of the chords based on a presumed symmetry around the x axis. P1P3 = P1P4, P2P4 = P3P5. None of these constraints eliminate the case where the pentagon degenerates into a triangle.
Additional interesting factors: 8*b^2-a^2, 12*b^2-a^2. So, a complete list of interesting factors: (3bb-aa), (8bb-aa), (12bb-aa), (15bb-aa), (27bb-aa). It would be interesting to understand why the ratios a/b=sqrt(3), sqrt(8), sqrt(12), sqrt(15), sqrt(27) are so special. At 09:19 AM 11/21/2010, Henry Baker wrote:
I did some more work on embedding an equilateral pentagon (cyclicly, P1,P2,P3,P4,P5; P1=(x1,y1), etc.) into a standard ellipse (x/a)^2+(y/b)^2=1.
Once again, I assumed that (x1,y1)=(a,0).
The problem with the polynomial equations is that they incorporate all sorts of degenerate cases. For example, P4=P2, where the links P2P3 and P3P4 are the same line segment. We then have an equilateral triangle with an extra link hanging off of one corner.
Another degenerate case has P1-P5 all distinct, but the links P2P3 and P4P5 cross each other to the other side.
I don't know how to easily eliminate these degenerate cases from the set of equations.
For this pentagonal case, there are three interesting factors that enter in: (3*b^2-a^2), (15*b^2-a^2), and (27*b^2-a^2). I assume that these again define critical ratios of b/a. For example, as b gets larger relative to a, the degenerate case where P2P3 crosses P4P5 collapses to where P3P4,P2P3,P4P5 all become the same link, and we have just an equilateral triangle. So this behavior can only happen when b/a is small enough.
In order to eliminate these degenerate solutions, we may have to put in additional constraints. For example, in the case of the pentagon with P1=(a,0), the (squared) distances P1P3 and P1P4 are equal. I don't know how many degenerate solutions we can eliminate, but we can add equality constraints for many of the chords based on a presumed symmetry around the x axis.
P1P3 = P1P4, P2P4 = P3P5.
None of these constraints eliminate the case where the pentagon degenerates into a triangle.
participants (5)
-
Bill Thurston -
David Wilson -
Henry Baker -
Michael Kleber -
Veit Elser