for those who hava mathematica, there is a really neat interactive demo of this theorem at: http://demonstrations.wolfram.com/MardensTheorem/ bob --- Mike Stay wrote:

On Mon, Nov 15, 2010 at 7:35 AM, Henry Baker <hbaker1@pipeline.com> wrote:

...

I just ran across this "Most Marvelous Theorem in Mathematics":

http://www.maa.org/joma/Volume8/Kalman/index.html

It is well-known/well-taught that the roots of p'(x) lie within the convex hull of the roots of p(x).

However, in the case of a cubic, we can say a lot more: the roots of p'(x) are the _foci_ of the inscribed _ellipse_ that passes through the midpoints of the sides of the triangle formed by the roots of p(x). This is Marden's/Siebeck's theorem.

If we assume the n roots of an nth degree polynomial are algebraically independent, is the obvious generalization of this theorem true, i.e. do the roots of the derivative suffice to describe some kind of surface tangent to the sides of the n-simplex?

On 11/22/10, Thomas Colthurst <thomaswc@gmail.com> wrote:

I don't know the answer to Mike's question, but I do have two observations about the roots of p'(x) that may be of interest. ... One consequence of this is that for any point z and multiset of n-1 points Y, there is another multiset of n-1 points Z such that C(z union Z) = Y. Just coordinate shift so that z is the origin, then integrate p'(x) to get p(x) + k. But p(0) = 0, so k = 0.

"multiset of n points Y", I think ... ? Interesting points you made there, TC. WFL

Interesting perspective! Thanks! However, you _did_ prove one type of uniqueness: Choose an equilateral triangle with its inscribed circle and a particular angle of rotation of the triangle around its center (<120 degrees). Then the mapping of the inscribed circle into a given ellipse is unique. (Do I need to worry about a reflected mapping here in order to further reduce the angle parameter to <60 degrees?) At 01:37 PM 11/15/2010, rcs@xmission.com wrote:

answering one question ...

...

Given an ellipse, is the triangle that the ellipse is inscribed within unique?

No. Start with an equilateral triangle, and the inscribed circle. Choose any direction, and stretch the diagram in that direction, while leaving perpendicular distances alone. This operation preserves some geometric properties, and mangles others. Tangency and midpoints are preserved. The inscribed circle becomes an inscribed ellipse. Foci are not preserved. We can get an ellipse of prescribed shape, in any direction. Working backwards, knowing an ellipse alone doesn't determine the orientation of the circumscribed "midpoint triangle".

Rich

--------- Quoting Henry Baker <hbaker1@pipeline.com>:

...

I just ran across this "Most Marvelous Theorem in Mathematics":

http://www.maa.org/joma/Volume8/Kalman/index.html

It is well-known/well-taught that the roots of p'(x) lie within the convex hull of the roots of p(x).

However, in the case of a cubic, we can say a lot more: the roots of p'(x) are the _foci_ of the inscribed _ellipse_ that passes through the midpoints of the sides of the triangle formed by the roots of p(x). This is Marden's/Siebeck's theorem.

So in the case of (my) standardized cubic p(x)=x^3-(3/4)*x-cos(3*a)/4 [roots are cos(a), cos(a+2*pi/3), cos(a-2*pi/3), even when a is complex], the roots of p'(x) are +-1/2. So even as the triangle becomes enormous (a is complex), the ellipse becomes closer & closer to a circle, and the triangle becomes closer & closer to an equilateral triangle.

Marden's/Siebeck's Theorem is one property of ellipses that Newton was almost certainly not aware of. Are there any consequences of this theorem that show up in Newton's 2-body problem? Given an ellipse, is the triangle that the ellipse is inscribed within unique? If so, are there any interesting physical/mechanical consequences of those triangle vertices?

My conjecture is that this inscribed ellipse may describe an maximum speed trajectory that lies within the triangle, subject to certain acceleration constraints, but I don't know enough calculus of variations to be able to get any further.

I should have explained my reasoning. If every configuration of triangles with inscribed ellipses through its midpoints can be produced by a sequence of translations, rotations, stretches and shears from an equilateral triangle with its incircle, then that same sequence of transformations will take its circumcircle into a similar ellipse twice as big. At 04:48 PM 11/16/2010, Henry Baker wrote:

It would appear that the triangle vertices must appear on a second similar ellipse exactly twice as large as the first one & centered & oriented the same as the given ellipse. This answers the question about why, given a random ellipse and a point outside that ellipse, it isn't always possible to construct a triangle with that point as one of its vertices that satisfies all of the constraints. However, given an ellipse and a random point along the _doubled_ ellipse, it _is_ always possible to construct such a triangle.

In the case of the triangle vertices cosh(b),(-cosh(b)+-i*sqrt(3)*sinh(b))/2, the foci of the inscribed ellipse are at +-1/2, the major axis is cosh(b), the minor axis is |sinh(b)|, the eccentricity is 1/cosh(b), and the area is pi*|sinh(2b)|/8.

At 02:37 PM 11/15/2010, Henry Baker wrote:

...

Interesting perspective! Thanks!

However, you _did_ prove one type of uniqueness: Choose an equilateral triangle with its inscribed circle and a particular angle of rotation of the triangle around its center (<120 degrees). Then the mapping of the inscribed circle into a given ellipse is unique. (Do I need to worry about a reflected mapping here in order to further reduce the angle parameter to <60 degrees?)

At 01:37 PM 11/15/2010, rcs@xmission.com wrote:

...

answering one question ...

...

Given an ellipse, is the triangle that the ellipse is inscribed within unique?

No. Start with an equilateral triangle, and the inscribed circle. Choose any direction, and stretch the diagram in that direction, while leaving perpendicular distances alone. This operation preserves some geometric properties, and mangles others. Tangency and midpoints are preserved. The inscribed circle becomes an inscribed ellipse. Foci are not preserved. We can get an ellipse of prescribed shape, in any direction. Working backwards, knowing an ellipse alone doesn't determine the orientation of the circumscribed "midpoint triangle".

Rich

--------- Quoting Henry Baker <hbaker1@pipeline.com>:

...

I just ran across this "Most Marvelous Theorem in Mathematics":

http://www.maa.org/joma/Volume8/Kalman/index.html

It is well-known/well-taught that the roots of p'(x) lie within the convex hull of the roots of p(x).

However, in the case of a cubic, we can say a lot more: the roots of p'(x) are the _foci_ of the inscribed _ellipse_ that passes through the midpoints of the sides of the triangle formed by the roots of p(x). This is Marden's/Siebeck's theorem.

So in the case of (my) standardized cubic p(x)=x^3-(3/4)*x-cos(3*a)/4 [roots are cos(a), cos(a+2*pi/3), cos(a-2*pi/3), even when a is complex], the roots of p'(x) are +-1/2. So even as the triangle becomes enormous (a is complex), the ellipse becomes closer & closer to a circle, and the triangle becomes closer & closer to an equilateral triangle.

Marden's/Siebeck's Theorem is one property of ellipses that Newton was almost certainly not aware of. Are there any consequences of this theorem that show up in Newton's 2-body problem? Given an ellipse, is the triangle that the ellipse is inscribed within unique? If so, are there any interesting physical/mechanical consequences of those triangle vertices?

My conjecture is that this inscribed ellipse may describe an maximum speed trajectory that lies within the triangle, subject to certain acceleration constraints, but I don't know enough calculus of variations to be able to get any further.