Gary
GaryVasco
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Post by Gary on Mar 16, 2020 0:41:13 GMT
Hi The note I refer to is nh.ch7.ex4.pdf which is linked to above in your post of Apr 8, 2017 at 11:52pm. I've found that the ellipse result is standard (Marden's Theorem) and follows immediately from the expressions in my above post, in that if you have a cubic function whose roots are say, a,b,c then when expressed in the form f(z) = z^3 +pz +q , then the foci of the "inellipse" are at +/- sqrt(-p/3). Exactly what your result ties into as per my earlier post. If you are prepared to accept Marden's theorem www.su.se/polopoly_fs/1.229312.1426783194!/menu/standard/file/marden.pdf then I think we have proved what was needed. It took a fair bit of blind kludging around and I couldn't have gone any further than the serendipitous simplifications I got from translating to the origin ...so finding this doc was a godsend. Do I feel a bit like I've "cheated" by relying on Marden for the final bit...yes I guess so. Let me know your view. Anyway a fun day on Q4...onto Q5! Regards Telemeter To any readers,
I haven't quite digested Marden's theorem yet, but perhaps I have made a bit of progress in understanding the problem. "Blind kludging around" sounds like a fair description of my process. I attach a stab at part (iv).
Gary
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Post by telemeter on Mar 16, 2020 15:01:38 GMT
Hi Gary,
I would imagine (intuition talking here) that to go from an equilateral triangle to any other and see the symmetry of the original "in circle" transformed to that of an "in ellipse" you would need to stretch in the y direction as you did, but shear (and not stretch) in the x-direction with the shear leaving the bottom edge of the triangle unaffected. (I am assuming that the triangle is aligned with bottom edge on or parallel to the real axis). In this way the centroid would automatically remain the centroid and you retain full ability to manipulate the triangle to any desired ratio of sides because the bottom edge is invariant under these transformations.
Unfortunately, I haven't got to p539 yet to appreciate the transformation you applied, but at first glance that looks to be a very specific transformation to an ellipse. I would guess we need the triangle deformation to drive the ellipse rather than the other way around..as you found. ie not any old ellipse will do.
telemeter
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Gary
GaryVasco
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Post by Gary on Mar 16, 2020 16:25:54 GMT
Hi Gary, I would imagine (intuition talking here) that to go from an equilateral triangle to any other and see the symmetry of the original "in circle" transformed to that of an "in ellipse" you would need to stretch in the y direction as you did, but shear (and not stretch) in the x-direction with the shear leaving the bottom edge of the triangle unaffected. (I am assuming that the triangle is aligned with bottom edge on or parallel to the real axis). In this way the centroid would automatically remain the centroid and you retain full ability to manipulate the triangle to any desired ratio of sides because the bottom edge is invariant under these transformations. Unfortunately, I haven't got to p539 yet to appreciate the transformation you applied, but at first glance that looks to be a very specific transformation to an ellipse. I would guess we need the triangle deformation to drive the ellipse rather than the other way around..as you found. ie not any old ellipse will do. telemeter telemeter,
May I assume that the statement "the bottom edge is invariant under these transformations" actually refers only to the y-value of the edge? I like the use of "shear" for the manipulation of the triangle, but I am not used to the idea that the shear applies in two directions at the same time. My only experience with the concept is in the context of wind blowing on a house frame and shifting all the vertical timbers in the same rotational direction. My notion of how this works with a given ellipse is that letting $\theta$ vary in the negative (clockwise) direction in (a cos $\theta$ b sin $\theta$) slides the touch-points (tangent points) down the shoulders of the ellipse, so that the shear of the triangle sides (their rotation about the moving tangent point) raises the apex and shortens the base (which remains at y = -i). For the ellipse a = 2, b = 1 (major, minor axes) the midpoint occurs at $\pi/6$ on the right side and $5\pi/6$ on the left, assuming my Figure 3, but they show up at near half these values on the plot itself.
I don't yet understand how the centroid fits in, but that was to be my next question.
I regret that I named the vertices a, b, and c, because there is confusion with major and minor axes and foci.
Gary
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Gary
GaryVasco
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Post by Gary on Mar 16, 2020 23:19:49 GMT
telemeter, I have attached a study sheet which I wrote for myself in an effort to understand Carlson's proof. I have a few questions referenced to the numbers on the sheet. Most of the numbers refer to expressions or equations in the proof in roughly the order that Carlson has them.
7. What role does this sum of products play in the proof? 8. What role does this sum of squares play in the proof? 11. Should this be $2a$ or $a^2$ rather than $(2a)^2$? 12. Same as 9. 13. Same as 9. 14. I wonder if this is another instance of the same problem as in 9, 10, and 11. 19. This is possibly another instance.
21, 21b. The numbers don't add up. Beyond that, I'm afraid I don't yet see the underlying structure of the proof. What purpose is served by introducing the general cubic $P(z) = z^3 + pz + q$ and it's derivative? Is it just to show that the foci coalesce to the origin? Gary
P.S. revised work sheet, which had numerous mistakes, on 03/19/20. The numbers add up now, allowing only for inversion of major/minor axes.
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Post by telemeter on Mar 19, 2020 11:10:03 GMT
Gary
On the shear
Assume you have equilateral triangle ABC and you want to transform it to be similar to a target triangle DEF. To do this you have to move apex C to C' such that ABC' will be similar to your target DEF.
Place base AB on the x axis with C above. Apply a stretch in y direction to get the apex to the required y coordinate:
Thus, y'=Yy, where Y is the stretch factor and so each y coordinate is moved to y'. So if C starts at (x,y) it will move to (x,y')
In this way, AB is unaffected.
Then apply a shear in the x direction to move C to its target shear is x'=Xyx. Where X is the amount of shear (per unit of 'height') and x and y are the coordinate of the point being sheared.
In this way AB is again unaffected (y=0) and C will coincide with target position C' at (x',y'). Also the centroid of ABC (important because it is the centre of your target ellipse) maps on to the centroid of ABC' and the initial circle will have been transformed to the inscribed ellipse we seek.
Thus I am advocating just one shear and not two.
I agree with your proof of uniqueness.
On Marden and the proof structure
The way I see it this is all an example of Macdonald's Theorem referred to in Chapter 7 Ex 13 part iv) p373. This is that f has one more root inside contour gamma than f'. Consequently, f' has one more root inside the contour than f''. We know in the case of Ex 4 that for a contour around the triangle f has three roots. Therefore by Macdonald f' has two roots and f'' has one root.
In your answer you found the two roots of f' (ie the ciritical points of f). And your expression for f' can be used to show f'' has one root.
I noted in my playing with this that the root of f'' was the centroid (as mentioned in the thread above). To answer Ex 4 we just need to prove that the roots of f' are the foci which is exactly Marden's Theorem. Needham, is effectively asking us either to prove Marden or recognise when we have got close enough to just use it.
The theorem applies to any cubic ie with term x^2 say, x^3+Ax^2+Bx+C. The proof (and statements of Marden's Theorem that I have seen) operates on a standard cubic (no x^2) x^3 + px + q. Thus the first part of the proof is to get the general cubic to a standard form so that it can be shown that whatever applies to a standard cubic is generally true of all cubics.
As we saw earlier in the other thread moving the centroid to the origin is the same as reducing the cubic to standard form.
Purpose of derivative
You ask about P and P' and its purpose.
The very last bit of the proof looks at the derivative P'. This is the key moment, the roots of P' are shown to be the foci of the ellipse at +/- c = sqrt (-p/3). Here the 'p' is the coefficient of x in the standard form. This is what we need. These roots are exactly the critical points of P which you found in your answer...after the coordinate translation in my earlier post on the other thread.
It seems this ellipse is called the 'Steiner inellipse' (I find on wiki) and there is more on it on wiki.
telemeter
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Post by telemeter on Mar 19, 2020 12:18:20 GMT
Gary
As to your detailed questions.
7. The vertices of the equilateral triangle lie on a circumscribed circle, centre the origin. They are by definition 2*pi/3 apart and so form the cube roots of some number as in 6. By considering the cube roots of unity it is easy to see that the sum of the products will come to zero as each product of two roots just maps on to the third root. I can't quite see their use yet!
8. As the vertices are of an the equilateral as in 7, the sum of the squares will equal zero. (Again consider the roots of unity as an example). This is a critical result. It leads to 14, and that the sum of the cross terms is zero. The sum of the cross products coming to zero is used later in 21. In 21 the term summing over x(i)y(i) is zero. The x(i)y(i) are proportional to the imaginary term from 8 and this is zero.
11. Consider an equilateral triangle. If the distance from the centroid to a side is 'a', then the distance to a vertex is 2a. That, is if the inscribed circle has radius a, the circumscribed circle has radius 2a. Thus each vertex is 2a from the centroid/ origin in this case. Therefore by Pythagoras the sum of the squares of the cords of each vertex = (2a)^2.
12. 13. Ditto. I think you may have gone astray by using a and b differently from the proof. In the proof a is the minor axis (y direction), in your workings it is the major axis (x direction).
If you switch them around all your problems disappear!
To apply the proof to the exercise you need to generate the equation of the cubic whose roots are the vertices of the equilateral triangle.
telemeter
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Gary
GaryVasco
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Post by Gary on Mar 19, 2020 15:49:24 GMT
telemeter,
Thank you for the detailed replies. I will go through it and try to resolve the details for myself. I see the place of the standard cubic more clearly now, but I will have to work through some of those products again in light of the geometry of the equilateral triangle that you pointed out. In an early study, I attempted to put an ellipse into an arbitrary triangle touching at the midpoints. When that didn't work out well, I reversed direction, started with the equilateral triangle, and just amplified the y-axis. That produced the desired ellipse, but the triangle was isosceles rather than arbitrary. I will try the shear to create an arbitrary triangle with the inscribed ellipse.
I will play with the major and minor axes. I think I see the issue, but I note that Carlson's proof only describes "semi-axes" and doesn't specify major and minor. I think when I stretched with a/b rather than b/a, I should have also swapped b for a in subsequent equations. Or I could stand the ellipse on its head.
Thanks to the corona virus, one has plenty of time to revise.
Gary
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Gary
GaryVasco
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Post by Gary on Mar 19, 2020 18:32:21 GMT
telemeter,
I found numerous mistakes in the original worksheet. It is revised and reposted in the same place now. It retains the same semi-axes, which are switched from those in Carlson's proof. I think one needs only to invert the b/a in $y_k = (b/a)\eta_k$ and everything works as in the proof.
Gary
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Gary
GaryVasco
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Post by Gary on Mar 20, 2020 6:59:53 GMT
telemeter,
I'm still chewing away at this. I added a figure to the document and reposted. It now has both compression on the y-axis and shear on the x-axis. I notice that shear moves the centroid but the compression or amplification on the y-axis does not.
I think the equation in the last line must be x' = Xy. At least that was the only thing that worked for me in the new figure.
Gary
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Post by telemeter on Mar 29, 2020 15:02:43 GMT
Looks good, yes that must be what I meant! I take it you're happier with the use of f' now in the proof. I'm back to Needham after a short break because its got colder outside again and my permitted one daily walk is unappealing!
telemeter
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Gary
GaryVasco
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Post by Gary on Mar 30, 2020 14:02:11 GMT
Looks good, yes that must be what I meant! I take it you're happier with the use of f' now in the proof. I'm back to Needham after a short break because its got colder outside again and my permitted one daily walk is unappealing! telemeter telemeter,
Thank you. After completely revising the figures in part (i), I'm still working on parts (ii) and (iii) of the problem.
Gary
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Gary
GaryVasco
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Post by Gary on Apr 6, 2020 23:09:37 GMT
telemeter
It’s at least a month now since the topic of the triangle with an inscribed ellipse touching the midpoints of the sides was broached here. As you have seen, I found it necessary to be able to create arbitrary ellipses with those qualities before attempting a solution. The amplification and shear did the trick, but the problem showing that an ellipse created in that manner is unique still seemed elusive. The foci do not seem to shear in the same way that the points of the ellipse, the triangle, and the centroid do. I probably should have written the transformations in the format $re^{i\theta} + b$, but I doubt that would have made everything transparent. I say this because I found Kalman’s original paper. It seems this problem was actually asking us to rediscover Kalman’s proof, which occupies 8 and 1/2 pages of the journal of the MAA 155: 330-338, 2008. I see from the paper that I was a few lemmas short of the proof. I still don’t quite see why a root that is tangent to the midpoint of one side of T implies that it is also tangent to the other two sides. It was instructive and I may continue to whittle at it.
Gary
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Post by telemeter on Apr 8, 2020 11:34:55 GMT
Gary,
It seems to me you have proved uniqueness geometrically by showing that any such Steiner in-ellipse can be transformed to the incircle of an equilateral triangle and vice versa. There is only one transformation to take the equilateral triangle to the triangle of your choice. That unique transformation when applied to the incircle of the equilateral triangle generates the in-ellipse. Therefore the Steiner in-ellipse is unique. Also the nature of the transformation is such that the in-ellipse has to be tangent to the sides of the triangle, because the incircle was. Are you not looking for too much?
Or am I missing something?
Not sure what "why a root that is tangent to the midpoint of one side of T implies that it is also tangent to the other two sides" refers to.
telemeter
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Gary
GaryVasco
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Post by Gary on Apr 8, 2020 14:25:32 GMT
Gary, It seems to me you have proved uniqueness geometrically by showing that any such Steiner in-ellipse can be transformed to the incircle of an equilateral triangle and vice versa. There is only one transformation to take the equilateral triangle to the triangle of your choice. That unique transformation when applied to the incircle of the equilateral triangle generates the in-ellipse. Therefore the Steiner in-ellipse is unique. Also the nature of the transformation is such that the in-ellipse has to be tangent to the sides of the triangle, because the incircle was. Are you not looking for too much? Or am I missing something? Not sure what "why a root that is tangent to the midpoint of one side of T implies that it is also tangent to the other two sides" refers to. telemeter telemeter,
Thank you. Perhaps I am looking for too much. And possibly I misinterpreted Kalman's paper. It appeared to me that he was attempting to prove two things: there is a unique ellipse with tangents at the midpoints, and the foci are the roots. Perhaps he was actually assuming the former and proving the latter. What you say in the first paragraph makes sense to me.
Gary
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Gary
GaryVasco
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Post by Gary on Apr 15, 2020 23:08:14 GMT
My answer to Part (v)
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