Very nice observation. Does this follow from the Vieta's theorem as we vary only the constant term, or is the proof more complicated?

midpoint F1 - C, A +P(x^2) when quadrilateral perimeter is equal to circle circumference

Point P is called the "4th point", the 4th intersection of the ABC circumconic (here the ellliptic billiard with equation in trilinears 1/apha + 1/ beta + 1/gamma =0) with the ABC circumcircle centerd at O. From ETC, P = X(100) lies on the circumcircle and the circumconic with center Mittenpunkt X(9). Fix point C0 is X(3035) = 3*X(2) + X(100).

The coefficients of x^3 and x^4 in the degree 4 polynomial obtained by computed the intersection of the ellipse and the circle only depends on the center of the circle and the parameters of the ellipse. Thus the sum of its roots P(t), A, B, C is constant when the radius is moving.

Let O_1=a, O_2=b, O_3=c. Then the equation of the ellipse shall be [conj(a)*(b-c)+conj(b)*(c-a)+conj(c)*(a-b)]^2-4*(|a|^2-1)(|b|^2-1)(|c|^2-1)*cos(pi/n)^2=0 for n sided polygons, when replacing a by x+iy.

This is indeed an ellipse, with equation x^2+y^2*(1+16*c^2/(c^2-1)^2)=1 I guess, where c=O_3. Great insight by the way !

Case N=4 is due to the fact that degree 4 Blaschke products are associated to an ellipse iff it is the composition of two degree 2 Blaschke products. Case N>4 should be the consequence of such a decomposition too and somehow is a generalization of Brianchon's theorem.

Funny ! One can easily find three Möbius transformations f,g,h such that P_(i+1)=f(P(i)) for i = 0 mod 3 P_(i+1)=g(P(i)) for i = 1 mod 3 P_(i+1)=h(P(i)) for i = 2 mod 3. Then the question is : when is the nth composition of h(g(f))(x) the identity ? But if I'm not wrong, this happen when O_1 (or O_2 or O_3) is a solution of 4*cos(pi/n)^4*(x^2-1)^3+27*x^4=0. When n=2, we get x=0 which is case N=6. Thus the phenomenon because non trivial for multiple of 3 greater than 8 as you thought.

2:13 Hypnotisant ! I think the only way this could work (err=0 for every polygons) is by setting the centers of the inner circles equal to +/-(sqrt(4-3*m)-sqrt(4-m))/(sqrt(4-3*m)+sqrt(4-m)) with radius = (2*sqrt(2 - m))/(sqrt(4 - 3*m) + sqrt(4 - m)) for any 0<=m<1. This is interesting as it would gives a way to construct, from two circles, a "poncelet curve" that is not a conic but that keeps poncelet porism properties. Thus, somehow, it should exist two elliptic curves (E,+) and (E'',++) and a point S such that for an infinite number of points P (all?) we have P+(-S)++(S)=P++(-S)+S.

How about 🥚?

Hello, For Poncelet triangles, area is constant iff the ellipse is centered at 0 or one of the foci is 0. Take a degree 3 Blaschke product, B(z) with zeroes a and b. Solving the equation B(z)=lambda for some |lambda|=1 gives the vertices of some triangle circumbscribed to some 3-ellipse with foci a and b (any ellipse inscribed in such triangles can be obtained like that, any triangle circumbscribed to this ellipse can be obtained like that too). Multiply both sides by (1-conj(a)z)(1-conj(b)z) where conj(u) is the complex conjugate. Extract the coefficient of z^2: conj(a)*conj(b)*lambda+(a+b). This is equal to minus the sum of the roots of the polynomial B(z)-lambda*(1-conj(a)z)*(1-conj(b)*z) which are the vertices of the triangle. Now the sum of the areas of the power circles is constant iff A=conj(z_1)*(z_2+z_3)+conj(z_2)*(z_1+z_3)+conj(z_3)*(z_1+z_2) is (as shown in the article in source). Now set z_1+z_2+z_3=S. Then A=conj(z_1)(S-z_1)+conj(z_2)(S-z_2)+conj(z_3)*(S-z_3). Thus (after simplifying and using |z_i|=1) A is constant iff |S| is constant. But |conj(a)*conj(b)*lambda+(a+b)| is constant iff a=-b (ellipse is centered at 0) or one of the foci is 0 (because here lambda is also equal to the norm 1 product z_1z_2z_3 which always moves when we do move the vertices of the triangle). For n-ellipses I think it it will depends on the foci of the ellipse but also on the foci of the ellipses obtained by considering the polygons with sides (z_{i}z_{i+k}) for k>=1 mod n.

Hello, Using the fact that a n sided polygon inscribed in the circle centered at 0 with radius 1 and circumbscribed to an ellipse centered at 0 has its jth vertex z_j = exp(i*am(4K(m)j/n+t,m)) (modulo multiplication by some fixed rotation depending on the axis of the foci), one should be able to deduce that the sum of |z_i-1/2(z_(i+k)+z_(i+k+1))|^2 for all i=0..n-1 is constant for any non zero integer k where indexes are reduced mod n. One can indeed see for example that it is true for n=4 and with recursive proof, show that it is true for n=2^s, s>1.

...and the value of this is?

Excellent work ! 💯

As X74 and X3 are both on the Jerabek hyperbola there may be some interesting phenomena related to points at this hyperbola. X125 will have a circular locus when the locus of X74 is frozen as X125 is the midpoint of X74 and X4, and X4 has a circular locus.

I understand this well enough for ellipses. I had some trouble with parabolas and hyperbolae though. Especially the latter. There is no major or minor axes; just a transverse axis. Can find vertices but dont know how to find foci. Much less conjugate axis, asymptotes, directrices, etc. of these conics. The construction for hyperbolae works identically to that of ellipses, right up to vertices, where it starts to deviate.

There is a much easier tangent construction that doesnt require any "infinitesimal limit" argument.

Is this animation made using Geogebra?

this obs is not related to Poncelet. for any triangle, the radius of the inverse image of the circumcircle wrt incircle only depends on the inradius.

What software are you using to visualize? I'm looking for something better than Geogebra. I find that once diagrams get beyond easy concepts, Geogebra is an impediment, not an aid.

X80 is the reflection of X1 into X11. As X1 is stationary and X11 is on the incircle X80 is on a circle centered at X1 with twice the radius of the incircle.

Note: further investigation reveals the r-isocurves are higher degree curves that "fool the eye" as ellipses. Too bad!

Nice and interesting ;-)

J'ai une démonstration algébrique avec mes outils en coordonnées trilinéaires. On part d'un triangle isosceles ABC inscrit dans le cercle avec BC = a, AB = AC = b, et on calcule les coordonnées paramétriques d'une 3-orbite générale P1P2P3 (notamment du fait que le centroid G est fixe, car le Nine Point center et l'orthocenter le sont)...à la fin on obtient que la somme des carrés des longueurs des cotés de P1P2P3 est invariante égale à a^2 + 2*b^2. Quand le centroid est fixe, c'est plus facile, car je paramétrise une droite P2P3 tangente à l''ellipse. Le rapport de proportion des longueurs des semi-axes de l'ellipse est une fonction directe de b/a.

So glad I stumbled upon this channel

Like in orher Poncelet cases, the inradius r is a function of circumradius R and semi-perimeter s. Here for MacBeath Poncelet s^2 - r*(4*R + r) is a constant. I am confident to prove your invariant with Cayley condition (for Poncelet 3-orbit) and my parametrization of elliptic billiard (I have the trilinear coordinates of the foci for an ABC isosceles 3 orbit with A on minor axis plus axis lengths).

Hi, the reason the sum of squares of side lengths of the quadrilaterals is constant 8 is because the condition is equivalent to the circumcenter having an isogonal conjugate with respect to the quadrilateral, which is equivalent to <AOB+<COD=180 if we call the quadrilateral ABCD. thus if we consider the midpoints M and N of AB and CD we get right triangles AMO and ONC, that must be similar since <AOM+<CON=90. Then they share a hypotenuse so they are congruent, so AM^2+CN^2=AM^2+MO^2=AO^2=1, so AB^2+CD^2=4, also BC^2+DA^2=4 so the sum of squares of sides is 8

When you let P move on any line (not necessarily through the centroid) all the other points will also move on lines. These six lines circumscribe a permutation ellipse, as the six points at which these lines touch this ellipse form a permutation set. This still holds in any type of homogenous coordinates, such as trilinear.

Neat way to think about it.

Are these results algebraically proved or just approximations? Maybe it is true for a pencil of coaxal circles, but I am not sure.

is a porism. (3 degenerate caustics)

In a Poncelet (general) porism, the outer conic is unique, nevertheless there can be more caustics (all in the same pencil).

como pode??

Bravo!

This is quite remarkable. It is non trivial, not intuitive and numerically remarcable. Someday, someone will bring it up!

sorry audio is screwed up

Accidental ASMR? btw, whenever I say "complete quadrangle" I meant "complete quadrilateral"

this is a visual pattern

P.S. -- forgot to say: strophoid passes through the 6 vertices in the quadrilateral, so inverting 5 of them wrt incircle would have been "dayeinu" for the rectangular hyperbola.

Correction: line QQ' of the complete quadrilateral is the polar of Li wrt *any* inconic.

My intermediary results : Poncelet n=4 for d = sqrt(2+sqrt(5)) and for d=1 too + M13,M24,O' colinear .... computations for Q,Q' and unproven phenoma to be done soon ;-)

research wtc 7 9/11 was an inside job

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Hi. Beautyfull. The center of Jerabek hyperbola is x125, and describes a circle of 9 points. Which hyperbola is this?

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