Hexagons whose vertices lie on a conic and whose sides touch a conic

Question

Yesterday I generalized the idea of ​​a bicentric quadrilateral

Here is the following definition:

A biconcentric hexagon is a hexagon whose vertices pass through a conic section and at the same time touch a conic section on its sides.

We know from Brianchon's theorem that the red lines converge at a point, but what I found is that in a biconcentric hexagon, the blue lines converge at the same point, and the opposite is also true. This means that if the main diagonals of a hexagon tangent to a conic section converge at one point, the hexagon will be biconcentric.

Is this feature already discovered?

How do we prove that anyway?

Answer 1

A partial answer, but with important features.

In fact, up to a projective transform, your issue can be placed in the following framework where the external conic is a parabola and the internal one an ellipse, in fact a circle.

Let us consider parabola $(P)$ with generic point (parametric equations) :

$$P_t = (x=4t, \ \ y=t^2) \tag{1}$$

We are going to "jump" on this parabola from points $P_t$ to points $P_{f(t)}$ where $f$ is the following fractional linear transform (FLT in short) :

$$f(t)=\frac{at-1}{t+a} \ \text{where} \ a:=\sqrt{3}\tag{2}$$

Essential lemma : Line $(L_t)$ joining point $P_t=(4t,t^2)$ and point $P_{f(t)}=(4f(t),f(t)^2)$ is tangent to the circle $(C)$ with parametric equations :

$$\left(x_t=4a\frac{t^2+2at-1}{t^2+2at+7}, \ \ \ y_t=\frac{7t^2-2at+1}{t^2+2at+7}\right)\tag{3}$$

or under an equivalent implicit equation :

$$x^2+y^2-14y+1=0\tag{3'}$$

(i.e., with center $(0,7)$ and radius $4\sqrt{3}$ ; proof left to the reader...)

Let us consider the red hexagon with its vertices on the parabola. If we start for example with $t=1$ (first red point in $(1,1/16)$), the next point will be associated with $t_2=f(t_1)=2(2-a)$ giving point $\approx(2.14, 0.29)$ ; then we take $t_3=f(t_2)$; etc.

An important feature of function $f$ is that

$$f(f(f(t)))=- \frac{1}{t}$$

giving the essential property :

$$\underbrace{f(f(f(f(f(f}_{6 \ \text{times}}(t)))))))=t\tag{4}$$

(for any possible $t$).

In this way, terating function six times, we are back in the initial point, giving rise to an inscribed-circumscribed hexagon !

What has been done with initial value $t=1$ can be done with any initial value of $t$ : in all cases, an hexagon is formed, which isn't surprising : it is the classical Poncelet theorem.

But a natural question we can ask is : how has this transform (1) been found ? For example, If I want a pentagon instead of an hexagon to be formed, which transform should I choose ?

Explanation : It is a consequence of the correspondence between FLTs and the $2 \times 2$ matrices of their coefficients :

$$f(t)=\frac{at+b}{ct+d}$ \ \ \leftrightarrow \ \ \pmatrix{a&b\\c&d}$$

This correspondence is important ; here is why.

Composition $f \circ f'$ of two FLT given by $f(t)=\frac{at+b}{ct+d}$ and $f'(t)=\frac{a't+b'}{c't+d'}$, is clearly again a FLT $f''(t)=\frac{a''t+b''}{c''t+d''}$. The remarkable fact is that these coefficients $a'',b'',c'',d''$ can be obtained as the matrix product :

$$\pmatrix{a&b\\c&d}\pmatrix{a'&b'\\c'&d'}=\pmatrix{a''&b''\\c''&d''}$$

(composition of functions on one side, product of matrices on the other ; this kind of correspondence is called a homomorphism).

Now, the explanation : why have we taken $f$ as it is given by (2) ? Because this FLT can be written :

$$f(t)=\frac{\tfrac12 \sqrt{3}t-\tfrac12}{\tfrac12t+\tfrac12 \sqrt{3}}=\frac{\cos(\pi/6)t-\sin(\pi/6)}{\sin(\pi/6)t+\cos(\pi/6)} \ \text{associated with} \ \pmatrix{\cos(\pi/6)&-\sin(\pi/6)\\ \sin(\pi/6)&\cos(\pi/6)},$$

a rotation matrix which, when taken at the power 6, gives the opposite of the identity matrix which is in fact identical to the identity matrix, because of a simplification of the two minus signs in the numerator and the denominator.

Fig. 1 : Parabola $(P)$ with parametric equations (1), and circle $(C)$ with equations (3) or (3'), with five different incribed/circumscribed hexagons. Please note the very narrow "tunnel" between $(P)$ and $(C)$.

Can you take profit of this representation for your issue ? (use in particular Remark 2 below).

Remarks :

1. Equations (3) give the exact point of tangency of line $(L_t)$ with circle $(C)$.

2. For further calculations, the parametric equation of line $(L_t)$ (defined above in the lemma) :

$$(t^2+2at-1)x-(4t+4a)y+(-4at^2+4t)=0 \tag{5}$$

3. Lines $(L_t)$ have circle $(C)$ as their envelope. This aspect is developed in specialized books on projective geometry.

SAGE program for the algebraic results (3), (3') and (5) and plotting of the figure :

 var('x y t');
 a=sqrt(3);f(t)=(a*t-1)/(t+a);
 g=line([[-18,0],[18,0],[0,0],[0,24]],color='black');
 g+=parametric_plot((4*t,t^2),(t,-4.7,4.7),color='blue');# parabola
 for q in range(5) :
    t0=1+0.05*(q-2);
    for p in range(6) :
       t1=f(t0);
       if q==4 :
              g+=plot(point([4*t0,t0^2],size=30));
    LL=line([[4*t0,t0^2],[4*t1,t1^2]],rgbcolor=(q/5,1-q/5,0));
    g+=plot(LL);
    t0=t1;
 # equation of line :
 M=matrix([[4*t,t^2,1],
   [4*f(t)*(t+a)^2,(t+a)^2*f(t)^2,(t+a)^2],
   [x,y,1]]);
 N=(M.determinant())/(1+t^2);
 R=numerator(N);show(4*R);
 dR=diff(R,t);show(4*dR);
 s=solve(dR,t);
    P=solve([R==0,dR==0],x,y);show(P[0]); # parametric equ. of circle
    # envelope obtained by solving system L_t=0 and d(L_t)/dt=0
 C=numerator(R.subs(t=s[0].rhs()));show(C); # cartesian equ. of circle
    g+=plot(point([0,7],color='black'));
    g+=implicit_plot(x^2+y^2-14*y+1,(x,-8,8),(y,0,16),color='blue');
    g

Answer 2

HINT:(corrected)

So what you want to show is this: take an arbitrary conic ( the inside one), and three chords of this conic passing through the same point. You want to show that the hexagon formed by the tangents to the conic at the ends of the chords is inscribed in a conic. Now this is a projective condition. So we may reduce to the case: the conic is a circle, and also, the common point of the three chords is the center of the circle ( since the group of projective transforms invarianting the circle acts transitively on points not on the conic). So now we have to prove the result: take three diameters of a circle, take the tangents at the end, the hexagon formed is inscribed in a bigger conic. I think it is easier now.

Note: it is not an easy condition to see when there is a hexagon inscribed in a (bigger conic) and tangent to the (inside) conic. Some invariants are $0$, and they seem complicated ( even for $n=3$ it is).

For facts on conics, the book of Salmon ( from late 1800's) on conics seems nice.

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Still, a beautiful result! ( not clear it is (widely) known)

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$\bf{Added:}$

Back to the particular case, the hexagon will have parallel opposite sides, so there is a conic through its vertices. Now parallel means intersecting at the line at infinity, which is the polar of the center of the circle with respect to the circle. So I think one can prove this fact directly:

Consider a non-degenerate conic, and three chords. Draw tangents to the conic at the ends of chords. We get a hexagon. The intersection of its opposite sides will be poles of corresponding chords ( out of the three). They will lie on the same line if and only if the chords are concurrent.