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In my course "Projective Algebraic Geometry" I was asked to solve following problem:

Prove that if two triangles are each self-polar relative to a nondegenerate conic $\Gamma$, then their six vertices all lie on another conic and their six edges are all tangent to the third conic.

Projective plane is considered to be over the field $\mathbb k$ with $char \mathbb k \neq 2$.

I can prove that if two triangles are inscibed in some conic, then their edges are tangnet to another conic. It also seems that proving other direction is also not a problem.

I just have no idea how to prove that vertices of self-polar triangles lie on conic or that their edges are tangent to conic.

So any hints or solutions are appreciated.

Thanks!

Update: I came up with proof.

Let $ABC$ and $PQR$ be self-polar triangles with respect to conic $\Gamma$. Without loss of generality we can assume that $A=[1; 0; 0]$, $B=[0; 1; 0]$, $C=[0; 0; 1]$ and $\Gamma$ is writen in canonic form $ax^2+by^2+cz^2=0$ (just let $ABC$ be a reference triangle).

Now let $P=[x_1; y_1; z_1]$, $Q=[x_2; y_2; z_2]$ and $R=[x_3; y_3; z_3]$. Denote new conic $\Gamma_1=(ax_1x_2x_3)yz+(by_1y_2y_3)xz+(cz_1z_2z_3)xy$. It is clear that $A$, $B$ and $C$ lie on $\Gamma_1$.

Substitude $P$ in $\Gamma_1$: $x_1y_1z_1(ax_2x_3+by_2y_3+cz_2z_3)$. $(ax_2x_3+by_2y_3+cz_2z_3)=0$ because $Q$ lies on the polar of $R$ with respect to $\Gamma$. Repeating this argument for $Q$ and $R$ we will insure that all six points lie on $\Gamma_1$.

Now we will see that $\Gamma_1$ is nonsingular. $Det(\Gamma_1)=\frac{1}{4}abcx_1x_2x_3y_1y_2y_3z_1z_2z_3$. $\Gamma$ is nondegenerate by assumption, thus $a, b$ and $c \neq 0$. Other terms are also different from zero due to neither three points from $A,B,C,P,Q,R$ are collinear. Therefore $Det(\Gamma_1) \neq 0$ and $\Gamma_1$ is nonsingular.

Then we can apply Pascal's and Brianchon's theorems to prove that edges of triangles are tangent to another conic, like here: Two triangles are inscribed in conic iff their edges are tangent to another conic.

Is everything fine here?

Thanks!

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  • $\begingroup$ Looks good to me. Nice! I'm happy someone else besides me is doing coordinate-based proofs with a suitable wlog in front. $\endgroup$ – MvG Dec 14 '17 at 18:56
  • $\begingroup$ Did you solve the question about infinite dimensional closed subspace of $L^1(\mathbb{R})$ consisting of only continuous functions? Do you still need a hint? $\endgroup$ – orole Jan 18 '18 at 22:01

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