Given an arbitrary triangle as 3 Feuerbach points, can we always reconstruct the original triangle?

Question

There are 4 tangency points $F,F_A,F_B,F_C$ of the nine-point circle with the incircle or excircles. Given an arbitrary triangle as 3 of these points, we can always reconstruct the original triangle (non-uniquely).

(It is difficult to distinguish the incircle with the excircles so we won't distinguish $F$ with $F_A,F_B,F_C$. The only requirement is that the triangle has the given points as three points among $F,F_A,F_B,F_C$.)

Analogous to "The incenter $I$ and excenters $J_A,J_B,J_C$ form an orthogonal system", I try to deduce relations of the 4 points $F,F_A,F_B,F_C$:

Let the Nine-point circle be the unit circle, let the incircle touches the sides at $X$, $Y$, and $Z$.

From this post, the center of the nine-point circle is $N=\frac{(X Y+Y Z+ZX)^2}{(X+Y)(Y+Z)(Z+X) }$, and the Feuerbach point is $F=\frac{X Y + Y Z + Z X}{X + Y + Z}$, the tangency point of the Nine-Point Circle and the $A$-excircle is $F_1=\frac{-X^2 Y^2+X^2 Y Z+X^3 Y-X^2 Z^2+X^3 Z-2 X Y^2 Z-2 X Y Z^2-Y^2 Z^2}{(X+Y) (X+Z) (X-Y-Z)}$, and the radius of the Nine-Point Circle is $-\frac{X Y Z}{(X + Y) (Y + Z) (Z + X)}$.

$z_0=F-N=-\frac{X Y Z (X Y+X Z+Y Z)}{(X+Y) (X+Z) (Y+Z) (X+Y+Z)}$

$z_1=F_1-N=\frac{X Y Z (X Y+X Z-Y Z)}{(X+Y) (X+Z) (Y+Z) (X-Y-Z)}$ and by cyclic permutation $z_2=\dots,z_3=\dots$

We eliminate $X$, $Y$, $Z$ from $z_0,z_1,z_2,z_3$ and get $$2 p^2 q s-4 p^2 r^2+p q^2 r+4 p r s+2 q r^2=0\tag1\label1$$ where $p, q, r, s$ are elementary symmetric polynomials in $z_0,z_1,z_2,z_3$.

So \eqref{1} is a symmetric homogeneous polynomial of degree $4$ in $z_0,z_1,z_2,z_3$. Write it as a polynomial $p(z_0)$.

Let $F(s)=p(\frac{s}{z_1z_2z_3})$ be a polynomial with coefficients in $z_1,z_2,z_3$.

Surprisingly we can verify that, $F(s)=s^4\overline{F(1/\overline{s})}$, so $F(s)$ is a Conjugate reciprocal polynomial.

So $F(s)=0$ iff $F(1/\overline{s})=0$.

Let $f(x)=(1+ix)^4F(\frac{1-ix}{1+ix})$, then $f(x)$ is a polynomial with coefficients in $\mathbb{R}$.

There is a bijection between roots of $f(x)$ and roots of $p(z_0)$. $$\frac{1-ix}{1+ix}\frac1{z_1z_2z_3}=z_0$$ There is a bijection between real roots of $f(x)$ and roots of $p(z_0)$ on $|z_0|=1$.

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Returning to the original question, since the Feuerbach point $z_0$ is on the Nine-point circle, so $|z_0|=1$.

Given unit complex numbers $z_1,z_2,z_3$, then $z_0$ is a root of \eqref{1} on $|z_0|=1$, corresponding to a real root of $f(x)$.

I try to find example of unit complex numbers $z_1,z_2,z_3$ such that there is no root $z_0$ of \eqref{1} with $|z_0|=1$, so that there does not exist triangle with $z_1,z_2,z_3$ as the Feuerbach Triangle.

Mathematica code to generate a random sample of unit complex numbers $z_1,z_2,z_3$:

Block[{z1=Exp[I RandomReal[2Pi]],z2=Exp[I RandomReal[2Pi]],z3=Exp[I RandomReal[2Pi]]},Chop[Abs[z0]]/.NSolve[{0==2 p^2 q s-4 p^2 r^2+p q^2 r+4 p r s+2 q r^2/. {p->SymmetricPolynomial[1,{z0,z1,z2,z3}],q->SymmetricPolynomial[2,{z0,z1,z2,z3}],r->SymmetricPolynomial[3,{z0,z1,z2,z3}],s->SymmetricPolynomial[4,{z0,z1,z2,z3}]}}]]

Running the test a few times, there is always a root $z_0$ of \eqref{1} with $|z_0|=1$. So I didn't find a counterexample to the original problem.

Equivalently, to find unit complex numbers $z_1,z_2,z_3$ such that $f(x)$ has no real roots. Here I used this condition for the existence of real root of the degree 4 polynomial $f(x)$ but it is taking too long to compute.

Mathematica code to generate a random sample to test this:

Block[{z1=Exp[I RandomReal[2Pi]],z2=Exp[I RandomReal[2Pi]],z3=Exp[I RandomReal[2Pi]]},CountRoots[Chop[Simplify[(2 p^2 q s-4 p^2 r^2+p q^2 r+4 p r s+2 q r^2/. {p->SymmetricPolynomial[1,{z0,z1,z2,z3}],q->SymmetricPolynomial[2,{z0,z1,z2,z3}],r->SymmetricPolynomial[3,{z0,z1,z2,z3}],s->SymmetricPolynomial[4,{z0,z1,z2,z3}]}/.z0->((1-I x)/(1+I x))/(z1 z2 z3))(1+I x)^4]],x]]

Running the test a few times, $f(x)$ always have a real root. Continue to construct the original triangle, I see that sometimes $z_1,z_2,z_3$ are all on the excircles of the original triangle, sometimes one of them is on the incircle of the original triangle.

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Footnote:

Here is an alternative proof of \eqref{1}, using the property $LFe+MFe=NFe$ proven here.

Let the incircle be the unit circle, $l,m,n,z$ be unit complex numbers corresponding to points $L,M,N,Fe$.

Substitute $LFe=|l-z|,MFe=|m-z|,NFe=|n-z|$ into the equation $(LFe+MFe+NFe)(LFe+MFe-NFe)(LFe-MFe-NFe)(LFe-MFe+NFe)=0$

(Mathematica code:

FullSimplify/@CoefficientList[Expand[(l^2 m^2 n^2)/(l^2 (m-n)^2+m^2 n^2-2 l m n (m+n)) z^2 (LFe+MFe+NFe)(LFe+MFe-NFe)(LFe-MFe-NFe)(LFe-MFe+NFe)/.{LFe->Sqrt[(l-z)(1/l-1/z)],MFe->Sqrt[(m-z)(1/m-1/z)],NFe->Sqrt[(n-z)(1/n-1/z)]}],z]

) we get \begin{align*}\frac{l^2 m^2 n^2 \left(l^2-2 l (m+n)+(m-n)^2\right)}{l^2 (m-n)^2-2 l m n (m+n)+m^2 n^2}\\+\frac{4 l^2 m^2 n^2 (l+m+n)}{l^2 (m-n)^2-2 l m n (m+n)+m^2 n^2}x\\-\frac{2 l m n (l+m+n) (l (m+n)+m n)}{l^2 (m-n)^2-2 l m n (m+n)+m^2 n^2}x^2\\+\frac{4 l m n (l (m+n)+m n)}{l^2 (m-n)^2-2 l m n (m+n)+m^2 n^2}x^3\\+x^4=0\end{align*} Let $p, q, r, s$ be elementary symmetric polynomials in the 4 roots $z_0,z_1,z_2,z_3$. Then \begin{align*}s=\frac{l^2 m^2 n^2 \left(l^2-2 l (m+n)+(m-n)^2\right)}{l^2 (m-n)^2-2 l m n (m+n)+m^2 n^2}\\r=-\frac{4 l^2 m^2 n^2 (l+m+n)}{l^2 (m-n)^2-2 l m n (m+n)+m^2 n^2}\\q=-\frac{2 l m n (l+m+n) (l (m+n)+m n)}{l^2 (m-n)^2-2 l m n (m+n)+m^2 n^2}\\p=-\frac{4 l m n (l (m+n)+m n)}{l^2 (m-n)^2-2 l m n (m+n)+m^2 n^2}\end{align*} Substitute $p,q,r,s$ into LHS of \eqref{1} and simplify, we get $0$.

(Mathematica code:

2 p^2 q s-4 p^2 r^2+p q^2 r+4 p r s+2 q r^2/.{p->-((4 l m n (m n+l (m+n)))/(l^2 (m-n)^2+m^2 n^2-2 l m n (m+n))),q->-((2 l m n (l+m+n) (m n+l (m+n)))/(l^2 (m-n)^2+m^2 n^2-2 l m n (m+n))),r->-((4 l^2 m^2 n^2 (l+m+n))/(l^2 (m-n)^2+m^2 n^2-2 l m n (m+n))),s->(l^2 m^2 n^2 (l^2+(m-n)^2-2 l (m+n)))/(l^2 (m-n)^2+m^2 n^2-2 l m n (m+n))}//Simplify

) So we proved \eqref{1}.