I came up with this guess about three years ago, but I wrote it down for the first time about two months ago, and I don’t know how it can be proven, nor whether it was previously discovered or not. Whoever can help, please do so, thank you.
It is impossible for the six points to lie on a conic section if the triangle is not isosceles
On
In fact, this solution is very similar to the solution given @Blue. I just take a different parametrization, and the determinantal condition I use deals with a $4 \times 4$ determinant. But I use as well a CAS and get also a final condition where the determinant has a nice factorization, the different factors accounting for the different ways triangle $ABC$ is isosceles.
Consider the following figure
Fig. 1 : A particular case where an ellipse through the given points $A_1,A_2,B_1,B_2,C_1,C_2$ exists because $AB=AC$.
where, without loss of generality, we have taken the midpoint of $AB$ as the origin, the other vertices of the square built on $AB$ being $C_1=(-1,-2)$ and $C_2=(1,-2)$. Let
$$\vec{AC}=\pmatrix{u\\v}, \ \ \ \vec{BC}=\pmatrix{u-2\\v}$$
($v$ assumed $>0$, of course). As a consequence, the coordinates of the four other vertices are :
$$A_1=\pmatrix{1+v\\2-u}, \ A_2=\pmatrix{u+v-1\\2-u+v}\tag{1a}$$
$$B_1=\pmatrix{-1-v\\u}, \ B_2=\pmatrix{-1+u-v\\u+v}\tag{1b}$$
The general equation of a conic curve being ;
$$ax^2+by^2+2cxy+2dx+2ey+f=0$$
Taking into account the fact that the conic curve passes through $C_1$ and $C_2$, this equation can be reduced to :
$$(4e-4b-f)x^2+by^2+2cxy+4cx+2ey+f=0 \tag{2}$$
with four unknowns $b,c,e,f$. If we plug in (2) the different values $x=1+v, y=2-u$, etc... issued from (1a) and (1b), we get a linear system of four equations with four unknowns, which has a non-trivial solution iff its determinant is $0$.
Using a CAS (see below), this determinant can be obtained under a factorized form :
$$16(u-1)(u^2+v^2-4)(u^2+v^2-4u)\underbrace{(u^2+v^2-2u+8v+4)}_{(u-1)^2+3+v^2+8v > 0}\tag{3}$$
Its different factors can be placed in correspondence with the 3 different cases where triangle $ABC$ is "isosceles" :
$$\begin{cases}u&=&1\ \leftrightarrow \ AC=BC\\ u^2+v^2&=&4 \ \leftrightarrow \ AB=AC\\u^2+v^2-4u&=&0 \ \leftrightarrow \ AB=BC\end{cases}$$
SAGE program which has given formula (3)
A.<u,v>=QQ[]
B.<b,c,e,x,y>=A[]
ex=(4*e-4*b-1)*x^2+b*y^2+2*c*x*y+4*c*x+2*e*y+1;
A1=ex.subs(x=1+v,y=2-u);a1=A1.coefficients();
A2=ex.subs(x=u+v-1,y=2-u+v);a2=A2.coefficients();
B1=ex.subs(x=-1-v,y=u);b1=B1.coefficients();
B2=ex.subs(x=-1+u+v,y=u+v);b2=B2.coefficients();
M=matrix([a1,a2,b1,b2]);d=M.determinant();df=factor(d);df
(where in particular "subs" means "substitution").
You can execute this code yourself : just
copy this program into the edit window
then press "Evaluate" ;
On
For the case of an isosceles triangle, will there be a conic section that passes through the six points? Yes, this can be easily proven, more generally according to the inverse of Pascal's theorem
Note that any six points with an axis of symmetry that does not pass through any of them will lie on a conic section
Defining $\triangle ABC$ by $$A=(0,0) \qquad B=(c,0) \qquad C=(b\cos A, b\sin A) \tag{1}$$ we can express the coordinates of the six points as $$\begin{align} A+(A-B)^\perp \quad B+(A-B)^\perp \\ B+(B-C)^\perp \quad C+(B-C)^\perp \\ C+(C-A)^\perp \quad A+(C-A)^\perp \end{align} \tag{2}$$ where $v^\perp:= (-v_y,v_x)$ is the $90^\circ$ counter-clockwise rotation of vector $v$. Then, we recall that points $P$, $Q$, $R$, $S$, $T$, $U$ are co-conicial if and only if $$\left| \begin{array}{cccccc} P_x^2 & P_y^2 & P_x P_y & P_x & P_y & 1 \\ Q_x^2 & Q_y^2 & Q_x Q_y & Q_x & Q_y & 1 \\ R_x^2 & R_y^2 & R_x R_y & R_x & R_y & 1 \\ S_x^2 & S_y^2 & S_x S_y & S_x & S_y & 1 \\ T_x^2 & T_y^2 & T_x T_y & T_x & T_y & 1 \\ U_x^2 & U_y^2 & U_x U_y & U_x & U_y & 1 \end{array} \right| = 0 \tag{3}$$
A computer algebra system makes light work of expanding and simplifying the determinant condition to $$\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)\left(a^2+b^2+c^2+8bc\sin A\right) = 0 \tag{4}$$ Since the last factor is strictly positive, the condition is satisfied when, and only when, at least two of $a$, $b$, $c$ are equal: the triangle must be isosceles, so your guess was correct!
I don't doubt that this result exists in the literature somewhere, but that wouldn't diminish your insight.
Note: If the squares are attached "the other way", so that each overlaps the triangle itself, then we change the "$+$"s in $(2)$ to "$-$"s, and the counterpart of $(4)$ becomes: $$\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)\left(a^2+b^2+c^2-8bc\sin A\right) = 0 \tag{5}$$ The sign change in that last factor is crucial, as it allows for a continuum of non-isosceles solution triangles.