<s>[INST] Given $\triangle ABC$ can we construct point $O$ such that $AO\times BC=BO\times AC=CO\times AB$? Given a $\triangle ABC$, is it possible to construct, with compass and straightedge, a point $O$ such that
$$AO\cdot BC=BO\cdot AC=CO\cdot AB$$
Does that point exist?
 [/INST] These points are known
in ETC
as 1st and 2nd isodynamic points,
the triangle centers $X_{15}$ and $X_{16}$.
Isodynamic point:

In Euclidean geometry, the isodynamic points of a triangle are points
associated with the triangle, with the properties that  ... the
distances from the isodynamic point to the triangle vertices are
inversely proportional to the opposite side lengths of the triangle.

The barycentric coordinates of these points are
\begin{align} 
X_{15}:\quad&
a\sin(\alpha + \tfrac\pi3) &: b\sin(\beta + \tfrac\pi3) : c \sin(\gamma + \tfrac\pi3)
,\\
X_{16}:\quad&
a\sin(\alpha - \tfrac\pi3) &: b\sin(\beta - \tfrac\pi3) : c \sin(\gamma - \tfrac\pi3)
.
\end{align}
As a linear combination of the vertices,
\begin{align} 
X_{15}&=\frac{u\cdot A+v\cdot B+w\cdot C}{u+v+w}
,\\
u&=a\sin(\alpha + \tfrac\pi3)
,\\
v&=b\sin(\beta + \tfrac\pi3)
,\\
w&=c \sin(\gamma + \tfrac\pi3)
, 
\end{align}
and the invariants are
\begin{align} 
a\cdot |AX_{15}|&=b\cdot |BX_{15}|=c\cdot |CX_{15}|
\\
&=
\frac{\sqrt2\,abc}{\sqrt{a^2+b^2+c^2+4\sqrt3 S}}
,\\
a\cdot |AX_{16}|&=b\cdot |BX_{16}|=c\cdot |CX_{16}|
\\
&=
\frac{\sqrt2\,abc}{\sqrt{a^2+b^2+c^2-4\sqrt3 S}}
,
\end{align}
where $S$ is the area of $\triangle ABC$.
Example: for the the nominal $6-9-13$ triangle,
\begin{align} 
a&=6,\quad b=9,\quad c=13,\quad S=4\sqrt{35}
,\\
a\cdot |AX_{15}|&=b\cdot |BX_{15}|=c\cdot |CX_{15}|
=\frac{702 \sqrt2}{\sqrt{286+16\sqrt{105}}}
\approx 46.80
,\\
a\cdot |AX_{16}|&=b\cdot |BX_{16}|=c\cdot |CX_{16}|
=\frac{702 \sqrt2}{\sqrt{286-16\sqrt{105}}}
\approx 89.86
.
\end{align}
Construction.

Points $A_b.A_e$ and $B_b,B_e$
are the feet of the internal and external bisectors
of the angles $CAB=\alpha$
and $ABC=\beta$, respectively.
Points
$O_a=\tfrac12(A_b+A_e)$,
$O_b=\tfrac12(B_b+B_e)$
are the centers of the circles
$\mathcal{C_a}$ and $\mathcal{C_b}$
through points
$A,A_b,A_e$
and $B,B_b,B_e$, respectively.
Intersection of the circles
$\mathcal{C_a}$ and $\mathcal{C_b}$
gives the pair of isodynamic points,
1st, $X_{15}$ inside $\triangle ABC$
and 2nd, $X_{16}$, outside of $\triangle ABC$.
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