<s>[INST] Find the area of the triangle defined in the figure provided Find the area of the triangle defined in the figure below

This question appeared in a math olympiad contest and was considered invalid later on without any specific reason. Is it solvable? If yes, provide the answer! 
 [/INST] 
In $\triangle ABC$,
$|BC|=a$, $|CA|=b$, $|AB|=c$,
points $D,E,F$ are midpoints,
$|D_1D_2|=|E_1E_2|=|F_1F_2|=2\,R$.
Using the power of a points $D,E$ and $F$ with respect to
the circumscribed circle with radius $R$, we have
\begin{align}
\tfrac{a}2\cdot\tfrac{a}2&=2\cdot(2\,R-2)
,\\
\tfrac{b}2\cdot\tfrac{b}2&=1\cdot(2\,R-1)
,\\
\tfrac{c}2\cdot\tfrac{c}2&=3\cdot(2\,R-3)
,\\
a&=4\sqrt{R-1}
,\\
b&=2\sqrt{2\,R-1}
,\\
c&=2\sqrt{6\,R-9}
,\\
S&=\frac{a\,b\,c}{4\,R}
=\frac{4\sqrt3}R\,\sqrt{(R-1)(2\,R-1)(2\,R-3)}\tag{1}\label{1}
,\\
S&=\tfrac14\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}
=4\sqrt{R\,(2\,R-3)}\tag{2}\label{2}
.
\end{align}
From $\eqref{1}=\eqref{2}$
we have
\begin{align}
3\,(R-1)(2\,R-1)(2\,R-3)
&=
R^3
,
\end{align}
and as it was noted 
in the comments, this results in a cubic equation 
\begin{align}
R^3-6\,R^2+9\,R-3&=0 \tag{3}\label{3}
.
\end{align}
Using substitution $R=2\,t+2$ and dividing the resulting equation by 2,
we get
\begin{align}
4\,t^3-3\,t-\tfrac12&=0
.
\end{align}
Using identity 
\begin{align}
4\,\cos^3\phi-3\,\cos\phi&=\cos3\phi
\end{align}
for $\cos3\phi=\tfrac12$, 
\begin{align}
t&=\cos\phi
=\cos\left(
\tfrac13(\arccos\tfrac12+2\pi k)
\right)
=\cos
\frac{\pi\,(1+6 k)}9
,\quad k=0,1,2
.
\end{align}
Thus there are three real solutions for \eqref{3},
\begin{align}
R_0&=2+2\cos\tfrac\pi9\approx 3.87938524157182
,\\
R_1&=2+2\cos\tfrac{7\pi}9\approx 0.467911113762044
,\\
R_2&=2+2\cos\tfrac{13\pi}9\approx 1.65270364466614
,
\end{align}
and the only valid solution that
satisfies condition $R>3$
is $R=2+2\cos\tfrac\pi9$, which gives the area
\begin{align}
S&=4\sqrt{R\,(2\,R-3)}
=4\sqrt2\sqrt{(\cos\tfrac\pi9+1)(4\,\cos\tfrac\pi9+1)}
\approx 17.1865547357625
.
\end{align}
</s>