, 2016

Solution:
Consider a triangle $A B C$ with Fermat point $P$ such that $A P=a$, $B P=b$, $C P=c$. Then
$$A B^{2}=A P^{2}+B P^{2}-2 A P \cdot B P \cos \left(120^{\circ}\right)$$
by the Law of Cosines, which becomes
$$A B^{2}=a^{2}+a b+b^{2}$$
and hence $A B=3$. Similarly, $B C=\sqrt{52}$ and $A C=7$.
Furthermore, we have
$$\begin{aligned} B C^{2}=52 & =A B^{2}+B C^{2}-2 A B \cdot B C \cos \angle B A C \\ & =3^{2}+7^{2}-2 \cdot 3 \cdot 7 \cos \angle B A C \\ & =58-42 \cos \angle B A C \end{aligned}$$
And so $\cos \angle B A C=\frac{1}{7}$.
Invert about $A$ with arbitrary radius $r$. Let $B'$, $P'$, $C'$ be the images of $B$, $P$, $C$ respectively. Since $\angle A P B=\angle A B' P'=120^{\circ}$ and $\angle A P C=\angle A C' P'=120^{\circ}$, we note that $\angle B' P' C'=120^{\circ}-\angle B A C$, and so
$$\begin{aligned} \cos \angle B' P' C' & =\cos \left(120^{\circ}-\angle B A C\right) \\ & =\cos 120^{\circ} \cos \angle B A C-\sin 120^{\circ} \sin \angle B A C \\ & =-\frac{1}{2}\left(\frac{1}{7}\right)+\frac{\sqrt{3}}{2}\left(\frac{4 \sqrt{3}}{7}\right) \\ & =\frac{11}{14} \end{aligned}$$
Furthermore, using the well-known result
$$B' C'=\frac{r^{2} B C}{A B \cdot A C}$$
for an inversion about $A$, we have
$$\begin{aligned} B' P' & =\frac{B P r^{2}}{A B \cdot A P} \\ & =\frac{b r^{2}}{a \cdot 3} \\ & =\frac{b r^{2}}{3 a} \end{aligned}$$
and similarly $P' C'=\frac{c r^{2}}{7 a}$, $B' C'=\frac{r^{2} \sqrt{52}}{21}$. Applying the Law of Cosines to $B' P' C'$ gives us
$$\begin{aligned} B' C'^{2} & =B' P'^{2}+P' C'^{2}-2 B' P' \cdot P' C' \cos \left(120^{\circ}-\angle B A C\right) \\ \Longrightarrow \frac{52 r^{4}}{21^{2}} & =\frac{b^{2} r^{4}}{9 a^{2}}+\frac{c^{2} r^{4}}{49 a^{2}}-\frac{11 b c r^{4}}{147 a^{2}} \\ \Longrightarrow \frac{52}{21^{2}} & =\frac{b^{2}}{9 a^{2}}+\frac{c^{2}}{49 a^{2}}-\frac{11 b c}{147 a^{2}} \\ \Longrightarrow \frac{52}{21^{2}} & =\frac{49 b^{2}-33 b c+9 c^{2}}{21^{2} a^{2}} \end{aligned}$$
and so $\frac{49 b^{2}-33 b c+9 c^{2}}{a^{2}}=52$.