<s>[INST] Centroid of a Triangle and Cosine Law In $\triangle ABC$, $M$ and $N$ are midpoints of $BC$ and $CA$ respectively such that $AM=14$ and $BN=8$. If $\angle C= 60^{\circ}$, find the length of $AB$.
For simplicity sake, let $x=AB$, $y=BM=MC$ and $z=AN=NC$. Then by Cosine Law, we have
$$8^2=z^2+4y^2-4yz\cos 60^{\circ},$$
$$14^2=4z^2+y^2-4yz\cos 60^{\circ},$$ and 
$$x^2=4z^2+4y^2-8yz\cos 60^{\circ}.$$ 
Subtracting the 2nd equation from 1st gives $z^2-y^2=(14^2-8^2)/3=44,$ and subtracting the 1st and the 2nd from the 3rd gives $x^2+y^2+z^2=14^2+8^2=260.$ 
But then I couldn't get further. I observed that the intersection of $BN$ and $AM$ is the centroid of the triangle but I don't know if this information would help, please help, thanks!
 [/INST] $$z^2-y^2=44 \tag{1}$$ Adding first two equations we get
$$5(y^2+z^2)-260=4yz \tag{2}$$ ans third equation can be written as
$$4(y^2+z^2)-x^2=4yz \tag{3} $$ Now subtracting $(3)$ from $(2)$ we get
$$y^2+z^2=260-x^2 \tag{4}$$
From $(1)$ and $(4)$ we get
$$2z^2=304-x^2 \tag{5}$$ and
$$2y^2=216-x^2 \tag{6}$$
Now substitute $y^2+z^2$ from $(4)$ in $(2)$ we get
$$1040-5x^2=4yz$$ squaring above equation both sides we get
$$(1040-5x^2)^2=4(2y^2)(2z^2)$$ $\implies$
$$(1040-5x^2)^2=4(304-x^2)(216-x^2)$$ from which you will get quadratic in $x^2$
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