<s>[INST] Parallelogram with an angle and the midpoint of a side $M$ is the midpoint of the side $AB$ of a parallelogram $ABCD$. Find the perimeter of the parallelogram if $MD=4,MC=6$ and $\measuredangle BAD=60^\circ$.

Let $M$ be the midpoint of $AB\Rightarrow AM=BM=x$ and $AD=BC=y$. We are supposed to find $P_{ABCD}=2(AB+AD)=2(2x+y)=4x+2y$. Two cosine rules for $\triangle AMD$ and $\triangle MBC$ give $$DM^2=AM^2+AD^2-2AM.AD\cos60^\circ\\x^2+y^2-xy=16\\MC^2=MB^2+BC^2-2MB.BC\cos120^\circ\\x^2+y^2+xy=36$$ This seems useless for the perimeter (we can find the area, though).
 [/INST] There are two possible values of perimeter.
$x^2+y^2-xy=16$, $x^2+y^2+xy=36$ $\Rightarrow x^2+y^2=26$, $xy=10$ $\Rightarrow (x+y)^2=46$, $(x-y)^2=6$ $\Rightarrow$
$x+y=\sqrt{46}$, $x-y=\pm\sqrt{6}$ $\Rightarrow 4x+2y=3(x+y)+(x-y)=3\sqrt{46}\pm\sqrt{6}$.
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