In [convex](https://artofproblemsolving.com/wiki/index.php/Convex) [quadrilateral](https://artofproblemsolving.com/wiki/index.php/Quadrilateral) $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The [perimeter](https://artofproblemsolving.com/wiki/index.php/Perimeter) of $ABCD$ is $640$. Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest [integer](https://artofproblemsolving.com/wiki/index.php/Integer) that is less than or equal to $x.$)
Solution 1

By the [Law of Cosines](https://artofproblemsolving.com/wiki/index.php/Law_of_Cosines) on $\triangle ABD$ at angle $A$ and on $\triangle BCD$ at angle $C$ (note $\angle C = \angle A$),
\[180^2 + AD^2 - 360 \cdot AD \cos A = 180^2 + BC^2 - 360 \cdot BC \cos A\]
\[(AD^2 - BC^2) = 360(AD - BC) \cos A\]
\[(AD - BC)(AD + BC) = 360(AD - BC) \cos A\]
\[(AD + BC) = 360 \cos A\]
We know that $AD + BC = 640 - 360 = 280$. $\cos A = \dfrac{280}{360} =  \dfrac{7}{9} = 0.777 \ldots$
$\lfloor 1000 \cos  A \rfloor = \boxed{777}$.

Solution 2
Notice that $AB = CD$, and $BD = DB$, and $\angle{DAB} \cong \angle{BCD}$, so we have side-side-angle matching on triangles $ABD$ and $CDB$.  Since the problem does not allow $\triangle{ABD} \cong \triangle{CDB}$, we know that $\angle{ADB}$ is not a right angle, and there is a unique other triangle with the matching side-side-angle.
Extend $AD$ to $C'$ so that $\triangle{ABC'}$ is isosceles with $AB = C'B$.  Then notice that $\triangle{DC'B}$ has matching side-side-angle, and yet $\triangle{ADB} \not\cong \triangle{C'DB}$ because $\angle{ADB}$ is not right.  Therefore $\triangle{C'DB}$ is the unique triangle mentioned above, so $\triangle{CDB}$ is congruent, in some order of vertices, to $\triangle{C'DB}$.  Since $\triangle{CDB} \cong \triangle{C'DB}$ would imply $\triangle{CDB} = \triangle{C'DB}$, making quadrilateral $ABCD$ degenerate, we must have $\triangle{CDB} \cong \triangle{C'BD}$.
Since the perimeter of $ABCD$ is $640$, $AD + BC = 640 - 180 - 180 = 280$.  Hence $280 = AD + BC = AD + DC'$.  Drop the altitude of $\triangle{ABC'}$ from $B$ and call the foot $P$.  Then right triangle trigonometry on $\triangle{APB}$ shows that $\cos{A} = AP/AB = 140/180 = 7/9$, so $\lfloor 1000 \cos A \rfloor = \boxed{777}$.