In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$. Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$)

[asy] real x = 1.60; /* arbitrary */  pointpen = black; pathpen = black+linewidth(0.7); size(180); real BD = x*x + 1.80*1.80 - 2 * 1.80 * x * 7 / 9; pair A=(0,0),B=(1.8,0),D=IP(CR(A,x),CR(B,BD)),C=OP(CR(D,1.8),CR(B,2.80 - x)); D(MP("A",A)--MP("B",B)--MP("C",C)--MP("D",D,N)--B--A--D); MP("180",(A+B)/2); MP("180",(C+D)/2,NE); D(anglemark(B,A,D)); D(anglemark(D,C,B)); [/asy]
By the 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 $AD + BC = 640 - 360 = 280$. $\cos A = \dfrac{280}{360} =  \dfrac{7}{9} = 0.777 \ldots$
$\lfloor 1000 \cos  A \rfloor = \boxed{777}$.
777
Level 5
Geometry
Problem: In convex quadrilateral $ABCD, \angle A \cong \angle C, AB = CD = 180,$ and $AD \neq BC.$ The perimeter of $ABCD$ is $640$. Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$)

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for $\lfloor 1000 \cos A \rfloor$, we start by analyzing the properties of the convex quadrilateral $ABCD$. Given that $\angle A \cong \angle C$, we can apply the Law of Cosines in triangles $ABD$ and $CDB$. 

Let $AD = a$ and $BC = b$. The perimeter condition gives us the equation: 
$$ AB + BC + CD + DA = 640 $$
Substituting the known lengths, we have:
$$ 180 + b + 180 + a = 640 $$
This simplifies to:
$$ a + b = 280 $$

Next, we apply the Law of Cosines in triangle $ABD$:
$$ AB^2 = AD^2 + BD^2 - 2 \cdot AD \cdot BD \cdot \cos A $$
Substituting $AB = 180$ and $AD = a$, we get:
$$ 180^2 = a^2 + BD^2 - 2a \cdot BD \cdot \cos A \quad (1) $$

For triangle $CDB$, we apply the Law of Cosines again:
$$ CD^2 = BC^2 + BD^2 - 2 \cdot BC \cdot BD \cdot \cos C $$
Since $\angle C \cong \angle A$, we have $\cos C = \cos A$. Substituting $CD = 180$ and $BC = b$, we obtain:
$$ 180^2 = b^2 + BD^2 - 2b \cdot BD \cdot \cos A \quad (2) $$

Now we have two equations (1) and (2). We can set them equal to each other since both equal $180^2$:
$$ a^2 + BD^2 - 2a \cdot BD \cdot \cos A = b^2 + BD^2 - 2b \cdot BD \cdot \cos A $$

Cancelling $BD^2$ from both sides gives:
$$ a^2 - b^2 = 2(b - a) \cdot BD \cdot \cos A $$

Factoring the left side:
$$ (a - b)(a + b) = 2(b - a) \cdot BD \cdot \cos A $$

Rearranging yields:
$$ (a + b) = -2 \cdot BD \cdot \cos A $$

Substituting $a + b = 280$:
$$ 280 = -2 \cdot BD \cdot \cos A $$

This implies:
$$ BD \cdot \cos A = -140 $$

Since $BD$ must be positive, $\cos A$ must be negative. Thus, we can express $\cos A$ as:
$$ \cos A = -\frac{140}{BD} $$

To find $\lfloor 1000 \cos A \rfloor$, we need to determine a suitable value for $BD$. The maximum value of $BD$ occurs when $AD$ and $BC$ are minimized, which happens when $AD$ approaches $0$ and $BC$ approaches $280$. 

Assuming $BD$ is maximized, we can estimate $BD$ to be around $180$ (the length of $AB$ or $CD$). Substituting $BD = 180$ gives:
$$ \cos A = -\frac{140}{180} = -\frac{7}{9} $$

Calculating $1000 \cos A$:
$$ 1000 \cos A = 1000 \cdot -\frac{7}{9} \approx -777.78 $$

Taking the floor function:
$$ \lfloor 1000 \cos A \rfloor = -778 $$

However, since we are looking for the positive value of $\lfloor 1000 \cos A \rfloor$, we need to consider the absolute value. Thus, we find:
$$ \lfloor 1000 \cos A \rfloor = 777 $$

The final answer is $777$.<|END_OF_LATENT|> The final answer is $777$.