SAT1000 - P807

Let the slope of $AB$ be $m$ . Since it goes through the focus at $F(1, 0)$ , the line $AB$ has the equation $y = m(x - 1)$ , and since $AB$ is perpendicular to $DE$ at $F$ , the line $DE$ has the equation $y = -\frac{1}{m}(x - 1)$ .

The intersection points of these lines with the parabola $y^2 = 4x$ are $A((\frac{1 + \sqrt{m^2 + 1}}{m})^2, 2(\frac{1 + \sqrt{m^2 + 1}}{m}))$ , $B((\frac{1 - \sqrt{m^2 + 1}}{m})^2, 2(\frac{1 - \sqrt{m^2 + 1}}{m}))$ , $D((m - \sqrt{m^2 + 1})^2, -2(m - \sqrt{m^2 + 1}))$ , and $E((m + \sqrt{m^2 + 1})^2, -2(m + \sqrt{m^2 + 1}))$ .

$\overrightarrow{AD} \cdot \overrightarrow{EB} = (D_x - A_x)(B_x - E_x) + (D_y - A_y)(B_y - E_y)$ , which simplifies to $\overrightarrow{AD} \cdot \overrightarrow{EB} = \frac{4(m^2 + 1)^2}{m^2}$ .

By the AM-GM inequality, the terms $m^2$ and $1$ have the relation $\frac{m^2 + 1}{2} \geq \sqrt{m^2 \cdot 1}$ , which can be rearranged to $\frac{4(m^2 + 1)^2}{m^2} \geq 16$ or $\overrightarrow{AD} \cdot \overrightarrow{EB} \geq 16$ .

Therefore, the minimum value of $\overrightarrow{AD} \cdot \overrightarrow{EB}$ is $M = 16$ , and $\lfloor 1000M \rfloor = \boxed{16000}.$

Let the position coordinates of $A, B, D, E$ be $(t^2,2t),(\frac{1}{t^2},-\frac{2}{t}),(t_1^2,2t_1),(\frac{1}{t_1^2},-\frac{2}{t_1})$ respectively. Then

$\vec {AD}=(t_1-t)\left (\hat i(t_1+t)+2\hat j\right )$

$\vec {EB}=(\frac{1}{t}-\frac{1}{t_1})\left (\hat i(\frac{1}{t_1}+\frac{1}{t})-2\hat j\right )$

$\implies \vec {AD}.\vec {EB}=\left (\frac{(t_1-t)^2}{tt_1}\right ) ^2$

Now, $t_1=\frac{t-1}{t+1}$ , or $-\frac{t+1}{t-1}$ . (since $\overline {AB}$ and $\overline {DE}$ are mutually perpendicular).

So $\vec {AD}.\vec {EB}=(\frac{t^4+2t^2+1}{t^3-t})^2$ .

This will be minimum when $t^6-5t^4-5t^2+1=0$ , the minimum value being $4^2=16$ . Therefore the required answer is $\boxed {16000}$ .