SAT1000 - P808

Equation of the tangent to the circle is $y=\dfrac{m+x}{\sqrt {m^2-1}}$ .

$x$ -coordinates of $A$ and $B$ are the roots of the equation $(m^2+3)x^2+8mx+4=0$ . Assuming them to be $x_1,x_2, x_1-x_2=\dfrac{4\sqrt {3(m^2-1)}}{m^2+3}$ .

The difference between the $y$ -coordinates of these points is

$y_1-y_2=\dfrac{x_1-x_2}{\sqrt {m^2-1}}=\dfrac{4\sqrt 3}{m^2+3}$

So, $|\overline {AB}|^2=(x_1-x_2)^2+(y_1-y_2)^2=\dfrac{48m^2}{m^4+6m^2+9}\leq \dfrac{48}{6+2\times 3}$ .

Hence the maximum value of $|\overline {AB}|$ is $2$ , and the answer is $\boxed {2000}$ .

Just an observation. C is a focus of the ellipse. Define D(-m,0). Nagel point for triangle ABD is (0,0).

Let the tangent point be $P$ with an $x$ -coordinate of $p$ . Since it is on the circle $x^2 + y^2 = 1$ , its $y$ -coordinate is $\sqrt{1 - p^2}$ .

The slope between $P$ and the origin is $\frac{\sqrt{1 - p^2}}{p}$ , so the slope perpendicular to that is $-\frac{p}{\sqrt{1 - p^2}}$ . Therefore, the equation of the line $AB$ is $y = -\frac{p}{\sqrt{1 - p^2}}(x - p) + \sqrt{1 - p^2}$ .

The intersection of $y = -\frac{p}{\sqrt{1 - p^2}}(x - p) + \sqrt{1 - p^2}$ and $\frac{x^2}{4} + y^2 = 1$ are at $A$ and $B$ with coordinates that solve to $(\frac{4p \pm 2\sqrt{3}\sqrt{p^2 - p^4}}{3p^2 + 1}, \frac{\sqrt{1 - p^2} \pm 2\sqrt{3}p^2}{3p^2 + 1})$ .

The distance $|AB|$ is then $\sqrt{(\frac{4p + 2\sqrt{3}\sqrt{p^2 - p^4}}{3p^2 + 1} - \frac{4p - 2\sqrt{3}\sqrt{p^2 - p^4}}{3p^2 + 1})^2 + (\frac{\sqrt{1 - p^2} + 2\sqrt{3}p^2}{3p^2 + 1} - \frac{\sqrt{1 - p^2} - 2\sqrt{3}p^2}{3p^2 + 1})^2} = \frac{4\sqrt{3p^2}}{3p^2 + 1}$ .

By the AM-GM inequality, the terms $3p^2$ and $1$ have the relation $\frac{3p^2 + 1}{2} \geq \sqrt{3p^2 \cdot 1}$ , which can be rearranged to $2 \geq \frac{4\sqrt{3p^2}}{3p^2 + 1}$ or $2 \geq |AB|$ .

Therefore, the maximum value of $|AB|$ is $M = 2$ , and $\lfloor 1000M \rfloor = \boxed{2000}$ .