SAT1000 - P801

Circle is $X^2+Y^2=2$ , where $X=x-2,Y=y$ .

Let the equation of a tangent to the circle from $P$ be $Y=mX+\sqrt {2(1+m^2)}$ . Then

$m^2(X^2-2)-2mXY+(Y^2-2)=0$ .

From the given condition, the product of the two roots of this equation for $m$ is $\dfrac {1}{2}$ .

So $2Y^2-X^2=2$ .

Substituting values of $X, Y$ and simplifying we get $5x^2-8x-36=0$ .

Therefore, $x_1=-2,y_1=-3,x_2=-2,y_2=3,x_3=3.6, y_3\approx -1.5099,x_4=3.6,y_4\approx 1.5099,N=4$ ,

$M=\displaystyle \sum_{k=1}^4 k(x_k+y_k)\approx 23.7096, M+N\approx 27.7096$ , and $\lfloor 1000(M+N)\rfloor =\boxed {27709}$ .

Let $B$ and $D$ be the tangent points, and draw $\triangle ABC$ and $\triangle ADC$ .

Since the equation for circle $C$ , $x^2 + y^2 -4x + 2 = 0$ , can be rearranged to $(x - 2)^2 + y^2 = 2$ , its center is $C(2, 0)$ and its radius is $BC = CD = \sqrt{2}$ .

Since $\triangle ABC$ and $\triangle ADC$ are congruent by HL congruence, let $\theta = \angle BAC = \angle DAC$ .

Let the $x$ -coordinate of $A$ be $p$ . Since $A$ is on the equation $\frac{x^2}{16} + \frac{y^2}{12} = 1$ , its $y$ -coordinate is $\frac{1}{2}\sqrt{48 - 3p^2}$ .

The slope of $AC$ is $m = \frac{\frac{1}{2}\sqrt{48 - 3p^2}}{p - 2}$ or $m = \frac{\sqrt{48 - 3p^2}}{2p - 4}$ .

By the distance equation, $AC = \sqrt{(p - 2)^2 + (\frac{1}{2}\sqrt{48 - 3p^2})^2} = \frac{1}{2}|p - 8|$ , and by the Pythagorean Theorem, $AB = \sqrt{AC^2 - BC^2} = \sqrt{(\frac{1}{2}|p - 8|)^2 + (\sqrt{2})^2} = \frac{1}{2}\sqrt{p^2 - 16p + 56}$ . Therefore, $\tan \theta = \frac{BC}{AB} = \frac{\sqrt{2}}{\frac{1}{2}\sqrt{p^2 - 16p + 56}}$ or $\tan \theta = \frac{2\sqrt{2}}{\sqrt{p^2 - 16p + 56}}$ .

Let $m_1$ be the slope of $AB$ and $m_2$ be the slope of $AD$ . From the equation for the angle between two slopes, $\tan \theta = \frac{m_1 - m}{1 + m_1m}$ and $\tan \theta = \frac{m - m_2}{1 + mm_2}$ . These equations rearrange to $m_1 = \frac{m + \tan \theta}{1 - m\tan \theta}$ and $m_2 = \frac{m - \tan \theta}{1 + m\tan \theta}$ .

Since $m_1m_2 = \frac{1}{2}$ , $(\frac{m + \tan \theta}{1 - m\tan \theta})(\frac{m - \tan \theta}{1 + m\tan \theta}) = \frac{1}{2}$ or $\frac{m^2 - \tan^2 \theta}{1 - m^2\tan^2 \theta} = \frac{1}{2}$ . Substituting $m = \frac{\sqrt{48 - 3p^2}}{2p - 4}$ and $\tan \theta = \frac{2\sqrt{2}}{\sqrt{p^2 - 16p + 56}}$ and simplifying gives $(p - 8)^2(p + 2)(5p - 18) = 0$ , which solves to $p = -2$ and $p = \frac{18}{5}$ for $-4 < p < 4$ .

Each $p$ -value has two possible $y$ -coordinate values for the ellipse, for a total of $N = 4$ possible solutions. Using $y = \pm \frac{1}{2}\sqrt{48 - 3x^2}$ , the possible coordinates for $A$ are $(-2, -3)$ , $(-2, 3)$ , $(\frac{18}{5}, -\frac{\sqrt{57}}{5})$ , and $(\frac{18}{5}, \frac{\sqrt{57}}{5})$ .

Therefore, $M = 1(-2 + -3) + 2(-2 + 3) + 3(\frac{18}{5} + -\frac{\sqrt{57}}{5}) + 4(\frac{18}{5} + \frac{\sqrt{57}}{5}) = \frac{111 + \sqrt{57}}{5}$ , $M + N = \frac{131 + \sqrt{57}}{5}$ , and $\lfloor 1000(M + N) \rfloor = \boxed{27709}$ .