SAT1000 - P802

Let $D$ be the point of tangency to the circle on $AP$ , and $E$ be the point of tangency to the circle on $BP$ .

Since $P$ is on the parabola $x^2 = y$ , let its coordinates be $P(p, p^2)$ . Since $M$ is the center of circle $x^2 + (y - 4)^2 = 1$ , its coordinates are $M(0, 4)$ and the radius is $DM = EM = 1$ .

The slope of $MP$ is then $m_{MP} = \frac{p^2 - 4}{p}$ , and the distance $MP$ is $d = \sqrt{p^2 + (p^2 - 4)^2} = \sqrt{p^4 - 7p^2 + 16}$ .

By the Pythagorean Theorem on $\triangle DMP$ , $DP = \sqrt{MP^2 - DM^2} = \sqrt{(\sqrt{p^4 - 7p^2 + 16})^2 + 1^2} = \sqrt{p^4 - 7p^2 + 15}$ . Let $\theta = \angle DPM = \angle EPM$ . Then $\tan \theta = \frac{DM}{DP} = \frac{1}{\sqrt{p^4 - 7p^2 + 15}}$ .

By the angle between two slopes equation, $\tan \theta = \frac{m_{AP} - m_{MP}}{1 + m_{AP}m_{MP}}$ and $\tan \theta = \frac{m_{MP} - m_{BP}}{1 + m_{MP}m_{BP}}$ . Substituting $m_{MP} = \frac{p^2 - 4}{p}$ and $\tan \theta = \frac{1}{\sqrt{p^4 - 7p^2 + 15}}$ and solving gives $m_{AP} = \frac{(p^2 - 4)\sqrt{p^4 - 7p^2 + 15} + p}{p\sqrt{p^4 - 7p^2 + 15} - (p^2 - 4)}$ and $m_{BP} = \frac{(p^2 - 4)\sqrt{p^4 - 7p^2 + 15} - p}{p\sqrt{p^4 - 7p^2 + 15} + (p^2 - 4)}$ .

The equation of line $AP$ is $y = m_{AP}(x - p) + p^2$ and the equation of line $BP$ is $y = m_{BP}(x - p) + p^2$ . Both $A$ and $B$ are also on $x^2 = y$ , and these equations solve to $A(\frac{\sqrt{p^4 - 7p^2 + 15} - 3p}{p^2 - 1}, \frac{p^4 + 2p^2 + 15 - 6p\sqrt{p^4 - 7p^2 + 15}}{(p^2 - 1)^2})$ and $B(\frac{\sqrt{p^4 - 7p^2 + 15} + 3p}{p^2 - 1}, \frac{p^4 + 2p^2 + 15 + 6p\sqrt{p^4 - 7p^2 + 15}}{(p^2 - 1)^2})$ .

The slope of $AB$ is then $m_{AB} = \frac{B_y - A_y}{B_x - A_x} = \frac{-6p}{p^2 - 1}$ . Since $MP \perp AB, m_{MP}m_{AB} = -1$ or $\frac{p^2 - 4}{p}\frac{-6p}{p^2 - 1} = -1$ , which solves to $p = \pm \frac{\sqrt{115}}{5}$ .

That means $m_{MP} = \frac{p^2 - 4}{p} = \frac{(\pm \frac{\sqrt{115}}{5})^2 - 4}{\pm \frac{\sqrt{115}}{5}} = \pm \frac{3\sqrt{115}}{115}$ , so that the line $l$ has an equation of $y = \pm \frac{3\sqrt{115}}{115}x + 4$ , so $k = \frac{3\sqrt{115}}{115}$ , $b = 4$ , and $\lfloor 1000(k + b) \rfloor = \boxed{4279}$ .

Let the position coordinates of $P$ be $(t, t^2)$ . Then the slopes $m_1,m_2$ of the tangents to the circle from $P$ are

$\dfrac{t(t^2-4)\pm \sqrt {t^4-7t^2+15}}{t^2-1}$

Points $A, B$ will have position coordinates $(m_1-t,(m_1-t)^2)$ and $(m_2-t,(m_2-t)^2)$ .

Let the mid point of $\overline {AB}$ be $D(\frac{m_1+m_2-2t}{2},\frac{(m_1-t)^2+(m_2-t)^2}{2})$ .

Then $t^2-\dfrac{t^4+2t^2+15}{(t^2-1)^2}=\dfrac{t(t^2-1)}{6t}+\dfrac{1}{2}\implies 5t^6-18t^4-3t^2-92=0$ .

There are two real solutions to this equation : $t\approx \pm 2.145$ .

Hence the equation of $\overline {PD}$ is

$y=\dfrac {t^2-1}{6t}x+\dfrac{t^4+2t^2+15}{(t^2-1)^2}+\dfrac{1}{2}=4\pm 0.2797x$ .

So, $k=4,b=0.2797, k+b=4.2797$ and $\lfloor 1000(k+b)\rfloor =\boxed {4729}$ .