SAT1000 - P845

Let us call the coordinates of $A$ , $B$ , $C$ and $D$ as $(x_A, y_A)$ , $(x_B, y_B)$ , $(x_C, y_C)$ and $(x_D, y_D)$ , respectively.

To find the $y$ -values one just substitutes the parabola equation $x=y^2$ in the circle equation

$\large \displaystyle (y^2-4)^2 + y^2 = r^2$

$\large \displaystyle y^4-7y^2 + (16-r^2) = 0$

$\large \displaystyle y^2 = \frac{7 \pm \sqrt{4r^2-15}}{2}$

So, since $x = y^2$ , we have:

$\large \displaystyle (x_A,y_A) = \left( \frac{7 - \sqrt{4r^2-15}}{2}, \sqrt{\frac{7 - \sqrt{4r^2-15}}{2}} \right )$

$\large \displaystyle (x_B,y_B) = \left( \frac{7 - \sqrt{4r^2-15}}{2}, -\sqrt{\frac{7 - \sqrt{4r^2-15}}{2}} \right )$

$\large \displaystyle (x_C,y_C) = \left( \frac{7 + \sqrt{4r^2-15}}{2}, -\sqrt{\frac{7 + \sqrt{4r^2-15}}{2}} \right )$

$\large \displaystyle (x_D,y_D) = \left( \frac{7 + \sqrt{4r^2-15}}{2}, \sqrt{\frac{7 + \sqrt{4r^2-15}}{2}} \right )$

The quadrilateral $ABCD$ is actually a trapezoid, so its area (which I'll denote by $S)$ will be:

$\large \displaystyle S = \frac12 \left [ (y_D-y_C) + (y_A-y_B) \right ] \cdot (x_C - x_A)$

$\large \displaystyle S = \frac12 \left [ 2 \sqrt{\frac{7 + \sqrt{4r^2-15}}{2}} + 2 \sqrt{\frac{7 - \sqrt{4r^2-15}}{2}} \right ] \cdot \sqrt{4r^2-15}$

Since $S$ is an area and, so, always a positive value, maximize $S$ is the same as maximizing $S^2$ , which I'll call $T$ :

$\large \displaystyle T = S^2 = (4r^2-15) \cdot (7 + \sqrt{64-4r^2} )$

Maximizing:

$\large \displaystyle \frac{dT}{dr} = 8r(7 + \sqrt{64-4r^2}) - (4r^2-15) \frac{8r}{2\sqrt{64-4r^2}} = 0$

$\large \displaystyle 12r^2 - 143 = 14 \sqrt{64-4r^2}$

Squaring both sides:

$\large \displaystyle 144r^4 + 784r^2 + 7905 = 0$

$\large \displaystyle r^2 = \frac{527}{36}$ or $\large \displaystyle r^2 = \frac{15}{4}$

The condition $r^2 = \frac{15}{4}$ will actually make $S = 0$ , so it's the minimizing condition. The maximizing condition is, then:

$\color{#20A900} \boxed{\large \displaystyle r^2 = \frac{527}{36} }$

We will focus on points $B$ and $D$ , because $P$ is the intersection of $\overline{BD}$ with the $x$ -axis (the same could be made with the line $\overline{AC}$ ). So:

$\large \displaystyle (x_B,y_B) = \left( \frac{21-14\sqrt{2}}{6}, -\sqrt{\frac{21-14\sqrt{2}}{6}} \right )$

$\large \displaystyle (x_D,y_D) = \left( \frac{21+14\sqrt{2}}{6}, \sqrt{\frac{21+14\sqrt{2}}{6}} \right )$

The line $\overline{BD}$ has equation:

$\color{#20A900} \boxed{ \large \displaystyle x = \sqrt{ \frac{14}{3}} y + \frac76 }$

Just as a confirmation, points $A$ and $C$ have coordinates:

$\large \displaystyle (x_A,y_A) = \left( \frac{21-14\sqrt{2}}{6}, \sqrt{\frac{21-14\sqrt{2}}{6}} \right )$

$\large \displaystyle (x_C,y_C) = \left( \frac{21+14\sqrt{2}}{6}, -\sqrt{\frac{21+14\sqrt{2}}{6}} \right )$

And the line $\overline{AC}$ has equation:

$\color{#20A900} \boxed{ \large \displaystyle x = - \sqrt{ \frac{14}{3}} y + \frac76 }$

So:

$\color{#3D99F6} \large \displaystyle x_0 = \frac76 \rightarrow \boxed{ \large \displaystyle \lfloor 1000 x_0 \rfloor = 1166 }$