Dynamic Geometry: P30

Note that the red loci are parabolas (see Dynamic Geometry: P28) . The area bounded by a parabola is $\frac 23$ of the area of the smallest rectangle inscribing the parabola.

Let radius of the circle be $1$ . Then we side length of the square is $\sqrt 2$ . The height of the circle above the top side of the square is $1-\dfrac {\sqrt 2}2 = \dfrac {2-\sqrt 2}2$ . Then the height of the parabola $h = \dfrac {2-\sqrt 2}4$ , a half of the height of circle above the top side of the square. Then the area of the rectangle is $\sqrt 2 h = \dfrac {\sqrt 2 -1}2$ and the area of a parabola is $\dfrac 23 \times \dfrac {\sqrt 2 - 1}2 = \dfrac {\sqrt 2 -1}3$ . The area of four parabolas, $A_p = \dfrac {4(\sqrt 2 -1)}3$ .

The area of the green rectangle is maximum, when it is a rectangle. Then its vertex is $\dfrac {\sqrt 2}2 + h = \dfrac {2+\sqrt 2}4$ away from its center. Its side length is $\dfrac {2+\sqrt 2}4 \times \sqrt 2 = \dfrac {\sqrt 2+1}2$ and its area $A_r = \left(\dfrac {\sqrt 2+1}2 \right)^2 = \dfrac {2\sqrt 2+3}4$ . And

$\frac {A_r}{A_p} = \frac {2\sqrt 2+3}4 \times \frac 3{4(\sqrt 2 -1)} = \frac {15\sqrt 2+21}{16}$

Therefore $\sqrt[3]{a-b-c+d} + 1 = \sqrt[3]8 + 1 = \boxed 3$ .

Let the center of the square be at $(0, 0)$ and the sides of the square be $2$ , so that its vertices are at $(\pm 1, \pm 1)$ , and the radius of the black circle is $\sqrt{2}$ .

Then consider the following right triangle:

whose hypotenuse is $\sqrt{2} - r$ and whose vertical leg is $y = 1 + r$ . By the Pythagorean Theorem, the horizontal leg is $x^2 = (\sqrt{2} - r)^2 - (1 + r)^2 = 1 - 2(1 + \sqrt{2})r$ .

Combining $y = 1 + r$ and $x^2 = (\sqrt{2} - r)^2 - (1 + r)^2 = 1 - 2(1 + \sqrt{2})r$ by eliminating $r$ gives $y = -\frac{1}{2}(\sqrt{2} - 1)x^2 + \frac{1}{2}(\sqrt{2} + 1)$ , a parabola with a vertex at $(0, \frac{1}{2}(\sqrt{2} + 1))$ , displayed by the top red curve in the diagram.

The area inside a parabola is $\frac{2}{3}$ the rectangle it is bounded in, so the area bounded by the blue square and the red curves is $A_1 = 4 \cdot \frac{2}{3} \cdot \frac{1}{2}(\sqrt{2} - 1) \cdot 2 = \frac{8}{3}(\sqrt{2} - 1)$ .

The area of the green rectangle will be a maximum when it is a square, and its vertices will be at the vertices of each parabola which are at $(0, \pm \frac{1}{2}(\sqrt{2} + 1)))$ and $(\pm \frac{1}{2}(\sqrt{2} + 1)), 0)$ , for an area of $A_2 = (\sqrt{2} \cdot \frac{1}{2}(\sqrt{2} + 1)))^2 = \frac{1}{2}(3 + 2\sqrt{2})$ .

The ratio of these areas is then $\cfrac{A_2}{A_1} = \cfrac{\frac{1}{2}(3 + 2\sqrt{2})}{\frac{8}{3}(\sqrt{2} - 1)} = \cfrac{15\sqrt{2} + 21}{16}$ , so $a = 15$ , $b = 2$ , $c = 21$ , $d = 16$ , and $\sqrt[3]{a - b - c + d} = \boxed{2}$ .