Maximal Quadrangle

Rotate the diagram and place on the coordinate plane so that $E$ is at the origin and $O$ is at $(\frac{1}{\sqrt{2}}, 0)$ .

The unit circle will then have the equation $(x - \frac{1}{\sqrt{2}})^2 + y^2 = 1$ .

Let the line through $AC$ have the equation $y = mx$ and the line through $BD$ have the equation $y = nx$ . Without loss of generality, let $n > m$ .

As the intersections of $(x - \frac{1}{\sqrt{2}})^2 + y^2 = 1$ and $y = mx$ , $A$ and $C$ have the coordinates of $A\bigg(\cfrac{1-\sqrt{m^2+2}}{\sqrt{2}(m^2+1)}, \cfrac{m(1-\sqrt{m^2+2})}{\sqrt{2}(m^2+1)}\bigg)$ and $C\bigg(\cfrac{1+\sqrt{m^2+2}}{\sqrt{2}(m^2+1)}, \cfrac{m(1+\sqrt{m^2+2})}{\sqrt{2}(m^2+1)}\bigg)$ .

Likewise, $B$ and $D$ have the coordinates of $B\bigg(\cfrac{1-\sqrt{n^2+2}}{\sqrt{2}(n^2+1)}, \cfrac{n(1-\sqrt{n^2+2})}{\sqrt{2}(n^2+1)}\bigg)$ and $D\bigg(\cfrac{1+\sqrt{n^2+2}}{\sqrt{2}(n^2+1)}, \cfrac{n(1+\sqrt{n^2+2})}{\sqrt{2}(n^2+1)}\bigg)$ .

The area of $\triangle AED$ is $A_{\triangle AED} = \frac{1}{2}|\overrightarrow{EA}_x \cdot \overrightarrow{ED}_y - \overrightarrow{ED}_x \cdot \overrightarrow{EA}_y| = \cfrac{1}{2}\left|\cfrac{1-\sqrt{m^2+2}}{\sqrt{2}(m^2+1)} \cdot \cfrac{n(1+\sqrt{n^2+2})}{\sqrt{2}(n^2+1)} - \cfrac{1+\sqrt{n^2+2}}{\sqrt{2}(n^2+1)} \cdot \cfrac{m(1-\sqrt{m^2+2})}{\sqrt{2}(m^2+1)}\right| = \cfrac{(n - m)(\sqrt{m^2 + 2} - 1)(\sqrt{n^2 + 2} + 1)}{4(m^2 + 1)(n^2 + 1)}$ .

Likewise, the area of the other triangles are $A_{\triangle DEC} = \cfrac{(n - m)(\sqrt{m^2 + 2} + 1)(\sqrt{n^2 + 2} + 1)}{4(m^2 + 1)(n^2 + 1)}$ , $A_{\triangle BEC} = \cfrac{(n - m)(\sqrt{m^2 + 2} + 1)(\sqrt{n^2 + 2} - 1)}{4(m^2 + 1)(n^2 + 1)}$ , and $A_{\triangle AEB} = \cfrac{(n - m)(\sqrt{m^2 + 2} - 1)(\sqrt{n^2 + 2} - 1)}{4(m^2 + 1)(n^2 + 1)}$ .

The area of quadrilateral $ABCD$ is then $A_{ABCD} = A_{\triangle AED} + A_{\triangle DEC} + A_{\triangle BEC} + A_{\triangle AEB} = \cfrac{(n-m)\sqrt{m^2+2}\sqrt{n^2+2}}{(m^2+1)(n^2+1)}$ .

The maximum area of quadrilateral $ABCD$ for a given $m$ -value will be when $\cfrac{dA_{ABCD}}{dn} = \cfrac{\sqrt{m^2 + 2}(mn^3 + 3mn + 2)}{\sqrt{n^2 + 2}(m^2 + 1)(n^2 + 1)^2} = 0$ , or when $mn^3 + 3mn + 2 = 0$ , which is when $n = \cfrac{\sqrt[3]{(\sqrt{m^2+1}-1)^2}-\sqrt[3]{m^2}}{\sqrt[3]{m(\sqrt{m^2+1}-1)}}$ .

The angle between $y = mx$ and $y = nx$ is $\tan \theta = \cfrac{n - m}{1 + mn}$ , which has a minimum for $m < 0$ when $\cfrac{d(\tan \theta)}{dm} = 0$ , which after substituting $n = \cfrac{\sqrt[3]{(\sqrt{m^2+1}-1)^2}-\sqrt[3]{m^2}}{\sqrt[3]{m(\sqrt{m^2+1}-1)}}$ solves to $m = -\cfrac{\sqrt{2}}{5}$ .

If $m = -\cfrac{\sqrt{2}}{5}$ , then $n = \cfrac{\sqrt[3]{(\sqrt{m^2+1}-1)^2}-\sqrt[3]{m^2}}{\sqrt[3]{m(\sqrt{m^2+1}-1)}} = \cfrac{\sqrt[3]{(\sqrt{(-\frac{\sqrt{2}}{5})^2+1}-1)^2}-\sqrt[3]{(-\frac{\sqrt{2}}{5})^2}}{\sqrt[3]{(-\frac{\sqrt{2}}{5})(\sqrt{(-\frac{\sqrt{2}}{5})^2+1}-1)}} = \sqrt{2}$ , and the area of the quadrilateral is $A_{ABCD} = \cfrac{(n-m)\sqrt{m^2+2}\sqrt{n^2+2}}{(m^2+1)(n^2+1)} = \cfrac{(\sqrt{2} + \frac{\sqrt{2}}{5})\sqrt{(-\frac{\sqrt{2}}{5})^2+2}\sqrt{(\sqrt{2})^2+2}}{((-\frac{\sqrt{2}}{5})^2+1)((\sqrt{2})^2+1)} = \cfrac{8\sqrt{26}}{27}$ .

Therefore, $u = 8$ , $v = 26$ , $w = 27$ , and $u + v + w = \boxed{61}$ .