BE smart

Let $x$ = $AB = BC = CD = AD$ . By Pythagoras, $AC$ *= $x\sqrt { 2 }$ . Then, $EC = BC$ since $EC$ is a radius, and since $AC$ = $x\sqrt { 2 }$ , so $AE$ = $x\sqrt { 2 } - x$ . Let $M$ be a point in $AB$ such that $AB$ $\bot$ $ME$ as shown in the figure. Note that $\triangle AME$ is an isosceles triangle with $\angle EAM$ = $\angle AME$ = ${45}^{\circ}$ . This implies, by Thales, that $ME = AM = x - \frac { \sqrt { 2 } }{ 2 } x$ and $MB = \frac { \sqrt { 2 } }{ 2 } x$ . Once again, by Pythagoras in $\triangle MEB$ , $BE = \sqrt { { x }^{ 2 }(2-\sqrt { 2 } ) }$ . In order to find $x$ , it is known that area of $ABCD = {x}^{2}$ , but from a different perspective, note that shaded area is 121 and the whole area outside the circumference is 242. Area of quadrant of the circumference is $\frac { { x }^{ 2 }\pi }{ 4 }$ , we have the following equation: ${ x }^{ 2 }\quad=\quad 242\quad+\quad \frac { { x }^{ 2 }\pi }{ 4 }$ Hence, ${x}^{2} = \frac { 968 }{ 4-\pi }$ . Replacing this in $BE$ , we have that $BE \approx 25.7016...$ . Thus, $\left\lceil BE \right\rceil = 26$ .

Let $x$ be equal to the side of the square. Due to symmetry, we can say that the area of the square minus the area of the quadrant of the circle is equal to double the area of the shaded section, i.e.

${x}^{2}-\frac { \pi }{ 4 } {x}^{2}=242 \quad \Rightarrow \quad {x}^{2}(1-\frac { \pi }{ 4 } )=242$ $x=\sqrt { \frac { 242 }{ 1-\frac { \pi }{ 4 } } } \approx 33.581$

Consider $\triangle BEC$ . Line segments $EC$ and $BC$ are both radii of length $x$ so the triangle is isosceles. Using the sine rule we have

$\frac { x }{ \sin { 67.5° } } =\frac { \overline { BE } }{ \sin { 45° } }$ $\overline { BE } \approx 25.702$ $\left\lceil \overline { BE } \right\rceil =\large\color{#20A900}{\boxed{26}}$