A geometry problem by Mohammad Hamdar

First thing, Let $I=2\sin ^{ 4 }{ x } +2\cos ^{ 2 }{ x } +\frac { \sin ^{ 2 }{ 2x } }{ 2 }$ . Adding $cos ^{ 4 }{ x } -\cos ^{ 4 }{ x }$ to $I$ we get, $I=(\sin ^{ 4 }{ x } +\cos ^{ 4 }{ x } +\frac { \sin ^{ 2 }{ 2x } }{ 2 } )+(\sin ^{ 4 }{ x } -\cos ^{ 4 }{ x } +2\cos ^{ 2 }{ x } )$ $={ (\sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } ) }^{ 2 }+({ (1-\cos ^{ 2 }{ x } ) }^{ 2 }-\cos ^{ 4 }{ x } +2\cos ^{ 2 }{ x } )$ $=1+1=2$ . Therefore, $A=\sum _{ k=1 }^{ 99 }{ { 2 }^{ k } } ={ 2 }^{ 100 }-2$ and so, $\left\lceil \frac { A }{ { 4 }^{ 49 } } \right\rceil =4$ . Looking through the answers we notice that $\sqrt { 6+\sqrt { 32 } } +\sqrt { 6-\sqrt { 32 } } =4$

Let $r= \sqrt { 6+\sqrt { 32 } } +\sqrt { 6-\sqrt { 32 } }$ then, ${ r }^{ 2 }=(6+\sqrt { 32 } )+(6-\sqrt { 32 } )+2\sqrt { 6+\sqrt { 32 } } \sqrt { 6-\sqrt { 32 } } =12+2\sqrt { { 6 }^{ 2 }-32 } =16$ and as $r\ge 0,then\quad r=4$