Consider two squares as in the figure to the right. Let the coordinates of their vertexes be
$$\begin{array} { l l } \mathrm { A } ( 2 t , 0 ) , \quad \mathrm { B } ( 0,2 t ) , & \mathrm { C } ( - 2 t , 0 ) , \quad \mathrm { D } ( 0 , - 2 t ) , \\ \mathrm { P } \left( 4 - t ^ { 2 } , 4 - t ^ { 2 } \right) , & \mathrm { Q } \left( - 4 + t ^ { 2 } , 4 - t ^ { 2 } \right) , \\ \mathrm { R } \left( - 4 + t ^ { 2 } , - 4 + t ^ { 2 } \right) , & \mathrm { S } \left( 4 - t ^ { 2 } , - 4 + t ^ { 2 } \right) , \end{array}$$
where $0 < t < 2$. Denote the areas of the two squares ABCD and PQRS by $S _ { 1 }$ and $S _ { 2 }$, respectively.
Then we have
$$S _ { 1 } = \mathbf { M } t ^ { 2 } \text { and } S _ { 2 } = \mathbf { N } \left( t ^ { 2 } - \mathbf { O } \right) ^ { 2 } .$$
(1) $S _ { 1 } + S _ { 2 }$ is minimized at $t = \sqrt { \mathbf { P } }$, and the minimum value is $\mathbf { Q } \mathbf { R }$.
(2) For $\mathbf { W }$ and $\mathbf { X }$ below, choose the correct answer from among (0) $\sim$ (9), and for the other $\square$, enter the correct numbers.
We are to find the range of $t$ such that $S _ { 1 } < S _ { 2 }$. If $S _ { 1 } < S _ { 2 }$, then $t$ satisfies the inequality
$$t ^ { 4 } - \mathbf { ST } t ^ { 2 } + \mathbf { UV } > 0 .$$
From the above inequality, a condition on $t ^ { 2 }$ is $\mathbf { W }$. Hence, $S _ { 1 } < S _ { 2 }$ if and only if $t$ satisfies $\mathbf { X }$.
(0) $t ^ { 2 } < 4$ or $6 < t ^ { 2 }$ (1) $4 < t ^ { 2 } < 6$ (2) $t ^ { 2 } < 2$ or $8 < t ^ { 2 }$ (3) $2 < t ^ { 2 } < 8$ (4) $t ^ { 2 } \neq 4$ (5) $0 < t < 2$ (6) $0 < t < \sqrt { 2 }$ (7) $\sqrt { 2 } < t < 2$ (8) $2 < t < \sqrt { 6 }$ (9) $t \neq 2$