The [polygon](https://artofproblemsolving.com/wiki/index.php/Polygon) enclosed by the solid lines in the figure consists of 4 [congruent](https://artofproblemsolving.com/wiki/index.php/Congruent) [ squares](https://artofproblemsolving.com/wiki/index.php/Square_(geometry)) joined [edge](https://artofproblemsolving.com/wiki/index.php/Edge)-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a [ cube](https://artofproblemsolving.com/wiki/index.php/Cube_(geometry)) with one face missing? 
[2003amc10a10.gif](https://artofproblemsolving.com/wiki/index.php/File:2003amc10a10.gif)
$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$
Solution 1
[2003amc10a10solution.gif](https://artofproblemsolving.com/wiki/index.php/File:2003amc10a10solution.gif)
Let the squares be labeled $A$, $B$, $C$, and $D$.
When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$, and the "bottom" edge of square $A$ becomes adjacent to the "bottom" edge of square $D$. 
So, any "new" square that is attatched to those edges will prevent the polygon from becoming a cube with one face missing. 
Therefore, squares $1$, $2$, and $3$ will prevent the polygon from becoming a cube with one face missing.
Squares  $4$, $5$, $6$, $7$, $8$, and $9$ will allow the polygon to become a cube with one face missing when folded. 
Thus the answer is $\boxed{\mathrm{(E)}\ 6}$.

Solution 2
Another way to think of it is that a cube missing one face has $5$ of its $6$ faces.  Since the shape has $4$ faces already, we need another face.  The only way to add another face is if the added square does not overlap any of the others.  $1$,$2$, and $3$ overlap, while squares $4$ to $9$ do not. The answer is $\boxed{\mathrm{(E)}\ 6}$

Solution 3
If you're good at visualizing, you can imagine each box and fold up the shape into a 3D shape. This solution is only recommended if you are either in a hurry or extremely skilled at visualizing. We find out that $4,5,6,7,8$ and $9$ work. Therefore, the answer is $\boxed{\mathrm{(E)}\ 6}$. ~Sophia866