2003 AMC 12A Problems/Problem 13

The following problem is from both the 2003 AMC 12A #13 and 2003 AMC 10A #10, so both problems redirect to this page.

Problem

The polygon enclosed by the solid lines in the figure consists of 4 congruent squares joined 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 with one face missing?

2003amc10a10.gif

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$

Solution

Solution 1

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 $\boxed{\mathrm{(E)}\ 6}$

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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