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2003 AMC 12A Problems/Problem 13

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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 attatched 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?

Image:2003amc10a10.gif

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

Solution


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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 6 \Rightarrow E.

See Also

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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