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| − | == Problem ==
| + | #REDIRECT[[2003 AMC 12A Problems/Problem 13]] |
| − | 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?
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| − | [[Image:2003amc10a10.gif]]
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| − | <math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6 </math>
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| − | == Solution ==
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| − | [[Image:2003amc10a10solution.gif]] | |
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| − | Let the squares be labeled <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>.
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| − | When the polygon is folded, the "right" edge of square <math>A</math> becomes adjacent to the "bottom edge" of square <math>C</math>, and the "bottom" edge of square <math>A</math> becomes adjacent to the "bottom" edge of square <math>D</math>.
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| − | So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.
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| − | Therefore, squares <math>1</math>, <math>2</math>, and <math>3</math> will prevent the polygon from becoming a cube with one face missing.
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| − | Squares <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math> will allow the polygon to become a cube with one face missing when folded.
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| − | Thus the answer is <math>6 \Rightarrow E</math>.
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| − | Another way to think of it is that a cube missing one edge has 5 of it's 6 faces. Since the shape has 4 faces already, we need another face. The only way to add anopther face is if the added square does not overlap any of the others. 1,2, and 3 overlap, while 4 <math>\Rightarrow</math> 9 do not. The answer is 6 <math>\Rightarrow E</math>
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| − | == See Also ==
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| − | {{AMC10 box|year=2003|ab=A|num-b=9|num-a=11}}
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| − | [[Category:Introductory Geometry Problems]]
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