Difference between revisions of "2003 AMC 12A Problems/Problem 13"

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{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #13]] and [[2003 AMC 10A Problems|2003 AMC 10A #10]]}}
 
== Problem ==
 
== 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?  
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The [[polygon]] enclosed by the solid lines in the figure consists of 4 [[congruent]] [[square (geometry) | 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 (geometry) | cube]] with one face missing?  
  
 
[[Image:2003amc10a10.gif]]
 
[[Image:2003amc10a10.gif]]
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== Solution ==
 
== Solution ==
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===Solution 1===
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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>.
 
Let the squares be labeled <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>.
  
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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.  
 
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.  
  
Thus the answer is <math>6 \Rightarrow E</math>.
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Thus the answer is <math>\boxed{\mathrm{(E)}\ 6}</math>.
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===Solution 2===
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Another way to think of it is that a cube missing one face has <math>5</math> of its <math>6</math> faces.  Since the shape has <math>4</math> 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.  <math>1</math>,<math>2</math>, and <math>3</math> overlap, while squares <math>4</math> to <math>9</math> do not. The answer is <math>\boxed{\mathrm{(E)}\ 6}</math>
  
 
== See Also ==
 
== See Also ==
*[[2003 AMC 12A Problems]]
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{{AMC10 box|year=2003|ab=A|num-b=9|num-a=11}}
*[[2003 AMC 12A/Problem 12|Previous Problem]]
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{{AMC12 box|year=2003|ab=A|num-b=12|num-a=14}}
*[[2003 AMC 12A/Problem 14|Next Problem]]
 
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:10, 22 April 2017

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

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