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Difference between revisions of "2003 AMC 12A Problems/Problem 13"
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===Solution 2===
 
===Solution 2===
 
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>
 
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>
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===Solution 3===
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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 <math>4,5,6,7,8</math> and <math>9</math> work. Therefore, the answer is <math>\boxed{\mathrm{(E)}\ 6}</math>. ~Sophia866
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==Video Solution==
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https://www.youtube.com/watch?v=PTA8_vkOekc&t=5s  ~David
  
 
== See Also ==
 
== See Also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 14:43, 19 August 2023

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

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

Video Solution

https://www.youtube.com/watch?v=PTA8_vkOekc&t=5s ~David

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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