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Difference between revisions of "1986 AJHSME Problems/Problem 21"

(New page: ==Problem== Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a ...)
 
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==Solution==
 
==Solution==
  
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{{Solution}}
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The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these two places.
 +
 
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* <math>A</math> would be the top side - OK
 +
* <math>B</math> would be the left side - OK
 +
* <math>C</math> would cause that the figure would not be foldable at all
 +
* <math>D</math> would be the left side - OK
 +
* <math>E</math> would be the top side - OK
 +
* <math>F</math> is the same case as <math>B</math> - OK
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* <math>G</math> is the same case as <math>C</math>
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* <math>H</math> is the same case as <math>A</math> - OK
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In total, there are <math>\boxed{6}</math> good possibilities.
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 19:56, 25 January 2009

Problem

Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?

[asy] draw((1,0)--(2,0)--(2,5)--(1,5)--cycle); draw((0,1)--(3,1)--(3,4)--(0,4)--cycle); draw((0,2)--(4,2)--(4,3)--(0,3)--cycle); draw((1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(2,4)--(1,4)--cycle,linewidth(.7 mm)); label("A",(1.5,4.2),N); label("B",(.5,3.2),N); label("C",(2.5,3.2),N); label("D",(.5,2.2),N); label("E",(3.5,2.2),N); label("F",(.5,1.2),N); label("G",(2.5,1.2),N); label("H",(1.5,.2),N); [/asy]

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

Solution

The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these two places.

  • $A$ would be the top side - OK
  • $B$ would be the left side - OK
  • $C$ would cause that the figure would not be foldable at all
  • $D$ would be the left side - OK
  • $E$ would be the top side - OK
  • $F$ is the same case as $B$ - OK
  • $G$ is the same case as $C$
  • $H$ is the same case as $A$ - OK

In total, there are $\boxed{6}$ good possibilities.

See Also

1986 AJHSME Problems

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