Difference between revisions of "1999 AMC 8 Problems/Problem 8"

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==Problem==
 
==Problem==
  
Six squares are colored, front and back, (R=red, B=blue,
+
Six squares are colored, front and back, (R = red, B = blue, O = orange, Y = yellow, G = green, and W = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is
O=orange, Y=yellow, G=green, and W=white). They
+
 
are hinged together as shown, then folded to form a cube.
+
<asy>
The face opposite the white face is
+
draw((0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,2)--(2,2)--(2,3)--(0,3)--cycle);
(A) B (B) G (C) O (D) R (E) Y
+
draw((1,3)--(1,2)--(2,2)--(2,1)--(3,1)--(3,2));
 +
label("R",(.5,2.3),N);
 +
label("B",(1.5,2.3),N);
 +
label("G",(1.5,1.3),N);
 +
label("Y",(2.5,1.3),N);
 +
label("W",(2.5,.3),N);
 +
label("O",(3.5,1.3),N);
 +
</asy>
 +
 
 +
<math>\text{(A)}\ \text{B} \qquad \text{(B)}\ \text{G} \qquad \text{(C)}\ \text{O} \qquad \text{(D)}\ \text{R} \qquad \text{(E)}\ \text{Y}</math>
  
 
==Solution==
 
==Solution==
  
(A) B: When G is arranged to be
+
===Solution 1===
the base, B is the back face and W is
+
When G is arranged to be the base, B is the back face and W is
the front face. Thus, B is opposite W .
+
the front face. Thus, <math>\boxed{\text{(A)}\ B}</math> is opposite W.
OR
+
 
 +
===Solution 2===
 
Let Y be the top and fold G, O, and W down.
 
Let Y be the top and fold G, O, and W down.
Then B will fold to become the back face and be
+
Then <math>\boxed{\text{(A)}\ B}</math> will fold to become the back face and be
opposite W .
+
opposite W.
  
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=1999|num-b=7|num-a=9}}
 
{{AMC8 box|year=1999|num-b=7|num-a=9}}

Revision as of 12:33, 23 December 2012

Problem

Six squares are colored, front and back, (R = red, B = blue, O = orange, Y = yellow, G = green, and W = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is

[asy] draw((0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,2)--(2,2)--(2,3)--(0,3)--cycle); draw((1,3)--(1,2)--(2,2)--(2,1)--(3,1)--(3,2)); label("R",(.5,2.3),N); label("B",(1.5,2.3),N); label("G",(1.5,1.3),N); label("Y",(2.5,1.3),N); label("W",(2.5,.3),N); label("O",(3.5,1.3),N); [/asy]

$\text{(A)}\ \text{B} \qquad \text{(B)}\ \text{G} \qquad \text{(C)}\ \text{O} \qquad \text{(D)}\ \text{R} \qquad \text{(E)}\ \text{Y}$

Solution

Solution 1

When G is arranged to be the base, B is the back face and W is the front face. Thus, $\boxed{\text{(A)}\ B}$ is opposite W.

Solution 2

Let Y be the top and fold G, O, and W down. Then $\boxed{\text{(A)}\ B}$ will fold to become the back face and be opposite W.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 AJHSME/AMC 8 Problems and Solutions
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