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

 
Line 30: Line 30:
  
 
{{AMC8 box|year=1999|num-b=7|num-a=9}}
 
{{AMC8 box|year=1999|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Latest revision as of 23:34, 4 July 2013

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS