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== | ||
| − | + | ===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. |
| − | + | ||
| + | ===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]](http://latex.artofproblemsolving.com/7/9/c/79c831b7f1e842c1127b2c37447a942e1ffff08b.png)
Solution
Solution 1
When G is arranged to be the base, B is the back face and W is
the front face. Thus, 
is opposite W.
Solution 2
Let Y be the top and fold G, O, and W down.
Then will fold to become the back face and be
opposite W.
See Also
| 1999 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
