Difference between revisions of "1995 AHSME Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | The marked side is the side with the x. We imagine it folding up. First, we fold the x upwards. Now we fold the A upwards, and thus x is touching B on it's left side. We now fold B up, and we realize that x won't be touching <math>\boxed{\mathrm{(C)}}</math> at all. | + | The marked side is the side with the <math>x.</math> We imagine it folding up. First, we fold the <math>x</math> upwards. Now we fold the <math>A</math> upwards, and thus <math>x</math> is touching <math>B</math> on it's left side. We now fold <math>B</math> up, and we realize that <math>x</math> won't be touching <math>\boxed{\mathrm{(C)}}</math> at all. |
==See also== | ==See also== | ||
{{AHSME box|year=1995|num-b=5|num-a=7}} | {{AHSME box|year=1995|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 23:49, 13 October 2013
Problem
The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked x?
![[asy] defaultpen(linewidth(0.7)); path p=origin--(0,1)--(1,1)--(1,2)--(2,2)--(2,3); draw(p^^(2,3)--(4,3)^^shift(2,0)*p^^(2,0)--origin); draw(shift(1,0)*p, dashed); label("$x$", (0.3,0.5), E); label("$A$", (1.3,0.5), E); label("$B$", (1.3,1.5), E); label("$C$", (2.3,1.5), E); label("$D$", (2.3,2.5), E); label("$E$", (3.3,2.5), E);[/asy]](http://latex.artofproblemsolving.com/c/d/2/cd26443e7d52459f1c3554f595e6832511d1d1f7.png)
Solution
The marked side is the side with the 
We imagine it folding up. First, we fold the 
upwards. Now we fold the 
upwards, and thus is touching 
on it's left side. We now fold up, and we realize that won't be touching 
at all.
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
