Difference between revisions of "2002 AMC 8 Problems/Problem 22"
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==Video Solution== | ==Video Solution== | ||
| − | https://www.youtube.com/watch?v=PyvyO5JMgHI | + | https://www.youtube.com/watch?v=PyvyO5JMgHI ~David |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=21|num-a=23}} | {{AMC8 box|year=2002|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:03, 15 April 2023
Problem
Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.
![[asy] /* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy]](http://latex.artofproblemsolving.com/6/5/2/6529094e601d11c16feb8a8834704d6fc4d4cfc0.png)
Solution
Count the number of sides that are not exposed, where a cube is connected to another cube and subtract it from the total number of faces. There are 
places with two adjacent cubes, covering 
sides, and 
faces. The exposed surface area is 
.
Solution 2
We can count the number of showing faces from each side. One thing that we notice is that the front face has the same number of squares as the back face, the side faces have the same surface area, etc. Therefore, we are looking for 
front surface area 
side surface area top surface area
. We find that this is 
.
Solution by ILoveMath31415926535
Video Solution
https://www.youtube.com/watch?v=PyvyO5JMgHI ~David
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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. 
