Problem:
A three-dimensional rectangular box with dimensions X,Y, and Z has faces whose surface areas are 24,24,48,48,72, and 72 square units. What is X+Y+Z?
Answer Choices:
A. 18
B. 22
C. 24
D. 30
E. 36
Solution:
Without loss of generality, assume that X≤Y≤Z. Then the geometric description of the problem can be translated into the system of equations, XY=24,XZ=48, and YZ=72. Dividing the second equation by the first yields YZ=2, so Z=2Y. Then 72=YZ=2Y2, so Y2=36. Because Y is positive, Y=6. It follows that X=24÷6=4 and Z=72÷6=12, so X+Y+Z=(B)22.
OR
With X,Y, and Z as above, multiply the three equations to give
X2Y2Z2=24⋅48⋅72=24⋅24⋅2⋅24⋅3=242⋅144=(24⋅12)2
Therefore XYZ=24⋅12, and dividing successively by the three equations gives Z=12,Y=6, and X=4, so X+Y+Z=(B)22.
The problems on this page are the property of the MAA's American Mathematics Competitions