Four cubes of volumes 1cm^3, 8cm^3, 27cm^3, and 125cm^3 are glued together at their faces. What is the number of square centimeters in the smallest possible surface area of the resulting solid figure?
I want a clear explanation with the answer. :):) thanks!
Yep, Michael is correct, but it would have been better if he didn't copy/paste the AoPS solution.
HACK
BLA BLA BLA BLA BLA
Michael is correct
From the volumes, we deduce that the side lengths of the cubes are 1 cm, 2 cm, 3 cm, and 5 cm. We position the cubes as follows:
[asy]
unitsize(0.5 cm);
draw((0,0)--(5*dir(-30))--(5*dir(-30) + 5*dir(30))--(10*dir(-30))--(5*dir(-30) + 5*dir(-90))--(5*dir(-90))--(0,0));
draw((5*dir(-30))--(5*dir(-30) + 5*dir(-90)));
draw((0,0)--(0,2)--((0,2) + 2*dir(-30))--(2*dir(-30)));
draw((0,2)--((0,2) + 2*dir(30))--((0,2) + 2*dir(30) + 2*dir(-30))--(2*dir(30)));
draw((2*dir(-30))--(2*dir(-30) + dir(30))--(2*dir(-30) + dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,2)));
draw((2*dir(-30) + dir(30))--(3*dir(-30) + dir(30))--(3*dir(-30) + dir(30) + (0,1))--(2*dir(-30) + dir(30) + (0,1)));
draw((3*dir(-30) + dir(30) + (0,1))--(3*dir(-30) + 2*dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,1)));
draw((2*dir(30) + (0,2))--(2*dir(30) + (0,3))--(2*dir(30) + 3*dir(-30) + (0,3))--(2*dir(30) + 3*dir(-30))--(dir(30) + 3*dir(-30)));
draw((2*dir(30) + (0,3))--(5*dir(30) + (0,3))--(5*dir(30) + 3*dir(-30) + (0,3))--(5*dir(30) + 3*dir(-30))--(5*dir(30) + 5*dir(-30)));
draw((3*dir(-30) + 2*dir(30))--(3*dir(-30) + 5*dir(30)));
draw((3*dir(-30) + 2*dir(30) + (0,3))--(3*dir(-30) + 5*dir(30) + (0,3)));
[/asy]
The surface area of a cube with side length $s$ is $6s^2$, so the total surface area of the cubes is $6 \cdot 1^2 + 6 \cdot 2^2 + 6 \cdot 3^2 + 6 \cdot 5^2 = 234$.
Note that every pair of cubes touches, and furthermore, they have maximum contact. (This is why this solid has the smallest possible area.) The area of contact of the 1-cube and the 2-cube is 1 square centimeter, so we must subtract this twice from 234 (because this portion of the area from both the 1-cube and 2-cube is not seen anymore).
Doing this for every pair of cubes, we find that the surface area of this solid is $234 - 2 \cdot 1^2 - 2 \cdot 1^2 - 2 \cdot 1^2 - 2 \cdot 2^2 - 2 \cdot 2^2 - 2 \cdot 3^2 = \boxed{194}$.
[asy]
unitsize(0.5 cm);
draw((0,0)--(5*dir(-30))--(5*dir(-30) + 5*dir(30))--(10*dir(-30))--(5*dir(-30) + 5*dir(-90))--(5*dir(-90))--(0,0));
draw((5*dir(-30))--(5*dir(-30) + 5*dir(-90)));
draw((0,0)--(0,2)--((0,2) + 2*dir(-30))--(2*dir(-30)));
draw((0,2)--((0,2) + 2*dir(30))--((0,2) + 2*dir(30) + 2*dir(-30))--(2*dir(30)));
draw((2*dir(-30))--(2*dir(-30) + dir(30))--(2*dir(-30) + dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,2)));
draw((2*dir(-30) + dir(30))--(3*dir(-30) + dir(30))--(3*dir(-30) + dir(30) + (0,1))--(2*dir(-30) + dir(30) + (0,1)));
draw((3*dir(-30) + dir(30) + (0,1))--(3*dir(-30) + 2*dir(30) + (0,1))--(2*dir(-30) + 2*dir(30) + (0,1)));
draw((2*dir(30) + (0,2))--(2*dir(30) + (0,3))--(2*dir(30) + 3*dir(-30) + (0,3))--(2*dir(30) + 3*dir(-30))--(dir(30) + 3*dir(-30)));
draw((2*dir(30) + (0,3))--(5*dir(30) + (0,3))--(5*dir(30) + 3*dir(-30) + (0,3))--(5*dir(30) + 3*dir(-30))--(5*dir(30) + 5*dir(-30)));
draw((3*dir(-30) + 2*dir(30))--(3*dir(-30) + 5*dir(30)));
draw((3*dir(-30) + 2*dir(30) + (0,3))--(3*dir(-30) + 5*dir(30) + (0,3)));
[/asy]
Yep Thanks Michael
thanks michael
Thanks for actually trying steve, not like Micheal who just copy and pasted.
the cubes are of side 1,2,3,5
Their areas are 6,24,54,150
If we glue the 3-cube to the 5-cube, it will cover 9 cm^2 of both cubes, leaving 150+54-18=186cm^2 exposed
If we then glue the 2-cube to the 5-cube next to the 3-cube, it will cover 4cm^2 on the 5-cube and 3-cube, and 8cm^2 of the 2-cube
Now the exposed area is 186+24-4-4-8 = 194
Now if we glue the 1-cube to the 5-cube so it is against the 3-cube and 2-cube, we will cover 1+1+1 of its 6 faces, leaving 194+6-3 = 197 exposed.
Can't think of a way to cover more area.