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?

Question

I want a clear explanation with the answer. :):) thanks!

Answer 1

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.

Answer 2

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 $234 - 2 \cdot 1^2 - 2 \cdot 1^2 - 2 \cdot 1^2 - 2 \cdot 2^2 - 2 \cdot 2^2 - 2 \cdot 3^2 = \boxed{194}$.

Answer 3

Michael is correct

Answer 4

Yep Thanks Michael

Answer 5

thanks michael

Answer 6

Yep, Michael is correct, but it would have been better if he didn't copy/paste the AoPS solution.

Answer 7

HACK

Answer 8

BLA BLA BLA BLA BLA

Answer 9

Thanks for actually trying steve, not like Micheal who just copy and pasted.

Answer 10

[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]