2016 AMC 12A Problem 14

Question

Each vertex of a cube is to be labeled with an integer from

1

through

8,

with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?

1

3

6

12

24

Answer: C

Concepts:cube geometry casework symmetry

Difficulty rating: 1730

Small hint:

Each vertex lies on

3

faces, so the six face-sums total

3(1+2+\cdots+8)

Big hint:

The face-sum is

18;

show

1

and

8

must be on adjacent vertices

Solution:

Each vertex belongs to

3

faces, so

6S=3(1+2+\cdots+8)=108,

giving each face-sum

S=18.

The four-element subsets containing

1

with sum

18

are

\{1,2,7,8\},

\{1,3,6,8\},

\{1,4,5,8\},

and

\{1,4,6,7\}.

Only one omits

8,

so at least two of the three distinct faces through

1

contain

8.

Two vertices of a cube lie on two common faces exactly when they are adjacent; hence

1

and

8

are adjacent.

Rotate the cube so that

1

is at the lower-left-front vertex and

8

at the lower-right-front vertex. The unique face through

1

that does not contain

8

must use

4,6,7,

and these can be placed in

3!=6

orders. Each order forces

5,3,2

at the three opposite vertices, and the remaining face sums are then

18.

Hence there are

6

arrangements.

Thus, the correct answer is C.