2016 AMC 10A Problems/Problem 18

Problem

Each vertex of a cube is to be labeled with an integer $1$ 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?

$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$

Solution 1

Note that the sum of the numbers on each face must be 18, because $\frac{1+2+\cdots+8}{2}=18$ .

So now consider the opposite edges (two edges which are parallel but not on same face of the cube); they must have the same sum value too. (It is not too hard to see this through making one side $a$ , the side on the same face and parallel to it $b.$ Then, the side diametrically opposite to $a$ must also have a side length of $a$ in order to maintain the constant face sum $a+b.$ ) Now think about the points $1$ and $8$ . If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$ . However, this scenario would yield no solution for $[7,2]$ , which is a contradiction. (Try drawing out the cube if it doesn't make sense to you.)

The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9$ .

Now we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$ . Thinking about $4$ endpoints, we realize they need to sum to $18$ . It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.

So if we fix one direction $1-8 ($ or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$

Now, the problem is the same as arranging $4$ points in a two-dimensional square, which is $\frac{4!}{4}=\boxed{\textbf{(C) }6.}$

Solution 2

Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to $1$ , then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x<y<z.$ $3,7,8\text{: Does not work.}$ $4,6,8\text{: Works.}$ $4,7,8\text{: Works.}$ $5,6,7\text{: Does not work.}$ $5,6,8\text{: Does not work.}$ $5,7,8\text{: Does not work.}$ $6,7,8\text{: Works.}$

Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with $[x, y, z]$ and one with $[z, y, x]$ ) for each combination of $x$ , $y$ , and $z$ from above. So, our answer is $3\cdot 2=\boxed{\textbf{(C) }6.}$

Solution 3

We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$ , it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same edge, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same edge, the only way to get the sum is with $1$ and $2$ . This means that $6$ and $7$ are not on the same edge as $8$ , or in other words they are diagonally across from it on the same face, or on the other end of the cube.

Now we look at three cases, each yielding two solutions which are reflections of each other:

1) $6$ and $7$ are diagonally opposite $8$ on the same face.

2) $6$ is diagonally across the cube from $8$ , while $7$ is diagonally across from $8$ on the same face.

3) $7$ is diagonally across the cube from $8$ , while $6$ is diagonally across from $8$ on the same face.

This means the answer is $3\cdot 2=\boxed{\textbf{(C) }6.}$

Video Solution

https://www.youtube.com/watch?v=enRv3Z4Mv5k

Video Solution by TheBeautyofMath

https://youtu.be/yJFQ2lsh6wg

~IceMatrix

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

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