Hi, sorry, I haven't been able to make a post for a few days.
Yesterday and today I did the 2016 AMC 10A, untimed. I'm doing all of the AMC 10s untimed, and I'm going to do timed mocks later, once I'm more prepared.
The hardest (and best, in my opinion) problem out of all of them was #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?
Solution:
Consider two faces opposite to each other. These two faces, combined, cover all of the vertices of the cube. Since the sum of both opposite faces is the same, this sum is $\frac{1+2+3+4+5+6+7+8}{2}=18$. Using this logic, we know that the sum of the vertices on each face is $18$. Now, consider two adjacent faces. They share an edge (with two vertices), and each face has one other edge, which has the remaining two vertices on the face. Since these two adjacent faces have the same sum, and they have a common edge, the sum of the two vertices on their uncommon edge must be the same. This implies that the sum of the vertices on an edge is equal to the sum of the vertices of the edge diagonally opposite to it. The sum of all of the vertices is $1+2+3+4+5+6+7+8=36$, so the sum on each edge must be $\frac{36}{4}=9$. The possibilities for this are $(1,8), (2,7), (3,6), (4,5)$. The 4 parallel edges must be one of these possibilities each. We fix the edge $(1,8)$ (let's say on left edge of the front face), where $1$ is on the top and $8$ is on the bottom. We still require that the sum of the vertices on the top faces be $18$ as well. Considering our $4$ parallel edges, by experimentation, we realize that the only way for this is to have $1,7,6,4$ on the top face and $2,3,5,8$ on the bottom face (not vice versa, since we have fixed the $(1,8)$ edge). So there's only one way for this. The only choices we still have are about how we arrange the $4$ parallel edges. This is just $\frac{4!}{4}$, where we divide by $4$ to account for rotations. The answer is thus $\boxed{6}$.
Phew! I didn't get this problem fully on my own, but I was really, really close. I did everything until the part where I realized that the sum on each parallel edge must be the same. For some reason, I couldn't think of the $\frac{36}{4}$ part after that.
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