Difference between revisions of "2016 AMC 10A Problems/Problem 18"
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So if we fix one direction <math>1-8 (</math>or <math>8-1)</math> all other <math>3</math> parallel sides must lay in one particular direction. <math>(1-8,7-2,6-3,4-5)</math> or <math>(8-1,2-7,3-6,5-4)</math> | So if we fix one direction <math>1-8 (</math>or <math>8-1)</math> all other <math>3</math> parallel sides must lay in one particular direction. <math>(1-8,7-2,6-3,4-5)</math> or <math>(8-1,2-7,3-6,5-4)</math> | ||
− | Now, the problem is same as arranging <math>4</math> points in a two-dimensional square, which is <math>\frac{4!}{4}=\boxed{\textbf{(C) }6.}</math> | + | Now, the problem is the same as arranging <math>4</math> points in a two-dimensional square, which is <math>\frac{4!}{4}=\boxed{\textbf{(C) }6.}</math> |
== Solution 2 == | == Solution 2 == |
Revision as of 10:38, 23 February 2020
Problem
Each vertex of a cube is to be labeled with an integer through , 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 1
Note that the sum of the numbers on each face must be 18, because . 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. Now think about the points and . If they are not on the same edge, they must be endpoints of opposite edges, and we should have , but no solution for , which is a contradiction.
The points and are therefore on the same side and all edges parallel must also sum to .
Now we have parallel sides . thinking about endpoints number need to have a sum of . It is easy to notice only and would work.
So if we fix one direction or all other parallel sides must lay in one particular direction. or
Now, the problem is the same as arranging points in a two-dimensional square, which is
Solution 2
Again, all faces sum to If are the vertices next to , then the remaining vertices are Now it remains to test possibilities. Note that we must have Without loss of generality, let
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 and one with ) for each combination of , , and from above. So, our answer is
Solution 3
We know the sum of each face is If we look at an edge of the cube whose numbers sum to , it must be possible to achieve the sum in two distinct ways, looking at the two faces which contain the edge. If and were on the same edge, it is possible to achieve the desired sum only with the numbers and since the values must be distinct. Similarly, if and were on the same edge, the only way to get the sum is with and . This means that and are not on the same edge as , 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) and are diagonally opposite on the same face. 2) is diagonally across the cube from , while is diagonally across from on the same face. 3) is diagonally across the cube from , while is diagonally across from on the same face.
This means the answer is
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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