Difference between revisions of "2016 AMC 10A Problems/Problem 18"
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== Solution 2 == | == Solution 2 == | ||
| − | Again, all faces sum to <math>18.</math> If <math>x,y,z</math> are the vertices next to one, then the remaining vertices are <math>17-x-y, 17-y-z, 17-x-z, x+y+z-16.</math> Now it remains to test possibilities. Note that we must have <math>x+y+z>17.</math> | + | Again, all faces sum to <math>18.</math> If <math>x,y,z</math> are the vertices next to one, then the remaining vertices are <math>17-x-y, 17-y-z, 17-x-z, x+y+z-16.</math> Now it remains to test possibilities. Note that we must have <math>x+y+z>17.</math> Without loss of generality, let <math>x<y<z.</math> |
<math>3,7,8:</math> Does not work. | <math>3,7,8:</math> Does not work. | ||
Revision as of 17:50, 8 January 2017
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
First of all, the adjacent faces have same sum 
, because 
, 
,
consider the 
(the two sides which are parallel but not in same face of the cube)
they must have same sum value too.
Now think about the extreme condition 1 and 8 , if they are not sharing the same side, which means they would become end points of ,
we should have 
, but no solution for ![$[2,7]$](http://latex.artofproblemsolving.com/b/b/5/bb5c51ab95361af0710c72615ecf364221682321.png)
, contradiction.
Now we know and must share the same side, which sum is 
, the 
also must have sum of , same thing for the other two parallel sides.
Now we have 
parallel sides 
.
thinking about end points number need to have sum of 
.
It is easy to notice only 
vs 
would work.
So if we fix one direction 
or 
all other 
parallel sides must lay in one particular direction. 
or 
Now, the problem is same as the problem to arrange points in a 
square. which is 
=
Solution 2
Again, all faces sum to 
If 
are the vertices next to one, then the remaining vertices are 
Now it remains to test possibilities. Note that we must have 
Without loss of generality, let 
Does not work.
Works.
Does not work.
Works.
Does not work.
Works.
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 face, 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 face, 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
