Difference between revisions of "1973 IMO Problems/Problem 2"
(→Solution) |
(→Solution) |
||
| Line 23: | Line 23: | ||
The new 4 longer diagonals will cross the diagonal of two of the cubes and will have a parallel line on one of the other cubes. | The new 4 longer diagonals will cross the diagonal of two of the cubes and will have a parallel line on one of the other cubes. | ||
| − | So, we found a set at least one finite set <math>M</math> that we can define as <math>M=\{(x,y,z)\}</math> where | + | So, we found a set at least one finite set <math>M</math> that we can define as <math>M=\{(x,y,z)\}</math> where <math>x,y,z \in {-k,0,k}</math> giving 27 points |
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 18:29, 12 September 2023
Problem
Determine whether or not there exists a finite set 
of points in space not lying in the same plane such that, for any two points 
and 
of ; one can select two other points 
and 
of so that lines 
and 
are parallel and not coincident.
Solution
In order to solve this problem we can start by finding at least one finite set that satisfies the condition.
We start by defining our first set 
with the vertices of a cube of side 
as follows:
Since all the faces of this cube have a parallel face, then any two points on one face will have corresponding 2 points on the opposite face that is parallel. However we have four diagonals on this cube that do not have two points that are parallel to any of these diagonals.
By doing a reflection of the points on the 
plane along the 
these four diagonals will have their respective parallel diagonals on the 
space.
But now we have four more diagonals on the set of two cubes that do not have a parallel line. That is, diagonal 
does not have a parallel line and neither do the other three.
By doing a reflection of the points on the 
plane along the 
these new four diagonals will have their respective parallel diagonals on the 
space.
But now we have four more diagonals on the set of 4 cubes that do not have a parallel line. That is, diagonal 
does not have a parallel line and neither do the other three.
By doing a reflection of the points on the 
plane along the 
these new four diagonals will have their respective parallel diagonals on the space.
The new 4 longer diagonals will cross the diagonal of two of the cubes and will have a parallel line on one of the other cubes.
So, we found a set at least one finite set that we can define as 
where 
giving 27 points
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1973 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
