1973 IMO Problems/Problem 2
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
If set of points in space consist of 3 points or less, then we can't satisfy the condition because we would need at least 4 points.
If set of points in space consist of 4 points, then we can't satisfy the condition because for the condition of lines and to be parallel the 4 points would need to be co-planar. But the points in set shall not be lying in the same plane. So, a finite set with 4 points would not satisfy the condition.
If set of points in space consist of 5 points, then we can't satisfy the condition either because even though we can construct a parallelogram in space with 4 co-planar vertices in the set with a 5th point outside of the parallelogram plane, the condition is to select any two points. So if one of the $A& or$ (Error compiling LaTeX. Unknown error_msg)B
C
D
ABCD$parallel because any of those combinations of lines will be skewed.
be parallel because those other two points will provide skew lines.
If set$ (Error compiling LaTeX. Unknown error_msg)M
ABCDEF
ABC
DEF
AE$, such diagonal can't be parallel with anything else.
If set$ (Error compiling LaTeX. Unknown error_msg)M
ABCDEFG
M
AG
H% and 
parallel to the diagonals, but then the two new points have the lines 
not parallel to anything else. ...and you keep adding points until infinity at which time the condition will be satisfied. But that would make the set infinite and not finite.
Therefore the finite set of points in space for this problem does not exist.
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 | ||
