Difference between revisions of "1973 IMO Problems/Problem 2"
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So, a finite set <math>M</math> with 4 points would not satisfy the condition. | So, a finite set <math>M</math> with 4 points would not satisfy the condition. | ||
| − | If set <math>M</math> 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 <math>M</math> with a 5th point outside of the parallelogram plane, the condition is to select any two points. So if one of the <math>A | + | If set <math>M</math> 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 <math>M</math> with a 5th point outside of the parallelogram plane, the condition is to select any two points. So if one of the <math>A</math> or <math>B</math> points is that 5th point, there would be no other two other points <math>C</math> and <math>D</math> for which will make lines <math>AB</math> and <math>CD</math> parallel because any of those combinations of lines will be skewed. |
be parallel because those other two points will provide skew lines. | be parallel because those other two points will provide skew lines. | ||
| − | If set < | + | If set <math>M</math> of points in space consist of 6 points, then let's consider a the 6 vertices of a pentahedron <math>ABCDEF</math> with triangular faces <math>ABC</math>, and <math>DEF</math> of the same size parallel to each other with the other faces being parallelograms. Here one might be tempted to think that this set complies with the condition because all lines of the solid are parallel to at least another line. But when we take a diagonal on that solid like <math>AE</math>, such diagonal can't be parallel with anything else. |
| − | If set < | + | If set <math>M</math> of points in space consist of 8 points, then let's consider the 8 vertices of parallelepiped <math>ABCDEFG</math> where all faces are parallelograms. Like in the example of the set <math>M</math> despite almost all combinations of two vertices having another two points in parallel even the diagonal to the faces, when we take a diagonal of this solid like diagonal <math>AG</math> that doesn't lie on any of the faces, such line does not have a parallel two points. So, then you add more points <math>H% and </math>I<math> parallel to the diagonals, but then the two new points have the lines </math>AH<math> 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 <math>M | + | Therefore the finite set </math>M$ of points in space for this problem does not exist. |
Revision as of 19:51, 11 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
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 or points is that 5th point, there would be no other two other points and for which will make lines and parallel because any of those combinations of lines will be skewed.
be parallel because those other two points will provide skew lines.
If set of points in space consist of 6 points, then let's consider a the 6 vertices of a pentahedron 
with triangular faces 
, and 
of the same size parallel to each other with the other faces being parallelograms. Here one might be tempted to think that this set complies with the condition because all lines of the solid are parallel to at least another line. But when we take a diagonal on that solid like 
, such diagonal can't be parallel with anything else.
If set of points in space consist of 8 points, then let's consider the 8 vertices of parallelepiped 
where all faces are parallelograms. Like in the example of the set despite almost all combinations of two vertices having another two points in parallel even the diagonal to the faces, when we take a diagonal of this solid like diagonal 
that doesn't lie on any of the faces, such line does not have a parallel two points. So, then you add more points $H% and$ (Error compiling LaTeX. Unknown error_msg)I
AH$not parallel to anything else. ...and you keep adding points until infinity at which time the condition will be satisfied. But that would make the set infinite and not finite.
Therefore the finite set$ (Error compiling LaTeX. Unknown error_msg)M$ 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 | ||
