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Difference between revisions of "1973 IMO Problems/Problem 2"

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Another way to define this set of points is let <math>M</math> be:
 
Another way to define this set of points is let <math>M</math> be:
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Let <math>V</math> be a solid cube or parallelepiped
 
Let <math>V</math> be a solid cube or parallelepiped
 +
 
Let <math>M_{v}</math> be the set of all 8 vertices <math>V</math>
 
Let <math>M_{v}</math> be the set of all 8 vertices <math>V</math>
 +
 
Let <math>M_{me}</math> be the set of all 12 midpoints of the edges of <math>V</math>
 
Let <math>M_{me}</math> be the set of all 12 midpoints of the edges of <math>V</math>
 +
 
Let <math>M_{me}</math> be the set of all 6 midpoints of the faces of <math>V</math>
 
Let <math>M_{me}</math> be the set of all 6 midpoints of the faces of <math>V</math>
 +
 
Let <math>M_{c}</math> be the center of <math>V</math>
 
Let <math>M_{c}</math> be the center of <math>V</math>
 +
 
<math>M</math>=<math>M_{v} \cup M_{me} \cup M_{me} \cup M_{c}</math>
 
<math>M</math>=<math>M_{v} \cup M_{me} \cup M_{me} \cup M_{c}</math>
  

Revision as of 13:27, 13 September 2023

Problem

Determine whether or not there exists a finite set $M$ of points in space not lying in the same plane such that, for any two points $A$ and $B$ of $M$; one can select two other points $C$ and $D$ of $M$ so that lines $AB$ and $CD$ are parallel and not coincident.

Solution

In order to solve this problem we can start by finding at least one finite set $M$ that satisfies the condition.

We start by defining our first set $M_{8}$ with the vertices of a cube of side $k$ as follows: $M_{8} = \{ (0,0,0), (k,0,0), (k,k,0), (0,k,0), (0,0,k), (k,0,k), (k,k,k), (0,k,k) \}$

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 $z=k$ plane along the $xy-plane$ these four diagonals will have their respective parallel diagonals on the $z \le 0$ space.

But now we have four more diagonals on the set of two cubes that do not have a parallel line. That is, diagonal $(0,0,-k) \rightarrow (k,k,k)$ does not have a parallel line and neither do the other three.

By doing a reflection of the points on the $x=k$ plane along the $yz-plane$ these new four diagonals will have their respective parallel diagonals on the $x \le 0$ space.

But now we have four more diagonals on the set of 4 cubes that do not have a parallel line. That is, diagonal $(-k,0,-k) \rightarrow (k,k,k)$ does not have a parallel line and neither do the other three.

By doing a reflection of the points on the $y=k$ plane along the $xz-plane$ these new four diagonals will have their respective parallel diagonals on the $y \le 0$ 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 $M$ that we can define as $M=\{(x,y,z)\}$ where $x,y,z \in \{-k,0,k\}$ giving a total of 27 points. Therefore such a set exists.

Another way to define this set of points is let $M$ be:

Let $V$ be a solid cube or parallelepiped

Let $M_{v}$ be the set of all 8 vertices $V$

Let $M_{me}$ be the set of all 12 midpoints of the edges of $V$

Let $M_{me}$ be the set of all 6 midpoints of the faces of $V$

Let $M_{c}$ be the center of $V$

$M$=$M_{v} \cup M_{me} \cup M_{me} \cup M_{c}$

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
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