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Difference between revisions of "1973 IMO Problems/Problem 2"
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==Problem==
 
==Problem==
 +
 
Determine whether or not there exists a finite set <math>M</math> of points in space not lying in the same plane such that, for any two points <math>A</math> and <math>B</math> of <math>M</math>; one can select two other points <math>C</math> and <math>D</math> of <math>M</math> so that lines <math>AB</math> and <math>CD</math> are parallel and not coincident.
 
Determine whether or not there exists a finite set <math>M</math> of points in space not lying in the same plane such that, for any two points <math>A</math> and <math>B</math> of <math>M</math>; one can select two other points <math>C</math> and <math>D</math> of <math>M</math> so that lines <math>AB</math> and <math>CD</math> are parallel and not coincident.
 +
  
 
==Solution==
 
==Solution==
Line 44: Line 46:
  
 
~Tomas Diaz.  orders@tomasdiaz.com
 
~Tomas Diaz.  orders@tomasdiaz.com
 +
 +
 +
==Remarks (added by pf02, June 2025)==
 +
 +
1. Very closely related to the solution above, it is easy
 +
to verify that if we delete the center (the point <math>(0, 0, 0)</math>)
 +
from the set <math>M</math>, we obtain a set with <math>26</math> points which still
 +
satisfies the requirements of the problem.  In the second way
 +
of defining this set, we don't need <math>M_c</math>.  In other words, we
 +
just take <math>M = M_v \cup M_{me} \cup M_{mf}</math>.
 +
 +
2. The last paragraph of the solution above ought to be deleted.
 +
The solution would gain from deleting it.  Stating that maybe
 +
one can find a set by "using regular tetrahedrons with some
 +
reflections ... or by translating or rotating or skewing all
 +
the the points simultaneously of the finite set <math>M</math> that we
 +
defined" is vague and useless.
 +
 +
3. Below I will give another solution, inspired by rhombic
 +
dodecahedrons.  Note that there are several non-equivalent
 +
rhombic dodecahedrons.  See for example
 +
https://en.wikipedia.org/wiki/Rhombic_dodecahedron and
 +
https://en.wikipedia.org/wiki/Bilinski_dodecahedron
 +
(By non-equivalent, I mean that one can not be obtained from
 +
the other by an affine transformation.)  Not all of them suggest
 +
solutions to this problem.
 +
 +
 +
==Solution 2==
 +
 +
 +
 +
 +
 +
 +
TO BE CONTINUED.
 +
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 18:41, 9 June 2025

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 right angled parallelepiped

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

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

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

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

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

It is possible that one can construct many other sets of $M$ using regular tetrahedrons with some reflections and with less points than 27, or by translating or rotating or skewing all the the points simultaneously of the finite set $M$ that we defined here. But that is not necessary for this problem because it asks to prove whether there exist a set with the described conditions. By showing that at least one set exists with those conditions, the problem is proved.

IMO 1973 P2 01.png

~Tomas Diaz. orders@tomasdiaz.com


Remarks (added by pf02, June 2025)

1. Very closely related to the solution above, it is easy to verify that if we delete the center (the point $(0, 0, 0)$) from the set $M$, we obtain a set with $26$ points which still satisfies the requirements of the problem. In the second way of defining this set, we don't need $M_c$. In other words, we just take $M = M_v \cup M_{me} \cup M_{mf}$.

2. The last paragraph of the solution above ought to be deleted. The solution would gain from deleting it. Stating that maybe one can find a set by "using regular tetrahedrons with some reflections ... or by translating or rotating or skewing all the the points simultaneously of the finite set $M$ that we defined" is vague and useless.

3. Below I will give another solution, inspired by rhombic dodecahedrons. Note that there are several non-equivalent rhombic dodecahedrons. See for example https://en.wikipedia.org/wiki/Rhombic_dodecahedron and https://en.wikipedia.org/wiki/Bilinski_dodecahedron (By non-equivalent, I mean that one can not be obtained from the other by an affine transformation.) Not all of them suggest solutions to this problem.


Solution 2

TO BE CONTINUED.


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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