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

Latest revision as of 20:33, 21 November 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 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.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 4: / Line 4: @@
 ==Solution==
-In order to solve this problem we can start by finding at least one finite set M that satisfies the condition.
+In order to solve this problem we can start by finding at least one finite set <math>M</math> that satisfies the condition.
-Let point in space <math> P_{xyz} = [x,y,z]*k </math> where <math>k</math> is a positive constant
+We start by defining our first set <math>M_{8}</math> with the vertices of a cube of side <math>k</math> as follows:
+<math>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) \}</math>
-If set <math>M</math> 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.
+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.
-If set <math>M</math> of points in space consist of 4 points, then we can't satisfy the condition because for the condition of lines <math>AB</math> and <math>CD</math> to be parallel the 4 points would need to be co-planar.  But the points in set <math>M</math> shall not be lying in the same plane.
+By doing a reflection of the points on the <math>z=k</math> plane along the <math>xy</math>-plane these four diagonals will have their respective parallel diagonals on the <math>z \le 0</math> space.
-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</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.
+But now we have four more diagonals on the set of two cubes that do not have a parallel line.  That is, diagonal <math>(0,0,-k) \rightarrow (k,k,k)</math> does not have a parallel line and neither do the other three.
-be parallel because those other two points will provide skew lines.
-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.
+By doing a reflection of the points on the <math>x=k</math> plane along the <math>yz</math>-plane these new four diagonals will have their respective parallel diagonals on the <math>x \le 0</math> space.
-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</math> 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.
+But now we have four more diagonals on the set of 4 cubes that do not have a parallel line.  That is, diagonal <math>(-k,0,-k) \rightarrow (k,k,k)</math> does not have a parallel line and neither do the other three.
-Therefore the finite set <math>M</math> of points in space for this problem does not exist.
+By doing a reflection of the points on the <math>y=k</math> plane along the <math>xz</math>-plane these new four diagonals will have their respective parallel diagonals on the <math>y \le 0</math> space.
-NOTE: I made a mistake. I found a case that it does exist. I'm working on a solution to update this.
+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 <math>x,y,z \in \{-k,0,k\}</math> giving a total of 27 points.  Therefore such a set exists.
+Another way to define this set of points is let <math>M</math> be:
+Let <math>V</math> be a solid cube or right angled parallelepiped
+Let <math>M_{v}</math> be the set of all 8 vertices 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_{mf}</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>
+<math>M</math>=<math>M_{v} \cup M_{me} \cup M_{mf} \cup M_{c}</math>
+It is possible that one can construct many other sets of <math>M</math> 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 <math>M</math> 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.
+[[File:IMO_1973_P2_01.png]]
+~Tomas Diaz.  orders@tomasdiaz.com
 {{alternate solutions}}

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

Latest revision as of 20:33, 21 November 2023

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