Difference between revisions of "1972 IMO Problems/Problem 6"
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==Problem== | ==Problem== | ||
+ | |||
Given four distinct parallel planes, prove that there exists a regular tetrahedron | Given four distinct parallel planes, prove that there exists a regular tetrahedron | ||
with a vertex on each plane. | with a vertex on each plane. | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
+ | |||
Let our planes be <math>\pi_1,\pi_2,\pi_3,\pi_4</math>, which we assume to be parallel to the <math>xy</math>-plane, listed in the increasing order of their <math>z</math>-coordinates. First take a plane <math>\pi</math> orthogonal to <math>\pi_i</math>, which cuts <math>\pi_1,\pi_2,\pi_3</math> along three lines <math>d_1,d_2,d_3</math>. On these three lines, take three vertices <math>A_1,A_2,A_3</math> respectively of an equilateral triangle (it is well-known that this is possible; in fact, the problem here is the <math>3</math>-dimensional version of this), and then complete the two regular tetrahedra <math>A_1A_2A_3P_1,A_1A_2A_3P_2</math> having <math>A_1A_2A_3</math> as one of their faces. Both <math>P_i</math> lie below <math>\pi_4</math>. | Let our planes be <math>\pi_1,\pi_2,\pi_3,\pi_4</math>, which we assume to be parallel to the <math>xy</math>-plane, listed in the increasing order of their <math>z</math>-coordinates. First take a plane <math>\pi</math> orthogonal to <math>\pi_i</math>, which cuts <math>\pi_1,\pi_2,\pi_3</math> along three lines <math>d_1,d_2,d_3</math>. On these three lines, take three vertices <math>A_1,A_2,A_3</math> respectively of an equilateral triangle (it is well-known that this is possible; in fact, the problem here is the <math>3</math>-dimensional version of this), and then complete the two regular tetrahedra <math>A_1A_2A_3P_1,A_1A_2A_3P_2</math> having <math>A_1A_2A_3</math> as one of their faces. Both <math>P_i</math> lie below <math>\pi_4</math>. | ||
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The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p390035] | The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p390035] | ||
+ | |||
==Solution 2== | ==Solution 2== | ||
+ | |||
Let's denote the (directed) distance between two parallel planes p and p' by d (p; p'), and the (directed) distance between two parallel lines g and g' by d (g; g'). (Directed distances are defined as follows: If <math>p_1</math>, <math>p_2</math>, <math>p_3</math>, ... is a family of parallel planes in space, then we choose a unit vector <math>\overrightarrow{v}_p</math> perpendicular to all of these planes (there are two such unit vectors, and we have to choose one of them), and then, by the directed distance between two of these planes <math>p_i</math> and <math>p_j</math>, we denote the real number k such that the translation with translation vector <math>k\cdot\overrightarrow{v}_p</math> maps the plane <math>p_i</math> to the plane <math>p_j</math>. Similarly, we define the directed distance between two of a family of parallel lines in a plane.) | Let's denote the (directed) distance between two parallel planes p and p' by d (p; p'), and the (directed) distance between two parallel lines g and g' by d (g; g'). (Directed distances are defined as follows: If <math>p_1</math>, <math>p_2</math>, <math>p_3</math>, ... is a family of parallel planes in space, then we choose a unit vector <math>\overrightarrow{v}_p</math> perpendicular to all of these planes (there are two such unit vectors, and we have to choose one of them), and then, by the directed distance between two of these planes <math>p_i</math> and <math>p_j</math>, we denote the real number k such that the translation with translation vector <math>k\cdot\overrightarrow{v}_p</math> maps the plane <math>p_i</math> to the plane <math>p_j</math>. Similarly, we define the directed distance between two of a family of parallel lines in a plane.) | ||
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The above solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [https://aops.com/community/p399688] | The above solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [https://aops.com/community/p399688] | ||
+ | |||
+ | |||
+ | --Remarks (added by pf02, April 2025)-- | ||
+ | |||
+ | 1. The first "solution" is based on a good idea, but it is so | ||
+ | incomplete (it contains so mach hand-waiving) that it can not | ||
+ | be called a solution. Here are the issues with this solution: | ||
+ | |||
+ | The first problem (minor) is the statement "On these three lines, | ||
+ | take three vertices <math>A_1, A_2, A_3</math> respectively of an equilateral | ||
+ | triangle (it is well-known that this is possible...)". This is *not* | ||
+ | well known, it should be proven. | ||
+ | |||
+ | The second problem (a more serious one) is the statement that "If | ||
+ | <math>\pi</math> makes a small enough angle with the <math>\pi_i</math>'s, one of the | ||
+ | <math>P_i</math>'s we get this time must lie above <math>\pi_4</math>". This is probably | ||
+ | true, but not obvious at all, and it should have a proof. | ||
+ | |||
+ | The third problem (this is so serious that it is a "show stopper") | ||
+ | is the assumption (not stated explicitly) that the <math>z</math>-coordinate | ||
+ | of <math>P_1, P_2</math> as functions of the angle <math>\theta</math> between <math>\pi</math> and | ||
+ | <math>pi_i</math> are continuous functions. This is not even true, it depends | ||
+ | on the choice of <math>A_1, A_2, A_3</math>. | ||
+ | |||
+ | As I said, the idea is good, and this could be turned into a solution | ||
+ | to the problem. I leave this task to the diligent reader. | ||
+ | |||
+ | 2. Below, I will give another solution to the problem. It is similar | ||
+ | to solution 2, but different enough to warrant writing it down. Both | ||
+ | this solution, and solution 2, show that in fact, the tetrahedron | ||
+ | does not have to be regular. The problem could be restated as | ||
+ | "Given four distinct parallel planes and a tetrahedron, prove that | ||
+ | there exists a tetrahedron similar to the given one with a vertex on | ||
+ | each plane." | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | As I remarked, this solution is similar with the previous one. Let | ||
+ | <math>P_1, P_2, P_3, P_4</math> be the four planes. We can assume they are | ||
+ | horizontal (in some <math>x, y, z</math> coordinate system) and that their | ||
+ | <math>z</math> coordinates are in increasing order. Let <math>d_1</math> be the distance | ||
+ | between <math>P_1, P_2</math>, <math>d_2</math> be the distance between <math>P_2, P_3</math>, and | ||
+ | <math>d_3</math> be the distance between <math>P_3, P_4</math>. | ||
+ | |||
+ | |||
+ | |||
+ | Now consider the tetrahedron <math>ABCD</math> | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
== See Also == {{IMO box|year=1972|num-b=5|after=Last Question}} | == See Also == {{IMO box|year=1972|num-b=5|after=Last Question}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
[[Category:3D Geometry Problems]] | [[Category:3D Geometry Problems]] |
Revision as of 19:23, 10 May 2025
Problem
Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane.
Solution 1
Let our planes be , which we assume to be parallel to the
-plane, listed in the increasing order of their
-coordinates. First take a plane
orthogonal to
, which cuts
along three lines
. On these three lines, take three vertices
respectively of an equilateral triangle (it is well-known that this is possible; in fact, the problem here is the
-dimensional version of this), and then complete the two regular tetrahedra
having
as one of their faces. Both
lie below
.
Now take another plane and repeat the construction above. If
makes a small enough angle with the
's, one of the
's we get this time must lie above
. Now, if we move the initial position of
towards the new one continuously and record the
-coordinates of
, these will be continuous functions of the angle that
makes with
, and for one of the points
the
-coordinate will move continuously from being smaller than that of
to being larger than it, meaning that at some point, one of the points
will lie on
, and this is what we want.
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]
Solution 2
Let's denote the (directed) distance between two parallel planes p and p' by d (p; p'), and the (directed) distance between two parallel lines g and g' by d (g; g'). (Directed distances are defined as follows: If ,
,
, ... is a family of parallel planes in space, then we choose a unit vector
perpendicular to all of these planes (there are two such unit vectors, and we have to choose one of them), and then, by the directed distance between two of these planes
and
, we denote the real number k such that the translation with translation vector
maps the plane
to the plane
. Similarly, we define the directed distance between two of a family of parallel lines in a plane.)
The problem can be rewritten as follows: Given four distinct parallel planes ,
,
,
in space, prove that there exists a regular tetrahedron XYZW such that
,
,
,
.
In order to do this, it is enough to find a regular tetrahedron ABCD somewhere in space and four parallel planes ,
,
,
such that
,
,
,
and
. In fact, once we have found such a tetrahedron ABCD and such planes
,
,
,
, then, because of
,
and
, there exists a similitude transformation which maps the planes
,
,
,
to the planes
,
,
,
; this similitude transformation will then obviously map the regular tetrahedron ABCD with
,
,
,
to a regular tetrahedron XYZW with
,
,
,
; hence, the existence of such a tetrahedron XYZW will be proven, and the problem will be solved.
So consider a regular tetrahedron ABCD lying arbitrarily in space; we try to find four parallel planes ,
,
,
such that
,
,
,
and
.
In fact, we start working in the plane ABC. Let T be the point on the line AC such that (where the segments AT and TC are directed). Let
be the line BT, and let
and
be the parallels to the line
through the points A and C, respectively. Then, the lines
,
,
are parallel and, by Thales,
. Thus,
. Now, denote by
the line in the plane ABC which is parallel to the lines
,
,
and satisfies
.
Now, let be the plane passing through the line
and the point D. Let
,
,
be the planes parallel to
and passing through the lines
,
,
, respectively (of course, we can construct such planes since the lines
,
,
are parallel to
). Thus, we have found four parallel planes
,
,
,
such that
,
,
,
, and these planes obviously satisfy
. Since
, we thus have
. Hence, according to the above, the problem is solved.
The above solution was posted and copyrighted by darij grinberg. The original thread for this problem can be found here: [2]
--Remarks (added by pf02, April 2025)--
1. The first "solution" is based on a good idea, but it is so incomplete (it contains so mach hand-waiving) that it can not be called a solution. Here are the issues with this solution:
The first problem (minor) is the statement "On these three lines,
take three vertices respectively of an equilateral
triangle (it is well-known that this is possible...)". This is *not*
well known, it should be proven.
The second problem (a more serious one) is the statement that "If
makes a small enough angle with the
's, one of the
's we get this time must lie above
". This is probably
true, but not obvious at all, and it should have a proof.
The third problem (this is so serious that it is a "show stopper")
is the assumption (not stated explicitly) that the -coordinate
of
as functions of the angle
between
and
are continuous functions. This is not even true, it depends
on the choice of
.
As I said, the idea is good, and this could be turned into a solution to the problem. I leave this task to the diligent reader.
2. Below, I will give another solution to the problem. It is similar to solution 2, but different enough to warrant writing it down. Both this solution, and solution 2, show that in fact, the tetrahedron does not have to be regular. The problem could be restated as "Given four distinct parallel planes and a tetrahedron, prove that there exists a tetrahedron similar to the given one with a vertex on each plane."
Solution 3
As I remarked, this solution is similar with the previous one. Let
be the four planes. We can assume they are
horizontal (in some
coordinate system) and that their
coordinates are in increasing order. Let
be the distance
between
,
be the distance between
, and
be the distance between
.
Now consider the tetrahedron
See Also
1972 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |