Is it p ossible to cut a cub e into pieces
and to assemble a tetrahedron
Hilb ert s problem and Dehn s theorem
Dmitry FUCHS
Hilb ert s Problem Three
Is it p ossible to cut a cub e by nitely many planes and assemble out of the
p olyhedral pieces obtained a regular tetrahedron of the same volume
This is a slight mo di cation of one of the problems presented by David
Hilb ert in his famous talk at the Congress of Mathematicians in Paris on
August it go es under the numb er Hilb ert s problems had a tremen
dous impact on Mathematics Most of them were solved during XX century
and each has a very sp ecial history Still Problem Three stands exceptional
in many resp ects
First this was the rst of Hilb ert s problems to b e solved The solution
b elonged to a years old German geometer Hilb ert s student Max Dehn
His article app eared two years after the Paris Congress but the solution
existed earlier mayb e even b efore Hilb ert stated the problem
or less the same as the one presented b elow was short Dehn s pro of more
for p opular lectures and clear and it b ecame one of the favorite sub jects
articles and b o oks in geometry like the one you are holding in your hands I
can recommend the b o ok by Boltianskii for a more extended exp osition
But among working mathematicians it was almost forgotten
Certainly the name of Dehn was not forgotten He b ecame one of the few
top exp erts in top ology of three dimensional manifolds and his work of
has b een never regarded as his main achievement it is not even mentioned
in Dehn s biography available on the web
In American Mathematical So ciety published a two volume collec
tion of articles under the title Mathematical Developments Arising from
Hilb ert Problems It was a very solid account of the three quarters of
century of history of the problems solutions full and partial generaliza
tions similar problems and so on This edition contains a thorough analysis
of of Hilb ert s problems And only Problem Three is not discussed
there The opinion of the editors is obvious no developments no in uence
on Mathematics nothing to discuss
How strange it seemed just a couple of years later Dehn s theorem
Dehn s theory Dehn s invariant b ecame one of the hottest sub jects in geom
an exciting domain etry This was stimulated by then new b orn K theory
develop ed along the b orderline b etween algebra and top ology We will not
follow this development but will just revise the theorem and its pro of
For a similar problem in the plane the an
swer is yes
Let P P be two polygonal domains in the plane having the Theorem
1 2
same ar ea Then it is possible to cut P into pieces by straight lines and to
1
assemble Of these pieces P
2
Pro of First it is clear that it is su cient to consider the case when P
2
is a rectangle with the sides and area P in doing this we can shorten the
1
notation of P to just P
1
Second since any p olygonal domain can b e cut into triangles we can
reduce the general case to that of a triangle see Figure
P ! ! !
P area
Figure
Third we need to remake by cutting and pasting a triangle into a rect
angle with one if the sides having length one This is done in four steps on
Figure
!
!
Step Step
rational length
p q
p
!
q
!
p
Step Step
q
Figure
We do this in four steps First we make a parallelogram out of our triangle
Step Then we cut a small triangle on one side of the parallelogram and
attach it to the other side in such a way that the length of one of the sides of
Step On Step we make this the parallelogram b ecomes rational p q
parallelogram a rectangle the numb er of horizontal cuts needed dep ends on
the shap e of the parallelogram On the nal step we cut the rectangle into
pq equal pieces by p horizontal lines and q vertical lines with the
understanding that it is the vertical side of the rectangle that has the length
p q then we rearrange these pq pieces into a rectangle with the length of
the vertical side b eing
A planar problem which do es not lo ok sim
ilar to Hilb ert s Problem Three but has a
similar solution
rectangle into nitely many smaller rectangles with Is it p ossible to cut a
p p
sides parallel to the sides of the given rectangle and to assemble a
square
The answer is NO The pro of is more algebraic than geometric but still
unlike the Hilb ert Problem it requires a small geometric preparation
A geometric preparation
Let us given two rectangles with vertical and horizontal sides b elow we will
call such rectangles brie y V H rectangles and supp ose that is is p ossible
to cut them into smaller V H rectangles such then the pieces of the rst are
equal congruent to the pieces of the second
2 6
1 1
5 2 7 5
3 4 3 4
(C )
6 7
A) (
a b m n o
e f g h
a b c d c d p
i j k l
e f g h
) (D
i j k l
m n o p
(B )
(E )
Figure
still smaller V H rectangles such Then there exists a collection of N
that each of the given rectangles can b e obtained by a sequence of N
admissible moves An admissible move we take two of our small rectangles
having equal widths or equal heights and attach them to each other vertically
or horizontally creating one rectangle of the same width or height Thus
our pro cess of cutting is replaced by a more strict pro cess of attaching of
rectangles How to do this is shown on Figure
Supp ose that two rectangles are cut into equal pieces as requested by
Problem rectangles A and C on Figure equal pieces are marked there
by the same Arabic numb ers Then we extend the sides of the pieces to the
whole width or length of the rectangle see rectangle B of Figure Some
of the pieces of division are cut into smaller pieces marked by Roman letters
b e and f b ecomes a union in rectangle B so b ecomes a union of a
of c and d etc Then we divide in the same way the pieces of the second
given rectangle see rectangle D of Figure we break the rectangle of
rectangle C into pieces congruent to a b e f the rectangle into pieces
c d and so on We obtain a new division of the second given rectangle
into smaller rectangles and again extend the sides of these smaller pieces
to the whole width or length of the rectangle see rectangle E of Figure
These last pieces form our collection Obviously we can assemble the
second rectangle C from these pieces using the admissible moves Other
admissible moves pro duce out of our small rectangles the parts of the ner
division of the rectangle A that is a b c o p and out of this part
we can assemble using admissible moves the rectangle A The geometric
preparation is over
An algebraic pro of
Let us have a nite collection of V H rectangles of the total area Then no
more than one of the following two is p ossible
to comp ose out of these rectangles a rectangle using only admissible
moves
p p
to comp ose out of these rectangles a square using only admis
sible moves
This is what we need to answer negatively the question of these section
w b e the widths of the rectangles of our collection N b eing Let w
1 N
the numb er of these rectangles and h h b e their heights
1 N
Consider the sequence
p
w w
1 N
remove a memb er of this sequence if it is a linear combination with rational
co e cients of the preceding memb ers Thus we do not remove we do not
p p
remove since it is irrational we remove w if and only if w r r
1 1 1 2
with rational r r and so on Let a a b e the remaining numb ers
1 2 1 m
p
thus a a It is imp ortant that each of the numb ers can
1 2
b e presented as a rational linear combination of the numb ers a a in a
1 m
1
unique way
No w do the same with the sequence
p
h h
1 N
p
with b b such that each of We will get the numb ers b b
1 2 1 n
the numb ers can b e presented as a rational linear combination of the
numb ers b b in a unique way
1 n
Call a rectangle admissible if its width is a rational linear combination
of a a and its height is a rational linear combination of b b
1 m 1 n
Let P b e an admissible rectangle of the width w and the length h and
P P
n m
s r a and h b with rational r s an d s s We de let w
i i j j i j
=1 =1 j i
ne the symbol Symb P of the rectangle P as the rational m n matrix
k k with S r s We usually will use for the symb ols the notation S
ij ij i j
P
Symb P r s a b which is simply the alternative notation for
i j i j
i j
the matrix ab ove Thus we regard the symb ols as formal rational linear
combination of the expressions a b Such formal linear combinations
i j
cab b e added in the obvious way we consider two formal rational linear
P P
combinations t a b j t a b equal if t t for all i
i j i j
ij ij ij ij
i j i j
Let P and P b e two admissible rectangles of equal heights or equal
widths Then we can merge these two rectangles into one rectangle P using
an admissible move see ab ove Obviously P is also an admissible rectangle
and Symb P Symb P Symb P Indeed if P and P have widths
1
This is a standard theorem from linear algebra but for the sake of completeness let us
p
is a rational linear combination of a give a pro of a a so is a Assume
1 1 m 2
by induction that all the numb ers preceding w are rational linear combinations
k
of a a If w is not a rational linear combination of preceding numb ers then
1 m k
it is one of a s and hence is a rational linear combination of a a if w is a
j 1 m k
rational linear combination of preceding numb ers then it is a rational linear combination of
a a since all the preceding numb ers are rational linear combinations of a a
1 m 1 m
If two di erent rational linear combinations of a a It remains to prove uniqueness
1 m
P P
m m
0 0 0 0 0 0
r is the largest of m for which r r then are equal a r a and s
i j
s s
i j
=1 j =1 i
1 s
0 0 0
X
r r
i i
a a which shows that a is a rational linear combination of preceding a s
s i s j
00 0
r r
s s
i=1
in contradiction to the choice of a a
1 m
P P P
n m m
s b then r a and the same height h r a and w w
j j i i
i i
j =1 i=1 i=1
P
m
r r a and the height h and P has the width w w
i
i i
i=1
P
Symb P r r s a b
j i j
i i
i j
P P
r s a b r s a b
j i j j i j
i i
i j i j
Symb P Symb P
Thus if we have a collection of admissible rectangles P P and can
1 N
assemble out of them by N admissible moves a rectangle P then
P
N
t Symb P If we can assemble in this way two di eren Symb P
i
i=1
rectangles P and P then Symb P Symb P This proves our theorem
p p
since the symb ol of a rectangle is a b and the symb ol of a
1 1
square is a b which is di erent
2 2
Pro of of Dehn s Theorem
We want to prove the following
Theorem Let C and T be a cube and a regular tetrahedron of the same
volume Suppose that each of them is cut into the same number of pieces by
is we cut our polyhedron into two pieces then cut one of the planes That
two pieces into two pieces then cut one of the three pieces into two pieces
and so on It is not possible that the two col lection of polyhedral pieces are
the same
Pro of Let b e the lengths of all edges of all p olyhedra involved in
1 N
the two cutting pro cesses Let are corresp onding dihedral angles
1 N
we supp ose that for all i Take the sequence el l
i 1 N
and remove from it any term which is a rational linear combination of the
previous terms we obtain a sequence a a such that each of the s
1 m k
is equal to a unique rational linear combination of a s Then do the same
i
with the sequence the resulting sequence is denoted as
1 N 0
and each of s is equal to a unique linear combination of s
1 n k j
Call a convex p olyhedron admissible if the length of every edge is a rational
linear combination of a a and each dihedral angle is a rational linear
1 m
combination of
0 1 n
b e the lengths of edges of an admissible convex p oly Let m m
1 q
hedron P and let b e the corresp onding dihedral angles Let
1 q
P P
n m
s De ne the symb ol of P by the r a and m
k j j k i i k k
j =0 i=1
formula
q
n m
X X X
a r s Symb P
i j k i k j
i=1 j =1
k =1
I mp ortant remark it is not a misprint that the second summation is
taken from j to n not from j to n we do not include into the
symb ol the summand s Thus if one changes an angle by a rational
k 0
multiple of then the symb ol is not a ected if some dihedral angle is a
rational multiple of then the corresp onding edge do not app ear in the
expression for the symb ol at all
E xample the symb ol of a cub e or of a rectangular b ox is zero Indeed
all the angles are
E xercise the symb ol of any rectangular prism with a p olygonal base is
zero
Lemma Let P be a convex polyhedron Suppose that it is cut by a plane L
into two pieces P and P Then provided that P P and P are admissible
Symb P Symb P Symb P
e e g b e the set of all edges of P let Pro of of Lemma Let S f
1 q k
b e the length of the edge e and b e the corresp onding dihedral angle We
k k
divide the set S into four subsets S consists of edges which have no interior
1
p oints in L and lie on the P side of L S is the similar set with P instead
2
of P S consists of edges e cut by L into an edge e of P and an edge
3 k
k
e of P and S consists of edges which are totally contained in L for each
4
k
e S the dihedral angle is divided by L into two parts and
k 4 k
k k
P This is a convex p olygon each e S Consider also the intersection L
k 4
is its side let T ff f g b e the set of all other sides of R Each f
1 p k
is a side b oth of P and P let m b e the length of f and b e the
k k
k k
corresp onding dihedral angles in P and P Obviously
k k
Edges of P
the edges e S the lengths are the angles are
k 1 k k
for e S the lengths are the angles are the edges e
k 3 k
k k
the edges e S the lengths are the angles are
k 4 k
k
the edges f T the lengths are m the angles are
k k
k
Edges of P
the edges e S the lengths are the angles are
k 2 k k
the edges e for e S the lengths are the angles are
k 3 k
k k
the edges e S the lengths are the angles are
k 4 k
k
the edges f T the lengths are m the angles are
k k
k
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Figure
The symb ols of each of the p olyhedra P P and P consists of four groups
of summands for P and P these groups corresp ond to the four groups
of edges as listed ab ove for P they corresp ond to the sets S S S S
1 2 3 4
The rst group of summands in Symb P is the same as the rst group
of summands in Symb P The rst group of summands in Symb P is
the same as the second group of summands in Symb P The sum of the
second groups of summands in Symb P and Symb P is the third group
of summands in Symb P b ecause The sum of the third groups
k
k k
of summands in Symb P and Symb P is the fourth group of summands
in Symb P b ecause At last the sum of the fourth groups of
k
k k
summands in Symb P and Symb P is zero since Thus
k k
Symb P Symb P Symb P as stated by Lemma
An example is shown on Figure A p olyhedron P a four gonal prism
with non parallel bases shown at the left of the rst row is cut into two
p olyhedra by a plane the cut is shown in the rst row the p olyhedra P and
P are shown in the second row The edges of P are e e the sets S
1 12 i
are S fe e e e g S fe e e e e e g S fe g S fe g
1 1 3 4 5 2 6 7 9 10 11 12 3 8 4 2
Bach to Theorem If two p olyhedra can b e cut into the same collection
of p olyhedral parts then their symb ols are b oth equal to the sum of the
symb ols of the part and hence the symb ols of the given two p olyhedra are
equal to each other But the symb ol of a cub e is equal to zero since all the
angles are see Example ab ove The symb ol of a regular tetrahedron
is equal to where is the length of the edge and is the dihedral
angle All we need to check is that is not a rational multiple of
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Figure
The dihedral angle of a regular tetrahedron is the largest angle of an
p p
isosceles triangle whose sides are see Figure The cosine
theorem shows that
p p
2 2
3 3
2
2 2
cos
p p
3 3
2 2
Lemma If cos then is irrational
Pro of of Lemma Otherwise cos n for some n However it is known
from trigonometry that
cos n P cos
n
n 1
where P is a p olynomial of degree n with the leading co e cient
n
Pro of by induction Statement for all n
cos sin cos n P cos sin n Q
n n
where deg P n deg Q n and the leading co e cients of b oth P
n n n
n 1
and Q are equal to For n this is true P t t Q t
n 1 1
assume that the statement is true for some n Then
cos n cos n sin sin n sin
2
P cos cos Q cos sin
n n
2
P cos cos Q cos cos
n n
sin n sin n cos cos n sin
Q cos sin cos P cos sin
n n
Q cos cos P cos sin
n n
Hence
2
t P t t Q t t P
n n n+1
Q t Q t t P t
n+1 n n
and the statement for the degrees and leading terms follows
This shows that
n 1
an integer
cos n P
n
n n 1
which cannot b e an integer in particular
This proves Lemma and completes the pro of of Dehn s theorem
Some further results
In the language of algebra which may technically not familiar to the reader
but the formulas b elow seem to me self explanatory the construction of the
previous section assigns to every convex actually not necessarily convex
p olyhedron a certain invariant Dehn s symb ol
Symb P R R Q
Q
and Dehn s theorem states that if two p olyhedra P and P are e quipartite
1 2
that is can b e cut by planes into identical collections of parts then
Symb P Symb P
1 2
This is precisely the result of the previous section
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Figure
Certainly this may b e applied not only to cub es and tetrahedra The
initial Hilb ert s problem by the way dealt with a di erent example Hilb ert
conjectured that two tetrahedra with equal bases and equal heights like
those on Figure are not equipartite
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Figure
The origin of this question b elongs to the foundations of geometry The
whole theory of volumes of solids is based on the lemma stating that the
volumes of tetrahedra in Figure are the same The similar planar lemma
involving the areas of triangles has a direct geometric pro of based on cut
ting and pasting But the three dimensional fact requires a limit stair con
struction involving pictures like Figure you can nd a gure like this in
textb o oks in the spatial geometry The question is is this really necessary
and the answer is yes Dehn s theorem easily implies that the tetrahedra
like those in Figure are not in general equipartite
More than years after Dehn s work Sydler proved that p olyhedra
with equal volumes and equal Dehn s invariant are equipartite There are
similar results in spherical and hyp erb olic geometries
Dehn s symb ol may b e generalized to p olyhedra of any dimension for an
n dimensional p olyhedron P
X
dihedral
volume s Symb P R R Q
Q
angle at s
n dimensional
faces s of P
the angle is formed by the two n dimensional faces of P attached to s
In dimension like in dimension two p olyhedra are equipartite if and only
if their volumes and their symb ols are the same But in dimension it is not
true any longer there arises a new invariant a secondary Dehn s symb ol
involving a summation over the edges for an n dimensional p olyhedron
over n dimensional faces of P There is a conjecture I do not know its
current status that an equipartite typ e of an n dimensional p olyhedron
n
invariants the volume Dehn s is characterized by a sequence of
symb ol secondary Dehn s symb ol and so on taking values in more and
more complicated tensor pro ducts k th Dehn symb ol involves a summation
over n k dimensional faces In particular for one and two dimensional
p olyhedra segments and p olygons only the volume the length and the
area counts in dimensions and we also have Dehn s symb ol and so on
If you want to know more ab out this you can read in addition to the
p opular b o ok of Boltianskii the article of Cartier in Pro ceedings of the Bour
baki Seminar But I am not sure that it has b een ever translated from
French into English so if you read this article you have a chance to study a
b eautiful language in addition to a b eautiful geometry
References
V G Boltianskii Hilb ert s Third Problem Scripta Series in Math
John Wiley and Sons Wash N Y
osition des p oly edres le p oint sur le troisi eme P Cartier Decomp
probl eme de Hilb ert Ast erisque
M Dehn Ub er den Rauminhalt Math Ann
n ecessaires et su santes p our l equivalence J P Sydler Conditions
des p oly edres de l espace euclidien a trois dimension Comment
Math Helv
developments arising from Hilb ert s problems F Mathematical
Browder ed Pro c Symp Pure Math XXVI II