11
Area in neutral geometry
11.1
Equidecomposable and equicomplementable figures
Definition 11.1. A rectilinear figure, or simply figure, is a finite union of non-overlapping
triangles. To express that the figure P is the union of non-overlapping triangles Ti for
i = 1 . . . n, we write
n
P =
Ti
i=1
We call the union Ti a dissection of the figure P .
Two triangles (or figures) are non-overlapping if their intersection is at most a finite
union of points and segments.
Definition 11.2. Two rectilinear figures P and Q are equidecomposable means they
are finite unions of non-overlapping, pairwise congruent triangles. Too, we say that the
figures P and Q can be dissected into congruent triangles, or simply, have congruent
dissections.
In this case, we write
P =
n
Ti
and Q =
i=1
n
Si
with Ti ∼
= Si
for i = 1 . . . n
i=1
Definition 11.3. Two rectilinear figures P and Q are equicomplementable means there
exist figures R and S such that
(1) R and S are equidecomposable
(2) P and R are non-overlapping
(3) Q and S are non-overlapping
(4) P R and Q S are equidecomposable
Remark. As a short hand, I write
P ∼ Q for figures P and Q being equidecomposable .
P Q to indicate that the figures P and Q are equicomplementable .
Remark. More in everyday language, one may explain as follows: Two figures are
equicomplementable means that one can obtain equidecomposable figures from them,
by adding to each one an appropriately chosen extra figure. These two extra figures
need to be equidecomposable and may not overlap with the two originally given figures.
304
Remark. In German, Hilbert uses the words ”zerlegungsgleich” for equidecomposable ,
and ”ergänzungsgleich” for equicomplementable .
Lemma 11.1. Any two dissections of the same figure have a common refinement.
Proof.
Question. What kinds of polygons occur by intersecting two triangles. How can they
be dissected into triangles.
Answer. The intersection of two overlapping triangles is a convex polygon with at most
six vertices. In the cases of four, five and six vertices, it can be dissected into triangles
by one, two, or three diagonals, respectively.
Question. Explain how to construct a common refinement of two arbitrary dissections
of a figure P .
Answer. Given are the two dissections P = ni=1 Ti and P = m
j=1 Sj of the same figure.
A common refinement is easily obtained by dissecting all overlapping intersection Sj ∩ Ti
into triangles, since these are convex polygons with at most six vertices.
Theorem 11.1. Any two congruent figures are equidecomposable . Any two equidecomposable figures are equicomplementable .
All four notions—equality, congruence, being equidecomposable as well as being equicomplementable —are equivalence relations among figures. Indeed these are four different
equivalence relations.
Problem 11.1. Prove that two figures which are both equidecomposable to a third figure
are equidecomposable . You need to use that any two dissections of the same figure have
a common refinement, as shown in Lemma 11.1 above.
Problem 11.2. Prove that two figures which are both equicomplementable to a third figure are equicomplementable . You can use that the relation of being ”equidecomposable ”
is a equivalence relation.
Remark. Euclid just uses in all four cases the same word ”equal”, and tells in his common
notions that two things equal to a third are equal. That is a good postulate, but not
enough to prove the Theorem 11.1 above.
Given is a triangle ABC, a side AB of which has been chosen as base. I call the line
through the midpoints of the two remaining sides a midline.
Proposition 11.1 (A neutral version of Euclid I.37). Two triangles with the same
base and the same midline are equicomplementable .
Two triangles with the same base, and the midpoints of the four remaining sides
lying on one line, are equicomplementable .
305
Remark. Astonishing enough, proposition 11.1 is even valid in neutral geometry. On
the other hand, Euclid’s original proposition I.37 is not valid in hyperbolic geometry.
Proof. The proof exploits the correspondence between a triangle and a Saccheri quadrilateral given in Hilbert’s Proposition 39. Complete details are given in Proposition 7.2
from the section about Legendre’s geometry.
For a given triangle ABC and base AB, we obtain the corresponding Saccheri
quadrilateral F GBA by the construction given in the figure on page 306. We have
Figure 11.1: To every triangle corresponds a Saccheri quadrilateral.
shown by SAA congruence that
AF D ∼
= CHD
and
BGE ∼
= CHE
Problem 11.3. Prove that the segment DE which the triangle cuts out of the middle
line has length half the base F G of the corresponding Saccheri quadrilateral.
Question. Take the case of an acute triangle and convince yourself that the triangle
ABC and the Saccheri quadrilateral F GBA are equidecomposable .
We can leave a short answer to this question to the reader. The Saccheri quadrilateral
depends only on the base AB and the midline DE of the triangle. Hence we obtain the
same Saccheri quadrilateral from another triangle with the same base and same midline.
Using the positive answer to the question above, by transitivity (see Theorem 11.1) we
conclude that two acute triangles with the same base and midline are equidecomposable .
In the case of an obtuse triangle, it can happen that the triangle ABC and the
Saccheri quadrilateral F GBA are only equicomplementable . The details are explained
below. By transitivity, now used for the relation of equicomplementarity, we conclude
that any two triangles with the same base and midline are equicomplementable .
Question. Take the case of an obtuse triangle, as given in the figure on page 307. Explain why the triangle ABC and the Saccheri quadrilateral F GBA are equicomplementable . Let the numbers denote non-overlapping regions.
306
Figure 11.2: Even an obtuse triangle is equicomplementable to its corresponding Saccheri
quadrilateral.
Answer. Triangle congruences yield
1 5∼
=4 6
and
The Saccheri quadrilateral and the triangle
ABGF ∼ 1 2 and
3
5∼
=6
ABC ∼ 2
3 4
are complemented by the triangular region 5 as additional figure. We conclude
5∼2
4 6
ABGF
5∼ 1 2
1 5 ∼2
6∼ 2 4
5
∼ 2 4
3 5 ∼ 2 4 3
∼ ABC
5
Hence the Saccheri quadrilateral and the triangle are equicomplementable .
11.2
The winding number
In every ordered incidence plane, it is possible to define an orientation. The orientation
is fixed, as soon as one has agreed which are the left and right half plane for one fixed
ray. We denote the congruence class of right angles by R.
Definition 11.4 (Winding number function). Given a triangle ABC with positive (counterclockwise) orientation, we define the winding number function ABC (X)
307
around any point X to be
⎧
4R
⎪
⎪
⎪
⎪
⎪
2R
⎪
⎪
⎪
⎨α
ABC (X) =
⎪
β
⎪
⎪
⎪
⎪
⎪
γ
⎪
⎪
⎩
0
if
if
if
if
if
if
X
X
X
X
X
X
lies inside ABC.
= A, B, C lies on a side of ABC.
= A.
= B.
= C.
lies outside the ABC.
which is greater or equal zero. The winding number function for a triangle ABC with
negative (clockwise) orientation is the negative of the above. If the three points A, B, C
lie on a line, the winding number function is identical zero.
Lemma 11.2. The winding number function ABC is identically zero if and only if
the three points A, B, C lie on a line.
The winding number functions satisfy OAB + OBC = OAC if and only if
point B lies between the points A and C.
Proposition 11.2. For any point X = O, A, B, C the winding number functions satisfy
the permutation property
(11.1)
ABC = BCA = CAB = − BAC = − ACB = − CBA
and the cyclic additivity properties:
(11.2)
ABC = OAB + OBC + OCA
Problem 11.4. Check that
ABC = P AB + P BC + P CA
still holds for an arbitrary vertex P = O.
11.3
Area of rectilinear figures
As well known from everyday life, it is more practical to measure the area of figures,
instead of only comparing different figures. As a first step in the discussion of area, we
state the properties which need to hold for a useful notion of area. Secondly, we give a
construction of the area function and thus prove its existence.
Definition 11.5. The area or measure is a function assigning to any rectilinear figure
P a number area(P ) such that the following properties hold:
(1) area(T ) > 0 for all triangles T
308
(2) T ∼
= S implies area(T ) = area(S) for any two triangles
(3) area(P Q) = area(P ) + area(Q) for any non-overlapping figures P and Q.
More generally, the value area(P ) can be an element of any ordered Abelian group.
Step 1. As the first step, we define the area for any triangle ABC. The is done quite
differently for Euclidean geometry on the one side and for semi-hyperbolic or semi-elliptic
geometry on the other side.
In any Pythagorean plane the area of a triangle is defined to be half of base times
height.
1
1
1
area(ABC) = aha = bhb = chc
2
2
2
In any semi-hyperbolic plane the area of a triangle is defined to be the deviation
of its angle sum from two right angles
δ(ABC) = 2R − α − β − γ
Recall that δ(ABC) is the defect of the triangle, according to definition 7.4.
In any semi-elliptic plane the area of a triangle is defined to be the excess of its
angle sum over two right angles α + β + γ − 2R
Remark. We see from this definition that the values of area are in different Abelian
groups depending on the geometry and are justified in different ways.
• The former definition satisfies the positivity requirement (1) because, as guaranteed by Theorem 4.1, the segment lengths in any Pythagorean plane are an ordered
field. Moreover we have to use the similarly of triangles with opposite orientation.
• The latter two definitions satisfy the positivity requirement (1) because of the
Uniformity Theorem 15.
Step 2. We define the left- and right half-planes for any ray, and the clockwise or counterclockwise orientation for any triangle. Note that the orientation depends on the order
of the vertices of the triangle. We define the signed area [ABC] of ABC as
+area(ABC)
if ABC has counterclockwise orientation
[ABC] =
−area(ABC)
if ABC has clockwise orientation
Lemma 11.3. For any base point O, the signed triangle areas satisfy
(11.3)
[ABC] = [OAB] + [OBC] + [OCA]
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Lemma 11.4. [P AB] + [P BC] + [P CA] = 0 if and only the three points A, B and C
lie on a line.
Lemma 11.5. If the points X and Y lie on the same side of line EF , the two signed
areas [XEF ] and [Y EF ] have the same signs. If the points X and Y lie on opposite
sides of line EF , the two signed areas [XEF ] and [Y EF ] have the opposite signs.
Step 3. We define the area of a figure as the sum of the areas of the triangles in any
dissection. We show that such a definition cannot lead to a contradiction.
Lemma 11.6 (Main Additivity Lemma). If any triangle is dissected into finitely
many non-overlapping triangles, its area is the sum of the triangles of the dissection.
T =
n
Ti
implies area(T ) =
i=1
n
area(Ti )
i=1
Proof. We give the triangle T = ABC and all triangles of the dissection the positive
orientation.
Any reference point O is chosen. We use the formula (11.3) for each term
of the sum ni=1 area(Ti ). There occur contributions [OEF ] of two different types:
• from a side of some triangle Ti in the dissection, which does not entirely lie on one
of the sides AB, BC or CA of the big triangle T
• from a side of some triangle Ti lying entirely on a side of the triangle T .
Assume that for side EF from the dissection, the first case applies. There exist two
triangles in the dissection having side EF , say T1 = EF X and T2 = F EY , lying in
the two half-planes of EF . Hence they do not overlap. They have the same orientation,
which we have assumed to be positive. The total area of the quadrilateral formed by
the two triangles is
area(EY F X) = area(EF X) + area(F EY ) = [EF X] + [F EY ]
= [OEF ] + [OF X] + [OXE] + [OF E] + [OEY ] + [OY F ]
= [OF X] + [OXE] + [OEY ] + [OY F ]
We see that the two contributions from the common side EF cancel. The same cancelation occurs for all interior sides of the dissection.
The contributions from sides of some triangle Ti lying entirely on one of the sides of
the big triangle T add up to
[OAB] + [OBC] + [OCA] = area(ABC)
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Theorem 11.2. Assume that the area for triangles has the properties
(1) area(T ) > 0 for all triangles T
(2) T ∼
= S implies area(T ) = area(S) for any two triangles
(3t) If two non-overlapping
triangles P and Q are joined to a larger triangle, this has
the area(P Q) = area(P ) + area(Q).
Then the area of any rectilinear figure is well defined. Indeed, one gets the same sum
n
area(Ti ) =: area(P )
i=1
independent of the dissection
P = Ti one has used.
Furthermore, area(P Q) = area(P )+area(Q) holds for any non-overlapping figures
P and Q.
Proof of Theorem 11.2. Given are the two dissections P = ni=1 Ti and P = m
j=1 Sj of
the same figure. Let Rk with k = 1 . . . K ≤ 3nm be a common refinement. By applying
the additivity lemma to each Ti , and each Sj , we obtain
n
i=1
area(Ti ) =
K
area(Rk ) =
m
area(Sj )
j=1
k=1
Hence any two dissections yield the same value of sum of area for the given figure, which
is thus well defined.
The proof of the second item can be left to the reader.
Corollary 30. The interior domain R of a simple closed polygon P1 , P2 , . . . Pn , Pn+1 =
P1 has the area
n
area(R) =
[OPi Pi+1 ]
i=1
We have put Pn+1 = P1 , and assumed positive orientation.
We can dissect the interior region R into triangles, and let all triangles have positive
orientation. Then the circumference curves of these triangles add up to the positively
oriented polygon surrounding them all.
Theorem 11.3. (a) Any two equidecomposable figures have the same area.
(b) Any two equicomplementable figures have the same area.
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Proof. Let P and Q be two equidecomposable figures. By definition, they have dissections
P =
n
i=1
Ti
and Q =
Ti ∼
= Si
n
Si
such that
i=1
for i = 1 . . . n
From the definition of area, we get
area(P ) =
n
area(Ti ) =
n
i=1
area(Si ) = area(Q)
i=1
as to be shown. Now let P and Q be two equicomplementable figures. By definition,
there exist figures R and S such that
(1) R and S are equidecomposable
(2) the non-overlapping unions P R and Q S are equidecomposable
Hence part (a) and the additivity from theorem 11.2 imply
area(R) = area(S)
area(P
R) = area(Q S)
area(P ) + area(R) = area(Q) + area(S)
Since substraction is defined in the Abelian group from which the value of area are
taken, we get area(P ) = area(Q), as to be shown.
Proposition 11.3. If the figures Pi are equidecomposable to the figures Qi for i = 1 . . . n,
n
and neither the figures P
nor
the
figures
Q
overlap,
then
the
union
i
i
i=1 Pi is equiden
composable to the union i=1 Qi .
The corresponding statement holds in the case of equicomplementable figures.
Proposition 11.4 (The postulate of de Zolt and Stolz). Let Q ⊂ P be two figures
such that there exists a triangle T inside P which does not overlap with the smaller
figure Q. Then the figures P and Q are not equicomplementable .
Proof. From dissections of the figures P and Q and Q T , we construct a common
refinement Rk such that
Q=
K
k=1
Rk ,
T =
L
k=K+1
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Rk ,
P =
M
k=1
Rk
Figure 11.3: deZolts postulate tells that P and Q are not equicomplementable .
with K < L ≤ M . By the additivity of area (Theorem 11.2), we get
area(Q) =
K
area(Rk ) <
M
k=1
area(Rk ) = area(P )
k=1
Hence area(Q) = area(P ). By part (b) of Theorem 11.3, the figures P and Q are not
equicomplementable .
11.4
A standard equicomplementable form for a figure
There remains the question—are any two figures with the same area equicomplementable ?
To prove this converse of Theorem 11.3, one needs a standard equicomplementable form
for any given figure. In Euclidean geometry, a rectangle with one side unit length, the
other side the given area, is a convenient standard form. In hyperbolic geometry, no
rectangles exist. Moreover, the area of Saccheri quadrilaterals with a unit base turns
out to be bounded above. A way out is to use for the purpose of comparison a finite
set of non-overlapping congruent triangles. We present this approach in a way valid for
any Hilbert plane for which the circle-line intersection property holds.
Lemma 11.7. To any given triangle and a segment longer than its shortest side, there
exists an equicomplementable triangle which has the given segment as one of its sides.
Proof. Let the given triangle ABC have side AB < c where c is the given segment.
We bisect sides AB and AC and draw the middle line DE connecting the midpoints D
and E, respectively. Let D be an intersection point of the middle line with the circle
313
−−→
around B of radius c /2. We extend the ray BD to get the point A such that D is
the midpoint of segment BA . The triangle A BC and the given triangle have the
common base BC and the common middle line DE. The triangles ABC A BC
are equicomplementable by Proposition 11.1, and obviously the triangle A BC has the
side A B of length c as required.
Lemma 11.8. Any triangle is equicomplementable to an isosceles triangle with the same
base and the same midline.
Proof. For a given triangle ABC and side AB, we obtain the corresponding Saccheri
quadrilateral F GBA by the construction given in the figure on page 306. Let DE
be the middle line and p be the perpendicular bisector of side AB. The top of the
isosceles triangle ABC is the point on p which has the same distance from line DE
as the points A and B, but lies on the other side. The two triangles ABC ABC have the common side AB and the same midline DE. By Proposition 11.1 they are
equicomplementable .
Lemma 11.9. For any two given triangles T1 and T2 there exists a pair of non-overlapping
congruent triangles T ∼
= T such that the unions
T1 T2 T T are equicomplementable .
Proof. If Ti are two congruent equilateral triangles, we are ready. Suppose this is not
the case and that triangle T2 has a side a longer than one side of T1 . By Lemma 11.7
there exists a triangle T1 equicomplementable to T1 and having the side a . We have
obtained two equicomplementable triangles T1 = A BC and T2 = A2 BC which we
can put into the two opposite half planes of their common side a = BC.
By Lemma 11.8 there exist isosceles triangles T ∗ = A∗ BC and T ∗∗ = A∗∗ BC
on opposite sides of their common base BC such that T ∗ T1 and T ∗∗ T2 . The kite
A∗ BCA∗∗ has be constructed as the non-overlapping union of T ∗ with T ∗∗ but clearly
A∗ BCA∗∗ = T ∗ T ∗∗ = BA∗ A∗∗
CA∗ A∗∗
and the latter are two congruent triangles. We put T := BA∗ A∗∗ and T :=
CA∗ A∗∗ , which are congruent triangles and are ready.
Lemma 11.10. Any given figure is equicomplementable to a disjoint union 2n T of 2n
congruent non-overlapping triangles T .
Proof. By further subdivisions, any figure P is equicomplementable to a union of 2n
non-overlapping triangles T1 , T2 , . . . T2n . Applying the Lemma 2n−1 times we construct
triangles Ti ∼
= Ti for i = 1 . . . 2n−1 such that
n
2
i=1
Ti n−1
2
i=1
314
Ti
Ti
In an inductive process for s = 1 . . . n, we get a union of 2n non-overlapping triangles
which
• together are still equicomplementable to the originally given figure P ;
• consists of 2n−s non-overlapping subsets,
• each of which is the non-overlapping union of 2s congruent triangles.
In the end, we see that the given figure P is equicomplementable to a disjoint union of
2n congruent non-overlapping triangles.
Lemma 11.11. For any two given triangles33 S and T there exists isosceles triangles
P and Q with a common base such that S P and T Q.
Proof. If S and T are two congruent equilateral triangles, we are ready. Suppose this is
not the case and that triangle T has a side a longer than one side of S. By Lemma 11.7
there exists an equicomplementable triangle S S having the side a . We have obtained
two triangles S and T with the common side a = BC. By Lemma 11.8 there exist
isosceles triangles P S and Q T , all on the common side a .
Proposition 11.5 (Figures with same area are equicomplementable). Assume
in the given Hilbert plane that
• the circle-line intersection property holds;
• an area has been defined in accordance to Definition 11.5. The area takes values
in an ordered and hence free Abelean group such that
(1) area(T ) > 0 for all triangles T
(2) T ∼
= S implies area(T ) = area(S) for any two triangles
(3) area(P Q) = area(P ) + area(Q) for any non-overlapping figures P and Q.
Then any two figures P and Q for which area(P ) = area(Q) are equicomplementable .
Remark. An additive group is called free if ng = 0 implies g = 0 for any integer n = 0
and any group element g. An ordered Abelean group is always free.
Proof. By Lemma 11.10, the given figure F is equicomplementable to a disjoint union
2n S of 2n congruent non-overlapping triangles S and the given figure G is equicomplementable to a disjoint union 2n T of 2n congruent non-overlapping triangles T . The
integer n can be chosen to be equal for both figures, by means of further subdivisions.
By assumption, both figures have the same area. Hence
2n area(T ) = area(F ) = area(G) = 2n area(S)
33
they need not have the same content
315
Since the area takes values in an ordered and hence free Abelean group, we conclude
that area(T ) = area(S).
By Lemma 11.11 there exists isosceles triangles P and Q with a common base such
that S P and T Q. Hence
area(P ) = area(S) = area(T ) = area(Q)
We put the isosceles triangles P = XBC and Q = Y BC on the same side of their
common base BC. Let M be the midpoint of BC. We may assume that the altitude of
XM ≥ Y M .
Because of the assumption that the area of any triangle is positive, it is impossible
that X = Y . Indeed in that case
XBC = Y BC
XY B
XY C
would imply that area(P ) > area(Q).
Now since X = Y , we see that actually P = Pt BC = Qt BC = Q. Wee that the
originally given figures are equicomplementable since
SP =QT
2n T G
F 2n S 11.5
The role of the Archimedean axiom
Problem 11.5. We assume that the Archimedean axiom holds. Explain why any triangle T even equidecomposable to a corresponding Saccheri quadrilateral.
Proof of Theorem ??. The assertion holds for every acute triangle. For an obtuse triangle, we reverse the process used in the first Legendre Theorem. Thus we obtain a
sequence of triangles by bisection of the longest side and having all the same base. They
have a common middle line. According to Problem 11.3 and the figure on page 306
, the middle line cuts these triangles in segments DE which have the length half the
base F G of the Saccheri quadrilateral. By the Archimedean axiom, it takes only a finite
number of steps to obtain a triangle which cuts the midline in two point between or
on the base of the Saccheri quadrilateral and hence is equidecomposable to the Saccheri
quadrilateral.
Lemma 11.12. We assume that circle-line intersection property and the Archimedean
axiom hold. To any given triangle and a segment longer than its shortest side, there
exists an equidecomposable triangle which has the given segment as one of its sides.
Lemma 11.13. We assume that circle-line intersection property and the Archimedean
axiom hold. Then any two equicomplementable triangles are even equidecomposable .
316
Theorem 11.4. Assume that the Archimedean axiom (V.1) holds, any two equicomplementable figures are even equidecomposable .
We now turn the case that the Archimedean axiom is not satisfied. We refer to
the figure on page 307 with the equicomplementable Saccheri quadrilateral and obtuse
triangle
AF BG ABC
Note that AG is the diameter of the Saccheri quadrilateral. Let M be the midpoint of
its base F G.
Lemma 11.14. Suppose these two figures have congruent dissections into t triangles.
Then
3t + 4
|HM | ≤
|AG|
2
Proof. Let C be the sum of the circumferences of all t triangles of the dissection.
|AB| + |BC| + |AB| ≤ C ≤ 3t |AG|
For the circumference of triangle ABC, we need to get a the lower bound from the
triangle inequality.
2|HM | = |HF | + |HG| ≤ 4|F A| + |AB| + |BC| ≤ 4|AG| + 3t |AG|
Lemma 11.15. Suppose that the Saccheri quadrilateral and obtuse triangle from the
figure on page 307 are even equidecomposable
AF BG ∼ ABC
and that
|HM | ≥ N · |AG|
for some large integer N . Then the congruent dissections contain at least
t≥
2N − 4
3
triangles.
Proof. The assumption and Lemma 11.14 imply
N · |AG| ≤ |HM | ≤
3t + 4
· |AG|
2
and solving for t yields the estimate to be shown.
317
Proposition 11.6. Assume the Archimedean axiom (V.1) does not hold. Then there
exist a triangle and a Saccheri quadrilateral that are equicomplementable but not equidecomposable .
Proof.
The logical relations between the different concepts of content gained in this section
are summed up in the following diagram:
clear
P and Q are equidecomposable
⏐
⏐
Th. 11.3
−−−→
P and Q are equicomplementable
⏐
⏐
Th. 11.3
area(Q)
area(P )
⏐
⏐
Archimedes and Th. 6.4
⏐
⏐
Th. 6.2
P and Q are equidecomposable ←−−−−−−−−−−− P and Q are equicomplementable
Archimedean axiom
Corollary 31. Assuming the Archimedean axiom (V.1) holds, any two equicomplementable figures are even equidecomposable . Assuming the Archimedean axiom (V.1)
does not hold, there exist two equicomplementable figures that are not equidecomposable .
318