11 Area in neutral geometry

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 T i for i =1...n, we write n P = We call the union T i 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 T i and Q = T i n S i with T i = Si for i =1...n 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 = n T i and P = m j=1 S j of the same figure. A common refinement is easily obtained by dissecting all overlapping intersection S j T i 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 FGBA 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 FG of the corresponding Saccheri quadrilateral. Question. Take the case of an acute triangle and convince yourself that the triangle ABC and the Saccheri quadrilateral FGBA 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 FGBAare 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 FGBA 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 The Saccheri quadrilateral and the triangle 1 5 = 4 6 and 3 5 = 6 ABGF 1 2 and ABC 2 3 4 are complemented by the triangular region 5 as additional figure. We conclude ABGF ( 5 1 ) 2 5 2 ( 1 ) 5 2 ( 4 ) 6 ( 2 ( 4) 6 2 ) ( 4 3 ( 5) 2 4 3) 5 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 if X lies inside ABC. 2R if X A, B, C lies on a side of ABC. α if X = A. ABC (X) = β if X = B. γ if X = C. 0 if X 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 still holds for an arbitrary vertex P O. ABC = PAB + PBC + PCA 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 anumberarea(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 height. theareaofatriangleisdefinedtobehalfofbasetimes area( ABC) = 1 2 ah a = 1 2 bh b = 1 2 ch c In any semi-hyperbolic plane theareaofatriangleisdefinedtobethedeviation of its angle sum from two right angles δ(abc) =2R α β γ Recall that δ(abc) isthedefect 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] 309

Lemma 11.4. [PAB]+[PBC]+[PCA]=0if 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 [YEF] have the same signs. If the points X and Y lie on opposite sides of line EF, the two signed areas [XEF] and [YEF] 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. n T = T i implies area(t )= n area(t i ) 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 n area(t i). There occur contributions [OEF] of two different types: from a side of some triangle T i 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 T i 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, sayt 1 = EFX and T 2 = FEY, 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 FX)=area( EFX)+area( FEY)=[EFX]+[FEY] =[OEF]+[OFX]+[OXE]+[OFE]+[OEY ]+[OY F] =[OFX]+[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 T i lying entirely on one of the sides of the big triangle T add up to [OAB]+[OBC]+[OCA] =area( ABC) 310

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(t i )=:area(p) independent of the dissection P = T i 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 = n T i and P = m j=1 S j of the same figure. Let R k with k =1...K 3nm be a common refinement. By applying the additivity lemma to each T i,andeachs j,weobtain n area(t i )= K area(r k )= k=1 m area(s j ) j=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 P 1,P 2,...P n,p n+1 = P 1 has the area n area(r) = [OP i P i+1 ] We have put P n+1 = P 1, 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. 311

Proof. Let P and Q be two equidecomposable figures. By definition, they have dissections P = n T i and Q = From the definition of area, we get n S i T i = Si for i =1...n such that area(p )= n area(t i )= n area(s i )=area(q) 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 P i are equidecomposable to the figures Q i for i =1...n, and neither the figures P i nor the figures Q i overlap, then the union n P i is equidecomposable to the union n Q i. 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 R k such that Q = K R k, T = k=1 L k=k+1 R k, P = M k=1 R k 312

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(r k ) < k=1 M area(r k )=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 FGBA 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 T 1 and T 2 there exists a pair of non-overlapping congruent triangles T = T such that the unions are equicomplementable. T 1 T2 T T Proof. If T i are two congruent equilateral triangles, we are ready. Suppose this is not the case and that triangle T 2 has a side a longer than one side of T 1. By Lemma 11.7 there exists a triangle T 1 equicomplementable to T 1 and having the side a. We have obtained two equicomplementable triangles T 1 = A BC and T 2 = A 2 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 T 1 and T T 2. 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 2 n T of 2 n congruent non-overlapping triangles T. Proof. By further subdivisions, any figure P is equicomplementable to a union of 2 n non-overlapping triangles T 1,T 2,...T 2 n. Applying the Lemma 2 n 1 timesweconstruct triangles T i = T i for i =1...2 n 1 such that 2 n T i 2 n 1 ( 314 T i ) T i

In an inductive process for s =1...n, we get a union of 2 n non-overlapping triangles which together are still equicomplementable to the originally given figure P ; consists of 2 n s non-overlapping subsets, each of which is the non-overlapping union of 2 s congruent triangles. In the end, we see that the given figure P is equicomplementable to a disjoint union of 2 n congruent non-overlapping triangles. Lemma 11.11. For any two given triangles 33 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 =0foranyintegern 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 2 n S of 2 n congruent non-overlapping triangles S and the given figure G is equicomplementable to a disjoint union 2 n T of 2 n 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 33 they need not have the same content 2 n area(t )=area(f )=area(g) =2 n area(s) 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 = YBC on the same side of their common base BC. LetM be the midpoint of BC. We may assume that the altitude of XM YM. Because of the assumption that the area of any triangle is positive, it is impossible that X Y. Indeed in that case XBC = YBC XY B XY C would imply that area(p ) > area(q). Now since X = Y, we see that actually P = P t BC = Q t BC = Q. Wee that the originally given figures are equicomplementable since S P = Q T F 2 n S 2 n T G 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 FG 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 FG. 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 FA + 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 and that AF BG ABC HM N AG for some large integer N. Then the congruent dissections contain at least triangles. t 2N 4 3 Proof. The assumption and Lemma 11.14 imply 3t +4 N AG HM 2 and solving for t yields the estimate to be shown. 317 AG

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: P and Q are equidecomposable clear Th. 11.3 Th. 11.3 area(p ) P and Q are equicomplementable area(q) Archimedes and Th. 6.4 Th. 6.2 P and Q are equidecomposable Archimedean axiom P and Q are equicomplementable 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