6 Area in Euclidean Geometry

6.1 Equidecomposable and equicomplementable figures

Problem 6.1. Draw two parallelograms on the same base with congruent altitudes, andexplain why they are equidecomposable . Clearly draw the dissections you use.

Figure 6.1: A square and a parallelogram that are easily seen to be equidecomposable .

Answer. In the figure on page 454, the square is decomposed as 1⊎

2. The parallelogramon the same base is decomposed as 2

⊎3. Since the triangles 1 ∼= 3 are congruent, the

square and the parallelogram as equidecomposable .

Problem 6.2. Draw two parallelograms on the same base with congruent altitudes thatare not easily seen to be equidecomposable . Explain why they are equicomplementable .Clearly draw the dissections and mark the additional figures used.

Figure 6.2: A square and a parallelogram that are only shown to be equicomplementable .

Answer. In the figure on page 454, the square is decomposed as 1⊎

2. The parallelogramon the same base is decomposed as 2

⊎3. But this time 1 and 3 are not congruent.

To see that the square and the parallelogram are equicomplementable , we use thetriangle 4 as additional figure. Since the unions

1⊎

4 ∼= 3⊎

4

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are indeed congruent, we conclude

square⊎

4�∼(

1⊎

2)⊎

4�∼ 2⊎(

1⊎

4)�∼ 2⊎(

3⊎

4)

�∼(

2⊎

3)⊎

4�∼ parallelogram

⊎4

Hence the square and the parallelogram—with the same base and congruent altitudes—are equicomplementable . Similarly, we prove that any two parallelograms with the samebase and congruent altitudes are equicomplementable .

We can now give the exact modern version for the propositions about area in Euclid’sbook one.

Proposition 6.1 (Instead of Euclid I.35). Two parallelograms on the same base withcongruent altitudes are equicomplementable .

Proposition 6.2 (Instead of Euclid I.36). Two parallelograms with congruent bases andcongruent altitudes are equicomplementable .

Problem 6.3. Explain why two parallelograms with the same base and congruent alti-tudes are always equicomplementable .

Proposition 6.3 (Instead of Euclid I.37). Two triangles with the same base and con-gruent altitudes are equicomplementable .

Remark. Euclid talks about triangles ”in the same parallels”. Euclid I.37 says literallythat "Triangles on the same base in the same parallels are equal."

Proposition 6.4 (Instead of Euclid I.38). Two triangles with congruent bases and con-gruent altitudes are equicomplementable .

Problem 6.4. Convince yourself by an informal argument that two triangles with thesame base and congruent altitudes are equicomplementable .

Problem 6.5. Use the neutral version of Euclid I.37 given by proposition 11.1 aboveto prove in Euclidean geometry that two triangles with the same base and congruentaltitudes are equicomplementable .

Answer. In Euclidean geometry, but only in Euclidean geometry, two triangles with thesame base and congruent altitudes have the same midline. (Work out further details!)Hence by proposition 11.1 above they are equicomplementable , too.

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Figure 6.3: Euclid’s leg theorem tells the square over the leg is equicomplementable to therectangle as indicated.

Figure 6.4: Euclid’s proof of the leg theorem.

6.2 The Theorem of Pythagoras and related results

As a first step towards the Pythagorean Theorem, Euclid proves the following versionof the leg theorem:

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Proposition 6.5 (Euclid’s leg theorem). The square on a leg of a right triangle isequicomplementable to the rectangle one side of which is the hypothenuse and the otherside is the projection of this leg onto the hypothenuse.

Modern proof following Euclid. For a given right triangle �ABC, we have constructedthe square �ACED over the leg AC and the square �ABIH over the hypothenuseAB. We have dropped the altitude from vertex C onto the hypothenuse and got theprojection AF of the leg AC. Finally we have obtained the rectangle �AFGH, one sideof which is the hypothenuse and the other side is the projection AF of the leg AC ontothe hypothenuse AB.

To show that the square �ACDE and the rectangle �AFGH are equicomple-mentable , we use as intermediate steps the two parallelograms �ABJD and �AHKC.Here point J can be obtained by transferring the leg AD ∼= BJ , and point K can beobtained by transferring the altitude FC ∼= GK.

Question. Convince yourself that both quadrilaterals �ABJD and �AHKC are indeedparallelograms.

Question. Convince yourself that these two parallelograms are congruent. How can oneeasily obtain parallelogram �AHKC from parallelogram �ABJD.

Question. Why is the square�ACED equicomplementable to the parallelogram�ABJD.

Answer. These two parallelograms have the same base AD and congruent altitudes.Hence they are equicomplementable by Euclid I.37 (see proposition 6.3).

Question. Why is the parallelogram�ACKH equicomplementable to the rectangle�AFGH.

Answer. These two parallelograms have the same base AH and congruent altitudes.Hence they are equicomplementable by Euclid I.37 (see proposition 6.3).

By these steps, we have obtained the following chain of equicomplementable figures:

�ACED � �ABJD ∼= �ACKH � �AFGH

By Theorem 11.1, any two congruent figures are equicomplementable . Furthermore,the relation ”equicomplementable ” is an equivalence relation among figures. Hence weconclude that

�ACED � �AFGH

as to be shown.

Proposition 6.6 (Euclid’s (weaker) Pythagorean theorem). The sum of thesquares on the legs of a right triangle is equicomplementable to the square over the hy-pothenuse.

Modern proof following Euclid. We use Euclid’s leg theorem 6.5 twice, for both legs:

�ACED � �AFGH and �BCKJ � �BFGI

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Figure 6.5: Euclid’s proof of the Pythagorean Theorem.

Since these figures do not overlapping we conclude

�ACED⊎�BCKJ � �AFGH

⊎�BFGI � �ABIH

as to be shown.

Theorem 6.1 (The Pythagorean Theorem). The sum of the squares on the legs ofa right triangle is equidecomposable to the square over the hypothenuse.

Problem 6.6. We want to prove the Theorem of Pythagoras by dissection. What wentwrong in the above figure on page 459 to get an appropriate dissection. Draw a figurewith a more useful dissection.

Problem 6.7. Prove the Theorem of Pythagoras using the figure on page 459. For aproof by dissection, we need to

• give a construction of this diagram which explains in which order it was obtained.

• Based on this construction, we need to prove the congruence of the pieces to whichthe square on the hypothenuse is dissected to the pieces of which the two squareserected on the legs are made up.

Proof. Here is a way to do these steps: We construct the center J of the square �BCFEon the larger leg of the given right triangle �ABC. Drop the perpendicular from J

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Figure 6.6: What went wrong?

Figure 6.7: A dissection proof of the Theorem of Pythagoras.

onto the hypothenuse AB and erect the ”double perpendicular” at point J . The square�BCFE cuts the segments NM and KL out of these two perpendicular lines.

We need to determine the segment lengths |CK| = x and |BL| = y. We see that

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y+x = |BL|+|LE| = a. The parallelogram �ABLK has two pair of opposite congruentsides. Hence |KL| = |AB| = c and b + x = |AC| + |CK| = |AK| = |BL| = y.

Since |KJ | = |JL| = c2, the four quadrilaterals into which the square �BCFE

is dissected, are congruent. They are transferred by several different parallel shiftsto the square �ABIH on the hypothenuse. We do this in such a manner that thefour right angles at vertex J are transferred to the four vertices A,B, I,H. Hence thetransferred quadrilaterals have vertices at the midpoints P,Q,W,Z of the sides of thesquare �ABIH. Furthermore, they have each one vertex A′, C ′, G′ or D′ lying insidethe square.

We get right angles at these four points. Since A′C ′ ∼= C ′G′ ∼= G′D′ ∼= D′A′ =y − x = b, we see that the square �A′C ′G′D′ is congruent to the square �ACGDerected over the shorter leg AC.

Hence the square over the shorter leg, and the four congruent quadrilaterals intowhich the square over the longer leg has been dissected are put together without over-lapping to form the square over the hypothenuse.

Figure 6.8: Dissection proof of the Pythagorean theorem in a special case

Problem 6.8. Given is a right triangle �ABC with angle α = 60◦ and side a = 2.Squares are erected on its three sides. As done in the figure on page 459, one canconstruct dissections of them which prove the Theorem of Pythagoras. What kind ofquadrilateral is �ALKB. Determine the segment lengths x = |LE| and y = |BL|.

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Answer. The quadrilateral �ABLK is a parallelogram. Its opposite sides are congruent.Hence y = |AK| = |AC| + |CK| = |AC| + |LE| = b + x. Secondly, a = |BE| =|BL| + |LE| = y + x. From the equations

x + y = a , y − x = b one concludes

x =a− b

2= 1 −

√3

3and y =

a + b

2= 1 +

√3

3

6.3 Dudeney’s dissection problem

How can one dissect a unit square into a few polygons and rearrange them into anequilateral triangle? This is a problem from ”The Canterbury Puzzles” by H.E. Dudeney[?], first published in 1908. Dudeney gave the solution indicated in the drawing below,which uses just four polygons. A less ingenious solution is given in the figure on page 462.

Figure 6.9: Dudeney’s dissection of an equilateral triangle and a square of the same area.

It uses five pieces some of which occur as mirror images.

Problem 6.9. We assume the side of the square to be equal to one. Calculate segmentsa and x + y and the angle α in the accompanying drawing for Dudeney’s dissection toconfirm its existence.

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Figure 6.10: Another dissection of an equilateral triangle and a square of the same area.

Answer. The side a is obtained from the total area 1 =√34

a2 to be

a = 2 (3−1/4) = 2

√√3

3

The two congruent quadrilaterals on the left side

�HAY Q ∼= �HCY ′Q

are obtained rotating by 180◦ about H. Hence y = |AY | = |CY ′|. The two congruentquadrilaterals on the right side

�GBXL ∼= �GCX ′P

are obtained rotating by 180◦ about G. Hence x = |BX| = |CX ′|. To complete thedissection, the triangles

�XLY ∼= �X ′P ′Y ′

have to be congruent. Hence a− x− y = |XY | = |X ′Y ′| = x + y and

x + y =a

2

Since |P ′X ′| = |LX| = |X ′P | = 12, we get from the right triangle �X ′P ′Y ′

cosα =|P ′X ′||X ′Y ′| =

1

a=

31/4

2<

√2

2and α ≈ 48.85◦ > 45◦

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Problem 6.10. Find four segments of the length z = |QY ′|. Calculate the length |Q′L|in terms of z.

Problem 6.11. Give a Euclidean construction for Dudeney’s dissection beginning witha unit square.

Answer. The three figures below on page 463, page 464, and page 465 give stages ofsuch a construction.

Figure 6.11: A construction of Dudeney’s dissection starting with the square I.

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Figure 6.12: A construction of Dudeney’s dissection starting with the square II.

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Figure 6.13: A construction of Dudeney’s dissection starting with the square III.

465

Problem 6.12. Calculate the segment x and the ratios xaand y

a. You can use the cos

Theorem for the triangle �Y GB.

Answer. The sides of the triangle �Y GB have the lengths

|Y G| = 1 , |GB| =a

2, |BY | = x +

a

2

Using the 60◦ angle at vertex B, the cos Theorem yields

12 =a2

4+(x +

a

2

)2− 2

a

2

(x +

a

2

)cos 60◦ = x2 + xa +

a2

2− a

2

(x +

a

2

)x2 +

ax

2+

a2

4− 1 = 0

x =−a +

√16 − 3a2

4

x

a=

−1 +√

4√

3 − 3

4

y

a=

3 −√

4√

3 − 3

4

Hencex ≈ .24549 and y ≈ .25451

Problem 6.13. Give a Euclidean construction for Dudeney’s dissection beginning witha given equilateral triangle.

Answer. The figure on page 467 shows such a construction.

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Figure 6.14: A construction of Dudeney’s dissection starting with the triangle.

467

6.4 Euclidean area

As the first step, we define the area for any triangle �ABC. In Euclidean geometry,the area of a triangle is defined to be half of base times height.

area(�ABC) =1

2aha =

1

2bhb =

1

2chc

Problem 6.14. Prove that it does not matter in this definition, which side of the triangleis chosen as base. 40

Figure 6.15: For the calculation of its area, any side of a triangle may be used as its base.

Answer. For the triangle �ABC, we can take side BC as base. The correspondingaltitude is AD, were D is the foot-point of the perpendicular dropped from vertex Aonto side BC.

As a second possibility, we can take side AC as base. The corresponding altitudeis BE, were E is the foot-point of the perpendicular dropped from vertex B onto sideAC.

The two triangles �CAD and �CBE are equiangular, and hence similar. We getthe proportion

|AD||AC| =

|BE||BC| = sin γ

40If one wants to define the area of rectangle as width time length, one get a similar difficulty—whichagain needs to be resolved by means of similar triangles.

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By multiplication with the denominators, we obtain

|AD| · |BC| = |BE| · |AC| = sin γ|AC| · |BC|

which is both the double area. We see that it does not matter which side one choses asbase.

Remark. We see from this definition that the area takes values the field of segmentarithmetic. This is indeed an Abelian group, as required. Note that this step uses thesimilarly of triangles with different orientation.

Proposition 6.7. The area of the triangle is given by one half product of two sides timeand the sin of the angle between them:

area(�ABC) =1

2ab sin γ =

1

2bc sinα =

1

2ca sin β

Proposition 6.8 (Instead of Euclid I.39). Two equicomplementable triangles with thesame base have congruent altitudes.

Figure 6.16: Two triangles with same base and different altitudes are not equicomple-mentable .

Proof inspired by Euclid. We assume that �ABC and �ABD are equicomplementable .Let their altitudes be CF ⊥ AB and DG ⊥ AB. To check whether they are congruent,

we transfer GD onto the ray−→FC and get the congruent segment FE ∼= GD. The

triangles �ABD and �ABE have the same base and congruent altitudes. By theproposition 6.3, which Euclid has already proved, they are equicomplementable . Henceby theorem 11.1, the triangles �ABC and �ABE are equicomplementable . Too, theiraltitudes dropped from C and E have the same foot point F , and lie on the same line.

The question is now, whether C = E. If not, there are two possible cases left. EitherE lies between F and C, or C lies between F and E. It is enough to rule out the firstcase, the argument in the second case is similar.

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Assume—towards a contradiction—that F ∗E ∗C. In that case the triangle �ABEis a subset of triangle �ABC, to which it is equicomplementable . Too, the triangle�ECB lies inside �ABC, but does not overlap the smaller triangle �ABE. Hence,by deZolt’s postulate 11.4, the triangles �ABE and �ABC cannot be equicomple-mentable . This contradiction rules out the case F ∗ E ∗ C. The case F ∗ C ∗ E can beruled out similarly.

Hence E = C, and the originally given triangles �ABC and �ABD have congruentaltitudes CF = EF ∼= DG ⊥ AB, as to be shown.

Euclid does not succeed to prove deZolt’s postulate in his treatment of area. Buthe did see that an additional assumptions is needed to get besides the simpler EuclidI.37 its converse Euclid I.39. To this end, he suggests an even stronger very generalpostulate about any magnitudes:

”The whole is more than its parts”.Such a general principle is clearly not justifiable. Indeed, it is not valid if dissectionconstructed from infinite sets are allowed. This is exemplified by the Banach-Tarskiparadox.

As we have seen from the proof above, deZolt’s postulate would have been a workableversion for a new axiom that Euclid could have used to complete his theory of area.

Short simple proof of Proposition 6.8. Let �ABC and �ABD be equicomplementable .By theorem 11.3, they have same area. The area is given by half the product of basetimes height. Let CF ⊥ AB and DG ⊥ AB be their altitudes. Hence we know that

1

2|AB| · |CF | =

1

2|AB| · |DG|

Since segment arithmetic is a field, we can multiply both sides by 2 and divide by|AB| = 0 and get

|CF | = |DG|from which follows CF ∼= DG, as to be shown.

Already in the segment arithmetic, it was necessary to introduce a unit segment.Areas are naturally measured by taking the square over the unit segment as unit forarea.

Lemma 6.1. A right triangle T with legs of lengths a and b is equicomplementable to aright triangle with legs 1 and ab, and to a rectangle with legs 1 and area(T ).

Problem 6.15. Prove lemma 6.1.

Proposition 6.9. Every figure P is equicomplementable to a rectangle one side of whichis the unit segment. The second side of this rectangle has the length area(P ).

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Proof. The given figure P can be dissected into triangles. Each of them is either a righttriangle or can be dissected into two right triangles. Hence the figure P can be dissectedinto finitely many right triangles Ti for i = 1...n. Each of them is equicomplementable toa rectangle Ri with sides 1 and area(Ti). By proposition 11.3,

P =n⊎

i=1

Ti �n⊎

i=1

Ri = R

were the latter is a rectangle R with sides 1 and

n∑i=1

area(Ti) = area(P )

as to be shown.

Theorem 6.2. Any two figures with same area are equicomplementable .

Proof. Let P and Q be two figures with same area area(P ) = area(Q). By proposi-tion 6.9, figure P � R, where R is a rectangle, one side of which is the unit segment,and the other side has the length area(P ). Similarly, figure Q � R′, where R′ is a rect-angle with lengths of sides 1 and area(Q). But two rectangles with the same lengths ofsides are congruent, and hence clearly R � R′. By theorem 11.1, ”equicomplementable ”is an equivalence relation. Hence

P � R � R′ � Q

implies the originally given figures P and Q are equicomplementable , as to be shown.

Corollary 42. Any two figures are equicomplementable if and only if they have equalareas.

6.5 The role of the Archimedean axiom

Theorem 6.3. Assuming the Archimedean axiom (V.1) holds, any two figures with samearea are even equidecomposable .

Problem 6.16. We assume that the Archimedean axiom holds. Explain why a righttriangle T is legs of lengths a and b is even equidecomposable to a rectangle with legs 1and area(T ).

Proof of Theorem 6.4. As explained in the problem 6.16 above, with the assumption ofthe Archimedean axiom, a right triangle T with legs of lengths a and b and a rectanglewith legs 1 and area(T ) are shown even to be equidecomposable .

In the same way as in proposition 6.9, we show that any figure P is even equide-composable to a rectangle one side of which is the unit segment. The second side of thisrectangle has the length area(P ).

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Finally, as in the proof of theorem 6.2 that any two figures P and Q with same areaare equidecomposable . (We do not use the Archimedean axiom in the two last steps, butthe reasoning is done now with the stronger equivalence relation ”equidecomposable ”.)

Problem 6.17. Between two parallels AB ‖ CD of distance 5 units, there are given twoparallelograms �ABCD and �ABEF . Let the lengths of side be |AB| = 1, |AC| = 10and |AE| = 20.

Construct dissections into triangles to shown that the two parallelograms are equide-composable . How many triangles do you need?

Problem 6.18. Let n be a natural number. What is the number t of triangles whichare needed to get congruent dissections of a square of side n and a rectangle of sides 1and n2?

(a) Prove that at least �√

2n� triangles are needed.

(b) Give a dissection using 2n triangles.

Can one improve the lower or upper bound?

Answer. Suppose we have congruent dissections of the square of side n and rectangle ofsides 1 and n2 into t triangles. Let L be the length of the longest side of these triangles.

The two opposite long sides of the rectangle are dissected into sides of differenttriangles. Since each triangle fits into the square of side n, the longest side L can havelength at most

√2n. Hence

2n2 ≤ t · L ≤ t√

2n

which implies t ≥ �√2n� as claimed.

Proposition 6.10. Assume the Archimedean axiom (V.1) does not hold. Then thereexist a square and a rectangle that are equicomplementable but not equidecomposable .

Proof. We assume the Archimedean axiom (V.1) does not hold. Hence there existsegments OB and OA such that

OA > n ·OB for all natural numbers n

Now OB takes the role of the unit segment. We put the right triangle �OAB and theright isosceles triangle �OAA′ on the same side of their common leg OA. We draw theparallel to the hypothenuse AB through point A′ and get an intersection point P with

the ray−→OA. In this way, we have constructed a segment OP such that

|OP | · |OB| = |OA|2

The right triangle �OBP and the right isosceles triangle �OAA′ are equicomple-mentable . By doubling these figures, we get a rectangle with sides OB and OP , aswell as a square with side OA, which, too, are equicomplementable .

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We claim these two figures are not equidecomposable . Suppose—towards a contradiction—we have congruent dissections of the square and the rectangle into t triangles. Let L bethe length of the longest side of these triangles.

The two opposite long sides of the rectangle together have total length 2|OP |. Theyare dissected into sides of different triangles. Since each triangle fits into the square, thelongest side L can have length at most

√2 |OA|. Hence

2|OP | ≤ t · L ≤ t√

2 |OA|√2|OP | · |OA| ≤ t|OA|2 = t|OP | · |OB|√

2|OA| ≤ t|OB|On the other hand, we have assumed OA > n ·OB for all natural numbers n, and hencewith n := t we get

t|OB| < |OA|Thus we arrive at the contradiction

√2|OA| < |OA|. This confirms the claim the square

and the rectangle are not equidecomposable .

The logical relations between the different concepts of content gained in this sectionare summed up in the following diagram:

P and Q are equidecomposableclear−−−→ P and Q are equicomplementable⏐⏐Th. 11.3 Th. 11.3

⏐⏐area(P ) area(Q)⏐⏐Archimedes and Th. 6.4 Th. 6.2

⏐⏐P and Q are equidecomposable ←−−−−−−−−−−−

Archimedean axiomP and Q are equicomplementable

Corollary 43. Assuming the Archimedean axiom (V.1) holds, any two equicomple-mentable figures are even equidecomposable . Assuming the Archimedean axiom (V.1)does not hold, there exist two equicomplementable figures that are not equidecomposable .

Problem 6.19. Convince yourself once more that two parallelograms with the same baseand congruent altitudes are always equicomplementable . Why can one not claim theyare always equidecomposable ?

6.6 Some uniqueness results for justification

Given is a Pythagorean plane. Let UQ denote a square the sides of unit length. Doesassigning the area 1 to this figure determine the area function uniquely?

Proposition 6.11. Take any Pythagorean plane. Let G be any ordered Abelian group.Suppose that μ is an measure function having the following properties:

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(1) the measure μ(P ) ≥ 0 of any figure P is a nonnegative value in the ordered Abeliangroup G;

(2) μ(T ) = μ(S) for any two congruent triangles T ∼= S;

(3) μ(P⊎

Q) = μ(P ) + μ(Q) for any non-overlapping figures P and Q.

Then there exists an order preserving homomorphism φ : F �→ G from field F of segmentarithmetic to the group G such that μ = φ ◦ area. In other word, each measure is ahomomorphic image of the area:

(6.1) μ(P ) = φ(area(P )) for all figures P

and we get the diagramfigure P

area

⏐⏐ ⏐⏐measure μ

area(P ) −−−→φ

measure μ(P )

Proposition 6.12. Under the additional assumption that

(1*) the measure μ(T ) > 0 is strictly positive for any triangle T .

the mapping φ is injective, hence an endomorphism.

Proof of Proposition 6.11. By the same reasoning as used in the proof of Theorem 11.3,we derive from the assumptions (2) and (3) the following more general properties of themeasure μ:

(4) Any two equidecomposable figures P�∼ Q have the same measure μ(P ) = μ(Q).

(5) Any two equicomplementable figures P � Q have the same measure μ(P ) = μ(Q).

We go on to construct the homomorphism φ : F �→ G with the property (6.1). For anysegment length r, we define φ(r) = μ(R) to be the measure of the rectangle R, one sideof which is the unit segment and the second side has the length r.

Given is any figure P . By Proposition 6.9, the figure P is equicomplementable toa rectangle R � P , one side of which is the unit segment and the second side has thelength area(P ). Hence we get both area(R) = area(P ), as well as μ(R) = μ(P ) fromproperty (5). In the end that means

μ(P ) = μ(R) = φ(r) = φ(area(R)) = φ(area(P ))

holds for any figure P , as required.It remains to check that the mapping φ preserves the addition and the order. Take

two rectangles R with sides r× 1, and S with sides s× 1. We may assume they are non

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overlapping and have a common side of unit length. In that case R⊎

S is a rectanglewith sides r + s and 1, and thus

φ(r + s) = μ(P⊎

R) = μ(R) + μ(S) = φ(r) + φ(s)

confirms the additivity. Now assume that r ≤ s, and both R and S lie on the same sideof their common side of unit length. In this case R ⊆ S, and hence

φ(r) = μ(R) ≤ μ(S) = φ(s)

confirms that φ preserves the order.

Proof of Proposition 6.12. In order to check whether the homomorphism φ is injective,we suppose that r and s are two segment lengths and φ(r) = φ(s). The rectangles R withsides r×1 and S with sides s×1 have the same measure since μ(R) = φ(r) = φ(s) = μ(S).Since any two segments are comparable, we may assume r ≤ s and hence R ⊆ S.

We use the equality μ(R) = μ(S) to rule out the R is strictly less that S. In thatcase, the difference S \ R would be the disjoint union of two congruent right triangles.By assumption (1*), they would have a positive measure and hence additivity wouldimply μ(R) < μ(S).

Thus we have seen that φ(r) = φ(s) implies μ(R) = μ(S), secondly R = S andfinally r = s, as to be checked for injectivity of φ.

Theorem 6.4. In a Pythagorean and Archimedean plane, any measure function μ isequal to the common known area, provided it has the following properties:

(0) the measure μ(UQ) = 1 for a unit square;

(1) the measure μ(P ) ≥ 0 of any figure P is a nonnegative value in the segment arith-metic;

(2) μ(T ) = μ(S) for any two congruent triangles T ∼= S;

(3) μ(P⊎

Q) = μ(P ) + μ(Q) for any non-overlapping figures P and Q.

It is interesting to note that this uniqueness proof uses really Eudoxus idea to getequality from rational upper and lower bounds. This principle appear in Euclid’s bookV as a definition for the equality of ratios of magnitudes. We reformulate this result asa proposition about Archimedean fields.

Proposition 6.13 (Eudoxus principle). Given are any four elements a, b, c, d > 0 inan Archimedean field. The ratios a : b and c : d are equal, if and only if for any wholenumbers m and n either one of these three cases occurs:

both na > mb and nc > md,

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both na = mb and nc = md,

both na < mb and nc < md.

Proof. By the the Archimedean axiom there exists a natural number n such that (n +1) · a > m · b. By the basic properties of the natural numbers, there exists a smallestsuch n. In this case both n · a ≤ m · b as well as (n+ 1) · a > m · b hold. By assumption,the same inequalities hold for a and b replaced by c and d. Hence

n · a ≤ m · b < (n + 1) · a and n · c ≤ m · d < (n + 1) · c

Division implies

n

m≤ b

a<

n + 1

mand

n

m≤ c

d<

n + 1

m

We have squeezed both fractions b/a and c/dbetween the same upper and lower rational bounds. By subtraction we conclude

− 1

m≤ b

a− c

d≤ 1

m

Since m is an arbitrary natural number, and the Archimedean axiom has been assumedto hold, we conclude b

a= c

d. This last step also called the principle of exhaustion.

Remark. Eudoxus and Archimedes’ principles are used by Euclid in book XII.2 to provethat the ”The areas of circles are in the same ratio as the squares on their diameters”.Too, they are essential for Euclid’s theory of volume. Lateron, the same prinicples appearagain in Archimedean proof of the law of levels, and many times more in mathematicalphysics.

Proof of Proposition 6.4 . By Proposition 6.11, there exists a homomorphism φ : F �→F from the field of segment arithmetic to itself such that

(6.1) μ(P ) = φ(area(P )) for all figures P

Moreover normalization for the unit square implies

φ(1) = φ(area(UQ)) = μ(1) = 1

The additivity of the area and the measure of any n non overlapping congruent rectanglesR of dimensions r × 1 imply, by means of an easy induction on n that

φ(n · r) = φ(area(n ·R)) = μ(n ·R)

= n · μ(R) = n · φ(area(R)) = n · φ(r)

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For any natural n and m we conclude

m · φ( r

m

)= φ

(m · r

m

)= φ(r)

φ

(1

m· r)

=1

m· φ(r)

φ( n

m· r)

=n

m· φ(r)

proving φ(qr) = q φ(r) for any rational q. We need the corresponding property for anyeven irrational segment length s.

Given is any segment s > 0 and denominator m = 0, by the Archimedean axiomthere exists a natural number n such that (n + 1) · 1 > m · s. By the basic properties ofthe natural numbers, there exists a smallest such n. In this case, we get

n · 1 ≤ m · s < (n + 1) · 1 and n · 1 ≤ m · φ(s) < (n + 1) · 1

The second inequalities hold since φ(qr) = qφ(r) for any rational q and r = 1, andφ(1) = 1, and finally the homomorphism φ preserves the order. We have squeezedboth s and φ(s) between the same upper and lower rational bounds, and conclude bysubtraction

− 1

m· 1 ≤ φ(s) − s ≤ 1

m· 1

Since m is an arbitrary natural number, and the Archimedean axiom has been assumedto hold, we conclude φ(s) = s. Once more, we have used Archimedes’ principle ofexhaustion. Hence it turns out that φ is the identity and the only additive positivemeasure μ is the common area.

6.7 About the volume of polyhedra

Already Gauss brought it to the attention of mathematicians, whether a correspondingjustification of the three-dimensional volume based on dissections is possible. In aletter to Gerling in 1844, Gauss says that it is too bad that the equality of volume oftwo symmetrical, but not congruent figures, can be proved only using the method ofexhaustion. Gerling could solve the dissection problem for this example. In his reply,Gerling gives a direct proof that any triangular pyramid can be dissected into twelvepieces that are congruent to those of a dissection of its mirror image. Gauss respondsthat this is a nice result. But it is still unfortunate that the proof of Euclid’s proposition(XII.5)

"Pyramids of the same height on triangular bases are in the same ratio

as their bases"

needs the method of exhaustion. Hilbert conjectured that the method of exhaustionis really necessary. At the International Congress of Mathematicians in Paris 1900, he

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presented his famous list of the 23 most important problems facing mathematics in thetwentieth century. The third problem of this list addresses the present question:

Give two tetrahedra with equal bases and equal heights, which in no waycan be dissected into congruent tetrahedra, and which, even by addingcongruent tetrahedra, cannot be complemented into polyhedra, which inturn can be dissected into congruent tetrahedra.

In the same year, Max Dehn has given such a counterexample. Max Dehn proves that itis not possible to dissect a regular tetrahedron into a finite number of pieces that can bereassembled into a cube, even after possibly adding on other figures that are equivalentby dissection. The important consequence is that the method of exhaustion is reallynecessary in Euclid’s proof of (XII.5).

Figure 6.17: Herons’ formula for the area of triangle.

6.8 Heron’s formula for the area of a triangle

Problem 6.20. Derive Heron’s formula

area(�ABC) =1

4

√(a + b + c)(a + b− c)(a− b + c)(b + c− a)

for the area a triangle in terms of its three sides a, b and c.

Answer. Take the triangle �ABC with vertices at the centers of the two circles andtheir intersection point A. We begin with the basic fact that the area of a triangle ishalf the product of base and altitude. We use the base CB and the altitude AP and get

Area �ABC =1

2|CB| · |AP |

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Let |CP | = p and |BP | = q. The height is calculated with the Pythagorean theorem inthe right triangle �CPA

h2 = |AP |2 = |CA|2 − |CP |2 = b2 − p2 = (b− p)(b + p)

Similarly, the Pythagorean theorem in the right triangle �BPA yields

h2 = |AP |2 = |BA|2 − |BP |2 = c2 − q2

Subtracting yields

b2 − c2 = p2 − q2 = p2 − (a− p)2

a2 + b2 − c2 = a2 + p2 − (a− p)2 = 2ap

2a(b + p) = (a + b)2 − c2 = (a + b + c)(a + b− c)

2a(b− p) = −(a− b)2 + c2 = (−a + b + c)(a− b + c)

Hence we get

Area �ABC =1

2|CB| · |AP | =

1

4

√2a(b + p)

√2a(b− p)

=1

4

√(a + b + c)(a + b− c)(−a + b + c)(a− b + c)

6.9 Algebraic relations for the pieces of a triangle

Problem 6.21. Let r be the radius of the in-circle of triangle �ABC. Prove the areais

area(�ABC) =(a + b + c) r

2

Answer. It is explained in an earlier problem 9.3 in the context of neutral trianglegeometry, how to construct the in-circle of a triangle. The perpendiculars are droppedfrom its center I onto the three sides. The foot points F,G,H are the touching pointsof the in-circle to the three sides. The six segments from the center of the in-circle tothe vertices and the touching points partition the given triangle into six right triangles.All of them have the radius of the in-circle as one of their legs. The other legs havelengths x = |AH|, y = |BF |, and z = |CG|, each one occurring twice for two congruenttriangles. Since the circumference is a+ b+ c = 2x+ 2y + 2z, the total area of the giventriangle �ABC is

area(�ABC) = 2 · x r

2+ 2 · y r

2+ 2 · z r

2= (x + y + z) r =

(a + b + c) r

2

Problem 6.22. Let R is the radius of the circum-circle of triangle �ABC. Prove thearea is

area(�ABC) =abc

4R

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Figure 6.18: An arbitrary triangle is partitioned into six right triangles.

Answer. As already stated in the Proposition 6.7, the area of a triangle is one halfproduct of two sides time and the sin of the angle between them:

area(�ABC) =1

2ab sin γ

On the other hand, the extended sin theorem tells that

c

sin γ= 2R

Hence sin γ = c2R

. Plugging into the first formula yields the required formula

area(�ABC) =abc

4R

Problem 6.23. Can the three sides of a triangle be determined from the radius of thecircum-circle, the radius of the in-circle and the circumference of the triangle?

Answer. Yes, this is possible, but one needs to solve a third order equation. Hence theproblem cannot be solved by a Euclidean construction— one needs to use Cardanos’formula.

We assume as given: the radius r of the in-circle, the radius R of the circum-circle,and the circumference 2s. From the two previous problems, we get

area(�ABC) =abc

4R= s r

Hence we can calculate the sum a + b + c = 2s and the product abc = 4Rrs. We definethe polynomial

P (x) := (x− a)(x− b)(x− c) = x3 − (a + b + c)x2 + (ab + ac + bc)x− abc

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To get all coefficients of this polynomial, we have still to use Heron’s formula, given inProblem 6.20 below in the form

area(�ABC) =1

4

√(a + b + c)(a + b− c)(a− b + c)(b + c− a)

This can also be written with the semi-perimeter s as

area(�ABC) =√

s(s− a)(s− b)(s− c)

Hence we get one further value of the polynomial P :

P (s) =area(�ABC)2

s= rs2

One now can calculate the polynomial and obtain

P (x) = x3 − 2sx2 + (rs + s2 + 4Rr)x− 4Rrs

From this information, it is possible to calculate the three sides, for example by Car-danos’ formula or by some numerical method.

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