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18 The Lunes of Hippocrates

18.1 The historic lunes

Figure 18.1: Just a lune.

Definition 18.1 (Lune). Aluneconsists of two circular arcs having a common chordand lying on the same side of this chord. The interior of the lune is the the crescentshaped area formed by the difference of the interiors of the corresponding circles. It isbounded by the lunes two arcs.

Hippocrates of Chios (ca. 430 B.C.) posed the problem:

(Hippocrates Problem). Find the lunes which are constructible and squarable withstraightedge and compass.

He gave three examples of constructible lunes. They are obtained beginning with

the following two assumptions:

(a) The two circular sectors corresponding to the lunes arcs have the same area.

(b) The central angles of the two circular arcs are commensurable.

Hippocrates of Chios is credited with discovering three such lunes; two more were dis-covered in the 18th century. In the 20th century Tschebatorev and Dorodnov (1947)proved these fives are the only ones.

Lemma 18.1. Assumption(a) implies the lune is squarable.

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Proof. Let CD be the common chord of the two arcs of the lune. LetSA be thecircular sector corresponding to the lunes longer arc and have centerA, andSB be thecircular sector corresponding to the lunes shorter arc and have center B . Both sectorsare delimited by the radiuses from their respective center to the endpoints of commonchord CD. The sectorSB lies on the other side of the lunes shorter arc and does notintersect its interior area. We add to the lune the latter sector and then subtract thesectorSA, and obtain the kite ACBD. Since by assumption (a) the two sectors havethe same area, the lune has the same area as the kite and hence is squarable.

Lemma 18.2. Assumption(a) and(b) together imply

(18.1)

= n

m

= b2

a2

with integersn, m1. Indeedn < m becomes possible only if one or both or anglesandare more than360.

Proof. We now use assumption (b). Let the angle be the (greatest) common measureof the central angles and of the lunes longer and shorter arcs around centers Aand B . Hence = n and =m with integers n > m1. Subdivision of the sectorSA yieldsn congruent sectors with center A, and similarly, subdivision of the sectorSByieldsm congruent sectors with center B. All sectors we have obtained have the samecentral angle . Hence they are similar.

By assumption (a), the two circular sectorsS

A andS

A have equal area, hence the nsmall sectors of radius a around center A with central angle have together the samearea as m similar sectors with central angle and radius b around center B. Sincethe areas of similar figures are proportional to the square of their linear dimension, weconcludena2 =mb2, and finally get equation (18.1).

Definition 18.2 (Circular segment). A circular segmentis bounded by an arc and achord. A segment of central angle is obtained from the circular sector with the sameangle and arc by subtraction or addition of the triangle with vertices at the endpointsof its circular arc and the center of the circle. For a short arc 0 <

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Figure 18.2: The constructible and squarable 3 : 1 lune of Hippocrates.

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Figure 18.3: Hippocrates 3 : 1 lune is greater than a semicircle.

Proposition 18.2(Hippocrates 430 B.C.). The external arc of Hippocrates 3 : 1luneis greater than a semicircle.

Proof. We drop the perpendiculars fromA and Conto lineB D. Since by construction|BD| = 3|AB| >|AC|, the foot points F and G of the perpendiculars lie insidethe segment BD. Thus we obtain a rectangle and two right triangles, one of which is

CDFwith the acute angles F DCand F CD.(i) The angle ACD is the sum of the right angle ACFand angle F CD, and hence

obtuse.

(ii) The triangleACD has the obtuse angle . Hence the Pythagorean comparisonimplies|AC|2 + |CD|2

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Figure 18.4: The 3 : 2 lune of Hippocrates.

Construction 18.1 (Hippocrates construction of a squarable 3 : 2 lune). Drawa circleCwith diameterAB around centerK, and the perpendicular bisectorp of radiusKB . Construct a segment DH such that|DH|2/|KB |2 = 3/2 and mark a congruentsegmentEF= DH on your ruler. Finally, one places the marked ruler such that the

following three requirements are met:

1. The ruler line goes through the pointB.

2. The point markedEon the ruler lies on the circleC.3. The second point markedFon the ruler lies on the perpendicular bisectorp.

LetG be the reflection image ofEacross the perpendicular bisector. We draw a circular

arc

EKBG with centerL, and a circular arc

EF G with centerI.

Result: We claim that the luneLbetween the two circular arcs, the pentagonEF GBK,and the kite IGLEhave the same area.

Problem 18.2..

(a) The circle aroundLcontains three congruent circular segments, which are similar totwo congruent circular segments of the circle aroundI. Which are these segments.

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(b) Prove the claim in (a) by means of circumference angles and the axial symmetry

across the perpendicular bisectorp.(c) Prove by means of(b) that the luneLhas the same area as the pentagonEF GBK.(d) Similarly to items(a) and(b), we get three congruent circular sectors of the circle

aroundL, which are similar to two congruent circular sectors of the circle aroundI. Prove that the first three segments have the same area as the latter two.

(e) Prove by means of(d) that the luneL between the two circular arcs has the samearea as the kite IGLE.

Answer. (a) The circle around L contains three congruent circular segments since |EK|=

|KB

|by construction and

|EK

|=

|BG

|by reflection across the bisector p. Let

be their central angle.

(b) We see that ELG = = 3is the central angle of the longer arc of the lune. Thecircumference angle EGK=/2 is half of the central angle ELK= .

The neusis used in the construction puts points E, FandB on a line. By reflectionacross the bisectorp, we see that the points G, F andKare on a line, too. HenceEGK = EGF is a circumference angle of the inner arcof the lune, too. Weget the corresponding central angle EI F = . By reflection across the bisector

p, we get GIF = . We see that EI G = = 2 is the central angle of theinner arc of the lune.

(c) We choose a scale such that|EK| =|KB| = 1 is a unit segment. Hence|CD| =|CH| = (3)/2 and|EF| =|HD| =

(3/2). From similar trianglesELK

EI Fwe get the scaling factorb

a=

|EI||EL| =

|EF||EK| =

3

2(18.2)

b2

a2 =

3

2=

(18.3)

confirming the requirement (18.1).

The n = 3 circular segments around center Lhave together the same area as them = 2 similar circular segments around center I. We add to the lune the twolatter segmentsEF andF Gonto the other side of the lunes inner arc , and cutalong the longer arc the three circular segments EK, KBandBG. We obtain the

pentagonEF GBKwhich has the same area as the nearby lune

EF G

GBKE.

(d) Similarly to items (a) and (b), we get three congruent circular sectors of the circlearoundL, which are similar to two congruent circular sectors of the circle aroundI. Because of equation (18.3) the first three segments have the same area as thelatter two.

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(e) This time we add the two circular sectorsEIF andF IGto the lune onto the other

side of the lunes inner arc, and cut along the longer arc the three circular sectorsELK,KLB and BLG. We obtain the kite IGLEwhich has the same area asthe lune.

Figure 18.5: Hippocrates 3 : 2 lune is smaller than a semicircle.

Proposition 18.3 (Hippocrates 430 B.C.). The external arc of Hippocrates 3 : 2 luneis less than a semicircle.

Proof. By construction|EF| =

3/2|EK| >|EK|. In triangleEF K the greaterangle lies across the longer side, and hence = E K F > EF K = . Since atriangle can have only one obtuse or right angle, we conclude that is acute. Finally

the supplementary angle KF B is obtuse.(i) 2 |KF|2

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(iv) Since the three points K, F, G lie on a line = EK G is obtuse by item (iii).

Since the vertex of angle EK G lies on the arc

EK G, Corollary 36 implies thatthis arc is less than a semicircle.

Figure 18.6: Hippocrates squared a suitable union of a circle and a lune.

According to the account of van der Waerden, Hippocrates squared a union of a circleand a lune in a special case, constructed in the following way. A hexagon is inscribedinto a circle of radius 1, and into a concentric smaller circleC of radius 1/6 with thesame centerO. A circular arc which is tangent at the endpoints AandCof two adjacent

sidesAB andB Cof the larger hexagon become the inner arc, and the arc

ABCof theunit circle the external arc of the luneL. The two arcs of the lune intersect at 30.

Proposition 18.4 (Hippocrates). The disjoint unionL C has the same area as theunion of the triangleABCand the hexagon inscribed into the circleC.Proof. LetS, S andS be three congruent circular segments between the unit circleand the inscribed hexagon. The first two of them, segmentsS, S, have the verticesA, BandB , C, respectively. LetT be the large circular segment with verticesA andC.

The circular segmentsS andTare similar, since both have arcs intersecting at anangle of 30. Since twice the altitude of the equilateral triangleABO is|AC| =

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3 |AB|, we conclude that

area(T)area(S) = 3

area(S S S) = area(T)

Only two circular segmentsS andS are cut away from the external arc of the luneL. The third segmentS has the same area as the six circular segments chopped awayfrom the small circleC.

The big circular segmentT is added into the inner arc of the lune, to obtain thetriangleABC. Meanwhile, the smaller hexagon is left from the smaller circle. Thedisjoint unionL C has the same area as the union of the triangleABC and thehexagon inscribed into the smaller circleC.

18.2 Historic remark

B.L. van der Waerdens book Science Awakening [39], contains a detailed account ofHippocrates work, as far as it has been reconstructed. The book [25] The AncientTradition of Geometric Problemsby Wilbur Richard Knorr contains interesting historicinformation about Hippocrates work, too.

Hippocrates of Chios (ca. 430 B.C.) is credited with discovering three squarablelunes; two more were discovered in the 18th century. The manner in which Hippocratessquared his lunes can be learned from a famous fragment by Simplicius (ca. 530 A.D.).

According to his own statement, Simplicius had copied word by word from the His-tory of Mathematics of Eudemus (ca. 335 B.C.). Hippocrates proofs were preservedthrough the History of Geometry compiled by Eudemus of Rhodes, which has also notsurvived, but which was excerpted by Simplicius of Cilicia in his commentary on Aris-totles Physics. Many scholars of history have attempted to reconstruct this lost workof Eudemus. (Obviously, no such attempt of any historic reconstruction is intended inthe present notes.)

Another line of information about Hippocrates of Chios comes from Alexander ofAphrodisias. Alexander was the teacher of Simplicius, the most learned and reliableamong the commentators of Aristotle.

Heath concludes that, in proving his result, Hippocrates was also the first to provethat the area of a circle is proportional to the square of its diameter. Hippocrates bookon geometry with the title Elements in which this result appears, has been lost, butmay have formed the model for Euclids Elements.

Van der Waerden given the following judgement of Hippocrates:

Looking at all these developments as a whole, we see in the first place thatHippocrates mastered a considerable number of propositions from ele-mentary geometry. The Elements of Geometry, which he has writtenaccording to the Catalogue of Proclus, must have contained a large part

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of Books III and IV of Euclid, as well as the contents of Books I and II,

which were baby food for the Pythagoreans.Hippocrates knows the relation between inscribed angles and arcs, the con-

struction of the regular hexagon; he knows how to circumscribe a circleabout a triangle and he knows that a circle can be circumscribed aboutan isosceles trapezoid. He is familiar with the concept of similarity andhe knows that the areas of similar figures are proportional to the squaresof homologous sides. He knows not only the Theorem of Pythagoras forthe right triangle (Euclid I.47) but also its generalization for obtuse- andacute-angled triangles (Euclid II.12 and II.13). Furthermore, he is ableto square an arbitrary rectilinear figure, i.e. to construct a square withthe same area. By means of this, he knows how to construct lines whosesquares have the ratio 3 : 2 or 6 : 1 to the square on a given line.

Still more important for the evaluation of the mathematical level, reachedin Athens during the second half of the fifth century and of Hippocratesin particular, is the excellent demonstrative technique and the high re-quirements of rigor demanded in the proofs. Hippocrates is not satisfiedmerely to construct the lunules and to conclude from the drawings thatthe external boundary is greater than or less than a semicircle; he wantsand succeeds to prove this rigorously. One has to remember that theoperation with inequalities is a very late achievement of modern mathe-

matics, about which even Euler did not worry much.

18.3 Some historic and less historic exercises

Lemma 18.5. The areas of similar figures are proportional to the square of their ho-mologous sides.

Proposition 18.5. Three similar figures are attached to the sides of a right triangle,with scales proportional to the sides of the triangle. The sum of the areas of the figuresput onto the legs of the triangle equals the area of the figure put onto the hypothenuse.

Problem 18.3. Provide drawings with four different examples for this proposition. Con-vince yourself that the Proposition follows from the Lemma above and the PythagoreanTheorem.

Answer.

Biography of Alhazen, the polymath

Abu Ali al-Hasan ibn al-HaythamBorn: 965 in (possibly) Basra, Persia (now Iraq)Died: 1040 in (possibly) Cairo, Egypt

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Figure 18.7: On a generalization of the Pythagorean Theorem.

Alhazen was born in Basra, in the Iraq province of the Buyid Empire. He probablydied in Cairo, Egypt. During the Islamic Golden Age, Basra was a key beginningof learning, and he was educated there and in Baghdad, the capital of the AbbasidCaliphate, and the focus of the high point of Islamic civilization. During his time inBuyid Iran, he worked as a civil servant and read many theological and scientific books.

One account of his career has him called to Egypt by Al-Hakim bi-Amr Allah, rulerof the Fatimid Caliphate, to regulate the flooding of the Nile, a task requiring an earlyattempt at building a dam at the present site of the Aswan Dam. After his field workmade him aware of the impracticality of this scheme, and fearing the caliphs anger,

he feigned madness. He was kept under house arrest from 1011 until al-Hakims deathin 1021. During this time, he wrote his influential Book of Optics. After his housearrest ended, he wrote scores of other treatises on physics, astronomy and mathematics.He later traveled to Islamic Spain. During this period, he had ample time for hisscientific pursuits, which included optics, mathematics, physics, medicine, and practicalexperiments.

http://www-history.mcs.st-andrews.ac.uk/Biographies/Al-Haytham.html

http://en.wikipedia.org/wiki/Alhazen%27s_problem

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In elementary geometry, Alhazen attempted to solve the problem of squaring the circle

using the area of lunes (crescent shapes), but later gave up on the impossible task. Thetwo lunes formed from a right triangle by erecting a semicircle on each of the trianglessides, inward for the hypotenuse and outward for the other two sides, are known as thelunes of Alhazen; they have the same total area as the triangle itself.

Problem 18.4. Provide a drawing of the two lunes of Alhazen. Explain and prove howthe lunes can be squared.

Figure 18.8: The lunes of Alhazen can be squared.

Answer. Because of the generalization of the Pythagorean Theorem stated as Proposi-tion 18.5, the unionS of the semicircles erected onto the legs of the triangle has thesame area as the semicircleT erected onto the hypothenuse.

The intersectionS Tof these figures is the sum of the two circular segments overthe legs. We subtract the intersection and conclude that the figuresS \ T, which is theunion of the two lunes, and

T \ S=

ABC, which is a triangle, have the same area.

Problem 18.5. In the figure on page 709 is shown another lune construction. Thistime three circular arcs have been used, but two of them fit together with a commontangent. Explain how this lune is squared, and why squaring the lune is possible.

Construction 18.2 (Another constructible and squarable lune). Into the semi-circle with diameter AB, we inscribe the right triangleABC. Let point R be themidpoint of circular arc

ARB opposite to the semicircle

BC A. We draw the segment

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Figure 18.9: Can you square this lune?

RC. The perpendiculars dropped onto the segmentRC from verticesA andB have thefoot-pointsQ andP, respectively.

Onto the legs of triangleABCare put quartercircles

BCaroundP, and

CAaround

Q. Onto the hypothenuse is put the quartercircle

BA aroundR. These three quartercir-

cles form a luneL.Squaring the lune. The circular segments erected onto the three sides of triangle ABCare similar since all three are bounded by a quartercircle.

According to Proposition 18.5, the unionS of the circular segments erected ontothe legs of the triangle has the same area as the circular segment T erected onto thehypothenuse. We can obtain the luneLfrom the triangleABCby adding the unionSand subtracting circular segmentT . Hence the the luneLhas the same area as thetriangleABC.

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Figure 18.10: Can you square this lune?

Another line of information about Hippocrates of Chios comes from Alexander ofAphrodisias. According to Alexander, Hippocrates began with an isosceles right triangle.Two congruent lunes are formed by the semicircle on the legs as external arcs, and thesemi-circle circumscribed about the triangle as the inner arc. He proved that the sumof the areas of these two lunes is equal to the area of the triangle.

Problem 18.6. Provide a drawing with named points, and explain the reasoning.

Answer. The reasoning how to square this lune is the same as for the lunes of Alhazen.

Here is how Hippocrates, again according to Alexander, tried to find a secondsquarable and constructible lune. Take an isosceles trapezoid formed by the diame-ter of a circle and three consecutive sides of an inscribed regular hexagon.

Problem 18.7. Look at the graph on page 711. Prove that the sum of the areas of asemicircle on a side of the hexagon and the three lunes formed by the semicircles on

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Figure 18.11: One can square this simple lune.

Figure 18.12: Can one square the circle?

three sides of the hexagon and by the semicircle circumscribing the trapezoid, is equal tothe area of the trapezoid.

Remark. Now there is a tempting speculation: if it were possible to square the threelunes, it would be possible to square the semicircle and hence the circle!

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Answer. Since the side of an inscribed hexagon equals the radius of the circle, and the

diameter is double the radius, the unionSof four semicircles over sides of the hexagonhas the same area as the big semicircleT. We place one of the four semicircles aside.Thus the intersectionS Tof the figures consists of the circular segments over threeside of the hexagon, or trapezoid ABCD.

We subtract the intersection and conclude that the figuresS \ T, which is the unionof three lunes together with the semicircle put aside, andT \ S = ABCD, which isthe trapzoid, have the same area.

18.4 The lune equation

Proposition 18.6. Any lune satisfying assumption(a) and(b) satisfies the lune equa-

tion

(18.4) sin(n/2)

sin(m/2)=

n

m

with integersn, m1.Proof. The common chord of the two arcs of the lune has the length

|CD|= 2a sin(/2) = 2b sin(/2)

Since = n and =m and na2 =mb2, the equation is easy to confirm.

Problem 18.8. The figure on page 713 is a numerical production of Hippocrates3 : 1lune, obtained from the lune equation

sin(3/2)

sin(/2) =

3

Convince yourself that the area of the lune equals the area of the kiteACBD.In the figure on page 714 you see the graph for the second solution of the same lune

equation. Convince yourself that in this case, the area of the lune equals the area of thekiteACBD plus twice the area of the smaller circle aroundA.

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Figure 18.13: A numerical production of Hippocrates 3 : 1 lune.

Answer. The figure on page ?? is a numerical production of Hippocrates 3 : 1 lune.The unionSof the three congruent circular sectors ADD, ADD and ADChas thesame area as the circular sector BDC=T. The luneL= ADD DCis obtained fromthe kite BDACby adding the unionSand subtracting the circular sectorT . Hencethe the lune has the same area as the kite BDAC.

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Figure 18.14: The area of the lune equals twice the area of smaller circle plus the area ofthe kite.

Answer. The figure on page 714 is the graph for the second solution of the same luneequation. In order to translate the explanation above to this case, one has to count areapositive or negative according to the orientation of their boundary.

The unionSof the three congruent circular sectors ADD, ADD and ADC hasthe area of two small circlesCA around A plus the left sector smallADC. The area ofS equals the area of the circular sector BDC =T. The area of the kite BDAC isnegative because of the clockwise orientation.

The big lune L= bigADCsmallADCis obtained from the negative kite BDACby adding the union

Sand subtracting the circular sector

T .

area(CADB) =area(CADB) + area(S) area(T)= (area(CADB) + area(T)) + area(S)=area(big ADC) + 2 area(CA) + area(small ADC)=area(L) + 2 area(CA)

area(L) = area(CADB) + 2 area(CA)Hence the big lune has the same area as the kite CADB plus twice the area of thesmaller circle .

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Remark. If one or both or angles = n and= m are more than 360, we have really

not squared the lune as drawn. Instead we have squared combined figure consisting ofthe lune plus q=n/360 circles of radius a minus p =m/360 circles

of radius b.

We get

|CD|= 2a sin(/2) = (1)q 2a sin(/2) = (1)q 2a sin(n/2)|CD|= 2b sin(/2) = (1)p 2b sin(/2) = (1)p 2b sin(m/2)

where the primed angles are in the range (0, 360). The sign in the lune equation turnsout to be

sin(n/2)sin(m/2)= (1)p+q

nm

Indeedn < mor the minus sign in the lune equation both become possible, but only insuch a situation.

Problem 18.9. Solve the lune equation

sin(n/2)

sin(m/2)=

n

m

forn= 3, m= 2 and0<

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Figure 18.15: Another 3 : 2 lune.

There are two real root, and both correspond to real angles. The numerical values arecos .5930703308 or .8430703308 and 53.6 or 147.5 in degrees. The lunesare drawn in the figures on page 701 and page 716.

Problem 18.10. The figure on page 716 shows another3 : 2 lune construction.

(a) The circle aroundL contains three congruent circular segments, which together turn

more than a full circle, and which are similar to two congruent circular segmentsof the circle aroundI. Which are these segments.

(b) Prove the claim in (a) by means of circumference angles and the axial symmetryacross the perpendicular bisectorp.

(c) Similarly to items(a) and(b), we get three congruent circular sectors of the circlearoundL. Together they are a full circle plus a sector. There are two congruentcircular sectors of the circle around I which are similar to former ones. Provethat the first three segments have the same area as the latter two.

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(d) Prove by means of (c) that lune minus circle have the same difference of areas as

the kite

IGLE.

Figure 18.16: In this case, the 3 : 2 lune minus the lower circle is squarable.

Problem 18.11. Find a lune and a circle in the figure on page 717 with squarabledifference of areas. Convince yourself that they have the same difference of areas as

the self-crossing pentagonEF GBK, were the area of the upper quadrilateral is countedpositive, and the area of the lower triangle is counted negative.

Convince yourself directly that the signed area of the self-crossing pentagonEF GBK,which is reallyEF GNBKN, and the area of the kite IGLEare equal.

Answer. The luneL is bounded by the long arc

GF Eand the short arc

EM G. Thelune minus the lower circleKaround L is squarable.

Indeed, the two circular segments

F G and

F E inside the upper circle have the

same area as the three circular segments

EK,

KB , and

BG inside the lower circle.

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All these five segments have congruent central angles 147. We subtract the twofirst segments from the luneL, and the three latter segments from the circleK. Theintersection

EMG N of the circular segments

EKand

BG appears twice. Thus we getthe differences

L

F G+

F E

=

EMG F

K

EK+

KB+

BG

=BN K

EMG N

L K=L

F G+

F E

K +

EK+

KB +

BG

=

GMEF BN K+

EMG N = GNEF BN K

The last difference has the signed area of the self-crossing pentagon EF GBK.We can also see directly that this difference equal in area to the kite IGLE.

Indeed, we subtract the two congruent trianglesGF I andF EI inside the uppercircle and add the three congruent trianglesEK L,KBL, andBGL inside thelower circle. The sum of the areas of the former two is equal the sum of the areas of thelatter three triangles.

Subtraction of the former two and addition of the latter three figures from the dif-ference GNEF BN Kyields the kite IGLE, which has hence the same area.

18.5 Vietas and Eulers lunes

Problem 18.12. Solve the lune equation

sin(n/2)

sin(m/2)=

n

m

forn= 4, m= 1 and0<

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Figure 18.17: The 4 : 1 lune of Vieta.

There is only one real root. The numerical value is 2 cos(/2) 1.769292354 and55.58 in degrees. The lune is draw in the figure on page 719.Problem 18.13. Solve the lune equation

sin(n/2)

sin(m/2)=

n

m

forn= 5, m= 1 and0<

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Figure 18.18: The constructible 5 : 1 lune of Euler.

From the lune equation we see that x = cos satisfies the quadratic equation

4x2 + 2x 1 =5

The solutions are:

cos =1

5 45

4There are two real roots and two complex roots. Only the root with both plus signscorresponds to a real angle. The numerical value is cos .6835507455 and 46.9in degrees.

Problem 18.14. Solve the lune equation

sin(n/2)

sin(m/2)=

n

m

forn= 5, m= 3 and0<

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Figure 18.19: The constructible 5 : 3 lune of Euler.

Answer.

sin(5/2)

sin3(/2)=

4cos2 + 2 cos 12cos + 1

From the lune equation we see that t = cos satisfies the quadratic equation

3(4xt + 2t 1) =

5(2t+ 1)

The solutions are with =1:

cos =

5 3

20 + 2

15

4

3

For = +1, there are two real roots, both correspond to a real angle. The numericalvalues are cos .8330386705 or .6875414461 and33.6 or 133.4 in degrees.

For =1, there are two real roots, only one corresponds to a real angle. Thenumerical values are cos .0674839681 or 1.078013256 and93.9 in degrees.

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Remark. Vieta found around 1593 the 4 : 1 lune leading to a cubic equation. The

constructible and squarable 5 : 1 and 5 : 3 lunes were found in 1766 by Martin JohanWallenius. Too, Leonard Euler made the same discovery around 1771, published in So-lutio problematis geometrici circa lunules a circulis formatas. The two lunes from Eulerand the two not so obvious ones by Hippocrates are once more obtained algebraicallyfrom the lune equation in an article by Th. Clausen [11] published in 1840. The resultsof Hippocrates were apparently not known at that time. Except for Hippocrates, noneof these authors consider a combination of lune and circle to be squared.

Remark. A popular account of some of the material from this section is contained inWilliam Dunhams book Journey through Genius. A short introduction and somebiographic information is given by

http://en.wikipedia.org/wiki/Quadrature_of_the_lune

Figure 18.20: A constructible 5 : 3 lune minus circle around A.

18.6 About transcendental numbers

A number is calledalgebraicif there exists a nonzero integer polynomial pZ[x] suchthatp() = 0. In that case is called a root of the polynomialp.

The set of all real or complex algebraic numbers is denoted by A. As stated inTheorem 17.1, the set of algebraic numbers is both a countable and a field. Obviously,

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Figure 18.21: The overlapping polygon has the same signed area as the kite ADBC.

all rational numbers are algebraic, but the converse is not true. We know that

2 and

i= 1 are two important examples of irrational but algebraic numbers.The real or complex numbers, which are not algebraic are called transcendental.Definition 18.3. The numbers 1, 2, . . . n are called Z-linearly independent if thedependence relation

k11+k22+ +knn= 0with integers k1, k2, . . . kn holds only for allk1= k2 = = kn= 0.

The numbers1, 2, . . . nare called linearly independent over the algebraic numbersif the dependence relation

11+22+ +nn = 0

with algebraic numbers 1, 2, . . . n holds only for all 1= 2= = n= 0.Obviously, independence over the integers is equivalent to linear independence of the

rationalsQ.

Theorem 18.1 (Lindemann-Weierstrass Theorem). Let the algebraic numbers1, 2, . . . n be distinct. Then the exponentialse

1, e2, . . . en are linearly independentover the algebraic numbers. Hence

ni=1

iei = 0

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Figure 18.22: A constructible 5 : 3 lune plus circle aroundB

minus circle aroundA

.

for any algebraic numbers1, 2, . . . n, unless1 = 2= = n= 0.The Lindemann-Weierstrass Theorem immediately yields the classical results that

e,, ln 2 are transcendental.

Assume towards a contradiction that Eulers number e is algebraic. We put 1=0, 2= 1, 1 =e, 2= 1 and get the contradictione e0 + 1 e1 = 0. (Hermite1873)

Assume towards a contradiction that is algebraic. We put1 = 0, 2= i, 1=

2= 1 and get the contradiction 1 e0 + 1 e = 0. (Lindemann 1882) Assume towards a contradiction that ln 2 is algebraic. We put 1 = 0, 2 =

ln 2, 1 =2, 2 = 1 and get the contradiction2 e0 + 1 eln 2 = 0.Corollary 55. Let the distinct algebraic numbers 1= 0, 2= 0, . . . n= 0 be allnonzero and the algebraic numbers1= 0, 2= 0, . . . n= 0 be nonzero. Then the sum

ni=1

iei is transcendental.

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18.7 Which lunes are algebraic, which constructible?

We look once more at the generic figure 18.1. I call a lunealgebraicfor which the radiusa and b, the angle functions sin , cos , sin , cos of the central angles, and the areaof the lune are algebraic numbers.

Theorem 18.5. All algebraic lunes satisfy Hippocrates two basic assumptions

(a) The two circular sectors corresponding to the lunes arcs have the same area.

(b) The central angles of the two circular arcs are commensurable.

Proof. Since the area of the kite ACDB is algebraic, and the area of the luneL isassumed to be algebraic, their difference is algebraic. Since

area(L) = area(SA) + area(ACDB) area(SB)

we conclude that the difference of the areas of the circular sectorsSA andSB:

area(SA) area(SB) =a2 b2

is algebraic, too. The angles and at the centers A and B , corresponding to the twoarcs of the lune have to be measured in radian measure.

The assumptions imply thatei, ei anda, band the areaa2 b2 are all algebraic.We use Alan Bakers Corollary 56 with 1 :=i, 2 :=iand 1 := a

2, 2 :=

b2. We

conclude that the the sum

11+22 = a2 b2 = 0

Hence Hippocrates assumption (a) holds.We now use the Theorem 18.4 of Gelfond Schneider and M. Waldschmidt with x=

/ and z = i. Assume towards a contradiction that x is irrational. We wouldconclude that at least one of the numbers /, ei, ei is transcendental. The latter twoare algebraic by assumption and

=

a2

b2

is algebraic, too, by the first part of the proof. Hence / is rational, confirmingHippocrates assumption (b).

Main Theorem 29 (N.G. Tschebatorev and A.W. Dorodnov 1947). There existonly five squarable and constructible lunes, corresponding to the cases in whichn: m is2 : 1 , 3 : 1 , 3 : 2 , 5 : 1 , 5 : 3.

Remark. The final result was obtained by A. W. Dorodnow, after many earlier partialresults.

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E. Landau showed in 1903 that the p : 1 lune is not constructible ifp is a primebut not a Fermat prime.

L. Tschakaloff showed in 1926 that the 17 : 1 lune is not constructible, neither anyp: m lune where p is a prime andp > m.

N. Tschebotarov [37] showed in 1934 that no n: m lune with n, mboth odd andm >5 is constructible.

18.8 Irreducibility of the lune equation

The lune equation (18.4) can be written with the complex variable z= ei. Since

sin(n/2)sin(/2)

=z(n1)/2 zn

1z 1

we get in the new variable

m sin(n/2)) (n) sin(m/2)) = 0

m zn 1

z 1 (

n)z(nm)/2 zm 1

z 1 = 0

After taking the square, we get an integer polynomial equation:

Pn,m(z) :=m zn 1

z 1 2

nznm z

m 1z 1

2

= 0(18.5)

Proposition 18.7. We assume that p = n be an odd prime and p > m. Then thepolynomial Pn,m(z) in the squared lune polynomial equation (18.5) is irreducible overthe integers.

Theorem 18.6 (L. Tschakaloff 1926). Ifp is a prime, but not a Fermat prime, andm < p, thep: m lune is not constructible.

Proof. We proceed similarly as in the proof of Proposition 17.19. We use the substitutionz = x + 1. Recall that the binomial formula implies that is the new variable x, allcoefficients of

Rp(x) :=(1 +x)p 1

x =

pk=1

p

k

xk1

=p+

p

2

x+

p

3

x2 + +

p

p 2

xp3 +pxp2 +xp1

except the leading ones are divisible by p. Similarly, we substitutez= 1 + x into thesquared lune polynomial

Pn,m(z) =mR2p(x) p(1 +x)pmR2m(x) =Qn,m(x)

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and use the Eisenstein criterium, given as Proposition 17.18, to show that the resulting

polynomial is irreducible.Indeed, all coefficients of Qn,m(x) except the leading one are divisible by p. Theleading coefficient ismpwhich is not divisible by the primep. The constant coefficientismp2pm2 =mp(pm) which is divisible byp, but not byp2. Hence all assumptions ofthe Eisenstein criterium are satisfied and the polynomialsQandPp,mare irreducible.

Remark. Similar results are contained in the article [30] which is translated from Post-nikovs 1963 Russian book on Galois theory. Note that we have only dealt with themost easy special case of the main theorem of N.G. Tschebatorev and A.W. Dorodnov.Biographic information is given in

http://en.wikipedia.org/wiki/N._G._Chebotarev

18.9 Tschebychev polynomials

Definition 18.4(Tschebychev polynomials). The Tschebychev polynomials of firstand second kind are defined by

Tn(cos t) = cos nt and Un(cos t) =sin(n+ 1)t

sin t

for integersn= 0, 1, 2, . . . .

Proposition 18.8. Both polynomialsTn andUn satisfy the same recursion formula

Tn+1= 2xTn Tn1 Un+1= 2xUn Un1with the initial dataT0 = U0= 1 andT1= x butU1= 2x, respectively. BothTn andUnare integer polynomials of degreen. They are even for evenn, odd for oddn. MoreoverTn(1) = 1 andUn(1) =n+ 1 for alln.

TheTn satisfy the composition formulaTn Tm = Tnm. The polynomials2Tn(v/2)andUn(v/2) are integer polynomials of the variablev.

n Tn Un0 1 11 x 2x2 1 + 2x2 1 + 4x23 3x+ 4x3 4x+ 8x34 1 8x2 + 8x4 1 12x2 + 16x45 5x 20x3 + 16x5 6x 32x3 + 32x56 1 + 18x2 48x4 + 32x6 1 + 24x2 80x4 + 64x67 7x+ 56x3 112x5 + 64x7 8x+ 80x3 192x5 + 128x78 1 32x2 + 160x4 256x6 + 128x8 1 40x2 + 240x4 448x6 + 256x89 9x 120x3 + 432x5 576x7 + 256x9 10x 160x3 + 672x5 1024x7 + 512x9

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Lemma 18.6 (binomial expansion).

Tn(x) =n/2m=0

n

2m

xn2m(x2 1)m(18.6)

Un(x) =

n/2m=0

n+ 1

2m+ 1

xn2m(x2 1)m(18.7)

Proof. Let x = cos t. Use eint = cos nt+ i sin nt = (cos t+ i sin t)n and separate thebinomial formula into real and imaginary parts:

cos nt+i sin nt=n

k=0

n

k

(cos t)nk (i sin t)k

Tn(x) = cos nt=

n/2m=0

n

2m

(1)m(cos t)n2m (sin t)2m =

n/2m=0

n

2m

xn2m(x2 1)m

sin nt=

(n1)/2m=0

n

2m+ 1

(1)m(cos t)n2m1 (sin t)2m+1

Un1(x) =(n1)/2

m=0

n

2m+ 1

xn12m(x2 1)m

Lemma 18.7 (generating function).n0

Tn(x) rn =

1 xr1 +r2 2xr(18.8)

n0Un(x) r

n = 1

1 +r2 2xr(18.9)

We see once more that both polynomialsTn andUn satisfy the same recursion formula.

Proof. Letx = cos t. One separates a complex geometric series into real and imaginaryparts:

n0 e

int

r

n

=

1

1 eit r = 1

eit r

|1 eit r|2n0

rn cos nt+irn sin nt= 1 r cos t+ir sin t(1 r cos t)2 +r2 sin2 t

n0rn Tn(cos t) =

1 r cos t1 +r2 2r cos t

r sin tn1

rn1 Un1(cos t) = r sin t

1 +r2 2r cos t

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The Tschebychev polynomials of second kind are for even and odd index are evenand odd, respectively.

U2n(x) =Pn(4x2 4) and U2n+1(x) = 2x Qn(4x2 4)

Lemma 18.8. The polynomials P and Q with the independent variable s = 4x2 4have the generating functions

k0Pk(s) r

n = 1 +s

1 (2 +s)r+r2(18.10)

l0

Ql(s) rn =

1

1

(2 +s)r+r2

(18.11)

Hence both satisfy the same the recursion formula:

Pn+2= (2 +s)Pn+1 Pn and Qn+2= (2 +s)Qn+1 Qnn Pn0 11 3 +s2 5 + 5s+s2

3 7 + 14s+ 7s2 +s3

4 9 + 30s+ 27s2 + 9s3 +s4

5 11 + 55s+ 77s2

+ 44s3

+ 11s4

+s5

6 13 + 91s+ 182s2 + 156s3 + 65s4 + 13s5 +s6

7 15 + 140s+ 378s2 + 450s3 + 275s4 + 90s5 + 15s6 +s7

8 17 + 204s+ 714s2 + 1122s3 + 935s4 + 442s5 + 119s6 + 17s7 +s8

n Qn0 11 2 +s2 3 + 4s+s2

3 4 + 10s+ 6s2 +s3

4 5 + 21s+ 21s2 + 8s3 +s4

We putn = 2kor n = 2l +1 into the binomial formula forU2k(x) =Pk(s) orU2l+1(x) =2x Ql(s) and get

Lemma 18.9.

Pk(s) = 4k

km=0

2k+ 1

2m+ 1

(4 +s)kmsm(18.12)

2Ql(s) = 4l

lm=0

2l+ 2

2m+ 1

(4 +s)lmsm(18.13)

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Lemma 18.10. With the new independent variables = 4x2 4, the polynomialsP andQare monic integer polynomials with positive coefficients. The constant coefficients arePk(0) = 2k+ 1 andQl(0) =l+ 1.

Assume now thatp is an odd prime. In this case, all coefficients of the polynomialP(p1)/2(s) except the leading one are divisible byp. The constant coefficient equalsp.

The polynomialPk is related to the construction of a regular 2k + 1-gon. Indeed, itszeros are

zi=4sin2 i 180

2k+ 1 fori = 1, 2, . . . , k

Problem 18.15. Confirm thatP7 is divisible byP1 P2. Calculate the quotientP7/(P1 P2). Find its zeroszi and the corresponding central anglesi 24 for the15-gon. Which

four integersi do you obtain, and why?

Answer. The quotient

P7P1 P2 = 1 + 8s+ 14s

2 + 7s3 +s4

have zeros4sin2 i 18015 for i = 1, 2, 4, 7. These are the integers less than 8 relativelyprime to 15.

Proposition 18.9. Letp an odd prime. Then the polynomialP(p1)/2(s), the Tscheby-chev polynomialUp1(x), and the cyclotomic polynomialp(z) are irreducible.

Proof. We know by Lemma 18.10 that all coefficients of the polynomial P(p1)/2(s), ex-cept the leading one, are divisible byp. The constant coefficient equalsp. By the Eisen-stein criterium, given as Proposition 17.18, we conclude that the polynomial P(p1)/2(s)is irreducible.

We put the substitutionz= eit into the cyclotomic polynomial and get

p(z) =zp 1

z 1 =z(p1)/2 z(1p)/2

z1/2 z1/2=

sin((p 1)t/2)sin(t/2)

=Up1(cos(t/2)) =P(p1)/2(4sin2(t/2))

We see that any factoring of p(z) into integer polynomials entails an integer factoringof the Tschebychev polynomial Up1(s) with s = cos(t/2), and an integer factoring ofthe polynomialP(p1)/2(s) with independent variable s =4sin2(t/2).

We have ruled out such a factoring. Hence the Tschebychev polynomialUp1(s) andfinally the cyclotomic polynomial p(z) are irreducible, too.

The lune equationsin(n/2)

sin(m/2)=

n

m

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can be written with the Tschebychev polynomials of second kind

Un(cos t) =sin(n+ 1)t

sin t

We use instead ofx = cos(/2) the new independent variable

s= 4x2 4 =4sin2(/2) = 2 cos 2

The Tschebychev polynomials of second kind are expressed by the new polynomials PandQ using the independent variable s.

Lemma 18.11 (Some reduction of the lune equation). Fornandmboth odd, we

get the lune polynomial equation

(18.14)

m P(n1)/2(2cos 2) (

n) P(m1)/2(2cos 2) = 0

which is of half the degree, and easier to solve. Forn odd andm even, we get the originallune polynomial equation

(18.15)

m P(n1)/2(4+4 cos2(/2))(

n)2cos(/2)Q(m2)/2(4+4 cos2(/2)) = 0

The half angle needs to be used and there is no reduction of the degree.

An interesting peculiarity occurring in the case n = 9, m = 1 is mentioned by

Postnikov in the survey article [30]. We get the lune polynomial equations

s4 + 9s3 + 27s2 + 30s+ 9 + 30= 0

again for the variables= 2 cos 2 and0 =1. The case with the minus sign0=1can indeed be solved by square roots! The factoring

s4 + 9s3 + 27s2 + 30s+ 12 = ((s+ 2)2 +(s+ 2) 2) ((s+ 2)2 (s+ 2) 2)

with = (1 +3)/2 leads to the four roots

si=9 +13 +2

30 2134

with 1=1 and 2=1. Only because these turn out to be complex roots does onenot obtain another constructible lune.

The polynomial s4 + 9s3 + 27s2 + 30s+ 3 for the case with the plus sign 0 = 1can be factored into quadratic polynomials by Descartes method, too. One obtains two

l l ti d t l l ti f hi h di t l B t