constructions in ancient greek spherics: *1em …constructions in ancient greek spherics:...

Constructions inSpherics

Outline

IntroductionInstruments

Theorems and Problems

Structural Elements

Using Constructions

Constructions in practice

TheodosiusInternal Constructions,Book I

External Constructions,Book I

External Constructions,Book II

Menelaus

Final Remarks

Constructions in ancient Greek Spherics:

Mathematical spheres and solid globes

Nathan Sidoli(joint work with Ken Saito)

Paris, February 2010


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Menelaus

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IntroductionInstrumental practicesTheorems and ProblemsStructural Elements of Greek geometryUsing ConstructionsExamples of constructions in Practice

Theodosius’s SphericsInternal Constructions, Book IExternal Constructions, Book IExternal Constructions, Book II

Menelaus’s Spherics

Final Remarks


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Instruments in Greek mathematics

Aside from some abacus boards and papyri containingcalculations, we have no material evidence for practicesin Greek mathematics. Nevertheless, there is textualevidence for various instruments. Some examples:

I Plato’s and Euratothenes’ devices for analogcalculation of two mean proportionals.

I Nicomedes’ device for producing a conchoid for usein neusis constructions.

I Diocles description of a bone ruler for drafting aparabola.

I Ptolemy’s “analemma board” and Heron’s “localhemisphere,” for analog calculations.


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The geometry of instrumental practice

Certain domains of geometric theory may have been, insome sense, delimited by certain instrumental practices.

I The classic example is Euclid’s elementary geometry,which is based on the postulation of operationsequivalent to the use of an unmarked ruler and acollapsing compass.

I Pappus’s Collection VIII gives constructions using anunmarked ruler and a fixed compass.

I The analemma techniques rely on the use of a setsquare and a non-collapsing compass.

I I will argue in this talk that ancient spherics was ageometry constructed with an unmarked ruler and anon-collapsing compass.


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Instruments in Greek astronomy

We have no surviving observational instruments, but wehave many descriptions.

I Mostly from Ptolemy: the meridian quadrant, theequatorial ring, the armillary sphere, etc.

I And others as Archimedes, Geminus and Proclus:the dioptra, the angular size dioptra, the armillarysphere, the gnomon, etc.

We have a fair number of objects related to the generalcultural interest in astronomy and astrology.

I Sundials (fixed, portable, geared), the Antikytheramechanism, the parapegmata and otherastronomical inscriptions, etc.

I Astrologer’s boards and markers, inscribedhoroscopes, etc.


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Evidence for Greek globes

We have both material and textual evidence for globes inGreek antiquity. Some examples:

I Ptolemy (Almagest) describes the construction of anelaborate star globe, Leontius describes ademonstration globe.

I Geminus refers to “inscribed globes” and Ptolemy(Plansiphere) talks about the lines that are found on“struck globes.”

I We know of three ornamental globes – the FarneseAtlas and two simple celestial spheres – inscribedwith the principle circles and decorated with imagesof the constellations (all 1st or 2nd century).


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The propositions of elementary treatises, like the Elementsor the Spherics, are divided into two types:

Theorems Given some set of initial objects, a theoremasserts some property that is true of theseobjects. (“If... then...”)

Problems Given some set of initial objects, a problemshows how to do something (say, to find,draw, set out) and then demonstrates thatwhat has been done is satisfactory. (“To dosuch-and-such...”)

Some theorems can be intelligibly expressed as problemsand the converse. (We will see that theorem Spher. I 8might very well have been a problem.)


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The components of a Greek proposition

Proclus (5th CE) put forward the following six parts of aGreek proposition. (In fact, they are only ever found socomplete and clearly divided in Euclid’s Elements.)

Introductory components1

1 Enunciation (prìtasic): A general statement of whatis to be shown (done).

2 Exposition (êkjesic): A statement setting out thegiven objects with letter names.

3 Specification (diorismìc): A restatement in terms ofthe specific objects of (1) what is to be shown(theorem), or of (2) what is to be done, including anyconditions of solvability (problem).

1Introductory division not distinguished by Proclus.


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The components of a Greek proposition

Proclus (5th CE) put forward the following six parts of aGreek proposition. (In fact, they are only ever found socomplete and clearly divided in Euclid’s Elements.)

Core components1

4 Construction (kataskeuă): Statements about theproduction of new objects that will be required in theproof. (Relies on posts. 1-3 and problems.)

5 Proof (Ćpìdeixic): A logical argument that theproposition holds (has been done). (Relies on theother assumptions and theorems.)

6 Conclusion (sumpèrasma): A restatement, in generalterms, of what has been shown (done).2

1Core division not distinguished by Proclus.2Very rare, except in the Elements.


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Using constructions to prove theorems

In a geometric theorem, it is sometimes the case that theproperties of the objects stated in the enunciation aresufficient to demonstrate the proposition, but more oftenthan not we have to introduce new objects and use theirproperties in the argument.

These new objects are introduced using constructions. (Infact, the geometer asserts that these objects must havebeen produced, using the same grammaticalconstructions, and sometimes even the same verbs as areused when the initially given objects are set out.3)

3In Spher. I 8, we will see a case where the exposition isundistinguishable from the construction, except on the basis of theproof structure.


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Using constructions to solve problems

A problem is solved by producing a specific geometricobject. A problem shows how to produce the object usingconstructions and then uses deductive argumentation toshow why this object is correct.

This deductive argumentation sometimes requires newconstructions in the same way as a theorem. That is, thegeometric object that solves the problem together withthe initial objects stipulated in the problem are sometimesnot sufficient to show that this object is a solution. In suchcases, we must construct new objects for the proof.


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Example theorem, Spherics I 1

D

A

B

G

A

G

BE

E

D

Upper: manuscript figure

Lower: reconstruction

Introduction

1 “If a plane cuts a sphericalsurface, the curved line in thespherical surface is a circle.”

2 Let a plane cut a sphere andmake curved line ABG.

3 “I say,” ABG is a circle.


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D

A

B

G

A

G

BE

E

D



Core [Case 1]

51 If the plane goes through thecenter of the sphere, thenevery line from the center ofthe sphere to ABG is equal(definition of a sphere), so thatcurved line ABG is a circle.


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D

A

B

G

A

G

BE

E

D



Core [Case 2]

42 If the plane does not gothrough the center of thesphere, let the center of thesphere be D [?],a draw DE ⊥ toplane ABG [El.11.11], join AE,BE, AD and BD (P.1).

52 Use El.1.47 to show thatEB = EA. This could be donefor any other two points oncurved line ABG. Hence, ABGis a circle with center E.

aHow we locate D is passed over with asimple use of êstw. Compare with thenext proposition.


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Example problem, Spherics I 2

DA B

G

A G

BE

D

Z

H

KK

C

E

Z

C

K

H



Introduction

1 “To find the center of a givensphere.”

2 Let there be a given sphere.a

3 “I say,” it is necessary to findits center.

aNotice there are no letter names.


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DA B

G

A G

BE

D

Z

H

KK

C

E

Z

C

K

H



Core [solving the problem]

4 Let some plane cut it, makingcircle ABG [?;Sph.1.1].a Find Das the center of ABG [El.3.1],draw ZDE ⊥ circle ABG tomeet the sphere at Z, E[El.11.12;P.2]. Bisect EZ at H[El.1.10].

3const “I say,” H is the center of thesphere.

aWe are not told how this is done.Perhaps we can imagine some sort ofpostulate, “To pass a plane through a solidfigure.” At any rate, such a constructionwas regularly performed.


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DA B

G

A G

BE

D

Z

H

KK

C

E

Z

C

K

H



Core [verifying the solution]

5 Assume, if possible, that thereis some other center, say C.

4proof Let a perpendicular bedrawn from C to circle ABGat K [El.11.11].

Then, K is a center of ABG[Sph.1.1], and so is D, which isimpossible.


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Components of a Geometry Problem

1p Enunciation: General statement of what is to be done.2p Exposition: Statement of what is given, usually using

specific, letter names.3p First Specification: Specific statement of what is to be

done, often with qualifications.4p Solution: Construction of the geometric object which

satisfies the requirements of the problem.5p Second Specification: Specific statement of what is to

be shown.1

6p Construction: Construction of any new objectsnecessary to the proof.1

7p Proof: Argument that the solution meets therequirements of the proposition; if necessary, usingthe new objects of the construction.

1May be absent.


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The role of constructions in Spherics I

The book opens with a number of propositions that uselines internal to the sphere to demonstrate the propertiesof great circles and sets of parallel circles and theirrelations.

For most of the book, constructions serve the role ofsupplying objects which are necessary to proofs.

We then have two problems that show us how to drawthese internal lines outside of the sphere, into a spacewhere we can use them. (From this point on, we can solveproblems by actually drawing some object.)

The book ends with two problems that show how todraw a great circle through two points and how to findthe pole of a circle working entirely outside of the sphere.


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Using internal constructions, Spherics I 8

DA

BG

A

GB

E

D

Z

H

C

E

Z

C

H



Introduction

1 “If a circle is in a sphere and aperpendicular is drawn fromthe center of the sphere andextended in both directions, itwill fall on the poles of thecircle.”

2 Let ABG be a sphere, let itscenter, D, be taken [Sph.1.2],let DE be drawn ⊥ ABG[El.11.11], and extended tomeet the sphere at Z and H[El.P.2].

3 “I say,” Z and H are the polesof ABG.


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Using internal constructions, Spherics I 8

DA

BG

A

GB

E

D

Z

H

C

E

Z

C

H



Core

4 Let AEG, BEC, AZ, GZ, AHand HG be drawn [El.P.2].

5 AE = EG, the ∠s at E are rightand ED is common, soAZ = GZ. We could make thesame argument for any otherpoint on ABC. Likewise withrespect to H. So Z and H arethe poles of circle ABC.

6 If a circle is in a sphere, fromthe center of the sphere, “andso on.”


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Bringing out a diameter, Spherics I 18

A

B

C

Introduction

1p “To set out (âkjèsjai) thediameter of a given circle in asphere.”

2p Let the given circle be ABC.3p So, it is necessary to set out its

diameter.


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A

B

C


4p Let three random points, A, Band C, be taken on thecircumference.

Let 4DEZ beput together from (sunestĹtw)three lines (such that DEequals that from A to B, etc.)[?]. At point E, draw EH ⊥ ED[El.1.11], At point Z, drawZH ⊥ ZD [El.1.11], and joinDH [El.P.1].

5p <I say, line DH is equal to thediameter of circle ABC.>


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A

B

C

D

FE

Upper: a plane diagram,outside the sphere

Lower: a solid sphere


4p Let three random points, A, Band C, be taken on thecircumference. Let 4DEZ beput together from (sunestĹtw)three lines (such that DEequals that from A to B, etc.)[?].

At point E, draw EH ⊥ ED[El.1.11], At point Z, drawZH ⊥ ZD [El.1.11], and joinDH [El.P.1].



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A

B

C

D

FE

H




4p Let three random points, A, Band C, be taken on thecircumference. Let 4DEZ beput together from (sunestĹtw)three lines (such that DEequals that from A to B, etc.)[?]. At point E, draw EH ⊥ ED[El.1.11],

At point Z, drawZH ⊥ ZD [El.1.11], and joinDH [El.P.1].



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A

B

C

D

FE

H




4p Let three random points, A, Band C, be taken on thecircumference. Let 4DEZ beput together from (sunestĹtw)three lines (such that DEequals that from A to B, etc.)[?]. At point E, draw EH ⊥ ED[El.1.11], At point Z, drawZH ⊥ ZD [El.1.11],

and joinDH [El.P.1].



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A

B

C

D

FE

H




4p Let three random points, A, Band C, be taken on thecircumference. Let 4DEZ beput together from (sunestĹtw)three lines (such that DEequals that from A to B, etc.)[?]. At point E, draw EH ⊥ ED[El.1.11], At point Z, drawZH ⊥ ZD [El.1.11], and joinDH [El.P.1].



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A

B

C

G

D

FE

H




6p Draw AG the diameter ofcircle ABC [?; El.3.1&P.1],

andjoin AB, BG, GA and GC[El.P.1].

7p Now, AB = ED, BC = DE andAC = DZ, ∴ ∠ABC = ∠DEZ[El.1.4]. And ∠ABC = ∠AGCwhile ∠DEZ = ∠DHZ[El.3.27], so ∠AGC = ∠DHZ.But ∠ACG = ∠DZH = R andAC = ED, so AG = DH.Therefore, DH is equal to thediameter of the circle.


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A

B

C

G

D

FE

H


6p Draw AG the diameter ofcircle ABC [?; El.3.1&P.1], andjoin AB, BG, GA and GC[El.P.1].

7p Now, AB = ED, BC = DE andAC = DZ, ∴ ∠ABC = ∠DEZ[El.1.4]. And ∠ABC = ∠AGCwhile ∠DEZ = ∠DHZ[El.3.27], so ∠AGC = ∠DHZ.But ∠ACG = ∠DZH = R andAC = ED, so AG = DH.Therefore, DH is equal to thediameter of the circle.


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Summary of the solutionLet there be a sphere.

We take tworandom points, A and B. With A aspole and AB as distance, we drawcircle ABC [?]. Then it is possibleset out the diameter of this circle asEF [Sph.1.18], so that we can puttogether 4EFH [?;Ele.1.22*]. Wedraw HG ⊥ EH and FG ⊥ EF[Ele.1.11], meeting at point G. Wejoin EG [Ele.P.1].

We draw the internal lines andshow that line EG is equal to thediameter of sphere ABC.


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AB

Summary of the solutionLet there be a sphere. We take tworandom points, A and B.

With A aspole and AB as distance, we drawcircle ABC [?]. Then it is possibleset out the diameter of this circle asEF [Sph.1.18], so that we can puttogether 4EFH [?;Ele.1.22*]. Wedraw HG ⊥ EH and FG ⊥ EF[Ele.1.11], meeting at point G. Wejoin EG [Ele.P.1].



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AB

C

Summary of the solutionLet there be a sphere. We take tworandom points, A and B. With A aspole and AB as distance, we drawcircle ABC [?].

Then it is possibleset out the diameter of this circle asEF [Sph.1.18], so that we can puttogether 4EFH [?;Ele.1.22*]. Wedraw HG ⊥ EH and FG ⊥ EF[Ele.1.11], meeting at point G. Wejoin EG [Ele.P.1].



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F

AB

C

H



Summary of the solutionLet there be a sphere. We take tworandom points, A and B. With A aspole and AB as distance, we drawcircle ABC [?]. Then it is possibleset out the diameter of this circle asEF [Sph.1.18],

so that we can puttogether 4EFH [?;Ele.1.22*]. Wedraw HG ⊥ EH and FG ⊥ EF[Ele.1.11], meeting at point G. Wejoin EG [Ele.P.1].



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F

E

AB

C

H



Summary of the solutionLet there be a sphere. We take tworandom points, A and B. With A aspole and AB as distance, we drawcircle ABC [?]. Then it is possibleset out the diameter of this circle asEF [Sph.1.18], so that we can puttogether 4EFH [?;Ele.1.22*].

Wedraw HG ⊥ EH and FG ⊥ EF[Ele.1.11], meeting at point G. Wejoin EG [Ele.P.1].



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F

E

AB

C

G

H



Summary of the solutionLet there be a sphere. We take tworandom points, A and B. With A aspole and AB as distance, we drawcircle ABC [?]. Then it is possibleset out the diameter of this circle asEF [Sph.1.18], so that we can puttogether 4EFH [?;Ele.1.22*]. Wedraw HG ⊥ EH and FG ⊥ EF[Ele.1.11], meeting at point G.

Wejoin EG [Ele.P.1].



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F

E

AB

C

G

H

Summary of the solutionLet there be a sphere. We take tworandom points, A and B. With A aspole and AB as distance, we drawcircle ABC [?]. Then it is possibleset out the diameter of this circle asEF [Sph.1.18], so that we can puttogether 4EFH [?;Ele.1.22*]. Wedraw HG ⊥ EH and FG ⊥ EF[Ele.1.11], meeting at point G. Wejoin EG [Ele.P.1].



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D

F

E

AB

C

G

H

K

Summary of the solutionLet there be a sphere. We take tworandom points, A and B. With A aspole and AB as distance, we drawcircle ABC [?]. Then it is possibleset out the diameter of this circle asEF [Sph.1.18], so that we can puttogether 4EFH [?;Ele.1.22*]. Wedraw HG ⊥ EH and FG ⊥ EF[Ele.1.11], meeting at point G. Wejoin EG [Ele.P.1].



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An interlude

F

E

AB

C

G

H

Spherics I 19 is never explicitlyused in the text.

In a number of propositions,however, it is required to let acircle be drawn with a given pointas pole and a pole-distance equalto the side of a square described ina great circle. This construction canbe effected using Spherics I 19.

We set out the diameter of thesphere, as EG. We construct asquare on GE as diameter, and useone of its sides, say EH, as thepole-distance of the great circle.


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Drawing a great circle through two points,Spherics I 20

AB

Summary of the solution, 1Let A and B be two given points.


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AB Summary of the solution, 1

Let A and B be two given points.[Case 1:] If A and B are theendpoints of a diameter of thesphere, there are an infinitenumber of great circles throughthem [?].


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AB

C

D

E

Summary of the solution, 2Let A and B be two given points.[Case 2:] With A as pole, we drawgreat circle CDE [Sph.1.19 (*inter.)].

With B as pole, we draw greatcircle EFH [Sph.1.19 (*inter.)],intersecting great circle CDE atpoints D and E. With D, or E, aspole, we draw a circle passingthrough either A or B [Ele.P.3*].

We argue that AD = BD are equalto the side of a square inscribed ina great circle.


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AB

C

D

E

F

H

Summary of the solution, 2Let A and B be two given points.[Case 2:] With A as pole, we drawgreat circle CDE [Sph.1.19 (*inter.)].With B as pole, we draw greatcircle EFH [Sph.1.19 (*inter.)],intersecting great circle CDE atpoints D and E.

With D, or E, aspole, we draw a circle passingthrough either A or B [Ele.P.3*].



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AB

C

D

E

F

H

G

Summary of the solution, 2Let A and B be two given points.[Case 2:] With A as pole, we drawgreat circle CDE [Sph.1.19 (*inter.)].With B as pole, we draw greatcircle EFH [Sph.1.19 (*inter.)],intersecting great circle CDE atpoints D and E. With D, or E, aspole, we draw a circle passingthrough either A or B [Ele.P.3*].



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Finding the pole of a given circle, Spherics I 21

A B

C

Summary of the solution, 1Let ABC be a given circle.

We takea random point, D. We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].[Case 1:] If circle ABC is not a greatcircle, we draw great circle AFG[Sph.1.20]. We bisect arc AHF atpoint H [?].

We argue that all the lines drawnfrom H to circle ABC are equal.


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D

A B

C

Summary of the solution, 1Let ABC be a given circle. We takea random point, D.

We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].[Case 1:] If circle ABC is not a greatcircle, we draw great circle AFG[Sph.1.20]. We bisect arc AHF atpoint H [?].



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D

A B

C

E

Summary of the solution, 1Let ABC be a given circle. We takea random point, D. We cut off arcsAD = AE [Ele.P.3*].

We bisect arcDE at F [?].[Case 1:] If circle ABC is not a greatcircle, we draw great circle AFG[Sph.1.20]. We bisect arc AHF atpoint H [?].



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D

A B

C

E

F

Summary of the solution, 1Let ABC be a given circle. We takea random point, D. We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].a

[Case 1:] If circle ABC is not a greatcircle, we draw great circle AFG[Sph.1.20]. We bisect arc AHF atpoint H [?].


aIf we set out the diameter of ABC, wecan transfer the arcs AD = AE and bisectthe arc using Ele.3.30.


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D

A BE

F

G

C

Summary of the solution, 1Let ABC be a given circle. We takea random point, D. We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].[Case 1:] If circle ABC is not a greatcircle, we draw great circle AFG[Sph.1.20].

We bisect arc AHF atpoint H [?].



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D

A BE

F

G

H

C

Summary of the solution, 1Let ABC be a given circle. We takea random point, D. We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].[Case 1:] If circle ABC is not a greatcircle, we draw great circle AFG[Sph.1.20]. We bisect arc AHF atpoint H [?].a


aWe set out the diameter of the sphere[Sph.1.19] and draw a great circle. Then weset out the diameter of circle ABC[Sph.1.18], fit it into the great circle [Ele.4.1]and bisect the resulting cord [Ele.3.30].


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D

AB

E

FC

Summary of the solution, 2Let ABC be a given circle. We takea random point, D. We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].[Case 2:] If circle ABC is a greatcircle, we bisect arc AF at C [?].a

With C as pole, we draw a greatcircle through A and F [Sph.1.19(*inter.)]. We bisect arc AF at H [?].


aWe set out the diameter of the sphere[Sph.1.19], draw a great circle, and bisect asemicircle [Ele.3.30].


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D

AB

E

FG C

Summary of the solution, 2Let ABC be a given circle. We takea random point, D. We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].[Case 2:] If circle ABC is a greatcircle, we bisect arc AF at C [?].With C as pole, we draw a greatcircle through A and F [Sph.1.19(*inter.)].

We bisect arc AF at H [?].



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D

AB

E

FG C

H

Summary of the solution, 2Let ABC be a given circle. We takea random point, D. We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].[Case 2:] If circle ABC is a greatcircle, we bisect arc AF at C [?].With C as pole, we draw a greatcircle through A and F [Sph.1.19(*inter.)]. We bisect arc AF at H [?].a


aAgain, we reproduce the arc outsidethe sphere, bisect it and then transfer thedistance back to the sphere.


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D

AB

E

FG C

H

Summary of the solution, 2Let ABC be a given circle. We takea random point, D. We cut off arcsAD = AE [Ele.P.3*]. We bisect arcDE at F [?].[Case 2:] If circle ABC is a greatcircle, we bisect arc AF at C [?].With C as pole, we draw a greatcircle through A and F [Sph.1.19(*inter.)]. We bisect arc AF at H [?].



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Menelaus

Final Remarks

Drawing a tangent great circle, Spherics II 14

A

B

Summary of the solutionLet there be a given lesser circle ABwith point B given on it.

We findthe pole of circle AB, as point C[Sph.1.21]. We draw a great circlethrough C and B [Sph.1.21], andcut off arc BD equal to a quadrant[Sph.1.19 (*inter.)]. With pole D, wedraw great circle BEF [Ele.P.3*].

The proof follows immediatelyfrom the properties of tangency[Sph.2.3-5].


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Menelaus

Final Remarks


A

B

C

Summary of the solutionLet there be a given lesser circle ABwith point B given on it. We findthe pole of circle AB, as point C[Sph.1.21].

We draw a great circlethrough C and B [Sph.1.21], andcut off arc BD equal to a quadrant[Sph.1.19 (*inter.)]. With pole D, wedraw great circle BEF [Ele.P.3*].



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Menelaus

Final Remarks


A

B

C

Summary of the solutionLet there be a given lesser circle ABwith point B given on it. We findthe pole of circle AB, as point C[Sph.1.21]. We draw a great circlethrough C and B [Sph.1.21],

andcut off arc BD equal to a quadrant[Sph.1.19 (*inter.)]. With pole D, wedraw great circle BEF [Ele.P.3*].



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Final Remarks


D

A

B

C

Summary of the solutionLet there be a given lesser circle ABwith point B given on it. We findthe pole of circle AB, as point C[Sph.1.21]. We draw a great circlethrough C and B [Sph.1.21], andcut off arc BD equal to a quadrant[Sph.1.19 (*inter.)].

With pole D, wedraw great circle BEF [Ele.P.3*].



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Final Remarks


D

A

B

C

EF

Summary of the solutionLet there be a given lesser circle ABwith point B given on it. We findthe pole of circle AB, as point C[Sph.1.21]. We draw a great circlethrough C and B [Sph.1.21], andcut off arc BD equal to a quadrant[Sph.1.19 (*inter.)]. With pole D, wedraw great circle BEF [Ele.P.3*].



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Menelaus

Final Remarks


A

BC

Summary of the solution, 1Let AB be a given lesser circle andC a given point, not on it.a

We findthe pole of circle AB, as point D[Sph.1.21]. With D as pole, wedraw lesser circle CF [Ele.P.3*]. Wedraw great circle DC, meetingcircle AB at point B [Sph.1.20], andcut off quadrant BG [Sph.1.19(*inter.)]. With G as pole, we drawgreat circle EBH [Ele.P.3*].

aPoint C must be between AB and thecircle equal and parallel to it.


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Menelaus

Final Remarks


AD

BC

Summary of the solution, 1Let AB be a given lesser circle andC a given point, not on it. We findthe pole of circle AB, as point D[Sph.1.21].

With D as pole, wedraw lesser circle CF [Ele.P.3*]. Wedraw great circle DC, meetingcircle AB at point B [Sph.1.20], andcut off quadrant BG [Sph.1.19(*inter.)]. With G as pole, we drawgreat circle EBH [Ele.P.3*].


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Menelaus

Final Remarks


A

F

D

BC

Summary of the solution, 1Let AB be a given lesser circle andC a given point, not on it. We findthe pole of circle AB, as point D[Sph.1.21]. With D as pole, wedraw lesser circle CF [Ele.P.3*].

Wedraw great circle DC, meetingcircle AB at point B [Sph.1.20], andcut off quadrant BG [Sph.1.19(*inter.)]. With G as pole, we drawgreat circle EBH [Ele.P.3*].


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Menelaus

Final Remarks


A

F

D

BC

Summary of the solution, 1Let AB be a given lesser circle andC a given point, not on it. We findthe pole of circle AB, as point D[Sph.1.21]. With D as pole, wedraw lesser circle CF [Ele.P.3*]. Wedraw great circle DC, meetingcircle AB at point B [Sph.1.20],

andcut off quadrant BG [Sph.1.19(*inter.)]. With G as pole, we drawgreat circle EBH [Ele.P.3*].


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Menelaus

Final Remarks


A

F

D

BC

G

Summary of the solution, 1Let AB be a given lesser circle andC a given point, not on it. We findthe pole of circle AB, as point D[Sph.1.21]. With D as pole, wedraw lesser circle CF [Ele.P.3*]. Wedraw great circle DC, meetingcircle AB at point B [Sph.1.20], andcut off quadrant BG [Sph.1.19(*inter.)].

With G as pole, we drawgreat circle EBH [Ele.P.3*].


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Menelaus

Final Remarks


A

F

D

BC

H

E

G

Summary of the solution, 1Let AB be a given lesser circle andC a given point, not on it. We findthe pole of circle AB, as point D[Sph.1.21]. With D as pole, wedraw lesser circle CF [Ele.P.3*]. Wedraw great circle DC, meetingcircle AB at point B [Sph.1.20], andcut off quadrant BG [Sph.1.19(*inter.)]. With G as pole, we drawgreat circle EBH [Ele.P.3*].


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Menelaus

Final Remarks


A

F

D

BC

H

E

M

G

N

Summary of the solution, 2We draw two great circles, DH andDE, meeting circle AB at points Nand M [Sph.1.20].

We cut off arcsHL and EK equal to arc CG[Ele.P.3*]. With L as pole anddistance LN, we draw circle XNC,and with K as pole and KM asdistance, we draw circle OMC[Ele.P.3*].

We draw in the internal lines andshow that the are all equal to theside of a great square.


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Menelaus

Final Remarks


A

F

D

BC

H

E

M

KG

L

N

Summary of the solution, 2We draw two great circles, DH andDE, meeting circle AB at points Nand M [Sph.1.20]. We cut off arcsHL and EK equal to arc CG[Ele.P.3*].

With L as pole anddistance LN, we draw circle XNC,and with K as pole and KM asdistance, we draw circle OMC[Ele.P.3*].



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Menelaus

Final Remarks


A

F

D

BC

H

X

E

M

O

KG

L

N

Summary of the solution, 2We draw two great circles, DH andDE, meeting circle AB at points Nand M [Sph.1.20]. We cut off arcsHL and EK equal to arc CG[Ele.P.3*]. With L as pole anddistance LN, we draw circle XNC,and with K as pole and KM asdistance, we draw circle OMC[Ele.P.3*].



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Menelaus

Final Remarks

Al-Harawı̄ on Menelaus

“One sees he considers Theodosius inadequate in histreatise On Spheres and thinks that the way he followed isother than satisfactory, since there is difficulty in it, andthe setting out of many lines, and he did not adhere, in it,to the properties of the figures that occur on the sphere,namely the conditions of the angles that arise from theintersection of the circles.

Upon my life, Menelaus has easily shown everything thatTheodosius proved in that book and he intends that theproof be straightforward ( �

éÓA�®�J�B

@

��KQ¢�.)4 without using

straight lines.”

4That is “by the correct way,” the “by the straight route.” Maybedirect, as opposed to indirect.


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To construct equal angles, Menelaus Spher. I 1

BA

G

D

E

Introduction

1p “We want to make an angle ata given point on thecircumference of a great circlethat is equal to a given angle.”

2p Let B be the given point ongreat-circle arc AB and let thegiven angle be angle GDE.

3p So, it is necessary to make anangle at B equal to angle GDE.


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BA

G

D

E Core [solving the problem]

4p With D as pole, we draw arcGE with “whatever distance”(YªK.

�ø

AK.).

With B as pole, wedraw arc AZ with the samedistance, and we cut off arcAZ = arc GE. We joingreat-circle arc BZ.

5p I say, angle ABZ is equal toangle GDE.


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Final Remarks


BA

Z

G

D

E



�ø

AK.). With B as pole, wedraw arc AZ with the samedistance,

and we cut off arcAZ = arc GE. We joingreat-circle arc BZ.



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Menelaus

Final Remarks


BA

Z

G

D

E



�ø

AK.). With B as pole, wedraw arc AZ with the samedistance, and we cut off arcAZ = arc GE.

We joingreat-circle arc BZ.



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Menelaus

Final Remarks


BA

Z

G

D

E



�ø

AK.). With B as pole, wedraw arc AZ with the samedistance, and we cut off arcAZ = arc GE. We joingreat-circle arc BZ.



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Menelaus

Final Remarks


BA

Z

G

D

E


6p Two great circles, GD & DE,pass through the pole of circleGE, so their planes are ⊥ tocircle GE [TSph.1.15]. The twointersections of great circlesGE & DE with circle GE, passthrough the center of circleGE, while intersection ofcircles GD & DE is ⊥ to circleGE at its center, hence theintersections of circles GE &DE with circle GE are each ⊥the intersection of circles GD& DE...


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Final Remarks


BA

Z

G

D

E


6p ... The same argument can bemade for the internal objectsof triangle ABZ. But circleAZ = circle GE and arc AZ =arc GE, so the inclination ofplane BZ on plane AB is equalto the inclination of plane EDon plane GD. Therefore,∠ABZ = ∠GDE [def. of equalinclination].


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Final Remarks

Final Remarks

I All of the constructions used to solve problems inancient spherics can be carried out through physicalmanipulations of an unmarked ruler and anon-colapsing compass.

I In the texts on spherics, an instrument like a compassmust be used to transfer distances.

I The use of practical constructions appears to haveguided not only the solution to problems but also tohave influenced the overall, theoretical approach.

I Theodosius moves from considerations of internallines to considerations of the objects on the surface ofthe sphere.

I Menelaus attempts to restrict his attention to theobjects on the surface. (Of course, he still usesinternal lines, but he does not name them; moreover,he uses Theodosius’s theorems that make use ofinternal lines.)

constructions in ancient greek spherics: *1em …constructions in ancient greek spherics:...

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