Lesson 00

How to use this course

Pappus wrote in the early fourth century AD, some five hundred years after Euclid and Apollonius. He is often called a commentator, and much of his Collection explains, supplements and corrects older books. Yet the Collection also contains results that appear nowhere earlier: the hexagon theorem, the invariance of the cross-ratio, the focus-directrix property of all three conics, and the rule for volumes of revolution later credited to Paul Guldin. It also preserves our only descriptions of several lost works of Apollonius and Euclid.

Prerequisites for Part II

Euclidean geometry at the level of similar triangles, the power of a point and areas of parallelograms.
The conic sections and their equations in Cartesian coordinates.
For Lesson 08, homogeneous coordinates are helpful but not required.
For Lesson 10, the definition of a centroid as an integral.

Part I sets out what is documented about Pappus and where each fact comes from. The evidence is thin: his own writings, a short entry in the tenth-century Byzantine lexicon called the Suda, a marginal note in a chronological table, and mentions by later commentators. Part II follows the Collection book by book, with modern statements and proofs placed beside what Pappus wrote. Part III follows the book from a single medieval manuscript to the printed Latin edition of 1588, and then into analytic and projective geometry.

Reference

Timeline

Dates before 1500 are approximate unless stated. Pappus’s birth and death years are modern estimates.

c. 300 BC	Euclid writes the Elements, Data and the lost Porisms, all later discussed by Pappus.
c. 200 BC	Apollonius of Perga writes the Conics and the shorter works Pappus summarizes in Book VII.
c. 150 AD	Claudius Ptolemy writes the Almagest.
c. 290	Approximate birth of Pappus (estimate in MacTutor).
320	18 October: partial solar eclipse at Alexandria, computed and described in Pappus’s commentary on the Almagest.
c. 320–340	Composition of the Collection; Adolphe Rome proposed c. 340, others a date soon after 320.
c. 350	Approximate death (estimate).
c. 360–380	Theon of Alexandria writes his own Almagest commentary, drawing on Pappus.
5th–6th c.	Proclus and Eutocius cite Pappus.
10th c.	The Suda lists his works and wrongly places him under Theodosius I.
12th c.	Codex Vaticanus Graecus 218 is copied, the archetype of all surviving manuscripts of the Collection.
1588	Federico Commandino’s Latin translation is printed at Pesaro, thirteen years after his death.
1600	François Viète’s Apollonius Gallus reconstructs the lost Tangencies from Pappus’s description.
1631–37	Jacob Golius sends René Descartes the locus problem of Book VII; Descartes solves it in La Géométrie (1637).
1640	Paul Guldin publishes the centroid rule for solids of revolution.
1687	Isaac Newton solves the four-line locus synthetically in the Principia (Book I, Lemmas 17–19).
1876–78	Friedrich Hultsch’s Greek–Latin critical edition, Berlin.
1899	David Hilbert’s Grundlagen der Geometrie uses the hexagon theorem to build an arithmetic of segments.
1931	Adolphe Rome identifies the eclipse of 320 in the Almagest commentary.
1933	Paul Ver Eecke’s complete French translation.
1986	Alexander Jones’s edition and English translation of Book VII.
1999	Thomas Hales proves the honeycomb conjecture, the modern form of the bee problem of Book V.

Part I · Sources

Lesson 01

Alexandria and the eclipse of 320

Nothing is known of Pappus’s birth, parents or education. His books and the Suda both call him an Alexandrian, and he seems to have spent his working life in that city. Alexandria in the early fourth century was the administrative centre of Roman Egypt and still the main seat of Greek mathematical learning. The age of original research was long past. Teachers worked from a canon of classical texts by Euclid, Archimedes, Apollonius and Ptolemy, and much of their writing took the form of commentary.

Three conflicting dates

The ancient witnesses disagree. The Suda makes Pappus a contemporary of Theon of Alexandria and dates both to the reign of Theodosius I (379–395). A marginal note in a manuscript of a chronological table, placed beside the reign of Diocletian (284–305), puts Pappus there instead. Thomas Heath accepted the Diocletian date.

The question was settled from Pappus’s own text. In his commentary on Book VI of Ptolemy’s Almagest he works through the calculation of a solar eclipse in the Egyptian month Tybi of the year 1068 of the era of Nabonassar. In 1931 the Belgian scholar Adolphe Rome showed that this is the eclipse of 18 October 320, visible at Alexandria, and that Pappus wrote as one who had seen it recently. This is the only firm date in his life. It shows that the Suda is wrong by about half a century.

Bounds from other authors

Pappus quotes Ptolemy, who died around 170. Proclus, born around 411, quotes Pappus. Within these bounds the eclipse pins his activity to the 320s and after. MacTutor’s estimates of c. 290 for his birth and c. 350 for his death are reasonable guesses, not evidence.

Documented: Alexandrian origin; the 320 eclipse; the dedications in the Collection (Lesson 02). Unreliable: the Suda’s Theodosian date and the Diocletian marginal note. Unknown: family background, schooling, dates of birth and death.

Lesson 02

The Collection as a book

The Synagoge or Mathematical Collection survives in eight books, of which Book I and the first part of Book II are lost. Every surviving manuscript descends from a single twelfth-century copy, Codex Vaticanus Graecus 218. It is not a textbook. Pappus meant it to be read alongside the classical works, filling gaps in their proofs, giving alternative solutions and recording lemmas a reader would need.

The books

II (fragment): Apollonius’s system for writing and multiplying very large numbers in powers of the myriad, \(10^4\).
III: the two mean proportionals, the theory of means, a set of geometrical paradoxes, and the inscription of the five regular solids in a sphere.
IV: a generalization of the Pythagorean theorem, the arbelos, the spiral of Archimedes, the conchoid of Nicomedes, the quadratrix, and angle trisection.
V: isoperimetric problems, opening with the bees’ honeycomb; the thirteen semiregular solids attributed to Archimedes.
VI: difficulties in the “Little Astronomy”, the short works of Theodosius, Autolycus, Aristarchus and Euclid read between the Elements and the Almagest.
VII: the “Treasury of Analysis”, with summaries of and lemmas to 33 books by Euclid, Apollonius, Aristaeus and Eratosthenes.
VIII: mechanics, the centre of gravity, the inclined plane, toothed wheels, and the conic through five points.

Dedications and named people

Book III is addressed to Pandrosion, Book V to Megethion, and Books VII and VIII to Pappus’s son Hermodorus. The reference to his son is the only fact we have about his family. A philosopher friend, Hierius, had pressed him about the problem of the two mean proportionals.

Pandrosion deserves a closer look. Book III opens by correcting the work of “some people” who taught geometry under Pandrosion’s supervision, including a solution of the mean proportional problem that Pappus judged unproven. The manuscript addresses Pandrosion with a feminine adjective. Hultsch, in his 1878 edition, emended it to the masculine. Alexander Jones argued in 1986 that the manuscript reading should stand, and most historians now accept that Pandrosion was a woman who ran a mathematical school in Alexandria, two generations before Hypatia. Pappus’s tone suggests a rival teacher.

Exercise 2.1Book II multiplies out the numerical values of the Greek letters in a line of verse and writes the result in Apollonius’s notation. Write \(N=2\times10^{17}\) in “myriads”: as a sum \(\sum a_k M^k\) with \(M=10^4\) and \(0\le a_k\lt M\).

Show solution

\(10^{17}=10\cdot 10^{16}=10\,M^4\), so \(N=20\,M^4\). In Apollonius’s language this is 20 “quadruple myriads”. The system works like base \(10^4\) place notation, with the powers of the myriad named rather than written by position.

Lesson 03

Teacher and commentator

A passage in Proclus has been read as evidence that Pappus headed a school, and the dedication of two books to his son suggests that he taught Hermodorus himself. The Collection reads like the work of a teacher who knew the classical texts thoroughly and cared about the gaps students would meet.

Other works

Commentary on the Almagest. The parts on Books V and VI survive in Greek. They follow Ptolemy’s lunar theory, parallax and eclipse computation step by step. Otto Neugebauer judged this work much weaker than the geometry. Theon of Alexandria’s later commentary drew on it.
Commentary on Euclid’s Elements. Proclus and Eutocius cite it. A two-part commentary on Book X, on irrational lines, survives in Arabic under Pappus’s name, though some scholars doubt the attribution.
Geography. The Suda lists a description of the inhabited world. An Armenian geography attributed to Moses of Khoren appears to depend on it.
Lost or doubtful works listed by the Suda include a book on the rivers of Libya and one on dream interpretation. A commentary on Ptolemy’s Harmonics and a hydrostatic instrument described by al-Khazini have also been linked to his name, without certainty.

Pappus as a writer

His prefaces are the liveliest Greek mathematical prose of late antiquity. Book V opens with a meditation on the intelligence of bees. Book VII opens with the clearest ancient account of method in geometry. Elsewhere he is sharp with predecessors: in Book IV he counts it a real fault in geometers to solve a plane problem with conics. Historians disagree on how original he was. Heath valued him mainly as a preserver; Jones and others have shown that many lemmas in Book VII go beyond anything in his sources.

Exercise 3.1The eclipse was dated in Nabonassar’s era, which starts on 26 February 747 BC and counts Egyptian years of exactly 365 days. Show that year 1068 of the era began in AD 320, as Rome’s identification requires.

Show solution

Year 1068 begins after 1067 Egyptian years, \(1067\times365=389\,455\) days, which is \(389\,455/365.25\approx1066.27\) Julian years. In astronomical numbering 747 BC is the year \(-746\), and late February is about \(0.15\) of a year in, so year 1068 begins near \(-746+0.15+1066.27\approx320.42\): late May of AD 320. Tybi is the fifth Egyptian month, days 121–150 of the year, which runs from late September to late October 320. In fact 18 October 320 is Tybi 17.

Part II · The mathematics

Lesson 04

Plane, solid and linear problems

For the general reader. Some constructions need only a ruler and compass. Others, like doubling a cube or trisecting an angle, cannot be done that way. Pappus sorted problems by the tools they genuinely require, and insisted that a geometer use the simplest tool that works. This sorting anticipates a question that was only answered with algebra in the nineteenth century.

Goals

State Pappus’s classification of problems.
Reduce cube duplication to two mean proportionals.
Prove the hyperbola trisection of Book IV.

The classification

Pappus’s three kinds of problem (Books III and IV)

A problem is plane if it can be solved with straight lines and circles; solid if it needs conic sections; linear if it needs more complicated curves such as spirals, quadratrices, conchoids or cissoids. It is a fault to solve a problem by a higher class of curve than it requires.

In modern terms a plane problem leads to an equation solvable by square roots, and a solid problem to a cubic or quartic. Pierre Wantzel proved in 1837 that cube duplication and general trisection are not plane, confirming what Pappus and his predecessors suspected without proof.

Mean proportionals

Hippocrates of Chios had reduced doubling the cube to inserting two mean proportionals \(x,y\) between segments \(a\) and \(b\): \(a:x=x:y=y:b\). Book III collects several solutions, all solid or mechanical, including one of Pappus’s own, and criticises the approximate method from Pandrosion’s school.

Trisection by a hyperbola

Theorem (Book IV)

Let \(B,C\) be fixed points. The locus of points \(P\) such that \(\angle PCB=2\angle PBC\) is one branch of a hyperbola with focus \(C\), directrix the perpendicular bisector of \(BC\), and eccentricity 2.

Proof. Put \(B=(0,0)\), \(C=(2,0)\), \(\angle PBC=\theta\), \(\angle PCB=2\theta\). The angle at \(P\) is \(\pi-3\theta\), so by the sine rule \(PC=2\sin\theta/\sin3\theta\). Since \(\sin3\theta=\sin\theta\,(3-4\sin^2\theta)=\sin\theta\,(1+2\cos2\theta)\), we get \(PC=2/(1+2\cos2\theta)\). The \(x\)-coordinate of \(P\) is \(2-PC\cos2\theta\), so its distance from the line \(x=1\) is \(1-PC\cos2\theta\). A short computation gives

\[ 2\,(1-PC\cos 2\theta)=PC\,(1+2\cos2\theta)-2PC\cos2\theta = PC. \]

So \(PC=2\cdot\operatorname{dist}(P,\ x=1)\). To trisect an angle \(\varphi\), draw on \(BC\) the arc of points from which \(BC\) subtends \(\pi-\varphi\). Its intersection \(P\) with the hyperbola has \(\pi-3\theta=\pi-\varphi\), so \(\theta=\varphi/3\). The problem is solid, as Pappus says.

Exercise 4.1Show that if \(x,y\) are two mean proportionals between \(a\) and \(2a\), then \(x^3=2a^3\).

Show solution

From \(a:x=x:y\) we get \(y=x^2/a\). From \(x:y=y:2a\) we get \(y^2=2ax\). Substituting, \(x^4/a^2=2ax\), so \(x^3=2a^3\). The cube on \(x\) has twice the volume of the cube on \(a\).

Exercise 4.2Check the theorem numerically: with \(B=(0,0)\), \(C=(2,0)\) and \(\theta=30^\circ\), find \(P\) and verify \(PC=2(x_P-1)\).

Show solution

\(2\theta=60^\circ\), so the angle at \(P\) is \(90^\circ\). \(PC=2/(1+2\cos60^\circ)=1\). Then \(P=(2-\cos60^\circ,\ \sin60^\circ)=(1.5,\ 0.866)\). Indeed \(2(x_P-1)=1=PC\). Also \(PB=\sqrt{3}\), and \(\angle PBC=\arctan(0.866/1.5)=30^\circ\).

Exercise 4.3Explain, using field degrees, why trisecting \(60^\circ\) is not a plane problem.

Show solution

Trisecting \(60^\circ\) means constructing \(c=\cos20^\circ\). From \(\cos3t=4\cos^3t-3\cos t\) with \(\cos60^\circ=\tfrac12\), \(c\) satisfies \(8c^3-6c-1=0\). This cubic has no rational root (test \(\pm1,\pm\tfrac12,\pm\tfrac14,\pm\tfrac18\)), so it is irreducible and \([\mathbb Q(c):\mathbb Q]=3\). Ruler and compass only produce numbers of degree a power of 2. Hence the problem is at least solid.

Lesson 05

Book IV: areas and the arbelos

For the general reader. Book IV opens with a version of Pythagoras’s theorem that works for any triangle, using parallelograms instead of squares. It then studies the arbelos, or shoemaker’s knife, a figure bounded by three semicircles, and proves a striking fact about an infinite chain of circles packed inside it.

Goals

Prove Pappus’s area theorem and recover Pythagoras.
State the Pappus chain theorem.
Prove it with inversion, a tool Pappus did not have.

Theorem (Pappus’s area theorem, Book IV, Prop. 1)

Let \(ABC\) be any triangle. On \(AB\) and \(BC\) draw arbitrary parallelograms \(ABDE\) and \(BCFG\) outward. Extend \(DE\) and \(FG\) to meet at \(H\). On \(AC\) draw a parallelogram \(ACKL\) whose sides \(AL\) and \(CK\) are equal and parallel to \(HB\). Then \(\operatorname{area}(ACKL)=\operatorname{area}(ABDE)+\operatorname{area}(BCFG)\).

Proof sketch. Let \(HB\) meet \(AC\) at \(M\) and \(LK\) at \(N\), splitting \(ACKL\) into two parallelograms \(AMNL\) and \(MCKN\). Slide parallelogram \(ABDE\) along the direction \(AB\) (same base, same parallels) into a parallelogram with sides \(AB\) and \(BH\). That one in turn shears, keeping the side \(AL\parallel BH\) of equal length, onto \(AMNL\). The same two shears carry \(BCFG\) onto \(MCKN\). Shears preserve area.

The arbelos and the Pappus chain

Let \(A,C,B\) lie on a line with \(C\) between \(A\) and \(B\). Draw semicircles on \(AB\), \(AC\) and \(CB\) on the same side. The region inside the large semicircle and outside the two small ones is the arbelos. Let \(\omega_1\) be the circle tangent to all three semicircles, and let \(\omega_{n+1}\) be tangent to \(\omega_n\), to the large semicircle and to the semicircle on \(AC\).

Theorem (Pappus chain, Book IV)

The centre of \(\omega_n\) lies at distance \(n\,d_n\) from the line \(AB\), where \(d_n\) is the diameter of \(\omega_n\).

Pappus proves it with a long chain of similar triangles. The modern proof uses inversion, introduced in the nineteenth century.

Proof by inversion. Invert in the circle centred at \(A\) that is orthogonal to \(\omega_n\). This inversion fixes \(\omega_n\). The two semicircles through \(A\) (on \(AB\) and \(AC\)) become two lines perpendicular to \(AB\), both tangent to \(\omega_n\). The chain \(\omega_1,\dots,\omega_{n-1}\) and the semicircle on \(CB\) become circles of the same diameter \(d_n\) stacked between these parallel lines, each tangent to the next, with the image of the semicircle on \(CB\) centred on the line \(AB\). Counting \(n\) diameters up from \(AB\) to the centre of \(\omega_n\) gives the height \(n\,d_n\). Since \(\omega_n\) is its own image, this is the height of the original centre.

Exercise 5.1Show that Pappus’s area theorem contains the Pythagorean theorem.

Show solution

Let the angle at \(B\) be right and take squares on \(AB\) and \(BC\). Then \(H\) is the fourth corner of the rectangle on \(B\) and the two outer square sides, so \(HB\) is a diagonal of a rectangle with sides \(AB,BC\): \(|HB|=|AC|\), and \(HB\) is perpendicular to \(AC\) (it is \(AC\) rotated by \(90^\circ\)). The parallelogram on \(AC\) is therefore a square, and the theorem reads \(AC^2=AB^2+BC^2\).

Exercise 5.2Verify the case \(n=1\) in the symmetric arbelos \(A=(-2,0)\), \(C=(0,0)\), \(B=(2,0)\).

Show solution

The semicircles have radii 2 (centre 0), 1 (centre \(-1\)) and 1 (centre 1). By symmetry \(\omega_1\) has centre \((0,h)\) and radius \(r\). Tangency to the big circle: \(h+r=2\). Tangency to the circle centred at \((1,0)\): \(\sqrt{1+h^2}=1+r\). Substituting \(h=2-r\): \(1+(2-r)^2=(1+r)^2\), so \(5-4r=1+2r\), \(r=\tfrac23\), \(h=\tfrac43\). The diameter is \(\tfrac43\), equal to \(h\), as the theorem claims for \(n=1\).

Lesson 06

Book V: bees and isoperimetry

For the general reader. Why are honeycomb cells hexagons? Pappus answered that bees, given a fixed amount of wax, choose the shape that tiles the plane and encloses the most honey. Only three regular polygons tile without gaps, and of those the hexagon holds the most for a given perimeter. The claim took until 1999 to prove in full generality.

Goals

Compute and compare areas of regular polygons with equal perimeter.
State the isoperimetric results Pappus took from Zenodorus.
Compare the sphere with the regular solids.

Tiling around a point

Regular \(n\)-gons meeting at a point have interior angle \((1-\tfrac2n)\pi\). For \(k\) of them to fill the angle around a point, \(k(1-\tfrac2n)=2\), that is \((n-2)(k-2)=4\). The solutions are \((n,k)=(3,6),(4,4),(6,3)\): triangles, squares and hexagons. Pappus states these three possibilities in his preface.

Theorem (Zenodorus, reported in Book V)

(a) Of regular polygons with equal perimeter, the one with more sides has greater area, and the circle exceeds them all. (b) Of all \(n\)-gons with given perimeter, the regular one has the greatest area.

For part (a), a regular \(n\)-gon of perimeter \(L\) splits into \(n\) isosceles triangles with base \(L/n\) and height \(\frac{L}{2n}\cot\frac\pi n\), so

\[ A_n=\frac{L^2}{4n\tan(\pi/n)}, \qquad \lim_{n\to\infty}A_n=\frac{L^2}{4\pi}. \]

The function \(n\tan(\pi/n)\) decreases to \(\pi\) because \(\tan x/x\) increases on \((0,\pi/2)\). Zenodorus proved this with Euclidean inequalities on chords and angles. Pappus extends the comparison to solids: among the five regular solids of equal surface, the one with more faces has greater volume, and the sphere exceeds them all. He also describes the thirteen semiregular solids, which he credits to Archimedes; Archimedes’ own treatise is lost.

The modern bee problem

Pappus compared only regular tilings. The question whether any partition of the plane into equal areas has less total boundary than the hexagonal one, allowing curved walls, is the honeycomb conjecture. Thomas Hales proved it in 1999.

Exercise 6.1Take perimeter 12. Compute the areas of the equilateral triangle, square, regular hexagon and circle.

Show solution

Triangle, side 4: \(\tfrac{\sqrt3}{4}\cdot16=4\sqrt3\approx6.93\). Square, side 3: 9. Hexagon, side 2: \(\tfrac{3\sqrt3}{2}\cdot4=6\sqrt3\approx10.39\). Circle: \(144/(4\pi)=36/\pi\approx11.46\). The hexagon beats the other tilings by about 15%.

Exercise 6.2A cube and a sphere each have surface area 6. Compare their volumes.

Show solution

The cube has edge 1 and volume 1. The sphere has \(4\pi r^2=6\), \(r=\sqrt{3/(2\pi)}\approx0.691\), and volume \(\tfrac43\pi r^3\approx1.382\). The sphere holds about 38% more.

Lesson 07

Analysis and synthesis

For the general reader. How do you find a proof or a construction? Pappus gave the classic answer: suppose the thing done, and reason backward until you reach something you already know. Then reverse the chain. This short passage at the start of Book VII is the most cited ancient text on mathematical method.

Goals

Distinguish theoretical and problematical analysis.
Know the contents of the Treasury of Analysis.
Carry out an analysis and synthesis in full.

Pappus’s definitions (Book VII, preface)

Analysis takes what is sought as granted and proceeds through its consequences to something already known or admitted. Synthesis starts from that known thing and reaches the sought result by reversing the steps. In theoretical analysis the sought thing is a theorem: if the endpoint is true, the theorem is true, provided each step reverses. In problematical analysis it is a construction: if the endpoint is possible, so is the construction, and the reversal also gives the diorismos, the conditions under which a solution exists.

The weak point is reversibility. Backward steps often go from a statement to a consequence, and the converse may fail. Synthesis is the check.

The Treasury of Analysis

Pappus lists the books a student of analysis should master: Euclid’s Data (1 book), Porisms (3) and Surface Loci (2); Apollonius’s Cutting off of a Ratio (2), Cutting off of an Area (2), Determinate Section (2), Tangencies (2), Inclinations (2), Plane Loci (2) and Conics (8); Aristaeus’s Solid Loci (5); and Eratosthenes’ On Means (2). That makes 33 books. Of the minor works only Cutting off of a Ratio survives, in Arabic. For the others, Pappus’s summaries and lemmas are our main source, and they guided every later reconstruction (Lesson 11).

Exercise 7.1Inscribe a square in a given acute triangle \(ABC\) with one side on \(BC\). Give the analysis, then the synthesis.

Show solution

Analysis. Suppose \(PQRS\) is the square, with \(P\) on \(AB\), \(Q\) on \(AC\) and \(R,S\) on \(BC\). The homothety centred at \(B\) carries any square with one side on \(BC\) and a vertex on \(BA\) to another such square, and its fourth vertex moves along a line through \(B\). So \(Q\) is where that line meets \(AC\). Synthesis. Draw any square \(P'Q'R'S'\) with \(S'R'\) on \(BC\) and \(P'\) on \(BA\). Draw \(BQ'\) and let it meet \(AC\) at \(Q\). Drop perpendiculars and complete the square \(PQRS\) by the homothety centred at \(B\) with ratio \(BQ/BQ'\). Diorismos. One solution exists when angles \(B\) and \(C\) are acute.

Exercise 7.2A student “proves” \(1=2\) by analysis: assume \(1=2\); multiply by 0; get \(0=0\), which is true. Which requirement of Pappus’s method fails?

Show solution

The step “multiply by 0” does not reverse. Theoretical analysis proves a theorem only when each step is an equivalence, so that synthesis can run the chain back. Here synthesis would need to divide \(0=0\) by 0.

Lesson 08

Cross-ratio and the hexagon

For the general reader. Among the lemmas Pappus wrote for Euclid’s lost Porisms are two results that look modern: a quantity built from four points on a line that survives any projection, and a theorem about nine points and nine lines that needs no measurement at all. Both became foundations of projective geometry.

Goals

Prove invariance of the cross-ratio under central projection.
State and prove the hexagon theorem.
See why the theorem is equivalent to commutativity of multiplication.

For four points \(A,B,C,D\) on a line, with signed lengths, define the cross-ratio

\[ (A,B;C,D)=\frac{\overline{CA}\cdot\overline{DB}}{\overline{CB}\cdot\overline{DA}}. \]

Theorem (Book VII, Prop. 129)

If four lines through a point \(O\) meet one transversal at \(A,B,C,D\) and another at \(A',B',C',D'\), then \((A,B;C,D)=(A',B';C',D')\).

Proof. Compare triangle areas: \(\overline{CA}/\overline{CB}=[OCA]/[OCB]=\dfrac{OA\sin\angle COA}{OB\sin\angle COB}\), and similarly for \(D\). In the quotient the lengths \(OA,OB\) cancel, leaving only sines of the angles at \(O\). These are the same for both transversals.

Theorem (Pappus’s hexagon theorem, Book VII, Prop. 139)

Let \(A,B,C\) be distinct points on a line \(\ell\) and \(A',B',C'\) distinct points on another line \(\ell'\), none at the intersection of \(\ell\) and \(\ell'\). Then the three points \(AB'\cap A'B\), \(AC'\cap A'C\) and \(BC'\cap B'C\) are collinear.

Proof (affine reduction). In the projective plane, send the line through two of the three crossing points to infinity. Then \(AB'\parallel A'B\) and \(BC'\parallel B'C\), and we must show \(AC'\parallel A'C\). If \(\ell,\ell'\) meet at \(O\), place coordinates \(a,b,c\) on \(\ell\) and \(a',b',c'\) on \(\ell'\) measured from \(O\). By similar triangles, \(AB'\parallel A'B\) means \(a/b=b'/a'\), that is \(a\,a'=b\,b'\). Likewise \(b\,b'=c\,c'\). Hence \(a\,a'=c\,c'\), which says \(AC'\parallel A'C\). (If \(\ell\parallel\ell'\), the same argument works with translations.)

The step from \(a/b=b'/a'\) to \(a\,a'=b\,b'\) silently commutes \(a'\) past \(b^{-1}\). This is not an accident. Over a division ring the theorem holds exactly when multiplication is commutative (Lesson 12).

Exercise 8.1Compute \((A,B;C,D)\) for \(A=0\), \(B=\infty\), \(C=1\), \(D=x\) on the real line.

Show solution

With \(B\) at infinity the factors involving \(B\) cancel in the limit: \((A,B;C,D)=\overline{CA}/\overline{DA}\cdot\lim\overline{DB}/\overline{CB}=\frac{0-1}{0-x}\cdot1=\frac1x\). Every cross-ratio value occurs, so four points have one projective invariant.

Exercise 8.2Verify the hexagon theorem with \(A=(0,0),B=(1,0),C=(3,0)\) on the \(x\)-axis and \(A'=(0,1),B'=(2,1),C'=(3,1)\) on \(y=1\).

Show solution

\(AB'\): \(y=x/2\); \(A'B\): \(y=1-x\). They meet at \((\tfrac23,\tfrac13)\). \(AC'\): \(y=x/3\); \(A'C\): \(y=1-x/3\). Meet at \((\tfrac32,\tfrac12)\). \(BC'\): \(y=(x-1)/2\); \(B'C\): \(y=3-x\). Meet at \((\tfrac73,\tfrac23)\). Slopes between consecutive points: \(\frac{1/6}{5/6}=\tfrac15\) and \(\frac{1/6}{5/6}=\tfrac15\). Collinear.

Exercise 8.3Count points and lines in the Pappus configuration and show it is self-dual in its counts.

Show solution

Points: 6 given plus 3 crossing points, 9 in all. Lines: \(\ell,\ell'\), the six cross-joins and the Pappus line, 9 in all. Each point lies on 3 of the lines and each line contains 3 of the points, a \((9_3)\) configuration.

Lesson 09

Focus, directrix and the locus problem

For the general reader. Ellipses, parabolas and hyperbolas look different, but Pappus proved they share a single description in terms of a point and a line. He also posed a locus problem with several lines that the Greeks could not finish. Descartes made it the test case of his new coordinate geometry.

Goals

Prove the focus-directrix property in coordinates.
State Pappus’s locus problem precisely.
See why three or four lines give a conic.

Theorem (Book VII, lemmas to Euclid’s Surface Loci)

Given a point \(F\), a line \(d\) not through \(F\), and \(e>0\), the locus of points \(P\) with \(PF=e\cdot\operatorname{dist}(P,d)\) is a conic: an ellipse if \(e<1\), a parabola if \(e=1\), a hyperbola if \(e>1\).

This is the earliest surviving proof of the property for all three conics. Apollonius had used foci for the ellipse and hyperbola, but no directrix.

The locus to three, four or more lines

Pappus’s problem (Book VII)

Given lines \(\ell_1,\dots,\ell_{2m}\) (or \(2m-1\)), angles \(\alpha_i\), and a ratio \(k\), let \(d_i(P)\) be the length of the segment from \(P\) to \(\ell_i\) drawn at angle \(\alpha_i\). Find the locus of \(P\) with \(d_1d_2\cdots d_m=k\,d_{m+1}\cdots d_{2m}\) (for an odd number, multiply the shorter side by a given segment).

Each \(d_i\) is a constant multiple of the perpendicular distance, an affine-linear function of \((x,y)\). So for three or four lines the condition is a polynomial equation of degree 2, and the locus is a conic. Pappus reports that Apollonius treated the three- and four-line loci incompletely and that for more than four lines the loci were unknown. For more than six lines he noted that a product of more than three lengths has no geometric meaning, and recast the condition as a compound of ratios. Descartes’s Géométrie (1637) solves the problem for any number of lines: with \(2m\) lines the locus is a curve of degree at most \(m\).

Exercise 9.1With focus at the origin and directrix \(x=p\), write the equation of the locus and classify it by \(e\).

Show solution

\(x^2+y^2=e^2(x-p)^2\), so \((1-e^2)x^2+2e^2px+y^2=e^2p^2\). If \(e=1\) the \(x^2\) term vanishes: \(y^2=p^2-2px\), a parabola. If \(e<1\) both squares have positive coefficients: an ellipse. If \(e>1\) they have opposite signs: a hyperbola.

Exercise 9.2Three-line locus: lines \(x=0\), \(y=0\), \(x+y=1\), perpendicular distances, condition \(d_1d_2=k\,d_3^2\). Find the equation and identify the conic for \(k=\tfrac12\).

Show solution

\(d_3=|x+y-1|/\sqrt2\), so \(|xy|=\tfrac k2(x+y-1)^2\). Taking the branch \(xy\ge0\) with \(k=\tfrac12\): \(4xy=(x+y-1)^2\), i.e. \(x^2-2xy+y^2-2x-2y+1=0\). The quadratic part \((x-y)^2\) is degenerate, so the conic is a parabola. It is tangent to both axes, at \((1,0)\) and \((0,1)\).

Lesson 10

Centroids and machines

For the general reader. Spin a flat shape around an axis and it sweeps out a solid. Pappus stated that the volume equals the shape’s area times the distance its balance point travels. The rule, stated without proof in Book VII, was rediscovered in the seventeenth century. Book VIII then gathers Greek mechanics, from the centre of gravity to geared lifting machines.

Goals

State and prove the centroid theorems with calculus.
Use them to find volumes and centroids.
Know the contents of Book VIII.

Theorem (Pappus, Book VII; Guldin, 1640)

Let a plane region \(F\) of area \(A\) lie on one side of a line \(\ell\) in its plane, with centroid at distance \(\bar r\) from \(\ell\). Rotating \(F\) about \(\ell\) through a full turn produces a solid of volume \(V=2\pi\bar r\,A\). Likewise a plane curve of length \(s\) whose centroid is at distance \(\bar r\) sweeps a surface of area \(S=2\pi\bar r\,s\).

Proof. Take \(\ell\) as an axis and let \(r\) be distance from it. A small element \(dA\) at distance \(r\) sweeps a ring of volume \(2\pi r\,dA\). So \(V=\int_F2\pi r\,dA=2\pi\int_Fr\,dA=2\pi\bar rA\), by the definition \(\bar r=\frac1A\int_Fr\,dA\). The curve case is the same with \(ds\) for \(dA\).

Pappus states the volume rule in general terms in the preface to Book VII and says it contains many theorems of every kind. No proof survives. Johannes Kepler used related ideas in 1615, and Guldin published the rule in 1640; whether Guldin knew Pappus’s statement is debated.

Book VIII

Book VIII defines the centre of gravity, the only surviving Greek definition of it, and treats the problem of moving a weight up an inclined plane. Pappus’s answer to that problem is wrong by later standards, since he assumed a force is needed even to move the weight on the level. The book then covers Heron’s five simple machines (wheel and axle, lever, pulley, wedge, screw) and gear trains for lifting a given weight with a given force. It also contains a construction of an ellipse through five given points (Prop. 14) and of the axes of an ellipse from two conjugate diameters (Prop. 15).

Exercise 10.1Find the volume and surface area of a torus made by rotating a disc of radius \(r\) about a line at distance \(R>r\) from its centre.

Show solution

The centroid of the disc and of its boundary circle is the centre, at distance \(R\). Volume \(=2\pi R\cdot\pi r^2=2\pi^2Rr^2\). Surface \(=2\pi R\cdot2\pi r=4\pi^2Rr\).

Exercise 10.2Use the theorem backwards to find the centroid of a half-disc of radius \(r\), and of a semicircular arc.

Show solution

Rotating the half-disc about its diameter gives a ball: \(\tfrac43\pi r^3=2\pi\bar y\cdot\tfrac12\pi r^2\), so \(\bar y=\frac{4r}{3\pi}\). Rotating the arc gives a sphere: \(4\pi r^2=2\pi\bar y\cdot\pi r\), so \(\bar y=\frac{2r}{\pi}\).

Part III · Life and legacy

Lesson 11

From one manuscript to Descartes

Pappus had readers in late antiquity. Theon of Alexandria used his Almagest commentary, and Proclus and Eutocius cited him on Euclid and on the mean proportionals. The Arabic world knew his commentary on Elements X and parts of Book VIII. The Latin Middle Ages knew almost nothing of him. The Collection survived through one Byzantine copy, now in the Vatican.

Commandino and the Renaissance

Federico Commandino of Urbino, translator of Archimedes, Apollonius and Euclid, made a Latin version of the Collection. He died in 1575, and the translation was printed at Pesaro in 1588. It reached readers at the moment when European mathematicians were trying to recover Greek methods of discovery. Pappus’s Book VII was their main witness to what had been lost.

Viète drew on Pappus’s account of analysis in his Isagoge (1591) and reconstructed Apollonius’s Tangencies in Apollonius Gallus (1600).
Fermat reconstructed Apollonius’s Plane Loci from Pappus’s summary, and his work on Determinate Section fed his method of maxima and minima.
Descartes received Pappus’s locus problem from Jacob Golius in 1631. Its solution became the centrepiece of La Géométrie (1637), where he presented it as proof that his method surpassed the ancients.
Newton answered with a purely geometric treatment of the four-line locus in the Principia (1687), partly to show that the ancients’ methods were adequate.
Edmond Halley translated the Arabic Cutting off of a Ratio into Latin (1706) and restored Cutting off of an Area using Pappus’s description.

Robert Simson, Michel Chasles and others later tried to reconstruct Euclid’s Porisms from Pappus’s lemmas. Chasles recognised in them the cross-ratio and involution, ideas then being developed by Jean-Victor Poncelet and Jakob Steiner.

DiscussionDescartes chose a problem from Pappus to show the power of algebra over Greek geometry; Newton answered with geometry alone. What does each solution make easy, and what does it hide?

Lesson 12

Pappus in modern mathematics

The scholarly recovery of Pappus continued with Hultsch’s critical edition (1876–78), Ver Eecke’s French translation (1933), Jones’s edition of Book VII (1986) and Heike Sefrin-Weis’s translation of Book IV (2010). Each made clearer that Book VII is more than a reader’s guide. Its lemmas include harmonic division, the complete quadrangle and a version of pole and polar.

The hexagon theorem and the foundations of geometry

Theorem (Hilbert 1899; Hessenberg 1905)

Let \(\Pi\) be the projective plane over a division ring \(D\). Pappus’s theorem holds in \(\Pi\) if and only if \(D\) is commutative. Moreover, in any projective plane Pappus’s theorem implies Desargues’s theorem.

Hilbert’s Grundlagen der Geometrie used Pappus’s theorem, which he called Pascal’s theorem, to define multiplication of segments and prove it commutative without numbers. Gerhard Hessenberg showed in 1905 that Pappus implies Desargues; his proof had a gap, closed in 1953. The quaternion plane is the standard example in which Desargues holds and Pappus fails. Pascal’s theorem on a hexagon inscribed in a conic contains Pappus’s as the case of a conic degenerated into two lines.

Other descendants

The centroid theorems are taught in every calculus course.
The bee problem became the honeycomb conjecture, proved by Hales in 1999.
George Pólya’s How to Solve It (1945) has an entry on Pappus and presents working backward as a heuristic in his tradition.
The Pappus graph, the incidence graph of the \((9_3)\) configuration, is a standard example in algebraic graph theory.

Exercise 12.1In the quaternions \(\mathbb H\), find \(a,b\) with \(ab\ne ba\), and point to the step of the Lesson 08 proof that needs commutativity.

Show solution

\(ij=k\) but \(ji=-k\). In Lesson 08 the parallel condition is a proportion \(a/b=b'/a'\), which over a division ring must be written with one-sided inverses, say \(b^{-1}a=b'a'^{-1}\). Passing to \(a\,a'=b\,b'\) and then chaining two such equalities requires moving factors past one another. With \(ab\ne ba\) the two given parallelisms no longer force the third, and one can choose coordinates in \(\mathbb H\) for which the three crossing points are not collinear.

Final projectChoose one lost work from the Treasury of Analysis. Using Pappus’s summary (Jones’s translation of Book VII), write a two-page account of what it contained, which of Pappus’s lemmas belong to it, and how a later mathematician tried to reconstruct it.

Reference

Sources and further reading

Biography and context

J. J. O’Connor and E. F. Robertson, Pappus of Alexandria, MacTutor History of Mathematics.
Pappus of Alexandria, Wikipedia.
I. Bulmer-Thomas, Pappus of Alexandria, Dictionary of Scientific Biography.
A. Rome, Commentaires de Pappus et de Théon d’Alexandrie sur l’Almageste, vol. 1, Vatican, 1931.
Pandrosion: man or woman?, MacTutor; and Pandrosion, Wikipedia.
S. Cuomo, Pappus of Alexandria and the Mathematics of Late Antiquity, Cambridge University Press, 2000.
T. L. Heath, A History of Greek Mathematics, vol. 2, Oxford, 1921, chapter on Pappus.

Texts and translations

F. Hultsch (ed.), Pappi Alexandrini Collectionis quae supersunt, 3 vols., Berlin, 1876–78.
F. Commandino, Pappi Alexandrini Mathematicae Collectiones, Pesaro, 1588.
P. Ver Eecke, Pappus d’Alexandrie: La Collection mathématique, 2 vols., Paris and Bruges, 1933.
A. Jones, Pappus of Alexandria: Book 7 of the Collection, 2 vols., Springer, 1986.
H. Sefrin-Weis, Pappus of Alexandria: Book 4 of the Collection, Springer, 2010.
Pappus on the wisdom of bees, MacTutor extract.

Mathematics

Pappus’s hexagon theorem and Pappus’s centroid theorem, Wikipedia.
D. Hilbert, Grundlagen der Geometrie, Teubner, 1899.
H. S. M. Coxeter, Projective Geometry, 2nd ed., Springer, 1987.
T. C. Hales, The honeycomb conjecture, arXiv:math/9906042; Discrete Comput. Geom. 25 (2001).
H. J. M. Bos, Redefining Geometrical Exactness: Descartes’ Transformation of the Early Modern Concept of Construction, Springer, 2001.
G. Pólya, How to Solve It, Princeton, 1945.