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Girard Desargues

Girard Desargues, “le S.G.D.L.” (sieur Girard Desargues, Lyonnais)

A course in two strands: the life of a Lyon engineer and architect who wrote for painters and stonecutters, and the geometry of projection, involution and infinity that his contemporaries ignored and the nineteenth century rediscovered.

Born 21 February 1591, Lyon Died 1661, Lyon Signature work Brouillon project on conics, 1639 Level advanced undergraduate → graduate
O ABC A′B′C′ XYZ centre of perspective axis of perspective
Desargues’s theorem. The lines AA′, BB′, CC′ meet at O, so the side pairs AB, A′B′ and BC, B′C′ and CA, C′A′ meet at three points X, Y, Z on one line.
Lesson 00

How to use this course

Girard Desargues published a handful of short pamphlets, some only a few pages long, in Paris between 1636 and 1643. Most were addressed to craftsmen: painters, engravers, stonecutters, sundial makers. One of them, a 30-page draft on conic sections printed in 1639, contained the core of projective geometry two centuries before Poncelet. It was nearly lost. This course tells how a gentleman engineer from Lyon came to write it, explains what it contains in modern language, and follows the theorem that bears his name into the foundations of geometry.

Prerequisites for Part II
  • Euclidean plane geometry: similar triangles, the theorems of Menelaus and Ceva, power of a point.
  • Analytic geometry of conics and some linear algebra.
  • For Lesson 10, the definition of a field and of a division ring (skew field).

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proofs or proof sketches, and exercises with folded solutions. Desargues wrote in an invented vocabulary of trees, trunks and branches. The lessons translate it into modern terms and show the original word where it helps.

On sources: few documents about Desargues’s life survive. Where a story rests only on a later biographer, the lesson says so.

Reference

Timeline

1591Born 21 February in Lyon; baptised 2 March at Sainte-Croix. Father Girard Desargues, royal notary and tithe collector; mother Jeanne Croppet.
1626In Paris; on 9 September proposes, with François Villette, a machine to raise water from the Seine for the city.
c. 1630Settles in Paris; joins the circle of Marin Mersenne.
1636Method of teaching music printed in Mersenne’s Harmonie universelle. First treatise: a universal method of perspective.
1639Brouillon project d’une atteinte aux evenemens des rencontres du cone avec un plan: the projective theory of conics.
1640Pamphlets on stonecutting and on sundials. Blaise Pascal’s Essay pour les coniques credits Desargues. Jean de Beaugrand attacks the Brouillon project.
1642Six erreurs, against a pirated perspective book. Quarrel with Jean Du Breuil.
1643Abraham Bosse publishes Desargues’s methods for stonecutting and sundials.
1644Jacques Curabelle’s attack and a pamphlet war.
c. 1645Turns to architecture.
1648Bosse’s Manière universelle de M. Desargues pour pratiquer la perspective prints the theorem on triangles in perspective. Desargues returns to Lyon.
1649–50In Paris again.
1657–60In Paris; designs private houses.
1661Dies in Lyon; his will is opened on 8 October.
1822Poncelet’s Traité draws attention to Desargues as a forerunner.
1845A manuscript copy of the Brouillon project by Philippe de La Hire comes to light.
1864N.-G. Poudra’s two-volume Œuvres de Desargues.
1951A printed copy of the 1639 Brouillon project is found. René Taton’s L’œuvre mathématique de G. Desargues.
Part I · Formation
Lesson 01

A Lyon family, 1591–1630

Girard Desargues was born in Lyon on 21 February 1591 and baptised on 2 March in the parish of Sainte-Croix. A horoscope found by the historian René Taton fixes the date; for a long time biographers had placed his birth in 1593, following a remark in Adrien Baillet’s 1691 life of Descartes. His father, also called Girard Desargues, was a royal notary and an investigating officer of the seneschal’s court of Lyon, and later collected ecclesiastical tithes for the diocese. His mother was Jeanne Croppet. The family was wealthy and had served the crown for generations as lawyers and magistrates in Lyon and Paris. They owned several houses in the city and property at Vourles, to the south.

Nothing is documented about his schooling. When he first appears in the records, in his mid-thirties, he is a man of independent means with practical interests: machines, pumps, staircases, and the geometry behind them. On 9 September 1626 he and François Villette proposed to the city of Paris a machine to raise water from the Seine. The city replied with conditions a week later; nothing more is known, and the project was probably dropped.

Baillet also claimed that Desargues served as an engineer at the siege of La Rochelle in 1628 and met Descartes there. No contemporary document supports this, and modern historians treat it as unproven.

Lesson 02

Paris and the Mersenne circle, 1630–1638

By about 1630 Desargues lived in Paris. He joined the informal academy that met around the Minim friar Marin Mersenne, who corresponded with every natural philosopher in Europe. The group included René Descartes (by letter, from the Netherlands), Pierre Gassendi, Gilles Personne de Roberval, Étienne Pascal and, later, Étienne’s son Blaise. Pierre de Fermat and Descartes both held his opinions in regard. By 1635 Mersenne counted him as a regular member.

His first known publication was a short method for learning and teaching music, which Mersenne printed in his Harmonie universelle (1636). The same year Desargues printed a twelve-page pamphlet on perspective, signed only with initials: Exemple de l’une des manières universelles du S.G.D.L. touchant la pratique de la perspective sans emploier aucun tiers point, de distance ny d’autre nature, qui soit hors du champ de l’ouvrage. The title states the selling point: a method for drawing in perspective that never needs a construction point outside the picture.

This was the pattern of his career. He wrote brief, dense broadsheets, often printed in small numbers and handed to friends, offering a single “universal manner” to replace the many rules of thumb used by a trade. His style was compressed, his vocabulary new, and his readers were few.

Lesson 03

The Brouillon project, 1639

In 1639 Desargues printed his most important work: Brouillon project d’une atteinte aux evenemens des rencontres du cone avec un plan, a “rough draft of an attempt to deal with the outcomes of the meeting of a cone with a plane”. It ran to about thirty pages.

The classical theory of conics, from Apollonius of Perga, treated the ellipse, parabola and hyperbola separately and proved their properties with long chains of proportions. Desargues treated all conics as plane sections of a cone over a circle. A property of the circle that survives projection from the vertex then holds for every conic at once. To make this work he had to treat parallel lines as lines meeting at a point infinitely far away, and he had to find properties that survive projection. He found one, the involution of six points (Lesson 06), and built the theory on it.

He wrote in a vocabulary drawn from botany. A line carrying marked points is a tronc (trunk); the points are nœuds (knots); segments from them are rameaux (branches); a configuration of a trunk with three pairs of knots is an arbre (tree); a family of concurrent or parallel lines is an ordonnance. Only a few of his words survived. Involution is one of them.

Readers

Descartes read the work and responded, but he thought algebra a better route to conics. Jean de Beaugrand attacked it in a letter of August 1640. The one reader who took it up fully was sixteen-year-old Blaise Pascal, who in 1640 printed a one-page Essay pour les coniques containing the hexagon theorem and acknowledged Desargues as the source of a key idea. Pascal’s fuller treatise on conics was later lost.

Part II · The mathematics
Lesson 04

Perspective, sundials and stonecutting

For the general reader. A painter who wants a tiled floor to look right must make the lines of tiles converge toward a vanishing point, and must space the rows correctly as they recede. Renaissance artists did this with special construction points, often outside the canvas. Desargues found a single rule that worked inside the picture. The geometry behind it is central projection.

Goals
  • Model perspective as central projection and compute vanishing points.
  • See why parallel lines in space meet in the picture.
  • Compute the hour lines of a sundial, one of Desargues’s practical subjects.

Central projection

Place the eye at the origin and the picture on the plane \(z=f\). A point \((X,Y,Z)\) with \(Z>0\) is drawn at

\[ (x,y)=\Big(\frac{fX}{Z},\ \frac{fY}{Z}\Big). \]
Proposition (vanishing points)

Let a family of parallel lines in space have direction \(\mathbf d=(d_1,d_2,d_3)\) with \(d_3\neq0\). The images of all these lines pass through the single point \(V=(fd_1/d_3,\ fd_2/d_3)\), the image of the line through the eye parallel to \(\mathbf d\). If \(d_3=0\), the images are parallel lines in the picture. The vanishing points of all horizontal directions lie on one line, the horizon.

Proof. A point of the line \(\mathbf P+t\mathbf d\) projects to \(\big(f\frac{P_1+td_1}{P_3+td_3},\ f\frac{P_2+td_2}{P_3+td_3}\big)\), which tends to \(V\) as \(t\to\infty\). The image of a line is a line, so the image line passes through \(V\). For horizontal directions (\(d_2=0\), with \(y\) vertical), \(V\) has \(y=0\): all such vanishing points lie on the line \(y=0\) at eye height.

This is where points at infinity come from. In space, parallel lines never meet. In the picture, they meet at \(V\). Desargues took the step of saying that they meet in space too, at a point infinitely far away, which the projection makes visible.

Practical geometry

In 1640 Desargues issued two more pamphlets. One gave a general method of trait for stonecutting, the drawing of templates for cutting the stones of vaults and arches; the other gave a method for placing the style of a sundial and drawing its hour lines. The originals are almost entirely lost. Their content is known from the books of his disciple, the engraver Abraham Bosse, published in 1643.

Exercise 4.1The eye is at height \(h\) above a flat floor, the picture plane is at distance \(f\), and the floor has square tiles of side 1 whose rows lie at distances \(Z=f,f+1,f+2,\dots\) from the eye. Where are the rows drawn? Show that they crowd toward the horizon.

Show solution
With \(y\) vertical and the eye at height \(h\), the floor is \(Y=-h\). A row at depth \(Z\) is drawn at \(y=-fh/Z\). So the rows are at \(-h,\ -\frac{fh}{f+1},\ -\frac{fh}{f+2},\dots\), tending to \(0\), the horizon. The gaps \(\frac{fh}{Z}-\frac{fh}{Z+1}=\frac{fh}{Z(Z+1)}\) shrink like \(1/Z^2\).

Exercise 4.2In Exercise 4.1, the diagonals of the tiles are horizontal lines with direction \((1,0,1)\). Find their vanishing point, and explain how a painter can use it to space the rows without a point outside the picture when \(f\) is small compared with the picture width.

Show solution
\(V=(f\cdot1/1,\ 0)=(f,0)\), on the horizon at distance \(f\) from the centre. This is the classical “distance point”. Drawing one diagonal toward it cuts each receding line of tiles at the successive rows. If \(f\) exceeds the half-width of the picture, the distance point lies outside the canvas. Desargues’s method replaced it by scales drawn inside the picture that encode the same ratios.

Exercise 4.3On a horizontal sundial at latitude \(\varphi\), the style is parallel to the Earth’s axis, and the shadow line for hour angle \(H\) (15° per hour from noon) makes an angle \(\theta\) with the noon line, where \(\tan\theta=\sin\varphi\tan H\). Compute \(\theta\) for 3 p.m. at Lyon (\(\varphi\approx45.8^\circ\)) and at Paris (\(\varphi\approx48.9^\circ\)).

Show solution
\(H=45^\circ\), so \(\tan\theta=\sin\varphi\). Lyon: \(\sin45.8^\circ\approx0.717\), \(\theta\approx35.6^\circ\). Paris: \(\sin48.9^\circ\approx0.753\), \(\theta\approx37.0^\circ\). The hour lines are the central projection, from the style, of 24 equally spaced half-planes through the Earth’s axis onto the dial plane. A dial drawn for Lyon is slightly wrong in Paris, which is why a method valid for any latitude and any wall orientation was worth selling.
Lesson 05

Ordonnances and points at infinity

For the general reader. A bundle of lines through one point and a bundle of parallel lines behave alike in a picture. Desargues gave them one name, ordonnance, and said the parallel bundle also has a meeting point, at infinite distance. With this convention, any two lines in a plane meet, and the exceptions that cluttered Greek geometry disappear.

Goals
  • Construct the projective plane by adding points at infinity.
  • Work with homogeneous coordinates.
  • See how a cone projects one conic onto another.

Desargues defined an ordonnance de droites as a family of lines in a plane all passing through one point, its but (goal). Parallel lines form an ordonnance whose goal is at infinite distance. In the same way the planes through a line, or parallel to it, form an ordonnance of planes. He also said that a straight line can be thought of as a circle of infinite radius.

Definition (projective plane)

Add to the Euclidean plane one new point for each direction of lines, the common point of all lines with that direction, and one new line containing exactly these points. In the resulting projective plane, any two distinct points lie on exactly one line, and any two distinct lines meet in exactly one point. In homogeneous coordinates, points are triples \((x:y:z)\neq(0:0:0)\) up to scaling, ordinary points are \((x:y:1)\), and the points at infinity form the line \(z=0\).

The two lines \(ax+by+cz=0\) and \(a'x+b'y+c'z=0\) meet at the point given by the cross product \((a,b,c)\times(a',b',c')\). For parallel lines, \((a,b)\) and \((a',b')\) are proportional, and the cross product has third coordinate \(ab'-a'b=0\): they meet at infinity, as Desargues said.

Conics as shadows of a circle

Let a cone have vertex \(S\) and base a circle. A plane not through \(S\) cuts it in a conic. Projection from \(S\) maps the circle to the conic, sends lines to lines, and preserves incidence. The section is an ellipse, a parabola or a hyperbola according as the conic meets the line at infinity of its plane in zero points, one point (tangentially) or two points. For Desargues the asymptotes of a hyperbola are its tangents at its two points at infinity.

Exercise 5.1Find the intersection of the lines \(x+2y=3\) and \(2x+4y=1\) in homogeneous coordinates.

Show solution
The lines are \((1,2,-3)\) and \((2,4,-1)\). Cross product: \((2\cdot(-1)-(-3)\cdot4,\ (-3)\cdot2-1\cdot(-1),\ 1\cdot4-2\cdot2)=(10,-5,0)\sim(2:-1:0)\). The point at infinity in the direction \((2,-1)\), the common direction of both lines.

Exercise 5.2Find the points at infinity of the hyperbola \(x^2-y^2=1\), and check that the tangents there are the asymptotes.

Show solution
Homogenize: \(x^2-y^2=z^2\). At \(z=0\): \(x=\pm y\), giving \((1:1:0)\) and \((1:-1:0)\). The tangent to \(F=x^2-y^2-z^2\) at \(P\) is \(\nabla F(P)\cdot(x,y,z)=0\). At \((1:1:0)\), \(\nabla F=(2,-2,0)\), so the tangent is \(x-y=0\). At \((1:-1:0)\) it is \(x+y=0\). These are the asymptotes \(y=\pm x\).
Lesson 06

Involution of six points

For the general reader. Distances change under projection, so most of Euclid’s measurements are useless for a theory of perspective. Desargues found a relation between three pairs of points on a line, expressed with products of distances, that projection does not destroy. He called it an involution, and it became the foundation of his treatment of conics.

Goals
  • State Desargues’s definition of an involution and its modern equivalent.
  • Prove that involutions are preserved by projection.
  • Classify involutions by their fixed points.

Desargues’s definition

Three pairs of points \(\{B,H\},\{C,G\},\{D,F\}\) on a line are in involution if there is a point \(A\) on the line (Desargues’s souche, the stump) such that, in signed lengths,

\[ AB\cdot AH=AC\cdot AG=AD\cdot AF. \]

If the common product is positive, there are two points \(E\) with \(AE^2\) equal to it; Desargues called them the nœuds moyens, the middle knots. The stump may be at infinity, in which case the three pairs share a common midpoint.

Theorem (modern form)

Three pairs of points are in involution in Desargues’s sense if and only if there is a projective involution of the line (a projective map \(\sigma\neq\mathrm{id}\) with \(\sigma^2=\mathrm{id}\)) exchanging the points of each pair. If the stump is at the origin and the common product is \(k\), this involution is \(x\mapsto k/x\). The middle knots are its fixed points, and each pair is harmonic with respect to them. An involution is determined by two of its pairs.

Proof. If \(A=0\), the relation says that each pair has the form \(\{x,k/x\}\), and \(x\mapsto k/x\) is projective with square the identity. Conversely, a projective involution \(x\mapsto\frac{px+q}{rx+s}\) satisfies \(\sigma^2=\mathrm{id}\) iff \(s=-p\) (excluding the identity). If \(r\neq0\), translate so that \(p=0\); then \(\sigma(x)=q/(rx)\), which has the required form with \(A\) the point sent to infinity. If \(r=0\), \(\sigma(x)=-x+\text{const}\), a reflection, and the stump is at infinity. Two pairs \(\{x_1,y_1\},\{x_2,y_2\}\) give two linear equations for the coefficients of the symmetric relation \(r\,xy-p(x+y)-q=0\), which determine it up to scale.

Theorem (Desargues, 1639)

If three pairs of points on a line are in involution, their projections from any point onto any other line are in involution.

Desargues proved this by a long argument with Menelaus’s theorem. With the modern form it is immediate: a projection between lines is a projective map \(\pi\), and \(\pi\sigma\pi^{-1}\) is a projective involution exchanging the projected pairs.

Exercise 6.1Are the pairs \(\{1,8\},\{2,4\},\{-1,-8\}\) in involution? Find the stump, the common product and the middle knots.

Show solution
With stump \(A=0\): \(1\cdot8=2\cdot4=(-1)(-8)=8\). So they are in involution with \(k=8\). The middle knots are \(\pm\sqrt8=\pm2\sqrt2\). Check that \(\{2,4\}\) is harmonic with respect to them, i.e. \((2\sqrt2,-2\sqrt2;2,4)=\frac{(2-2\sqrt2)(4+2\sqrt2)}{(2+2\sqrt2)(4-2\sqrt2)}=-1\). Indeed \((2-2\sqrt2)(4+2\sqrt2)=8+4\sqrt2-8\sqrt2-8=-4\sqrt2\) and \((2+2\sqrt2)(4-2\sqrt2)=8-4\sqrt2+8\sqrt2-8=4\sqrt2\). The ratio is \(-1\).

Exercise 6.2Find the involution that exchanges \(0\leftrightarrow3\) and \(1\leftrightarrow2\). Where is its stump? Does it have real fixed points?

Show solution
Both pairs have midpoint \(\tfrac32\), so the involution is the reflection \(x\mapsto3-x\). Its stump is at infinity. Its fixed points are \(\tfrac32\) and the point at infinity, and each pair is harmonic with respect to them, which is the statement that \(\tfrac32\) is the midpoint of each pair.

Exercise 6.3Show that an involution with negative common product \(k\) has no real fixed points, and that its pairs then separate each other.

Show solution
Fixed points satisfy \(x^2=k<0\): none are real. If \(k<0\), each pair \(\{x,k/x\}\) has one positive and one negative member, so every pair contains the stump \(0\) between its points. For two pairs \(\{a,k/a\},\{b,k/b\}\) with \(0<a<b\): \(k/b>k/a\) (both negative, \(|k/b|<|k/a|\)), so the order on the line is \(k/a<k/b<0<a<b\). Going round the projective line, the pairs alternate: they separate each other. Desargues treated both cases, with and without middle knots.
Lesson 07

The involution theorem for conics

For the general reader. Pick four points and draw every conic through them. A straight line crosses each of these conics twice. Desargues proved that all these pairs of crossing points, from infinitely many curves, belong to one involution. This was the central theorem of the 1639 draft, and it lets one transfer properties from a circle to any conic.

Goals
  • State and prove Desargues’s involution theorem.
  • Recover the power of a point for a circle as a special case.
  • Connect it to Pascal’s hexagon theorem.
Theorem (Desargues’s involution theorem)

Let \(P_1,P_2,P_3,P_4\) be four points in the plane, no three collinear, and let \(\ell\) be a line through none of them. Each conic through the four points meets \(\ell\) in a pair of points (possibly complex or coincident). All these pairs, together with the three pairs in which \(\ell\) meets the opposite sides \(P_1P_2,P_3P_4\); \(P_1P_3,P_2P_4\); \(P_1P_4,P_2P_3\) of the quadrangle, are pairs of one involution of \(\ell\).

Proof. Conics through the four points form a pencil \(\lambda F+\mu G=0\), where \(F\) and \(G\) are two of them; the three pairs of opposite sides are degenerate members. Parametrize \(\ell\) by \(t\). Restricted to \(\ell\), \(F\) and \(G\) become quadratics \(f(t)=a_0+a_1t+a_2t^2\) and \(g(t)=b_0+b_1t+b_2t^2\). The pair cut by the member \((\lambda:\mu)\) is the root pair of \(\lambda f+\mu g\). A pair \(\{t,t'\}\) is the root pair of \(c_0+c_1t+c_2t^2\) iff \(c_2tt'=c_0\) and \(c_2(t+t')=-c_1\). Root pairs of a two-dimensional linear family of quadratics are exactly the pairs of a fixed symmetric relation \(\alpha\,tt'+\beta(t+t')+\gamma=0\), namely the one orthogonal to both coefficient vectors. Such a relation defines an involution (or, in a degenerate case, all pairs share a point, which cannot happen here because \(\ell\) avoids the base points).

Desargues proved the theorem for the circle first, where it follows from the power of a point, and then projected. That is the method of the whole Brouillon project.

Pascal’s hexagon

Pascal’s 1640 Essay pour les coniques stated that for six points on a conic, the three intersection points of opposite sides of the hexagon lie on a line. Pascal proved it for the circle and extended it by projection, the method he learned from Desargues. Pascal named Desargues as the inventor of one of the key properties he used.

Exercise 7.1For a circle and a line \(\ell\) through a point \(A\), let \(\ell\) meet the circle at \(X,X'\). The power of a point says \(AX\cdot AX'\) is the same for all circles of a pencil through two fixed points \(P,Q\) whenever \(A\) lies on the line \(PQ\) (and \(\ell\) is any line through \(A\)). Show that this is a special case of the involution theorem.

Show solution
Every circle of the pencil passes through \(P\) and \(Q\) and through the circular points \(I,J\) at infinity, so these circles are conics through four fixed points. The line \(PQ\) is a side of the quadrangle \(PQIJ\), and its opposite side is the line at infinity \(IJ\). So the involution induced on \(\ell\) pairs \(A=\ell\cap PQ\) with \(\ell\cap IJ\), the point at infinity of \(\ell\). An involution that pairs \(A\) with \(\infty\) has \(A\) as stump. Hence \(AX\cdot AX'\) is constant over the pencil, equal to \(AP\cdot AQ\).

Exercise 7.2Take the four points \((\pm1,\pm1)\) and the line \(y=0\). Find the pairs cut on the line by the conics \(x^2-1=0\), \(x^2+y^2-2=0\), \(x^2-y^2=0\) and \(x^2+3y^2-4=0\), all of which pass through the four points, and identify the involution.

Show solution
On \(y=0\): \(x^2-1=0\) gives \(\{1,-1\}\); \(x^2-2=0\) gives \(\{\sqrt2,-\sqrt2\}\); \(x^2=0\) gives \(\{0,0\}\); \(x^2-4=0\) gives \(\{2,-2\}\). Every pair has the form \(\{x,-x\}\): the involution is \(x\mapsto-x\), with stump at infinity and fixed points \(0\) and \(\infty\). The fixed point \(0\) appears as the double pair cut by the degenerate member \(x^2-y^2=0\), the two diagonals of the square. The member \(y^2-1=0\) (the sides \(y=\pm1\)) is parallel to the line and cuts it in the fixed point \(\infty\) taken twice.
Lesson 08

The traversale: poles and polars

For the general reader. Apollonius knew that the midpoints of parallel chords of a conic lie on a line, a diameter. Desargues asked what happens when the chords are not parallel but pass through a common point. The answer is again a line, which he called the traversale. Today it is the polar of the point, one of the basic objects of projective geometry.

Goals
  • Define the traversale (polar) of a point by harmonic division.
  • Recover Apollonius’s diameters as polars of points at infinity.
  • Compute polars in coordinates.
Theorem (Desargues, 1639; pole and polar)

Let \(\Gamma\) be a non-degenerate conic and \(P\) a point not on \(\Gamma\). For each line through \(P\) meeting \(\Gamma\) at \(U,V\), let \(W\) be the harmonic conjugate of \(P\) with respect to \(U,V\). Then all such points \(W\) lie on one line \(p\), the polar of \(P\). If \(\Gamma\) is given by \(X^{\mathsf T}MX=0\), then \(p\) is the line \(P^{\mathsf T}MX=0\).

Proof. Points of the line are \(P+tQ\). They lie on \(\Gamma\) when \(P^{\mathsf T}MP+2t\,P^{\mathsf T}MQ+t^2\,Q^{\mathsf T}MQ=0\). If \(Q\) satisfies \(P^{\mathsf T}MQ=0\), the roots are \(\pm t_0\), and the points with parameters \(\pm t_0\) are harmonic with respect to the parameters \(0\) (the point \(P\)) and \(\infty\) (the point \(Q\)). So \(W=Q\) lies on the line \(P^{\mathsf T}MX=0\). The harmonic conjugate is unique, so this line is the full locus.

Desargues proved the theorem for a circle and transferred it by projection. When \(P\) is a point at infinity, the chords through \(P\) are parallel; the harmonic conjugate of the point at infinity with respect to \(U,V\) is the midpoint of \(UV\). So the polar of a point at infinity is a diameter in Apollonius’s sense. Desargues called it a diamétrale. The polar of the centre is the line at infinity. The terms “pole” and “polar” came later, from François-Joseph Servois and Joseph Gergonne in the 1810s.

Exercise 8.1Find the polar of \(P=(3,1)\) with respect to the ellipse \(\frac{x^2}{4}+y^2=1\).

Show solution
For \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) the polar of \((x_0,y_0)\) is \(\frac{x_0x}{a^2}+\frac{y_0y}{b^2}=1\). Here \(\frac{3x}{4}+y=1\). It meets the ellipse where \(\frac{x^2}4+\big(1-\frac{3x}4\big)^2=1\), i.e. \(\frac{13x^2}{16}-\frac{3x}2=0\): \(x=0\) or \(x=\frac{24}{13}\). The points \((0,1)\) and \(\big(\frac{24}{13},-\frac5{13}\big)\) are where the tangents from \(P\) touch.

Exercise 8.2Use the polar of a point at infinity to find the locus of midpoints of chords of the parabola \(y=x^2\) with slope \(m\).

Show solution
Homogeneous form: \(yz-x^2=0\), matrix \(M\) with \(X^{\mathsf T}MX=yz-x^2\). The point at infinity of slope \(m\) is \(P=(1:m:0)\). Its polar is \(P^{\mathsf T}MX=-x+\tfrac{m}{2}z=0\), i.e. the vertical line \(x=m/2\). Check: chords \(y=mx+c\) meet the parabola where \(x^2-mx-c=0\), whose roots have mean \(m/2\).
Lesson 09

Desargues’s theorem on triangles

For the general reader. Two triangles seen from a single point, as when one is the shadow of the other, are said to be in perspective from that point. Desargues showed that their corresponding sides, extended, then meet in three points that lie on a single line. The statement uses only points, lines and incidence, and it is the most famous theorem bearing his name.

Goals
  • State Desargues’s theorem and its converse.
  • Prove it in space by a two-plane argument, and in the plane by Menelaus.
  • Study the Desargues configuration.
Theorem (Desargues; printed by Bosse, 1648)

Let \(ABC\) and \(A'B'C'\) be triangles in the projective plane (or projective space) with distinct vertices and sides. If the lines \(AA',BB',CC'\) pass through a common point \(O\), then the points \(X=AB\cap A'B'\), \(Y=BC\cap B'C'\), \(Z=CA\cap C'A'\) are collinear. Conversely, if \(X,Y,Z\) are collinear, then \(AA',BB',CC'\) are concurrent.

The theorem was first printed in Bosse’s 1648 book on Desargues’s perspective method, among several propositions stated there as Desargues’s.

Proof in space

Suppose first that the triangles lie in distinct planes \(\pi\) and \(\pi'\). The lines \(AB\) and \(A'B'\) both lie in the plane through \(O,A,B\) (which contains \(A'\) and \(B'\)), so they meet at a point \(X\). This point lies on \(AB\subset\pi\) and on \(A'B'\subset\pi'\), hence on the line \(\pi\cap\pi'\). The same holds for \(Y\) and \(Z\). So all three lie on the line \(\pi\cap\pi'\).

Proof in the plane

For two triangles in the same plane, lift: choose a point \(S\) outside the plane and a point \(B''\) on the line \(SB\), different from \(S\) and \(B\). Since \(O,B,B'\) are collinear, the lines \(OB''\) and \(SB'\) lie in one plane and meet at a point \(B^*\). Then \(ABC\) and \(A'B^*C'\) are in perspective from \(O\) and lie in different planes, so by the space proof their side pairs meet on a line. Projecting from \(S\) back to the original plane sends \(B^*\) to \(B'\) and that line to the line through \(X,Y,Z\). A purely planar proof uses Menelaus’s theorem three times (Exercise 9.2).

The plane theorem needs either an excursion into space or a metric tool like Menelaus. Lesson 10 shows this is unavoidable.

The configuration

The ten points \(O,A,B,C,A',B',C',X,Y,Z\) and ten lines \(AA',BB',CC',AB,A'B',BC,B'C',CA,C'A',XYZ\) form the Desargues configuration: each point lies on three of the lines and each line contains three of the points. Any of the ten points can play the role of the centre.

Exercise 9.1Take the centre \(O=(0,0)\) and the triangles \(A=(1,0),B=(0,1),C=(1,1)\), \(A'=(2,0),B'=(0,3),C'=(4,4)\). Verify that \(O\) is a centre of perspective, find \(X,Y,Z\), and check collinearity.

Show solution
\(A'=2A,\ B'=3B,\ C'=4C\), so \(AA',BB',CC'\) pass through \(O\). \(AB: x+y=1\); \(A'B': \frac x2+\frac y3=1\). Solving: \(y=1-x\), \(3x+2-2x=6\), \(x=4\), \(X=(4,-3)\). \(BC: y=1\); \(B'C'\): through \((0,3)\) and \((4,4)\), \(y=3+\frac x4\); \(Y=(-8,1)\). \(CA: x=1\); \(C'A'\): through \((2,0),(4,4)\), \(y=2x-4\); \(Z=(1,-2)\). Collinearity: slope \(XZ=\frac{-2+3}{1-4}=-\frac13\), slope \(XY=\frac{1+3}{-8-4}=-\frac13\). The axis is \(x+3y=-5\).

Exercise 9.2Prove Desargues’s theorem in the Euclidean plane using Menelaus’s theorem: points \(L,M,N\) on the sides \(BC,CA,AB\) of a triangle (extended) are collinear iff \(\frac{BL}{LC}\cdot\frac{CM}{MA}\cdot\frac{AN}{NB}=-1\) in signed ratios.

Show solution
Apply Menelaus to triangle \(OAB\) with transversal \(A'B'X\): \(\frac{OA'}{A'A}\cdot\frac{AX}{XB}\cdot\frac{BB'}{B'O}=-1\). To triangle \(OBC\) with transversal \(B'C'Y\): \(\frac{OB'}{B'B}\cdot\frac{BY}{YC}\cdot\frac{CC'}{C'O}=-1\). To triangle \(OCA\) with transversal \(C'A'Z\): \(\frac{OC'}{C'C}\cdot\frac{CZ}{ZA}\cdot\frac{AA'}{A'O}=-1\). Multiply the three. The factors involving \(O,A',B',C'\) cancel in pairs (e.g. \(\frac{OA'}{A'A}\cdot\frac{AA'}{A'O}=1\)), leaving \(\frac{AX}{XB}\cdot\frac{BY}{YC}\cdot\frac{CZ}{ZA}=-1\). By Menelaus for triangle \(ABC\), \(X,Y,Z\) are collinear.

Exercise 9.3Show that in the Desargues configuration, the point \(A\) is also a centre of perspective for two triangles formed from the remaining points. Identify them.

Show solution
The three lines through \(A\) are \(AA'\) (containing \(O\)), \(AB\) (containing \(X\)) and \(AC\) (containing \(Z\)). Pair the other points on each line: \(\{O,A'\}\), \(\{B,X\}\), \(\{C,Z\}\). Triangles \(OBC\) and \(A'XZ\) are in perspective from \(A\). Check the axis: \(OB\cap A'X=B'\), \(BC\cap XZ=Y\), \(CO\cap ZA'=C'\), and \(B',Y,C'\) are collinear on the line \(B'C'\). The configuration is symmetric under a group of order 120 (isomorphic to \(S_5\)) acting on its ten points.
Lesson 10

Desargues and the foundations of geometry

For the general reader. Around 1900, mathematicians asked which axioms geometry really needs. Desargues’s theorem turned out to be a watershed. In three dimensions it follows from the simplest axioms of incidence. In a plane it does not, and the planes where it holds are exactly those that can be described by coordinates from a number system. A seventeenth-century theorem about drawing became a test for algebra.

Goals
  • Define abstract projective planes.
  • State the coordinatization theorem for Desarguesian planes.
  • Construct a plane where Desargues’s theorem fails.

A projective plane is a set of points and a set of lines, with an incidence relation, such that any two points lie on exactly one line, any two lines meet in exactly one point, and there are four points with no three collinear. For any division ring \(D\) (a ring in which every non-zero element has a multiplicative inverse, commutativity not required), the one-dimensional and two-dimensional subspaces of \(D^3\) form a projective plane \(\mathbb P^2(D)\).

Theorem (Hilbert 1899, and the coordinatization theorem)
  1. Every projective plane contained in a projective space of dimension at least 3 satisfies Desargues’s theorem.
  2. A projective plane satisfies Desargues’s theorem if and only if it is isomorphic to \(\mathbb P^2(D)\) for some division ring \(D\).
  3. \(\mathbb P^2(D)\) satisfies Pappus’s theorem if and only if \(D\) is commutative, that is, a field. Pappus’s theorem implies Desargues’s theorem (Hessenberg, 1905).

In his Grundlagen der Geometrie (1899), David Hilbert showed that the plane Desargues theorem cannot be derived from the plane axioms of incidence and order alone, and that it is exactly what is needed to introduce an algebra of segments with an associative multiplication. Part (1) is the space proof of Lesson 09. For (2), one defines addition and multiplication of points on a line by geometric constructions; Desargues’s theorem in its special forms is what makes these operations associative and distributive.

A non-Desarguesian plane

The Moulton plane (1902) is the Euclidean plane with modified lines. Vertical lines and lines of non-positive slope are unchanged. A line of positive slope \(m\) is bent where it crosses the \(y\)-axis: it has slope \(m\) for \(x\le0\) and slope \(2m\) for \(x>0\). With points at infinity added, one checks that any two points still lie on exactly one line, and this gives a projective plane. A pair of triangles in perspective whose sides cross the \(y\)-axis appropriately violates Desargues’s theorem. Finite non-Desarguesian planes also exist; the smallest have order 9, and every projective plane of order at most 8 is Desarguesian.

Exercise 10.1Count the points and lines of \(\mathbb P^2(\mathbb F_q)\), and the points on each line.

Show solution
Non-zero vectors in \(\mathbb F_q^3\): \(q^3-1\). Each point is a line through the origin with \(q-1\) non-zero vectors, so there are \(\frac{q^3-1}{q-1}=q^2+q+1\) points. By duality there are as many lines. Each line is a two-dimensional subspace containing \(\frac{q^2-1}{q-1}=q+1\) points. For \(q=2\) this is the Fano plane: 7 points, 7 lines, 3 points per line.

Exercise 10.2In \(\mathbb P^2(D)\), use coordinates to see why Desargues holds: show that two triangles in perspective from \(O\) can be written \(A'=A\alpha,\ B'=B\beta,\ C'=C\gamma\) up to adding multiples of \(O\), and find \(X\).

Show solution
Choose vectors (right \(D\)-modules) \(o,a,b,c\) for \(O,A,B,C\). Since \(A'\) is on \(OA\), a vector for it is \(a'=o+a\alpha\); similarly \(b'=o+b\beta\), \(c'=o+c\gamma\) (rescaling \(a,b,c\) if needed one can even take \(\alpha=\beta=\gamma=1\)). Then \(a'-b'=a\alpha-b\beta\) lies in both the span of \(a,b\) and of \(a',b'\), so it represents \(X\). Likewise \(Y\) is represented by \(b\beta-c\gamma\) and \(Z\) by \(c\gamma-a\alpha\). These three vectors sum to zero, so they are linearly dependent and \(X,Y,Z\) are collinear. Only associativity and inverses were used, never commutativity.

Exercise 10.3In the Moulton plane, find the line through \((-1,0)\) and \((1,3)\).

Show solution
An ordinary line through these points has slope \(\tfrac32>0\), so it is not a Moulton line. Try a bent line with slope \(m\) for \(x\le0\) and \(2m\) for \(x>0\): from \((-1,0)\), at \(x=0\) it reaches \(y=m\), and at \(x=1\) it reaches \(m+2m=3m\). So \(3m=3\), \(m=1\). The Moulton line is \(y=x+1\) for \(x\le0\) and \(y=2x+1\) for \(x>0\).
Part III · Life and legacy
Lesson 11

Quarrels, architecture and death, 1640–1661

Desargues’s pamphlets made enemies. In 1642 an anonymous Perspective pratique, by the Jesuit Jean Du Breuil, borrowed from his method with errors. Desargues answered with placards posted around Paris and a pamphlet listing Six erreurs. His opponents replied that his ideas had been published earlier by others. The same year a collection of hostile Advis charitables gathered criticisms of his work, including Beaugrand’s letter of 1640 on the conics. In 1644 the stonemason Jacques Curabelle published an 81-page Examen accusing him of errors and of knowing nothing of practice. The pamphlet war that followed ended with a proposal for a formal debate before judges that seems never to have taken place.

His defender was the engraver Abraham Bosse. Bosse attended Desargues’s lessons for craftsmen from about 1641 and published accessible versions of his methods: stonecutting and sundials in 1643, and in 1648 the Manière universelle de M. Desargues pour pratiquer la perspective, which contains the theorem on triangles. Bosse later taught perspective at the Royal Academy of Painting and Sculpture and was expelled after disputes over Desargues’s method. The painter Laurent de La Hire also adopted the method, and his son Philippe de La Hire became a geometer.

Architecture

From about 1645 Desargues worked mainly as an architect and published little. He returned to Lyon in 1648, at the time of the Fronde, and designed buildings there; his work there included private houses. He was in Paris in 1649–50 and again from 1657 to about 1660, designing private houses. He is also credited with a water-raising machine near Paris using epicycloidal gear teeth, a profile later studied by Philippe de La Hire.

He died in Lyon in 1661. The exact date is unknown; his will was opened in Lyon on 8 October 1661, a few days after his death.

DiscussionDesargues addressed his geometry to craftsmen and wrote in an invented vocabulary. Bosse and Curabelle fought over whether his methods worked in practice. What did he gain and lose by writing for artisans rather than for the learned, and why did Pascal and La Hire, rather than his intended readers, preserve his ideas?

Lesson 12

Oblivion and rediscovery

After 1661 Desargues’s geometry nearly vanished. Analytic geometry in the style of Descartes dominated the next century, and the few printed copies of the Brouillon project disappeared. Philippe de La Hire had made a manuscript copy, and his own books on conics (1673, 1685) used projective methods, but he did not make Desargues widely known.

The revival came with nineteenth-century projective geometry. Jean-Victor Poncelet, in his 1822 Traité des propriétés projectives des figures, drew attention to Desargues as a forerunner, and Michel Chasles gave him a prominent place in his 1837 history of geometry. In 1845 La Hire’s manuscript copy of the Brouillon project came to light. In 1864 Noël-Germinal Poudra published the Œuvres de Desargues in two volumes. A printed copy of the 1639 original was found in 1951; it is now in the Bibliothèque nationale de France. The same year René Taton published L’œuvre mathématique de G. Desargues, the basis of modern scholarship, and in 1987 Judith Field and Jeremy Gray published an English translation and commentary.

Legacy

  • Projective geometry. Points at infinity, the unified treatment of conics by projection, involution, and polarity all appear in the 1639 draft.
  • Desargues’s theorem. It underlies the coordinatization of projective planes, the classification of finite planes, and the theory of incidence geometries and buildings.
  • Perspective and practical geometry. Through Bosse, his methods entered the teaching of perspective and stereotomy in France.
  • Honours. A lunar crater and a street in Paris’s 11th arrondissement bear his name.

Final projectRead the opening pages of the Brouillon project in Taton’s or Field and Gray’s edition. Translate one proposition into modern projective language, give a coordinate proof, and explain what the original vocabulary (tree, trunk, knot, ordonnance) expresses that modern terms leave out.

Reference

Connections

People linked to Girard Desargues across this collection. A name without a link has no page here yet. See Girard Desargues in the network of all 337 lives →

Students

  • Abraham Bosse 1641-1661

Influenced by

Influenced

Colleagues and collaborators

Rivals

  • Jacques Curabelle 1644
  • Jean de Beaugrand 1640-1642
  • Jean Du Breuil 1642

Also mentioned in this life

Fields

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Girard Desargues, MacTutor History of Mathematics.
  2. Girard Desargues, Wikipedia; French edition.
  3. Girard Desargues, Encyclopaedia Britannica.
  4. D. R. Strick, Girard Desargues, MacTutor.
  5. R. Taton, L’œuvre mathématique de G. Desargues, Presses universitaires de France, 1951.

Mathematics

  1. G. Desargues, Brouillon project d’une atteinte aux evenemens des rencontres du cone avec un plan, Paris, 1639.
  2. N.-G. Poudra (ed.), Œuvres de Desargues, 2 vols, Leiber, Paris, 1864.
  3. A. Bosse, Manière universelle de M. Desargues pour pratiquer la perspective, Paris, 1648.
  4. J. V. Field and J. J. Gray, The Geometrical Work of Girard Desargues, Springer, 1987.
  5. M. Anglade and J.-Y. Briend, Le diamètre et la traversale : dans l’atelier de Girard Desargues, arXiv:1812.08445.
  6. Desargues and the “trait à preuves”, arXiv:2504.08477.
  7. D. Hilbert, Grundlagen der Geometrie, Teubner, 1899.
  8. H. S. M. Coxeter, Projective Geometry, 2nd ed., Springer, 1987.
  9. R. Hartshorne, Foundations of Projective Geometry, Benjamin, 1967.