Lesson 00

How to use this course

Jean-Victor Poncelet led two careers at once. To geometers he is the founder of modern projective geometry: the man who named the subject, made points at infinity and imaginary points into working tools, and found the closure theorem that still drives research. To engineers he is the professor who made work the central quantity of machine theory and built a waterwheel twice as efficient as the old ones. He was also a soldier who rose to general.

Prerequisites for Part II

Analytic geometry of lines and conics in the plane; homogeneous coordinates are introduced as needed.
Linear algebra: matrices, determinants, eigenvectors of \(2\times2\) and \(3\times3\) matrices.
For Lesson 08, some familiarity with complex numbers; elliptic curves are mentioned but not required.
For Lesson 10, first-year mechanics: force, kinetic energy, momentum flux.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proofs or proof sketches, and exercises with folded solutions. Lessons 04 to 09 follow the logic of the 1822 Traité. Lesson 10 covers his mechanics.

Reference

Timeline

1788	Born 1 July in Metz, son of Claude Poncelet, a lawyer at the Parlement of Metz, and Anne-Marie Perrein. Raised by the Olier family in Saint-Avold.
1804	Returns to Metz and studies at the lycée.
1807	Enters the École polytechnique, where the descriptive geometry of Gaspard Monge shapes the teaching.
1810	Leaves Polytechnique for the engineering school at Metz; commissioned lieutenant of engineers.
1812	Russian campaign. Left for dead at Krasnoi in November and taken prisoner.
1813–14	Prisoner at Saratov on the Volga. Rebuilds geometry from memory in the “Saratov notebooks”.
1814	Released after the peace; back in France in September.
1815–25	Captain of engineers at Metz: arsenal workshops, fortifications, teaching.
1820	Cauchy’s report to the Académie criticizes the principle of continuity.
1822	Traité des propriétés projectives des figures.
1824–25	The curved-blade undershot waterwheel; Montyon Prize of the Académie des sciences.
1825–35	Professor of mechanics applied to machines at the École d’application, Metz.
1826–29	Priority dispute with Joseph Gergonne over the principle of duality.
1829	Introduction à la mécanique industrielle.
1834	Elected to the mechanics section of the Académie des sciences; moves to Paris.
1837–48	Chair of physical and experimental mechanics at the Paris Faculty of Sciences.
1842	Foreign Member of the Royal Society of London. Marries Louise Palmyre Gaudin.
1848	General of brigade; commandant of the École polytechnique; member of the Constituent Assembly.
1850	Retires from the army and from Polytechnique.
1851	Heads the French scientific commission at the Great Exhibition in London.
1862–64	Applications d’analyse et de géométrie, including the Saratov notebooks.
1865–66	Second edition of the Traité.
1867	Dies in Paris on 22 December, aged 79.
1868	His widow endows the Prix Poncelet of the Académie des sciences.

Part I · Formation

Lesson 01

Metz and the École polytechnique, 1788–1810

Jean-Victor Poncelet was born in Metz, in Lorraine, on 1 July 1788. His father, Claude Poncelet, was a lawyer at the Parlement of Metz and a landowner; his mother was Anne-Marie Perrein. He was born before his parents married, and they later legitimated him. As a young child he was sent to live with the Olier family at Saint-Avold, a small town east of Metz, where he stayed until about 1804.

Back in Metz at fifteen, he attended the lycée and prepared for the École polytechnique, the engineering school founded during the Revolution. He entered in 1807. Polytechnique was then the best school of mathematics in Europe. Gaspard Monge taught descriptive geometry there, the art of representing three-dimensional objects by projections onto two planes, and his lectures trained a whole generation to think about space through projections. Poncelet’s teachers and examiners included Monge, Charles-Julien Brianchon, Jean Hachette, Sylvestre Lacroix, Louis Poinsot and André-Marie Ampère. He also absorbed Lazare Carnot’s Géométrie de position (1803), which tried to make geometric theorems independent of the particular arrangement of a figure.

Illness held him back in his third year. He left in 1810 for the École d’application of artillery and engineering at Metz, and was commissioned a lieutenant in the engineering corps.

Lesson 02

Russia and Saratov, 1812–1814

His first post, in early 1812, was on fortifications on the island of Walcheren in the Netherlands. By June he had been called to the Grande Armée for the invasion of Russia. He took part in the crossing of the Dnieper near Smolensk and in the campaign that led to Borodino and Moscow. The retreat began in October. At the battle of Krasnoi in November 1812 his horse was killed under him, and he was left for dead on the field. He survived and was taken prisoner.

With other captured officers he was marched east across the frozen steppe for several months, a journey of well over a thousand kilometres, to a prison camp at Saratov on the Volga. He arrived in early 1813 and stayed until June 1814.

The Saratov notebooks

Poncelet had no books. To pass the time and to keep his mind alive, he set out to reconstruct what he had learned from Monge and Carnot, starting from first principles. Working without references, he went beyond what he had been taught. In the notebooks he studied properties of conics that survive central projection, and he began to use points at infinity and imaginary points as ordinary elements of a figure. He brought the notebooks home after the peace of 1814 and arrived in France in September. Half a century later he published them in the first volume of his Applications d’analyse et de géométrie (1862).

The story of a theory conceived in a prison camp became part of the legend of projective geometry. The documented core is clear: the notebooks exist, they date from the captivity, and they contain the first form of the ideas of the 1822 Traité.

Lesson 03

Metz and the Traité, 1815–1822

From 1815 Poncelet served at Metz as a captain of engineers. He ran workshops at the arsenal, worked on fortifications and machinery, and taught. In his spare time he turned the Saratov notes into a treatise. He sent a memoir on the projective properties of conics to the Académie des sciences in 1820. The report, written by Augustin-Louis Cauchy, praised the results but rejected their foundation, the “principle of continuity” (Lesson 06), as an unproved induction. Poncelet never forgave the slight, and he saw it as the opinion of analysts who did not value pure geometry.

The Traité des propriétés projectives des figures appeared in Paris in 1822. Its title gave the subject its name. The book defined a projective property as one preserved under central projection, and set out to derive the theory of conics from the simplest case, the circle, by projection. It contained the theory of poles and polars, the treatment of points and lines at infinity, the two circular points at infinity, the method of reciprocal polars that leads to duality, and the closure theorem now called Poncelet’s porism. It also stated that every ruler-and-compass construction can be carried out with a ruler alone once one circle and its centre are drawn.

Poncelet also recognized his predecessors. He drew attention to Girard Desargues and to Blaise Pascal, whose seventeenth-century projective ideas had been largely forgotten, and helped bring them back into view.

Part II · Geometry and mechanics

Lesson 04

Central projection and the cross-ratio

For the general reader. Hold a picture frame up to a window and trace the landscape on the glass. Lengths and angles change, circles become ellipses, and parallel railway tracks meet at the horizon. Poncelet asked which properties survive this kind of projection. The answer starts with a single number built from four points on a line.

Goals

Define central projection and the projective plane in homogeneous coordinates.
Prove that the cross-ratio of four collinear points is invariant.
Use projective maps of a line to simplify computations.

The projective plane

Central projection from a point \(O\) maps a plane \(\pi\) to a plane \(\pi'\): a point \(P\in\pi\) goes to the intersection of the line \(OP\) with \(\pi'\). Points of \(\pi\) on the plane through \(O\) parallel to \(\pi'\) have no image, and points on a certain line of \(\pi'\) have no preimage. Poncelet repaired this by adding to each plane a line at infinity. In coordinates, the real projective plane is \(\mathbb{RP}^2=(\mathbb R^3\setminus\{0\})/\mathbb R^\times\). A point is written \((x:y:z)\); the affine point \((x,y)\) is \((x:y:1)\) and the points \((x:y:0)\) form the line at infinity. Lines are the sets \(ax+by+cz=0\). Central projections, composed, are the maps \(v\mapsto Av\) with \(A\in\mathrm{GL}_3(\mathbb R)\), called projective transformations or homographies.

The cross-ratio

For four distinct points with coordinates \(a,b,c,d\) on a line, define

\[ (A,B;C,D)=\frac{(c-a)(d-b)}{(c-b)(d-a)}. \]

Theorem (invariance of the cross-ratio)

Let \(f(x)=\dfrac{px+q}{rx+s}\) with \(ps-qr\neq0\), a projective transformation of the line. Then \((f(a),f(b);f(c),f(d))=(a,b;c,d)\). Consequently, if four collinear points are projected from a centre \(O\) onto another line, their cross-ratio is unchanged. Conversely, a bijection of the projective line that preserves cross-ratios is a projective transformation.

Proof. A direct computation gives \(f(x)-f(y)=\dfrac{(ps-qr)(x-y)}{(rx+s)(ry+s)}\). In the cross-ratio, the factor \(ps-qr\) appears twice in the numerator and twice in the denominator, and each factor \((rx+s)\) appears once above and once below, so everything cancels. Projection from \(O\) between two lines, in suitable coordinates on each line, is such a map \(f\). For the converse, given three points there is a unique projective map sending them to \(\infty,0,1\), and the cross-ratio \((A,B;C,X)\) then equals the coordinate of the image of \(X\), up to the chosen convention.

Poncelet did not treat the cross-ratio as a primary notion; he worked mostly with harmonic sets and with the ratio relations he inherited from Desargues and Carnot. The cross-ratio as the fundamental invariant became central with August Möbius and Michel Chasles in the 1820s and 1830s. It is the cleanest modern summary of what his projective properties preserve.

Exercise 4.1Compute \((A,B;C,D)\) for the points \(0,1,2,3\), and check that the map \(x\mapsto 1/x\) preserves the cross-ratio of \(1,2,3,4\).

Show solution

For \(0,1,2,3\): \(\frac{(2-0)(3-1)}{(2-1)(3-0)}=\frac43\). For \(1,2,3,4\): \(\frac{(3-1)(4-2)}{(3-2)(4-1)}=\frac43\). The images are \(1,\tfrac12,\tfrac13,\tfrac14\): \(\frac{(\frac13-1)(\frac14-\frac12)}{(\frac13-\frac12)(\frac14-1)}=\frac{(-\frac23)(-\frac14)}{(-\frac16)(-\frac34)}=\frac{1/6}{1/8}=\frac43.\)

Exercise 4.2Permuting four points changes the cross-ratio. Show that if \((A,B;C,D)=\lambda\) then \((B,A;C,D)=1/\lambda\) and \((A,C;B,D)=1-\lambda\). Deduce that the 24 orderings give at most six values.

Show solution

Swapping \(a\) and \(b\) exchanges numerator and denominator, giving \(1/\lambda\). For the second identity, use invariance to put \(a=0\), \(d=1\) and let \(b\to\infty\). Then \((A,B;C,D)=\frac{c\,(1-b)}{(c-b)\cdot1}\to c\), so \(\lambda=c\), and \((A,C;B,D)=\frac{(b-0)(1-c)}{(b-c)(1-0)}\to1-c=1-\lambda\). The maps \(\lambda\mapsto1/\lambda\) and \(\lambda\mapsto1-\lambda\) generate a group of order 6 (isomorphic to \(S_3\)): \(\lambda,\ 1/\lambda,\ 1-\lambda,\ 1/(1-\lambda),\ (\lambda-1)/\lambda,\ \lambda/(\lambda-1)\). The Klein four-group of double transpositions acts trivially, and \(24/4=6\).

Exercise 4.3Show that a projective transformation of the plane is determined by the images of four points, no three collinear, and that any such four points can be sent to any other four.

Show solution

Choose representatives \(v_1,v_2,v_3\) of the first three points; they form a basis because the points are not collinear. Write \(v_4=\alpha v_1+\beta v_2+\gamma v_3\) with all coefficients non-zero (no three collinear). Rescale so that \(v_4=v_1+v_2+v_3\). Do the same for the targets \(w_i\). The linear map with \(Av_i=w_i\) for \(i\le3\) then sends \(v_4\) to \(w_4\). Any other such map must send each \(v_i\) to a multiple \(\mu_iw_i\) with \(\mu_1w_1+\mu_2w_2+\mu_3w_3\) proportional to \(w_1+w_2+w_3\), forcing \(\mu_1=\mu_2=\mu_3\). So \(A\) is unique up to scale.

Lesson 05

Harmonic sets and involutions

For the general reader. Some arrangements of four points on a line are special: they can be constructed with a ruler alone, from a quadrilateral, and they survive every projection. These harmonic sets, and the pairings of points called involutions, were the working tools of Desargues and Poncelet long before the cross-ratio had a name.

Goals

Construct harmonic conjugates with a straightedge and prove the construction.
Understand involutions on a line and their fixed points.
State Desargues’s involution theorem as Poncelet used it.

Four collinear points form a harmonic set, and \(D\) is the harmonic conjugate of \(C\) with respect to \(A,B\), when \((A,B;C,D)=-1\). If \(C\) is the midpoint of \(AB\), then \(D\) is the point at infinity of the line.

Theorem (complete quadrilateral)

Let \(A,B,C\) be distinct points on a line \(\ell\). Choose a point \(S\notin\ell\) and a point \(T\) on \(SC\), distinct from \(S\) and \(C\). Let \(P=AT\cap SB\) and \(Q=BT\cap SA\). Then the line \(PQ\) meets \(\ell\) at the harmonic conjugate \(D\) of \(C\) with respect to \(A,B\). In particular \(D\) does not depend on the choices of \(S\) and \(T\).

Proof. Apply a projective transformation sending the line \(PQ\) to the line at infinity. Lines through \(P\) become parallel, and so do lines through \(Q\). Since \(P=AT\cap SB\), we get \(AT\parallel SB\); since \(Q=BT\cap SA\), we get \(BT\parallel SA\). So \(ASBT\) is a parallelogram with diagonals \(AB\) and \(ST\). The diagonals bisect each other, so \(C=ST\cap AB\) is the midpoint of \(AB\), and \(D\) is the point at infinity of \(AB\). For the midpoint, \((A,B;C,\infty)=\frac{c-a}{c-b}=-1\). Cross-ratios are invariant, so the original four points are harmonic.

Involutions

An involution of a projective line is a projective map \(f\neq\mathrm{id}\) with \(f\circ f=\mathrm{id}\). In a suitable coordinate every involution has the form \(x\mapsto k/x\). If \(k>0\) it has two real fixed points \(\pm\sqrt k\) (a hyperbolic involution); if \(k<0\) it has none (an elliptic involution), and its fixed points are the complex pair \(\pm\sqrt{k}\). Each pair \(\{x,f(x)\}\) is harmonic with respect to the two fixed points. Desargues called a set of three such pairs an involution of six points.

Theorem (Desargues’s involution theorem)

Let four points in general position in the plane be given, and let \(\ell\) be a line through none of them. The conics through the four points meet \(\ell\) in pairs of points, and these pairs (together with the pairs cut on \(\ell\) by the three pairs of opposite sides of the quadrangle) are pairs of a single involution of \(\ell\).

Poncelet used this theorem, and the language of involutions, throughout the Traité. For instance, a pencil of circles through two fixed points cuts any line in an involution, which is a standard way to locate the two points where a coaxal system of circles is tangent to a given line.

Exercise 5.1On a number line, find the harmonic conjugate of \(C=1\) with respect to \(A=0\), \(B=3\).

Show solution

Solve \(\frac{(1-0)(d-3)}{(1-3)(d-0)}=-1\): \(d-3=2d\), so \(d=-3\). Check: the pairs are separated, and \(\frac{2}{AB}=\frac1{AC}+\frac1{AD}\) in signed lengths: \(\frac23=\frac11+\frac1{-3}\). This is the classical harmonic-mean characterization.

Exercise 5.2Show that the involution \(x\mapsto k/x\) with \(k>0\) pairs every \(x\) with a point harmonic to it with respect to \(\pm\sqrt k\).

Show solution

Put \(s=\sqrt k\) and compute \((s,-s;x,k/x)=\frac{(x-s)(k/x+s)}{(x+s)(k/x-s)}\). Multiply top and bottom by \(x\): \(\frac{(x-s)(k+sx)}{(x+s)(k-sx)}=\frac{(x-s)\,s(s+x)}{(x+s)\,s(s-x)}=-1\).

Exercise 5.3Prove that a projective map of a line that swaps two distinct points is an involution.

Show solution

Choose coordinates with the two points at \(0\) and \(\infty\). The map \(f(x)=\frac{px+q}{rx+s}\) sends \(0\mapsto\infty\), so \(s=0\), and \(\infty\mapsto0\), so \(p=0\). Then \(f(x)=\frac{q}{rx}=\frac{k}{x}\), and \(f(f(x))=x\).

Lesson 06

Infinity, continuity and the circular points

For the general reader. Two circles that cross meet in two points, and the line through those points is easy to draw. When the circles are pulled apart, the points vanish, yet a line with the same algebraic description remains. Poncelet insisted that the vanished points still exist, as imaginary points, and that the theorem survives. This boldness made geometry more general, and it alarmed the analysts.

Goals

State the principle of continuity and see what algebra justifies it.
Locate the circular points at infinity and characterize circles by them.
Explain why two circles meet in only two finite points although Bézout’s theorem predicts four.

The principle of continuity

Poncelet stated the principle roughly as follows. If a property holds for a figure in a general position, it continues to hold when the figure is moved continuously, even if some of its elements become imaginary or go off to infinity, provided the property is interpreted suitably. Cauchy objected in 1820 that this was an argument by analogy. In modern terms the principle is justified when the property is expressed by polynomial identities in the coordinates: an identity that holds on a non-empty open set of real parameters holds identically, also for complex values.

Example (radical axis)

For circles \(\Gamma_1: x^2+y^2+D_1x+E_1y+F_1=0\) and \(\Gamma_2: x^2+y^2+D_2x+E_2y+F_2=0\) with distinct centres, the radical axis is the line \((D_1-D_2)x+(E_1-E_2)y+(F_1-F_2)=0\). It is the locus of points with equal power with respect to both circles. When the circles meet, it is their common chord; when they do not, it still exists and passes through their two complex intersection points.

The circular points

Write a circle in homogeneous coordinates: \(x^2+y^2+Dxz+Eyz+Fz^2=0\). At \(z=0\) this reduces to \(x^2+y^2=0\), whose solutions are the two points

\[ I=(1:i:0),\qquad J=(1:-i:0). \]

Theorem (Poncelet)

Every circle in the plane passes through the two points \(I\) and \(J\) on the line at infinity. Conversely, a non-degenerate real conic through \(I\) and \(J\) is a circle.

The proof is the computation above, and its converse: a conic \(ax^2+bxy+cy^2+\dots\) contains \(I\) and \(J\) exactly when \(a-c\pm ib=0\), so \(a=c\) and \(b=0\). Two distinct circles are two conics, so by Bézout’s theorem they meet in four points counted with multiplicity. Two of these are always \(I\) and \(J\), which leaves at most two finite intersections. Edmond Laguerre showed in 1853 that angles can be expressed through cross-ratios involving \(I\) and \(J\), and Arthur Cayley and Felix Klein built metric geometry on this idea.

Exercise 6.1Find the radical axis of \(x^2+y^2=1\) and \((x-4)^2+y^2=4\). Do the circles meet? Find their complex intersection points.

Show solution

Subtract: \(x^2-(x-4)^2=1-4\), so \(8x-16=-3\) and \(x=\tfrac{13}{8}\). Since \(\tfrac{13}{8}>1\), the circles do not meet in real points. On \(x=\tfrac{13}8\), \(y^2=1-\tfrac{169}{64}=-\tfrac{105}{64}\), so the intersections are \(\big(\tfrac{13}8,\pm\tfrac{i\sqrt{105}}{8}\big)\), a complex-conjugate pair on a real line.

Exercise 6.2Show that every parabola \(y=ax^2+bx+c\) (\(a\neq0\)) touches the line at infinity, and find the point of contact.

Show solution

Homogenize: \(yz=ax^2+bxz+cz^2\). At \(z=0\) this becomes \(ax^2=0\), a double root \(x=0\). So the conic meets the line at infinity in the single point \((0:1:0)\) with multiplicity 2: it is tangent there, at the point at infinity of the vertical direction (the axis direction).

Exercise 6.3Use the circular points to prove that two concentric circles are tangent to each other at \(I\) and at \(J\).

Show solution

\(x^2+y^2=r_1^2z^2\) and \(x^2+y^2=r_2^2z^2\). Subtracting gives \((r_1^2-r_2^2)z^2=0\), so all intersections lie on \(z=0\), at \(I\) and \(J\). Bézout requires total multiplicity 4, so each of \(I,J\) has multiplicity 2: the circles are tangent at both. Their common tangents there are the lines \(x\pm iy=0\) through the common centre, called isotropic lines.

Lesson 07

Poles, polars and duality

For the general reader. A circle sets up a dictionary between points and lines. Every point has a partner line, its polar, and every line a partner point. Translating a theorem through this dictionary produces another theorem, often a new one. Poncelet turned this trick into a method, and others saw that it reflects a perfect symmetry between points and lines in the projective plane.

Goals

Define the polar of a point with respect to a conic and prove reciprocity.
Derive Brianchon’s theorem from Pascal’s by reciprocal polars.
Distinguish Poncelet’s polar transformation from the abstract principle of duality.

Let a non-degenerate conic be given by a symmetric invertible matrix \(M\), as the set \(X^{\mathsf T}MX=0\) in homogeneous coordinates. The polar of a point \(P\) is the line \(\{X: P^{\mathsf T}MX=0\}\). For the circle \(x^2+y^2=r^2\), the polar of \((x_0,y_0)\) is \(x_0x+y_0y=r^2\).

Theorem (pole and polar)

(Reciprocity, La Hire) \(P\) lies on the polar of \(Q\) if and only if \(Q\) lies on the polar of \(P\).
If \(P\) is outside the conic, its polar is the line through the two points of contact of the tangents from \(P\). If \(P\) is on the conic, its polar is the tangent at \(P\).
If a line through \(P\) meets the conic at \(U,V\) and the polar of \(P\) at \(W\), then \((U,V;P,W)=-1\).

Proof. (1) is the symmetry \(P^{\mathsf T}MQ=Q^{\mathsf T}MP\). For (2), the tangent at a point \(T\) of the conic is its polar; \(T\) lies on the polar of \(P\) iff \(P\) lies on the polar of \(T\), the tangent at \(T\). For (3), parametrize the line as \(P+tW\). Since \(W\) is on the polar, \(P^{\mathsf T}MW=0\), and the intersections satisfy \(P^{\mathsf T}MP+t^2\,W^{\mathsf T}MW=0\). The two parameters are \(\pm t_0\), while \(P\) and \(W\) have parameters \(0\) and \(\infty\). Points with parameters \(t_0,-t_0\) are harmonic with respect to \(0,\infty\).

Reciprocal polars and duality

Fix a conic. Sending each point to its polar and each line to its pole turns a configuration of points and lines into one of lines and points, preserving incidence. A curve, seen as the envelope of its tangents, goes to the curve formed by their poles. Applied to Pascal’s theorem, this gives a theorem first found by Brianchon in 1806:

Pascal (1640) and Brianchon (1806)

Pascal. If a hexagon is inscribed in a conic, the three intersection points of pairs of opposite sides are collinear.
Brianchon. If a hexagon is circumscribed about a conic, the three diagonals joining opposite vertices are concurrent.

Joseph Gergonne, editor of the Annales de mathématiques, argued from 1825 that the correspondence between points and lines is a general law of plane geometry, independent of any conic, and printed theorems in two columns side by side. Poncelet answered in 1826 that duality was just his method of reciprocal polars, which he had published first. The quarrel lasted until 1829. Julius Plücker, who gave duality its algebraic form through line coordinates, was drawn into it. In modern terms both men were right: the abstract principle comes from the fact that lines of \(\mathbb P^2\) form the dual projective plane, and Poncelet’s polarity is a concrete isomorphism between the two planes.

Exercise 7.1Find the polar of \(P=(2,0)\) with respect to the unit circle, and the points of contact of the tangents from \(P\).

Show solution

The polar is \(2x=1\), the line \(x=\tfrac12\). It meets the circle where \(y^2=\tfrac34\), at \(\big(\tfrac12,\pm\tfrac{\sqrt3}2\big)\). Check: the tangent at \(\big(\tfrac12,\tfrac{\sqrt3}2\big)\) is \(\tfrac12x+\tfrac{\sqrt3}2y=1\), which passes through \((2,0)\).

Exercise 7.2Write down the dual of Desargues’s theorem (“if two triangles are in perspective from a point, they are in perspective from a line”). What do you notice?

Show solution

Swapping points and lines: “if the lines joining corresponding vertices are concurrent, then the intersections of corresponding sides are collinear” becomes “if the intersections of corresponding sides are collinear, then the lines joining corresponding vertices are concurrent”. The dual is the converse. So, given duality, Desargues’s theorem and its converse prove each other.

Exercise 7.3Using item (3) of the theorem, show that the polar of the centre of a circle is the line at infinity.

Show solution

For the centre \(O\), each line through \(O\) meets the circle at \(U,V\) with \(O\) the midpoint. The harmonic conjugate of the midpoint with respect to \(U,V\) is the point at infinity of the line. So every point of the polar is at infinity. Algebraically, with \(M=\mathrm{diag}(1,1,-r^2)\) and \(O=(0:0:1)\), the polar is \(-r^2z=0\), the line \(z=0\).

Lesson 08

The closure theorem (Poncelet’s porism)

For the general reader. Draw one circle inside another. Start at a point of the outer circle, draw a tangent to the inner one, follow it to the outer circle, draw the next tangent, and keep going. Usually the zigzag never closes. Poncelet proved that if it closes once, after some number of steps, then it closes after the same number of steps wherever you start.

Goals

State the closure theorem precisely.
Derive the Chapple–Euler and Fuss conditions for triangles and quadrilaterals.
See the modern proof through an elliptic curve and Cayley’s criterion.

Theorem (Poncelet, 1822)

Let \(C\) and \(D\) be smooth conics in the complex projective plane meeting in four distinct points. Suppose there is a polygon with \(n\) vertices on \(C\) whose \(n\) sides are tangent to \(D\). Then for every point \(P\in C\) the polygon obtained by starting at \(P\) and alternately drawing tangents to \(D\) and meeting \(C\) again closes after \(n\) steps.

Such a statement, where a problem has either no solution or infinitely many, is a porism in the old Greek sense, hence the name. For two real circles with \(D\) inside \(C\), the zigzag is a real polygon.

Triangles and quadrilaterals

Let \(C\) have radius \(R\), \(D\) radius \(r\), and let \(d\) be the distance between the centres.

Chapple (1746), Euler (1765), Fuss (1798)

A triangle inscribed in \(C\) and circumscribed about \(D\) exists if and only if \(d^2=R^2-2Rr\). A quadrilateral exists if and only if

\[ \frac{1}{(R-d)^2}+\frac{1}{(R+d)^2}=\frac{1}{r^2}. \]

Combined with the closure theorem, the first condition says: if the circumcircle and incircle of one triangle are given, every point of the circumcircle is a vertex of a triangle with the same two circles. The triangle formula is often attributed to Leonhard Euler alone; William Chapple published it earlier.

Why it is true: an elliptic curve

Consider the set \(E=\{(P,\ell): P\in C,\ \ell \text{ tangent to } D,\ P\in\ell\}\). Each point \(P\) lies on two tangents, and each tangent meets \(C\) in two points, so \(E\) is a double cover of the conic \(C\cong\mathbb P^1\) branched over the four points of \(C\cap D\). It is therefore a curve of genus 1, an elliptic curve. Two involutions act on \(E\): \(\sigma(P,\ell)=(P',\ell)\), where \(P'\) is the other point of \(\ell\cap C\), and \(\tau(P,\ell)=(P,\ell')\), where \(\ell'\) is the other tangent through \(P\). One step of the zigzag is \(\tau\circ\sigma\). On an elliptic curve, an involution with fixed points has the form \(x\mapsto a-x\) in the group law, so the composition of two of them is a translation \(x\mapsto x+c\). The polygon closes after \(n\) steps from one starting point iff \(nc=0\), a condition independent of the start.

Poncelet’s own proof was synthetic. He embedded the two circles in a pencil of conics and used the fact that tangents from a moving point to the members of the pencil vary in a controlled way. Carl Jacobi gave a proof in 1828 using elliptic functions, which is the analytic form of the argument above.

Cayley’s criterion (1853)

Let \(C\) and \(D\) have symmetric matrices of the same names, and expand \(\sqrt{\det(tC+D)}=A_0+A_1t+A_2t^2+\cdots\) near \(t=0\). A Poncelet \(n\)-gon inscribed in \(C\) and circumscribed about \(D\) exists iff

\[ \begin{aligned} &\det\begin{pmatrix}A_2&\cdots&A_{m+1}\\ \vdots&&\vdots\\ A_{m+1}&\cdots&A_{2m}\end{pmatrix}=0 \quad (n=2m+1),\\ &\det\begin{pmatrix}A_3&\cdots&A_{m+1}\\ \vdots&&\vdots\\ A_{m+1}&\cdots&A_{2m-1}\end{pmatrix}=0 \quad (n=2m). \end{aligned} \]

For \(n=3\) the condition is \(A_2=0\); for \(n=4\) it is \(A_3=0\).

Exercise 8.1A triangle has circumradius \(R=5\) and inradius \(r=2\). How far apart are the circumcentre and incentre? What happens when \(R=2r\)?

Show solution

\(d^2=R^2-2Rr=25-20=5\), so \(d=\sqrt5\). Since \(d^2\ge0\), every triangle satisfies \(R\ge2r\) (Euler’s inequality). Equality \(R=2r\) gives \(d=0\): the centres coincide, which happens only for the equilateral triangle. The closure theorem then says every triangle with these two concentric circles is equilateral, which is clear by symmetry.

Exercise 8.2Check Fuss’s condition for concentric circles, and identify the polygon.

Show solution

With \(d=0\), the condition reads \(\frac2{R^2}=\frac1{r^2}\), so \(r=R/\sqrt2\). This is the incircle of a square inscribed in the circle of radius \(R\): the half-side of the square is \(R/\sqrt2\). Every starting point gives a rotated square.

Exercise 8.3Show that for two concentric circles of radii \(R>r\), a Poncelet \(n\)-gon exists iff \(\arccos(r/R)\) is a rational multiple of \(\pi\) with the right denominator. Connect this with the elliptic-curve proof.

Show solution

A chord of the outer circle tangent to the inner one subtends a central angle \(2\theta\) with \(\cos\theta=r/R\). Each step rotates the vertex by \(2\theta\). The polygon closes after \(n\) steps (winding \(k\) times) iff \(2n\theta=2\pi k\), i.e. \(\theta=\pi k/n\). Here the “translation” \(c\) of the general proof is literally a rotation by \(2\theta\), and closing means it has finite order \(n\). For \(n=3,k=1\): \(\theta=\pi/3\), \(r=R/2\).

Lesson 09

The straightedge and one circle

For the general reader. Euclid built figures with a ruler and a compass. Which of these tools is really needed? A ruler alone cannot even find the midpoint of a segment. Poncelet saw that one circle, drawn once with its centre marked, gives the ruler all the power of the compass. Jakob Steiner proved it in 1833.

Goals

Show that straightedge constructions are projectively invariant, and what that rules out.
Construct parallels and midpoints with a straightedge, given suitable data.
Understand the Poncelet–Steiner theorem and why the centre is essential.

Theorem (Poncelet 1822, proved by Steiner 1833)

Every point constructible with straightedge and compass from a given set of points can be constructed with a straightedge alone, provided one circle and its centre are drawn in the plane. Here a circle counts as “constructed” when its centre and one point on it are known.

Proof outline. It suffices to construct, with the straightedge and the fixed circle \(\Gamma\) (centre \(O\)), (a) the parallel to a given line through a given point, (b) the intersections of a line with a circle given by centre and point, and (c) the intersections of two such circles. For (a), every diameter of \(\Gamma\) is a segment with a known midpoint \(O\), so Exercise 9.1 gives parallels to any diameter. Fix two diameters with directions \(\delta_1,\delta_2\). For two points \(A,B\) on an arbitrary line \(\ell\), draw through \(A\) the parallels in directions \(\delta_1,\delta_2\) and through \(B\) the parallels in directions \(\delta_2,\delta_1\). They form a parallelogram with diagonal \(AB\); its other diagonal meets \(AB\) at the midpoint. Exercise 9.1 then gives the parallel to \(\ell\) through any point, and perpendiculars follow by the same method from perpendicular diameters. For (b), a homothety centred at a suitable point maps the given circle onto \(\Gamma\); homotheties can be carried out with parallels; intersect the image line with \(\Gamma\) and map back. For (c), reduce to (b) by drawing the radical axis of the two circles, which is constructible with parallels and perpendiculars.

Limits of the straightedge

A straightedge construction only intersects lines through known points. If all given points have coordinates in a field \(K\), so does every constructed point, because solving two linear equations stays in \(K\). Moreover, a projective transformation maps a straightedge construction to a straightedge construction. This is why the centre must be given: a projective map can fix a circle while moving its centre, so no straightedge construction from the circle alone can find the centre.

Exercise 9.1Given a segment \(AB\), its midpoint \(M\), and a point \(P\) off the line \(AB\), construct the parallel to \(AB\) through \(P\) with a straightedge alone.

Show solution

Choose \(S\) on line \(AP\) beyond \(P\). Let \(T=SM\cap PB\) and \(U=AT\cap SB\). Then \(PU\parallel AB\). Reason: in triangle \(SAB\), the cevians \(AU\), \(BP\), \(SM\) are concurrent at \(T\). By Ceva’s theorem \(\frac{AM}{MB}\cdot\frac{BU}{US}\cdot\frac{SP}{PA}=1\); since \(AM=MB\), \(\frac{SP}{PA}=\frac{SU}{UB}\), and by Thales \(PU\parallel AB\). In projective terms, \(M\) is the harmonic conjugate of the point at infinity of \(AB\), and the construction of Lesson 05 finds that point.

Exercise 9.2Conversely, given two parallel lines \(\ell\) and \(AB\), construct the midpoint of \(AB\) with a straightedge.

Show solution

Pick \(S\) outside the strip. Let \(SA\), \(SB\) meet \(\ell\) at \(A'\), \(B'\). Let \(X=AB'\cap A'B\). The line \(SX\) meets \(AB\) at its midpoint. Reason: \(ABB'A'\) is a trapezoid; the line through the intersection of its diagonals and the intersection of its legs bisects both parallel sides (apply a homothety from \(S\), or use the harmonic construction with the point at infinity of \(AB\)).

Exercise 9.3Starting from the points \((0,0),(1,0),(0,1)\), show that \(\big(\sqrt2,0\big)\) cannot be constructed with a straightedge alone.

Show solution

Every line through two points with rational coordinates has an equation with rational coefficients, and the intersection of two such lines has rational coordinates (Cramer’s rule). By induction every constructed point is rational, and \(\sqrt2\notin\mathbb Q\). With a compass one can solve quadratic equations, which is exactly what the fixed circle supplies in the Poncelet–Steiner theorem.

Lesson 10

Work, waterwheels and turbines

For the general reader. To an engineer of the 1820s the useful output of a machine was the weight it lifted times the height. Poncelet, with Gaspard-Gustave Coriolis, made this quantity, work, the basis of a teachable science of machines. He then used it to redesign the waterwheel, the main power source of French industry, and doubled its efficiency.

Goals

State the work–energy theorem in the form Poncelet taught.
Compute the ideal efficiency of a flat-blade undershot wheel and of Poncelet’s curved-blade wheel.
Relate the poncelet unit of power to modern units.

Work and living force

In his Metz course, lithographed from 1826 and printed as Introduction à la mécanique industrielle (1829), Poncelet put the quantity of work at the centre: the product of a force and the displacement of its point of application along its direction, \(W=\int\mathbf F\cdot d\mathbf s\). Coriolis, in his 1829 book on the effect of machines, made the same choice and introduced the factor \(\tfrac12\) in the living force.

Theorem (work–energy)

For a particle of mass \(m\) moving under a resultant force \(\mathbf F\), \(\displaystyle \tfrac12 m v_2^2-\tfrac12 m v_1^2=\int_{1}^{2}\mathbf F\cdot d\mathbf s\). For a machine in steady operation, motive work equals useful work plus work lost to friction, shocks and the kinetic energy carried away by the working fluid.

The second sentence was Poncelet’s design principle. Losses come from shocks (sudden changes of velocity, as in inelastic collisions) and from the speed of the water leaving the wheel. A good hydraulic machine should take in the water without shock and release it with as little speed as possible.

Two waterwheels

In an undershot wheel, a jet of speed \(v\) and volume flow \(Q\) strikes blades moving at speed \(u\). With flat blades the water hits, loses its relative speed and moves on with the blade. The force is the momentum change per second, \(\rho Q(v-u)\), so the power is

\[ P_{\text{flat}}=\rho Q\,(v-u)\,u,\qquad \eta_{\text{flat}}=\frac{P_{\text{flat}}}{\tfrac12\rho Q v^2}=\frac{2u(v-u)}{v^2}\le\frac12. \]

Poncelet’s wheel, presented in 1824, has curved blades. The water enters tangentially, runs up the curved blade, stops relative to it, and runs back down, leaving with relative velocity \(-(v-u)\). The momentum change per second doubles:

\[ P_{\text{curved}}=2\rho Q\,(v-u)\,u,\qquad \eta_{\text{curved}}=\frac{4u(v-u)}{v^2}, \]

which reaches 1 at \(u=v/2\), when the water leaves with zero absolute speed. Real wheels lost energy to friction and leakage, but Poncelet’s design more than doubled the efficiency of the old undershot wheel. It won the Montyon Prize of the Académie des sciences in 1825 and was widely built. In 1826 he also proposed an inward-flow turbine, a design later realized in modern radial turbines.

Experiment

From 1831 Poncelet and Arthur Morin carried out friction experiments at Metz that confirmed and extended the laws of Charles-Augustin de Coulomb, and they later built rotating dynamometers to measure the work delivered by machines. The unit called the poncelet, used in France in the late nineteenth century, was 100 kilogram-force metres per second.

Exercise 10.1Show that the flat-blade wheel has maximal power at \(u=v/2\) with efficiency \(\tfrac12\), and verify that \(\eta_{\text{curved}}=1-\big(\tfrac{2u-v}{v}\big)^2\).

Show solution

\(\frac{d}{du}\,u(v-u)=v-2u=0\) gives \(u=v/2\) and \(\eta=2\cdot\frac v2\cdot\frac v2/v^2=\tfrac12\). For the curved blade, the absolute exit speed is \(u-(v-u)=2u-v\), so the kinetic energy lost per unit mass is \(\tfrac12(2u-v)^2\), and \(\eta=1-\frac{(2u-v)^2}{v^2}=\frac{4uv-4u^2}{v^2}\), matching \(4u(v-u)/v^2\). Half the flat-blade loss is the shock, which the curved blade removes.

Exercise 10.2Water falls through a head of \(h=1.5\) m to drive an undershot wheel, with a flow of \(0.5\ \mathrm{m^3/s}\). Ignoring losses in the channel, find the jet speed, the jet power, and the ideal outputs of the two wheel types, in kilowatts and in poncelets (\(g=9.81\ \mathrm{m/s^2}\)).

Show solution

\(v=\sqrt{2gh}=\sqrt{29.43}\approx5.42\) m/s. Mass flow \(500\) kg/s, so the jet power is \(\tfrac12\cdot500\cdot29.43\approx7.36\) kW. Flat blades, ideal: \(3.68\) kW. Curved blades, ideal: \(7.36\) kW. One poncelet is \(100\times9.80665=980.665\) W, so these are about \(3.75\) and \(7.5\) poncelets.

Exercise 10.3Compare the poncelet with the metric horsepower (75 kgf·m/s). Why might engineers have preferred a decimal unit?

Show solution

The metric horsepower is \(75\times9.80665\approx735.5\) W, so one poncelet is \(\tfrac43\) metric horsepower. The poncelet is a round multiple (100) of the metric unit of work per second then in use, which fits a decimal system; the horsepower kept a historical number from James Watt’s estimate of a horse’s output.

Part III · Life and legacy

Lesson 11

Paris, the Academy and 1848

From 1825 Poncelet was professor of mechanics applied to machines at the École d’application in Metz, a post François Arago had urged him to accept. He taught there for ten years and rose to chef de bataillon in 1831. His mechanics, more than his geometry, opened the doors of Paris. In 1834 the Académie des sciences elected him to its mechanics section, in succession to Hachette. He moved to the capital, served on the committee for the fortifications of Paris, and in 1837 the Faculty of Sciences created a chair of physical and experimental mechanics for him, which he held until 1848.

He married Louise Palmyre Gaudin in 1842. The same year the Royal Society of London elected him a Foreign Member. He continued to rise in the army: lieutenant-colonel in 1841, colonel in 1844.

The revolution of 1848

After the February revolution of 1848 he was promoted general of brigade in April and made commandant of the École polytechnique. He was elected to the Constituent Assembly. During the June Days uprising he led students to protect the seat of government, and he was given command of the National Guard of the Seine department. He retired from the army and from the school in 1850.

In 1851 he led the French scientific commission at the Great Exhibition in London and wrote long reports on machinery and the industrial uses of science. He also took part in the Paris Universal Exhibition of 1855.

DiscussionPoncelet was elected to the Academy for his mechanics, not for the geometry that made him famous. What does this say about how French science in the 1830s weighed pure and applied work, and about Cauchy’s influence on what counted as rigorous?

Lesson 12

Last years and legacy

In retirement Poncelet returned to geometry. He published the two volumes of Applications d’analyse et de géométrie (1862 and 1864), which reproduce the Saratov notebooks and his later reflections, with long polemical notes on priority. A second edition of the Traité followed in 1865–66. He received the Prussian order Pour le Mérite for sciences and arts in 1863 and was elected to academies in Berlin, St Petersburg and Turin. He died in Paris on 22 December 1867 after a long illness.

In 1868 his widow endowed the Prix Poncelet of the Académie des sciences, awarded for work in mathematics or mechanics. His name is one of the 72 engraved on the Eiffel Tower, and a lunar crater bears it. His unpublished manuscripts disappeared after the First World War.

Legacy

Projective geometry. The Traité launched the great nineteenth-century school of synthetic geometry: Chasles in France, Steiner and Karl von Staudt in Germany, and the analytic approach of Möbius and Plücker. Cayley and Klein later showed that Euclidean and non-Euclidean metrics are both projective geometry with a chosen absolute conic, extending Poncelet’s circular points.
Imaginary elements. His insistence that complex points belong to a figure made complex projective space the natural home of algebraic geometry.
Poncelet’s porism. The closure theorem connects to elliptic curves, integrable billiards, and dynamics. It is still an active research topic.
Engineering. Work as the measure of a machine’s output became standard in engineering education. The curved-blade wheel and the inward-flow turbine were steps toward modern hydraulic machines.

Final projectTake one theorem from Lessons 05–09. Prove it twice: once synthetically in Poncelet’s style, by projecting a special case (a circle, a parallelogram, a point at infinity), and once with homogeneous coordinates. Discuss which proof shows more about why the theorem is true.

Reference

Sources and further reading

Biography

J. J. O’Connor and E. F. Robertson, Jean-Victor Poncelet, MacTutor History of Mathematics.
Jean-Victor Poncelet, Wikipedia; French edition.
Jean-Victor Poncelet, Encyclopaedia Britannica.
École polytechnique, Jean-Victor Poncelet, 225 years of stories.
I. Didion, Notice sur la vie et les ouvrages du général J.-V. Poncelet, Paris, 1869.

Mathematics and mechanics

J.-V. Poncelet, Traité des propriétés projectives des figures, Paris, 1822; 2nd ed., 2 vols, Gauthier-Villars, 1865–66.
J.-V. Poncelet, Applications d’analyse et de géométrie, 2 vols, Paris, 1862–64.
J.-V. Poncelet, Introduction à la mécanique industrielle, physique ou expérimentale, Metz, 1829.
J. Gray, Worlds Out of Nothing: A Course in the History of Geometry in the 19th Century, Springer, 2007, chapters on Poncelet and duality.
L. Flatto, Poncelet’s Theorem, American Mathematical Society, 2009.
V. Dragović and M. Radnović, Poncelet Porisms and Beyond, Birkhäuser, 2011.
P. Griffiths and J. Harris, On Cayley’s explicit solution to Poncelet’s porism, L’Enseignement Mathématique 24 (1978).
H. S. M. Coxeter, Projective Geometry, 2nd ed., Springer, 1987.
A survey of the development of geometry up to 1870, arXiv:1409.1140.