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Jakob Steiner

Jakob (Jacob) Steiner

A farmer’s son who could not write until he was fourteen became the leading synthetic geometer of the nineteenth century. This course follows his road from Pestalozzi’s school to a Berlin chair, and works through the circles, conics, extremal figures and configurations that still carry his name.

Born 18 March 1796, Utzenstorf, Canton of Bern Died 1 April 1863, Bern Berlin chair of geometry created for him, 1834 Level advanced undergraduate
A Steiner chain of seven circles between two non-concentric circles. Steiner’s porism: if the chain closes for one starting circle, it closes for every one. The dots are the centres, which lie on an ellipse.
Lesson 00

How to use this course

Jakob Steiner believed that geometry should be done with figures and reasoning, without coordinates. He distrusted analysis, and he wanted to find the few principles from which the mass of known geometric theorems would follow. Many of his results are taught today in a different language, often with algebra he would have refused to use. This course states them in modern form and, where it helps, gives both the synthetic idea and an analytic check.

Prerequisites for Part II
  • Euclidean plane geometry: similar triangles, the inscribed angle theorem, the theorem of Ceva.
  • Analytic geometry of lines, circles and conics, and some linear algebra.
  • For Lesson 05, homogeneous coordinates and the cross-ratio. For Lesson 07, basic calculus. For Lesson 09, polynomials in several variables and Bézout’s theorem.

Each technical lesson begins with a paragraph for the general reader, then gives precise statements, proof sketches and exercises with folded solutions. The lessons run roughly in the order of his publications: circles (1826), projective forms (1832), constructions (1833), extremal problems (1838–1842), the Rome years and algebraic curves (1844–1848), and combinatorics (1853).

Reference

Timeline

1796Born 18 March in Utzenstorf, about 24 km north of Bern, the youngest of eight children of Niklaus Steiner and Anna Barbara Weber.
1814Spring: leaves home against his parents’ wishes to enter Johann Heinrich Pestalozzi’s institute at Yverdon. Within about eighteen months he teaches mathematics there.
1818Autumn: moves to Heidelberg, attends lectures and lives by private tutoring.
1821Moves to Berlin. Obtains only a restricted teaching licence.
1822Dismissed in the autumn from the Werder Gymnasium after a dispute over teaching methods.
1822–24Attends courses at the University of Berlin; friendship with Carl Gustav Jacob Jacobi.
1825Assistant master at the Berlin Gewerbeschule (technical school); senior master from 1829.
1826August Leopold Crelle founds his Journal für die reine und angewandte Mathematik. Steiner’s first papers appear in volume 1.
1832Systematische Entwickelung der Abhängigkeit geometrischer Gestalten von einander.
1833Die geometrischen Constructionen ausgeführt mittelst der geraden Linie und eines festen Kreises. Honorary doctorate from Königsberg on Jacobi’s recommendation.
18345 June: elected to the Berlin Academy of Sciences. 8 October: extraordinary professor of geometry at the University of Berlin, a post created for him.
1838Einfache Beweise der isoperimetrischen Hauptsätze in Crelle’s Journal.
1843–44Travels to Rome with Jacobi, Peter Gustav Lejeune Dirichlet and Ludwig Schläfli. Studies the surface now called the Steiner or Roman surface.
1848Work on general properties of algebraic curves; poses the problem of conics tangent to five conics.
1853Combinatorische Aufgabe: the question that leads to Steiner systems.
1854–55Winter in Paris; elected a corresponding member of the Académie des sciences.
1863Dies 1 April in Bern, aged 67, after years of kidney illness. Bequest founds the Steiner Prize of the Berlin Academy.
1867Vorlesungen über synthetische Geometrie, edited by Carl Friedrich Geiser and Heinrich Schröter.
1881–82Gesammelte Werke, edited by Karl Weierstrass for the Berlin Academy.
Part I · Formation
Lesson 01

Utzenstorf and Yverdon, 1796–1818

Jakob Steiner was born on 18 March 1796 in Utzenstorf, a village in the Canton of Bern. His parents, Niklaus Steiner and Anna Barbara Weber, kept a small farm and a business. Jakob was the youngest of their eight children. He helped with the family accounts and was good with numbers, but he had almost no schooling. He did not learn to read and write until he was about fourteen.

In the spring of 1814, at eighteen, he left home against his parents’ wishes and went to Yverdon, on Lake Neuchâtel, where Johann Heinrich Pestalozzi ran his famous institute. Pestalozzi taught through observation and questioning, letting children reach ideas by their own reasoning instead of by memorising rules. Steiner paid nothing for his schooling. He progressed so quickly that within about a year and a half he was teaching mathematics to other pupils.

Pestalozzi’s method marked Steiner for life. As a professor in Berlin he lectured on geometry without drawing figures, asking students to see the configurations in their minds, and he questioned them in something like a Socratic dialogue. His whole programme in geometry, to find the few principles from which the theorems grow, can be read as Pestalozzi’s pedagogy applied to a science.

Lesson 02

Heidelberg and Berlin, 1818–1826

In the autumn of 1818 Steiner moved to Heidelberg. He supported himself by private lessons and attended lectures on combinatorial analysis, calculus and algebra, notably those of Ferdinand Schweins. He found the teaching weak and learned mostly on his own. In these years he wrote several unpublished manuscripts on mechanics and geometry; his earliest papers and notes were edited much later by Fritz Bützberger.

In 1821 he moved to Berlin, again as a private tutor, and soon had a reputation as one of the best in the city. One of his pupils was a son of Wilhelm von Humboldt, a connection that would matter later. To teach in a state school he sat the Prussian examinations, but he received only a restricted licence: his mathematics was excellent, his history, literature and languages were not. He taught for a while at the Werder Gymnasium, quarrelled with its director over teaching methods, and was dismissed in the autumn of 1822.

From November 1822 to August 1824 he attended courses at the University of Berlin. There he became friends with Carl Gustav Jacob Jacobi, eight years younger, who would be his most effective supporter. In 1825 he became assistant master at the Berlin Gewerbeschule, a technical school directed by Karl Friedrich von Klöden, and in 1829 senior master. Relations with Klöden were poor. Among the school’s pupils in those years was the future novelist Theodor Fontane.

Lesson 03

Crelle’s Journal and the road to a chair, 1826–1834

Steiner’s career turned on a friendship with August Leopold Crelle, an engineer and civil servant who loved mathematics. In 1826 Crelle founded the Journal für die reine und angewandte Mathematik, encouraged by the talent of two young men in Berlin: Steiner and the Norwegian Niels Henrik Abel. The first volume contained four papers by Steiner. The first, Einige geometrische Betrachtungen, gave a systematic theory of the power of a point and the centres of similitude of circles (Lesson 04). Another found the maximum number of regions into which \(n\) planes divide space (Lesson 10). Over his life Steiner published about sixty papers in Crelle’s Journal.

In 1832 he gathered six years of ideas into his first book, the Systematische Entwickelung der Abhängigkeit geometrischer Gestalten von einander. Its preface announced the aim of uncovering the organism by which the phenomena of space are linked. The book built conics and quadric surfaces from projectively related pencils and ranges (Lesson 05). In 1833 a second book proved that every Euclidean construction can be made with a straightedge alone once a single circle and its centre are drawn (Lesson 06).

Jacobi, then professor at Königsberg, admired the Systematische Entwickelung and arranged an honorary doctorate for Steiner from Königsberg in 1833. Alexander and Wilhelm von Humboldt then pressed the Prussian ministry. On 5 June 1834 Steiner was elected to the Berlin Academy of Sciences, and on 8 October 1834 he became extraordinary professor of geometry at the University of Berlin, in a chair created for him. He never became an ordinary professor. Felix Klein later suggested that his lack of formal schooling, of Latin and of social polish all counted against him.

Part II · The mathematics
Lesson 04

Power of a point, similitude, inversion

For the general reader. Draw any line through a fixed point and let it cut a circle at two points. The product of the two distances from the point is the same for every line. Steiner made this number, the power of the point, into a tool, and with it organised the geometry of several circles at once.

Goals
  • Define the power of a point and prove its invariance.
  • Construct radical axes, the radical centre and the centres of similitude.
  • Use inversion to turn circles into lines.

Power and radical axis

Let \(\Gamma\) be a circle with centre \(O\) and radius \(r\). The power of a point \(P\) is \(\operatorname{pow}(P)=|PO|^2-r^2\). In coordinates, if \(\Gamma\) is \(x^2+y^2+ax+by+c=0\), then \(\operatorname{pow}(P)\) is the value of the left side at \(P\).

Theorem (power of a point)

If a line through \(P\) meets \(\Gamma\) at \(A\) and \(B\), then \(\overline{PA}\cdot\overline{PB}=\operatorname{pow}(P)\), with signed lengths along the line. In particular the product does not depend on the line.

Proof. Parametrise the line as \(P+t u\) with \(|u|=1\). Substituting into \(|X-O|^2=r^2\) gives \(t^2+2t\,u\cdot(P-O)+|P-O|^2-r^2=0\). The roots are the signed distances \(t_A,t_B\), and their product is the constant term \(|PO|^2-r^2\).

Theorem (radical axis and radical centre)

For two non-concentric circles, the set of points with equal power with respect to both is a line perpendicular to the line of centres. For three circles with non-collinear centres, the three radical axes meet in one point, the radical centre.

Subtracting the normalised equations of the two circles cancels \(x^2+y^2\) and leaves a linear equation. If \(P\) has equal power for circles 1 and 2 and for 2 and 3, it has equal power for 1 and 3, which proves concurrence.

Centres of similitude

Two circles with centres \(O_1,O_2\) and radii \(r_1\ne r_2\) are related by two homotheties: one with positive ratio \(r_2/r_1\), centred at the external centre of similitude \(E\), and one with negative ratio \(-r_2/r_1\), centred at the internal centre \(I\). As position vectors,

\[ E=\frac{r_2\,O_1-r_1\,O_2}{r_2-r_1},\qquad I=\frac{r_2\,O_1+r_1\,O_2}{r_1+r_2}. \]

For three circles, the six centres of similitude lie by threes on four lines; the collinearity of the three external centres is usually credited to Gaspard Monge. Steiner’s 1826 paper used such configurations systematically, together with the power of a point.

Inversion

Inversion in the circle of centre \(O\) and radius \(k\) sends \(X\ne O\) to the point \(X'\) on the ray \(OX\) with \(|OX|\cdot|OX'|=k^2\), that is \(X'=O+k^2(X-O)/|X-O|^2\). It is an involution, it preserves angles, and it maps circles and lines to circles and lines. A circle through \(O\) becomes a line not through \(O\), and conversely. Steiner used inversion in his 1826 paper; it is the key to Lesson 08.

Exercise 4.1Let \(\Gamma_1: x^2+y^2=1\) and \(\Gamma_2:(x-4)^2+y^2=4\). Find the radical axis and both centres of similitude.

Show solution
Equal powers: \(x^2+y^2-1=x^2-8x+16+y^2-4\), so \(8x=13\) and the radical axis is \(x=13/8\). With \(O_1=(0,0)\), \(O_2=(4,0)\), \(r_1=1\), \(r_2=2\): \(E=(2\cdot 0-1\cdot4)/(2-1)=(-4,0)\) and \(I=(0+4)/3=(4/3,0)\). Check: from \(E\) the distances to the centres are 4 and 8, in ratio \(1:2\); from \(I\) they are \(4/3\) and \(8/3\).

Exercise 4.2Show that inversion in the unit circle centred at the origin maps the line \(x=1/2\) onto the circle with centre \((1,0)\) through the origin.

Show solution
The image of \((x,y)\) is \((X,Y)=(x,y)/(x^2+y^2)\), and the map is its own inverse, so \(x=X/(X^2+Y^2)\). The condition \(x=1/2\) becomes \(X^2+Y^2=2X\), that is \((X-1)^2+Y^2=1\): the circle with centre \((1,0)\) and radius 1, which passes through the origin.

Exercise 4.3Two circles meet at \(A\) and \(B\). Show that their radical axis is the line \(AB\), and deduce that the common chords of three pairwise intersecting circles are concurrent.

Show solution
\(A\) and \(B\) lie on both circles, so both powers are 0 at each of them; hence both lie on the radical axis, which is therefore the line \(AB\). For three circles the common chords are the three radical axes, which meet at the radical centre (or are parallel when the centres are collinear).
Lesson 05

Projective generation of conics

For the general reader. Take two points, and turn a line about each of them according to a fixed rule. The two lines cross at a moving point. Steiner showed that when the rule is projective, the crossing point traces a conic, and that every conic arises this way. Conics, which the Greeks had defined by cutting a cone, became the simplest product of two related pencils of lines.

Goals
  • Define projectivities between pencils and the cross-ratio.
  • Prove Steiner’s theorem and its converse.
  • Recognise degenerate cases and the dual statement.

Pencils and projectivities

Work in the real or complex projective plane with homogeneous coordinates \(x=(x_0:x_1:x_2)\). The pencil of lines through a point \(P\) is a projective line: if \(a(x)=0\) and \(b(x)=0\) are two lines through \(P\), every line through \(P\) is \(a(x)+t\,b(x)=0\) for a unique \(t\in\mathbb P^1\). A projectivity between pencils is a map \(t\mapsto s=(\alpha t+\beta)/(\gamma t+\delta)\) with \(\alpha\delta-\beta\gamma\ne0\), equivalently one that preserves the cross-ratio of any four lines. A projectivity between the pencils at \(P\) and \(Q\) is a perspectivity if it maps the line \(PQ\) to itself.

Theorem (Steiner, 1832)

Let \(P\ne Q\) and let \(\varphi\) be a projectivity from the pencil at \(P\) to the pencil at \(Q\). If \(\varphi\) is not a perspectivity, the points \(\ell\cap\varphi(\ell)\) form a non-degenerate conic through \(P\) and \(Q\). Conversely, if \(C\) is a non-degenerate conic and \(P,Q\in C\), the map sending the line \(PX\) to the line \(QX\), for \(X\in C\), extends to a projectivity between the pencils that is not a perspectivity.

Proof

Write the lines through \(P\) as \(a+tb=0\) and those through \(Q\) as \(c+sd=0\). A projectivity is a bilinear relation \(p\,ts+q\,t+r\,s+u=0\) with \(pu-qr\ne0\). A point \(x\) lies on corresponding lines exactly when \(t=-a(x)/b(x)\) and \(s=-c(x)/d(x)\) satisfy this relation. Clearing denominators,

\[ p\,a(x)c(x)-q\,a(x)d(x)-r\,b(x)c(x)+u\,b(x)d(x)=0, \]

a quadratic equation in \(x\): a conic. It contains \(P\) (where \(a=b=0\)) and \(Q\) (where \(c=d=0\)). If \(\varphi\) maps \(PQ\) to itself, the line \(PQ\) is a component and the conic splits into two lines. Otherwise one checks the conic is irreducible. For the converse, the inscribed angle theorem is the special case of a circle: the angle between \(PX\) and \(QX\) is constant, so the map is a rotation of directions, which is projective. A projective transformation sends any non-degenerate conic to a circle and preserves cross-ratios, so the general case follows.

The dual statement, also in the 1832 book, says that the lines joining corresponding points of two projectively related ranges on two lines envelop a conic tangent to both lines. Steiner extended both statements to space, generating quadric surfaces from projectively related pencils of planes and ranges of points.

Exercise 5.1Let \(P=(0,0)\) and \(Q=(1,0)\). Pair the line \(y=tx\) through \(P\) with the line \(y=s(x-1)\) through \(Q\) by \(ts=-1\). Find the locus of intersections and explain the result geometrically.

Show solution
At an intersection point \(t=y/x\) and \(s=y/(x-1)\). Then \(ts=-1\) gives \(y^2=-x(x-1)\), that is \((x-\tfrac12)^2+y^2=\tfrac14\): the circle with diameter \(PQ\). Here \(ts=-1\) says the two lines are perpendicular, so this is Thales’ theorem as a case of Steiner’s.

Exercise 5.2With the same pencils, use the relation \(s=t\) (corresponding lines are parallel in the affine sense, but meet at infinity). What is the locus, and why is this projectivity a perspectivity?

Show solution
Lines \(y=tx\) and \(y=t(x-1)\) never meet in the affine plane: they meet at the point at infinity in direction \((1,t)\). The line \(PQ\) (\(t=0\)) maps to itself, so the projectivity is a perspectivity, and the degenerate conic is the line \(PQ\) together with the line at infinity. In the formula, \(p=0\), \(q=1\), \(r=-1\), \(u=0\) and \(a=y\), \(b=-x\), \(c=y\), \(d=1-x\) give \(-ad+bc=-y\). In homogeneous coordinates this is the quadratic \(y\,x_0=0\): the line \(y=0\) together with the line at infinity \(x_0=0\).
Lesson 06

Straightedge and one circle

For the general reader. Greek geometry used a straightedge and a pair of compasses. In 1797 Lorenzo Mascheroni showed the compasses alone suffice (Georg Mohr had found this in 1672). Jean-Victor Poncelet suggested in 1822 that the opposite also holds if one circle is drawn once and for all. Steiner proved it in 1833.

Goals
  • State the Poncelet–Steiner theorem and its algebraic meaning.
  • Carry out the key straightedge constructions: parallels, midpoints, perpendiculars.
  • Reduce line–circle intersections to the given circle.
Theorem (Poncelet–Steiner)

Suppose a circle \(\Gamma\) and its centre \(O\) are drawn in the plane. Then every point constructible from given points by straightedge and compass is constructible from the same points by straightedge alone, using \(\Gamma\) and \(O\).

Why something must be given

A straightedge alone can produce only intersections of lines through known points. In coordinates these are rational functions of the data, and every such construction commutes with projective transformations. A midpoint is not projectively invariant, so the straightedge cannot even bisect a segment. Compass constructions add square roots: the constructible points have coordinates in the smallest field containing the data and closed under \(\sqrt{\ }\) of positive elements. The theorem says that one circle with centre supplies exactly these square roots.

The construction in steps

  1. Parallels from a midpoint. If segment \(AB\) and its midpoint \(M\) are known, the straightedge can draw the parallel to \(AB\) through any point \(P\) (Exercise 6.1). Conversely, given two parallel lines it can bisect any segment on one of them.
  2. Parallels in general. Two diameters are the diagonals of a rectangle inscribed in \(\Gamma\), so they give two pairs of parallel sides. Through \(O\) draw the parallel to one pair (steps above); it is their midline. Any line \(\ell\) not parallel to that pair cuts the two sides at \(U,V\), and the midline cuts \(UV\) at its midpoint. With a known midpoint on \(\ell\), Exercise 6.1 draws the parallel to \(\ell\) through any point. If \(\ell\) is parallel to the first pair, use the second.
  3. Perpendiculars. Given a line \(\ell\), take any diameter \(CD\) of \(\Gamma\). Through \(C\) draw the parallel to \(\ell\); it meets \(\Gamma\) again at \(E\). By Thales \(DE\perp CE\), so \(DE\) is perpendicular to \(\ell\).
  4. Transferring lengths. Parallelograms move segments; radii of \(\Gamma\) rotated into any direction (via the perpendicular construction) let one lay off a given length along any line.
  5. Line meets circle. To intersect a line \(m\) with an undrawn circle of centre \(C\) and radius \(\rho\), construct the homothety that maps this circle to \(\Gamma\) (its centre of similitude is constructible with parallels), map \(m\) to a line \(m'\), intersect \(m'\) with \(\Gamma\), and map the points back.
  6. Circle meets circle. The intersection points of two circles lie on their radical axis, which is a line computable by rational operations and perpendiculars. This reduces the problem to step 5.

Hilbert later showed that the centre cannot be dropped: there are projective transformations that map \(\Gamma\) to itself but move \(O\), so no straightedge construction can locate the centre from the circle alone. Francesco Severi showed in 1904 that an arbitrarily small arc of \(\Gamma\), with its centre, is enough.

Exercise 6.1Given a segment \(AB\), its midpoint \(M\), and a point \(P\) off the line \(AB\), construct the parallel to \(AB\) through \(P\) with a straightedge, and prove the construction correct.

Show solution
Draw line \(AP\) and choose \(S\) on it beyond \(P\). Draw \(SM\) and \(BP\), meeting at \(X\). Draw \(AX\), meeting \(SB\) at \(Q\). Then \(PQ\parallel AB\). Proof: in triangle \(SAB\) the cevians \(SM\), \(AQ\), \(BP\) concur at \(X\), so by Ceva \(\frac{AM}{MB}\cdot\frac{BQ}{QS}\cdot\frac{SP}{PA}=1\). Since \(AM=MB\), \(\frac{SP}{PA}=\frac{SQ}{QB}\), and by the converse of the intercept theorem \(PQ\parallel AB\).

Exercise 6.2Given two parallel lines and a segment \(AB\) on one of them, bisect \(AB\) with a straightedge.

Show solution
Choose \(S\) not between the lines. Let \(SA\) and \(SB\) meet the other line at \(A'\) and \(B'\). The diagonals \(AB'\) and \(A'B\) of the trapezium meet at \(X\). The line \(SX\) meets \(AB\) at its midpoint. Reason: let \(M,M'\) be the midpoints of \(AB\) and \(A'B'\). The homothety at \(S\) maps \(A\mapsto A'\), \(B\mapsto B'\), hence \(M\mapsto M'\), so \(S,M,M'\) are collinear. The homothety at \(X\) with negative ratio maps \(A\mapsto B'\), \(B\mapsto A'\), hence again \(M\mapsto M'\), so \(X,M,M'\) are collinear. Thus \(M\) lies on \(SX\).

Exercise 6.3Only two points \(A\) and \(B\) are given, and auxiliary points may be chosen freely. Explain why no straightedge construction can always produce the midpoint of \(AB\).

Show solution
A straightedge construction uses only joins and intersections, so applying a projective transformation \(T\) to every point of a valid construction gives another valid construction, with auxiliary points \(T(P)\). Choose \(T\) fixing \(A\) and \(B\) and mapping the line \(AB\) to itself, acting on it as \(x\mapsto x/(2-x)\) in a coordinate with \(A=0\), \(B=1\). It fixes 0 and 1 but sends the midpoint \(\tfrac12\) to \(\tfrac13\). If a construction produced the midpoint \(M\), the transformed construction, starting from the same \(A\) and \(B\), would produce \(T(M)\ne M\). So the output depends on the auxiliary choices, and no construction always yields the midpoint.
Lesson 07

The isoperimetric problem and symmetrization

For the general reader. Of all closed curves of a given length, the circle encloses the greatest area. The Greeks knew this, but no one had proved it. In 1838 Steiner gave several beautifully simple arguments. Each showed how to improve any non-circular figure, and one introduced symmetrization, a tool still central in analysis. All of them shared one gap, which took decades to fill.

Goals
  • Follow Steiner’s four-hinge argument and locate the gap.
  • Define Steiner symmetrization and see why it decreases perimeter.
  • Use Steiner’s formula for parallel sets and the Steiner inellipse.
Theorem (isoperimetric inequality)

If a rectifiable simple closed plane curve of length \(L\) encloses a region of area \(A\), then \(4\pi A\le L^2\), with equality if and only if the curve is a circle.

Steiner’s argument

Suppose a curve \(K\) of length \(L\) encloses the largest possible area.

  1. \(K\) is convex. Otherwise reflect a dent across a supporting line: the length is unchanged and the area grows.
  2. A chord that halves the perimeter halves the area. If the chord \(AB\) splits \(K\) into two arcs of equal length but one side has more area, replace the smaller side by the mirror image of the larger. The length stays \(L\) and the area increases.
  3. Each half is a semicircle. Take such a chord \(AB\) and a point \(C\) on one arc. The region on that side consists of triangle \(ABC\) plus two caps attached to \(AC\) and \(CB\). Hinge the caps at \(C\) and slide \(A,B\) along the line: the caps and the arc length are unchanged, and the triangle area \(\tfrac12|CA||CB|\sin\angle C\) is largest when \(\angle C=90^\circ\). Maximality forces \(\angle ACB=90^\circ\) for every \(C\), so the arc is a semicircle on \(AB\).

The argument shows that if a maximal curve exists, it is a circle. It does not show that one exists. Oskar Perron later made the point with a parody: let \(N\) be the largest positive integer; if \(N>1\) then \(N^2>N\), a contradiction, so \(N=1\). Dirichlet is reported to have raised the existence question with Steiner himself. Karl Weierstrass, in his Berlin lectures on the calculus of variations, and later authors, including Hermann Amandus Schwarz in 1884 for the sphere, supplied the missing compactness arguments.

Symmetrization

Let \(K\) be a convex body and \(\ell\) a line. For each line \(m\perp\ell\) meeting \(K\), replace the chord \(m\cap K\) by the segment of the same length on \(m\) centred on \(\ell\). The result \(S_\ell K\) is the Steiner symmetral.

Proposition

\(S_\ell K\) is convex and symmetric in \(\ell\), has the same area as \(K\), and its perimeter is at most that of \(K\), with equality only if \(K\) is already symmetric about a line parallel to \(\ell\). The same holds in space for volume and surface area with symmetrization in a plane.

Area is preserved by the principle of Bonaventura Cavalieri: every chord keeps its length. For perimeter, write the upper and lower boundaries as graphs \(f(x)\ge g(x)\); the symmetral has boundary graphs \(\pm h(x)\) with \(h=(f-g)/2\). Since \(\sqrt{1+u^2}\) is convex, \(2\sqrt{1+((f'-g')/2)^2}\le\sqrt{1+f'^2}+\sqrt{1+g'^2}\). Repeated symmetrizations in suitable directions converge to a disc, a fact proved rigorously long after Steiner.

Parallel sets

Steiner also studied the region within distance \(t\) of a convex body. For a convex plane region \(K\) of area \(A\) and perimeter \(L\), the parallel region \(K_t=\{x:\operatorname{dist}(x,K)\le t\}\) has area

\[ A(K_t)=A+L\,t+\pi t^2 . \]

In space, for a convex body of volume \(V\) and surface area \(S\), \(V(K_t)=V+S\,t+M\,t^2+\tfrac43\pi t^3\), where \(M\) is an integral of mean curvature. This is the first instance of what are now called quermassintegrals and, via Hermann Minkowski, mixed volumes.

The Steiner inellipse

Among Steiner’s extremal results: of all ellipses inscribed in a triangle, the one of largest area touches the sides at their midpoints, and its area is \(\pi/(3\sqrt3)\) times that of the triangle. Since affine maps preserve ratios of areas, midpoints and tangency, it suffices to check the equilateral triangle, where the answer is the incircle.

Exercise 7.1Verify Steiner’s formula \(A(K_t)=A+Lt+\pi t^2\) for a square of side \(a\), and then use it to prove the formula for any convex polygon.

Show solution
For the square, \(K_t\) is the square, plus four rectangles \(a\times t\) on the sides, plus four quarter discs of radius \(t\) at the corners: \(a^2+4at+\pi t^2\), and \(L=4a\). For a convex polygon the rectangles on the sides contribute \(Lt\), and the circular sectors at the vertices have angles equal to the exterior angles, which sum to \(2\pi\), so together they form a full disc of area \(\pi t^2\).

Exercise 7.2Use Steiner’s formula to show that a convex region of perimeter \(L\) has area at most \(L^2/(4\pi)\), assuming Brunn–Minkowski in the form \(\sqrt{A(K_t)}\ge\sqrt{A}+\sqrt{\pi}\,t\).

Show solution
Squaring, \(A(K_t)\ge A+2\sqrt{\pi A}\,t+\pi t^2\). Comparing with \(A+Lt+\pi t^2\) for small \(t>0\) gives \(L\ge 2\sqrt{\pi A}\), that is \(4\pi A\le L^2\).

Exercise 7.3Check the inellipse ratio for an equilateral triangle, and find the area of the Steiner inellipse of the triangle with vertices \((0,0)\), \((4,0)\), \((0,3)\).

Show solution
For an equilateral triangle with inradius \(r\), the side is \(2\sqrt3\,r\) and the area is \(3\sqrt3\,r^2\). The incircle area is \(\pi r^2\), giving ratio \(\pi/(3\sqrt3)\approx0.605\). The right triangle has area 6, so its inellipse has area \(6\pi/(3\sqrt3)=2\pi/\sqrt3\approx3.63\).
Lesson 08

Steiner chains and the porism

For the general reader. Place a small circle inside a larger one. Now fit a ring of circles into the gap, each touching both and touching its neighbour. Usually the ring fails to close up. Steiner found that whether it closes does not depend on where you start: if one ring closes, every ring closes. Inversion makes the reason transparent.

Goals
  • Prove Steiner’s porism by inversion.
  • Compute the closing condition.
  • Find the locus of centres in a chain.
Lemma

Two disjoint circles, one inside the other, can be mapped by an inversion to two concentric circles.

Proof sketch. For nested disjoint circles the radical axis lies outside both, so every point \(X\) on it has the same positive power \(\pi(X)\) with respect to both. The circle centred at \(X\) with radius \(\sqrt{\pi(X)}\) is orthogonal to both given circles. These orthogonal circles form a pencil through two points \(F_1,F_2\) on the line of centres (the limiting points). Inverting in a circle centred at \(F_1\) sends every circle through \(F_1\) to a line, so the orthogonal circles become lines through a common point. The two given circles stay orthogonal to all these lines, so they become concentric circles centred at that common point.

Theorem (Steiner’s porism)

Let \(\Gamma_{\text{in}}\) lie inside \(\Gamma_{\text{out}}\), disjoint from it. Start with a circle \(C_1\) tangent to both, then choose \(C_2\) tangent to both and to \(C_1\), and so on. If \(C_{n+1}=C_1\) after the chain winds \(m\) times around \(\Gamma_{\text{in}}\), then the same holds for every starting circle \(C_1\).

Proof. Invert to concentric circles of radii \(r<R\). Inversion preserves tangency, so chains go to chains. In the concentric picture every chain circle has radius \((R-r)/2\) and centre on the circle of radius \((R+r)/2\). Consecutive tangent circles subtend an angle \(2\theta\) at the common centre, where

\[ \sin\theta=\frac{(R-r)/2}{(R+r)/2}=\frac{R-r}{R+r}. \]

The chain closes after \(n\) circles and \(m\) turns exactly when \(2n\theta=2\pi m\), that is \(\sin(m\pi/n)=(R-r)/(R+r)\). This condition does not mention \(C_1\), and rotating the concentric picture shows that every starting position closes or none does. Inverting back gives the theorem.

Steiner’s porism is the circle analogue of Poncelet’s closure theorem for polygons inscribed in one conic and circumscribed about another. In the same early papers Steiner gave an elegant construction of the Malfatti circles, three circles in a triangle each tangent to the other two and to two sides, stating it without proof.

Exercise 8.1For concentric circles of radii \(r<R\), find \(R/r\) so that a chain of 6 circles closes in one turn. What about 3 circles?

Show solution
For \(n=6\): \(\sin(\pi/6)=\tfrac12=(R-r)/(R+r)\), so \(R=3r\). For \(n=3\): \(\sin(\pi/3)=\sqrt3/2\), so \(R/r=(1+\sqrt3/2)/(1-\sqrt3/2)=(2+\sqrt3)/(2-\sqrt3)=(2+\sqrt3)^2=7+4\sqrt3\approx13.9\).

Exercise 8.2In a Steiner chain between non-concentric circles with centres \(O_1\) (outer, radius \(R\)) and \(O_2\) (inner, radius \(r\)), show that the centres of the chain circles lie on an ellipse with foci \(O_1,O_2\). What is its major axis?

Show solution
A chain circle with centre \(X\) and radius \(\rho\) touches the outer circle internally and the inner circle externally, so \(|XO_1|=R-\rho\) and \(|XO_2|=r+\rho\). Adding, \(|XO_1|+|XO_2|=R+r\), independent of \(\rho\). So \(X\) lies on the ellipse with foci \(O_1,O_2\) and major axis \(R+r\). The dots in the illustration at the top of this page lie on such an ellipse.
Lesson 09

Curves, surfaces and the conic problem

For the general reader. In the 1840s Steiner turned to curves and surfaces defined by higher-degree equations, still trying to reason without algebra. In Rome in 1844 he found a strange four-lobed surface that contains a conic in every tangent plane. In 1848 he asked how many conics touch five given conics, and gave an answer that was wrong in an instructive way.

Goals
  • Describe the Steiner (Roman) surface by equation and parametrisation.
  • Understand Steiner’s count 7776 and why the true number is 3264.
  • Compute simple characteristic numbers of conics.

The Roman surface

Map the sphere \(x^2+y^2+z^2=r^2\) to space by \((x,y,z)\mapsto(X,Y,Z)=(yz,\,zx,\,xy)\). Antipodal points have the same image, so the image is a model of the real projective plane, and it satisfies

\[ X^2Y^2+Y^2Z^2+Z^2X^2-r^2XYZ=0 . \]

This quartic surface is Steiner’s Roman surface. Its singular set consists of three double lines (segments of the coordinate axes) meeting in a triple point at the origin, and the double segments end in six pinch points. In projective terms the surface is a projection into 3-space of the Veronese surface, the image of \(\mathbb P^2\) under all quadratic monomials. Since plane sections of the Veronese surface come from conics of \(\mathbb P^2\), every plane section of the Roman surface is the image of a conic, and the tangent planes cut it in pairs of conics.

Algebraic curves

Steiner’s long memoir on general properties of algebraic curves (1848) studied polar curves, the curve now called the Steinerian, inflections and double tangents. He stated many results without proof, including facts about the 28 bitangents of a plane quartic; Luigi Cremona later supplied proofs. He also drew Ludwig Schläfli’s attention to the 27 lines on a cubic surface, recently found by Arthur Cayley and George Salmon, which led Schläfli to his double-six configuration.

The conic problem

A conic \(\sum_{i\le j}a_{ij}x_ix_j=0\) is a point of \(\mathbb P^5\). The conics tangent to a fixed smooth conic form a hypersurface of degree 6 in \(\mathbb P^5\). Steiner reasoned that five such hypersurfaces meet in \(6^5=7776\) points, by Bézout’s theorem. The flaw: every double line \(\ell^2\) is tangent to every conic, so all five hypersurfaces contain the 2-dimensional Veronese surface of double lines, and Bézout’s count includes an excess contribution from it.

Theorem (de Jonquières, Chasles 1864; rigorous: Fulton–MacPherson)

Five conics in general position in the complex projective plane are tangent to exactly 3264 smooth conics.

Ernest de Jonquières found 3264 around 1859 but did not publish it, out of respect for Steiner. Michel Chasles published it in 1864 with his method of characteristics (see the course on Chasles). A complete justification of the count required the intersection theory of the 1970s.

Exercise 9.1Verify that the image of the sphere \(x^2+y^2+z^2=r^2\) under \((yz,zx,xy)\) satisfies the quartic equation above.

Show solution
\(X^2Y^2=x^2y^2z^4\), \(Y^2Z^2=x^4y^2z^2\), \(Z^2X^2=x^2y^4z^2\). Their sum is \(x^2y^2z^2(x^2+y^2+z^2)=r^2x^2y^2z^2\). Also \(XYZ=x^2y^2z^2\). So the left side equals \(r^2x^2y^2z^2-r^2x^2y^2z^2=0\).

Exercise 9.2How many conics pass through four general points and are tangent to a general line?

Show solution
The conics through four points form a pencil \(C_\lambda=C_0+\lambda C_1\). Restricted to the line, \(C_\lambda\) is a binary quadratic form whose coefficients are linear in \(\lambda\). Tangency means its discriminant vanishes, a quadratic equation in \(\lambda\). So there are 2 such conics. (The full list of these characteristic numbers, for \(5-k\) points and \(k\) lines, is \(1,2,4,4,2,1\).)

Exercise 9.3Show that the double lines form a surface in \(\mathbb P^5\), and that every double line \(\ell^2\) satisfies the tangency condition with a given conic \(D\).

Show solution
Lines form a \(\mathbb P^2\), and \(\ell\mapsto\ell^2\) embeds it in \(\mathbb P^5\) as the Veronese surface, which has dimension 2. A conic \(C\) is tangent to \(D\) when \(C\cap D\) has a point of multiplicity at least 2. If \(C=\ell^2\), every point of \(\ell\cap D\) has multiplicity at least 2 in \(C\cap D\), so the condition always holds.
Lesson 10

Counting: regions, triple systems, trees

For the general reader. Some of Steiner’s shortest papers have had the longest lives. A page on cutting space by planes opened his first year in print. A two-page question in 1853 gave its name to a central object of combinatorial design. And a classic shortest-network problem now carries his name, although his own contribution to it was modest.

Goals
  • Derive the maximum number of regions cut by lines and planes.
  • Prove the divisibility condition for Steiner triple systems.
  • Solve small Steiner tree problems.

Dividing space

Theorem (Steiner, 1826)

\(n\) lines in general position divide the plane into \(\frac{n^2+n+2}{2}\) regions, and \(n\) planes in general position divide space into \(\frac{n^3+5n+6}{6}\) regions. These are the maxima.

The \(n\)-th plane meets the previous \(n-1\) planes in \(n-1\) lines in general position on it. These cut it into \(R_2(n-1)\) pieces, and each piece splits one old region in two. So \(R_3(n)=R_3(n-1)+R_2(n-1)\); the same reasoning gives \(R_2(n)=R_2(n-1)+n\).

Triple systems

A Steiner triple system on a set of \(v\) points is a collection of 3-element subsets (triples) such that every pair of points lies in exactly one triple. More generally a Steiner system \(S(t,k,v)\) is a family of \(k\)-subsets with every \(t\)-subset in exactly one. Thomas Kirkman had solved the existence question for triple systems in 1847, answering a problem posed by Wesley Woolhouse in 1844. Steiner, unaware of this, asked in his 1853 Combinatorische Aufgabe for which \(v\) such systems exist, and the systems came to bear his name.

Theorem (Kirkman, 1847)

A Steiner triple system on \(v\) points exists if and only if \(v\equiv1\) or \(3\pmod 6\).

The great Steiner systems \(S(5,6,12)\) and \(S(5,8,24)\), studied by Robert Carmichael and by Ernst Witt in the 1930s, have the Mathieu groups as automorphism groups and lead to the Golay codes and the Leech lattice.

The Steiner tree problem

Given points \(P_1,\dots,P_n\) in the plane, find the shortest network connecting them, allowing extra junction points. Steiner studied the problem of a single point minimising the sum of distances to several given points, which goes back to Pierre de Fermat and Evangelista Torricelli for three points. Richard Courant and Herbert Robbins attached Steiner’s name to the network problem in What is Mathematics? (1941). In an optimal network every extra junction has degree 3 with angles of \(120^\circ\).

Exercise 10.1Show that a Steiner triple system on \(v\) points requires \(v\equiv1\) or \(3\pmod6\). Exhibit one for \(v=7\).

Show solution
Fix a point \(x\). The other \(v-1\) points are paired off by the triples through \(x\), so \(v-1\) is even. Each triple covers 3 pairs and there are \(\binom v2\) pairs, so \(3\mid v(v-1)/2\), that is \(6\mid v(v-1)\). With \(v\) odd, \(v\equiv1,3,5\pmod 6\), and \(v\equiv5\) gives \(v(v-1)\equiv 20\equiv2\pmod 6\), excluded. For \(v=7\) take the lines of the Fano plane: \(\{1,2,4\},\{2,3,5\},\{3,4,6\},\{4,5,7\},\{5,6,1\},\{6,7,2\},\{7,1,3\}\).

Exercise 10.2Solve the recursion to obtain \(R_2(n)\) and \(R_3(n)\), and check \(R_3(4)\).

Show solution
\(R_2(0)=1\) and \(R_2(n)=R_2(n-1)+n\) give \(R_2(n)=1+n(n+1)/2=(n^2+n+2)/2\). Then \(R_3(n)=1+\sum_{k=0}^{n-1}R_2(k)=1+n+\tfrac12\sum_{k<n}(k^2+k)=1+n+\tfrac{(n-1)n(n+1)}{6}\), which simplifies to \((n^3+5n+6)/6\). For \(n=4\): \((64+20+6)/6=15\).

Exercise 10.3Find the shortest network joining the four corners of a unit square, and compare it with the shortest network using no extra points.

Show solution
Without extra points the best is three sides, length 3 (the diagonals give \(2\sqrt2\approx2.83\) but need the centre as a junction). The Steiner tree uses two junctions on the midline, each joined to two adjacent corners at \(120^\circ\). Each corner edge has length \(1/\sqrt3\) and the middle edge \(1-1/\sqrt3\), so the total is \(4/\sqrt3+1-1/\sqrt3=1+\sqrt3\approx2.732\).
Part III · Life and legacy
Lesson 11

The Berlin professor, 1834–1863

For almost thirty years Steiner taught geometry at the University of Berlin alongside Dirichlet and, later, Ernst Kummer and Weierstrass. His lectures were demanding. He rarely prepared them in detail, drew few figures, and asked students to think the configurations through. The British geometer Thomas Archer Hirst, who heard him in the 1850s, described a stout, anxious-looking man troubled by rheumatism. Others recalled a fiery temper, liberal politics and a bluntness that offended colleagues. Yet he gathered a loyal circle. His nephew Carl Friedrich Geiser, later professor at Zürich, and Heinrich Schröter edited his lectures; the student and later biographer Emil Lampe recorded his manner.

Steiner was the leading German voice of synthetic geometry, against the analytic method of Julius Plücker. Their relations were bad from the start. Steiner had just taken the Berlin position when Plücker spent 1833 in Berlin, and Plücker soon left for Halle and then Bonn. Historians have linked the conflict to competition for posts and to Crelle’s editorial favour, as well as to temperament. Steiner also fell out, in later years, with Jacobi and with Schläfli.

In 1843 Steiner travelled to Italy with Jacobi and Dirichlet. He took with him Ludwig Schläfli, a young Bernese schoolteacher he had met that year, as interpreter. The party spent about six months in Rome, where Steiner studied the surface of Lesson 09. In the winter of 1854–55 he was in Paris and was elected a corresponding member of the Académie des sciences.

Klein judged that Steiner’s creativity declined from about 1845, when he turned to higher algebraic curves. Synthetic methods were less effective there, and Steiner sometimes published results that the English algebraists had obtained by calculation, without saying so.

DiscussionSteiner refused analytic methods on principle, yet several of his best-known results (the conic count, the isoperimetric theorem) were incomplete precisely where analysis was needed. Was the synthetic programme a strength or a limitation? Use Lessons 05, 07 and 09.

Lesson 12

Death, the Steiner Prize, legacy

In his last decade Steiner suffered from kidney disease and spent much of each year in Switzerland, eventually unable to teach. He died in Bern on 1 April 1863, aged 67. He never married. He left a third of his considerable fortune to the Berlin Academy to found the Steiner Prize for geometry. The rest went to relatives and to the school at Utzenstorf, to help educate poor children.

The Steiner Prize honoured much of the best geometry of the following decades. Winners included Luigi Cremona (1866), Ludwig Schläfli (1870) for his work on the cubic surface, Rudolf Sturm, Henry John Stephen Smith, Georges Henri Halphen (1880) and Sigmund Gundelfinger.

Legacy

  • Publications: his Vorlesungen über synthetische Geometrie appeared in 1867, edited by Geiser and Schröter; a third edition by Rudolf Sturm followed in 1887–98. Weierstrass edited the Gesammelte Werke for the Academy in 1881–82.
  • Projective geometry: the projective generation of conics and quadrics became the standard foundation of synthetic projective geometry, continued by Karl von Staudt and taught for a century.
  • Analysis and convexity: symmetrization is a central tool in isoperimetric and rearrangement inequalities, and Steiner’s formula for parallel bodies began integral geometry.
  • Combinatorics and optimisation: Steiner systems, the Steiner tree problem and arrangements of hyperplanes.
  • Named results: the Steiner chain, Steiner surface, Steiner inellipse, Steiner–Lehmus theorem, Poncelet–Steiner theorem, the Steinerian and Steiner’s conic problem, among many others. His name is attached to more concepts than that of almost any other geometer.

Exercise 12.1The Steiner–Lehmus theorem states that a triangle with two equal internal angle bisectors is isosceles. Prove it.

Show solution
Let the bisectors from \(B\) and \(C\) meet the opposite sides at \(D\) and \(E\), with \(BD=CE\), and suppose \(\angle B>\angle C\). Triangles \(DBC\) and \(ECB\) share \(BC\) and have \(BD=CE\); since \(\angle DBC=\tfrac12\angle B>\tfrac12\angle C=\angle ECB\), we get \(CD>BE\). Complete the parallelogram \(BDFE\), so \(DF=BE\) and \(EF=BD=CE\). In isosceles triangle \(ECF\), \(\angle ECF=\angle EFC\). Now \(\angle DFC=\angle EFC-\tfrac12\angle B\) and \(\angle DCF=\angle ECF-\tfrac12\angle C\), so \(\angle DFC<\angle DCF\), hence \(CD<DF=BE\), a contradiction. By symmetry \(\angle B<\angle C\) is also impossible.

Final projectPick one of Steiner’s results from Part II. Write a two-page account that gives his synthetic argument, an analytic or modern proof, and an assessment of what each version shows that the other hides.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Jakob Steiner, MacTutor History of Mathematics.
  2. Jakob Steiner, Wikipedia; German article with further detail on his Berlin career.
  3. C. F. Geiser, Zur Erinnerung an Jakob Steiner (memorial address).
  4. E. Lampe, biographical writings on Jacob Steiner, cited in the German Wikipedia article.
  5. W. Eccarius, Der Gegensatz zwischen Julius Plücker und Jakob Steiner im Lichte ihrer Beziehungen zu August Leopold Crelle, Annals of Science 37 (1980).
  6. F. Klein, Vorlesungen über die Entwicklung der Mathematik im 19. Jahrhundert, vol. 1, Springer, 1926.
  7. J. J. O’Connor and E. F. Robertson, Ludwig Schläfli, MacTutor (the Rome journey).

Mathematics

  1. J. Steiner, Systematische Entwickelung der Abhängigkeit geometrischer Gestalten von einander, Berlin, 1832.
  2. J. Steiner, Die geometrischen Constructionen ausgeführt mittelst der geraden Linie und eines festen Kreises, Berlin, 1833.
  3. J. Steiner, Einfache Beweise der isoperimetrischen Hauptsätze, J. reine angew. Math. 18 (1838).
  4. J. Steiner, Combinatorische Aufgabe, J. reine angew. Math. 45 (1853).
  5. J. Steiner, Gesammelte Werke, ed. K. Weierstrass, 2 vols., Berlin, 1881–82.
  6. V. Blåsjö, The isoperimetric problem, American Mathematical Monthly 112 (2005).
  7. Article on the history of the isoperimetric problem, Notices of the AMS 64(9) (2017), p. 980.
  8. Steiner’s conic problem, Steiner system, Poncelet–Steiner theorem, Steiner surface, Wikipedia.
  9. D. Eisenbud and J. Harris, 3264 and All That: A Second Course in Algebraic Geometry, Cambridge, 2016.
  10. R. Courant and H. Robbins, What is Mathematics?, Oxford, 1941.
  11. H. S. M. Coxeter and S. L. Greitzer, Geometry Revisited, MAA, 1967 (power of a point, inversion, Steiner chains).