Part I, Chapter V — General Considerations on Algebraic Constructions

Part I, Chapter V (§§ 1–5; Ginn & Co., 1897, pp. 42–48)

§ 1. Paper Folding (Wiener and Sundara Row)

1. We now lay aside constructions restricted to straightedge and compasses. Before leaving elementary methods altogether, however, we mention a new and remarkably simple geometric procedure: paper folding. Hermann Wiener showed how by folding paper one may obtain the planar nets of all the regular polyhedra (see W. Dyck, Katalog der Münchener mathematischen Ausstellung von 1893, Nachtrag, p. 52). Singularly, around the very same time an Indian mathematician, T. Sundara Row of Madras, published a suggestive little treatise, Geometrical Exercises in Paper Folding (Madras: Addison & Co., 1893), developing the same idea systematically and showing how to construct by points curves such as the ellipse, the cissoid, and others.

§ 2. Solving Cubic and Biquadratic Equations by Conic Sections

2. Let us now inquire by what geometric instruments problems leading to equations of the third or higher degree can be solved, and in particular how the ancients succeeded in solving the duplication of the cube and the trisection of the angle.

The most natural auxiliaries—extensively cultivated in antiquity—are the conic sections. Expressed in modern analytic language, suppose we wish to solve graphically either the cubic equation

\[x^3 + a x^2 + b x + c = 0\]

or the biquadratic (quartic) equation

\[x^4 + a x^3 + b x^2 + c x + d = 0.\]

Setting $x^2 = y$ reduces the two equations respectively to

\[x y + a y + b x + c = 0 \qquad \text{and} \qquad y^2 + a x y + b y + c x + d = 0.\]

In both cases the roots $x$ are the abscissas of the intersections of two conics:

  • The fixed parabola $x^2 = y$ with vertical axis.

  • For the cubic (Fig. 10), the hyperbola $xy + ay + bx + c = 0$, whose asymptotes are parallel to the coordinate axes. Of the four intersections required by Bézout’s theorem, one lies at infinity on the $Y$-axis, leaving three finite intersections whose abscissas are the three roots of the cubic.

  • For the biquadratic (Fig. 11), the hyperbola $y^2 + axy + by + cx + d = 0$ retains the horizontal asymptote while its second asymptote is inclined; all four intersections with $x^2 = y$ now lie at finite distance.

(For the historical methods of the Greek geometers, see M. Cantor, Geschichte der Mathematik, Leipzig, 1894, 2nd ed.; H. G. Zeuthen, Die Kegelschnitte im Altertum, Copenhagen, 1886; and R. Baltzer, Analytische Geometrie, Leipzig, 1882.)

§ 3. The Cissoid of Diocles and the Delian Problem

3. Besides conic sections, the ancients invented higher algebraic curves expressly to solve solid problems, notably the cissoid and the conchoid.

The cissoid of Diocles (c. 150 B.C., Fig. 12) is traced from a circle of unit diameter tangent at $(1, 0)$ to the vertical line $x = 1$ and passing through the origin $O(0, 0)$. Through the vertex $O$ draw any ray at angle $\theta$ to the $X$-axis, meeting the circle at distance $\cos\theta$ and the vertical tangent $x = 1$ at distance $\frac{1}{\cos\theta}$, and lay off from $O$ along the ray the difference $r$ of those two segments:

\[r = \frac{1}{\cos\theta} - \cos\theta = \frac{\sin^2\theta}{\cos\theta}.\]

Passing to Cartesian coordinates ($x = r\cos\theta$, $y = r\sin\theta$) yields the cubic equation

\[(x^2 + y^2)x - y^2 = 0.\]

The cissoid is an algebraic curve of the third order, symmetric about the $X$-axis, with a cusp at the origin $O$, having the vertical line $x = 1$ as asymptote, and cutting the line at infinity in the two imaginary circular points.

Cissoid of Diocles: (x² + y²)x − y² = 0 Conchoid of Nicomedes: (x² + y²)(x − a)² = b²x² O x = 1 P (y/x = ∛2) O x = a
Classical Higher Algebraic Curves (Klein Ch. V, §§ 3–4, Figs. 12–13): Left — Cissoid of Diocles (x^2 + y^2)x - y^2 = 0 constructing the cube root of 2; Right — Conchoid of Nicomedes trisecting angle MOY with exact Bezout root count 8 - 2 - 2 - 1 = 3.

To solve the Delian problem by means of the cissoid, rewrite its equation in the form

\[\left(\frac{y}{x}\right)^3 = \frac{y}{1 - x}.\]

A ray $\frac{y}{x} = \lambda$ through the origin cuts off the ordinate $\lambda$ on the vertical asymptote $x = 1$, and meets the cissoid at a point satisfying $\frac{y}{1 - x} = \lambda^3$. Now $\frac{y}{1 - x} = \lambda^3$ is the equation of the straight line joining $(1, 0)$ to that cissoid point, which cuts off on the $Y$-axis ($x = 0$) the intercept $\lambda^3$! Reversing the steps constructs $\sqrt[3]{2}$ immediately:

  1. Lay off the intercept $2$ on the $Y$-axis at $(0, 2)$ and join $(0, 2)$ to $(1, 0)$.

  2. Mark the intersection $P$ of that segment with the cissoid.

  3. Draw the ray $OP$ from the origin to the vertical tangent $x = 1$: the ordinate cut off on $x = 1$ equals $\sqrt[3]{2}$.

§ 4. The Conchoid of Nicomedes and Angle Trisection

4. The conchoid of Nicomedes (c. 150 B.C., Fig. 13) is generated from a fixed pole $O(0, 0)$ at perpendicular distance $a$ from a fixed line $x = a$. Through $O$ pass a pencil of rays and lay off on each ray, in both directions from its intersection with $x = a$, a constant segment $b$. Depending on whether $b > a$, $b < a$, or $b = a$, the origin $O$ is a node (looped branch), an isolated conjugate point, or a cusp. From the proportion $r : x = \pm b : (x - a)$ we obtain the Cartesian equation

\[(x^2 + y^2)(x - a)^2 - b^2 x^2 = 0.\]

The conchoid is thus of the fourth order, has a double point at the origin $O$ and two branches asymptotic to $x = a$, and—by virtue of the factor $(x^2 + y^2)$—passes through the circular points at infinity.

To trisect any given angle $\varphi = \angle MOY$:

  1. On the inclined side $OM$, lay off an arbitrary length $OM = b$.

  2. With center $M$ and radius $b$, describe a circle, and through $M$ draw the vertical line $x = a$ ($a = b\sin\varphi$) perpendicular to the $X$-axis as asymptote of the conchoid with pole $O$ and interval $b$.

  3. Connect $O$ to the intersection $A$ of the circle and the outer branch of the conchoid: then $\angle AOY = \frac{1}{3}\varphi$! (Indeed, if $N$ is where $OA$ meets the vertical line $x = a$, then $NA = b = MA$ and $\triangle NMA$ is isosceles while $\angle OMN = 90^\circ$ gives the exterior-angle relation $\varphi = 3\,\angle AOY$.)

Algebraic Counting of Intersections: Our earlier Galois analysis showed that angle trisection is an irreducible problem of the third degree, admitting the three roots $\frac{\varphi}{3}, \frac{\varphi + 2\pi}{3}, \frac{\varphi + 4\pi}{3}$. Every algebraic curve construction must necessarily furnish all three roots simultaneously—otherwise the trisection cubic would split! Let us verify this in the figure:

  • The circle (degree $2$) and conchoid (degree $4$) intersect in $2 \times 4 = 8$ points.

  • Two intersections fall at the double point $O(0, 0)$, and two fall at the imaginary circular points at infinity—none of which solve our problem, leaving $8 - 2 - 2 = 4$ finite intersections.

  • Among these four finite points lies the collinear point $B$ on ray $OM$ extended ($OMB = 2b$), which is constructible rationally beforehand without the curve.

  • Removing $B$ leaves precisely $3$ intersections, corresponding to the three algebraic roots $\frac{\varphi}{3}, \frac{\varphi + 2\pi}{3}, \frac{\varphi + 4\pi}{3}$!

§ 5. Continuous Mechanical Instruments vs. Pointwise Approximation

5. Whenever higher algebraic curves are invoked for geometric construction, practical execution demands an instrument that traces the curve by a continuous movement; plotting isolated points and joining them freehand is merely an approximation method, not an exact construction. Nicomedes himself invented a simple mechanical trammel for tracing the conchoid—the oldest curve-drawing mechanism known after straightedge and compasses (M. Cantor, Vol. I, p. 302; compare Pappus, Collection, Book IV). Modern linkages and curve-tracers are catalogued in W. Dyck’s Katalog (pp. 227–230, 340; Nachtrag, pp. 42–43).