Solving the quartic by conics

Lorenz Halbeisen; Norbert Hungerbühler

doi:10.1515/math-2025-0132

Article Open Access

Solving the quartic by conics

Lorenz Halbeisen and Norbert Hungerbühler

Published/Copyright: April 22, 2025

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$Open Mathematics$

From the journal Open Mathematics Volume 23 Issue 1

Abstract

Two conic sections C 1 and C 2 in the Euclidean plane that pass through two given points can generally have two further points of intersection. It is shown how these can be constructed using compass and ruler. The idea is to construct the degenerate conics in the pencil of the two conics C 1 and C 2 . Their intersections are then the four intersection points of C 1 and C 2 . The same idea is then used to reduce a general quartic equation to a cubic equation and to solve it. This is performed by interpreting the solutions of the quartic as intersections of two complex conic sections. The degenerate complex conics in their pencil can then be found through a cubic equation.

Keywords: intersection of conics; pencils of conics; quartic equations; cubic equations; ruler and compass constructions

MSC 2010: 51M05; 51M15; 12D10

1 Introduction

We investigate the problem of finding the intersection points of two conic sections from a geometric and from an algebraic point of view. For the geometric solution, we work in the Euclidean plane, whereas for the algebraic solution, we work in the complex plane. Let us first have a look at the Euclidean plane where conics are given by quadratic equations in the variables x and y , in the form

(1) a x x x 2 + a x y x y + a y y y 2 + a x x + a y y + a = 0

with real coefficients. The resultant of two such quadratic equations with respect to x is a quartic equation in y , and the resultant with respect to y is a quartic equation in x . The coefficients of these two quartic equations are polynomials in the coefficients of the two quadratic equations. The solutions of the quartic in x are the x -coordinates of the intersections of the two conics. The corresponding y -coordinates are then the common solutions of the quadratic equations of the two conics for the given values of x . Conversely, we will see that we can interpret the solutions of a quartic equation in the variable x as x -coordinates of the intersections of two conics.

Alternatively, one can also argue as follows by considering the Euclidean plane as embedded in the projective plane. If a point P of one of two conics is known, one can apply a projective transformation to send P to a point at infinity, and we can therefore assume that one conic is a parabola. Note that, in general, the solution of a quadratic equation is required to find such a point P . We may then assume, by a suitable affine transformation, that one conic is the normal parabola y = x 2 , and the other conic is given by (1). If we then substitute y = x 2 into (1), we obtain a quartic equation in x .

This article is organized as follows. In Section 2, we present a construction of the intersections of two conics by ruler and compass, provided two intersections are known. In Section 3, we use the geometric idea to show how a quartic equation can be reduced to a cubic equation by considering the pencil of two associated conics.

2 Construction of the intersection of two conic sections

As discussed in Section 1, the problem of finding the intersection points of two conic sections corresponds to a quartic equation. It is therefore, in general, impossible to construct the intersections by ruler and compass. However, if two points of intersection are known, the problem reduces to a quadratic equation, and a construction should be feasible. In [1], such a construction was used to intersect a circle and a hyperbola in a special position. In this section, we want to present a construction, which works for two arbitrary conics for which two common points are known. We work in the framework of the real projective plane. The construction with ruler and compass takes place in the Euclidean plane which we consider as embedded in the projective plane. The compass will only be used in one single construction step.

Before we describe the construction, we will prove a lemma concerning quadruples of points forming harmonic ranges. Recall that four points A , B , X , and Y on a line in the real projective plane form a harmonic range, denoted ( A , B ; X , Y ) , if their cross-ratio satisfies cr ( A , B ; X , Y ) = − 1 .

Lemma

Let C 1 and C 2 be two conics that meet in the four points I 1 , I 2 , J 1 , and J 2 (Figure 1). Let r ≔ I 1 I 2 , let P 1 and Q 2 be the poles of the polar line r with respect to C 1 and C 2 , respectively, let p 2 be the polar line of P 1 with respect to C 2 , and let q 1 be the polar line of Q 2 with respect to C 1 . Furthermore, let s ≔ P 1 Q 2 ; R , T 1 , and T 2 be the intersection points of s with r , q 1 , and p 2 , respectively; and let A 1 , B 1 and A 2 , B 2 be the intersection points of s with C 1 and C 2 , respectively. Finally, let H 1 and H 2 be the intersection points of I 1 J 2 with I 2 J 1 , and I 1 J 1 with I 2 J 2 , respectively.

Then, we have the following:

The points A 1 , B 1 , A 2 , B 2 , H 1 , and H 2 are collinear.
cr ( P 1 , R ; A 1 , B 1 ) = cr ( Q 2 , T 1 ; A 1 , B 1 ) = − 1 .
cr ( P 1 , T 2 ; A 2 , B 2 ) = cr ( Q 2 , R ; A 2 , B 2 ) = − 1 .
cr ( A 1 , B 1 ; H 1 , H 2 ) = cr ( A 2 , B 2 ; H 1 , H 2 ) = − 1 .

$Figure 1 Illustration for the lemma. Using a projective mapping, we can assume that the conic sections C 1 and C 2 are in a symmetrical position.$

Figure 1

Illustration for the lemma. Using a projective mapping, we can assume that the conic sections C ₁ and C ₂ are in a symmetrical position.

Proof

By a projective transformation, we can assume that the four points I 1 , I 2 , J 1 , and J 2 form an isosceles trapezoid, where the lines I 1 I 2 and J 1 J 2 are parallel. In this situation, the entire configuration is mirror symmetric with respect to s . In particular, the lines r , p 2 , and q 1 are parallel (as in Figure 1).

(a) follows directly from the symmetry. (b), (c), and (d) follow from the following fact (see, e.g., [2, Section 6 ⋅ 4] or [3, Satz 4.9]): Let C be a conic, let P be a point not on C , let p be the polar line of P with respect to C , and let s be a line through P which intersects p at T and the conic C at the two points A and B . Then, the four points P , T , A , and B on s form a harmonic range ( P , T ; A , B ) .□

Now, we are ready to present a construction of the intersections of two conics C 1 and C 2 by ruler and compass, provided that the two conics have four intersection points and that two intersections I 1 and I 2 are known.

Construction

Since a conic is determined by five points, assume that C 1 and C 2 are given by the points I 1 , I 2 , C 1,1 , C 1,2 , C 1,3 and I 1 , I 2 , C 2,1 , C 2,2 , C 2,3 , respectively. Then, we construct the following points and lines.

Point P 1 : By Pascal’s theorem, we can construct by ruler alone the tangents to the conic C 1 at the points I 1 and I 2 (see Figure 2). The intersection of these two tangents is the point P 1 .
Point Q 2 : The point Q 2 is constructed as above as the intersection of the tangents in I 1 and I 2 with respect to the conic C 2 .
Line s : Joining the points P 1 and Q 2 , we obtain the line s .
Point R : Intersecting the line r = I 1 I 2 with s gives the point R .
Point T 2 : By Pascal’s theorem we can construct by ruler alone the intersection point I 1 ′ of C 2 with the line P 1 I 1 (see Figure 3). Then, T 2 is the intersection of I 2 I 1 ′ and s (see Figure 1 and [2, Section 6 ⋅ 4] or [3, Satz 4.10]). Note that I 1 ′ , T 2 , I 2 as well as I 2 ′ , T 2 , I 1 are collinear (see Figure 4 for the general situation).
Point T 1 : The point T 1 is constructed in the same way as T 2 , above, with the point Q 2 in place of P 1 and the conic C 1 in place of C 2 .
Points A 1 and B 1 : By item (b) of the aforementioned lemma, we have the harmonic ranges ( P 1 , R ; A 1 , B 1 ) and ( Q 2 , T 1 ; A 1 , B 1 ) . Hence, the points A 1 and B 1 are determined by the points P 1 , R , Q 2 , T 1 – which we have already constructed – and can be constructed using a compass and ruler (see Figure 5, and [3, p. 78]).
Points A 2 and B 2 : The points A 2 and B 2 are constructed as above with respect to the points P 1 , T 2 , Q 2 , and R and using item (c) of the lemma.
Points H 1 and H 2 : The points H 1 and H 2 are constructed as above with respect to the points A 1 , B 1 , A 2 , and B 2 and using item (d) of the lemma.
Points J 1 and J 2 : Finally, the other two intersection points J 1 and J 2 of C 1 and C 2 are obtained as follows: J 1 is the intersection of the lines I 1 H 2 and I 2 H 1 , and J 2 is the intersection of the lines I 1 H 1 and I 2 H 2 . Observe that the line pairs { I 1 J 2 , I 2 J 1 } and { I 1 J 1 , I 2 J 2 } are degenerate conics in the pencil generated by C 1 and C 2 .

$Figure 2 Construction of the tangent in I 1 {I}_{1} : X Y XY is the Pascal line in the hexagon I 1 I 2 C 1,1 C 1,2 C 1,3 I 1 {I}_{1}\hspace{0.16em}{I}_{2}\hspace{0.16em}{C}_{\mathrm{1,1}}\hspace{0.16em}{C}_{\mathrm{1,2}}\hspace{0.16em}{C}_{\mathrm{1,3}}\hspace{0.16em}{I}_{1} . Hence, Z I 1 Z{I}_{1} is the tangent to C 1 {C}_{1} in I 1 {I}_{1} .$

Figure 2

Construction of the tangent in I 1 : X Y is the Pascal line in the hexagon I 1 I 2 C 1,1 C 1,2 C 1,3 I 1 . Hence, Z I 1 is the tangent to C 1 in I 1 .

$Figure 3 With U 1 {U}_{1} and U 2 {U}_{2} , we can construct the Pascal line of the hexagon I 1 I 1 ′ I 2 C 2,1 C 2,2 C 2,3 {I}_{1}\hspace{0.16em}{I}_{1}^{^{\prime} }\hspace{0.16em}{I}_{2}\hspace{0.16em}{C}_{\mathrm{2,1}}\hspace{0.16em}{C}_{\mathrm{2,2}}\hspace{0.16em}{C}_{\mathrm{2,3}} . With the Pascal line and the line through C 2,2 {C}_{\mathrm{2,2}} and C 2,3 {C}_{\mathrm{2,3}} , we obtain U 3 {U}_{3} , and I 1 ′ {I}_{1}^{^{\prime} } is the intersection of U 3 I 2 {U}_{3}\hspace{0.16em}{I}_{2} and P 1 I 1 {P}_{1}\hspace{0.16em}{I}_{1} .$

Figure 3

With U 1 and U 2 , we can construct the Pascal line of the hexagon I 1 I 1 ′ I 2 C 2,1 C 2,2 C 2,3 . With the Pascal line and the line through C 2,2 and C 2,3 , we obtain U 3 , and I 1 ′ is the intersection of U 3 I 2 and P 1 I 1 .

$Figure 4 Line p p is the polar line with respect to the pole P P , and P I 1 P{I}_{1} and P I 2 P{I}_{2} are arbitrary secants.$

Figure 4

Line p is the polar line with respect to the pole P , and P I 1 and P I 2 are arbitrary secants.

$Figure 5 Construction of the points A 1 {A}_{1} and B 1 {B}_{1} . The circle c c has P 1 R {P}_{1}R as a diameter. The segments in T 1 {T}_{1} and Q 2 {Q}_{2} are orthogonal to s s . ( P 1 , R ; A 1 , B 1 ) \left({P}_{1},R;\hspace{0.33em}{A}_{1},{B}_{1}) is a harmonic range since A 1 {A}_{1} is a point of the polar of B 1 {B}_{1} with respect to c c , and ( Q 2 , T 1 ; A 1 , B 1 ) \left({Q}_{2},{T}_{1};\hspace{0.33em}{A}_{1},{B}_{1}) is a harmonic range because the points are part of the complete quadrangle given by the straight lines.$

Figure 5

Construction of the points A 1 and B 1 . The circle c has P 1 R as a diameter. The segments in T 1 and Q 2 are orthogonal to s . ( P 1 , R ; A 1 , B 1 ) is a harmonic range since A 1 is a point of the polar of B 1 with respect to c , and ( Q 2 , T 1 ; A 1 , B 1 ) is a harmonic range because the points are part of the complete quadrangle given by the straight lines.

3 Solving the quartic by conics

It is well known that a quartic equation in the variable z can be reduced to a cubic equation. The usual procedure is to first obtain rid of the cubic term in the quartic equation. This is performed by a substitution z = x − μ , leading to the depressed quartic in the variable x . Then, one can follow Ferrari’s solution, for example, which leads to a cubic equation (see, e.g., [4, §3.2], or [5] for a full deduction of the formulas). Another way is to employ Galois theory and factorization to reduce the quartic to a cubic problem (see, e.g., [6]). In the course of time, numerous renowned mathematicians have given solution methods for the cubic and quartic equations. Among them are Vieta, Descartes, van Schooten, Euler, Lagrange, Bézout, Lebesgue, and others (see [7] or [8] for an extensive overview of the known solution methods).

In this section, we want to show an alternative way to reduce the quartic to a cubic equation. The idea is inspired by the geometric construction in Section 2. To make this idea work, we will reformulate a given quartic equation as a system of two quadratic equations in two variables that represent two conics C 1 and C 2 . Observe that the pairs of straight lines I 1 J 1 , I 2 J 2 , and I 1 J 2 , I 2 J 1 that we used in the construction are degenerate conics in the pencil spanned by C 1 and C 2 . To compute these (in general complex) degenerate conics from the equations of C 1 and C 2 is indeed only a cubic problem. The intersection points of the straight lines, and thus the solutions of the original quartic, can then be calculated immediately. In terms of computational effort, this method is no better or worse than others. But since it is geometrically motivated, it can be visualized particularly well. Before we describe the general case, we illustrate the method with the following example.

Example

Suppose we want to solve the quartic

(2) z 4 + z 3 − 45 z 2 − 97 z + 140 = 0 .

To have it easier later on, we first remove the quadratic term in the equation. If we substitute z = x − μ , we find the coefficient 3 ( 2 μ 2 − μ − 15 ) for x 2 . This quadratic expression vanishes for μ = 3 . The resulting quartic equation in x is then

(3) x 4 − 11 x 3 + 92 x + 80 = 0 .

Now we consider the two conics

(4) C 1 : x 2 − y = 0 ,

(5) C 2 : y 2 − 11 x y + 92 x + 80 = 0 .

Clearly, if x and y is a solution of the system (4)–(5), then x solves (3), and vice versa, if x is a solution of (3), then x , y = x 2 is a solution of (4)–(5).

In what follows, it is convenient to consider the Euclidean plane as embedded in the real projective plane by

R 2 → P 2 ( R ) , x y ↦ x y 1 .

In this sense, we can write the first conic C 1 as ⟨ X , A X ⟩ = 0 , where X = ( x , y , 1 ) ⊤ , and A may, for convenience, be taken as

A = 2 0 0 0 0 − 1 0 − 1 0 ,

where ⟨ ⋅ , ⋅ ⟩ denotes the Euclidean inner product in R 3 . Similarly, the second conic C 2 is given by the equation ⟨ X , B X ⟩ = 0 with

B = 0 − 11 92 − 11 2 0 92 0 160 .

The pencil of the two conics, i.e., the conics whose equations are linear combinations of (4) and (5), is the set of conics given by equations of the form

⟨ X , ( λ A + μ B ) X ⟩ = 0 ,

for scalars λ and μ , not both zero. Observe that all conics of the pencil pass through the intersections of C 1 and C 2 . The idea is now to find the values of λ , and μ , not both zero, such that the matrix λ A + μ B is singular, and hence, the corresponding conics of the pencil degenerate to straight lines. Then, the intersections of the conics C 1 and C 2 can simply be computed by intersecting these lines, as indicated in Figure 6. In this case and later, A is nonsingular, so μ cannot be zero; it may be taken as 1, and λ may be required to satisfy det ( λ A + B ) = 0 .

The point is that this is a cubic equation for λ , which is in our example det ( λ A + B ) = − 2 λ 3 + 2664 λ − 36288 = 0 . Observe that there is no quadratic term, which makes it quite easy to find the roots in general. In our model case, we find the values λ 1 = 18 and λ 2 = 24 . For these values, the quadratic form

⟨ X , ( λ A + B ) X ⟩

factors in two straight lines as follows (see below for the general case):

⟨ X , ( 18 A + B ) X ⟩ = ⟨ X , g 1 ⟩ ⟨ X , h 1 ⟩ , ⟨ X , ( 24 A + B ) X ⟩ = ⟨ X , g 2 ⟩ ⟨ X , h 2 ⟩ ,

with

g 1 = 9 − 1 10 , h 1 = 4 − 2 16 , g 2 = 8 − 1 20 , h 2 = 6 − 2 8 .

The intersections of these lines can be computed by the respective cross-products. For readers who are less familiar with projective geometry, this can be explained briefly as follows: let, for example, g 1 and g 2 be normal vectors for planes through the origin in R 3 that represent the projective lines. Then, the common point of the projective lines is represented by the intersection of those planes, which is the line from the origin in R 3 perpendicular to both those normals, i.e., in the direction of their cross-product. Formally, this can also be seen by just noting that X = g 1 × g 2 is a solution of the equations ⟨ X , g 1 ⟩ = 0 and ⟨ X , g 2 ⟩ = 0 of the two lines. In our example, we find

g 1 × g 2 = − 10 100 1 , g 1 × h 2 = − 12 − 1 1 1 , h 1 × g 2 = 12 − 2 4 1 , h 1 × h 2 = 4 4 16 1 .

We read off the solutions x 1 = 10 , x 2 = − 1 , x 3 = − 2 , and x 4 = 4 of equation (3). Hence, the solutions of the original equation (2) are z 1 = 7 , z 2 = − 4 , z 3 = − 5 , and z 4 = 1 .

It is instructive to compare this geometrically motivated method with Ferrari’s classical solution. He first removes the cubic term by substituting z = x − μ . The coefficient of x 3 is then 1 − 4 μ . Hence, by choosing μ = 1 4 , we obtain the depressed equation

x 4 − 363 8 x 2 − 595 8 x + 41325 256 = 0 .

Ferrari rewrites this as

x 4 − 2 ⋅ 363 8 x 2 + 363 8 2 = − 363 8 x 2 + 363 8 2 + 595 8 x − 41325 256

such that the left-hand side is a square. To make also the right-hand side a square, Ferrari introduces the variable y to obtain

(6) x 2 − 363 8 + y 2 = − 363 8 + 2 y x 2 + 595 8 x + 363 8 2 − 41325 256 − 2 ⋅ 363 8 y + y 2 .

The right-hand side is a square, if the discriminant of the quadratic expression vanishes, i.e., if

595 8 2 = 4 − 363 8 + 2 y 363 8 2 − 41325 256 − 2 ⋅ 363 8 y + y 2 .

This is a cubic equation in y , and one of its solutions is y = 413 16 . With this choice, equation (6) becomes

x 2 − 313 16 2 = 24 4 x 2 + 595 8 x + 14161 64 = 1 8 ( 20 x + 119 ) 2 .

Taking roots on both sides, we obtain the two quadratic equations

x 2 − 313 16 = ± 1 8 ( 20 x + 119 ) .

For the plus sign, we find x 1 = − 19 4 and x 2 = 29 4 , and for the minus sign, we obtain x 3 = − 15 4 , x 4 = 5 4 . This leads to z 1 = − 5 , z 2 = 7 , z 3 = − 4 , and z 4 = 1 .

$Figure 6 The parabola C 1 {C}_{1} and the hyperbola C 2 {C}_{2} are shown in red. The degenerate conics in the pencil of C 1 {C}_{1} and C 2 {C}_{2} are the blue lines g 1 {g}_{1} and h 1 {h}_{1} and the green lines g 2 {g}_{2} and h 2 {h}_{2} .$

Figure 6

The parabola C 1 and the hyperbola C 2 are shown in red. The degenerate conics in the pencil of C 1 and C 2 are the blue lines g 1 and h 1 and the green lines g 2 and h 2 .

The general case of a quartic equation. Let us now solve the general quartic equation

a 0 + a 1 z + a 2 z 2 + a 3 z 3 + z 4 = 0 ,

by the geometrically inspired method. The coefficients are allowed to be complex numbers, and all subsequent computations are carried out in C , i.e., we work in P 2 ( C ) with embedded C 2 . Substituting z = x − μ yields for the quadratic term of x 2 the expression a 2 − 3 a 3 μ + 6 μ 2 . This is a quadratic equation, and we can choose one of its solutions to obtain a quartic equation without quadratic term. So, from now on, we assume that the quartic has the form

(7) a 0 + a 1 x + a 3 x 3 + x 4 = 0 .

Consider the two conics

(8) C 1 : x 2 − y = 0 ,

(9) C 2 : a 0 + a 1 x + a 3 x y + y 2 = 0 .

As in the aforementioned example, we have that if x and y are solutions of the system (8)–(9), then x solves (7), and vice versa, if x is a solution of (7), then x , y = x 2 is a solution of (8)–(9). As before, the first conic C 1 is given by ⟨ X , A X ⟩ = 0 , and the second conic C 2 by ⟨ X , B X ⟩ = 0 , where

A = 2 0 0 0 0 − 1 0 − 1 0 , B = 0 a 3 a 1 a 3 2 0 a 1 0 2 a 0 ,

with X = ( x , y , 1 ) ⊤ . The cubic equation to determine the degenerate conics in the pencil of C 1 and C 2 is

(10) − 1 2 det ( λ A + B ) = a 1 2 + a 0 a 3 2 ︸ ≕ p + ( a 1 a 3 − 4 a 0 ) ︸ ≕ q λ + λ 3 = 0 .

If p = 0 , then λ = 0 is a solution. This means that B is singular and the left-hand side of (9) is the product of two linear terms, namely, ( − a 1 ⁄ a 3 + a 3 x + y ) ( a 1 ⁄ a 3 + y ) if a 3 ≠ 0 and ( y + − a 0 ) ( y − − a 0 ) if a 3 = 0 . The problem is therefore reduced to determining the intersection points of the parabola C 1 and straight lines.

Let us assume now that p ≠ 0 . Then, as the product of the solutions of (10) is − p , they are all nonzero. If all were equal (with nonzero value λ 0 ), the coefficient of λ 2 in (10) would be − 3 λ 0 ≠ 0 , whereas it is 0; thus, there are at least two different solutions. The solutions can be found by Cardano’s formulas. However, these formulas are not easy to remember. We therefore follow a simple, memorizable solution method. Any nonzero κ ∈ C may be expressed as α + β , where α 3 + β 3 = − p and α ≠ 0 . Indeed, set β = κ − α . Then, α 3 + β 3 = κ ( κ 2 − 3 κ α + 3 α 2 ) ; one has a quadratic equation 3 α 2 − 3 κ α + κ 2 = − p ⁄ κ for α . If κ 3 = − p , the solutions are α = 0 and α = κ ; take α ≔ κ (so that α ≠ 0 ) and β ≔ 0 . If κ 3 ≠ − p , then the equation for α has nonzero solutions only, and either value gives κ = α + β (again with α ≠ 0 ) as desired.

Now, write the (nonzero) solution λ of (10) as α + β , where α 3 + β 3 = − p and α ≠ 0 . Then, (10) becomes

0 = p + q ( α + β ) + ( α + β ) 3 = ( α + β ) ( q + 3 α β ) = λ ( q + 3 α β ) ,

from which q + 3 α β = 0 (as λ ≠ 0 ) and β = − q ⁄ 3 α . But, therefore,

− p = α 3 + β 3 = α 3 − q 3 27 α 3 ,

which gives finitely many possible values for α , hence for β , and so for α + β ; among them will be the solutions for (10). Take two distinct solutions λ 1 and λ 2 .

For these values, each conic given by ⟨ X , ( λ j A + B ) X ⟩ = 0 is the union of two straight lines

(11) 0 = ⟨ X , ( λ j A + B ) X ⟩ = ⟨ X , g j ⟩ ⟨ X , h j ⟩ ,

say

g j = r j s j t j , h j = u j v j w j .

Writing out the matrix, we have

λ j A + B = 2 λ j a 3 a 1 a 3 2 − λ j a 1 − λ j 2 a 0 .

If we compare the coefficients in (11), we obtain

r j u j = 2 λ j , s j u j + r j v j = 2 a 3 , s j v j = 2 , t j u j + r j w j = 2 a 1 , t j w j = 2 a 0 , t j v j + s j w j = − 2 λ j .

The solution can be written as follows (plug in to check while keeping (10) in mind):

( u j , v j , w j ) = 1 2 ( 2 λ j , a 3 + μ j , a 1 + q j ν j ) , ( r j , s j , t j ) = 1 λ j ( 2 λ j , a 3 − μ j , a 1 − q j ν j ) .

Since we are dealing with projective lines, one can drop the factors 1 2 and 1 λ j later. Here,

μ j ≔ a 3 2 − 4 λ j and ν j ≔ a 1 2 − 4 a 0 λ j

are negative minors of λ j A + B . Depending on the choice of the square roots of μ 1 and ν j , the sign q j is

q j ≔ a 1 a 3 + 2 λ j 2 μ j ν j ∈ { − 1 , 1 } , if μ j ν j ≠ 0 .

Observe that

0 = − 2 λ j det ( λ j A + B ) = ( a 1 a 3 + 2 λ j 2 ) 2 − μ j ν j .

If μ j ν j = 0 , one can choose either q j = 1 or q j = − 1 . Now that the lines g j and h j are determined, their intersections and hence the solutions x j are easily computed by the respective cross-products, as in the example.

Of course, there are also other ways to find the factorization in (11).

4 The symmetric case

Let us assume that two conics C A and C B in the real ( x 1 , x 2 ) -plane are centrally symmetric with respect to a point P = ( p 1 , p 2 ) . We claim that in this case, the intersection points (real or complex) are given by quadratic equations. To see this, let C A and C B be given by equations ⟨ X , A X ⟩ = 0 and ⟨ X , B X ⟩ = 0 with matrices

A = a 11 a 12 a 13 a 12 a 22 a 23 a 13 a 23 a 33 and B = b 11 b 12 b 13 b 12 b 22 b 23 b 13 b 23 b 33 .

The coordinate transformation X = T X ˜ for

T = 1 0 p 1 0 1 p 2 0 0 1

corresponds to a translation of the coordinates in the ( x 1 , x 2 ) -plane. In the new coordinates, the conics are centrally symmetric with respect to the origin in the ( x ˜ 1 , x ˜ 2 ) -plane, and they are given by ⟨ X ˜ , A ˜ X ˜ ⟩ = 0 and ⟨ X ˜ , b ˜ X ˜ ⟩ = 0 , where

A ˜ = T ⊤ A T = a ˜ 11 a ˜ 12 0 a ˜ 12 a ˜ 22 0 0 0 a ˜ 33 and b ˜ = T ⊤ B T = b ˜ 11 b ˜ 12 0 b ˜ 12 b ˜ 22 0 0 0 b ˜ 33 .

The condition that a ˜ 13 = a ˜ 23 = 0 yields

p 1 = cof A 31 cof A 33 and p 2 = cof A 32 cof A 33 ,

where cof A i j denotes the cofactors of the matrix A , i.e.,

cof A 31 = a 12 a 23 − a 13 a 22 , cof A 23 = a 12 a 13 − a 11 a 23 , cof A 33 = a 11 a 22 − a 12 2 .

In other words, the two original conics C A and C B are centrally symmetric, if and only if

cof A 31 cof A 33 = cof B 31 cof B 33 and cof A 32 cof A 33 = cof B 32 cof B 33 .

The equations ⟨ X ˜ , A ˜ X ˜ ⟩ = 0 and ⟨ X ˜ , b ˜ X ˜ ⟩ = 0 are

(12) a ˜ 11 x ˜ 1 2 + 2 a ˜ 12 x ˜ 1 x ˜ 2 + a ˜ 22 x ˜ 2 2 + a ˜ 33 = 0 ,

(13) b ˜ 11 x ˜ 1 2 + 2 b ˜ 12 x ˜ 1 x ˜ 2 + b ˜ 22 x ˜ 2 2 + b ˜ 33 = 0 .

The linear combination b ˜ 33 (12)– a ˜ 33 (13) is

( b ˜ 33 a ˜ 11 − a ˜ 33 b ˜ 11 ) x ˜ 1 2 + 2 ( b ˜ 33 a ˜ 12 − a ˜ 33 b ˜ 12 ) x ˜ 1 x ˜ 2 + ( b ˜ 33 a ˜ 22 − a ˜ 33 b ˜ 22 ) x ˜ 2 2 = 0 ,

which clearly factors into two real or complex lines or a double line. Their intersection with the conic given by (12) or (13) corresponds to a quadratic problem, which yields the real or complex intersection of the two conics in ( x ˜ 1 , x ˜ 2 ) -coordinates, and by adding ( p 1 , p 2 ) in the original ( x 1 , x 2 ) -coordinates.

Since the intersection points in the centrally symmetric case are given by quadratic equations, it is possible to construct the intersections by ruler and compass. The reader is invited to find a construction, e.g., for the case of a circle and an ellipse. Note however, that the property “centrally symmetric” is not projectively invariant. In particular, the construction from Section 2 does not help in the symmetric case.

Acknowledgment

We are grateful to the referees for the valuable comments and suggestions that helped improve this article.

Funding information: The authors state no funding involved.
Author contributions: Both authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript. Both authors contributed equally to the article.
Conflict of interest: The authors state no conflicts of interest.

References

[1] N. Hungerbühler and J. Läuchli, Construction of a quadrilateral from its vertex angles and the diagonal angle, Elem. Math. 79 (2024), no. 4, 159–166, DOI: https://doi.org/10.4171/em/517. 10.4171/em/517Search in Google Scholar

[2] H. S. M. Coxeter, The Real Projective Plane, Springer-Verlag, New York, 1993. 10.1007/978-1-4613-9281-1Search in Google Scholar

[3] L. Halbeisen, N. Hungerbühler, and J. Läuchli, Mit Harmonischen Verhältnissen zu Kegelschnitten, Perlen Der Klassischen Geometrie, Springer Spektrum, Berlin, 2021. 10.1007/978-3-662-63330-4Search in Google Scholar

[4] J.-P. Tignol, Galois’ Theory of Algebraic Equations, World Scientific Publishing Co. Pte. Ltd., Hackensack, NJ, 2016. 10.1142/9719Search in Google Scholar

[5] M. Merriman, The deduction of final formulas for the algebraic solution of the quartic equation, Amer. J. Math. 14 (1892), no. 3, 237–245. 10.2307/2369666Search in Google Scholar

[6] I. Stewart, Galois Theory, CRC Press, Boca Raton, FL, 2015. Search in Google Scholar

[7] L. Matthiessen, Grundzüge Der Antiken und Modernen Algebra Der Litteralen Gleichungen, B. G. Teubner, Leipzig, 1878. Search in Google Scholar

[8] W. S. Burnside and A.W. Panton, The Theory of Equations: With an Introduction to the Theory of Binary Algebraic Forms, Dover Publications, Inc., New York, 1960. Search in Google Scholar

Received: 2024-08-08

Revised: 2024-12-19

Accepted: 2025-01-23

Published Online: 2025-04-22

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2025-0132

Keywords for this article

intersection of conics; pencils of conics; quartic equations; cubic equations; ruler and compass constructions

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