Pencils of Cubics

Hubert Shutrick

The general cubic form in three variables $x$, $y$, $z$ is

$a_{1} x^{3} + a_{2} x^{2}y + a_{3}x^{2}z + a_{4} xy^{2} + a_{5} xyz + a_{6} yz^{2} + a_{7} y^{3} + a_{8}y^{2}z + a_{9} yz^{2} + a_{10}z^{3}$

and a cubic curve is defined in homogeneous coordinates $(x, y, z)$ by equating a cubic form to $0.$ If the cubic is required to pass through $n$ given points, then the coordinates of the points can be inserted in the general form to give $n$ homogeneous linear equations for the ten coefficients. In general, nine points will determine the coefficients up to a multiplicative constant and so determine the cubic. When only eight points are given, they will, in general, define a one parameter family of cubics, a pencil, and any cubic that satisfies the equations can then be written

$s (a_{1} x^{3} + ... + a_{10} z^{3}) + t (a_{1} x^{3} + ... + a_{10} z^{3}) = 0.$

where $s$ and $t$ are not both zero and the cubic forms in the parentheses represent any pair of (independent) solutions.

Note that if a cubic form factorizes, the cubic degenerates as a line and a conic or as three lines.

If four of the eight points lie on a line, then the cubic must degenerate into the line and a, possibly degenerate, conic. To prove this assume that the line is $x + l = 0$ where $l$ is a homogeneous linear form in $y$ and $z$ with a change of coordinates if necessary. The cubic form $c$ can be considered as a polynomial in $x$ whose coefficients are homogeneous polynomials in $y$ and $z$ and the division algorithm gives that it can be written

$c = (x + l) q + r,$

where $r = r(x, y, z)$ is a homogeneous polynomial of degree three in $y$ and $z$. Inserting the four points gives that $r$ has four roots and so is identically zero. A consequence of a theorem by Bézout is that, if seven of the eight points lie on a non-degenerate conic, then the cubic form factorizes. Without appealing to Bézout, it can be proved as follows using Euclid's algorithm. Consider the cubic form as

$c = c_{0}x^{3} + c_{1}x^{2} + c_{2}x + c_{3}$

and the quadratic form as

$q = x^{2} + q_{1}x + q_{2}$

where the coefficients are homogeneous polynomials in $y$ and $z$ whose degree is given by the subscript.

$ c = (c_{0} x + b_{1}) q + ( r_{2} x + r_{3})\\ r_{2} q = (r_{2} x + r_{3})(s_{0} x + s_{1}) + (t_{3} x + t_{4})\\ r_{2}(t_{3} x + t_{4}) - t_{3}(r_{2} x + r_{3}) = r_{2}t_{4} - t_{3}r_{3}. $

Substituting the second equation in the third and then the first in the result gives

$uq - (r_{2}s_{0} x + r_{2}s_{1}) c = r_{2}t_{4} - t_{3}r_{3}.$

Inserting the points gives that the homogeneous polynomial of degree six in $y$ and $z$ has seven roots and so must be identically zero and, since the conic is non-degenerate, its form must divide that of the cubic.

Consequently, the cases where we get pencils that include non-degenerate cubics are given by the following theorem, which is sufficient for the proofs of Pascal's hexagon and the pivot theorem below.


The cubics that pass through eight points form a pencil if no four points lie on a line and no seven on a conic.

Proof. The method of proof is by induction on the number of points, that is, if $n - 1$ points give independent linear equations, then we show that $n$ must do also. The induction is attained by showing that there is a cubic through the $n - 1$ points that doesn't pass through the $n^{th}.$ In each case, the cubic will be degenerate.

  • $n = 2$: the one point gives a single equation to start the induction and a line through the point but not the second and two lines that don't pass through the second point is the required cubic.

  • $n = 3$: three lines none through the third point but two go through one of the first two.

  • $n = 4$: three lines one through each of the three points but none through the fourth.

  • $n = 5$: since no four of the five points are collinear, there are at least two distinct pairs of the four that are not collinear with the fifth so two lines through those pairs and a line not through the fifth works.

  • $n = 6$: if three of the six points are collinear, take the line through them and a line each through the other two points that doesn't pass through the sixth; otherwise take two lines though pairs of points and a line and a line through the fifth that doesn't pass through the sixth.

  • $n = 7$: if no three of the seven points are collinear, then three lines each through two of the six will work; if three are collinear, take the line through them, a line through two others and, if that line doesn't pass through another of the points, a line through the sixth not containing the seventh, otherwise any that doesn't contain the seventh.

  • $n = 8$: if six points lie on a conic take the conic along with a line through the seventh that doesn't contain the eighth; if there aren't six points on a conic, then there can't be more than one set of three collinear points, so a line through the three and lines through the two other pairs will do; if no six points are on a conic and no three on a line, take a conic through five of the seven and a line through the other two.

We now apply the theorem just proved to two well known results. The cases that will be used in the proofs will be when the two cubics defining the pencil are degenerate: they consist either of three lines or a line and a conic.

Pascal's Theorem

For a hexagon inscribed in a conic, the intersections of the three pairs of opposite sides are collinear.

Proof. Let the hexagon be $ABCDEF$ and let $AB$ meet $DE$ in $P,$ $BC$ meet $EF$ in $Q$ and $CD$ meet $FA$ in $R.$ Consider the two degenerate cubics that consist of the three lines $AB, CD, EF$ and $BC, DE, FA.$ Each of the lines contain just three of the eight points $A, B, C, D, E, F, P, Q$ so the two cubics determine a pencil. Another degenerate cubic in the pencil consists of the line $PQ$ and the conic through the other six points. Since the two base cubics pass through $R$ and the conic doesn't, the line $PQ$ must also pass through it.

Pivot Theorem

If $A',B', C'$ are points on the sides $BC, CA, AB,$ respectively, of a triangle $ABC,$ then the circles through $AB'C', BC'A', CA'B'$ are concurrent. Moreover, if the points $A',B',C'$ are collinear, the circumcircle of $ABC$ passes through the same point.

Proof. The equation of a circle in homogeneous coordinates is of the form

$x^{2} + y^{2} + axz + byz + z^{2} = 0.$

so all circles intersect the line at infinity $z = 0$ in the imaginary points $(1, i, 0)$ and $(1, -i, 0)$, the eight points in this proof are the six in statement of the theorem along with these two circular points at infinity. As the two base cubics in the pencil, take the circle through $AB'C'$ with the line $BC$ and the circle through $BC'A'$ with the line $AC.$ Since the cubic consisting of the circle through $CA'B'$ with the line $AB$ is in the pencil its circle must pass through the other intersection point of the other two circles.

If $A',$ $B'$ and $C'$ are collinear, the cubic consisting of the circumcircle of $ABC$ and the line $A'B'C'$ is also in the pencil so the circumcircle must also pass through the extra point.


  1. K. Kendig, Conics, MAA, 2005

Pascal and Brianchon Theorems

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