## Lecture Notes 20: Pascal's Theorem

In this section we shall use the following result which is a consequence of the Fundamental Theorem of Projective Geometry (which we will prove in a later section).

Theorem 20.1: In a Desarguesian projective plane, PG(2,F), any four points no three of which are collinear, can be mapped to any other four points, no three collinear. In particular, there is a collineation mapping any four points, no three collinear, to the points (1,0,0), (0,1,0), (0,0,1) and (1,1,1).

As we shall prove below, any five points in a Desarguesian plane PG(2,F) lie on a quadric. In the degenerate cases, there may be more than one quadric containing the points, however we have:

Theorem 20.2: In a Desarguesian plane, PG(2,F), any five points, no three of which are collinear lie on a unique conic.

Pf: By Theorem 20.1, we may assume wlog that four of the points are (1,0,0), (0,1,0), (0,0,1), and (1,1,1). The fifth point is not on any line determined by two of these. Its coordinates are of the form (a,b,1) with neither a nor b being 0 or 1, and a not equal to b. Consider the general quadric with equation Ax2 + Bxy + Cy2 + Dxz + Eyz + Fz2 = 0. Since the first four of our points must satisfy this equation, we see that A = C = F = 0 and B + D + E = 0. Since (a,b,1) is to satisfy this equation, we have Bab + Da + Eb = 0. Subtract ab times the first equation from the second to obtain, D(a - ab) + E(b - ab) = 0. So, D = -E(b-ab)/(a-ab). [Note that the denominator is not zero since a is not 0 and b is not 1] From the first equation we then have that B = -D - E = E(b-a)/(a-ab). So, the equation of the quadric is thus, E(b-a)/(a-ab)xy + E(ab-b)/(a-ab)xz + E yz = 0. Now, if E is not 0 we may divide by it to obtain the quadric with equation:

[(b-a)/(a-ab)]xy + [(ab-b)/(a-ab)]xz + yz = 0.
Consider the quadrics with equations Bxy + Dxz + Eyz = 0. If any of the coefficients is 0 then the quadric is degenerate since the quadratic form factors into two linear factors. However, no degenerate quadric can contain 5 points no three collinear, so none of these coefficients can be zero and we have that the 5 points lie on a unique conic.

Corollary 20.3: In a Desarguesian plane, PG(2,F), any five points lie on at least one quadric.

Pf: We will argue based on the maximum number of collinear points in the set of five. If all five points are collinear, then the only degenerate quadrics that contain them must contain lines. Thus, either a pair of lines or a repeated line where one of the lines of the quadric is the line containing the 5 points contains the set. If only 4 points are collinear, then the repeated line is no longer a possibility, and the second line of the pair can be any line through the 5th point. If there are 3, but no more, collinear points, the pair of lines containing the set is uniquely determined - the line containing the three collinear points and the line determined by the other two points. Finally, if no three points are collinear the quadric is nondegenerate and a unique conic passes through the five points by Theorem 20.2.

The next theorem, known as Pascal's Theorem, gives necessary and sufficient conditions for six points to lie on a conic.

Theorem 20.4 [Pascal's Theorem]: In a Desarguesian plane PG(2,F), let A,A',B,B',C and C' be six points no three of which are collinear. Let R be the intersection of lines AB' and A'B, Q the intersection of AC' and CA' and P the intersection of BC' and CB'. These six points lie on a conic if and only if the points P,Q and R are collinear.

Pf: By Theorem 20.1, we can assign coordinates A = (1,0,0), A' = (0,1,0), B = (0,0,1) and B' = (1,1,1). Points C = (a,b,1) and C' = (c,d,1) are such that none of a,b,c,or d is 0 or 1, a is not equal to b or c, and d is not equal to c or b, by the assumption that no three of the points are collinear. The line AB' has equation y = 1 and the line A'B has equation x = 0, so R = (0,1,1). The line AC' has equation y = d and the line A'C has equation x = a, so Q = (a,d,1). The line BC' has homogeneous equation -dx + cy = 0, and the line B'C has homogeneous equation (1-b)x + (a-1)y + (b-a)z = 0, so the point P has coordinates (c(b-a), d(b-a), d(1-a)+c(b-1) ). The condition that P, Q and R are collinear, using the determinant method, is that c(b-a) + a(d(b-a)) - dc(b-a) -a(d(1-a)+c(b-1)) = 0, which simplifies to ad(b+c-1) -bc(a+d-1) = 0.
The equation for the conic through the points A,A',B,B',C (from the proof of Theorem 20.2) is [(b-a)/(a-ab)]xy + [(ab-b)/(a-ab)]xz + yz = 0. Thus, C' is on this conic if and only if [(b-a)/(a-ab)]cd + [(ab-b)/(a-ab)]c + d = 0. This equation simplifies to ad(b+c-1) - bc(a+d-1) = 0. So, we see that C' is on the conic through the other five points if and only if P,Q and R are collinear.

Notice that if we relax the condition on the 6 points to no four are collinear, then a special case of Pascal's Theorem would deal with the degenerate quadric consisting of two lines having three points on each. This is of course Pappus' theorem.

Pascal's theorem was first proved in the real plane setting by Blaise Pascal (1623-1662) when he was 16 years old. The line PQR of the theorem is called a Pascal line. Given 6 points on a conic, there are 60 distinct ways to label the points, thus giving rise to 60 Pascal lines. The configuration of these 60 lines is called the Hexagrammum Mysticum and has many interesting properties.

One may view the 6 points of the theorem as the vertices of an ordered hexagon inscribed on a conic. The points P,Q, and R are then the intersections of the opposite sides of this hexagon. As Desarguesian planes are self-dual, the dual of Pascal's theorem is also valid. This dual is known as Brianchon's theorem and may be stated as:

Theorem 20.5 [Brianchon's Theorem]: Given six lines, no three concurrent, forming the sides of a hexagon, then the hexagon circumscribes a conic (that is, the sides are all tangents to a conic) if and only if the lines joining opposite vertices of the hexagon are concurrent.