A very simple proof of Pascal s hexagon theorem and some applications

Proc. Indian Acad. Sci. (Math. Sci.) Vol. 120, No. 5, November 2010, pp. 619 629. Indian Academy of Sciences A very simple proof of Pascal s hexagon theorem and some applications NEDELJKO STEFANOVIĆ and MILOŠ MILOŠEVIĆ Group for Intelligent Systems, Faculty of Mathematics, University of Belgrade, Studentski trg 16, 11000 Beograd, Serbia Serbian Academy of Sciences and Arts, Mathematical Institute, Kneza Mihaila 36, 11001 Beograd, Serbia E-mail: stenedjo@gmail.com; mionamil@eunet.rs MS received 3 February 2010; revised 28 August 2010 Abstract. In this article we present a simple and elegant algebraic proof of Pascal s hexagon theorem which requires only knowledge of basics on conic sections without theory of projective transformations. Also, we provide an efficient algorithm for finding an equation of the conic containing five given points and a criterion for verification whether a set of points is a subset of the conic. Keywords. Pascal s hexagon theorem; conic sections; algebraic curves. 1. Introduction In this article we present a simple and elegant algebraic proof of Pascal s hexagon theorem which requires only the knowledge of the basics on conic sections without using any results from the theory of projective transformations. The idea of the proof is very simple and natural. These are the main advances compared to the proof given in [4]. Also, Pascal s theorem is a corollary of Bezout s theorem for algebraic curves (see [3]). Bezout s theorem is a somewhat deeper result (see [1 3]), while our approach is comprehensible to anyone knowing the foundations of linear algebra and its applications on conic sections. In order to make this paper self-contained as much as possible, in the next two subsections we prove a theorem about intersections of two conics and introduce the notion of homogenous coordinates. The final subsection contains the proofs of Pascal s theorem and its converse, and we give the number of operations which have to be performed to find coefficients of a conic containing five given points using eq. (2), a criterion for verification whether six points lie on the same conic and illustrate this criterion. The applications announced in the title are based more on the ideas used in our proof of Pascal s theorem, than on the statement of that theorem. 2. Intersections of two conics It is important to emphasize that by the conic we mean both non-degenerate and degenerate cases. We remind the reader that matrix of the conic section given by the equation 619

620 Nedeljko Stefanović and Miloš Milošević Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 is A B/2 D/2 B/2 C E/2. D/2 E/2 F In order to prove Pascal s hexagon theorem we need the following theorem. Theorem 1. If C 1 and C 2 are different conics and at least one of them is non-degenerate, then they contain at most four common points. In other words, two different conics can contain five common points only if both of them are degenerate. Proof. Let C i be the matrix of the conic C i. First we consider the case when C 1 is a nondegenerate conic and C 2 is degenerate. There exist lines p and q such that C 2 p q. The fact that a line and a non-degenerate conic can have at most two common points implies that C 1 C 2 4. From now on, we assume that C 2 is non-degenerate too. It is important to notice that det(c 1 ), det(c 2 ) 0. In order to prove the theorem it is sufficient to find a degenerate conic section C 3 satisfying the property C 1 C 2 = C 1 C 3. The function f(t)= det((1 t)det(c 2 )C 1 t det(c 1 )C 2 ) fulfills f(1) = det(c 1 ) 3 det(c 2 ) 0 and f(0) = det(c 1 )det(c 2 ) 3 0, and, thus, there exists α (0, 1) such that f(α)= 0. For λ = (1 α)det(c 2 ), μ = α det(c 1 ) and C 3 = λc 1 + μc 2 hold det(c 3 ) = 0 and μ 0. So, C 1 C 2 = C 1 C 3 and C 3 is degenerate, but it differs from the whole plane since C 1 C 2. 3. Homogeneous coordinates Each point of the plane can be represented as an ordered triple of real numbers (x,y,z) (0, 0, 0), called homogeneous coordinates of a point. That triple is uniquely determined up to a multiplicative constant. In that way, the point with Cartesian coordinates (x, y) has got homogeneous coordinates (λx, λy, λ), where λ R\{0}. A triple (x, y, 0) (0, 0, 0) is not attached to any point in a Cartesian plane. However, we extend each line ax + by + c = 0 with a point at infinity (λb, λa, 0) (0, 0, 0), keeping the convention that linearly dependent triples of real numbers represent the same point. In doing so, we add the same point at infinity to parallel lines, and different points at infinity to nonparallel lines. In order to preserve the property that there exists a line containing a given pair of points we consider the set of all points at infinity as a line called the line at infinity. Moreover, we have secured that each pair of different lines has a unique common point. The line ax + by + cz = 0 is described by the triple (a,b,c), while (0, 0,λ) (0, 0, 0) describes the line at infinity. It is obvious that different lines correspond to linearly independent triples, while linearly dependent triples determine the same line, and the lines, just like the points, are described by the triples from R 3 \{(0, 0, 0)}. Besides, the point (x,y,z)is on the line (a,b,c)if and only if ax + by + cz = 0.

Proof of Pascal s hexagon theorem 621 If we think of triples representing points and lines as of vectors in the three-dimensional space, then the vectors corresponding to a line and a point lying on that line are orthogonal. Thus, the vector of a line containing two given points can be found as the cross-product of the vectors of these points, and similarly, the vector of the intersecting point of two lines is the cross-product of the vectors of these lines. The points A, B, C are collinear if and only if the mixed triple product of the corresponding vectors is equal to 0. Also, it is possible to describe the projective plane P 2 as P 2 = A 2 P 1, where A 2 is the affine plane and P 1 is the line at infinity, i.e. the set of directions of lines. When we say the origin of the coordinate system, then we think of the origin of A 2. DEFINITION 1 A set S of points in the projective plane is general if it does not contain three collinear points. 4. Pascal s hexagon theorem, its converse and some applications Now, everything is prepared for the proof of Pascal s theorem. Theorem 2 (Pascal s hexagon theorem). Let A 1,A 2,A 3,A 4,A 5,A 6 be different points on a non-degenerative conic C. Then the points are collinear. P = A 1 A 2 A 4 A 5, Q = A 2 A 3 A 5 A 6 and R = A 3 A 4 A 6 A 1 Proof. It is sufficient to prove that the vectors u = (A 1 A 2 ) (A 4 A 5 ), v = (A 2 A 3 ) (A 5 A 6 ), w = (A 3 A 4 ) (A 6 A 1 ) are linearly dependent. Bearing in mind that (a b) (c d) = [a,c,d]b [b, c, d] a = [a,b,d]c [a,b,c]d we obtain u = [A 1,A 4,A 5 ]A 2 [A 2,A 4,A 5 ]A 1, w = [A 3,A 6,A 1 ]A 4 [A 4,A 6,A 1 ]A 3, v = [A 2,A 3,A 6 ]A 5 [A 2,A 3,A 5 ]A 6. Computing [u, v, w] and observing that six of the eight terms cancel out, we find that the condition [u, v, w] = 0 is equivalent to the equality [A 1,A 3,A 5 ][A 2,A 4,A 5 ][A 4,A 6,A 1 ][A 6,A 2,A 3 ] = [A 2,A 4,A 6 ][A 1,A 4,A 5 ][A 3,A 6,A 1 ][A 5,A 2,A 3 ]. (1) In order to prove (1) we replace A 6 with the general point A(x,y,z): [A 1,A 3,A 5 ][A 2,A 4,A 5 ][A 4,A,A 1 ][A, A 2,A 3 ] = [A 2,A 4,A][A 1,A 4,A 5 ][A 3,A,A 1 ][A 5,A 2,A 3 ]. (2) In this way we obtain an equation in x,y,z of at most degree 2, whose solutions include A {A 1,...,A 5 }. So, eq. (2) can be reduced to 0 = 0 or it is an equation of a conic C 1 which contains A 1,...,A 5. The fact that C is a non-degenerate conic implies that C = C 1 and (1) is a consequence of A 6 C.

622 Nedeljko Stefanović and Miloš Milošević Notice that the points P, Q, R from the above theorem are uniquely determined. Namely, since a line and a non-degenerate conic do not contain more than two common points, the set {A 1,...,A 6 } is general and the lines A 1 A 2 and A 4 A 5 do not coincide. Similar reasoning can be applied to Q and R. PROPOSITION 1 Let S ={A 1,...,A 5 } be a set of five points in the projective plane. Then there exists a nondegenerate conic containing S if and only if S is a general set. That conic is unique and its equation is given by (2), where A(x,y,z)is a variable point. In order to find coefficients of eq. (2) of that conic one needs to perform no more than 62 multiplications and 32 additions or subtractions. If there is no point at infinity in S and we want to determine the equation of the conic in Cartesian coordinates then the numbers of multiplications and additions are not greater than, respectively,42and 32. Moreover, if a point A 6 / S, then the points A 1,...,A 5,A 6 are on the same conic if and only if the points P, Q, R from Pascal s theorem are collinear. Proof. We consider the following functions: f(x,y,z) = [A 4,A,A 1 ][A, A 2,A 3 ], g(x,y,z) = [A 2,A 4,A][A 3,A,A 1 ], A = A(x, y, z). Obviously, f(x,y,z) = 0 is an equation of the union of lines A 1 A 4 and A 2 A 3, and g(x,y,z) = 0 is an equation of the union of lines A 1 A 3 and A 2 A 4, and thus, the functions f and g can not be linearly dependent. The assumption that there are no three collinear points in S, implies that f(a 5 ), g(a 5 ) 0. Since for λ = [A 1,A 3,A 5 ][A 2,A 4,A 5 ], μ = [A 1,A 4,A 5 ][A 5,A 2,A 3 ] hold λf (A 5 ) = μg(a 5 ) and λ, μ 0, we conclude that (2) is an equation of the nondegenerate conic containing A 1,...,A 5, because this equation is not reduced to 0 = 0, since the functions f and g are linearly independent. From the proof of Pascal s theorem it is clear that the equality (1) is equivalent to the collinearity of P, Q and R. In (2) there are four pairs of mixed triple products of the forms [A i,a j,a] and [A i,a j,a k ], where the second mixed triple product is obtained by replacement of A by A k.[a i,a j,a]isa3 3 determinant containing one row which consists of variables and the result of computation of such determinant is a polynomial. For the computation of a 2 2 determinant whose components are real numbers one has to perform 2 multiplications and one addition (well, in fact subtraction, but we will not point out the difference between addition and subtraction here). So, to determine the coefficients corresponding to x,y and z, 6 multiplications and 3 additions are needed. In order to compute [A i,a j,a k ], we are going to use the already computed minors and we have to perform additional 3 multiplications and 2 additions. Thus, it makes in total 9 multiplications and 5 additions for computing [A i,a j,a] and [A i,a j,a k ], and repeating this procedure for all four pairs requires 36 multiplications and 20 additions. Let us focus for a moment on the left-hand side of (2). One more multiplication must be done to compute [A 1,A 3,A 5 ][A 2,A 4,A 5 ] and multiplication of the polynomial [A 4,A,A 1 ] by the number [A 1,A 3,A 5 ][A 2,A 4,A 5 ] requires 3 multiplications to determine the coefficients corresponding to the variables x,y,z. Multiplication of polynomials [A 1,A 3,A 5 ][A 2,A 4,A 5 ][A 4,A,A 1 ] and [A, A 2,A 3 ]

Proof of Pascal s hexagon theorem 623 adds 9 multiplications and 3 additions. Same must be done on the right-hand side of (2) and the subtraction of the right-hand side from the left-hand side requires 6 more additions for finding the coefficients of all monomials. Finally, it makes 62 multiplications and 32 additions. If we are finding coefficients in Cartesian coordinates then we have to notice that the numbers of multiplications and additions needed for computation of both determinants x y 1 a 1 a 2 1 D 1 = b 1 b 2 1 and D 2 = b 1 b 2 1 c 1 c 2 1 c 1 c 2 1 are, respectively, 4 and 5. We obtain these numbers of operations when we expand D 1 along the first row, and again we already use computed minors to obtain D 2. The computation of D 1 requires 2 multiplications and 3 additions, and to compute D 2 one has to perform additional 2 multiplications and 2 additions. There is a simple condition for checking whether six points lie on the same conic section, and in order to prove it, we need some preparatory results. PROPOSITION 2 If C is a non-degenerate conic section, and p 1 (x,y,z) = 0 and p 2 (x,y,z) = 0 are two equations of C, where p 1 and p 2 are polynomials of degree at most 2, then there exists nonzero constant C such that p 1 = C p 2. Proof. Suppose that there is no such constant and denote C by C i seen as the set of solutions of p i (x,y,z) = 0, and denote their matrices by M 1 and M 2. M 1 and M 2 are linearly independent and their determinants are not equal to 0. The proof is analogous to the proof of Theorem 1. The function h(t) = det((1 t)det(m 2 )M 1 t det(m 1 )M 2 ) fulfills h(1) h(0) <0, and, thus, there exists α (0, 1) such that h(α) = 0. For λ = (1 α)det(m 2 ), μ = α det(m 1 ), C 0 = λc 1 + μc 2, and matrix M 0 of C 0 hold det(m 0 ) = 0 and μ 0. If C 0 is the whole plane then C 2 = λ μ C 1. Otherwise, C = C 1 C 2 = C 1 C 0 C. In both cases the initial assumption leads to contradiction. Observe the following homogenous function of degree 2 in variables x,y,z: x1 2 y1 2 z 2 1 y 1 z 1 z 1 x 1 x 1 y 1 x 2 2 y2 2 z 2 2 y 2 z 2 z 2 x 2 x 2 y 2 x3 2 y3 2 z3 2 y 3 z 3 z 3 x 3 x 3 y 3 g(x,y,z) = x4 2 y4 2 z4 2. y 4 z 4 z 4 x 4 x 4 y 4 x 5 2 y5 2 z5 2 y 5 z 5 z 5 x 5 x 5 y 5 x 2 y 2 z 2 yz zx xy

624 Nedeljko Stefanović and Miloš Milošević This quite naturally comes to mind if we want to construct a homogenous function of degree 2 having roots A 1 (x 1,y 1,z 1 ),...,A 5 (x 5,y 5,z 5 ). Suppose that {A 1,...,A 5 } is a general set and that the unique conic section through A 1,...,A 5 has the equation c 1 x 2 + c 2 y 2 + c 3 z 2 + c 4 yz + c 5 zx + c 6 xy = 0, where for example c 6 0. Dividing by c 6 and using the fact that (x 1,y 1,z 1 ),...,(x 5,y 5,z 5 ) are roots, one obtains the system of five linear equations in variables d 1,...,d 5 : d 1 x 2 i + d 2 y 2 i + d 3 z 2 i + d 4y i z i + d 5 z i x i = x i y i,i= 1,...,5. The system has nonzero determinant because it has an unique solution by the previous proposition. This determinant is exactly the coefficient which corresponds to the monomial xy in g(x,y,z), and g(x,y,z) is not the zero polynomial. So, we conclude that g(x,y,z) = 0 is an equation of the conic containing A 1,...,A 5. Now, we are going to prove the announced criterion. PROPOSITION 3 The six points A 1,...,A 6 in the projective plane are on the same conic section if and only if (1) holds. Proof. Let the homogenous coordinates of the point A i be (x i,y i,z i ).If{A 1,...,A 6 } is a general set, then it is already proved that these points lie on the same conic if and only if the equality (1) holds. Using MATLAB it is possible to show that the polynomials x1 2 y1 2 z 2 1 y 1 z 1 z 1 x 1 x 1 y 1 x 2 2 y2 2 z 2 2 y 2 z 2 z 2 x 2 x 2 y 2 x g(x 1,y 1,z 1,...,x 6,y 6,z 6 ) = 3 2 y3 2 z 2 3 y 3 z 3 z 3 x 3 x 3 y 3 x 2 4 y4 2 z4 2. y 4 z 4 z 4 x 4 x 4 y 4 x5 2 y5 2 z5 2 y 5 z 5 z 5 x 5 x 5 y 5 x6 2 y6 2 z6 2 y 6 z 6 z 6 x 6 x 6 y 6 and f(x 1,y 1,z 1,...,x 6,y 6,z 6 ) = [A 1,A 3,A 5 ][A 2,A 4,A 5 ][A 4,A 6,A 1 ][A 6,A 2,A 3 ] [A 2,A 4,A 6 ][A 1,A 4,A 5 ][A 3,A 6,A 1 ][A 5,A 2,A 3 ] are equal. Now, we can use the fact that permutation of the rows of a determinant does not change the absolute value of the determinant. If there exist three collinear points in {A 1,...,A 6 }, then without loss of generality we can assume that this triple is (A 1,A 3,A 5 ). The left-hand side of [A 1,A 3,A 5 ][A 2,A 4,A 5 ][A 4,A 6,A 1 ][A 6,A 2,A 3 ] = [A 2,A 4,A 6 ][A 1,A 4,A 5 ][A 3,A 6,A 1 ][A 5,A 2,A 3 ], which is equal to 0. If [A 2,A 4,A 6 ] = 0, then A 2,A 4 and A 6 are collinear, otherwise, there is a point from {A 2,A 4,A 6 } which lies on the line A 1 A 3 A 5. In any of these possible cases the points A 1,...,A 6 lie on two lines.

Proof of Pascal s hexagon theorem 625 In the appendix, an interested reader can find another proof for the existence of a nonzero constant C for which holds f = C g, where f and g are functions defined in the previous proposition. That proof does not require any computation. PROPOSITION 4 A set S of more than five points lies on a conic if and only if for each choice of its six elements there exists a conic containing those six points. Proof. We prove the nontrivial implication from right to left. Suppose that there exist five points, say A 1,...,A 5, without three collinear points among them. Each point B S lies on the unique conic determined by A 1,...,A 5. Otherwise, we distinguish the following subcases: there exist five points A 1,...,A 5, such that A 1,A 2,A 3 are collinear, and A 4 and A 5 are not on the line A 1 A 2 A 3 ; then each point B S is contained on the union of lines A 1 A 2 A 3 A 4 A 5 ; there is a line containing all points from S except at most one. Then S is a subset of union of two lines. The final illustration of our method is the proof of Theorem 6.13. from [1]. Theorem 3. Let A 1,...,A 8 be eight points on a non-degenerate conic in the projective plane. Then the points B 1 = A 1 A 2 A 4 A 5,B 2 = A 2 A 3 A 5 A 6, B 3 = A 3 A 4 A 6 A 7,B 4 = A 4 A 5 A 7 A 8, B 5 = A 5 A 6 A 8 A 1,B 6 = A 6 A 7 A 1 A 2, B 7 = A 7 A 8 A 2 A 3,B 8 = A 8 A 1 A 3 A 4 lie on a conic. Proof. Let {A 1,A 2,A 3,A 4,A 5 } be a general set of five points, C the unique conic determined by A 1,...,A 5, and C 6,C 7,C 8 points such that none of the lines A 1 C 6,A 1 C 7,A 1 C 8 is a tangent line on C at A 1. We determine A 6,A 7 and A 8 as intersection points of, respectively, A 1 C 6,A 1 C 7, and A 1 C 8 with C, different from A 1.Fori = 6, 7, 8 it is possible to express A i using A 1,A 2,A 3,A 4,A 5,C i. Namely, for the hexagon A 1 A 2 A 3 A 4 A 5 A 6 let P,Q and R be points from Pascal s theorem. Then it holds that P = A 1 A 2 A 4 A 5,R= A 3 A 4 A 1 C 6, Q = A 2 A 3 PR, and A 6 = A 5 Q A 1 C 6. Thus, A 6 = (A 5 ((A 2 A 3 ) (((A 1 A 2 ) (A 4 A 5 )) ((A 3 A 4 ) (A 1 C 6 ))))) (A 1 C 6 ). We check whether B m1,b m2,b m3,b m4,b m5,b m6,m s {1,...,8} are on the same conic using equality (1) for each of the possible 28 combinations. The equality (1) is an equality of polynomials in variables a ij,i = 1,...,5,j = 1, 2, 3, and c kl, k = 6, 7, 8,l = 1, 2, 3, where a ij are homogenous coordinates of A i and c kl are homogenous coordinates of C k. Using MATLAB one can convince oneself that the polynomials on the left-hand side and the right-hand side of (1) are equal. Once again, it is important to notice that from the beginning there is no connection between A 1,A 2,A 3,A 4,A 5,C 6,C 7,C 8, and no initial equality conditions, since these points are arbitrary.

626 Nedeljko Stefanović and Miloš Milošević In the Appendix we give another proof of the previous theorem which does not require MATLAB or any other computational equipment. Appendix Lemma 1. Let K be a field and p, q K[x 1,...,x n ], where deg xi p, deg xi q d i, for each i = 1,...,n. If there exist sets S 1,...,S n such that S i >d i and p(a) = q(a) for each a S 1 S n, then p = q. Proof. We use induction on the number of variables n.forn = 1, the polynomial p(x 1 ) q(x 1 ) of degree at most d 1 has more than d 1 roots, which implies that this is a zero polynomial. Assume that the statement is true for n = k and prove it for n = k + 1. Polynomial p(x 1,...,x k+1 ) q(x 1,...,x k+1 ) can be written in the form p(x 1,...,x k+1 ) q(x 1,...,x k+1 ) = c k+1 (x 1,...,x k )x d k+1 k+1 + +c 0(x 1,...,x k ), where c i are polynomials in x 1,...,x k. For fixed a = (a 1,...,a k ) S 1 S k degree of the polynomial p(a 1,...,a k,x k+1 ) q(a 1,...,a k,x k+1 ) is at most d k+1. But this polynomial has value 0 for every element of S k+1, so it must be zero polynomial, and consequently c i (a 1,...,a k ) = 0, i= 0,...,k+1. Since c i,i= 0,...,k+1have value zero for arbitrarily chosen a S 1 S k, by the inductive hypothesis we conclude that c i = 0, i= 0,...,k+ 1, implying p = q. COROLLARY 1 Let p, q K[x 1,...,x n ], where K = R or K = C. Polynomials p and q are equal if and only if the interior of the set {(a 1...,a n ) K n : p(a 1,...,a n ) = q(a 1,...,a n )} is nonempty. Lemma 2. Let p, q R[x 1,...,x 18 ]. Assume that p and q have the following properties: (1) p and q are of the degree at most 2 in each variable; (2) for arbitrarily chosen six points A 1 (a 1,a 2,a 3 ),...,A 6 (a 16,a 17,a 18 ) without three collinear points among them the following statements are equivalent: p(a 1,...,a 18 ) = 0, q(a 1,...,a 18 ) = 0, A 1,...,A 6 lie on the same conic section. Then there exists a constant C 0 such that p = C q. Proof. Throughout this proof we emphasize that by the equation of the unique conic containing five given points, where each three of them are non-collinear, we mean eq. (2), and consequently the matrix of the conic and its determinant are unique functions in homogenous coordinates of these five points. Namely, if A is determined by the coordinates

Proof of Pascal s hexagon theorem 627 (a,b,c), then in eq. (2) will occur exactly a,b and c, although it holds that (a,b,c) = (ta, tb, tc) for t R\{0}. If we express A as (ta, tb, tc) for some t 0, 1, the equation obtained in that way will be the product of a nonzero constant and (2). Let k be a circle in the projective plane such that the origin of the coordinate system O lies in its interior, and let A 1 (a 1,a 2,a 3 ),...,A 6 (a 16,a 17,a 18 ) be six different points on k satisfying a j 0,j = 1,...,18, and such that tangent lines through A i,i = 1...,6onk do not intersect x-axis and y-axis at points at infinity. The points of the form A(x, a 2,a 3 ), x R lie on the line a 3 y a 2 z = 0, which contains A 1 and has the same point at infinity (1, 0, 0) as the x-axis. It means that this line and the circle k have two different common points. The same holds for the line {(a 1,x,a 3 ) R 3 : x R}. The line OA 1 consists of the points with homogenous coordinates (a 1,a 2, x), x R and intersects k at two points, too. So, if we fix all variables with x j = a j, except x 1, then the equations p(x 1,a 2,...,a 18 ) = 0, q(x 1,a 2,...,a 18 ) = 0, have two solutions each, and, moreover, these sets of solutions are equal. One of these solutions is a 1 and the other we denote as b 1. Thus, there exist nonzero constants C 1 and C 1 such that p(x 1,a 2,...,a 18) = C 1 (x 1 a 1 )(x 1 b 1 ), q(x 1,a 2,...,a 18 ) = C 1 (x 1 a 1 )(x 1 b 1 ), and consequently, for C 1 = C 1 /C 1 it holds that p(x 1,a 2,...,a 18 ) = C 1 q(x 1, a 2,...,a 18 ). It is possible to find ε 1 > 0 such that for all a i a i <ε 1 points A 1 (a 1,a 2,a 3 ),..., A 6 (a 16,a 17,a 18 ) fulfill the following requirements: {A 2,A 3,A 4,A 5,A 6 } is a general set and the unique conic C 1 section it determines is non-degenerate; that can be fulfilled since the coefficients of a conic are continuous functions of coordinates a 4,...,a 18 and the determinant of a conic is a continuous function of its coefficients, so we can secure that the determinant differs from 0; the line {(x, a 2,a 3 ) x R} intersects C 1 at two points with first coordinates α and β; the discriminant of the quadratic equation which determines x-coordinates of the intersection points of the above mentioned line and conic is a continuous function in a 2,...,a 18, and in neighbourhoods of a 2,...,a 18 of sufficiently small radius this discriminant will stay positive; that is why this condition can be satisfied as well. And again there exist nonzero C 1 (a 2,...,a 18 ) and C 1 (a 2,...,a 18 ) for which p(x 1,a 2,...,a 18 ) = C 1 (a 2,...,a 18 )(x 1 α)(x 1 β), q(x 1,a 2,...,a 18 ) = C 1 (a 2,...,a 18 )(x 1 α)(x 1 β), and p(x 1,a 2,...,a 18 ) = C 1(a 2,...,a 18 )q(x 1,a 2,...,a 18 ), for C 1 (a 2,...,a 18 ) = C 1 (a 2,...,a 18 )/C 1 (a 2,...,a 18 ) hold.

628 Nedeljko Stefanović and Miloš Milošević Similarly, one can find ε i,i= 2,...,18, such that for all a j a j <ε i,j i the equality p(a 1,...,a i 1,x i,a i+1,...,a 18 ) = C 1 (a 1,...,a i 1,a i+1,...,a 18 )q(a 1,...,a i 1,x i,a i+1,...,a 18 ) is true. We denote ε = min{ε i i = 1,...,18}. There exist a j, j = 1,...,18 such that a j a j < ε and q(a 1,...,a 18 ) 0. Namely, we can set a j = a j for j 1 and for a 1 we can pick any value from (a 1 ε, a 1 + ε)\{a 1,b 1 }. There is δ, such that 0 <δ<ε/2and that for x j a j <δ,j= 1,...,18 holds q(x 1,...,x 18 ) 0. It is proved that for x j a j <δ, j= 1,...,18 the quotient p(x 1,...,x 18 ) q(x 1,...,x 18 ) is not a function of x j, for any j. Thus, this quotient is constant in an 18-dimensional cube, and Proposition 3 follows from Corollary 1. Now, we present the proof of Theorem 3 which is not based on the computational brute force approach. The following lemma is a variant of the previously mentioned theorem with one additional assumption. Lemma 3. Let the points A 1,...,A 8,B 1,...,B 8 satisfy the conditions of Theorem 3 and let {B 1,...,B 8 } be a general set of eight points. Then B 1,...,B 8 lie on a conic. Proof. We remind the reader that B 1 = A 1 A 2 A 4 A 5 and B 6 = A 6 A 7 A 1 A 2. Thus, the points B 1 and B 6 are on the line A 1 A 2, and consequently B 1 B 6 and A 1 A 2 represent the same line, i.e. B 1 B 6 = A 1 A 2. Similarly, we obtain B 2 B 7 = A 2 A 3,B 3 B 8 = A 3 A 4,B 4 B 1 = A 4 A 5,B 5 B 2 = A 5 A 6, B 6 B 3 = A 6 A 7,B 7 B 4 = A 7 A 8,B 8 B 5 = A 8 A 1. The set {A 1,A 2,A 3,A 4,A 5,A 6 } lies on a non-degenerate conic, and thus, by the Pascal s theorem, the points B 1 = A 1 A 2 A 4 A 5,B 2 = A 2 A 3 A 5 A 6,B= A 3 A 4 A 1 A 6 are collinear. So we can conclude that A 3 A 4 B 1 B 2 = B. Using the above observations, we obtain B 3 B 8 B 1 B 2 = B, B 1 B 6 B 8 B 5 = A 1, B 5 B 2 B 6 B 3 = A 6, and, since B A 1 A 6, the set {B 1,B 2,B 5,B 8,B 3,B 6 } lies on the unique non-degenerate conic C determined by points B 1,B 3,B 5,B 6,B 8. For the conclusion the inverse Pascal s theorem (Proposition 1) was used as well by the fact that {B 1,...,B 8 } is a general set. In an analogous way, it can be demonstrated that the set {B 1,B 3,B 4,B 5,B 6,B 8 } lies on C. Similarly, the set {B 2,...,B 8 } lies on some C, and since conics C and C contain at least six common points, we can conclude C = C. Let M 1,...,M 8 be vertices of a regular octagon. A reader can think of M i,i = 1,...,8, as of A i,i = 1,...,8, in a very special position. The corresponding points B M 1,...,BM 8 from Theorem 3 are also vertices of some regular octagon, and thus lie on a circle. Since the coordinates of B 1,...,B 8 are polynomial functions, and thus continuous functions of the coordinates of A 1,...,A 8, there exist neighborhoods O i of M i,i = 1,...,8, such that if

Proof of Pascal s hexagon theorem 629 A i O i,i = 1,...,8, then none of the mixed triple products [A i,a j,a k ], [B i,b j,b k ], for different i, j, k, is equal to 0, and consequently {A 1,...,A 8 } and {B 1,...,B 8 } are general sets of eight points. For i = 6, 7, 8, let P i,q i,r i be the points from the Pascal s theorem correlated to M 1,...,M 5,M i, and C i an arbitrary point on the line P i Q i R i which differs from P i,q i and R i. We represent points P i,q i,r i,m i as intersections of lines: P i = M 1 M 2 M 4 M 5,Q i = M 2 M 3 P i C i, R i = M 3 M 4 P i Q i,m i = M 1 R i M 5 Q i, and by doing so we obtain the coordinates of M i as functions of the coordinates of points M 1,...,M 5,C i. We pick some A 1,...,A 5 and C i, and in the same way as above, we define A i in terms of A 1,...,A 5, C i.fora 1,...,A 5, C i from some neighborhoods O i 1,...,Oi 5, O i of, respectively, M 1,...,M 5,C i, the point A i O i and A 1,...,A 5,A i lie on the non-degenerate conic C determined by A 1,...,A 5. We denote O j = O j O 6 j O 7 j O8 j,j = 1,...,5. Further, the coordinates of B 1,...,B 8 can be represented as the polynomial functions of the coordinates of A 1,...,A 5, C 6, C 7, C 8. The set {B 1,...,B 8 } is general, and by Lemma 3 lies on a conic. Proposition 3 implies that [B i1,b i3,b i5 ][B i2,b i4,b i5 ][B i4,b i6,b i1 ][B i6,b i2,b i3 ] = [B i2,b i4,b i6 ][B i1,b i4,b i5 ][B i3,b i6,b i1 ][B i5,b i2,b i3 ] (3) holds for every choice of distinct elements i 1,...,i 6 {1,...,8}, and this equality can be rewritten as the equality of a polynomial in the coordinates of A 1,...,A 5, C 6, C 7, C 8 and the zero polynomial. Cause the equality of the corresponding polynomial functions holds on the set O 1 O 8, one concludes, by Corollary 1, that these polynomials are equal. So, B i1,...,b i6 are on the same conic. Finally, an application of Proposition 4 finishes the proof. References [1] Bix R, Conics and cubics (Springer) (2006) [2] Kunz E, Introduction to plane algebraic curves (Birkhauser) (2005) [3] Silverman J H and Tate J, Rational points on elliptic curves (Springer) (1994) [4] van Yzeren J, A simple proof of Pascal s hexagon theorem, The American Mathematical Monthly 100(10) (1993) 930 931