Pascal's Theorem, Homogeneous Coordinates
The theorem states that if a hexagon is inscribed in a conic, then the three points at which the pairs of opposite sides meet, lie on a straight line.
The theorem has been demonstrated on a circle and ellipse and proved based on Chasles' theorem and also on Menelaus' theorem.
Below we give an additional proof that makes use of homogeneous coordinates. This particular proof comes from a book by E. A. Maxwell, popular half a century ago. The proof reached me through the good services of Professors Hubert Shutrick and his friend Gordon Walsh.
The proof refers to the diagram and notational conventions of the previous proofs. Pascal's mystic hexagon ABCDEF is inscribed into a conic with an equation in homogeneous coordinates
f(x, y, z) = ax² + by² + cz² + fxy + gxz + hyz = 0.
Choose the coordinates so that
f(A) = f(1, 0, 0) = ax² = 0.
So that, with this choice of coordinates, a = 0. Similarly, we find that b = c = 0. After the division by xyz, the equation of the conic becomes
(*) | f / x + g / y + h / z = 0. |
Denote D, F and B by (x_{1}, y_{1}, z_{1}),
EF is x/x_{2} = y/y_{2}, BC is x/x_{3} = z/z_{3} so that R is ( 1 , y_{2}/x_{2}, z_{3}/x_{3});
AB is y/y_{3} = z/z_{3}, DE is x/x_{1} = y/y_{1} so that P is (x_{1}/y_{1}, 1 , z_{3}/y_{3});
AF is y/y_{2} = z/z_{2}, CD is x/x_{1} = z/z_{1} so that Q is (x_{1}/z_{1}, y_{2}/z_{2}, 1 ).
P, Q and R are collinear if the determinant of the matrix
1 | y_{2}/x_{2} | z_{3}/x_{3} | ||||||
x_{1}/y_{1} | 1 | z_{3}/y_{3} | ||||||
x_{1}/z_{1} | y_{2}/z_{2} | 1 |
is zero. After you divide the first column by x_{1}, second by y_{2} and third by z_{3}, you see that, by (*), f times row 1 plus g times row 2 plus h times row 3 is zero. Which means that the determinant is indeed 0 and the points P, Q, R are collinear. Q.E.D.
References
- E. A. Maxwell, Methods of Plane Projective Geometry Based on the Use of General Homogeneous Coordinates, Cambridge, 1960
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