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Projective Geometry Projective geometry Projective geometry is an extension of Euclidean geometry, endowed with many nice properties incurred by affixing an extra ‘line at infinity’. Certain theorems (such as Desargues’ and Pascal’s theorems) have projective geometry as their more natural setting, and the wealth of projective transformations can simplify problems in ordinary Euclidean geometry. The real projective plane In Euclidean geometry, we assign a coordinate pair x, y to each point in the plane. In projective geometry, we augment this with an extra coordinate, so three values are used to represent a point: x, y, z. Moreover, scalar multiples are considered equivalent; x, y, z and x, y, z represent the same point. R 0, 0, 0 is not part of the projective plane, but can be regarded as a ‘projector’, from which all points, lines, circles et cetera emanate. Since scalar multiples of points are considered equivalent, we can identify points in 2 (the real projective plane) with lines through the origin. Projective lines are identified with planes through the origin, and are of the form a x b y c z 0. Note that this equation is homogeneous : all terms are of first degree. In general, all algebraic curves are represented by homogeneous polynomials in x, y and z. In the diagram above, a triangle A B C is shown in the reference plane ( z 1). The point A is ‘projected’ from R along the orange line, intersecting the reference plane at A. Similarly, the ‘plane’ containing the yellow triangle represents the line B C . Horizontal ‘lines’ such as the blue one do not intersect the reference plane, so correspond to points ‘at infinity’. The horizontal ‘plane’ (parallel to the reference plane) through R represents the line at infinity ( z 0). Parallel lines on the projective plane can be considered to meet at a point on the line at infinity. 1. Prove that any two distinct lines intersect in precisely one point. x y z 2. Show that the equation of a line through points x1, y1, z1 and x2, y2, z2 is given by det x1 y1 z1 0. x2 y2 z2 You have shown that any two points share a common line, and any two lines share a common point. This suggests a fundamental interchangeability between lines and points, known as projective duality . We shall explore this later. 3. Let A B C and A! B! C! be two triangles. Let A B meet A' B' at C!! , and define A!! and B!! similarly. Show that A A !, B B ! and C C ! are concurrent if and only if A!! , B!! and C!! are collinear. [Desargues’ theorem] A B C P C' C'' A' B' B'' A'' If two triangles exhibit this relationship, they are said to be in perspective . The point P is the perspector , and the line A!! B!! C!! is the perspectrix . As Desargues’ theorem is the projective dual of its converse, you only need to prove the statement in one direction. Surprisingly, it is actually easier to prove Desargues in three dimensions (where the triangles are in different planes); the two-dimensional result then follows by projecting it onto the plane. 4. Let A B C be a triangle and X be a point inside the triangle. The lines A X , B X and C X meet the circle A B C again at P, Q and R, respectively. Choose a point U on X P which is between X and P. Suppose that the lines through U which are parallel to A B and C A meet X Q and X R at points V and W respectively. Prove that the points R, W, V and Q lie on a circle. [BMO2 2011, Question 1] Cross-ratio and harmonic ranges You may have encountered homogeneous projective coordinates before in the form of areal (or barycentric) coordinates. Even though qualitative properties such as collinearity and concurrency can be defined in terms of any projective coordinates, to compare distances they must first be normalised , i.e. projected onto the reference plane. Firstly, we define a vector, n, perpendicular to the reference plane. (For Cartesian coordinates, we generally allow z 1 to be the reference plane, so n 0, 0, 1 . For areal coordinates, x y z 1 is the reference plane, so n 1, 1, 1 . In general, if the reference plane is given by a x b y c z 1, the vector n a, b, c. For a given x vector x in the projective plane, it is normalised by the operation x - nVx 5. If A, B, C and D are collinear points represented by normalised vectors a, b, c and d, respectively, show that ab A B . C D cd P R S Q A B C D Suppose we have three points, A, B and C, which are collinear. We can select an arbitrary point P, and a further arbitrary point Q lying on the line P B . Let Q C meet P A at R, and Q A meet P C at S. Finally, let R S meet line A B C at D. 6. Prove that A B V C D 1, irrespectively of the locations of P and Q. (Hint: this can be done by applying two B C V D A similar theorems in quick succession.) It is a remarkable fact that the location of D does not depend on that of P and Q. B and D are described as projective harmonic conjugates with respect to the line segment A C . For any four collinear points, the quantity A B V C D A, C; B, D is known as the cross-ratio . If A, C; B, D 1, we say that they form a harmonic range . B C V D A ^ A, C; B, D 1 is equivalent to A and C being inverse points with respect to the circle on diameter B D , which is in turn equivalent to the circles on diameters A C and B D being orthogonal. [Equivalent definitions of harmonic range] There is another remarkable and useful fact concerning harmonic ranges. Let P be a point not on the line A B C D . Then any two of the following four properties implies the other two: O A P C ; 2 P A is an angle bisector of O B P D ; P C is the other angle bisector of O B P D ; A, C; B, D 1. P A B C D By Thales’ theorem, the first of these conditions is equivalent to P lying on the circle with diameter A C . Hence, if P lies on either of the intersection points of the (orthogonal) circles on diameters A C and B D , the four lines through P divide the plane into eight equal octants of angle . 4 Projective transformations Returning to the idea of representing points in the projective plane as three-element vectors, we can consider the group of operations represented by linear maps x - M x , where M is a non-singular matrix. These are known as projective transformations , or collineations . 7. Show that applying a projective transformation to a line results in another line, and thus that collinear points remain collinear. A generalisation is that degree- d algebraic curves are mapped to degree- d algebraic curves by projective transformations. Hence, conics are preserved. Moreover, it is possible to choose a projective transformation carefully to map a given conic to any other conic, with three real degrees of freedom remaining. 8. Show that applying a projective transformation with matrix M to a point x p q results in a point with M p M q normalised coordinates . n V M p n V M q 9. Hence show that the cross-ratio A, C; B, D of four collinear points is preserved under projective transformations. 10. For any four points A, B, C and D, no three of which are collinear, show that there exists a unique projective transformation mapping them to 1, 0, 0 , 0, 1, 0 , 0, 0, 1 and 1, 1, 1 , respectively. 11. Hence show that there exists a unique projective transformation mapping any four points (no three of which are collinear) to any other four points (no three of which are collinear). The last of these theorems enables one to simplify a projective problem by converting any quadrilateral into a parallelogram. This enables one to find all of the harmonic ranges in the complete quadrangle displayed below. Try to spot as many as you can! B C Y A D X Z 12. The diagonals of the quadrilateral A B C D meet at X . The circumcircles of A B X and C D X intersect again at Y ; the circumcircles of B C X and D A X intersect again at Z. The midpoints of the diagonals A C and B D are denoted M and N, respectively. Prove that M, N, X , Y and Z are concyclic. [Sherry Gong, Trinity 2012] Another configuration occurring in many instances is a quadrilateral with an inscribed conic. With a projective transformation, we can convert the quadrilateral into a parallelogram. To simplify matters even further, a (possibly imaginary) affine transformation is capable of turning the conic into a circle. The symmetry of the configuration implies that the diagonals of the quadrilateral are concurrent with the lines joining opposite points of tangency. When viewed as a complete quadrilateral (four lines in general position intersecting in six points), we have three of these concurrency points and plenty of harmonic ranges! The centre of the inscribed conic also happens to lie on a line passing through the three midpoints of pairs opposite vertices, by Newton’s theorem .
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