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Example 1 Let \( \\bigtriangleup {ABC} \) be a triangle in which \( \\angle {ABC} = \\angle {ACB} \) . Prove that \( {AB} = {AC} \) . | Solution First, reflect the triangle in the perpendicular bisector of \( {BC} \), so that the points \( B \) and \( C \) change places and the point \( A \) moves to some point \( {A}^{\prime } \), say. Since reflection preserves angles, it follows that \( \\angle {A}^{\prime }{BC} = \\angle {ACB} \) . Also, we are giv... | Yes |
Prove that if \( \bigtriangleup {ABC} \) and \( \bigtriangleup {DEF} \) are two triangles such that\n\n\[ \n{AB} = {DE},\;{AC} = {DF}\;\text{ and }\;\angle {BAC} = \angle {EDF},\n\]\n\nthen \( {BC} = {EF},\angle {ABC} = \angle {DEF} \) and \( \angle {ACB} = \angle {DFE} \) . | Solution It is sufficient to show that there is an isometry which maps \( \bigtriangleup {ABC} \) onto \( \bigtriangleup {DEF} \) . We construct this isometry in stages, starting with the translation\n\nwhich maps \( A \) to \( D \) . This translation maps \( \bigtriangleup {ABC} \) onto \( \bigtriangleup D{B}^{\prime ... | Yes |
Theorem 2 The set of Euclidean transformations of \( {\mathbb{R}}^{2} \) forms a group under the operation of composition of functions. | It is instructive to check the group axioms algebraically, for in the process of doing so we obtain formulas for the composites and inverses of Euclidean transformations.\n\nWe start by considering closure. Suppose that \( {t}_{1} \) and \( {t}_{2} \) are two Euclidean transformations given by\n\n\[ \n{t}_{1}\left( \ma... | Yes |
Problem 5 Prove that if \( {t}_{1} \) is a Euclidean transformation of \( {\mathbb{R}}^{2} \) given by\n\n\[ \n{t}_{1}\left( \mathbf{x}\right) = \mathbf{U}\mathbf{x} + \mathbf{a}\;\left( {\mathbf{x} \in {\mathbb{R}}^{2}}\right) ,\n\]\n\nthen:\n\n(a) the transformation of \( {\mathbb{R}}^{2} \) given by\n\n\[ \n{t}_{2}\... | The solution of Problem 5 shows that we can calculate the inverse of a Euclidean transformation by using the following result.\n\nThe inverse of the Euclidean transformation \( t\left( \mathbf{x}\right) = \mathbf{U}\mathbf{x} + \mathbf{a} \) is given by\n\n\[ \n{t}^{-1}\left( \mathbf{x}\right) = {\mathbf{U}}^{-1}\mathb... | No |
Theorem 3 Euclidean-congruence is an equivalence relation. | Proof We show that the three equivalence relation axioms E1, E2 and E3 hold.\n\nEl REFLEXIVE For all figures \( F \) in \( {\mathbb{R}}^{2} \), the identity transformation maps \( F \) This uses the existence of onto itself; so Euclidean-congruence is reflexive. an identity transformation.\n\nE2 SYMMETRIC Let a figure ... | Yes |
Problem 8 Prove that if two figures in \( {\mathbb{R}}^{2} \) are each Euclidean-congruent to a third figure, then they are Euclidean-congruent to each other. | Since Euclidean-congruence is an equivalence relation, it partitions the set of all figures into disjoint equivalence classes. Each class consists of figures which are Euclidean-congruent to each other, and hence share the same Euclidean properties (for example, one class consists of all circles of unit radius, another... | Yes |
Property 1 A parallel projection maps straight lines to straight lines. | Proof Let \( \ell \) be a line in the plane \( {\pi }_{1} \), and let \( p \) be a parallel projection mapping \( {\pi }_{1} \) onto the plane \( {\pi }_{2} \) . Now consider all the rays associated with \( p \) that pass through \( \ell \) . Since these rays are parallel, they must fill a plane. Call this plane \( \pi... | Yes |
Property 2 A parallel projection maps parallel straight lines to parallel straight lines. | Proof Let \( {\ell }_{1} \) and \( {m}_{1} \) be parallel lines in the plane \( {\pi }_{1} \), and let \( p \) be a parallel projection mapping \( {\pi }_{1} \) onto the plane \( {\pi }_{2} \) . Let \( {\ell }_{2} \) and \( {m}_{2} \) be the lines in \( {\pi }_{2} \) that are the images under \( p \) of \( {\ell }_{1} ... | Yes |
Property 3 A parallel projection preserves ratios of lengths along a given straight line. | Proof Let \( A, B, C \) be three points on a line in the plane \( {\pi }_{1} \), and let \( p \) be a parallel projection mapping \( {\pi }_{1} \) onto the plane \( {\pi }_{2} \) . Let \( P, Q, R \) be the points in \( {\pi }_{2} \) that are the images under \( p \) of \( A, B, C \) . We know from Property 1 that \( P ... | Yes |
Theorem 4 Given any ellipse, there is a parallel projection which maps the ellipse onto a circle. | A suitable parallel projection is illustrated below. Here the plane \( {\pi }_{1} \) (initially parallel to \( {\pi }_{2} \) ) has been tilted about the minor axis of the ellipse. Under the projection distances which are parallel to the minor axis remain unchanged, but distances parallel to the major axis are scaled by... | Yes |
Determine the image of the line \( y = {2x} \) under the affine transformation \( t\left( \mathbf{x}\right) = \left( \begin{array}{ll} 4 & 1 \\ 2 & 1 \end{array}\right) \mathbf{x} + \left( \begin{array}{r} 2 \\ - 1 \end{array}\right) \;\left( {\mathbf{x} \in {\mathbb{R}}^{2}}\right) . | Solution Let \( \left( {x, y}\right) \) be an arbitrary point on the line \( y = {2x} \), and let \( \left( {{x}^{\prime },{y}^{\prime }}\right) \) be the image of \( \left( {x, y}\right) \) under \( t \) . Then\n\n\[ \left( \begin{array}{l} {x}^{\prime } \\ {y}^{\prime } \end{array}\right) = \left( \begin{array}{ll} 4... | Yes |
Determine the affine transformation which maps the points \( \left( {0,0}\right) \) , \( \left( {1,0}\right) \) and \( \left( {0,1}\right) \) to the points \( \left( {3,2}\right) ,\left( {5,8}\right) \) and \( \left( {7,3}\right) \), respectively. | Let \( t \) be the affine transformation given by\n\n\[ t : \left( \begin{array}{l} x \\ y \end{array}\right) \mapsto \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \left( \begin{array}{l} x \\ y \end{array}\right) + \left( \begin{array}{l} e \\ f \end{array}\right) . \]\n\nSince \( t\left( {0,0}\right) = \... | Yes |
Theorem 1 Fundamental Theorem of Affine Geometry\n\nLet \( \mathbf{p},\mathbf{q},\mathbf{r} \) and \( {\mathbf{p}}^{\prime },{\mathbf{q}}^{\prime },{\mathbf{r}}^{\prime } \) be two sets of three non-collinear points in \( {\mathbb{R}}^{2} \) . Then:\n\n(a) there is an affine transformation \( t \) which maps \( \mathbf... | ## Proof\n\n(a) Let \( {t}_{1} \) be the affine transformation which maps \( \left( {0,0}\right) ,\left( {1,0}\right) \) and \( \left( {0,1}\right) \) to the points \( \mathbf{p},\mathbf{q} \) and \( \mathbf{r} \), respectively, and let \( {t}_{2} \) be the affine transformation which maps \( \left( {0,0}\right) ,\left... | Yes |
Determine the affine transformation which maps the points \( \left( {2,3}\right) \) , \( \left( {1,6}\right) \) and \( \left( {3, - 1}\right) \) to the points \( \left( {1, - 2}\right) ,\left( {2,1}\right) \) and \( \left( {-3,5}\right) \), respectively. | Solution You have already seen in Problem 3 that the affine transformation \( {t}_{1} \) which maps the points \( \left( {0,0}\right) ,\left( {1,0}\right) \) and \( \left( {0,1}\right) \) to the points \( \left( {2,3}\right) ,\left( {1,6}\right) \) and \( \left( {3, - 1}\right) \), respectively, is given by\n\n\[ \n{t}... | Yes |
Theorem 2 An affine transformation maps straight lines to straight lines. | Proof \n\nLet \( \ell \) be a line through a point with position vector \( \mathbf{p} \), and let the direction of \( \ell \) be that of some vector \( \mathbf{a} \) . Then\n\n\[ \ell = \{ \mathbf{p} + \lambda \mathb... | Yes |
Theorem 3 An affine transformation maps parallel straight lines to parallel straight lines. | Let \( {\ell }_{1} \) and \( {\ell }_{2} \) be parallel lines through the points with position vectors \( \mathbf{p} \) and \( \mathbf{q} \), respectively, and let the direction of the lines be that of the vector \( \mathbf{a} \) . Then\n\n\[ \n{\ell }_{1} = \{ \mathbf{p} + \lambda \mathbf{a} : \lambda \in \mathbb{R}\}... | Yes |
Theorem 4 An affine transformation preserves ratios of lengths along parallel straight lines. | Proof We begin by examining what happens to the length of a line segment under an affine transformation.\n\n\n\nLet \( \ell \) be a line through a point with position vector \( \mathbf{p} \), and let the direction of... | Yes |
Let \( P, Q \) and \( R \) be points, other than vertices, on the (possibly extended) sides \( {BC},{CA} \) and \( {AB} \) of a triangle \( \bigtriangleup {ABC} \), such that\n\n\[ \frac{AR}{RB} \cdot \frac{BP}{PC} \cdot \frac{CQ}{QA} = 1 \]\n\nThen the lines \( {AP},{BQ} \) and \( {CR} \) are concurrent. | Proof Let the lines \( {BQ} \) and \( {CR} \) intersect at a point \( X \), and let the line \( {AX} \) meet\n\n\n\n\( {BC} \) at some point \( {P}^{\prime } \) . It is sufficient to prove that \( P = {P}^{\prime } \... | Yes |
Example 2 The triangle \( \bigtriangleup {ABC} \) has vertices \( A\left( {1,3}\right), B\left( {-1,0}\right) \) and \( C\left( {4,0}\right) \), and the points \( P\left( {0,0}\right), Q\left( {\frac{8}{3},\frac{4}{3}}\right) \) and \( R\left( {-\frac{2}{3},\frac{1}{2}}\right) \) lie on \( {BC},{CA} \) and \( {AB} \), ... | ## Solution\n\n(a) Using the coordinate formulas for calculating ratios, we obtain\n\n\[ \frac{AR}{RB} = \frac{{x}_{R} - {x}_{A}}{{x}_{B} - {x}_{R}} = \frac{-\frac{2}{3} - 1}{-1 + \frac{2}{3}} = 5,\;\frac{BP}{PC} = \frac{{x}_{P} - {x}_{B}}{{x}_{C} - {x}_{P}} = \frac{0 + 1}{4 - 0} = \frac{1}{4}, \]\n\n\[ \frac{CQ}{QA} =... | Yes |
Theorem 5 Converse to Menelaus' Theorem\n\nLet \( P, Q \) and \( R \) be points other than vertices on the (possibly extended) sides \( {BC},{CA} \) and \( {AB} \) of a triangle \( \bigtriangleup {ABC} \), such that\n\n\[ \frac{AR}{RB} \cdot \frac{BP}{PC} \cdot \frac{CQ}{QA} = - 1. \]\n\nThen the points \( P, Q \) and ... | Proof Let the line \( \ell \) that passes through \( Q \) and \( R \) meet \( {BC} \) at some point \( {P}^{\prime } \) . The strategy of the proof It is sufficient to prove that \( P = {P}^{\prime } \) . is the same as that of\n\nIt follows from Menelaus' Theorem that Theorem 3.\n\n\[ \frac{AR}{RB} \cdot \frac{B{P}^{\... | Yes |
Example 4 Determine barycentric coordinates for the point \( \left( {2,1}\right) \) with respect to the triangle of reference \( \bigtriangleup {ABC} \) where \( A = \left( {1,0}\right), B = \left( {1, - 1}\right) \) and \( C = \left( {-1,1}\right) \) . | Solution The matrix \( \mathbf{M} \) for the triangle of reference \( \bigtriangleup {ABC} \) is\n\n\[ \mathbf{M} = \left( \begin{array}{rrr} 1 & 1 & - 1 \\ 0 & - 1 & 1 \\ 1 & 1 & 1 \end{array}\right) \]\n\nwhose inverse is We omit the details of the\n\n\[ \left( \begin{array}{rrr} 1 & 1 & 0 \\ - \frac{1}{2} & - 1 & \f... | No |
Theorem 6 The points \( P, Q \) and \( R \) with barycentric coordinates \( \left( {{\xi }_{1},{\eta }_{1},{\zeta }_{1}}\right) ,\left( {{\xi }_{2},{\eta }_{2},{\zeta }_{2}}\right) \) and \( \left( {{\xi }_{3},{\eta }_{3},{\zeta }_{3}}\right) \) are collinear if and only if\n\n\[ \left| \begin{array}{lll} {\xi }_{1} & ... | Proof Let the points \( P, Q \) and \( R \) have cartesian coordinates \( \left( {{x}_{1},{y}_{1}}\right) ,\left( {{x}_{2},{y}_{2}}\right) \) and \( \left( {{x}_{3},{y}_{3}}\right) \), respectively. It follows that, if the triangle of reference \( \bigtriangleup {ABC} \) has vertices \( A = \left( {{a}_{1},{a}_{2}}\rig... | Yes |
Let \( \\bigtriangleup {ABC} \) be a triangle, and let \( \\ell \) be a line that crosses the sides \( {BC},{CA} \) and \( {AB} \) at three distinct points \( P, Q, R \), respectively. Then\n\n\[ \n\\frac{AR}{RB} \\cdot \\frac{BP}{PC} \\cdot \\frac{CQ}{QA} = - 1 \n\] | Proof Define \( \\lambda ,\\mu \) and \( v \) as follows:\n\n\[ \n\\frac{BP}{PC} = \\frac{1 - \\lambda }{\\lambda },\\;\\frac{CQ}{QA} = \\frac{1 - \\mu }{\\mu },\\;\\frac{AR}{RB} = \\frac{1 - v}{v}.\n\]\n\nHence, by the Section Formula, \( P \) has barycentric coordinates\n\n\[ \nP = \\lambda \\left( {0,1,0}\\right) + ... | No |
Theorem 4 Affine transformations map ellipses to ellipses, parabolas to parabolas, and hyperbolas to hyperbolas. | Proof Consider the non-degenerate conic with equation\n\n\[ A{x}^{2} + {Bxy} + C{y}^{2} + {Fx} + {Gy} + H = 0, \]\n\n(6)\n\nHere we omit the details of the particular transformations involved, and concentrate instead on the principles underlying the successive mappings which are used.\n\nRemember that a circle is a spe... | Yes |
Theorem 5 Let \( t \) be an affine transformation, and let \( C \) be an ellipse or hyperbola with centre \( R \). Then \( t\left( C\right) \) has centre \( t\left( R\right) \). | Proof Let \( {C}^{\prime } \) and \( {R}^{\prime } \) be the images of \( C \) and \( R \) under \( t \). If \( {P}^{\prime } \) is any point on \( {C}^{\prime } \), then it must be the image of some point \( P \) on \( C \). Since \( R \) is the centre of \( C \), we can rotate \( P \) about \( R \) through an angle \... | Yes |
Theorem 6 Let \( t \) be an affine transformation, and let \( H \) be a hyperbola with asymptotes \( {\ell }_{1} \) and \( {\ell }_{2} \) . Then \( t\left( H\right) \) has asymptotes \( t\left( {\ell }_{1}\right) \) and \( t\left( {\ell }_{2}\right) \) . | Proof The hyperbola \( H \) possesses exactly two (distinct) families of parallel lines each of which fills the plane, with each member of each family meeting \( H \) exactly once - apart from one line in each family that is an asymptote of \( H \) , and so does not meet \( H \) .\n\nThe image of \( H \) under the affi... | Yes |
Theorem 7 Let \( t \) be an affine transformation, and let \( \ell \) be a tangent to a conic \( C \) . Then \( t\left( \ell \right) \) is a tangent to the conic \( t\left( C\right) \) . | Solution We shall use the fact that a tangent to a conic (whether it is an meets a parabola in ellipse, a hyperbola or a parabola) intersects the conic at exactly one point. exactly one point.\n\nFirst, the image of an ellipse \( E \) under an affine transformation \( t \) is an ellipse. A tangent to \( E \) is a line ... | Yes |
Example 1 \( {AB} \) is a diameter of an ellipse. Prove that the tangents to the ellipse Recall that the diameter at \( A \) and \( B \) are parallel to the diameter conjugate to \( {AB} \) . conjugate to \( {AB} \) is the set of midpoints of all the | Solution First, map the ellipse onto the unit circle, by an affine transforma- chords parallel to \( {AB} \) (see tion \( t \) . Since the centre \( O \) of the ellipse maps to the centre \( {O}^{\prime } \) of the circle, the Subsection 2.2.3). image of the diameter \( {AB} \) is a diameter \( {A}^{\prime }{B}^{\prime... | Yes |
Example 2 The tangent at the point \( P \) on a hyperbola meets the asymptotes at the points \( A \) and \( B \) . Prove that \( {PA} = {PB} \) . | Solution Let \( t \) be an affine transformation which maps the hyperbola onto the rectangular hyperbola \( H = \{ \left( {x, y}\right) : {xy} = 1\} \) in such a way that \( t\left( P\right) = \left( {1,1}\right) \) . Then, by Theorem 6 of Subsection 2.5.1, the asymptotes of the hyperbola map to the asymptotes of \( H ... | Yes |
Example 1 Which of the following homogeneous coordinates represent the same Point in \( {\mathbb{{RP}}}^{2} \) as \( \left\lbrack {6,3,2}\right\rbrack \) ?\n\n(a) \( \left\lbrack {{18},9,6}\right\rbrack \) (b) \( \left\lbrack {{12}, - 6,4}\right\rbrack \) (c) \( \left\lbrack {1,\frac{1}{2},\frac{1}{3}}\right\rbrack \) ... | ## Solution\n\nThroughout the solution we use equation (1):\n\n(a) This represents the same Point as \( \left\lbrack {6,3,2}\right\rbrack \), for if \( \lambda = 3 \), then\n\n\[ \left\lbrack {a, b, c}\right\rbrack = \left\lbrack {{\lambda a},{\lambda b},{\lambda c}}\right\rbrack \]\n\n\[ \left\lbrack {{18},9,6}\right\... | Yes |
Example 2 Determine homogeneous coordinates of the form \( \\left\\lbrack {a, b,1}\\right\\rbrack \) for the Points\n\n\[ \n\\left\\lbrack {2, - 1,4}\\right\\rbrack ,\\;\\left\\lbrack {4,2,8}\\right\\rbrack ,\\;\\left\\lbrack {{2\\pi }, - \\pi ,{4\\pi }}\\right\\rbrack ,\n\]\n\n\[ \n\\left\\lbrack {{200},{100},{400}}\\... | Solution According to equation (1), a Point of \( {\\mathbb{{RP}}}^{2} \) is unchanged if its homogeneous coordinates are multiplied (or divided) by any non-zero real\n\n---\n\nFor, dividing by a non-zero number \( \\lambda \) is equivalent to multiplying by the non-zero number \( 1/\\lambda \) .\n\n---\n\nnumber. Sinc... | Yes |
Example 3 For each of the following pairs of Points, write down an equation for the Line that passes through them.\n\n(a) \( \\left\\lbrack {3,2,0}\\right\\rbrack \) and \( \\left\\lbrack {3,4,0}\\right\\rbrack \; \) | ## Solution\n\n(a) Both the Points have a \( z \) -coordinate equal to 0, so the homogeneous coordinates must satisfy the equation \( z = 0 \) . This equation is of the form (2)\n\n---\n\nThe equation \( x = 3 \) is not of the form (2), and so is not the equation of a Line.\n\n---\n\nwith \( a = 0, b = 0 \) and \( c = ... | No |
Example 4 Find an equation for the Line that passes through the Points \( \\left\\lbrack {1,2,3}\\right\\rbrack \) and \( \\left\\lbrack {2, - 1,4}\\right\\rbrack \) . | Solution An equation for the Line is\n\n\[ \n\\left| \\begin{array}{rrr} x & y & z \\\\ 1 & 2 & 3 \\\\ 2 & - 1 & 4 \\end{array}\\right| = 0 \n\]\n\nNow\n\n\[ \n\\left| \\begin{matrix} x & y & z \\\\ 1 & 2 & 3 \\\\ 2 & - 1 & 4 \\end{matrix}\\right| = x\\left| \\begin{array}{rr} 2 & 3 \\\\ - 1 & 4 \\end{array}\\right| - ... | Yes |
Example 5 Determine whether the Points \( \\left\\lbrack {2,1,3}\\right\\rbrack ,\\left\\lbrack {1,2,1}\\right\\rbrack \) and \( \\left\\lbrack {-1,4, - 3}\\right\\rbrack \) are collinear. | Solution We have\n\n\[ \n\\left| \\begin{matrix} 2 & 1 & 3 \\\\ 1 & 2 & 1 \\\\ - 1 & 4 & - 3 \\end{matrix}\\right| = 2\\left| \\begin{array}{rr} 2 & 1 \\\\ 4 & - 3 \\end{array}\\right| - 1\\left| \\begin{matrix} 1 & 1 \\\\ - 1 & - 3 \\end{matrix}\\right| + 3\\left| \\begin{matrix} 1 & 2 \\\\ - 1 & 4 \\end{matrix}\\righ... | Yes |
Any two distinct Lines in \( {\mathbb{{RP}}}^{2} \) intersect in a unique Point of \( {\mathbb{{RP}}}^{2} \) . | We can determine the Point of intersection of two Lines simply by solving the equations of the two Lines as a pair of simultaneous equations. | Yes |
Determine the Point of intersection of the Lines in \( {\mathbb{{RP}}}^{2} \) with We know that there is a equations \( x + {6y} - {5z} = 0 \) and \( x - {2y} + z = 0 \) . unique Point of intersection, by | Solution At the Point of intersection \( \left\lbrack {x, y, z}\right\rbrack \) of the two Lines, we have Theorem 3.\n\n\[ x + {6y} - {5z} = 0, \]\n\n\[ x - {2y} + z = 0. \]\n\nSubtracting the second equation from the first, we obtain\n\n\[ {8y} - {6z} = 0, \]\n\nso that \( y = \frac{3}{4}z \) . Substituting this into ... | Yes |
Show that the function \( t : {\mathbb{{RP}}}^{2} \rightarrow {\mathbb{{RP}}}^{2} \) defined by\n\n\[ t : \left\lbrack {x, y, z}\right\rbrack \mapsto \left\lbrack {{2x} + z, - x + {2y} - {3z}, x - y + {5z}}\right\rbrack \]\n\nis a projective transformation, and find the image of \( \left\lbrack {1,2,3}\right\rbrack \) ... | Solution The transformation \( t \) has the form \( t : \left\lbrack \mathbf{x}\right\rbrack \mapsto \left\lbrack \mathbf{{Ax}}\right\rbrack \), where \( \mathbf{x} = \) \( \left( {x, y, z}\right) \) and\n\n\[ \mathbf{A} = \left( \begin{array}{rrr} 2 & 0 & 1 \\ - 1 & 2 & - 3 \\ 1 & - 1 & 5 \end{array}\right) \]\n\nNow\... | Yes |
Determine the projective transformations \( {t}_{2} \circ {t}_{1} \) and \( {t}_{1}^{-1} \). | The transformations \( {t}_{1} \) and \( {t}_{2} \) have associated matrices\n\n\[ \n{\mathbf{A}}_{1} = \left( \begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 3 \\ - 2 & 0 & 1 \end{array}\right) \;\text{ and }\;{\mathbf{A}}_{2} = \left( \begin{array}{lll} 2 & 0 & 0 \\ 1 & 1 & 1 \\ 4 & 2 & 0 \end{array}\right)\n\]\n\nrespective... | Yes |
Find the image of the Line \( {2x} + y - {3z} = 0 \) under the projective transformation \( {t}_{1} \) defined by Projective Transformations | Solution The equation of the Line can be written in the form \( \mathbf{{Lx}} = 0 \), where\n\n\[ \mathbf{L} = \left( \begin{array}{lll} 2 & 1 & - 3 \end{array}\right) .\n\nIn Example 2 we showed that \( {t}_{1}{}^{-1} \) has an associated matrix\n\n\[ \mathbf{B} = \left( \begin{array}{rrr} 1 & 0 & - 1 \\ - 7 & 3 & - 2... | Yes |
Problem 6 Let \( {t}_{1} \) and \( {t}_{2} \) be the projective transformations with associated matrices \[ {\mathbf{A}}_{1} = \left( \begin{array}{rrr} - 4 & - 1 & 1 \\ - 3 & - 2 & 1 \\ 4 & 2 & - 1 \end{array}\right) \text{ and }\;{\mathbf{A}}_{2} = \left( \begin{array}{rrr} - 8 & - 6 & - 2 \\ - 3 & 4 & 7 \\ 6 & 0 & -... | You should have found that both of the projective transformations \( {t}_{1} \) and \( {t}_{2} \) map the Points \( \left\lbrack {1, - 1,1}\right\rbrack ,\left\lbrack {1, - 2,2}\right\rbrack \) and \( \left\lbrack {-1,2, - 1}\right\rbrack \) to the Points \( \left\lbrack {-2,0,1}\right\rbrack \) , \( \left\lbrack {0,3,... | No |
Find a projective transformation \( t \) that maps the Points \( \left\lbrack {1,0,0}\right\rbrack \) , \( \left\lbrack {0,1,0}\right\rbrack \) and \( \left\lbrack {0,0,1}\right\rbrack \) to the non-collinear Points \( \left\lbrack {1, - 1,1}\right\rbrack ,\left\lbrack {1, - 2,2}\right\rbrack \) and \( \left\lbrack {-1... | Solution Let \( \mathbf{A} \) be a matrix associated with \( t \), and let the first column of \( \mathbf{A} \) be \( \left( \begin{array}{l} a \\ b \\ c \end{array}\right) \) . Then since\n\n\[ \left\lbrack {\left( \begin{array}{lll} a & * & * \\ b & * & * \\ c & * & * \end{array}\right) \left( \begin{array}{l} 1 \\ 0... | Yes |
Find the projective transformation \( t \) which maps the Points \( \left\lbrack {1,0,0}\right\rbrack ,\left\lbrack {0,1,0}\right\rbrack ,\left\lbrack {0,0,1}\right\rbrack \) and \( \left\lbrack {1,1,1}\right\rbrack \) to the Points \( \left\lbrack {1, - 1,1}\right\rbrack ,\left\lbrack {1, - 2,2}\right\rbrack \) , \( \... | If \( \mathbf{A} \) is the matrix associated with \( t \), then its columns must be multiples of the homogeneous coordinates \( \left\lbrack {1, - 1,1}\right\rbrack ,\left\lbrack {1, - 2,2}\right\rbrack ,\left\lbrack {-1,2, - 1}\right\rbrack \) ; that is,\n\n\[ \mathbf{A} = \left( \begin{array}{rrr} u & v & - w \\ - u ... | Yes |
Theorem 3 The Fundamental Theorem of Projective Geometry\n\nunique projective transformation \n\nLet \( {ABCD} \) and \( {A}^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime } \) be two quadrilaterals in \( {\mathbb{{R... | Proof According to the strategy above, there is a projective transformation \( {t}_{1} \) which maps the Points \( \left\lbrack {1,0,0}\right\rbrack ,\left\lbrack {0,1,0}\right\rbrack ,\left\lbrack {0,0,1}\right\rbrack ,\left\lbrack {1,1,1}\right\rbrack \) to the Points \( A, B, C, D \), respectively. Similarly, there ... | Yes |
Example 1 Let \( A = \left\lbrack {1,2,3}\right\rbrack, B = \left\lbrack {1,1,2}\right\rbrack, C = \left\lbrack {3,5,8}\right\rbrack, D = \left\lbrack {1, - 1,0}\right\rbrack \) be Points of \( {\mathbb{{RP}}}^{2} \) . Calculate the cross-ratio (ABCD). | Solution First, we have to find real numbers \( \alpha \) and \( \beta \) such that the following vector equation holds:\n\n\[ \left( {3,5,8}\right) = \alpha \left( {1,2,3}\right) + \beta \left( {1,1,2}\right) . \]\n\nComparing corresponding coordinates on both sides of this vector equation, we deduce that\n\n\[ 3 = \a... | Yes |
Theorem 1 The cross-ratio (ABCD) is independent of the homogeneous coordinates that are used to represent the collinear Points \( A, B, C, D \) . | Proof Suppose that \( A = \left\lbrack \mathbf{a}\right\rbrack, B = \left\lbrack \mathbf{b}\right\rbrack, C = \left\lbrack \mathbf{c}\right\rbrack, D = \left\lbrack \mathbf{d}\right\rbrack \), and let\n\n\n\n\[ \math... | Yes |
Calculate the cross-ratios (BACD) and (ACBD) for the four Points used in Problem 1(a). | When answering Problem 2 you may have noticed that (BACD) is the reciprocal of the value which we obtained for \( \left( {ABCD}\right) \) in Problem 1(a). Also, (ACBD) is equal to \( 1 - \left( {ABCD}\right) \) . The next result shows that this is not simply chance! | No |
Theorem 3 Let \( t \) be a projective transformation, and let \( A, B, C, D \) be any four collinear Points in \( {\mathbb{{RP}}}^{2} \) . If \( {A}^{\prime } = t\left( A\right) ,{B}^{\prime } = t\left( B\right) ,{C}^{\prime } = t\left( C\right) ,{D}^{\prime } = t\left( D\right) \), then\n\n\[ \left( {ABCD}\right) = \l... | Proof Let \( t \) be the projective transformation \( t : \left\lbrack \mathbf{x}\right\rbrack \mapsto \left\lbrack \mathbf{{Ax}}\right\rbrack \), where \( \mathbf{A} \) is an invertible \( 3 \times 3 \) matrix. If \( A = \left\lbrack \mathbf{a}\right\rbrack, B = \left\lbrack \mathbf{b}\right\rbrack, C = \left\lbrack \... | Yes |
Theorem 4 Let \( A, B, C, D \) be four distinct Points on a Line, and let \( {A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime } \) be four distinct Points on another Line such that \( A{A}^{\prime }, B{B}^{\prime }, C{C}^{\prime }, D{D}^{\prime } \) all meet at a Point \( U \) . Then\n\n\[ \left( {ABCD}\right) = \... | Proof By the Fundamental Theorem of Projective Geometry, there is a unique projective transformation \( t \) which maps \( B \) to \( {B}^{\prime }, C \) to \( {C}^{\prime },{B}^{\prime } \) to \( B \) , and \( {C}^{\prime } \) to \( C \) . We shall show that \( t\left( A\right) = {A}^{\prime } \) and \( t\left( D\righ... | Yes |
Theorem 5 Unique Fourth Point Theorem\n\nLet \( A, B, C, X, Y \) be collinear Points in \( {\mathbb{{RP}}}^{2} \) such that\n\n\[ \left( {ABCX}\right) = \left( {ABCY}\right) \text{.} \]\n\nThen \( X = Y \) . | Proof Let \( A = \left\lbrack \mathbf{a}\right\rbrack, B = \left\lbrack \mathbf{b}\right\rbrack, C = \left\lbrack \mathbf{c}\right\rbrack, X = \left\lbrack \mathbf{x}\right\rbrack, Y = \left\lbrack \mathbf{y}\right\rbrack \) . Since \( A, B, C, X, Y \) are collinear, it follows that there are real numbers \( \alpha ,\b... | Yes |
Theorem 6 Let \( A, B, C, D \) and \( A, E, F, G \) be two sets of collinear Points (on different Lines in \( {\mathbb{{RP}}}^{2} \) ) such that the cross-ratios (ABCD) and (AEFG) are equal. Then the Lines \( {BE},{CF} \) and \( {DG} \) are concurrent. | Proof Let \( P \) be the Point at which the Lines \( {BE} \) and \( {CF} \) meet, and let \( X \) be the Point at which the Line \( {PG} \) meets the Line \( {ABCD} \) . Then the Points \( A, B, C \) and \( X \) are in perspective from \( P \) with the Points \( A, E, F \) and \( \mathrm{G} \), so that \[ \left( {ABCX}... | Yes |
Theorem 7 Pappus' Theorem You met this Theorem\n\nLet \( A, B \) and \( C \) be three Points on a Line in \( {\mathbb{{RP}}}^{2} \), and let \( {A}^{\prime },{B}^{\prime } \) and \( {C}^{\prime } \) earlier, in Subsection be three Points on another Line. Let \( B{C}^{\prime } \) and \( {B}^{\prime }C \) meet at \( P, C... | Proof Let \( V \) be the Point of intersection of the two given Lines. Also let the Lines \( B{A}^{\prime } \) and \( A{C}^{\prime } \) meet at the Point \( S \), and the Lines \( B{C}^{\prime } \) and \( C{A}^{\prime } \) meet at the Point \( T \) .\n\nNow, the Points \( V,{A}^{\prime },{B}^{\prime },{C}^{\prime } \) ... | Yes |
Example 2 In an embedding plane, the points \( A, B, C, D \) lie in order along a line with the distances \( {AB},{BC},{CD} \) being 1 unit,3 units and 2 units, respectively. Determine the cross-ratios (ABCD), (BACD) and (ACBD). | Solution Using equation (10) and the sign convention for ratios, we have\n\n\[ \left( {ABCD}\right) = \frac{AC}{CB}/\frac{AD}{DB} = \left( {-\frac{4}{3}}\right) /\left( {-\frac{6}{5}}\right) = \frac{10}{9}, \]\n\n\[ \left( {BACD}\right) = \frac{BC}{CA}/\frac{BD}{DA} = \left( {-\frac{3}{4}}\right) /\left( {-\frac{5}{6}}... | Yes |
Problem 4 The points \( A, B, C, D \) lie in order along a line with the distances \( {AB},{BC},{CD} \) being 2 units,1 unit and 3 units, respectively. Determine the cross-ratios (ABCD) and (DBCA). | To be specific, suppose that the Points \( A, B, C, D \) are collinear, but that \( A \) is an ideal Point for the embedding plane \( \pi \), as shown in the margin. As before, we can let \( \mathbf{b},\mathbf{c},\mathbf{d} \) be the position vectors of the points \( B, C, D \) on \( \pi \), but we take a to be a unit ... | No |
Example 3 Determine (ABCD) for the collinear points \( A, B, C, D \) illustrated in the margin, where \( C \) is an ideal Point. | Solution Since \( C \) is an ideal Point, we have\n\n\[ \left( {ABCD}\right) = \frac{BD}{AD} = \frac{4}{1} = 4. \] | Yes |
Example 4 An aerial camera photographs a car travelling along a straight road on flat ground towards a junction. Before the junction there are two warning signs at distances of \( 4\mathrm{\;{km}} \) and \( 2\mathrm{\;{km}} \) from the junction. On the film the signs are \( 1\mathrm{\;{cm}} \) and \( 3\mathrm{\;{cm}} \... | Solution Let \( A \) and \( B \) denote the signs, \( C \) denote the car, and \( D \) denote the junction, and let \( {A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime } \) be their images on the film. Then \[ \left( {{A}^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }}\right) = \frac{{A}^{\prime }{C}^{\prime }}{... | Yes |
Find an equation for the projective figure in \( {\mathbb{{RP}}}^{2} \) which corresponds to the parabola \( \left\{ {\left( {x, y, z}\right) : y = {x}^{2}, z = 1}\right\} \) in the standard embedding plane. Which ideal Points should be associated with the parabola? | In general, any conic in the standard embedding plane can be expressed in the form\n\n\[ \left\{ {\left( {x, y, z}\right) : A{x}^{2} + {Bxy} + C{y}^{2} + {Fx} + {Gy} + H = 0, z = 1}\right\} .\n\]\n\nSince any Point \( \left\lbrack {{x}^{\prime },{y}^{\prime },{z}^{\prime }}\right\rbrack \) on the corresponding projecti... | Yes |
Theorem 1 Let \( t \) be a projective transformation, and let \( E \) be a nondegenerate projective conic. Then \( t\left( E\right) \) is a non-degenerate projective conic. | Proof Let \( E \) have equation\n\n\[ A{x}^{2} + {Bxy} + C{y}^{2} + {Fxz} + {Gyz} + H{z}^{2} = 0. \]\n\n(2)\n\nThen any Point that lies on \( E \) has homogeneous coordinates \( \left\lbrack {x, y, z}\right\rbrack \) which satisfy equation (2). If \( \left\lbrack {{x}^{\prime },{y}^{\prime },{z}^{\prime }}\right\rbrack... | Yes |
Theorem 2 Let \( t \) be a projective transformation, and let the Line \( \ell \) be a tangent to a non-degenerate projective conic \( E \) at a Point \( P \) . Then \( t\left( \ell \right) \) is a tangent to \( t\left( E\right) \) at \( t\left( P\right) \) . Also, if \( Q \) is a Point inside \( E \), then \( t\left( ... | Proof By the definition of tangent, \( P \) is the only Point that \( \ell \) and \( E \) have in common. Since \( t \) is a one-one map of \( {\mathbb{{RP}}}^{2} \) onto itself, it follows that \( t\left( P\right) \) is the only Point that \( t\left( \ell \right) \) and \( t\left( E\right) \) have in common. In other ... | Yes |
Theorem 6 Three Tangents Theorem\n\nLet a non-degenerate plane conic touch the sides \( {BC},{CA} \) and \( {AB} \) of a triangle \( \bigtriangleup {ABC} \) in \( {\mathbb{R}}^{2} \) at the points \( P, Q \) and \( R \), respectively. Then \( {AP},{BQ} \) and \( {CR} \) are concurrent. | Proof The theorem concerns a non-degenerate conic, its tangents, and con-\n\n\n\nSubsection 2.4.2\n\ncurrency of lines. Since all of these properties are projective properties, it is sufficient to prove the result fo... | Yes |
Determine \( {s}_{11},{s}_{22},{s}_{12} \) and \( {s}_{1} \) for the hyperbola with equation\n\n\[ 3{x}^{2} - {2xy} - {y}^{2} + {5x} - y - 4 = 0 \]\n\nat the points \( {P}_{1} = \left( {3,2}\right) \) and \( {P}_{2} = \left( {-5, - 2}\right) \). Hence determine whether either \( {P}_{1} \) or \( {P}_{2} \) lies on the ... | Solution The equation of the conic may be written in Joachimsthal's notation as \( s = 0 \), where\n\n\[ s = 3{x}^{2} - {2xy} - {y}^{2} + {5x} - y - 4. \]\n\nSince here we have \( {x}_{1} = 3,{y}_{1} = 2,{x}_{2} = - 5 \) and \( {y}_{2} = - 2 \), we deduce that\n\n\[ {s}_{11} = 3 \cdot {3}^{2} - 2 \cdot 3 \cdot 2 - {2}^... | Yes |
Determine the ratios in which the hyperbola with equation\n\n\[ \n3{x}^{2} - {2xy} - {y}^{2} + {5x} - y - 4 = 0 \n\]\n\ndivides the line segment from \( {P}_{1} = \left( {3,2}\right) \) to \( {P}_{2} = \left( {-5, - 2}\right) \) . | Solution First observe that the hyperbola and the points \( {P}_{1} \) and \( {P}_{2} \) are the same as those used in Example 1, so we can use the values \( {s}_{11} = {20},{s}_{22} = {24} \) and \( {s}_{12} = - {34} \) calculated there. It follows that we can rewrite Joachimsthal’s Section Equation in this case as\n\... | Yes |
Example 3 Determine the equation of the tangent at \( {P}_{1} = \left( {1,1}\right) \) to the hyperbola with equation\n\n\[ 3{x}^{2} - {2xy} - {y}^{2} + {5x} - y - 4 = 0. \] | Solution The equation of the conic may be written in Joachimsthal's notation as \( s = 0 \), where\n\n\[ s = 3{x}^{2} - {2xy} - {y}^{2} + {5x} - y - 4. \]\n\nSince here we have \( {x}_{1} = 1 \) and \( {y}_{1} = 1 \), we deduce that\n\n\[ {s}_{1} = 3 \cdot 1 \cdot x - 2\frac{1 \cdot y + x \cdot 1}{2} - 1 \cdot y + 5\fr... | Yes |
Example 4 Find the equations of the tangents from the point \( \left( {1,1}\right) \) to the ellipse with equation \( {x}^{2} + 2{y}^{2} = 1 \) . | Solution The equation of the ellipse may be written in Joachimsthal's notation as \( s = 0 \), where\n\n\[ s = {x}^{2} + 2{y}^{2} - 1 \]\n\nSince here we have \( {x}_{1} = 1 \) and \( {y}_{1} = 1 \), we deduce that\n\n\[ {s}_{11} = {1}^{2} + 2 \cdot {1}^{2} - 1 = 2 \]\n\nand\n\n\[ {s}_{1} = 1 \cdot x + 2 \cdot 1 \cdot ... | Yes |
Example 5 Determine the polar of \( {P}_{1} = \\left( {2,2}\\right) \) with respect to the hyperbola \( E \) with equation\n\n\[ 3{x}^{2} - {2xy} - {y}^{2} + {5x} - y - 4 = 0. \] | Solution The equation of the hyperbola \( E \) may be written in Joachimsthal’s\n\nnotation as \( s = 0 \), where\n\n\[ s = 3{x}^{2} - {2xy} - {y}^{2} + {5x} - y - 4. \]\n\nSince here we have \( {x}_{1} = 2 \) and \( {y}_{1} = 2 \), we deduce that\n\n\[ {s}_{1} = 3 \\cdot 2 \\cdot x - 2\\frac{2 \\cdot y + x \\cdot 2}{2... | Yes |
Theorem 3 La Hire's Theorem\n\nLet \( E \) be a non-degenerate plane conic, and let \( {p}_{1} \) be the polar of a point \( {P}_{1} \) in \( {\mathbb{R}}^{2} \) . Then each point of \( {p}_{1} \) has a polar which passes through \( {P}_{1} \) . | Proof Let \( {P}_{1} = \left( {{x}_{1},{y}_{1}}\right) \) be a point in \( {\mathbb{R}}^{2} \) . Then, by definition, the polar \( {p}_{1} \) of \( {P}_{1} \) with respect to \( E \) has the equation \( {s}_{1} = 0 \) .\n\nLet \( {P}_{2} = \left( {{x}_{2},{y}_{2}}\right) \) be any point on \( {p}_{1} \), so that in par... | Yes |
Example 6 Determine \( {s}_{1},{s}_{11} \) and \( {s}_{12} \) for the projective conic\n\n\[ 4{x}^{2} + {xy} - 2{y}^{2} - {8xz} - {2yz} + 4{z}^{2} = 0 \]\n\nat the Points \( \left\lbrack {{x}_{1},{y}_{1},{z}_{1}}\right\rbrack = \left\lbrack {1,0,2}\right\rbrack \) and \( \left\lbrack {{x}_{2},{y}_{2},{z}_{2}}\right\rbr... | Solution Using Joachimsthal’s notation with \( A = 4, B = 1, C = - 2 \), \( F = - 8, G = - 2 \) and \( H = 4 \), we deduce that\n\n\[ {s}_{1} = {4x} + \frac{1}{2}\left( {y + 0}\right) - 2 \cdot 0 - 4\left( {z + {2x}}\right) - \left( {0 + {2y}}\right) + 4 \cdot {2z} \]\n\n\[ = - {4x} - \frac{3}{2}y + {4z} \]\n\n\[ {s}_{... | Yes |
(a) Determine the equation of the tangent to \( E \) at the Point \( \left\lbrack {0,1,1}\right\rbrack \) . | Let \( s = 4{x}^{2} + {xy} - 2{y}^{2} - {8xz} - {2yz} + 4{z}^{2} \) and \( \left\lbrack {{x}_{1},{y}_{1},{z}_{1}}\right\rbrack = \left\lbrack {0,1,1}\right\rbrack \) .\n\nThen\n\n\[ {s}_{1} = 0 + \frac{1}{2}\left( {0 + x}\right) - {2y} - 4\left( {0 + x}\right) - \left( {z + y}\right) + {4z} \]\n\n\[ = - \frac{7}{2}x - ... | Yes |
Determine the equation of the projective conic which passes through the Points \( \\left\\lbrack {1,0,0}\\right\\rbrack ,\\left\\lbrack {0,1,0}\\right\\rbrack ,\\left\\lbrack {0,0,1}\\right\\rbrack ,\\left\\lbrack {1,1,1}\\right\\rbrack \) and \( \\left\\lbrack {1,2,3}\\right\\rbrack \) . | Let the projective conic have equation\n\n\[ \nA{x}^{2} + {Bxy} + C{y}^{2} + {Fxz} + {Gyz} + H{z}^{2} = 0.\n\]\n\nSince \( \\left\\lbrack {1,0,0}\\right\\rbrack \) lies on the projective conic, we must have \( A = 0 \) . Similarly, since \( \\left\\lbrack {0,1,0}\\right\\rbrack \) and \( \\left\\lbrack {0,0,1}\\right\\... | Yes |
There is a unique non-degenerate projective conic through any given set of five Points, no three of which are collinear. In particular, if the five Points are \( \left\lbrack {1,0,0}\right\rbrack ,\left\lbrack {0,1,0}\right\rbrack ,\left\lbrack {0,0,1}\right\rbrack ,\left\lbrack {1,1,1}\right\rbrack \) and \( \left\lbr... | Proof By the Fundamental Theorem of Projective Geometry, there is a projective transformation \( t \) which maps four of the Points to \( \left\lbrack {1,0,0}\right\rbrack ,\left\lbrack {0,1,0}\right\rbrack \) , \( \left\lbrack {0,0,1}\right\rbrack \) and \( \left\lbrack {1,1,1}\right\rbrack \) . Let \( \left\lbrack {a... | Yes |
Example 2 The Points \( \left\lbrack {1,1,1}\right\rbrack ,\left\lbrack {1,2,2}\right\rbrack ,\left\lbrack {1,2,1}\right\rbrack \) lie on the projective conic \( E \) with equation \( 2{x}^{2} + {2xy} - {y}^{2} + {yz} - {5xz} + {z}^{2} = 0 \) .\n\n(a) Verify that the projective transformation \( {t}_{1} : \left\lbrack ... | ## Solution\n\n(a) Let \( {\mathbf{x}}^{\prime } = \mathbf{{Ax}} \), so that\n\n\[ \left( \begin{array}{l} {x}^{\prime } \\ {y}^{\prime } \\ {z}^{\prime } \end{array}\right) = \left( \begin{array}{rrr} 2 & - 1 & 0 \\ - 1 & 0 & 1 \\ 0 & 1 & - 1 \end{array}\right) \left( \begin{array}{l} x \\ y \\ z \end{array}\right) . ... | Yes |
Problem 3 The Points \( \left\lbrack {-2,0,1}\right\rbrack ,\left\lbrack {0, - 3,2}\right\rbrack ,\left\lbrack {1, - 2,1}\right\rbrack \) lie on the projective conic \( E \) with equation \( {17}{x}^{2} + {47xy} + {32}{y}^{2} + {67xz} + {92yz} + \) \( {66}{z}^{2} = 0 \) . | (a) Verify that the projective transformation \( t \) with an associated matrix\n\n\[ \mathbf{A} = \left( \begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 6 \end{array}\right) \]\n\nmaps \( \left\lbrack {-2,0,1}\right\rbrack ,\left\lbrack {0, - 3,2}\right\rbrack ,\left\lbrack {1, - 2,1}\right\rbrack \) to \( \left\... | Yes |
Theorem 4 Pascal's Theorem\n\nLet \( A, B, C,{A}^{\prime },{B}^{\prime } \) and \( {C}^{\prime } \) be six distinct Points on a non-degenerate projective conic. Let \( {BC} \) and \( {B}^{\prime }C \) intersect at \( P, C{A}^{\prime } \) and \( {C}^{\prime }A \) intersect at \( Q \) , and \( A{B}^{\prime } \) and \( {A... | Proof By the Three Points Theorem we can let the equation of the projective conic be in the standard form \( {xy} + {yz} + {zx} = 0 \), with \( A = \left\lbrack {1,0,0}\right\rbrack, B = \left\lbrack {0,1,0}\right\rbrack, C = \left\lbrack {0,0,1}\right\rbrack \) . Also, by the Parametrization Theorem, we can let \( {A}... | Yes |
Theorem 6 Let \( E \) be the projective conic with equation\n\n\[{x}^{2} + {y}^{2} = {z}^{2}\]\n\nand let \( P = \left\lbrack {a, b, c}\right\rbrack \) be any Point in \( {\mathbb{{RP}}}^{2} \) . Then:\n\n(a) if \( P \in E \), then \( {ax} + {by} - {cz} = 0 \) is the equation of the tangent to \( E \) at \( P \) ;\n\n(... | Proof In Joachimsthal’s notation, the standard form becomes \( s = 0 \), where Subsection 4.2.2 \( s = {x}^{2} + {y}^{2} - {z}^{2} \) . So at any Point \( P = \left\lbrack {{x}_{1},{y}_{1},{z}_{1}}\right\rbrack \) of \( {\mathbb{{RP}}}^{2} \), we have \( {s}_{1} = \) \( x{x}_{1} + y{y}_{1} - z{z}_{1} \)\n\nNow recall t... | Yes |
Example 3 Determine whether each of the following Lines touches the projective conic with equation \( {x}^{2} + {y}^{2} = {z}^{2} \) . For each Line that does, state the Point of tangency.\n\n(a) \( {3x} - {5y} + {4z} = 0 \)\n\[ \n\text{(b)}{3x} - {4y} + {5z} = 0 \n\] | Solution If a Line is the tangent to the projective conic \( {x}^{2} + {y}^{2} - {z}^{2} = 0 \) at some Point \( P = \left\lbrack {a, b, c}\right\rbrack \), say, then its equation must be \( {ax} + {by} - {cz} = 0 \) (or some multiple of this).\n\n(a) Comparing the equations \( {3x} - {5y} + {4z} = 0 \) and \( {ax} + {... | Yes |
Theorem 8 Three Tangents and Three Chords Theorem\n\nLet a non-degenerate plane conic touch the sides of a triangle at the points \( P, Q \) and \( R \), respectively, and let the tangents at \( P, Q \) and \( R \) meet the extended chords \( {QR},{RP} \) and \( {PQ} \) at the points \( A, B \) and \( C \), respectivel... | Proof Since the theorem is concerned exclusively with projective properties, we shall prove it as a projective theorem about a projective conic. Since The result then follows as this projective conic is non-degenerate, we may assume that its equation is in an interpretation in an the standard form \( {x}^{2} + {y}^{2} ... | Yes |
Theorem 2). Also, the equation of the chord \( {QR} \) is \[ \left| \begin{matrix} x & y & z \\ 1 & 0 & - 1 \\ 0 & 1 & 1 \end{matrix}\right| = 0 \] which we can rewrite in the form \[ x - y + z = 0. \] | It follows that at the Point \( A \), both equations (11) and (12) must hold, so that \( x = z \) and \( y = x + z = {2z} \) . Hence \( A \) must have homogeneous coordinates \( \left\lbrack {z,{2z}, z}\right\rbrack \) or, equivalently, \( \left\lbrack {1,2,1}\right\rbrack \) . | Yes |
Pascal’s Theorem Let \( A, B, C,{A}^{\prime },{B}^{\prime } \) and \( {C}^{\prime } \) be six distinct Points on a non-degenerate projective conic. Let \( B{C}^{\prime } \) and \( {B}^{\prime }C \) intersect at \( P, C{A}^{\prime } \) and \( {C}^{\prime }A \) intersect at \( Q \), and \( A{B}^{\prime } \) and \( {A}^{\... | <table><thead><tr><th>Pascal’s Theorem</th><th>Its dual</th></tr></thead><tr><td>Let \( A, B, C,{A}^{\prime },{B}^{\prime } \) and \( {C}^{\prime } \)</td><td>Let \( a, b, c,{a}^{\prime },{b}^{\prime } \) and \( {c}^{\prime } \)</td></tr><tr><td>be six distinct Points</td><td>be six distinct tangents</td></tr><tr><td>o... | Yes |
Theorem 2 Inversion in the unit circle \( \mathcal{C} \) is the function\n\n\[ t : \left( {x, y}\right) \mapsto \left( {\frac{x}{{x}^{2} + {y}^{2}},\frac{y}{{x}^{2} + {y}^{2}}}\right) \;\left( {\left( {x, y}\right) \in {\mathbb{R}}^{2}-\{ O\} }\right) . \] | We may use this theorem to find the image of any non-zero point of \( {\mathbb{R}}^{2} \) under inversion in the unit circle \( \mathcal{C} \) . For example, the image of \( \left( {3, - 2}\right) \) is the point\n\n\[ \left( {\frac{3}{{3}^{2} + {\left( -2\right) }^{2}},\frac{-2}{{3}^{2} + {\left( -2\right) }^{2}}}\rig... | Yes |
Problem 2 Determine the image of each of the following points under inversion in \( \mathcal{C} \). (a) \( \left( {4,1}\right) \; \) (b) \( \left( {\frac{1}{2}, - \frac{1}{4}}\right) \) | Now let \( \left( {{x}^{\prime },{y}^{\prime }}\right) \) be the image of \( \left( {x, y}\right) \) under inversion in \( \mathcal{C} \). Since inversion is self-inverse, it follows that \[ \left( {x, y}\right) = \left( {\frac{{x}^{\prime }}{{\left( {x}^{\prime }\right) }^{2} + {\left( {y}^{\prime }\right) }^{2}},\fra... | No |
Example 1 Determine the image under inversion in \( \mathcal{C} \) of the line \( {2x} + {4y} = 1 \) . | Solution Let \( \left( {x, y}\right) \) be an arbitrary point on the line \( {2x} + {4y} = 1 \), and let \( \left( {{x}^{\prime },{y}^{\prime }}\right) \) be the image of \( \left( {x, y}\right) \) under inversion in \( \mathcal{C} \) . Then\n\n\[ \left( {x, y}\right) = \left( {\frac{{x}^{\prime }}{{\left( {x}^{\prime ... | Yes |
Example 2 Determine the image under inversion in \( \\mathcal{C} \) of the line \( y = x \) (with the origin removed). | Solution Replacing \( x \) by \( \\frac{x}{{x}^{2} + {y}^{2}} \) and \( y \) by \( \\frac{y}{{x}^{2} + {y}^{2}} \), we obtain\n\n\[ \n\\frac{y}{{x}^{2} + {y}^{2}} = \\frac{x}{{x}^{2} + {y}^{2}}.\n\]\n\nHence\n\n\[ \ny = x\\text{.}\n\]\n\nThis is the line we started with. Just as the origin had to be excluded from that ... | No |
Theorem 3 Images of Lines under Inversion\n\nUnder inversion in a circle with centre \( O \) :\n\n(a) a line that does not pass through \( O \) maps onto a circle punctured at \( O \) ;\n\n(b) a line punctured at \( O \) maps onto itself. | Proof First choose a pair of coordinate axes with origin at \( O \), and choose a unit of length equal to the radius of the circle in which we are inverting. Then the circle in which we are inverting becomes the unit circle \( \mathcal{C} \) .\n\n(a) If \( \ell \) is a line that does not pass through the origin, then i... | Yes |
Example 3 Use the above strategy to determine the image under inversion in \( \mathcal{C} \) of the circle \( C \) with centre \( \left( {2,0}\right) \) and radius 1 . | Solution The circle \( C \) has equation \( {\left( x - 2\right) }^{2} + {y}^{2} = 1 \), which we may rewrite in the form\n\n\[{x}^{2} + {y}^{2} - {4x} + 3 = 0.\]\n\nUsing the above strategy, we deduce that the image of \( C \) under inversion in \( \mathcal{C} \) Note that the origin does has equation not lie on \( C ... | Yes |
Example 4 Let \( C \) be the circle with centre \( \left( {-2,0}\right) \) and radius 2, punctured at the origin. Determine the image of \( C \) under inversion in \( \mathcal{C} \) . | Solution The circle \( C \) has equation \( {\left( x + 2\right) }^{2} + {y}^{2} = {2}^{2} \), which we may rewrite in the form\n\n\[ \n{x}^{2} + {y}^{2} + {4x} = 0.\n\]\n\nUsing the above strategy, we deduce that the image of \( C \) under inversion in \( \mathcal{C} \) has equation\n\n\[ \n{\left( \frac{x}{{x}^{2} + ... | Yes |
Problem 6 Let \( C \) be the circle with centre \( \\left( {0, - \\frac{1}{4}}\\right) \) and radius \( \\frac{1}{4} \) , punctured at the origin. Determine the image of \( C \) under inversion in \( \\mathcal{C} \) . | ## Theorem 4 Images of Circles under Inversion\n\nUnder inversion in a circle with centre \( O \) :\n\n(a) a circle that does not pass through \( O \) maps onto a circle;\n\n(b) a circle punctured at \( O \) maps onto a line that does not pass through \( O \) .\n\nProof As for Theorem 3, we choose a pair of coordinate ... | Yes |
Example 5 Determine the image of each of the following under inversion in the unit circle \( \mathcal{C} \) :\n\n(a) the line \( \ell \) with equation \( x = 2 \) ;\n\n(b) the circle \( C \) with centre \( \left( {0,2}\right) \) and radius 1 . | (a) From the summary, we know that \( \ell \) maps to a circle \( C \) punctured at the origin. This circle passes through the point \( \left( {\frac{1}{2},0}\right) \), since \( \left( {\frac{1}{2},0}\right) \) is the image of the point \( \left( {2,0}\right) \) on the line. Since \( \ell \) is symmetrical about the \... | Yes |
Lemma 1 Symmetry Lemma\n\nLet \( \ell \) be a line that does not pass through the point \( O \) . Then under inversion in a circle with centre \( O,\ell \) maps to a circle \( C \) (punctured at \( O \) ), and the tangent to \( C \) at \( O \) is parallel to \( \ell \) . | Now consider what happens to the angle between two lines \( {\ell }_{1} \) and \( {\ell }_{2} \) which\n\nintersect at some point \( A \) other than \( O \), as shown below. For the moment we We ask you to investigate shall assume that neither line passes through \( O \), so that under the inversion what happens when o... | No |
Example 6 A family of circles shares a common tangent at the origin \( O \), as shown in the margin. Describe the effect of inverting the family of circles (all punctured at \( O \) ) in the unit circle \( \mathcal{C} \) . Illustrate your answer with a sketch. | Solution Let \( \ell \) be the common tangent to the circles, as shown on the left below. If \( d \) is the punctured line through \( O \) perpendicular to \( \ell \), then each of the circles crosses \( d \) at right angles. Since all the circles are punctured at \( O \), their images under the inversion must be strai... | Yes |
Theorem 1 Each isometry \( t \) of the plane can be represented in the complex plane by one of the functions\n\n\[ t\left( z\right) = {az} + b\;\text{ or }\;t\left( z\right) = a\bar{z} + b, \]\n\nwhere \( a, b \in \mathbb{C},\;\left| a\right| = 1 \) . Conversely, all such functions represent isometries. | Proof The converse is easy to prove, because every function of the type described is a composite of the basic isometries described above.\n\n## Extending the Plane\n\nSo let \( t \) be an isometry of the complex plane, and let \( t\left( 0\right) = b, t\left( 1\right) = c \) . If we denote \( c - b \) by \( a \), then ... | Yes |
(a) Show that \( t \) represents an isometry. | (a) The coefficient of \( \bar{z} \) is \( i \), which has modulus \( \left| i\right| = 1 \) . By Theorem 1, it follows that \( t \) is an isometry. | Yes |
Theorem 3 An inversion in a circle \( C \) of radius \( r \) with centre \( \left( {a, b}\right) \) may be represented in the complex plane by the transformation\n\n\[ t\left( z\right) = \frac{{r}^{2}}{\overline{z - c}} + c\;\left( {z \in \mathbb{C}-\{ c\} }\right) ,\] \n\nwhere \( c = a + {ib} \) . | Proof We first consider the case where \( C \) is the unit circle \( \mathcal{C} \) . The image\n\nunder inversion in \( \mathcal{C} \) of the point \( \left( {x, y}\right) \) is the point \( \left( {\frac{x}{{x}^{2} + {y}^{2}},\frac{y}{{x}^{2} + {y}^{2}}}\right) \) . We may By Theorem 2,\n\nreformulate this expression... | Yes |
Problem 5 Determine, in Cartesian form, the image under inversion in the unit circle \( \mathcal{C} \) of each of the following points. | \[ \text{(a)} - \sqrt{3} + i\;\text{(b)} - 3 - {4i} \] | No |
Find the composite \( t = {t}_{2} \circ {t}_{1} \) where\n\n\( {t}_{1} \) is the inversion in the unit circle \( \mathcal{C} \) ,\n\n\( {t}_{2} \) is the extended conjugation function. | Solution Since\n\n\[ \n{t}_{1}\left( z\right) = \left\{ {\begin{array}{ll} \frac{1}{\bar{z}}, & \text{ if }z \in \mathbb{C}-\{ O\} , \\ \infty , & \text{ if }z = 0, \\ 0, & \text{ if }z = \infty , \end{array}\;\text{ and }\;{t}_{2}\left( z\right) = \left\{ \begin{array}{ll} \bar{z}, & \text{ if }z \in \mathbb{C}, \\ \i... | Yes |
Express \( t \) as a composite of inversions of the extended complex plane. | In addition to mapping \( \infty \) to \( \infty \), the transformation \( t \) scales the complex plane by the factor \( \left| {2\left( {-1 + \sqrt{3}i}\right) }\right| = 4 \), rotates it through the angle \( \operatorname{Arg}\left( {2\left( {-1 + \sqrt{3}i}\right) }\right) = \frac{2\pi }{3} \), and then translates ... | Yes |
Theorem 6 Let \( \pi \) denote the mapping of the Riemann sphere \( \mathbb{S} \) onto \( \widehat{\mathbb{C}} \) given by stereographic projection. Then the stereographic projection of the point \( \left( {X, Y, Z}\right) \) of \( \mathbb{S} \) onto the point \( z = x + {iy} \) of \( \widehat{\mathbb{C}} \) is given b... | Proof Let the point \( {P}^{\prime }\left( {X, Y, Z}\right) \) be a point on \( \mathbb{S} \) (other than \( N \) ), and \( P\left( {x, y}\right) \) the point in the plane that corresponds to \( {P}^{\prime } \) under stereographic projection. Project the line \( N{P}^{\prime }P \) perpendicularly onto the \( \left( {Y... | No |
Theorem 7 Under stereographic projection, circles on the Riemann sphere map onto generalized circles in \( \widehat{\mathbb{C}} \) . | Proof A circle on the sphere is the intersection of the sphere \( {X}^{2} + {Y}^{2} + {Z}^{2} = 1 \) with some plane \( {aX} + {bY} + {cZ} + d = 0 \), where \( a, b \) and \( c \) are not all zero. It follows from substituting the expressions\n\n\[ X = \frac{2x}{{x}^{2} + {y}^{2} + 1},\;Y = \frac{2y}{{x}^{2} + {y}^{2} ... | Yes |
Theorem 8 Stereographic projection preserves the magnitude of angles. | Proof First, we make the following crucial observation. Recall that if a point \( P \) moves along a line \( \ell \) in the complex plane, then the corresponding point \( {P}^{\prime } \) on the Riemann sphere moves round a circle through \( N \) . Indeed, as \( P \) moves out ’towards \( {\infty }^{\prime },{P}^{\prim... | Yes |
Theorem 1 The extended reciprocal function and the extended linear functions are inversive transformations. | Since every inversion preserves the magnitude of angles and maps generalized circles to generalized circles, the same must be true of all composites of inversions. | No |
Theorem 3 The set of inversive transformations forms a group under the operation of composition of functions. | Proof We check that the four group axioms hold.\n\nGl closure Let \( r \) and \( s \) be inversive transformations. Then we can write\n\n\[ r = {t}_{1} \circ {t}_{2} \circ \ldots \circ {t}_{k} \]\n\nand\n\n\[ s = {t}_{k + 1} \circ {t}_{k + 2} \circ \ldots \circ {t}_{n} \]\n\nwhere \( {t}_{1},{t}_{2},\ldots ,{t}_{n} \) ... | Yes |
Theorem 4 Every Möbius transformation is an inversive transformation. | Proof Let \( M \) be the Möbius transformation defined by the formula\n\n\[ M\left( z\right) = \frac{{az} + b}{{cz} + d}. \]\n\nIf \( c = 0 \), then \( M \) is an extended linear function, and is therefore an inversive transformation.\n\nIf \( c \neq 0 \), then for \( z \in \mathbb{C} - \{ - d/c\} \) we can write\n\n\[... | Yes |
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