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Let \( T \) be the linear transformation with matrix \( A \) shown. If \( X \) is a point on the line \( l\left\lbrack {1, - 1,0}\right\rbrack \), show that \( T\left( X\right) \) is also a point on \( l \) and show that \( T\left( P\right) = P \) where \( P \) is the point \( P\left( {-\frac{2}{3}, - \frac{2}{3},1}\ri... | To find the images of points on \( l \), we note that \( X\left( {{x}_{1},{x}_{2},1}\right) \) is on \( l \) iff \( {x}_{1} - {x}_{2} = 0 \), that is, if \( {x}_{2} = {x}_{1} \) . Thus we can find the images of any point \( X \) on \( l \) as follows:\n\n\[ \left\lbrack \begin{array}{lll} 1 & 3 & 2 \\ 3 & 1 & 2 \\ 0 & ... | Yes |
Theorem 3.5. A one-to-one linear transformation of \( {V}^{ * } \) preserves collinearity (i.e., the images of collinear points are collinear). | Proof. Let \( {X}^{\prime }\left( {{x}_{1}^{\prime },{x}_{2}^{\prime },1}\right) ,{Y}^{\prime }\left( {{y}_{1}^{\prime },{y}_{2}^{\prime },1}\right) \), and \( {Z}^{\prime }\left( {{z}_{1}^{\prime },{z}_{2}^{\prime },1}\right) \) be images of the points \( X \) , \( Y \), and \( Z \) under a given one-to-one linear tra... | Yes |
Theorem 3.6. If the image of a point under a one-to-one linear transformation of \( {V}^{ * } \) is given by the matrix equation \( {X}^{\prime } = {AX} \) then the image of a line under this same transformation is given by the matrix equation \( k{u}^{\prime } = u{A}^{-1} \) for some nonzero scalar \( k \) . | Proof. Consider the line \( u\left\lbrack {{u}_{1},{u}_{2},{u}_{3}}\right\rbrack \) with equation \( {u}_{1}{x}_{1} + {u}_{2}{x}_{2} + {u}_{3}{x}_{3} = 0 \) ; that is, \( {uX} = 0 \) . Under the linear transformation, \( u \) maps to \( {u}^{\prime }, X \) maps to \( {X}^{\prime } \) , and \( {uX} = 0 \) iff \( {u}^{\p... | Yes |
Theorem 3.10. Let \( {u}^{\prime } \) and \( {v}^{\prime } \) be the images of lines \( u \) and \( v \) under an isometry. If the isometry is direct then \( m\left( {\angle \left( {{u}^{\prime },{v}^{\prime }}\right) }\right) = m\left( {\angle \left( {u, v}\right) }\right) \) . If the isometry is indirect then \( m\le... | Proof. Let \( u\left\lbrack {{u}_{1},{u}_{2},{u}_{3}}\right\rbrack \) and \( v\left\lbrack {{v}_{1},{v}_{2},{v}_{3}}\right\rbrack \) be two lines and \( {u}^{\prime }\left\lbrack {{u}_{1}^{\prime },{u}_{2}^{\prime },{u}_{3}^{\prime }}\right\rbrack \) and \( {v}^{\prime }\left\lbrack {{v}_{1}^{\prime },{v}_{2}^{\prime }... | No |
Theorem 3.12. A direct isometry other than the identity, with matrix \( A = \left\lbrack {a}_{ij}\right\rbrack \) has exactly one invariant point iff \( {a}_{11} \neq 1 \) . | Proof. The point \( X\left( {{x}_{1}{x}_{2},1}\right) \) is an invariant point of the isometry iff \( {AX} = X \) ;\n\n\[ \left\lbrack \begin{matrix} {a}_{11} & {a}_{12} & {a}_{13} \\ - {a}_{12} & {a}_{11} & {a}_{23} \\ 0 & 0 & 1 \end{matrix}\right\rbrack \left\lbrack \begin{array}{l} {x}_{1} \\ {x}_{2} \\ 1 \end{array... | Yes |
Theorem 3.13. A translation \( T \) has matrix representation\n\n\[ \left\lbrack \begin{array}{lll} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{array}\right\rbrack \]\n\n\( {T}^{-1} \) is also a translation and has matrix representation\n\n\[ \left\lbrack \begin{array}{rrr} 1 & 0 & - a \\ 0 & 1 & - b \\ 0 & 0 & 1 \end{arr... | The verification of the second half of the previous theorem as well as the next two theorems involve calculations with elementary matrix algebra (see Exercises 9-11). | No |
Theorem 3.16. If a translation maps a line \( u \) to a line \( v \), then \( u \) and \( v \) are either identical or parallel. | Proof. If \( A \) is the matrix of the translations, then \( {kv} = u{A}^{-1} \) since \( v \) is the image of \( u \) . So\n\n\[ k\left\lbrack {{v}_{1},{v}_{2},{v}_{3}}\right\rbrack = \left\lbrack {{u}_{1},{u}_{2},{u}_{3}}\right\rbrack \left\lbrack \begin{array}{rrr} 1 & 0 & - a \\ 0 & 1 & - b \\ 0 & 0 & 1 \end{array}... | No |
Theorem 3.17. If a translation maps \( P \) to \( {P}^{\prime }\left( {P \neq {P}^{\prime }}\right) \), then the line \( P{P}^{\prime } \) as well as all lines parallel to \( P{P}^{\prime } \) are invariant. No other lines are invariant. | Proof. Let \( P \) be a point with coordinates \( \left( {{p}_{1},{p}_{2},1}\right) \) . If \( T \) is a translation with a matrix of the form given in Theorem 3.13, \( {P}^{\prime } = T\left( P\right) \) has coordinates \( \left( {{p}_{1} + a,{p}_{2} + b,1}\right) \) so the equation of line \( P{P}^{\prime } \) is giv... | Yes |
Theorem 3.18. If \( u \) and \( v \) are parallel lines, then there is a translation mapping \( u \) to \( v \) . | Proof. Let \( X \) be a point on \( u,{X}^{\prime } \) a point on \( v \) . Then by Theorem 3.15 there is a translation maping \( X \) to \( {X}^{\prime } \) . But this translation also maps line \( u \) to a line \( {u}^{\prime } \) through \( {X}^{\prime } \) (see Fig. 3.2). Since \( {u}^{\prime } \) is parallel to \... | Yes |
Theorem 3.19. A rotation \( R \) with center \( C\left( {{c}_{1},{c}_{2},1}\right) \) has matrix representation:\n\n\[ \left\lbrack \begin{matrix} \cos \theta & - \sin \theta & {c}_{1}\left( {1 - \cos \theta }\right) + {c}_{2}\sin \theta \\ \sin \theta & \cos \theta & - {c}_{1}\sin \theta + {c}_{2}\left( {1 - \cos \the... | Proof. By definition, \( R \) has a matrix representation\n\n\[ \left\lbrack \begin{matrix} {a}_{11} & {a}_{12} & {a}_{13} \\ - {a}_{12} & {a}_{11} & {a}_{23} \\ 0 & 0 & 1 \end{matrix}\right\rbrack \text{ where }{\left( {a}_{11}\right) }^{2} + {\left( {a}_{12}\right) }^{2} = 1. \]\n\nSince \( \left| {a}_{11}\right| \le... | Yes |
Theorem 3.21. Under a rotation with center \( C \) and angle \( \theta \), any point \( P \neq C \) is mapped to a point \( {P}^{\prime } \) such that \( d\left( {C, P}\right) = d\left( {C,{P}^{\prime }}\right) \) and \( m\left( {\angle \left( {{PC}{P}^{\prime }}\right) = \theta }\right. \) . | Proof. Since a rotation is an isometry and the images of points \( C \) and \( P \) under this isometry are \( C \) and \( {P}^{\prime } \), respectively, it follows from the definition of isometries that \( d\left( {C, P}\right) = d\left( {C,{P}^{\prime }}\right) \) . To verify that \( m\left( {\angle {PC}{P}^{\prime ... | Yes |
Theorem 3.23. If \( \bigtriangleup {PQR} \) and \( \bigtriangleup {P}^{\prime }{Q}^{\prime }{R}^{\prime } \) are two triangles with \( m\left( \overline{PQ}\right) = m\left( \overline{{P}^{\prime }{Q}^{\prime }}\right) \) , \( m\left( \overline{QR}\right) = m\left( \overline{{Q}^{\prime }{R}^{\prime }}\right), m\left( ... | Proof. To show the congruence of the two triangles, it is sufficient to show that there is an isometry mapping \( \bigtriangleup {PQR} \) to \( \bigtriangleup {P}^{\prime }{Q}^{\prime }{R}^{\prime } \) (see Definition 3.12). In the following paragraph we will outline a procedure for obtaining such an isometry.\n\n![a19... | Yes |
Theorem 3.24. The matrix representation of a reflection \( {R}_{x} \) with axis \( x\left\lbrack {0,1,0}\right\rbrack \) is\n\n\[ \left\lbrack \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \] | Proof. All points on \( x \) have coordinates of the form \( \left( {{x}_{1},0,1}\right) .{R}_{x} \) is then an indirect isometry that keeps each point \( \left( {{x}_{1},0,1}\right) \) fixed, that is, for all \( {x}_{1} \in R \)\n\n\[ \left\lbrack \begin{matrix} {a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{12} & - {a}_{11} ... | Yes |
Theorem 3.25. The only invariant points under a reflection with axis \( m,{R}_{m} \), are those on \( m \) . For any point \( P \) not on \( m \), if \( {P}^{\prime } = {R}_{m}\left( P\right) \), then \( {P}^{\prime } \neq P \) and \( m \) is the perpendicular bisector of segment \( P{P}^{\prime } \) . | Proof. Let \( P \) be a point with coordinates \( \left( {{p}_{1},{p}_{2},1}\right) \) . Then \( {P}^{\prime } = {R}_{x}\left( P\right) \) has coordinates \( \left( {{p}_{1}, - {p}_{2},1}\right) \) so the invariant points of \( {R}_{x} \) are precisely those for which \( {p}_{2} = - {p}_{2} \) or \( {p}_{2} = 0 \) . Th... | No |
Theorem 3.26. Every line perpendicular to \( m \) is invariant under \( {R}_{m} \), and conversely any line invariant under \( {R}_{m} \) is either \( m \) or a line perpendicular to \( m \) . | Proof. As noted earlier, we will assume that \( m = x \) . Let \( u \) be a line perpendicular to \( x \) . Since \( x \) has coordinates \( \left\lbrack {0,1,0}\right\rbrack \) it follows from Definition 3.4 that \( u \) has coordinates \( \left\lbrack {{u}_{1},0,{u}_{3}}\right\rbrack \) . If \( {u}^{\prime } \) is th... | Yes |
Theorem 3.28. The product of two reflections \( {R}_{m} \) and \( {R}_{n} \) is (a) a translation mapping any point \( P \) to a point \( {P}^{\prime } \) where \( {d}^{ * }\left( {P,{P}^{\prime }}\right) = 2{d}^{ * }\left( {m, n}\right) \) if \( n \) and \( m \) are parallel ( \( {d}^{ * } \) indicates directed distan... | Proof. (a) We shall assume that \( m \) is the line with coordinates \( \left\lbrack {0,1,0}\right\rbrack ;n \) must then have coordinates \( \left\lbrack {0,1,{n}_{2}}\right\rbrack \) . We can measure the (perpendicular) distance between \( m \) and \( n \) along the line \( t\left\lbrack {1,0,0}\right\rbrack \), whic... | Yes |
Theorem 3.29. An indirect isometry is the product of one or three reflections. | Proof. The proof follows easily after noting that any indirect isometry can be expressed as a product of a direct isometry and \( {R}_{x} \) (see Exercise 9). | No |
Theorem 3.30. An indirect isometry is either a reflection or a glide reflection. | Proof. By Theorem 3.29 we need only consider indirect isometries that can be written as the product \( {R}_{c}{R}_{b}{R}_{a} \) . We will examine this product for each of several cases.\n\nCase 1. \( a, b, c \) are all parallel (see Fig. 3.8). Then let \( {c}^{\prime } \) be a line parallel to \( a \) such that \( {d}^... | Yes |
Theorem 3.34. The set of all symmetries of a set of points forms a group. | To find the symmetry group of a line segment \( \overline{PQ} \), we must determine which isometries keep \( \overline{PQ} \) invariant. Other than the identity, the only direct isometry that has this property is the rotation with center at the midpoint of \( \overline{PQ} \) and angle \( {180}^{ \circ } \) . Such rota... | Yes |
Theorem 3.40. A dilation with center \( O\\left( {0,0,1}\\right) \) and ratio \( r \) has matrix representation\n\n\[ \n\\left\\lbrack \\begin{array}{lll} r & 0 & 0 \\\\ 0 & r & 0 \\\\ 0 & 0 & 1 \\end{array}\\right\\rbrack \n\]\n\nA dilation with center \( C\\left( {{c}_{1},{c}_{2},1}\\right) \) has matrix representati... | Proof (Outline). For the first case, the requirement that a direct similarity with matrix \( A = \\left\\lbrack {a}_{ij}\\right\\rbrack \) keep \( O\\left( {0,0,1}\\right) \) invariant implies that \( {a}_{13} = {a}_{23} = 0 \) . The requirement that any point \( X\\left( {x,0,1}\\right) \) on the line \( \\left\\lbrac... | Yes |
Theorem 3.41. If \( {D}_{C, r} \) is a dilation with \( r \neq 1 \) and \( m \) is a line not incident with \( C \) , then \( {D}_{C, r}\left( m\right) = {m}^{\prime } \) is a distinct line parallel to \( m \) . | Proof. The line equation of this dilation requires the matrix of \( {\left( {D}_{C, r}\right) }^{-1} \) . Since this transformation is also a dilation with center \( C \) and ratio \( {r}^{\prime } = 1/r \) (see Exercise 11), its matrix representation is given by Theorem 3.40.\n\nUsing this matrix in the line equation ... | Yes |
Theorem 3.45. If \( T \) is an affinity and \( m \) and \( n \) are parallel lines, then \( T\left( m\right) \) is parallel to \( T\left( n\right) \) . | Proof. Assume \( m\left\lbrack {{m}_{1},{m}_{2},{m}_{3}}\right\rbrack \) and \( n\left\lbrack {{n}_{1},{n}_{2},{n}_{3}}\right\rbrack \) are parallel lines. Then there is a nonzero real number \( t \) such that \( {n}_{1} = t{m}_{1} \) and \( {n}_{2} = t{m}_{2} \) . We can find \( T\left( m\right) \), the image of line ... | Yes |
Theorem 3.47. If \( T \) is an affinity and \( P \) , \( Q \) , and \( R \) are three collinear points with \( P \) , between \( Q \) and \( R \), then \( T\left( P\right) \) is between \( T\left( Q\right) \) and \( T\left( R\right) \) . | Proof. Since \( P \) is between \( Q \) and \( R, d\left( {Q, P}\right) + d\left( {P, R}\right) = d\left( {Q, R}\right) \) . Dividing each term of this equation by \( d\left( {Q, R}\right) \) and letting \( d\left( {Q, P}\right) /d\left( {Q, R}\right) = k \), gives \( d\left( {P, R}\right) /d\left( {Q, R}\right) = 1 - ... | Yes |
Theorem 3.48. The matrix representation of a shear with axis \( x\left\lbrack {0,1,0}\right\rbrack \) is\n\n\[ \left\lbrack \begin{array}{lll} 1 & j & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \] | Proof. Since \( {S}_{x} \) keeps each point on the line \( x\left\lbrack {0,1,0}\right\rbrack \) invariant, the following equation must be true for all real numbers \( {x}_{1} \)\n\n\[ \left\lbrack \begin{array}{lll} {a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ 0 & 0 & 1 \end{array}\right\rbrack ... | Yes |
Theorem 3.49. The matrix representation of a strain with axis \( x\left\lbrack {0,1,0}\right\rbrack \) is\n\n\[ \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \] | In general, the matrix representation of strain \( {T}_{m} \) can be found using \( {T}_{m} = S{T}_{x}{S}^{-1} \) where \( S \) is a direct isometry mapping \( x \) to \( m\left( {S\left( x\right) = m}\right) \) . | No |
Theorem 3.50. Any affinity can be written as the product of a shear, a strain, and a direct similarity. | Proof. We can verify this theorem by merely demonstrating that the following product does indeed yield the matrix of a general affinity as indicated:\n\n\[ \left\lbrack \begin{array}{lll} {a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ 0 & 0 & 1 \end{array}\right\rbrack = \left\lbrack \begin{matrix}... | Yes |
Theorem 3.51. Given two triangles, \( \bigtriangleup {PQR} \) and \( \bigtriangleup {P}^{\prime }{Q}^{\prime }{R}^{\prime } \), there is an affinity mapping \( \bigtriangleup {PQR} \) to \( \bigtriangleup {P}^{\prime }{Q}^{\prime }{R}^{\prime } \) . | Proof. We can show that there is an affinity mapping \( P, Q \), and \( R \) to \( {P}^{\prime },{Q}^{\prime } \), and \( {R}^{\prime } \), resectively, by finding a matrix \( A \) such that \( {P}^{\prime } = {AP},{Q}^{\prime } = {AQ} \), and \( {R}^{\prime } = {AR} \) . This involves six equations in six unknowns. Ho... | Yes |
Theorem 3.53. The image of a conic section under an affinity is a conic section of the same type. Furthermore, if \( A \) is the matrix of an affinity, then the matrix of the image conic section is \( {C}^{\prime } = {\left( {A}^{-1}\right) }^{t}C{A}^{-1} \) . | Proof. Under the affinity, \( X \) is mapped to \( {X}^{\prime } = {AX} \) . Solving for \( X \), gives \( X = {A}^{-1}{X}^{\prime } \) . Substituting this into the matrix equation \( {X}^{t}{CX} = 0 \) yields \( {\left( {A}^{-1}{X}^{\prime }\right) }^{t}C\left( {{A}^{-1}{X}^{\prime }}\right) = 0 \) or \( {X}^{\prime t... | No |
Theorem 4.1 (Dual of Axiom 3). There exist at least four lines, no three of which are concurrent. | Proof. Let \( A, B, C, D \) be four points, no three collinear, as guaranteed by Axiom 3. Then by Axiom 1, there exist the four lines \( {AB},{AC},{CD} \), and \( {BD} \) . If any three of these were concurrent, the dual of Axiom 1 would be contradicted. | Yes |
Theorem 4.2 (Dual of Axiom 4). The three diagonal lines of a complete quadrilateral are never concurrent. | Proof. Let \( {abcd} \) be an arbitrary complete quadrilateral. Let \( E = a \cdot b, F = b \cdot c \) , \( G = c \cdot d, H = a \cdot d, I = a \cdot c \), and \( J = b \cdot d \) . Then the diagonal lines are \( {EG},{FH} \) , and \( {IJ} \) . Assume these three lines are concurrent; that is, \( {EG},{FH} \), and \( {... | Yes |
Theorem 4.3 (Dual of Axiom 5). If two triangles are perspective from a line, they are perspective from a point. | Proof. Assume \( \bigtriangleup {ABC} \) and \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) are perspective from a line, that is, \( {AB} \cdot {A}^{\prime }{B}^{\prime } = P,{B}^{\prime }{C}^{\prime } \cdot {BC} = Q \), and \( {AC} \cdot {A}^{\prime }{C}^{\prime } = R \) are collinear (see Fig. 4.6). It ... | Yes |
Theorem 4.4. If \( A, B \), and \( C \) are three distinct, collinear points, then \( D \), the harmonic conjugate of \( C \) with respect to \( A \) and \( B \), is unique. | Proof. Let \( {EFGH} \) be a quadrangle used to find the point \( D \) . Assume a second quadrangle \( {E}^{\prime }{F}^{\prime }{G}^{\prime }{H}^{\prime } \) is also constructed so that \( {E}^{\prime }{H}^{\prime } \cdot {F}^{\prime }{G}^{\prime } = B,{E}^{\prime }{F}^{\prime } \cdot {G}^{\prime }{H}^{\prime } = A \)... | Yes |
Theorem 4.5. \( H\left( {{AB},{CD}}\right) \Leftrightarrow H\left( {{CD},{AB}}\right) \). | Proof. We assume \( H\left( {{AB},{CD}}\right) \) and show \( H\left( {{CD},{AB}}\right) \). A similar proof can be used to verify the second half of the equivalence.\n\nSince \( H\left( {{AB},{CD}}\right) \), there is a quadrangle \( {EFGH} \) such that \( A = {EF} \cdot {GH}, B = {EH} \cdot {FG}, C = {EG} \cdot n \),... | Yes |
Theorem 4.8 (Fundamental Theorem). A projectivity between two pencils is uniquely determined by three pairs of corresponding elements. | Proof. The existence of a projectivity has been demonstrated. The uniqueness follows from Axiom 6 as shown.\n\nCase 1. Two pencils of points. Assume that \( A, B, C \) are elements of a pencil of points with axis \( p \) and that \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) are the corresponding elements of a second... | Yes |
Theorem 4.9. The harmonic relation is invariant under a projectivity. So, for example, if \( H\left( {{AB},{CD}}\right) \) and \( {ABCD} \land {A}^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime } \), then \( H\left( {{A}^{\prime }{B}^{\prime },{C}^{\prime }{D}^{\prime }}\right) \) . | Proof. Since the projective plane possesses duality, and any projectivity is a product of perspectivities, it is sufficient to show that \( H\left( {{AB},{CD}}\right) \) implies \( H\left( {{ab},{cd}}\right) \) where \( {ABCD} \land {abcd} \) . Let \( O = a \cdot b \) thus \( a = {OA}, b = {OB} \), and so on. Since \( ... | Yes |
Theorem 4.10. If four elements of one pencil, A, B, C, D, form a harmonic set, \( H\left( {{AB},{CD}}\right) \), and four elements of a second pencil, \( {A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime } \), form a second harmonic set, \( H\left( {{A}^{\prime }{B}^{\prime },{C}^{\prime }{D}^{\prime }}\right) \), ... | Proof. By Theorem 4.8, there is a projectivity such that \( {ABC} \land {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . Let \( {D}^{ * } \) be the image of \( D \) under this projectivity. Then by Theorem 4.9, \( H\left( {{A}^{\prime }{B}^{\prime },{C}^{\prime }{D}^{ * }}\right) \) ; but by Theorem 4.4, the harmonic conju... | Yes |
Theorem 4.11. A projectivity between two distinct pencils of points determines a unique line called the axis of homology, which contains the intersections of the cross joins of all pairs of corresponding points. | Proof. Consider two distinct pencils of points with axes \( p \) and \( {p}^{\prime } \) . Assume \( {ABC} \land {A}^{\prime }{B}^{\prime }{C}^{\prime } \), where \( P = p \cdot {p}^{\prime } \) is none of the six points. Clearly \( {A}^{\prime }A,{A}^{\prime }B \) , \( {A}^{\prime }C \barwedge {ABC} \) and \( {A}^{\pr... | Yes |
Theorem 4.12. The centers of the pencils of lines in the projectivity defining a point conic are points of the point conic. | Proof. Let \( P \) and \( {P}^{\prime } \) be the centers of the pencils. Let \( m = P{P}^{\prime } \), and consider \( m \) as a line in the pencil with center \( P \) (see Fig. 4.27). Then there is a corresponding line \( {m}^{\prime } \) in the pencil with center \( {P}^{\prime } \) . Note that \( m \neq {m}^{\prime... | Yes |
Theorem 4.13. If \( A, B, C, D \) are four points on a point conic defined by projectively related pencils with centers \( P \) and \( {P}^{\prime } \), then the diagonal points of hexagon PBP'ACD are collinear, and conversely, if the diagonal points of hexagon \( {PB}{P}^{\prime }{ACD} \) are collinear, then \( A, B, ... | Proof. (a) The diagonal points for hexagon \( {PB}{P}^{\prime }{ACD} \) are \( {PB} \cdot {AC} = J \) , \( B{P}^{\prime } \cdot {CD} = L \), and \( {P}^{\prime }A \cdot {DP} = K \) . Let \( {AC} \cdot {PD} = M \) and \( A{P}^{\prime } \cdot {DC} = N \) (see Fig. 4.29). By using these and the definition of a point conic... | Yes |
Theorem 4.14. A point conic is uniquely determined by five distinct points, no three of which are collinear. | Proof. Let \( {P}_{1},{P}_{2},{P}_{3},{P}_{4},{P}_{5} \) be five points, no three collinear. Then there exists a point conic determined by the pencils with centers \( {P}_{1} \) and \( {P}_{2} \) and the projectivity \( {P}_{1}{P}_{3},{P}_{1}{P}_{4},{P}_{1}{P}_{5} \land {P}_{2}{P}_{3},{P}_{2}{P}_{4},{P}_{2}{P}_{5} \), ... | Yes |
Corollary 1 (Pascal’s Theorem). If a hexagon is inscribed in a point conic (i.e., the vertices of the hexagon are points of the point conic), its diagonal points are collinear (see Fig. 4.30). | By considering hexagon \( {P}_{1}^{\prime }{P}_{1}{P}_{2}{P}_{3}{P}_{4}{P}_{5} \) and letting point \( {P}_{1}^{\prime } \) approach \( {P}_{1} \) so that line \( {P}_{1}^{\prime }{P}_{1} \) becomes the tangent at \( {P}_{1} \), we can verify a second corollary that gives an efficient method for constructing tangents t... | No |
For any point \( A \) of a point conic, there is exactly one line tangent to the conic at \( A \) . (This tangent is the line corresponding to line \( {AB} \) considered as a line of the pencil through \( B \) when the conic is defined by projectively related pencils with centers \( A \) and \( B \) .) | Proof. Let \( B, C, D, E \) be four more points of the point conic. Then the point conic can be defined by projectively related pencils with centers \( A \) and \( B \) . Let \( h \) be the line in the pencil with center \( A \) that corresponds to line \( {AB} \) considered as a line in the pencil with center \( B \) ... | Yes |
Theorem 4.16. A line intersects a point conic in at most two points. | Proof. Assume line \( n \) intersects a point conic in three distinct points \( Q, R \), and \( S \) . Let \( P \) and \( {P}^{\prime } \) be two other points of the conic and consider the pencils with centers \( P \) and \( {P}^{\prime } \) . Then, as shown previously, the conic can be defined in terms of a projectivi... | Yes |
Theorem 4.19. If \( P\left( {{p}_{1},{p}_{2},{p}_{3}}\right) \) and \( Q\left( {{q}_{1},{q}_{2},{q}_{3}}\right) \) are two distinct points, any point \( R \) of the line \( {PQ} \) has homogeneous coordinates \( \left( {{r}_{1},{r}_{2},{r}_{3}}\right) \) where \( {r}_{i} = {\lambda }_{1}{p}_{i} + {\lambda }_{2}{q}_{i} ... | Proof. (a) Assume \( R \) has homogeneous coordinates \( \left( {{\lambda }_{1}{p}_{1} + {\lambda }_{2}{q}_{1},{\lambda }_{1}{p}_{2} + {\lambda }_{2}{q}_{2}}\right. \) , \( \left. {{\lambda }_{1}{p}_{3} + {\lambda }_{2}{q}_{3}}\right) \), then\n\n\[ \left| \begin{array}{lll} {r}_{1} & {p}_{1} & {q}_{1} \\ {r}_{2} & {p}... | Yes |
Theorem 4.20. A projectivity between the elements of two pencils can be represented by a real matrix equation of the form\n\n\[ s\left\lbrack \begin{array}{l} {\lambda }_{1}^{\prime } \\ {\lambda }_{2}^{\prime } \end{array}\right\rbrack = \left\lbrack \begin{array}{ll} {a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22} \end{ar... | Proof. We first show that a perspectivity from a pencil of points to a pencil of lines has this algebraic form.\n\nLet \( P \) and \( Q \) be base points of the pencil of points and let the lines \( m \) and \( n \) be base lines of the pencil of lines. Let \( X\left( {{\lambda }_{1},{\lambda }_{2}}\right) \) be any ot... | Yes |
Theorem 4.22. A projectivity on a pencil, other than the identity, with matrix\n\n\\[ \n\\left\\lbrack \\begin{array}{ll} {a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22} \\end{array}\\right\\rbrack \n\\]\n\nhas two distinct invariant elements, one invariant element, or no invariant elements according as \\( {\\left( {a}_{22... | Proof. Note \\( \\left( {{\\lambda }_{1},{\\lambda }_{2}}\\right) \\) is an invariant element iff\n\n\\[ \ns\\left\\lbrack \\begin{array}{l} {\\lambda }_{1} \\\\ {\\lambda }_{2} \\end{array}\\right\\rbrack = \\left\\lbrack \\begin{array}{ll} {a}_{11} & {a}_{12} \\\\ {a}_{21} & {a}_{22} \\end{array}\\right\\rbrack \\lef... | Yes |
Theorem 4.26. The cross ratio of four distinct elements of a pencil is invariant under a projectivity (so, e.g., if \( {ABCD} \land {A}^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime } \), then \( R\left( {A, B, C, D}\right) = \) \( \left. {R\left( {{A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime }}\right) }\righ... | Proof. Assume that distinct elements \( A, B, C, D \) of one pencil map to corresponding elements \( {A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime } \) of a second pencil under a projectivity with matrix \( A = \left\lbrack {a}_{ij}\right\rbrack \) . Then\n\n\[ \left| \begin{array}{ll} {\gamma }_{1}^{\prime } &... | Yes |
Find \( R\left( {A, B, C, D}\right) \) where \( A\left( {1,2,1}\right), B\left( {3,6,1}\right), C\left( {2,4,1}\right) \), and \( D\left( {1,2,0}\right) \) are points on \( l\left\lbrack {2, - 1,0}\right\rbrack \) . | Since \( Z\left( {0,0,1}\right) \) is clearly a point on \( l \), we cannot use the corollary to Theorem 4.26 directly. However, since \( X\left( {1,0,0}\right) \) is not incident with \( l \), we can use a comparable result; that is, we can use the last two homogeneous coordinates of each point in the role of homogene... | Yes |
Theorem 4.28. If \( A, B, C, D \) are four distinct elements of one pencil and \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) , \( {D}^{\prime } \) are four distinct elements of a second pencil, with \( R\left( {{A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime }}\right) = \) \( R\left( {A, B, C, D}\right) \), the... | Proof. By the fundamental theorem, there exists a projectivity such that \( {ABC} \land {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . Let \( {D}^{ * } \) be the unique image of \( D \) under this projectivity. By Theorem 4.26 \( R\left( {A, B, C, D}\right) = R\left( {{A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{ * }}\... | Yes |
Theorem 4.29. If \( A, B, C, D \) are four distinct elements of a pencil, then \( R\left( {A, B, C, D}\right) = - 1 \) iff \( H\left( {{AB},{CD}}\right) \) . | Proof. (a) Since \( H\left( {{AB},{CD}}\right) \), it follows that \( H\left( {{AB},{DC}}\right) \) and, by Theorem 4.10, there is a projectivity such that \( {ABCD} \land {ABDC} \) . Thus by Theorem 4.26 \( R(A \) , \( B, C, D) = R\left( {A, B, D, C}\right) \) ; but by Theorem 4.24, if \( R\left( {A, B, C, D}\right) =... | Yes |
Theorem 4.31. A collineation maps collinear points to collinear points. The image of a line \( u\left\lbrack {{u}_{1},{u}_{2},{u}_{3}}\right\rbrack \) under a collineation with matrix \( A \) is given by the equation \( k{u}^{\prime } = u{A}^{-1}, k \neq 0 \) . | Proof. Assume that \( P \) is a point on line \( {QR} \) . Then it suffices to show that \( {P}^{\prime } \), the image of \( P \) under the collineation, is collinear with the images of \( Q \) and \( R \) , namely, \( {Q}^{\prime } \) and \( {R}^{\prime } \) . Since \( P \) is on \( {QR} \), Theorem 4.19 implies that... | Yes |
Theorem 4.33. A collineation of the projective plane induces a projectivity between the elements of corresponding pencils. | Proof. Let \( P, Q, R \) be three collinear points, so \( R = {\lambda }_{1}P + {\lambda }_{2}Q \) . Let \( {P}^{\prime },{Q}^{\prime } \), and \( {R}^{\prime } \) be their images under a collineation with matrix \( A \) . Then \( {P}^{\prime },{Q}^{\prime } \), and \( {R}^{\prime } \) are also collinear so \( {R}^{\pr... | Yes |
Theorem 4.34. There exists a unique collineation that maps any four points, no three collinear, to any four points, no three collinear. | Proof. The verification of this theorem consists of algebraically finding a matrix \( A \) of the collineation that maps any four points \( P, Q, R, S \) (no three collinear) to any four points \( {P}^{\prime },{Q}^{\prime },{R}^{\prime },{S}^{\prime } \) (no three collinear) and noting that this matrix is uniquely det... | Yes |
Theorem 4.35. A collineation has at least one invariant point and one invariant line. | Proof. To show that a collineation with matrix \( A \) has at least one invariant point, note that there will be an invariant point \( X \) iff there is a nonzero scalar \( s \) such that \( {sX} = {AX} \) . But \( {sX} = {AX} \) iff \( {sIX} - {AX} = \left( {{sI} - A}\right) X = 0 \) where \( I \) is the identity matr... | Yes |
Theorem 4.36. Every perspective collineation has a linewise invariant point. (This point is called the center). | Proof. Let \( m \) be the axis of the perspective collineation.\n\nCase 1. There is an invariant point not on \( m \) . Let this invariant point be called \( C \) . Then any line through \( C \) intersects \( m \) in a second invariant point\n\n and center \( C \) that maps a given point \( P\left( {P \neq C\text{and}P\text{not on}m}\right) \) to a given point \( {P}^{\prime } \) on \( {PC} \) . | Proof. Case 1. \( C \) is not on \( m \) . Let \( {PC} \cdot m = D \) and let \( E \) and \( F \) be two additional points on \( m \) (Fig. 4.37). Then by Theorem 4.34 there exists a unique collineation that maps \( P \) to \( {P}^{\prime }, C \) to \( C, E \) to \( E \), and \( F \) to \( F \) . Clearly, this collinea... | Yes |
Theorem 4.38. \( \bigtriangleup {P}^{\prime }{Q}^{\prime }{R}^{\prime } \) is the image of \( \bigtriangleup {PQR} \) under a perspective collineation with center \( C \) and axis \( m \), iff the triangles are perspective from the point \( C \) and perspective from the line \( m \) . | Proof. (a) For the first half of the proof, see Exercise 11.\n\n(b) Now assume that \( \bigtriangleup {PQR} \) and \( \bigtriangleup {P}^{\prime }{Q}^{\prime }{R}^{\prime } \) are perspective from \( C \) and \( m \) . Since \( \bigtriangleup {PQR} \) is a triangle, the three points \( P, Q, R \) are not collinear. Thu... | No |
Theorem 4.39. Under a homology with center \( C \) and axis \( m \), any point \( P \) not on \( m \) \( \left( {P \neq C}\right) \) has an image \( {P}^{\prime } \) such that \( C, P \), and \( {P}^{\prime } \) are collinear and if \( m \cdot {CP} = Q \) , then \( R\left( {C, Q, P,{P}^{\prime }}\right) \) is constant ... | Proof. The fact that \( C, P \), and \( {P}^{\prime } \) are collinear for all perspective collineations has been noted previously.\n\nCase 1. \( X \) is a point not on \( {CP} \) or on \( m \) . Let \( {X}^{\prime } \) be its image under the homology, and let \( D = {CX} \cdot m, E = {PX} \cdot m \) . So \( {X}^{\prim... | Yes |
Theorem 4.43. There exists a unique correlation that maps any four points, no three collinear, to any four lines, no three concurrent. | Thus a given correlation that maps points to lines according to the equation \( s{u}^{t} = {AX} \) also maps lines to points according to the equation \( k{X}^{t} = u{A}^{-1} \) . (The transpose is used in both equations, since points are represented by column matrices and lines are represented by row matrices.) In gen... | No |
Theorem 4.44. A point \( P \) is on the polar of a point \( Q \) under a given polarity iff \( Q \) is on the polar of \( P \) under this same polarity. | Proof. Let \( C \) be the matrix of the polarity and let \( q \) and \( p \) be the polars of \( Q \) and \( P \), that is, \( {s}_{1}{q}^{t} = {CQ} \) and \( {s}_{2}{p}^{t} = {CP} \) . Since \( P \) is on the polar of \( Q,{qP} = 0 \) ; but \( {s}_{1}q = {Q}^{t}C \), so \( {Q}^{t}{CP} = 0 \) . Transposing gives \( {P}... | Yes |
Theorem 4.46. A collineation with matrix \( A \) maps a set of self-conjugate points with matrix \( C \) to a set of self-conjugate points with matrix \( {C}^{\prime } = {\left( {A}^{-1}\right) }^{t}C\left( {A}^{-1}\right) \) . | Proof. Let \( S \) be a set of self-conjugate points with equation \( {X}^{t}{CX} = 0 \) where \( C \) is a \( 3 \times 3 \) nonsingular, symmetric matrix. Let \( A \) be the matrix of an arbitrary collineation. Then \( A \) is also a \( 3 \times 3 \) nonsingular matrix, and the corresponding point equation is \( s{X}^... | Yes |
Corollary 1. A point conic has an equation of the form \( {X}^{t}{CX} = 0 \) and a line conic has an equation of the form \( u{C}^{-1}{u}^{t} = 0 \) where \( C \) is a symmetric, nonsingular \( 3 \times 3 \) matrix. | Therefore, any point conic corresponds to a symmetric matrix that is the matrix of a polarity. This polarity matrix is called the matrix of the point conic. Furthermore, if line \( p \) corresponds to point \( P \) under the polarity determined by the conic, \( P \) and \( p \) are said to be pole and polar with respec... | No |
Theorem 4.48. The tangents to a point conic are the lines of the line conic determined by the same polarity. | Proof. Let \( X \) be a point on a point conic with matrix \( C \) . By Corollary 3 to Theorem 4.47, \( u \), the tangent at \( X \), is given by \( s{u}^{t} = {CX} \) . Solving this equation for \( X \), gives \( X = s{C}^{-1}{u}^{t} \) .\n\nSince \( X \) is on the point conic, \( {X}^{t}{CX} = 0 \) by Corollary 1 of ... | Yes |
Theorem 4.49. The point of intersection of two tangents to a point conic is the pole of the line joining the points of tangency. | Proof. Let \( p \) and \( q \) be tangents to a point conic at points \( P \) and \( Q \), respectively; that is, \( p \) and \( q \) are the polars of \( P \) and \( Q \), respectively. Let \( R = p \cdot q \) (Fig. 4.41). Then \( R \) is on both the polar of \( P \) and the polar of \( Q \) so by Theorem 4.44, \( P \... | Yes |
Theorem 4.50. If \( A, B, C \), and \( D \) are four distinct points of a point conic then the diagonal triangle of quadrangle \( {ABCD} \) is self-polar. | Proof. Let \( P = {CD} \cdot {AB}, Q = {CB} \cdot {AD} \), and \( R = {AC} \cdot {BD} \) be the diagonal points of the quadrangle, and let\n\n\[ S = \tan B \cdot \tan A\;\text{ and }\;T = \tan C \cdot \tan D \]\n\n\[ U = \tan C \cdot \tan A\;\text{ and }\;V = \tan D \cdot \tan B \]\n\nThen by a corollary to Theorem 4.1... | No |
Theorem 4.51. Triangle \( \bigtriangleup {XYZ} \) [where \( X\left( {1,0,0}\right), Y\left( {0,1,0}\right) \), and \( Z\left( {0,0,1}\right) \) ] is a self-polar triangle with respect to a conic iff the matrix of the conic is diagonal. | Thus any point conic is equivalent, that is, can be mapped via a collineation, to a conic with an equation of the form \( a{\left( {x}_{1}\right) }^{2} + b{\left( {x}_{2}\right) }^{2} + c{\left( {x}_{3}\right) }^{2} = 0 \) . However, the next theorem shows that conics with equations of this form can, in turn, be mapped... | No |
Theorem 4.52. Any point conic is projectively equivalent to a conic with an equation of the form \( {\left( {x}_{1}\right) }^{2} + {\left( {x}_{2}\right) }^{2} \pm {\left( {x}_{3}\right) }^{2} = 0 \) (i.e., any conic can be mapped via a collineation to a conic with this equation). | Proof. Let \( \bigtriangleup {PQR} \) be a self-polar triangle with respect to a given point conic \( \mathcal{C} \) , and let \( T \) be a collineation that maps \( P, Q, R \), to \( X, Y, Z \), respectively. Then \( \bigtriangleup {XYZ} \) will be self-polar with respect to the conic \( T\left( \mathcal{C}\right) \),... | Yes |
Problem 2.1. Let \( {ABC} \) be a triangle. Prove that its medians \( {CD} \) , \( {AE} \) and \( {BF} \) are concurrent. | Solution. Let \( {AE} \cap {BF} = G \) . Let \( {CG} \cap {AB} = {D}^{\prime } \) . We have\n\n\[ \frac{{S}_{\bigtriangleup {AGC}}}{{S}_{\bigtriangleup {BGC}}} = \frac{{S}_{\bigtriangleup {AGC}}}{{S}_{\bigtriangleup {AGB}}} \cdot \frac{{S}_{\bigtriangleup {AGB}}}{{S}_{\bigtriangleup {BGC}}} \]\n\n\[ = \frac{CE}{EB} \cd... | Yes |
Problem 2.2. Let \( {ABC} \) be a triangle. Prove that its altitudes \( {AE} \) , \( {CF} \) and \( {BD} \) are concurrent. | Solution. Let \( {BD} \cap {CF} = H \) . Let \( {AH} \cap {BC} = {E}^{\prime } \) . Therefore, the quadrilaterals \( {AFHD} \) and \( {BFDC} \) are cyclic. Hence, \[ \angle {DAH} = \angle {DFC} = \angle {CBD}\text{.} \] Thus, the quadrilateral \( {AD}{E}^{\prime }B \) is cyclic. It follows that \( \angle A{E}^{\prime }... | Yes |
Prove that the internal angle bisectors of a given \( \\bigtriangleup {ABC} \) are concurrent. | Solution. Let the bisectors of \( \\angle {BAC} \) and \( \\angle {ABC} \) intersect at \( I \) . Then \( \\operatorname{dist}\\left( {I,{AC}}\\right) = \) \( \\operatorname{dist}\\left( {I,{AB}}\\right) \) and \( \\operatorname{dist}\\left( {I,{AB}}\\right) = \) \( \\operatorname{dist}\\left( {I,{BC}}\\right) \) .\n\n... | Yes |
Problem 2.4. Let \( {ABC} \) be a triangle. Its incircle touches \( {AB},{BC} \) and \( {CA} \) at the points \( {C}_{1},{A}_{1} \) and \( {B}_{1} \), respectively. Prove that the lines \( C{C}_{1}, B{B}_{1} \) and \( A{A}_{1} \) are concurrent. | Solution. We use the properties of tangent lines to the incircle to get that \( A{B}_{1} = A{C}_{1}, B{C}_{1} = B{A}_{1} \) and \( C{A}_{1} = C{B}_{1} \) . Thus,\n\n\[ \frac{A{C}_{1}}{{C}_{1} \cdot B} \cdot \frac{B{A}_{1}}{{A}_{1}C} \cdot \frac{C{B}_{1}}{{B}_{1}A} = 1 \]\n\nand Ceva's Theorem (Problem 4.9.15), applied ... | Yes |
Problem 2.5. Prove that the perpendicular bisectors of the sides of a given \( \bigtriangleup {ABC} \) are concurrent. | Solution. Let \( {S}_{AB} \cap {S}_{AC} = O \) . Since \( O \) lies on both perpendicular bisectors, \( {AO} = {BO} \) and \( {AO} = {CO} \) . Hence, \( {BO} = {CO} \) . Hence, the point \( O \) lies on the perpendicular bisector of \( {BC} \) . Thus, \( O \) is the intersection point of the three perpendicular bisecto... | Yes |
Problem 2.6. Let \( {ABC} \) be a triangle with circumcircle \( k \) . Let \( {l}_{A},{l}_{B} \) and \( {l}_{C} \) be the tangent lines to \( k \) at the points \( A, B \) and \( C \), respectively. If \( {l}_{A} \cap {l}_{B} = {C}_{1},{l}_{B} \cap {l}_{C} = {A}_{1} \) and \( {l}_{C} \cap {l}_{A} = {B}_{1} \), prove th... | Solution. Note that \( k \) is inscribed in \( \bigtriangleup {A}_{1}{B}_{1}{C}_{1} \), and also note that \( A, B \) and \( C \) are the points of tangency of \( k \) and \( {B}_{1}{C}_{1},{C}_{1}{A}_{1} \) and \( {A}_{1}{B}_{1} \) , respectively. Problem 2.4 yields that the lines \( A{A}_{1}, B{B}_{1} \) and \( C{C}_... | Yes |
Problem 2.7. Let \( {ABC} \) be a triangle. Let \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) be the points of tangency of the segments \( {BC},{CA} \) and \( {AB} \) and the excircles of \( \bigtriangleup {ABC} \) . Prove that the lines \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) are concurrent. | Solution. We have \( A{C}_{1} = \)\n\n\n\n\( {A}_{1}C,{B}_{1}A = B{A}_{1} \) and \( {C}_{1}B = \) \( C{B}_{1} \) . Thus,\n\n\[ \n\frac{A{C}_{1}}{{C}_{1}B} \cdot \frac{B{A}_{1}}{{A}_{1}C} \cdot \frac{C{B}_{1}}{{B}_{1}A}... | Yes |
Problem 2.8. Let \( {ABC} \) be a triangle and let \( N \) be its Nagel point. Let \( {AN},{BN} \) and \( {CN} \) intersect the incircle of \( \bigtriangleup {ABC} \) at the points \( {A}_{1} \) , \( {B}_{1} \) and \( {C}_{1} \) (as shown in the figure), and the sides \( {BC},{CA} \) and \( {AB} \) at the points \( {A}... | Solution. We will prove that \( C{C}_{1} = N{C}_{2} \), and the remaining two statements will follow analogously. Let \( P \) be the foot of the altitude of \( \bigtriangleup {ABC} \) through \( C \), and let \( I \) be the incenter of \( \bigtriangleup {ABC} \) . Let the incircle of \( \bigtriangleup {ABC} \) touch \(... | Yes |
Problem 2.9. Let \( {ABC} \) be a triangle. The equilateral triangles \( \bigtriangleup {AB}{C}_{1},\bigtriangleup A{B}_{1}C \) and \( \bigtriangleup {A}_{1}{BC} \) are constructed externally to its sides. Prove that the lines \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) are concurrent. | Solution. Let the circumcircles of \( \bigtriangleup {AB}{C}_{1} \) and \( \bigtriangleup {A}_{1}{BC} \) intersect at the point \( {T}_{1} \) . Then \( \angle A{T}_{1}B = {120}^{ \circ } = \angle B{T}_{1}C \) and hence\n\n\[ \angle C{T}_{1}A = {120}^{ \circ }\text{.}\]\n\nTherefore, \( A{T}_{1}C{B}_{1} \) is cyclic. Th... | Yes |
Let \( {ABC} \) be a triangle. The equilateral triangles \( \bigtriangleup {AB}{C}_{1},\bigtriangleup A{B}_{1}C \) and \( \bigtriangleup {A}_{1}{BC} \) are constructed internally to its sides. Prove that the lines \( A{A}_{1}, B{B}_{1} \) and \( C{C}_{1} \) are concurrent. | Let the circumcircles of \( \bigtriangleup {AB}{C}_{1} \) and \( \bigtriangleup {A}_{1}{BC} \) intersect at the point \( {T}_{2} \) . Then\n\n\[ \angle A{T}_{2}C = \angle B{T}_{2}C - \angle A{T}_{2}B \]\n\n\[ = {120}^{ \circ } - {60}^{ \circ } = \angle A{B}_{1}C\text{. } \]\n\nTherefore, the quadrilateral \( A{B}_{1}{T... | Yes |
Problem 2.13. Let \( {ABC} \) be a triangle. Prove that there exists a unique point \( S \) such that the equality \( {BC} + {AS} = {CA} + {BS} = {AB} + {CS} \) holds. | Solution. Let \( \omega \) be the incircle of \( \bigtriangleup {ABC} \) and let \( \omega \) touch the sides \( {AB},{BC} \) and \( {CA} \) at the points \( {C}_{1} \) , \( {A}_{1} \) and \( {B}_{1} \), respectively.\n\nConsider the circles \( {k}_{1}\left( {A, A{C}_{1}}\right) \) , \( {k}_{2}\left( {B, B{A}_{1}}\righ... | Yes |
Problem 2.14. Let \( {ABC} \) be a triangle. Prove that there exists a unique point \( S \) such that the equality \( {BC} - {AS} = {CA} - {BS} = {AB} - {CS} \) holds. | Solution. Let \( \omega \) be the incircle of \( \bigtriangleup {ABC} \) and let \( \omega \) touch the sides \( {AB},{BC} \) and \( {CA} \) at the points \( {C}_{1},{A}_{1} \) and \( {B}_{1} \), respectively.\n\nWe construct the circles \( {k}_{1}\left( {A, A{C}_{1}}\right) ,{k}_{2}\left( {B, B{A}_{1}}\right) \) and \... | Yes |
Problem 2.15. Three circles \( {k}_{1}\left( A\right) ,{k}_{2}\left( B\right) \) and \( {k}_{3}\left( C\right) \) are given, and they all touch each other externally. Let \( {C}_{1} \) and \( {B}_{1} \) be the points of tangency of \( {k}_{1} \) and \( {k}_{2} \), and of \( {k}_{1} \) and \( {k}_{3} \), respectively. L... | Solution. The points \( {A}_{1} \) ,\n\n\n\n\( {B}_{1} \) and \( {C}_{1} \) coincide with the points of tangency of the incircle of \( \bigtriangleup {ABC} \) and the sides of the triangle. Thus, the statement follows ... | No |
Problem 2.16. Three circles \( {k}_{1}\left( A\right) ,{k}_{2}\left( B\right) \) and \( {k}_{3}\left( C\right) \) are given, and they all touch each other externally. Let \( {C}_{1} \) and \( {B}_{1} \) be the points of tangency of \( {k}_{1} \) and \( {k}_{2} \), and of \( {k}_{1} \) and \( {k}_{3} \), respectively. L... | Solution. The points \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) are the points of tangency of the incircle of \( \bigtriangleup {ABC} \) and the sides of the triangle. Thus, the statement follows from Problem 2.14, in which we use \( {k}_{4} \) to prove the existence of the second Soddy center. We will refer to \( {k}_{4}... | No |
Problem 2.17. Let \( {ABC} \) be a triangle and let \( {S}_{1} \) and \( {S}_{2} \) be its first and second Soddy centers, respectively. Prove that the points \( A, B \) and \( C \) lie on an ellipse with foci \( {S}_{1} \) and \( {S}_{2} \) . | Solution. The definition\n\n\n\nof Soddy centers yields \( {S}_{1}A + \) \( {S}_{2}A = \)\n\n\[= \left( {{S}_{1}A + {BC}}\right) + \left( {{S}_{2}A - {BC}}\right)\]\n\n\[= \left( {{S}_{1}B + {AC}}\right) + \left( {{S}_... | Yes |
Problem 2.18. Let \( {ABC} \) be a triangle and let \( {S}_{1} \) be its first Soddy center. Prove that there exists a circle, inscribed in the convex quadrilateral, formed by the lines \( C{S}_{1}, B{S}_{1},{AC} \) and \( {AB} \) . | Solution. We use the notations from the figure to prove that the quadrilateral \( A{A}_{1}{S}_{1}{C}_{1} \) is circumscribed. Problem 2.13 yields \( B{S}_{1} + {AC} = \) \( C{S}_{1} + {AB} \), and the result follows. | No |
Problem 2.19. Let \( {ABC} \) be a triangle and let \( {S}_{2} \) be its second Soddy center. Prove that there exists a circle that touches the lines \( {BA} \) and \( {BC} \) , and the segments \( A{S}_{2} \) and \( C{S}_{2} \) . | Solution. Problem 2.14  yields \( C{S}_{2} + {BC} = A{S}_{2} + \) \( {AB} \), and the result follows from the concave quadrilateral \( {ABC}{S}_{2} \) . | No |
Let \( {ABC} \) be a triangle with excircles \( {\omega }_{a},{\omega }_{b} \) and \( {\omega }_{c} \) . Let \( {I}_{a},{I}_{b} \) and \( {I}_{c} \) be the centers of \( {\omega }_{a},{\omega }_{b} \) and \( {\omega }_{c} \), respectively. Let \( {A}_{1} \) be the point of tangency of \( {\omega }_{a} \) and the side \... | The lines \( {C}_{1}{I}_{c},{B}_{1}{I}_{b} \) and \( {A}_{1}{I}_{a} \) are perpendicular to \( {BA},{AC} \) and \( {CB} \), respectively. We will prove that the perpendiculars, drawn from \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) to the respective sides of \( \bigtriangleup {ABC} \), are concurrent.\n\nLet \( I \) be the... | Yes |
Problem 2.21. Let \( {ABC} \) be a triangle. The first Brocard point \( B{r}_{1} \) is defined as the point for which \( \angle {BAB}{r}_{1} = \angle {ACB}{r}_{1} = \angle {CBB}{r}_{1} \) . Prove that it always exists. | Solution. Construct the circle through \( A \) and \( C \) which touches \( {AB} \), and construct the circle through \( B \) and \( C \) which touches \( \overline{AC} \) . Let their second intersection point be \( B{r}_{1} \). The alternate segments theorem yields \[ \angle {BAB}{r}_{1} = \angle {ACB}{r}_{1} = \angle... | Yes |
Problem 2.22. Let \( {ABC} \) be a triangle. The second Brocard point \( B{r}_{1} \) is defined as the point for which \( \angle {ABB}{r}_{2} = \angle {CAB}{r}_{2} = \angle {BCB}{r}_{2} \) . Prove that it always exists. | Solution. Construct the circle through \( A \) and \( C \), which touches \( {CB} \) , and construct the circle through \( B \) and \( C \) which touches \( {AB} \) . Let their second intersection point be \( B{r}_{2} \) . The alternate segments theorem yields\n\n\[ \angle {ABB}{r}_{2} = \angle {BCB}{r}_{2} = \angle {C... | Yes |
Problem 2.23. Let \( {ABC} \) be a triangle. Let \( L \) be its Lemoine point and let \( B{r}_{1} \) and \( B{r}_{2} \) be its first and second Brocard points, respectively. Let \( {CL} \cap {AB} = F \) . Prove that \( \angle {AFB}{r}_{1} = \angle {BFB}{r}_{2} \) . | Solution. The definition of the Brocard points and Problems 2.21 and 2.22 yield\n\n\[ \angle {FAB}{r}_{1} = \angle {ACB}{r}_{1} = \angle B{r}_{2}{CB} = \angle B{r}_{2}{BF} = \varphi . \]\n\nSince \( {CL} \) is isogonally conjugate to the median through \( C \) in \( \bigtriangleup {ABC} \) (see Problem 4.4.1), we have ... | Yes |
Problem 2.24. Let \( {ABC} \) be a triangle with circumcenter \( O \) . Let \( L \) be its Lemoine point and let \( B{r}_{1} \) and \( B{r}_{2} \) be its first and second Brocard points, respectively. Prove that the equalities \( \angle {OB}{r}_{1}L = \angle {OB}{r}_{2}L = {90}^{ \circ } \) and \( B{r}_{1}L = B{r}_{2}L... | Solution. Let us denote the projections of \( O \) onto \( {AL},{BL} \) and \( {CL} \) by \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) . We will prove that the quadrilateral \( {AB}{C}_{1}O \) is cyclic. Let the tangent lines to the circumcircle of \( \bigtriangleup {ABC} \) at the points \( A \) and \( B \) intersect at \(... | Yes |
Problem 2.25. Let \( {ABC} \) be a triangle. Let \( A{p}_{1} \) and \( A{p}_{2} \) be the two isodynamic points. Prove that the pedal triangles with respect to these two points are equilateral. | Solution. Let \( X \) be any one of \( A{p}_{1} \) and \( A{p}_{2} \) . Problems 2.11 and 2.12 yield \( {AX}.{BC} = {BX}.{CA} = {CX}.{AB} \) . Therefore, there exists a real constant \( k > 0 \) such that \( {AX}.{BC} = {BX}.{CA} = {CX}.{AB} = k \) . Let \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) be the projections of \( ... | Yes |
Problem 2.26. Let \( {ABC} \) be a triangle and let \( E \) and \( D \) be the feet of the internal and external bisectors at \( C \), respectively. Prove that the two isodynamic points of \( \bigtriangleup {ABC} \) (defined as in Problems 2.11 and 2.12) lie on the circle with diameter \( {ED} \) . | Solution. We will prove that the circle with diameter \( {ED} \) is the locus of points \( X \) for which the foot of the bisector of \( \angle {AXB} \) is the point \( E \) . Since \( \frac{AE}{EB} = \frac{AC}{CB} = \frac{AD}{DB} \), then \( D \) is the foot of the external bisector of \( \angle {AX}\bar{B} \) . Thus,... | Yes |
Problem 2.27. Let \( {ABC} \) be a triangle and let \( {T}_{1} \) be its first Fermat-Torricelli point. Prove that \( \angle A{T}_{1}B = \angle B{T}_{1}C = {120}^{ \circ } \) . | Solution. See Problem 2.9. | No |
Problem 2.28. Let \( {ABC} \) be a triangle and let \( {T}_{2} \) be its second Fermat-Torricelli point. Prove that exactly one of the equalities \( \angle A{T}_{2}B = \) \( \angle A{T}_{2}C = {60}^{ \circ },\angle B{T}_{2}A = \angle B{T}_{2}C = {60}^{ \circ } \) and \( \angle C{T}_{2}B = \angle C{T}_{2}A = {60}^{ \cir... | Solution. See Problem 2.10. The position of \( {T}_{2} \) determines which one of the three equations holds. | No |
Problem 2.31. Let \( {ABC} \) be a triangle and let \( L \) be its Lemoine point. The points \( M, K \in {AB}, H, I \in {BC} \) and \( J, G \in {AC} \) are chosen such that \( {MI} \parallel {AC},{GH} \parallel {AB},{KJ} \parallel {BC} \) and \( {MI} \cap {KJ} \cap {GH} = L \) . Prove that the points \( M, K, H, I, J \... | Solution. We will prove that the quadrilaterals MKHI, HIJG and \( {IJGM} \) are cyclic. We have that \( {BL} \) is a symmedian in \( \bigtriangleup {MBI} \) because of the homothety with center \( B \) that sends \( {CA} \) to \( {IM} \) . Let \( N \) be the midpoint of \( {KH} \) . Then \( B, N \) and \( L \) are coll... | Yes |
Problem 2.32. Let \( {ABC} \) be a triangle and let \( L \) be its Lemoine point. The points \( M, K \in {AB}, H, I \in {BC} \) and \( J, G \in {AC} \) are chosen such that the quadrilaterals \( {MICA},{GHBA} \) and \( {KJCB} \) are cyclic and \( {MI} \cap {KJ} \cap {GH} = L \) . Prove that the points \( M, K, H, I, J ... | Solution. Since \( \bigtriangleup {MBI} \sim \bigtriangleup {CBA} \), then \( {BL} \) is a median in \( \bigtriangleup {MBI} \). Therefore, \( {ML} = {LI} \) and analogously, \( {HL} = {LG} \) and \( {JL} = {LK} \). But simple angle chasing leads us to the fact that \( \bigtriangleup {MLK} \), \( \bigtriangleup {LHI} \... | Yes |
Problem 2.33. Let \( {ABCD} \) be a convex quadrilateral such that \( {AB} \cap \) \( {CD} = F \) and \( {AD} \cap {BC} = E \) . Prove that the circumcircles of \( \bigtriangleup {BFC} \) , \( \bigtriangleup {AFD} \) , \( \bigtriangleup {DCE} \) and \( \bigtriangleup {ABE} \) pass through a single point. | Solution. Let the circumcircles of \( \bigtriangleup {BFC} \) and \( \bigtriangleup {CDE} \) intersect each other at the points \( C \) and \( M \) . We have\n\n\[ \angle {DMF} = \angle {DMC} + \angle {CMF} \]\n\n\[ = \angle {DEC} + \angle {CBA} \]\n\n\[ = {180}^{ \circ } - \angle {BAE} \]\n\nhence the quadrilateral \(... | Yes |
Problem 2.34. The construction of Problem 2.33 is given. Prove that point \( M \) and the respective centers \( {O}_{1},{O}_{2},{O}_{3} \) and \( {O}_{4} \) of the circumcircles of \( \bigtriangleup {AFD},\bigtriangleup {BFC},\bigtriangleup {ABE} \) and \( \bigtriangleup {DCE} \) lie on a circle. | Solution. We have that\n\n\[ \angle {O}_{4}{O}_{3}{O}_{2} = {180}^{ \circ } - \angle {EMB} \]\n\n\n\nbecause \( {\mathrm{O}}_{3}{\mathrm{O}}_{4} \bot \mathrm{{EM}} \) and \( {\mathrm{O}}_{2}{\mathrm{O}}_{3} \bot \mathr... | Yes |
Let \( {ABCDE} \) be a convex pentagon such that \( {AC} \cap {BE} = {D}_{1},{BD} \cap {AC} = {E}_{1},{BD} \cap {EC} = {A}_{1}, E\bar{C} \cap {AD} = {B}_{1} \) and \( {AD} \cap {BE} = {C}_{1} \) . Let \( \left( {XYZ}\right) \) denote the circumcircle of \( \bigtriangleup {XYZ} \) . Let \( \left( {A{D}_{1}{C}_{1}}\right... | We will prove that the points \( {A}_{2},{B}_{2},{D}_{2} \) and \( {E}_{2} \) lie on a circle. By analogous reasoning it will follow that the point \( {C}_{2} \) lies on the same circle. First, we will prove that \( B{C}_{1}{B}_{2}D \) is cyclic. We have\n\n\[ \angle {C}_{1}{BD} = \angle {EB}{A}_{1} = \angle D{A}_{1}{B... | Yes |
Problem 2.37. Let \( {ABCDE} \) be a convex pentagon such that \( {AC} \cap {BE} = {D}^{\prime },{BD} \cap {AC} = {E}^{\prime },{BD} \cap {EC} = {A}^{\prime },\bar{EC} \cap \bar{A}D = {B}^{\prime } \) and \( {AD} \cap {BE} = {C}^{\prime } \) . Let \( \left( {XYZ}\right) \) denote the circumcircle of \( \bigtriangleup {... | Solution. Due to the radical axes theorem, it is enough to show that the quadrilateral \( B{B}^{\prime \prime }{E}^{\prime \prime }E \) is cyclic. Let \( \left( {{AD}{E}^{\prime \prime }}\right) \cap \left( {{AC}{B}^{\prime \prime }}\right) = {A}_{1} \).\n\nWe have\n\n\[ \angle {BEC} = {180}^{ \circ } - \angle E{B}^{\p... | Yes |
Problem 3.1. Let \( {ABC} \) be a triangle and let \( O \) be its circumcenter. Let \( M \) be its centroid and let \( H \) be its orthocenter. Prove that the points \( H, O \) and \( M \) are collinear. | Solution. Let \( {B}^{\prime } \) and \( {C}^{\prime } \) be the feet of the altitudes from \( B \) and \( C \) , and let \( {B}_{1} \) and \( {C}_{1} \) be the midpoints of \( {AC} \) and \( {AB} \), respectively. Then we have that \( O{B}_{1}\parallel {BH} \) and \( O{C}_{1}\parallel {CH} \) . But \( {B}_{1}{C}_{1}\p... | Yes |
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