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Theorem 4.4.4. Triangles with the same defect have the same area. | Proof. Case 1. Suppose that triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) have the same defect and, further, they have a pair of congruent corresponding sides, \( \overline{BC} \cong \overline{{B}^{\prime }{C}^{\prime }} \) . We construct the associated Sac-cheri ... | Yes |
Theorem 4.4.5 (Restatement of Theorem 4.1.1). There is a real number \( k \) such that, for every triangle \( \bigtriangleup {ABC} \), its area is \( k \) (Defect of \( \bigtriangleup {ABC} \) ). | Proof. The defect and the area are functions of the triangle. First, Exercise 4.3.13 shows the defect function to be additive: for \( D \) between \( B \) and \( C \), the defects of \( \bigtriangleup {ABD} \) and \( \bigtriangleup {ACD} \) add to the defect of \( \bigtriangleup {ABC} \) . Postulate 19 shows that the a... | No |
Theorem 4.4.6. The area of a convex polygon is proportional to its defect. | ## Proof. See Exercise 4.4.7(b). | No |
Verify that neighboring points \( {P}_{i} \) in Figure 4.43 are equally spaced along the \( x \) -axis in either the Poincaré or Klein model. The \( x \) -coordinates of the points are \( {P}_{0} = 0,{P}_{1} = \frac{1}{3} \) , \( {P}_{2} = \frac{3}{5},{P}_{3} = \frac{7}{9},{P}_{4} = \frac{15}{17},{P}_{5} = \frac{31}{33... | Solution. The Euclidean distances between the points are the differences of their \( x \) -coordinates. Then \( \left( {{P}_{0}\sum /{P}_{1}\sum }\right) \div \left( {{P}_{0}\Pi /{P}_{1}\Pi }\right) = \left( {1/\left( {4/3}\right) }\right) \div \left( {1/\left( {2/3}\right) }\right) = 1/2 \) . Similarly, \( \left( {{P}... | No |
Theorem 4.5.2. In single elliptic geometry, all lines perpendicular to a given line intersect in one point. | Proof. From Theorem 4.5.1 we know that, in a Saccheri quadrilateral \( {ABCD},\overline{EF} \) is perpendicular to both \( \overline{AB} \) and \( \overline{CD} \) . By the characteristic property of single elliptic geometry, \( \overset{⏜}{AB} \) and \( \overset{⏜}{CD} \) intersect in a unique point, say, \( P \) . We... | No |
Theorem 4.5.3. In single elliptic geometry, the summit angles of a Saccheri quadrilateral are obtuse. | Proof. In a Saccheri quadrilateral \( {ABCD} \) construct \( \overset{\overleftrightarrow{} }{DG} \) perpendicular to \( \overline{AD} \) and let \( Q \) be the intersection of \( \overrightarrow{DG} \) and \( \overrightarrow{AB} \) (Figure 4.52). Then \( d\left( {A, P}\right) \) is less than or equal to \( d\left( {A,... | Yes |
On the Euclidean plane \( {\mathbb{R}}^{2} \) define \( \rho \left( {x, y}\right) = \left( {y + 2,2 - x}\right) \). Show that \( \rho \) is a transformation. | Each point \( \left( {x, y}\right) \) has a unique image \( \left( {y + 2,2 - x}\right) \), so \( \rho \) is a function. To show that it is one-to-one and onto, we start with a point \( \left( {u, v}\right) \) and show that there is a unique point \( \left( {x, y}\right) \) that \( \rho \) sends to \( \left( {u, v}\rig... | Yes |
On \( {\mathbb{R}}^{2} \) define \( \psi \left( {x, y}\right) = \left( {{e}^{x},\sin y}\right) \). Although \( \psi \) is a function, it is neither one-to-one nor onto. | No matter what value of \( x \) we choose, \( {e}^{x} \) is positive, so \( \psi \) cannot be onto all of \( {\mathbb{R}}^{2} \). Furthermore, the sine function is periodic, so two different points can map to the same point, such as \( \psi \left( {0,0}\right) = \left( {1,0}\right) = \psi \left( {0,\pi }\right) \), dem... | Yes |
Theorem 5.1.1. If \( \alpha \) and \( \beta \) are transformations on \( S \) then \( \alpha \circ \beta \) is a transformation on \( S \) . | Proof. Let \( \alpha \) and \( \beta \) be transformations on \( S \) and \( P \in S \) . For \( \alpha \circ \beta \) to be a function, there must be a unique \( R \in S \) such that \( \alpha \circ \beta \left( P\right) = R \) . Now \( \beta \) is a transformation, so there is a unique \( Q \in S \) such that \( \bet... | Yes |
For a set \( S \), define the identity transformation by \( \iota \left( P\right) = P \) for \( P \in S \) . For a transformation \( \alpha \) on \( S,\alpha \circ \iota = \alpha = \iota \circ \alpha \) . | Every point of \( S \) is fixed by \( \iota \), and every subset of \( S \) is stable. | No |
On \( {\mathbb{R}}^{2} \), if we compose \( \rho \left( {x, y}\right) = \left( {y + 2,2 - x}\right) \) and \( \sigma \left( {x, y}\right) = \left( {2 - y, x - 2}\right) \), then what is \( \rho \circ \sigma \left( {x, y}\right) \)? | then \( \rho \circ \sigma \left( {x, y}\right) = \rho \left( {2 - y, x - 2}\right) = \left( {\left( {x - 2}\right) + 2,2 - \left( {2 - y}\right) }\right) = \left( {x, y}\right) \) . Thus \( \rho \circ \sigma = \iota \), the identity transformation: \( \rho \) undoes what \( \sigma \) did. | Yes |
Theorem 5.1.2. Every transformation \( \alpha \) on a set \( S \) has a unique inverse, \( {\alpha }^{-1} \), the transformation satisfying \( {\alpha }^{-1}\left( Q\right) = P \) if and only if \( \alpha \left( P\right) = Q \) . | Proof. Parts (i) and (ii) of the definition of a transformation are closely related. From the relationship \( {\alpha }^{-1} \), as defined in the statement of Theorem 5.2.1, is a function because \( \alpha \) is one-to-one and onto. Similarly, \( {\alpha }^{-1} \) is one-to-one and onto because \( \alpha \) is a funct... | Yes |
Theorem 5.1.3. The set of all transformations on a set is a transformation group. | Proof. See Theorems 5.1.1 and 5.1.2 and Example 6. ∎ | No |
A rotation of the plane is one type of Euclidean isometry and has exactly one fixed point, except for rotations of a multiple of \( {360}^{ \circ } \) . In Figure \( {5.4A} \) is the fixed point or center of rotation of the rotation \( \rho \) . Most rotations have no stable lines. For the special case of a rotation of... | For example, in Figure 5.5 the line \( \overrightarrow{AB} \) through the center \( A \) also goes through \( \rho \left( B\right) \), the image of \( B \) . | No |
Example 2. As shown in Figure 5.6, \( \sigma \) doubles \( x \) -coordinates and halves \( y \) -coordinates. The points \( \left( {1,2}\right) \) and \( \left( {-1,2}\right) \) become \( \left( {2,1}\right) \) and \( \left( {-2,1}\right) \), respectively. | Since the distance \( d\left( {\left( {1,2}\right) ,\left( {-1,2}\right) }\right) = 2 \) does not equal \( d\left( {\sigma \left( {1,2}\right) ,\sigma \left( {-1,2}\right) }\right) = 4 \), it is not an isometry, even though it is a transformation. The only fixed point is \( \left( {0,0}\right) \) . The axes are the sta... | Yes |
Theorem 5.2.1. The isometries of a set form a transformation group. | Proof. For closure, we let \( \alpha \) and \( \beta \) be isometries on a set \( S \) and show that \( \alpha \circ \beta \) is an isometry. Let \( P \) and \( Q \) be points in \( S \) . Then \( d\left( {P, Q}\right) = d\left( {\beta \left( P\right) ,\beta \left( Q\right) }\right) = d\left( {\alpha \left( {\beta \lef... | Yes |
Theorem 5.2.2. A Euclidean plane isometry that fixes three noncollinear points is the identity. | Proof. Let \( \alpha \) be an isometry, and \( A, B \), and \( C \) be three noncollinear points fixed by \( \alpha \), and \( D \) be any other point. We must show that \( \alpha \left( D\right) = D \) . By definition of an isometry, \( d\left( {A, D}\right) = \) \( d\left( {\alpha \left( A\right) ,\alpha \left( D\rig... | Yes |
Theorem 5.2.3. A Euclidean plane isometry is determined by what it does to any three noncollinear points. | Proof. Let \( A, B \), and \( C \) be three noncollinear points, \( \alpha \) and \( \beta \) be isometries such that \( \alpha \left( A\right) = \) \( \beta \left( A\right) ,\alpha \left( B\right) = \beta \left( B\right) \), and \( \alpha \left( C\right) = \beta \left( C\right) \) . By Theorem 5.2.1, \( {\beta }^{-1} ... | Yes |
Theorem 5.2.4. For two distinct points \( P \) and \( Q \) in the Euclidean plane, there is exactly one mirror reflection that takes \( P \) to \( Q \) . | Proof. Given distinct points \( P \) and \( Q \) we can use Euclid I-10 and I-11 to construct a perpendicular bisector. It is unique by SMSG postulates 3 and 12. Then the definition of a mirror reflection gives a unique mirror reflection, taking one point to the other. - | Yes |
Theorem 5.2.5. Every Euclidean plane isometry can be written as the composition of at most three mirror reflections. | Proof. Let \( \alpha \) be a Euclidean plane isometry and \( A, B \), and \( C \) be noncollinear points in the plane. By Theorem 5.2.3 we only have to find a composition of mirror reflections that together take \( A, B \), and \( C \) to the same images as \( \alpha \) does, say, \( P, Q \), and \( R \), respectively ... | Yes |
Theorem 5.2.7. There are four types of Euclidean plane isometries: mirror reflections, translations, rotations, and glide reflections. | Proof. From Theorem 5.2.5 every isometry can be written as the composition of one, two, or three mirror reflections. Exercises 5.2.11 and 5.2.12, Theorem 5.2.6, and the preceding discussion cover all the possibilities for isometries. | No |
Theorem 5.2.8 (SAS). For \( \bigtriangleup {ABC} \) and \( \bigtriangleup {PQR} \) if \( \overline{AB} \cong \overline{PQ},\angle {ABC} \cong \angle {PQR} \), and \( \overline{BC} \cong \) \( \overline{QR} \), then \( \bigtriangleup {ABC} \cong \bigtriangleup {PQR} \), that is, \( \angle {BAC} \cong \angle {QRP},\overl... | Proof. Given \( \bigtriangleup {ABC} \) and \( \bigtriangleup {PQR} \), by Theorem 5.2.3 there is a mirror reflection \( {\mu }_{1} \) taking \( A \) to \( P \) . Then \( \bigtriangleup {ABC} \cong \bigtriangleup P{\mu }_{1}\left( B\right) {\mu }_{1}\left( C\right) \) by (i). If \( {\mu }_{1}\left( B\right) = Q \), ski... | No |
In Exercise 5.3.2 the translation \( T = \left\lbrack \begin{array}{lll} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right\rbrack \) takes a point \( \left( {x, y,1}\right) \) to \( (x + \) \( 3, y + 2,1) \) . In Figure 5.19 the line \( y = - x \) or \( \left\lbrack {-1, - 1,0}\right\rbrack \) goes to the line \( y... | Theorem 5.3.1 shows that we need to use the inverse of \( T \) . | Yes |
Theorem 5.3.1. The affine matrix \( M \) takes \( \left\lbrack {a, b, c}\right\rbrack \) to the line \( \left\lbrack {a, b, c}\right\rbrack {M}^{-1} \) . | Proof. We need to show that, for any point \( \left( {x, y,1}\right) \), its image \( M\left( {x, y,1}\right) \) is on the proposed image of \( \left\lbrack {a, b, c}\right\rbrack \), namely, \( \left\lbrack {a, b, c}\right\rbrack {M}^{-1} \) if and only if \( \left( {x, y,1}\right) \) is on \( \left\lbrack {a, b, c}\r... | Yes |
Theorem 5.3.2. An affine matrix\n\n\\[ \nM = \\left\\lbrack \\begin{array}{lll} a & b & c \\\\ d & e & f \\\\ 0 & 0 & 1 \\end{array}\\right\\rbrack \n\\]\n\nis an isometry if and only if for some angle \\( \\theta \\) ,\n\n\\[ \nM = \\left\\lbrack \\begin{matrix} \\cos \\theta & - \\sin \\theta & c \\\\ \\sin \\theta &... | Proof. \\( \\left( \\Rightarrow \\right) \\) If \\( M \\) is an isometry, then, for the three points \\( {O}^{\\prime },{X}^{\\prime } \\), and \\( {Y}^{\\prime } \\) of Exercise 5.3.6, \\( \\bigtriangleup {O}^{\\prime }{X}^{\\prime }{Y}^{\\prime } \\cong \\bigtriangleup {OXY} \\) (Figure 5.20). The distance \\( d\\lef... | No |
Example 3. Find the center of rotation of \( \left\lbrack \begin{matrix} 0 & - 1 & 2 \\ 1 & 0 & 3 \\ 0 & 0 & 1 \end{matrix}\right\rbrack \) . | Solution. The center of this \( {90}^{ \circ } \) rotation is a fixed point, say, \( \left( {u, v,1}\right) \), satisfying\n\n\[ \left\lbrack \begin{matrix} 0 & - 1 & 2 \\ 1 & 0 & 3 \\ 0 & 0 & 1 \end{matrix}\right\rbrack \left\lbrack \begin{array}{l} u \\ v \\ 1 \end{array}\right\rbrack = \left\lbrack \begin{array}{l} ... | Yes |
Find the line of reflection of\n\n\[ M = \left\lbrack \begin{matrix} {0.6} & {0.8} & 2 \\ {0.8} & - {0.6} & - 4 \\ 0 & 0 & 1 \end{matrix}\right\rbrack \] | Solution. We can find the fixed points of the mirror reflection \( M \) as in Example 3, obtaining the pair of equations \( - {0.4u} + {0.8v} + 2 = 0 \) and \( {0.8u} - {1.6v} - 4 = 0 \) . The second is a multiple of the first, so we get an infinite family of fixed points, \( \left( {u,\frac{1}{2}u - 2\frac{1}{2},1}\ri... | Yes |
Example 5. Find the matrix \( S \) representing a rotation of \( \theta \) around the point \( \left( {u, v,1}\right) \) . | Solution. We build the rotation around \( \left( {u, v,1}\right) \) from the translation \( T = \left\lbrack \begin{array}{lll} 1 & 0 & u \\ 0 & 1 & v \\ 0 & 0 & 1 \end{array}\right\rbrack \) that moves \( \left( {0,0,1}\right) \) to \( \left( {u, v,1}\right) \) and the rotation \( R = \left\lbrack \begin{matrix} \cos ... | Yes |
Theorem 5.4.1. The set of similarities forms a transformation group. | ## Proof. See Exercise 5.3.9. | No |
The matrix \( S = \left\lbrack \begin{matrix} r & 0 & 0 \\ 0 & r & 0 \\ 0 & 0 & 1 \end{matrix}\right\rbrack \) takes \( \left( {x, y,1}\right) \) to \( \left( {{rx},{ry},1}\right) \) . Thus the origin \( \left( {0,0,1}\right) \) is fixed, and all points expand or contract with respect to the origin by a scaling ratio o... | Consider the two points \( \left( {a, b,1}\right) \) and \( \left( {p, q,1}\right) \) . Then\n\n\[ d(\left( {{ra},{rb},1}\right) ,\left( {{rp},{rq},1}\right) = \sqrt{{\left( ra - rp\right) }^{2} + {\left( rb - rq\right) }^{2}} \]\n\n\[ = \sqrt{{r}^{2}{\left( a - p\right) }^{2} + {r}^{2}{\left( b - q\right) }^{2}} \]\n\... | Yes |
Theorem 5.4.2. An affine matrix \( M \) represents a similarity if and only if\n\n\[ M = \left\lbrack \begin{matrix} r\cos \theta & \mp r\sin \theta & a \\ r\sin \theta & \pm r\cos \theta & b \\ 0 & 0 & 1 \end{matrix}\right\rbrack \text{ for some }r > 0\text{. The determinant is } \pm {r}^{2}. \] | Proof. See Exercise 5.4.10. | No |
Theorem 5.4.3. Similarities preserve angle measures and the proportions of distances. If a similarity has a scaling ratio of \( r \), then the area of the image of a convex polygon is \( {r}^{2} \) times the area of the original polygon. | Proof. We use the theorems of Section 1.4. See Exercise 5.4.11. \( \blacksquare \) | No |
Theorem 5.4.4. A similarity with a scaling ratio of \( r \neq 1 \) has a unique fixed point. | Proof. Let \( M \) be the similarity matrix and \( \left( {u, v,1}\right) \) be a candidate for a fixed point. From Theorem 5.4.2 we have one of the following systems of two equations in the two unknowns \( u \) and \( v \) :\n\n\[ \begin{array}{l} r\cos \left( A\right) u - r\sin \left( A\right) v + a = u \\ r\sin \lef... | Yes |
Theorem 5.4.5. The set of affine transformations is a transformation group. Affine transformations preserve lines, parallelism, betweenness, and proportions on a line. | Proof. In Section 5.3 we noted that the image of a line under an affine transformation was a line. Here we prove betweenness and proportions on a line. (See Exercise 5.4.16 for proofs about the group and parallelism.) From Section 3.5, the segment \( \overline{PQ} \) contains a point \( R \) (considered as a vector) if... | No |
Verify that the affine matrix \( \left\lbrack \begin{array}{lll} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \) triples every \( x \) -coordinate, doubles every \( y \) -coordinate, has a determinant of 6, and increases the area of every triangle by a factor of 6. | Solution. We leave all but the last part to you. For the last part, note that every triangle \( \bigtriangleup {ABC} \) without a horizontal side can be split into two triangles, each with a horizontal side \( (\overline{BD} \) in Figure 5.28). Then the smaller triangles have their bases tripled and their heights doubl... | No |
Theorem 5.4.6. An affine transformation preserves convexity. | Proof. Suppose that \( \alpha \) is an affine transformation and \( A \) is a convex set. We must show that the image \( \alpha \left\lbrack A\right\rbrack \) is convex. That is, for points \( {X}^{\prime } \) and \( {Y}^{\prime } \) in \( \alpha \left\lbrack A\right\rbrack \) and \( {Z}^{\prime } \), a point between t... | Yes |
Example 7. Figure 5.33 shows a distorted Koch curve made from the matrices\n\n\\[ \n{A}^{\prime } = \\left\\lbrack \\begin{matrix} 1/3 & 1/6 & 0 \\\\ 0 & 1/3 & 0 \\\\ 0 & 0 & 1 \\end{matrix}\\right\\rbrack ,\\;{B}^{\prime } = \\left\\lbrack \\begin{matrix} \\cos {60}^{ \\circ }/3 & \\cos {60}^{ \\circ }/6 - \\sin {60}^... | Figure 5.34 shows the images of the unit square for each of them, which are products of the shears \\( \\left\\lbrack \\begin{matrix} 1 & \\pm 1/2 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{matrix}\\right\\rbrack \\) with the matrices of Example 6. Shears shift rectangles to parallelograms, as illustrated in Figure 5.29. ... | Yes |
Theorem 5.4.7. Every contraction mapping has a unique fixed point. An IFS fractal is a closed and bounded set. | Proof. See Barnsley [2] for a proof, which uses analysis. The reasoning of Exercise 5.4.12(a) applies to a contraction mapping to give a unique fixed point. As in Example 6, the fractal is bounded by the sequence of parallelograms. \( \blacksquare \) | No |
Example 2. The transformation \( R = \left\lbrack \begin{matrix} 0 & - 1 & 0 \\ 0 & 0 & - 1 \\ 1 & 0 & 0 \end{matrix}\right\rbrack \) is a rotation of the sphere (and all of \( {\mathbf{R}}^{3} \) ). | Points of the form \( \left( {a, - a, a}\right) = a\left( {1, - 1,1}\right) \) are fixed by \( \rho \) and so form the axis of rotation, which is the line through the origin and \( \left( {1, - 1,1}\right) \) . (See Section 3.5 for a general description of lines in more than two dimensions.) Furthermore, \( R\left( {a,... | Yes |
Example 3. The rotatory reflection\n\n\[ \left\lbrack \begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & - 1 \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}\right\rbrack \cdot \left\lbrack \begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right\... | is the composition of a rotation of \( {90}^{ \circ } \) around the \( z \) -axis followed by a mirror reflection over the plane \( z = 0 \) (the equator) (Figure 5.38). The eigenvalues of the matrix are \( - 1, i \), and \( - i \), which show that there is no fixed point. The antipodal North and South poles \( \left( ... | Yes |
Theorem 5.5.1. The following statements are equivalent.\n\n(i) \( {A3} \times 3 \) matrix \( M \) is an isometry of the unit sphere.\n\n(ii) \( {M}^{-1} = {M}^{T} \), the transpose of \( M \) .\n\n(iii) The columns of \( M \) form an orthonormal basis of \( {\mathbf{R}}^{3} \) . | Proof. We show the equivalence of the first two parts and leave their connection with part (iii) to Exercise 5.5.12. ( \( \Rightarrow \) ) First suppose that \( M = \left\lbrack \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right\rbrack \) is a spherical isometry. Then \( M\left( {1,0,0}\right) = \lef... | No |
Every isometry of the sphere has at least two antipodal points on it that are either fixed or are mapped to each other. The circle on the sphere midway between them is stable. | Proof. When we use eigenvalues to find fixed points (and stable lines), we obtain an equation in \( \lambda \) called the characteristic equation. The characteristic equation of a \( 3 \times 3 \) matrix involves a third-degree real polynomial. All odd-degree real polynomials have a real root, so every isometry of the ... | Yes |
Determine what type of isometry \( M = \left\lbrack \begin{matrix} 0 & 1 & 0 & 2 \\ 1 & 0 & 0 & - 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right\rbrack \) is and find its fixed points | Solution. The determinant of \( M \) is -1 and \( {M}^{2} = I \) . So, as in the two-dimensional case, \( M \) is a mirror reflection, which we confirm with the fixed points. For the fixed points we solve\n\n\[ \left\lbrack \begin{matrix} 0 & 1 & 0 & 2 \\ 1 & 0 & 0 & - 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\r... | Yes |
Theorem 5.5.3. Every three-dimensional Euclidean isometry can be written as the composition of at most four mirror reflections. There are three types of three-dimensional direct Euclidean isometries: translations, rotations, and screw motions and three indirect types: mirror reflections, glide reflections, and rotatory... | ## Proof. See Project 12 and Coxeter [4]. - | No |
Theorem 5.5.4. Let \( \alpha \) be an \( n \) -dimensional affine transformation and \( A \) be an \( \left( {n + 1}\right) \times k \) matrix whose columns \( {A}_{1},{A}_{2},\ldots ,{A}_{k} \) are \( k \) points in \( n \) -dimensional affine space. Then the columns of \( {\alpha A} \) are \( \alpha \left( {A}_{1}\ri... | ## Proof. See Exercise 5.5.14. | No |
Example 6. The matrix \( \left\lbrack \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}\right\rbrack \), called a projection mapping, sends a point onto the \( {xy} \) -plane, which has the equation \( z = 0 \) . Two points with the same \( x \) - and \( y \) -coordinates are ... | \( \diamond \) | No |
Theorem 5.6.1. Let \( k \) be a line not through the center of inversion \( O \) of \( {v}_{c} \) . Then the image of \( k \) is a circle through \( O \) . Conversely, the image of a circle through \( O \) is a line not through \( O \) . | Proof. Let the perpendicular from \( O \) to \( k \) intersect \( k \) at \( A \) and let \( {A}^{\prime } = {v}_{c}\left( A\right) \) . Figure 5.42 illustrates the two cases, depending on whether \( k \) intersects the circle of inversion \( C \) . We show that the circle \( D \) with diameter \( \overline{O{A}^{\prim... | No |
Theorem 5.6.2. Let \( D \) be a circle that does not pass through the center of inversion of \( {v}_{c} \) . Then the inversive image of \( D \) is another circle that does not pass through the center of inversion. | Proof. See Eves [6, 78]. \( \blacksquare \) | No |
Theorem 5.6.3. If a circle \( D \) is orthogonal to the circle of inversion \( C \), then \( D \) is its own image: \( {v}_{c}\left\lbrack D\right\rbrack = D \) . | Proof. With \( D \) orthogonal to \( C \), the radii of \( C \) that go to the intersections \( P \) and \( Q \) are tangents to \( D \) (Figure 5.43). Because \( P \) and \( Q \) are on \( C \), they are fixed by the inversion. Lines \( \overrightarrow{OP} \) and \( \overrightarrow{OQ} \) are stable. By Theorem 5.6.2,... | Yes |
Theorem 5.6.4. The inversion with respect to the circle of radius \( r \) and center \( w \) is given by \( v\left( z\right) = \overline{{r}^{2}/\left( {z - w}\right) } + w \), for \( z \neq w \) . | Proof. First, we show that the formula works in the special case when \( w = 0 + {0i} = 0 \) . Call the function in this case \( {v}_{0}\left( z\right) \), which reduces to \( \overline{{r}^{2}/z} \) . The product \( z \cdot \overline{{r}^{2}/z} \) has a length of \( {r}^{2} \), showing that \( \overline{{r}^{2}/z} \) ... | Yes |
Find the transformation of the extended complexes \( \mathbf{C}\# \) that is the composition of the inversions \( \overline{1/z} \) followed by \( \overline{4/z} \) . | When we replace the \( z \) of \( \overline{4/z} \) with \( \overline{1/z} \), we get \( \overline{4/\overline{\left( 1/z\right) }} = \overline{\left( \overline{\left( 4z\right) }\right) } = {4z} \), a similarity (dilation) with a scaling factor of 4. \( \diamond \) | Yes |
Theorem 5.6.6. Möbius transformations preserve angle measure. | Proof. A general Möbius transformation can be written as a composition of similarities and inversions. (See Exercise 5.6.12.) We already know from Theorem 5.4.3 that similarities preserve angles. Hence we only need to show that inversions preserve angles. To simplify Figure 5.47, we will start with the angle between tw... | Yes |
Theorem 6.1.1. The symmetries of a subset \( T \) form a transformation group. | Proof. Recall from Section 5.1 that a group of transformations needs to have closure, identity, and inverses. If \( \alpha \) is a symmetry of \( T \), then \( \alpha \left\lbrack T\right\rbrack = T \) . For closure, let \( \alpha \) and \( \beta \) be symmetries of \( T \) . Then \( \alpha \circ \beta \left\lbrack T\r... | Yes |
Theorem 6.2.1. The isometries of a finite plane symmetry group fix a point and so are rotations around it or mirror reflections over lines through it. | Proof. We adapt a proof from Gallian [6,404]. Let \( \mathbf{G} \) be a finite symmetry group of plane isometries and assume that the plane has coordinates. For a point \( A \), let \( S = \{ \gamma \left( A\right) : \gamma \in \mathbf{G}\} \) ; that is, \( S \) is the set of images of \( A \) under the isometries of \... | Yes |
Theorem 6.2.2. A finite symmetry group containing only Euclidean plane isometries is either a cyclic group or a dihedral group. | Proof. We need to show that the possible rotations and mirror reflections from Theorem 6.2.1 always fit exactly as cyclic and dihedral groups require. First, consider the rotations. If there is only one, it is the identity, a rotation of \( {0}^{ \circ } \) . Otherwise, let the smallest positive angle of rotation be \(... | No |
Example 1. Count the symmetries of a pentagonal bipyramid. | Solution. The pentagonal bipyramid shown in the left of Figure 6.11 has seven vertices, but they can't all be mapped to one another. The middle of Figure 6.11 pictures the polyhedron from the top vertex. The symmetries fixing the top vertex form the dihedral group \( {\mathbf{D}}_{5} \), so ten symmetries fix it. The o... | Yes |
Theorem 6.2.3. The number of symmetries of a figure, if finite, equals the product nk, where \( n \) is the number of symmetries of the entire figure that leave a point fixed and \( k \) is the number of points, including that point, to which it can be moved by symmetries. | Proof. Let \( P \) be a point of the figure, \( \mathbf{G} \) the symmetry group, \( {\mathbf{G}}_{P} \) be the set of symmetries fixing \( P \), and let \( {\mathbf{G}}_{p} \) have \( n \) elements. We collect the symmetries of \( \mathbf{G} \) into disjoint classes, show the classes to be the same size, \( n \), and ... | Yes |
Theorem 6.2.4. In a finite symmetry group, either all the isometries are direct or exactly half of them are direct. | Proof. Let \( \mathbf{D} \) be the set of the direct isometries and \( \mathbf{I} \) the set of the indirect isometries, if any, in the symmetry group. If \( \mathbf{D} \) is the entire symmetry group, we are done. If \( \gamma \in \mathbf{I} \), then \( \gamma \mathbf{D} = \{ \gamma \circ \delta : \delta \in \mathbf{D... | Yes |
Example 2. Describe the symmetries of a pentagonal bipyramid. | Solution. By Theorem 6.2.4, we know that half of the twenty symmetries are rotations, including the identity. Four rotations around the axis through the top and bottom vertices have angles of rotation that are multiples of \( {72}^{ \circ } \) . (See Figure 6.11(a).) The five remaining rotations, each of \( {180}^{ \ci... | Yes |
Theorem 6.3.1. The only symmetries of a frieze pattern with horizontal translations are horizontal translations, vertical mirror reflections, rotations of \( {180}^{ \circ } \) with centers on the midline of the frieze pattern, and glide reflections and mirror reflections over the midline of the frieze pattern. | Proof. We show that all other isometries map the midline to a different line and so cannot be symmetries of the frieze pattern. Translations in a direction other than horizontal lift or lower the midline, so they are eliminated. Rotations other than \( {180}^{ \circ } \) (or \( {0}^{ \circ } \), the identity) tilt the ... | Yes |
Explain why a translation, a vertical mirror reflection, and the horizontal mirror reflection generate the group of symmetries of the frieze pattern shown in Figure 6.16. | Solution. We use properties from Chapter 5 to analyze compositions. The smallest translation generates all the others. The compositions of a vertical mirror reflection with the translations give all the vertical mirror reflections. Similarly, the horizontal mirror reflection and the translations generate the glide refl... | No |
Theorem 6.3.2. There are exactly seven groups of symmetries for frieze patterns, up to geometric isomorphism. | Proof. Example 3 shows that there are at least seven frieze groups. To show there are no others, we consider the possible sets of generators for frieze groups chosen from the isometries described in Theorem 6.3.1. We use \( \tau \) for the smallest translation to the right, \( \rho \) for a rotation, \( \eta \) for the... | No |
Theorem 6.3.3. The Crystallographic Restriction. The minimal positive angles of rotations that can be symmetries of a wallpaper pattern are \( {60}^{ \circ },{90}^{ \circ },{120}^{ \circ },{180}^{ \circ } \), and \( {360}^{ \circ } \) . All other angles of rotation for a given wallpaper pattern are multiples of the min... | Proof. Let \( A \) be a center of rotation for a wallpaper pattern and let \( B \) be a point closest to \( A \) for which a symmetry takes \( A \) to \( B \) . Then \( B \) is, by symmetry, also a center of rotation for the wallpaper pattern with the same angles as at \( A \) . No two other images of \( A \) can be an... | No |
Theorem 6.3.4. There are exactly seventeen groups of symmetries for wallpaper patterns, up to isomorphism. | ## Proof. See Crowe [2]. \( \blacksquare \) | No |
Example 4. Classify the patterns shown in Figure 6.23. | Solution. Both patterns have \( {180}^{ \circ } \) rotations. Following that branch in Figure 6.22, we note that both have mirror reflections and perpendicular mirror reflections. In the Chinese design, the motifs stack like boxes, so from the flowchart it has the group pmm. However, the Japanese motifs stack like bric... | Yes |
Example 5. Classify the patterns shown in Figure 6.24. | Solution. Both designs have rotations of \( {120}^{ \circ } \) but not \( {60}^{ \circ } \) . Each has some mirror reflections that pass through centers of rotation. The design on the left of Figure 6.24 has mirror reflections through every center of rotation, such as the centers of hexagons and the centers of the \( \... | Yes |
Example 6. Classify the colored patterns shown in Figure 6.25. | Solution. First consider the color-preserving symmetries of the two-color Peruvian design in Figure 6.25. All the black staircases are upright, implying no color-preserving rotations. The rows of black staircases alternate facing left and right, indicating no mirror reflections, but indicating glide reflections that pr... | Yes |
Theorem 6.3.5. The color-preserving symmetries of a design form a subgroup of the color symmetries of the design. | ## Proof. See Exercise 6.3.16. | No |
Example 1. Find the group of symmetries of the great stellated dodecahedron (Figure 6.31). | Solution. This polyhedron has axes of rotation other than those of a prism. The axis facing out allows rotations of \( {72}^{ \circ } \), so the group can’t be \( \overline{\mathbf{T}} \) or \( \overline{\mathbf{W}} \) or their subgroups. Because the polyhedron has mirror reflections, by Theorem 6.4.2 below, it must ha... | No |
Theorem 6.4.1. The three-dimensional isometries in a finite symmetry group must fix a point and so form a subgroup of the symmetries of a sphere. | Proof. See Exercise 6.4.9. \( \blacksquare \) | No |
Theorem 6.4.2. A finite group of three-dimensional isometries must be one of the following or one of their subgroups: \( {\mathbf{D}}_{nh},\overline{\mathbf{T}},\overline{\mathbf{W}} \), and \( \overline{\mathbf{P}} \) . | Proof. See Weyl [23, 149ff]. | No |
Theorem 6.4.3. For a positive integer \( n \), the group \( {\mathbf{S}}_{n} \) has \( n \) ! elements. | Proof. See Exercise 6.4.10. | No |
Theorem 6.4.4. The group of isometries of the \( \left( {n - 1}\right) \) -dimensional regular simplex is \( {\mathbf{S}}_{n} \) . | Proof. See Exercise 6.4.12. (A regular simplex is defined in Section 3.5.) | No |
Example 1. Show the Koch curve is infinitely long. | Solution. Suppose that the original segment in Figure 6.47 has a length of 1 unit. The first iteration (the motif) has four segments of length \( 1/3 \), for a length of \( 4/3 \approx {1.3} \) units. The second iteration has sixteen segments and a length of \( {16}/9 = {\left( 4/3\right) }^{2} \approx {1.8} \) units. ... | Yes |
Find the Hausdorff dimension of the Koch curve. | Solution. In Example 1 a scaling factor of \( r = \frac{1}{3} \) gives \( n = 4 \) times as many units. Then (6.3) gives \( d = \log 4/\log 3 \approx {1.262} \) . Verify that a scaling ratio \( r = \frac{1}{9} = {\left( \frac{1}{3}\right) }^{2} \) gives the same value of \( d \) . The Koch curve is too convoluted to be... | Yes |
We verify that the Hausdorff dimension of a circle is 1. | This makes sense since the circumference of a circle is a length, which is a one-dimensional measure. In Figure 6.50, we can approximate the circumference of a circle with increasingly smaller units. The table below gives data for regular polygons, starting with the number of sides in the first column. The next column ... | Yes |
In the motif of the Koch curve (see Figure 6.47) let each segment have a length of \( {f}_{1} = \frac{1}{3} \) . Then the four segments give \( L\left( {f}_{1}\right) = 4/3 \) as an estimated length of the Koch curve. In the second iteration when \( {f}_{2} = \frac{1}{3} \cdot \frac{1}{3} \), the sixteen segments give ... | With \( {f}_{1} \) we get \( 4/3 = c{\left( \frac{1}{3}\right) }^{1 - d} \), and with \( {f}_{2} \) we get \( \frac{16}{9} = c{\left( \frac{1}{3} \cdot \frac{1}{3}\right) }^{1 - d} = \) \( c{\left( \frac{1}{3}\right) }^{1 - d}{\left( \frac{1}{3}\right) }^{1 - d} \) . Substituting the first equation into the second yiel... | Yes |
From three points \( P, Q \), and \( R \) on a line, the following construction determines the unique fourth point \( S \) to form a harmonic set, illustrated in Figure 7.4. Although not obvious, every such construction will give the same fourth point. (See Exercises 7.1.7 and 7.1.8.) Uniqueness is an axiom in Section ... | Let \( P, Q \), and \( R \) be three distinct points on a line \( m \) . Draw lines \( {k}_{1} \) and \( {k}_{2} \) through \( P \) and another line \( {k}_{3} \) through \( R \) . (The lines \( {k}_{i} \) are distinct and differ from \( m \) .) Let \( {T}_{1} \) and \( {T}_{2} \) be the intersections of \( {k}_{3} \) ... | No |
Theorem 7.1.1. (Desargues' theorem). If two triangles are perspective from a point, then they are perspective from a line. | ## Proof. See Project 4. ∎ | No |
Theorem 7.2.1. (i) Two distinct lines have exactly one point on both lines. | Proof. See Exercise 7.2.12 for parts (i) and (iii). For part (ii), let \( k \) be a line. By axiom (ii) there are four points \( A,{A}_{1},{A}_{2} \), and \( {A}_{3} \) and at least one of them, say \( A \), is not on \( k \) . Consider the lines \( {k}_{i} \) on \( A \) and \( {A}_{i} \), for \( i = 1,2,3 \) . By axio... | No |
Theorem 7.2.2. (i) If \( {PQ}//{RS} \), then \( {QP}//{RS},{QP}//{SR},{RS}//{QP},{SR}//{PQ} \), and \( {SR}//{QP} \) . | Proof. See Exercise 7.2.14. | No |
Theorem 7.2.3. If \( {X}_{p} \) and \( {X}_{q} \) are determined from the definition and Exercise 7.2.1 and \( p \neq q \), then \( {X}_{p} \) and \( {X}_{q} \) are distinct. | Partial Proof. Let \( a, b \), and \( c \) be nonnegative integers with \( a < b < c \) . Then \( {X}_{a},{X}_{b} \), and \( {X}_{c} \) are determined from the definition. We use induction on the size of \( c \) to prove that \( X{X}_{b}//{X}_{a}{X}_{c} \) .\n\nFor the initial case, let \( k = 1 \) and \( c \leq k + 1 ... | No |
Theorem 7.2.4. The duals of axioms (i), (ii), and (iii) hold. | Proof. See Exercise 7.2.15. | No |
Theorem 7.2.5. The duals of axioms (iv)-(x) hold. | Proof. We prove the dual of axiom (iv). (See Exercise 7.2.15 for the others, which are similar.)\n\nThe dual of axiom (iv) states,\ | No |
Theorem 7.2.6. A perspectivity preserves harmonic sets of points and the relation of separation. That is, if a perspectivity from \( O \) maps the collinear points \( P, Q, R \), and \( S \) to the collinear points \( {P}^{\prime },{Q}^{\prime },{R}^{\prime } \), and \( {S}^{\prime } \) and \( H\left( {{PQ},{RS}}\right... | Proof. See Exercise 7.2.18. | No |
Theorem 7.2.7. Fundamental Theorem of Projective Geometry. A projectivity of a line is completely determined by three points on the original line and their images. | Proof. Without loss of generality call the points on the original line \( X,{X}_{0} \), and \( {X}_{1} \) and suppose that the projectivity takes them to \( Y,{Y}_{0} \), and \( {Y}_{1} \), respectively. Theorem 7.2.6 ensures that a point \( {X}_{p} \) described in Theorem 7.2.3 goes to \( {Y}_{p} \) . Exercise 7.2.8 a... | No |
We can think of an ordinary Euclidean point \( \left( {x, y}\right) \) as the projective point \( \left( {x, y,1}\right) \) or in general \( \left( {{\lambda x},{\lambda y},\lambda }\right) \) . For this example we ignore the scalar for simplicity. Figure 7.20 illustrates several projective points and lines, with the g... | The ideal point on this line is \( \left( {1,1,0}\right) \), which is on \( \left\lbrack {1, - 1,0}\right\rbrack \) and also on all the Euclidean lines parallel to it, \( y = x + c \), which are of the form \( \left\lbrack {1, - 1, c}\right\rbrack \) . However, there is no reason in projective geometry to single out an... | Yes |
We can relate each point of the projective plane to two antipodal points on a sphere. For each projective point \( \left( {x, y, z}\right) \), there are scalars \( \lambda \) and \( - \lambda \) such that \( \left( {{\lambda x},{\lambda y},{\lambda z}}\right) \) and \( \left( {-{\lambda x}, - {\lambda y}, - {\lambda z}... | Figure 7.21 illustrates this initially nonintuitive fact and Example 3 provides an explanation. | No |
We claim that distinct points \( \left( {p, q, r}\right) ,\left( {s, t, u}\right) \), and \( \left( {v, w, x}\right) \) are collinear if and only if the determinant of \( \left\lbrack \begin{matrix} p & s & v \\ q & t & w \\ r & u & x \end{matrix}\right\rbrack \) is 0. | The points are on the line \( \left\lbrack {a, b, c}\right\rbrack \) if and only if the products \( \left\lbrack {a, b, c}\right\rbrack \left( {p, q, r}\right) \) , \( \left\lbrack {a, b, c}\right\rbrack \left( {s, t, u}\right) \), and \( \left\lbrack {a, b, c}\right\rbrack \left( {v, w, x}\right) \) are 0 . In matrix ... | Yes |
The points on \( \left\lbrack {1, - 1,2}\right\rbrack \) are of the form \( \left( {x, y, z}\right) \), where \( x - y + {2z} = 0 \) or \( y = \) \( x + {2z} \) . | The last equation enables us to eliminate the \( y \) -value, so the two coordinates \( \left( {x, z}\right) \) are sufficient to describe which point on this line we are considering. From the Euclidean point of view, the point \( \left( {x,1}\right) = \left( {{\lambda x},\lambda }\right) \) corresponds to the real num... | No |
For \( P = \left( {p,1}\right), S = \left( {s,1}\right), U = \left( {u,1}\right) \), and \( W = \left( {w,1}\right), R\left( {P, S, U, W}\right) = \frac{p - u}{p - w} \div \frac{s - u}{s - w} \). When we can replace homogeneous coordinates by cartesian coordinates, we’ll write \( R\left( {p, s, u, w}\right) \). For exa... | \( \diamondsuit \) | No |
Example 7. Let \( P = \\left( {1,1}\\right), S = \\left( {2,1}\\right), U = \\left( {3,1}\\right) \), and \( W = \\left( {w,1}\\right) \) . Investigate what happens to \( R\\left( {p, s, u, w}\\right) \) as \( w \) varies. | Solution. \( R\\left( {p, s, u, w}\\right) = \\frac{1 - 3}{1 - w} \\div \\frac{2 - 3}{2 - w} = \\frac{4 - {2w}}{1 - w} \) . Figure 7.22 gives the graph of \( f\\left( w\\right) = \\frac{4 - {2w}}{1 - w} \) . From it we see that 1 and 2 separate 3 and \( w \) if and only if \( 1 < w < 2 \) . Also, as \( w \\rightarrow \... | Yes |
Theorem 7.4.1. For distinct collinear points \( \left( {{u}_{1},{v}_{1}}\right) ,\left( {{u}_{2},{v}_{2}}\right) \), and \( \left( {{u}_{3},{v}_{3}}\right) \) and distinct collinear points \( \left( {{s}_{1},{t}_{1}}\right) ,\left( {{s}_{2},{t}_{2}}\right) \), and \( \left( {{s}_{3},{t}_{3}}\right) \), there is a uniqu... | Proof. First we show that a projectivity can map the collinear points \( \left( {1,0}\right) ,\left( {0,1}\right) \), and \( \left( {1,1}\right) \) to any distinct collinear points \( \left( {{s}_{1},{t}_{1}}\right) ,\left( {{s}_{2},{t}_{2}}\right) \), and \( \left( {{s}_{3},{t}_{3}}\right) \) . The matrix \( \left\lbr... | Yes |
Theorem 7.4.2. Cross ratios, harmonic sets, and separation are preserved under projectivities. | Proof. Let \( {X}_{i} = \left( {{u}_{i},{v}_{i}}\right) \) for \( i = 1,2,3,4 \) be collinear points and \( M = \left\lbrack \begin{array}{ll} a & c \\ b & d \end{array}\right\rbrack \) be a projectivity. We first show the cross ratios \( R\left( {{X}_{1},{X}_{2},{X}_{3},{X}_{4}}\right) \) and \( R\left( {M{X}_{1}, M{X... | Yes |
Theorem 7.4.3. The image of the line \( \left\lbrack {a, b, c}\right\rbrack \) under the collineation \( M \) is \( \left\lbrack {a, b, c}\right\rbrack {M}^{-1} \) . The image of the conic \( C \) under \( M \) is \( {M}^{-{1T}}C{M}^{-1} \) . | ## Proof. See Exercise 7.4.5. \( \blacksquare \) | No |
Theorem 7.4.4. The set of projectivities of a line to itself forms a group of transformations. The set of collineations forms a group of transformations. | Proof. See Exercise 7.4.6. \( \blacksquare \) | No |
Consider the Euclidean translation\n\n\[ T = \left\lbrack \begin{matrix} 1 & 0 & 3 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}\right\rbrack \]\n\nIn Chapter 5 we saw that a translation has no fixed affine points \( \left( {x, y,1}\right) \). However, we now have more points \( \left( {x, y, z}\right) \). Because \( \lambd... | We obtain three equations: \( x + {3z} = {\lambda x}, y - {2z} = {\lambda y} \), and \( z = {\lambda z} \). The last equation forces \( \lambda = 1 \) or \( z = 0 \). Then the first two equations force both \( \lambda = 1 \) and \( z = 0 \). Thus every ideal point \( \left( {x, y,0}\right) \) is fixed by a translation.... | Yes |
The ideal line \( \left\lbrack {0,0,1}\right\rbrack \) is stable for all similarities and affine transformations, for the bottom row of their inverses is \( \left\lbrack \begin{array}{lll} 0 & 0 & 1 \end{array}\right\rbrack \) . | Consider\n\n\[ A = \left\lbrack \begin{array}{lll} 0 & 2 & 3 \\ 2 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right\rbrack \]\n\nwhich reflects over the line \( y = x - z\left( \left\lbrack {1, - 1, - 1}\right\rbrack \right) \) and expands by a factor of 2 around the fixed point \( \left( {-1, - 2,1}\right) \) . There are two oth... | Yes |
Example 3. Figure 7.24 portrays the transformation of a checkerboard pattern into a perspective view using the matrix \( \\left\\lbrack \\begin{matrix} 9 & 1 & {20} \\\\ 7 & 7 & 4 \\\\ 1 & 1 & 4 \\end{matrix}\\right\\rbrack \) . The first column gives the image \( {X}^{\\prime } \) of \( X = \\left( {1,0,0}\\right) \),... | You can check that the matrix transforms \( T = \\left( {2,2,1}\\right) \) to \( {T}^{\\prime } = \\left( {5,4,1}\\right) \) and \( \\left( {0,4,1}\\right) \) to \( \\left( {3,4,1}\\right) \), among other points. The ideal point \( D = \\left( {1,1,0}\\right) \) on the line through \( Z, U \), and \( T \) becomes \( {D... | Yes |
Theorem 7.4.5. Let \( {P}_{1},{P}_{2},{P}_{3} \), and \( {P}_{4} \) be four points, no three of which are collinear; and \( {Q}_{1},{Q}_{2},{Q}_{3} \), and \( {Q}_{4} \) be any four points. Then a unique collineation takes \( {P}_{1},{P}_{2},{P}_{3} \), and \( {P}_{4} \) to \( {Q}_{1},{Q}_{2},{Q}_{3} \), and \( {Q}_{4}... | ## Proof. See Exercise 7.4.9. \( \blacksquare \) | No |
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