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Theorem 7.4.6. Every collineation of the projective plane has at least one fixed point and at least one stable line. | Proof. To solve the eigenvector problem \( A\left( {x, y, z}\right) = \lambda \left( {x, y, z}\right) \) we find the values of \( \lambda \) for which the determinant of \( A - {\lambda I} \) is zero. As \( A - {\lambda I} \) is a \( 3 \times 3 \) matrix, with \( \lambda \) appearing in the three diagonal entries, the ... | Yes |
Example 4. The matrix \( \left\lbrack \begin{matrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & - 1 \end{matrix}\right\rbrack \) is a rotary reflection when thought of as a spherical isometry or three-dimensional Euclidean isometry. It rotates the sphere \( {90}^{ \circ } \) around the \( z \) -axis and reflects it over the e... | As a spherical isometry, it fixes no Euclidean point. However, it maps \( \left( {0,0,1}\right) \) and \( \left( {0,0, - 1}\right) \) to each other, which are the same projective point. The fixed point is for the only real eigenvalue, -1 . The only stable line is \( \left\lbrack {0,0,1}\right\rbrack \), the sphere’s eq... | Yes |
Verify that the adjacent points \( {P}_{i} \) and \( {P}_{i + 1} \) shown in Figure 7.26 have the same distance between them. 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},{P}_{-i} = - {... | Solution. The Euclidean distances between the points are the differences of their \( x \) - coordinates. Then \( \left( {{P}_{0}\Omega /{P}_{0}\Lambda }\right) \div \left( {{P}_{1}\Omega /{P}_{1}\Lambda }\right) = \left( {1/1}\right) \div \left( {\frac{4}{3}/\frac{2}{3}}\right) = \frac{1}{2} \) . Similarly, \( \left( {... | Yes |
A matrix \( {X}_{x} \) analogous to a translation shifting \( \left( {0,0,1}\right) \) along the \( x \) -axis to \( \left( {x,0,1}\right) \) has for its last column \( \left( {x,0,1}\right) \), with \( - 1 < x < 1 \) . The translation must leave the omega points \( \left( {1,0,1}\right) \) and \( \left( {-1,0,1}\right... | One way to satisfy the equations is \( a = 1, d = 0 \), and \( g = x \) . Figure 7.27 suggests that \( {X}_{x} \) should shift \( \left( {0,1,1}\right) \) and \( \left( {0, - 1,1}\right) \) to the points on the unit circle directly above and below \( \left( {x,0,1}\right) \) . This condition forces \( b = 0 = h \) and ... | No |
Theorem 7.5.1. The hyperbolic isometries form a group of transformations. The set \( \left\{ {\lambda {X}_{x}}\right. \) : \( - 1 < x < 1,\lambda \neq 0\} \) forms a subgroup. | ## Proof. See Exercise 7.5.10. | No |
Find conditions on a \( 3 \times 3 \) invertible matrix \( M \) so that \( M \) is a hyperbolic isometry. | For \( M \) to be a hyperbolic isometry, we must have \( {M}^{-{1T}}C{M}^{-1} = {\lambda C} \) for some \( \lambda \neq 0 \) . Multiply by \( {M}^{T} \) on the left and \( M \) on the right to get \( C = {M}^{T}{\lambda CM} \) . We can factor out \( \lambda \) and move it to the other side to get \( \frac{1}{\lambda }C... | Yes |
Theorem 7.5.3. A collineation \( M \) is an isometry of single elliptic geometry if and only if \( {M}^{T} = \lambda {M}^{-1} \) for some nonzero real number \( \lambda \) . The isometries form a group of transformations. | Proof. See Exercise 7.5.10. | No |
Example 4. Show that Euclidean isometries map \( I \) and \( J \) to themselves. | Solution. From Section 5.3, Euclidean isometries are of the form\n\n\[ \left\lbrack \begin{matrix} \cos \theta & - \sin \theta & c \\ \sin \theta & \cos \theta & f \\ 0 & 0 & 1 \end{matrix}\right\rbrack \;\text{ or }\;\left\lbrack \begin{matrix} \cos \theta & \sin \theta & c \\ \sin \theta & - \cos \theta & f \\ 0 & 0 ... | Yes |
Two points in \( {\mathbf{P}}^{3} \) (or any \( {\mathbf{P}}^{n} \) ) have a unique line on them. | In terms of linear algebra, two distinct one-dimensional subspaces are spanned by a unique two-dimensional subspace. However, in \( {\mathbf{P}}^{3} \) two lines can fail to intersect, which corresponds to skew lines in Euclidean space. For example, the lines (two-dimensional subspaces) \( \{ \left( {x, y,0,0}\right) :... | No |
We illustrate the effects of the perspective entries in a collineation by projecting a cube using several related matrices. (For simplicity we ignore the translation and scaling entries.) By Theorem 5.5.4, we need only consider what the collineations do to the \( 4 \times 8 \) matrix\n\n\[ V = \left\lbrack \begin{array... | The matrix\n\n\[ P = \left\lbrack \begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right\rbrack \]\n\nrepresents a projection parallel to the \( x \) -axis; it isn’t a collineation because it collapses three-dimensional objects to two dimensions. For example, \( P \) maps... | Yes |
The oriented points between \( \left( \widetilde{0,0,1}\right) \) and \( \left( \widetilde{1,0,1}\right) \) are \( \left( \widetilde{x,0,1}\right) \), where \( 0 < x < 1 \) . | Proof. Let \( \left( {u, v, w}\right) \) be between \( \left( {0,0,1}\right) \) and \( \left( {1,0,1}\right) \) . Then there are positive reals \( a \) and \( b \) such that \( \left( \widetilde{u, v, w}\right) = a\left( \widetilde{0,0,1}\right) + b\left( \widetilde{1,0,1}\right) = \left( {b,\widetilde{0, a + b}}\right... | Yes |
Theorem 7.6.1. Collineations in oriented projective geometry preserve betweenness and convexity. | Proof. Let \( \gamma \) be a collineation and \( \widetilde{P} \) and \( \widetilde{Q} \) be two points. Then \( \widetilde{R} = a\widetilde{P} + b\widetilde{Q} \), for \( a \) and \( b \) positive, is between \( \widetilde{P} \) and \( \widetilde{Q} \) . Because \( \gamma \) is a linear transformation, \( \gamma \wide... | Yes |
Theorem 7.6.2. A \( 4 \times 4 \) nonsingular matrix represents a hyperbolic isometry if and only if any two of its columns are h-orthogonal, the first three have the same h-length, and the last column has the opposite h-length. | ## Proof. See Exercise 7.6.9. \( \blacksquare \) | No |
Corollary 7.6.3. The Lorentz transformations are collineations of \( {\mathbf{P}}^{4} \), where the bottom row is \( \left\lbrack \begin{array}{lllll} 0 & 0 & 0 & 0 & 1 \end{array}\right\rbrack \) and the upper left \( 4 \times 4 \) submatrix is a hyperbolic isometry, with the first three columns having \( h \) -length... | ## Proof. See Exercise 7.6.10. | No |
In 1850, the Rev. Thomas Kirkman posed and solved the “15-schoolgirl problem.” He supposed that the girls took walks every day in an artificially regimented style of five rows of three each. He asked for daily arrangements of them so that two girls walked in the same row just once a week. | The solution in Table 8.2 makes use of Fano's finite three-dimensional projective geometry, discussed in Section 8.4. The girls are points of the geometry and the groups of three girls are the lines.\n\nTable 8.2\n\n<table><tr><td>Sunday</td><td>1, 2, 3</td><td>5, 11, 14</td><td>6, 9, 15</td><td>7, 10, 13</td><td>4, 8,... | Yes |
Theorem 8.2.1. In an affine plane\n\n(i) two distinct lines have at most one point in common,\n\n(ii) for every point there is a line not on that point,\n\n(iii) for every line there is a point not on that line,\n\n(iv) every line has at least two points on it, and\n\n(v) every point is on at least three lines. | Proof. For parts (iii) and (iv), let \( k \) be a line. Let \( A,{A}_{1},{A}_{2} \), and \( {A}_{3} \) be the four points guaranteed by axiom (ii). Again by axiom (ii), at least one of the points, say \( A \), is not on \( k \), showing part (iii). From axiom (i) there are lines \( {k}_{i} \) that are on \( A \) and \(... | No |
Theorem 8.2.2. Parallelism is an equivalence relation for lines in an affine plane. That is, for lines \( k, l \), and \( m \), three properties hold: reflexive, \( k\parallel k \) ; symmetric, if \( k\parallel l \), then \( l\parallel k \) ; and transitive, if \( k\parallel l \) and \( l\parallel m \), then \( k\paral... | Proof. See Exercise 8.2.8. \( \blacksquare \) | No |
Theorem 8.2.3. If a line of an affine plane has \( n \) points on it, then each line has \( n \) points on it and each point has \( n + 1 \) lines on it. | Proof. Let \( k \) be a line with \( n \) points on it, say, \( {P}_{1},\ldots ,{P}_{n} \), where \( n \geq 2 \) . First, let \( l \) be a line not parallel to \( k \) . By axiom (ii), there is a point \( Q \) on neither \( l \) nor \( k \) . By axiom (i), \( Q \) has exactly \( n \) lines on it that intersect \( k \) ... | Yes |
Theorem 8.2.4. In an affine plane if some line has \( n \) points on it, then there are \( {n}^{2} \) points and \( {n}^{2} + n \) lines, and each line has \( n \) lines parallel to it, including itself. | Proof. Let \( P \) be a point. Then there are \( n + 1 \) lines on \( P \) and each has \( n - 1 \) points on it other than \( P \) by Theorem 8.2.3. Thus there are \( \left( {n + 1}\right) \left( {n - 1}\right) \) points besides \( P \), giving a total of \( \left( {n + 1}\right) \left( {n - 1}\right) + 1 = {n}^{2} \)... | No |
There is no affine plane of order six. | We show that an affine plane with six points on a line would provide a solution to Euler's 36-officer problem, which we assume to be impossible. So, for a contradiction, suppose that there were an affine plane of order 6 . We choose two nonparallel lines to determine the rows and columns of the officers and fix a point... | Yes |
## Theorem 8.2.5. In a projective plane\n\n(i) every two distinct lines have exactly one point on them both,\n\n(ii) there are at least four lines with no three on the same point,\n\n(iii) every line is on at least three points, and\n\n(iv) every point is on at least three lines. | ## Proof. See Exercise 8.2.13. | No |
Theorem 8.2.6. If one line of a projective plane has \( n + 1 \) points on it, then all lines have \( n + 1 \) points on them and all points have \( n + 1 \) lines on them. | Proof. See Exercise 8.2.14. \( \blacksquare \) | No |
Theorem 8.2.7. If one line of a projective plane has \( n + 1 \) points on it, there are \( {n}^{2} + n + 1 \) points and \( {n}^{2} + n + 1 \) lines. | ## Proof. See Exercise 8.2.15. | No |
Imagine that a food company wants a taste test to compare eleven products, some potential new ones and some current products. Research has shown that individual tasters can't reliably compare so many different tastes. So each taster will compare just five of the products. Yet every pair of products needs to be compared... | Table 8.3 , with eleven tasters (A to K) and eleven products (1 to 11), satisfies all the conditions. Each pair of products is compared by two tasters, who compare them with different combinations of products.\n\n<table><tr><td>A: \( 1,3,4,5,9 \)</td><td>B: 2, 4, 5, 6, 10</td><td>C: 3, 5, 6, 7, 11</td><td>D: 1, 4, 6, 7... | Yes |
Theorem 8.3.1. In a \( \left( {v, k,\lambda }\right) \) BIBD, each variety is in the same number \( r \) of blocks, \( r(k - 1) = \lambda \left( {v - 1}\right) \) and \( v \cdot r = b \cdot k \) . | Proof. Let \( W \) be a variety of a BIBD and \( {r}_{W} \) be the number of blocks containing \( W \) . Each block has \( k - 1 \) other varieties. So \( {r}_{W}\left( {k - 1}\right) \) counts all appearances of the other varieties in blocks containing \( W \) . The value \( \lambda \) counts the number of times anoth... | Yes |
Theorem 8.3.2. There is a Steiner triple system \( \left( {v,3,1}\right) \) if and only if \( v = {6n} + 1 \) or \( v = \) \( {6n} + 3 \) . | Proof. Exercise 8.3.5 shows \( v \) must equal either \( {6n} + 1 \) or \( {6n} + 3 \) . (See Anderson [1,112] for the existence of such systems.) \( \blacksquare \) | No |
A code can detect as many as \( k \) errors if the Hamming distance between two code words is at least \( k + 1 \) . A code can correct as many as \( k \) errors if the Hamming distance between any two code words is at least \( {2k} + 1 \) . | Proof. Suppose that all the code words are \( n \) -tuples. The number of errors in a received \( n \) -tuple is the number of places it differs from the code word sent, or the Hamming distance between them. Suppose that the Hamming distance between any two code words is at least \( k + 1 \) . If between 1 and \( k \) ... | Yes |
Let \( {\mathbb{Z}}_{3} \) be the set \( \{ 0,1,2\} \) together with addition and multiplication modulo 3 . That is, after doing the usual arithmetic, we subtract multiples of 3 until we get back to a number in \( {\mathbb{Z}}_{3} \) . For example, \( 2 + 2 = 4 \) becomes 1 (mod 3) because \( 4 - 3 = 1 \) . We write \(... | Tables 8.6 and 8.7 give additions and multiplications. | No |
Example 2. Let \( {\mathbb{Z}}_{5} \) be the set \( \{ 0,1,2,3,4\} \) with addition and multiplication modulo 5, given in Tables 8.8 and 8.9. For example, \( 3 \times 4 \equiv 2\left( {\;\operatorname{mod}\;5}\right) \) because \( 3 \times 4 = {12} \equiv {12} - 5 - 5 = \) 2 (mod 5). \( {\mathbb{Z}}_{5} \) is a field. | As in Example 1, we can place the numbers of \( {\mathbb{Z}}_{5} \) clockwise around a circle, as in Figure 8.13, and then addition corresponds to clockwise rotation. The numbers 1 and 4 are additive inverses or, more informally, negatives of each other. Adding 4 is the same as subtracting 1. Similarly, 2 and 3 are add... | No |
Theorem 8.4.1. \( {\mathbb{Z}}_{p} \) is a field if and only if \( p \) is a prime. There is, up to isomorphism, exactly one field with \( {p}^{k} \) elements, where \( p \) is a prime number. | Proof. See Gallian [6, 213 and 328]. ∎ | No |
Example 4. The 25 vectors of \( \mathbf{A}{\mathbb{Z}}_{5}^{2} \) form an affine plane of order 5. | Figure 8.14 illustrates the five lines of the form \( \left\lbrack {m,4,0}\right\rbrack \) through the origin, more familiarly known as \( y = {mx} \), as well as the vertical line \( x = 0 \) or \( \left\lbrack {1,0,0}\right\rbrack \) . The thin line segments indicate horizontal line \( \left\lbrack {0,4,0}\right\rbra... | Yes |
Theorem 8.4.2. The set of affine transformations for \( {\mathbf{{AF}}}^{2} \) and the collineations for \( {\mathbf{{PF}}}^{2} \) form a group. | Proof. We replace the specific field \( \mathbb{R} \) with the general field \( \mathbf{F} \) in the proofs of Theorem 5.4.5 and Theorem 7.4.4. \( \blacksquare \) | No |
Example 6. Find the tangent at each point of the oval \( 2{x}^{2} + {y}^{2} + {2xz} + {yz} + 4{z}^{2} = 0 \) in \( \mathbf{P}{\mathbb{Z}}_{5}^{2} \) . | Solution. From Figure 8.17 we can guess that the tangents to this oval through the points \( \left( {0,2,1}\right) \) and \( \left( {4,2,1}\right) \) are vertical. Indeed, the lines \( \left\lbrack {1,0,0}\right\rbrack \) and \( \left\lbrack {1,0,1}\right\rbrack \) are tangents at these points. For the point \( \left( ... | Yes |
Theorem 8.4.3. Every point on an oval in a projective plane of odd order has exactly one tangent to the oval. | Proof. Let \( {P}_{0} \) be a point on an oval and \( {P}_{1},\ldots ,{P}_{n} \) be the other points on it. The definition of an oval guarantees that the lines \( {P}_{0}{P}_{i} \), for \( i > 0 \), are distinct, accounting for \( n \) of the \( n + 1 \) lines through \( {P}_{0} \) . The remaining line cannot intersect... | Yes |
The solution in Example 2 of Section 8.1 to Kirkman's schoolgirl problem comes from the projective space of Example 7. We write the numbers from 1 to 15 in base 2 and convert them into coordinates, partially represented in Table 8.10. | Then the 35 rows with three girls each correspond to lines in the geometry. The five lines for each day are skew in this space - no two intersect. The collineation \[ \left\lbrack \begin{array}{llll} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right\rbrack \] takes the lines making up S... | No |
Theorem 8.4.4. For \( d \geq 2 \), an affine space \( \mathbf{A}{\mathbb{Z}}_{n}^{d} \) is a \( \left( {{n}^{d}, n,1}\right) \) BIBD. | ## Proof. See Project 14. - | No |
Example 9. Use an affine space to model the game of Set. | Solution. In the game of Set each card has objects on it characterized by four properties: the number of objects, their color, their shape, and their shading. For each property there are three choices, so a full deck with one card of each possibility has \( {3}^{4} = {81} \) cards. The four-dimensional affine space \( ... | Yes |
Consider the circle through the three points \( \left( {0,0}\right) ,\left( {a,{a}^{2}}\right) \), and \( \left( {-a,{a}^{2}}\right) \) on the parabola. From Exercise 1.2.15, the circle's center is on the perpendicular bisectors of the segments of Figure 9.8. Their equations are \( x = 0 \) and \( y = \pm \frac{x}{a} +... | In the limit as \( a \rightarrow 0 \), the center is \( \left( {0,\frac{1}{2}}\right) \) and the radius is \( \frac{1}{2} \) . Leibniz called this best approximating circle, circle \( C \) in Figure 9.7, the osculating circle. (\ | Yes |
For the parabola \( f\left( x\right) = {x}^{2} \), Newton’s formula gives \( r\left( x\right) = {\left( 1 + 4{x}^{2}\right) }^{3/2}/2 \) for the radius of curvature. | Even though \( {f}^{\prime }\left( 0\right) = 0 \), by Exercise 9.2.8 (c) the formulas are valid there and \( r\left( 0\right) = \frac{1}{2} \) and \( \kappa \left( 0\right) = 2 \) . In Figure 9.7, circle \( C \) has radius \( \frac{1}{2} \) . As \( x \) increases, the radius \( r\left( x\right) \) quickly increases an... | No |
We call the curve \( \overrightarrow{c\left( t\right) } = \left( {t\cos \left( t\right), t\sin \left( t\right) }\right) \) for \( 0 \leq t \) the spiral of Archimedes, depicted in Figure 9.11, because Archimedes proved several theorems about it. We use the formula \( {\cos }^{2}\left( t\right) + {\sin }^{2}\left( t\rig... | Figure 9.12 illustrates how well the circle with radius \( \frac{1}{2} \) fits the spiral at the origin. As \( t \) increases, the curvature goes to \( 0 \) . | Yes |
Theorem 9.2.2. For a curve \( \overrightarrow{c\left( t\right) },{T}^{\prime }\left( t\right) \) is orthogonal to \( T\left( t\right) \) . | Proof. By definition, \( \parallel T\left( t\right) \parallel = 1 \), and so \( T\left( t\right) \cdot T\left( t\right) = 1 \) . The product rule gives \( {\left( T\left( t\right) \cdot T\left( t\right) \right) }^{\prime } = \) \( {T}^{\prime }\left( t\right) \cdot T\left( t\right) + T\left( t\right) \cdot {T}^{\prime ... | Yes |
For a helix, as in Figure 9.14, \( \overrightarrow{c\left( t\right) } = \left( {R\cos \left( t\right), R\sin \left( t\right) ,{mt}}\right) \) . The arrows mark \( T\left( t\right) \) at various places on this curve. We have \( \overrightarrow{{c}^{\prime }\left( t\right) } = \left( {-R\sin \left( t\right), R\cos \left(... | All the helices lie on the surface of a vertical cylinder of radius \( R \) with equation \( {x}^{2} + {y}^{2} = {R}^{2} \) . We will see in Section 9.4 that they are geodesics for the cylinder. | Yes |
We generate a torus, or more informally an inner tube, by rotating the circle \( {\left( y - R\right) }^{2} + {z}^{2} = {r}^{2} \) in the \( {yz} \) -plane around the \( z \) -axis. (See Figures 9.17 and 9.18.) The rotation replaces the \( y \) -value in the equation with corresponding points \( \left( {x, y}\right) \)... | Figure 9.19 depicts the torus as a surface of revolution using cross sections from the parametric form \( \left( {\left( {r\cos \left( v\right) + R}\right) \cos \left( u\right) ,\left( {r\cos \left( v\right) + R}\right) \sin \left( u\right), r\sin \left( v\right) }\right) \), for \( 0 \leq u < {2\pi } \) and \( - \frac... | No |
Example 3. For the sphere of Example \( 1,\mathbf{s}\left( {u, v}\right) = \left( {R\cos \left( u\right) \cos \left( v\right), R\sin \left( u\right) \cos \left( v\right), R\sin \left( v\right) }\right) \) . Then | \[ {\mathbf{s}}_{u} = \left( {-R\sin \left( u\right) \cos \left( v\right), R\cos \left( u\right) \cos \left( v\right) ,0}\right) \] \[ \text{and}{\mathbf{s}}_{v} = \left( {-R\cos \left( u\right) \sin \left( v\right) , - R\sin \left( u\right) \sin \left( v\right), R\cos \left( v\right) }\right) \text{.} \] We find \[ {\... | Yes |
Exercise 9.3.16 shows that the normal to the torus in Example 2 is \( N\left( {u, v}\right) = \) \( \left( {\cos \left( u\right) \cos \left( v\right) ,\sin \left( u\right) \cos \left( v\right) ,\sin \left( v\right) }\right) \) . To determine the curvature of the surface at a point \( \left( {a, b}\right) \) we need to ... | Exercise 9.3.16 gives formulas for \( {N}_{u} \) and \( {N}_{v} \) and shows that \( {N}_{u} = \frac{\cos \left( v\right) }{R + r\cos \left( v\right) }{\mathbf{s}}_{u} \) and \( {N}_{v} = \frac{1}{r}{\mathbf{s}}_{v} \) . The ratios indicate how quickly the normal is changing in the \( u \) and \( v \) directions. We ca... | Yes |
The curvature of the torus of Examples 2 and 4 is \( \kappa \left( {u, v}\right) = \det M = \frac{\cos \left( v\right) }{r\left( {R + r\cos \left( v\right) }\right) } \) . | For \( - \frac{\pi }{2} < v < \frac{\pi }{2} \) the torus bulges out and so should have a positive curvature, matching \( \cos \left( v\right) > 0 \) and \( \kappa \left( {u, v}\right) > 0 \) for these values. For \( v = 0,\kappa \left( {u,0}\right) = \frac{1}{r\left( {R + r}\right) } \), the product of the curvatures ... | Yes |
The lines of longitude on a sphere run through the north and south pole and are, as proved here, geodesics. | A line of longitude has the form\n\n\[ \overrightarrow{c\left( t\right) } = \left( {R\cos \left( {u}_{0}\right) \cos \left( \frac{t}{R}\right), R\sin \left( {u}_{0}\right) \cos \left( \frac{t}{R}\right), R\sin \left( \frac{t}{R}\right) }\right) \;\text{ for a fixed }{u}_{0} \]\n\nand the sphere is\n\n\[ \mathbf{s}\left... | No |
Theorem 9.4.1. Let \( \mathbf{s}\left( {u, v}\right) = \left( {f\left( v\right) \cos \left( u\right), f\left( v\right) \sin \left( u\right), g\left( v\right) }\right) \) be a smooth surface of revolution with \( {f}^{\prime }{\left( v\right) }^{2} + {g}^{\prime }{\left( v\right) }^{2} = 1 \) for all \( v \) . Then for ... | Proof. From the assumptions,\n\n\[ \overrightarrow{{c}^{\prime }\left( t\right) } = T\left( t\right) = \left( {{f}^{\prime }\left( t\right) \cos \left( {u}_{0}\right) ,{f}^{\prime }\left( t\right) \sin \left( {u}_{0}\right) ,{g}^{\prime }\left( t\right) }\right) \]\n\n\[ \text{and}{T}^{\prime }\left( t\right) = \left( ... | Yes |
For the sphere of radius \( R,\mathbf{s}\left( {u, v}\right) = \left( {R\cos \left( u\right) \cos \left( \frac{v}{R}\right), R\sin \left( u\right) \cos \left( \frac{v}{R}\right), R\sin \left( \frac{v}{R}\right) }\right. \) . From Exercise 9.4.3, \( E\left( {u, v}\right) = {R}^{2}{\cos }^{2}\left( \frac{v}{R}\right), F ... | \[ s = {\int }_{0}^{2\pi }\sqrt{{R}^{2}{\cos }^{2}\left( \frac{{v}_{0}}{R}\right) \cdot 1 + 0 + 0}{dt} = {\int }_{0}^{2\pi }R\left| {\cos \left( \frac{{v}_{0}}{R}\right) }\right| {dt} = {2\pi R}\left| {\cos \left( \frac{{v}_{0}}{R}\right) }\right| ,\] the circumference of a circle with radius \( R\cos \left( \frac{{v}_... | Yes |
Theorem 9.4.3 (Gauss part of Gauss-Bonnet Theorem, 1827). Suppose a triangular region \( T \) of a smooth surface is bounded by geodesics and the measures of the angles made by the geodesics are \( \alpha ,\beta \), and \( \gamma \) . Assume that \( T \) can be shrunk continuously to a point on the surface. Then \( \al... | Proof. See Do Carmo [1, 264-270]. | No |
A sphere of radius \( R \) has constant curvature \( \frac{1}{{R}^{2}} \) . Then \( \int {\int }_{T}\kappa \left( {u, v}\right) {d\sigma } = \frac{1}{{R}^{2}}\operatorname{area}\left( T\right) \) . | Thus we can rearrange the equation of Theorem 9.4.3 to give Theorem 1.5.3 for spherical triangles: \( {R}^{2}\left( {\alpha + \beta + \gamma - \pi }\right) = \operatorname{area}\left( T\right) \) . In Chapter 1 we called \( \alpha + \beta + \gamma - \pi \) the spherical excess. | No |
Example 1. An imaginary town has five fire stations. Figure 10.5 depicts the regions closest to each fire station. To minimize response time to a fire, each fire station should respond to fires in its region. How do we determine the boundaries of the regions? | Solution. We first solve an easier problem: find the points equidistant from two points \( A \) and \( B \) . In Figure 10.6, the points equidistant from \( A \) and \( B \) form the perpendicular bisector of the points \( A \) and \( B \) . In Figure 10.6, the half plane determined by the bisector and containing \( A ... | Yes |
Theorem 10.2.1. For positive integers \( n \) and \( d,{D}_{d + 1}\left( n\right) \leq {D}_{d}\left( n\right) \leq {D}_{d}\left( {n + 1}\right) \) . | ## Proof. See Exercise 10.2.3. | No |
Theorem 10.2.2. On a Euclidean line, the number of nonzero distances n distinct points can determine can be any integer between \( n - 1 \) and \( \left( \begin{array}{l} n \\ 2 \end{array}\right) \) . Thus \( {D}_{1}\left( n\right) = n - 1 \) . | Proof. Consider each point \( {a}_{i} \) as a real number, assuming \( {a}_{1} < {a}_{2} < \cdots < {a}_{n} \) . Then \( d\left( {{a}_{i},{a}_{j}}\right) = \) \( \left| {{a}_{j} - {a}_{i}}\right| \) . Further \( \left| {{a}_{2} - {a}_{1}}\right| < \left| {{a}_{3} - {a}_{1}}\right| < \cdots < \left| {{a}_{n} - {a}_{1}}\... | No |
Theorem 10.2.3. For positive integers \( n \) and \( d, D\left( {n, d}\right) = 1 \) if and only if \( d \geq n - 1 \) . | Proof. We determine the maximum number of points in \( d \) dimensions equidistant from one another. Because of the homogeneity of space, we can place the points conveniently using any distance. For ease, start with \( d \) points \( {P}_{1} \) to \( {P}_{d} \) given by \( {P}_{1} = \left( {1,0,\ldots ,0}\right) \) , \... | Yes |
Theorem 10.2.4. The minimum number of distances determined by a convex set of \( n \) points in the Euclidean plane is \( \left\lfloor \frac{n}{2}\right\rfloor \), the greatest integer less than or equal to \( \frac{n}{2} \) . | ## Proof. See Altman [1]. \( \blacksquare \) | No |
Lemma 10.2.5. A polygon with at least four vertices has a diagonal. | Proof. Let \( A \) be the leftmost vertex of a polygon with at least four vertices and let \( B \) and \( C \) be those adjacent to \( A \) . (See Figure 10.16.) If two vertices are leftmost, pick the lower one of them. Draw the segment \( \overline{BC} \) . If \( \overline{BC} \) is entirely inside the polygon, we hav... | Yes |
Theorem 10.2.7. A triangulation of an \( n \) -gon has \( n - 3 \) diagonals and \( n - 2 \) triangles. | Proof. Again we use induction on \( n \), the number of vertices. For the initial step when \( n = 3 \) , a triangle has \( 0 = 3 - 3 \) diagonals and is \ | No |
Theorem 10.2.8. A convex \( n \) -gon has \( \frac{n\left( {n - 3}\right) }{2} \) diagonals. | Proof. See Exercise 10.2.9. | No |
Let \( T\left( n\right) \) be the number of triangulations of a convex \( n \) -gon. Quick inspection gives \( T\left( 3\right) = 1, T\left( 4\right) = 2 \), and \( T\left( 5\right) = 5 \) . A little more work yields \( T\left( 6\right) = {14} \), but drawing all the possibilities quickly becomes tedious. For \( T\left... | Edge \( \overline{12} \) in Figure 10.17 must be part of exactly one triangle in any triangulation. From the five other vertices we have five cases. For \( \bigtriangleup {123} \), the remaining polygon is the hexagon 134567 and there are \( T\left( 6\right) = {14} \) ways to triangulate it. Next, \( \bigtriangleup {12... | Yes |
Theorem 10.2.9. A convex \( n \) -gon has \( \frac{1}{n - 1}\left( \begin{matrix} {2n} - 4 \\ n - 2 \end{matrix}\right) \) triangulations. | Proof. See Devadoss and O'Rourke [3, 9-10]. | No |
Lemma 10.2.10. Every polygon with at least four vertices has at least two ears. | Proof. Given an \( n \) -gon, triangulate it. If any of the triangles contains two consecutive edges and so one diagonal of the polygon, we have an ear. (For example, triangles \( \bigtriangleup {P}_{1}{P}_{2}{P}_{3} \) and \( \bigtriangleup {P}_{4}{P}_{5}{P}_{6} \) in Figure 10.19.) Any triangle of the triangulation h... | Yes |
Lemma 10.2.11. Fix a triangulation of a polygon. Then the vertices of the polygon can be colored with three colors so that every triangle in the triangulation has a vertex of each color: | Proof. We use induction on \( n \), the number of vertices of the polygon. For \( n = 3 \), we have a triangle and we satisfy the theorem by coloring its vertices with three colors. For the induction step, assume that any triangulation of any \( k \) -gon can be colored as specified. Consider a triangulation of a \( k ... | Yes |
Theorem 10.2.12. (Art Gallery Theorem) An n-gon needs at most \( \left\lfloor \frac{n}{3}\right\rfloor \) guards to be able to observe all interior points. For each \( n \) there is an \( n \) -gon needing \( \left\lfloor \frac{n}{3}\right\rfloor \) guards. | Proof. Triangulate the \( n \) -gon. Now use Lemma 10.2.11 to color the vertices with three colors so that each triangle has one vertex of each color. If we pick as our guards the vertices of the same color, every triangle has a guard observing it and so the entire polygon is guarded. Choose the color used the fewest n... | No |
Theorem 10.2.13. (Fortress Theorem). An \( n \) -gon needs at most \( \left\lceil \frac{n}{2}\right\rceil \) guards to be able to observe all exterior points. For each \( n \) there is an \( n \) -gon needing \( \left\lceil \frac{n}{2}\right\rceil \) guards. | Sketch of Proof. Exercise 10.2.16 shows that a convex \( n \) -gon needs \( \left\lceil \frac{n}{2}\right\rceil \) guards, proving the second statement of the theorem. Figure 10.21 suggests an approach to the general situation. We add dashed segments to the polygon to make its convex hull, ABCDFGH. By Exercise 10.2.16 ... | No |
Theorem 10.3.1. A centrally symmetric hexagon gives a monohedral tiling of the plane. | Proof. Let \( C \) be the center of symmetry of hexagon \( {PQR}{P}^{\prime }{Q}^{\prime }{R}^{\prime } \) . That is, a \( {180}^{ \circ } \) rotation around \( C \) switches the pair \( P \) and \( {P}^{\prime } \) and the other pairs \( Q \) and \( {Q}^{\prime } \) and \( R \) and \( {R}^{\prime } \) . (See Figure 10... | No |
Corollary 10.3.2. Any quadrilateral gives a monohedral tiling of the plane. Any pentagon with two adjacent supplementary angles gives a monohedral tiling of the plane. | ## Proof. See Exercise 10.3.7. | No |
Corollary 10.3.3. Any centrally symmetric orthogonal octagon gives a monohedral tiling of the plane. | Proof. See Exercise 10.3.8 for an outline of the proof. - | No |
Theorem 10.3.4. If a convex polygon gives an edge-to-edge monohedral tiling of the plane, than the polygon has at most six vertices. | Proof. Let \( P \) be a convex \( n \) -gon giving a monohedral tiling and consider a finite part of the tiling with \( k \) connected copies of \( P \) . We count the number of edges \( \left( E\right) \) and vertices \( \left( V\right) \) in it. We count as faces \( \left( F\right) \) all the copies of \( P \) togeth... | Yes |
Example 1. Figure 10.33 depicts a tiling with two tiles, both squares of different sizes. Besides arranging squares in a more interesting way than a checkerboard pattern, the tiling suggests a proof of the Pythagorean theorem. Let the side of the small square be \( a \) and the side of the larger square be \( b \) . Ca... | The square with side \( c \) is divided into five pieces. Two of them fit together to give a small square with area \( {a}^{2} \) . The other three assemble into a larger square with area \( {b}^{2} \) . Putting these facts into one equation we get the formula for the Pythagorean theorem: \( {a}^{2} + {b}^{2} = {c}^{2}... | Yes |
Theorem 10.4.1. A Voronoi region is a convex set. | Proof. Let \( A \) be a site in a Voronoi diagram. Its Voronoi region is the intersection of the half planes containing \( A \) and bounded by the perpendicular bisectors of \( A \) and the other sites. Because of the separation axiom (Appendix B or C), any half plane is a convex set. By Exercise 10.4.11, the intersect... | No |
Theorem 10.4.2. A Voronoi diagram with at least three sites has an infinite line if and only if all the sites are collinear. | ## Proof. See Exercise 10.4.13. | No |
Theorem 10.4.3. A Voronoi diagram with \( n \) sites, where \( n \geq 3 \), has at most \( {3n} - 6 \) Voronoi edges and at most \( {2n} - 5 \) Voronoi vertices. | Proof. Given a Voronoi diagram, as in Figure 10.40, make a diagram by bending the infinite Voronoi edges of the diagram to meet at an extra added point, denoted \( P \) in Figure 10.41. The new diagram has \( F = n \) faces and the same number \( E \) of edges as the Voronoi diagram. Its number \( V \) of vertices is t... | Yes |
Lemma 2 The three lines through the symmedian point of a triangle parallel to each side cut the sides at 6 concyclic points. Also, the intersections closer to any vertex are antiparallel to the corresponding side. | Proof. Theorem 10 of this PDF by Cosmin Pohoata. | No |
The sum of the squares of the lengths of the sides of the triangle \( {A}_{0}{B}_{0}{C}_{0} \), determined by the intersection of the nedians of order \( i \) of the triangle \( {ABC} \) is given by the following relation:\n\n\[ {A}_{0}{B}_{0}^{2} + {B}_{0}{C}_{0}^{2} + {C}_{0}{A}_{0}^{2} = \frac{{\left( n - 2i\right) ... | Proof\n\nWe’ll apply the Menelaus ’theorem in the triangle \( A{A}_{i}C \) for the transversals \( B - {B}_{0} - {B}_{i} \), see Fig. 1.\n\n\[ \frac{B{A}_{i}}{BC} \cdot \frac{{B}_{i}C}{{B}_{i}A} \cdot \frac{{B}_{0}A}{{B}_{0}{A}_{i}} = 1 \]\n\n(3)\n\nBecause \( B{A}_{i} = \frac{ia}{n},{B}_{i}C = \frac{ib}{n},{B}_{i}A = ... | Yes |
Proposition 4\n\nThe coefficient of deformation \( k \) of triangle \( {ABC} \) ha the following formula:\n\n\[ k = {\left( \frac{{a}^{2} + {b}^{2} + {c}^{2} - {4s}\sqrt{3}}{{a}^{2} + {b}^{2} + {c}^{2} + {4s}\sqrt{3}}\right) }^{\frac{1}{2}} \]\n\n(10)\n\nwhere \( s \) is the aria of the triangle \( {ABC} \) . | Proof\n\nWe’ll apply the cosine theorem in the triangle \( {\mathrm{{CO}}}_{1}{}^{\prime }{\mathrm{O}}_{2}{}^{\prime } \) (see Fig. 3), in which\n\n\[ C{O}_{1}{}^{\prime } = \frac{a\sqrt{3}}{3}, C{O}_{2}{}^{\prime } = \frac{b\sqrt{3}}{3}\text{, and}m\left( { < {O}_{1}C{O}_{2}{}^{\prime }}\right) = C - {60}^{ \circ }\te... | Yes |
The necessary and sufficient condition for two triangles to have the same coefficient of deformation is to have the same Brocard angle. | If the triangles \( {ABC} \) and \( {A}_{1}{B}_{1}{C}_{1} \) have equal coefficients of deformation \( k = {k}_{1} \) then from relation 21 it results\n\n\[ \frac{\operatorname{ctg}\omega - \sqrt{3}}{\operatorname{ctg}\omega + \sqrt{3}} = \frac{\operatorname{ctg}{\omega }_{1} - \sqrt{3}}{\operatorname{ctg}{\omega }_{1}... | Yes |
Two triangles \( {ABC} \) and \( {A}_{1}{B}_{1}{C}_{1} \) have the same coefficient of deformation if and only if\n\n\[ \frac{{s}_{1}}{s} = \frac{{a}_{1}^{2} + {b}_{1}^{2} + {c}_{1}^{2}}{{a}^{2} + {b}^{2} + {c}^{2}} \]\n\n( \( {s}_{1} \) being the aria of triangle \( {A}_{1}{B}_{1}{C}_{1} \), with the sides \( {a}_{1},... | If \( \omega ,{\omega }_{1} \) are the Brocard angles of triangles \( {ABC} \) and \( {A}_{1}{B}_{1}{C}_{1} \) then, taking into consideration (17) and Proposition 6, we'll obtain (22). Also from (22) taking into consideration of (17) and Proposition 6, we’ll get \( k = {k}_{1} \). | No |
If in triangle ABC inscribed in a circle, the tangents to this circle in the points B and C intersect in a point \( \mathrm{S} \), then \( \mathrm{{AS}} \) is symmedian in the triangle \( \mathrm{{ABC}} \). | We'll note L the intersection point of the line AS with BC (see fig. 1).\n\n\n\nFig. 1\n\nWe have\n\n\[ \n\frac{\text{ Aria }\Delta \mathrm{{ABL}}}{\text{ Aria }\Delta \mathrm{{ACL}}} = \frac{\mathrm{{BL}}}{\mathrm{{LC... | Yes |
If in the triangle \( \mathrm{ABC},\mathrm{AL} \) is the interior symmedian \( \mathrm{L} \in \mathrm{BC} \), and \( \mathrm{AP} \) is the external median \( \mathrm{P} \in \mathrm{BC} \), then the points \( \mathrm{P},\mathrm{B},\mathrm{L},\mathrm{C} \) form a harmonic division. | It is known that the external symmedian AP in the triangle ABC is tangent in A to the circumscribed circle (see fig. 2), also, it can be proved that:\n\n\[ \frac{\mathrm{PB}}{\mathrm{PC}} = {\left( \frac{\mathrm{AB}}{\mathrm{AC}}\right) }^{2} \]\n\n(1)\n\nbut\n\n\[ \frac{\mathrm{LB}}{\mathrm{LC}} = {\left( \frac{\mathr... | Yes |
Theorem 2 (The reciprocal of the Desargues' theorem)\n\nIf two triangles \( {ABC} \) and \( {A}_{1}{B}_{1}{C}_{1} \) are such that\n\n\[ \n{AB} \cap {A}_{1}{B}_{1} = \{ N\}\n\]\n\n\[ \n{BC} \cap {B}_{1}{C}_{1} = \{ M\}\n\]\n\n\[ \n{CA} \cap {C}_{1}{A}_{1} = \{ P\}\n\]\n\nAnd the points \( N, M, P \) are collinear, then... | Proof\n\nWe'll use the reduction ad absurdum method .\n\nLet\n\n\[ \nA{A}_{1} \cap B{B}_{1} = \{ O\}\n\]\n\n\[ \nA{A}_{1} \cap C{C}_{1} = \left\{ {O}_{1}\right\}\n\]\n\n\[ \nB{B}_{1} \cap C{C}_{1} = \left\{ {O}_{2}\right\}\n\]\n\nWe suppose that \( O \neq {O}_{1} \neq {O}_{2} \neq {O}_{3} \) .\n\nThe Menelaus’ theorem ... | Yes |
If \( {ABCD} \) is a parallelogram, \( {A}_{1} \in \left( {AB}\right) ,{B}_{1} \in \left( {BC}\right) ,{C}_{1} \in \left( {CD}\right) ,{D}_{1} \in \left( {DA}\right) \) such that the lines \( {A}_{1}{D}_{1},{BD},{B}_{1}{C}_{1} \) are concurrent, then:\n\na) The lines \( {AC},{A}_{1}{C}_{1} \) and \( {B}_{1}{D}_{1} \) a... | Let \( \{ P\} = {A}_{1}{D}_{1} \cap {B}_{1}{C}_{1} \cap {BD} \) see Fig. 2. We observe that the sides \( {A}_{1}{D}_{1} \) and \( {B}_{1}{C}_{1};C{C}_{1} \) and \( A{D}_{1};{A}_{1}A \) and \( C{B}_{1} \) of triangles \( A{A}_{1}{D}_{1} \) and \( C{B}_{1}{C}_{1} \) intersect in the collinear points \( P, B, D \) . Apply... | Yes |
i) The triangle POR is homological with each of the triangles: GHI, JKL, MNQ, UVT. | i) when proving this problem we must observe that the \( {ABCDEF} \) is a complete quadrilateral and if \( {O}_{1},{O}_{2},{O}_{3} \) are the middle of the diagonals \( \left( {AC}\right) ,\left( {BD}\right) \) respective \( {EF} \), these point are collinear. The line on which the points \( {O}_{1},{O}_{2},{O}_{3} \) ... | Yes |
If \( {ABC} \) is a given triangle and \( {A}_{1}{B}_{1}{C}_{1} \) is its first triangle Brocard, then the triangles \( {ABC} \) and \( {A}_{1}{B}_{1}{C}_{1} \) are ortho-homological. | We'll perform the proof of this theorem in two stages.\n\nI. We prove that the triangles \( {A}_{1}{B}_{1}{C}_{1} \) and \( {ABC} \) are orthological.\n\nThe perpendiculars from \( {A}_{1},{B}_{1},{C}_{1} \) on \( {BC},{CA} \) respective \( {AB} \) are perpendicular bisectors in the triangle \( {ABC} \), therefore are ... | Yes |
In a triangle \( {ABC} \), the Brocard’s Cevian \( {B\Omega } \), symmedian from \( C \) and the median from \( A \) are concurrent. | It is known that the symmedian \( {CK} \) of triangle \( {ABC} \) intersects \( {AB} \) in the point \( {C}_{2} \) such that \( \frac{A{C}_{2}}{B{C}_{2}} = \frac{{b}^{2}}{{c}^{2}} \) . We had that the Cevian \( {B\Omega } \) intersects \( {AC} \) in \( {B}^{\prime \prime } \) such that \( \frac{B{C}^{\prime \prime }}{{... | Yes |
Give a triangle \( {ABC} \) and \( \omega \) the Brocard’s angle, then\n\n\[ \operatorname{ctg}\omega = \operatorname{ctg}A + \operatorname{ctg}B + \operatorname{ctg}C \] | Proof\n\nFrom the relation (8) we find:\n\n\[ \sin \left( {A - \omega }\right) = \frac{a}{b} \cdot \frac{\sin A}{\sin C} \cdot \sin \omega \]\n\n(10)\n\nFrom the sinus’ theorem in the triangle \( {ABC} \) we have that\n\n\[ \frac{a}{b} = \frac{\sin A}{\sin B} \]\n\nSubstituting it in (10) it results: \( \sin \left( {A ... | Yes |
The Cevians \( A{A}_{1}, B{B}_{1}, C{C}_{1} \) are the isotomics of the symmedians \( A{A}_{2}, B{B}_{2}, C{C}_{2} \) in the triangle \( {ABC} \) . | \n\nFig. 5\n\nIn figure 5 we note J the intersection point of the Cevians from the Lemma 2. Because \( K{A}_{1}\parallel {BC} \), we have that \( {A}_{1}{A}^{\prime } = K{K}_{1} = \frac{1}{2}\operatorname{atg}\omega \)... | Yes |
If through the vertexes \( A, B, C \) of a triangle are constructed the parallels to the sides \( {B}_{1}{C}_{1},{C}_{1}{A}_{1} \) respectively \( {A}_{1}{B}_{1} \) of the first triangle of Brocard of this triangle, then these lines are concurrent in a point \( \mathrm{S} \) (the Steiner point of the triangle) | We note with \( S \) the polar intersection constructed through \( A \) to \( {B}_{1}{C}_{1} \) with the polar constructed through \( B \) to \( {A}_{1}{C}_{1} \) (see Fig. 6).\n\nWe have\n\n\[ m\left( { < {ASB}}\right) = {180}^{ \circ } - m\left( { < {A}_{1}{C}_{1}{B}_{1}}\right) \text{(angles with parallel sides)} \]... | Yes |
In a triangle \( {ABC} \) the Tarry point \( T \), the center of the circumscribed circle \( O \), the third point of Brocard \( {\Omega }^{\prime \prime } \) and Steiner’s point \( S \) are collinear points | The P. Sondat's theorem relative to the orthological triangles (see [4]) says that the points \( T, O,{\Omega }^{\prime \prime } \) are collinear, therefore the points: \( T, O,{\Omega }^{\prime \prime }, S \) are collinear. | Yes |
Let \( I \) be the center of the circumscribe circle to triangle \( {ABC} \) and \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) its projections on the sides \( {BC},{CA},{AB} \) . We consider the points \( {A}^{\prime \prime },{B}^{\prime \prime },{C}^{\prime \prime } \) such that:\n\n\[ \overrightarrow{I{A}^{\prime \... | The barycentric coordinates of the point \( I \) are \( I\left( {\frac{a}{2p},\frac{b}{2p},\frac{c}{2p}}\right) \).\n\nEvidently:\n\n\[ {abc}\left( {\cos A - \cos B}\right) + {abc}\left( {\cos B - \cos C}\right) + {abc}\left( {\cos C - \cos A}\right) = 0 \]\nand\n\n\[ \cos A\left( {\cos B - \cos C}\right) + \cos B\left... | No |
Let \( O \) be the center of the circumscribed circle to the triangle \( {ABC} \) and \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) its projections on the sides \( {BC},{CA},{AB} \) . Consider the points \( {A}^{\prime \prime },{B}^{\prime \prime },{C}^{\prime \prime } \) such that \( \frac{O{A}^{\prime }}{O{A}^{\pri... | \[ O\left( {\frac{{R}^{2}}{2S}\sin {2A},\frac{{R}^{2}}{2S}\sin {2B},\frac{{R}^{2}}{2S}\sin {2C}}\right), P = N\text{, because}\frac{\sin {2B}\cos A}{\sin B} - \frac{\sin {2A}\cos B}{\sin A} = 0\text{.} \]\n\nSimilarly we find that \( R = Q \) and \( M = S \) .\n\nAlso \( {\alpha MP} = {\alpha NS},{\beta PR} = {\beta NQ... | Yes |
Theorem 2 (The theorem of orthological triangle of J. Steiner)\n\nIf the triangle \( {ABC} \) is orthological with the triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \), then the triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) is also orthological in rapport to triangle \( {ABC} \) . | For the proof of this theorem we recommend [1]. | No |
Let \( {ABC} \) and \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) two orthological triangles. The orthogonal projections of the vertexes \( B \) and \( C \) on the sides \( {A}^{\prime }{C}^{\prime } \) respectively \( {A}^{\prime }{B}^{\prime } \) are concyclic. | We note with \( E, F \) the orthogonal projections \( \mathrm{f} \) the vertexes \( B \) and \( C \) on \( {A}^{\prime }{C}^{\prime } \) respectively \( {A}^{\prime }{B}^{\prime } \) (see Fig. 2). Also, we’ll note \( O \) the common orthological center of the orthological triangles \( {ABC} \) and \( {A}^{\prime }{B}^{... | Yes |
The tangency point with the side \( {BC} \) of the \( A \) -mixt-linear circle adjointly ex-inscribed associated to the triangle is the leg of the of the external bisectrix of the angle \( {BAC} \) | Let \( {D}^{\prime } \) the contact point with the side \( {BC} \) of the \( A \) -mixt-linear circle adjointly ex-inscribed and let \( {A}^{\prime } \) the intersection of the tangent in the point \( A \) to the circumscribed circle to the triangle \( {ABC} \) with \( {BC} \) (see Fig. 1) We have\n\n\[ m\left( { \ll A... | Yes |
The \( A \) -mixt-linear circle adjointly ex-inscribed to triangle \( {ABC} \) intersects the sides \( {AB},{AC} \), respectively, in two points of a cord which is parallel to \( {BC} \) . | We’ll note with \( M, N \) the intersection points with \( {AB} \) respectively \( {AC} \) of the \( A \) -mixt-linear circle adjointly ex-inscribed. We have \( \because {BCA} \equiv \because {BA}{A}^{\prime } \) and \( \because {A}^{\prime }{AB} \equiv \because {A}^{\prime \prime }{AM} \) (see Fig.1).\n\nBecause \( < ... | Yes |
Theorem 1 (Blaise Pascal)\n\nThe opposite sides of a hexagon inscribed in a circle intersect in collinear points. | To prove the Pascal theorem one may use [1]. | No |
In a circumscribable quadrilateral its diagonals and the cords determined by the contact points of the opposite sides of the quadrilateral with the circumscribed circle are four concurrent lines. | \n\nFig. 1\n\n\n\nWe constructed the circles \( {O}_{1},{O}_{2},{O}_{3},{O}_{4} \) tangent to the extensions of the quadrilateral \( {ABCD} \) such that\n\n\[ \n{A}_{1}M = {A}_{1}N = {B}_{1}P = {B}_{1}Q = {C}_{1}R = {C... | Yes |
Two triangles are homological two by two and have a common homological center (their homological centers coincide) then their homological axes are concurrent. | Let’s consider the homological triangles \( {A}_{1}{B}_{1}{C}_{1},{A}_{2}{B}_{2}{C}_{2},{A}_{3}{B}_{3}{C}_{3} \) whose common homological center is \( O \) (see figure 1.)\n\n\n\nFig. 1\n\nWe consider the triangle fo... | Yes |
If three triangles are homological two by two and have their homological centers collinear, then these have the same homological axis. | We will use the triangles from figure 2. Let therefore \( {O}_{1},{O}_{2},{O}_{3} \) the three homological collinear points. We consider the triangles \( {B}_{1}{B}_{2}{B}_{3} \) and \( {C}_{1}{C}_{2}{C}_{3} \), we observe that these admit as homological axis the line \( {\mathrm{O}}_{1}{\mathrm{O}}_{2}{\mathrm{O}}_{3}... | Yes |
Theorem 4 (The Veronese theorem)\n\nIf the triangles \( {A}_{1}{B}_{1}{C}_{1},{A}_{2}{B}_{2}{C}_{2} \) are homological and\n\n\[ \left\{ {A}_{3}\right\} = {B}_{1}{C}_{2} \cap {B}_{2}{C}_{1},\left\{ {B}_{3}\right\} = {A}_{1}{C}_{2} \cap {A}_{2}{C}_{1},\left\{ {C}_{3}\right\} = {A}_{1}{B}_{2} \cap {A}_{2}{B}_{1} \]\n\nth... | Proof\n\nLet \( {O}_{1} \) be the homological center of triangles \( {A}_{1}{B}_{1}{C}_{1} \) and \( {A}_{2}{B}_{2}{C}_{2} \) (see figure 3) and \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) their homological axis.\n\nWe observe that \( {O}_{1} \) is a homological center also for the triangles \( {A}_{1}{B}_{1}{C}_{2... | Yes |
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