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Example 1. \( U = 1 : 1 : 1 \) . The characteristic polynomial is\n\n\[ \left| \begin{matrix} - t & 1 & 1 \\ 1 & - t & 1 \\ 1 & 1 & - t \end{matrix}\right| = \left( {2 - t}\right) {\left( t + 1\right) }^{2} \] | We have two cases: \( t = 2 \) and \( t = - 1 \) . For \( t = 2 \), we easily find the fixed point \( 1 : 1 : 1 \) . For \( t = - 1 \), the method of proof of Theorem 1 does not apply because \( {t}^{2} = {w}^{2} \) . Instead, the system to be solved degenerates to the single equation \( z = - x - y \) . The solutions,... | Yes |
Example 2. \( U = - 1 : 1 : 1 \), the \( A \) -excenter. The characteristic polynomial is\n\n\[ \left| \begin{matrix} - t & 1 & 1 \\ - 1 & - t & 1 \\ - 1 & 1 & - t \end{matrix}\right| = - \left( {t + 1}\right) \left( {{t}^{2} - t + 2}\right) . \] | For \( t = - 1 \), we find that every point on the line \( x + y + z = 0 \) is a fixed point. If \( {t}^{2} - t + 2 = 0 \), then \( t = \left( {1 \pm \sqrt{-7}}\right) /2 \), and the (nonreal) fixed point is \( 1 : 1 : t - 1 \) . Similar results are obtained for \( U \in \{ 1 : - 1 : 1,1 : 1 : - 1\} \). | Yes |
Example 3. \( U = \cos A : \cos B : \cos C \) . It can be checked using a computer algebra system that \( {X}_{6},{X}_{2574} \), and \( {X}_{2575} \) are fixed points. The first of these corresponds to the eigenvalue \( t = 1 \) | \[ x : y : z = {tv} + {uw} : {tu} + {vw} : {t}^{2} - {w}^{2} \] \[ = \cos B + \cos A\cos C : \cos A + \cos B\cos C : 1 - {\cos }^{2}C \] \[ = \sin A\sin C : \sin B\sin C : \sin C\sin C \] \[ = \sin A : \sin B : \sin C \] \[ = {X}_{6}\text{.} \] | Yes |
Example 4. \( U = a\left( {{b}^{2} + {c}^{2}}\right) : b\left( {{c}^{2} + {a}^{2}}\right) : c\left( {{a}^{2} + {b}^{2}}\right) = {X}_{39} \) . The three roots of \( {t}^{3} - {gt} - h = 0 \) are | \[ - {2abc},\;{abc} - \sqrt{3{a}^{2}{b}^{2}{c}^{2} + S\left( {2,4}\right) },\;{abc} + \sqrt{3{a}^{2}{b}^{2}{c}^{2} + S\left( {2,4}\right) }, \] where \[ S\left( {2,4}\right) = {a}^{2}{b}^{4} + {a}^{4}{b}^{2} + {a}^{2}{c}^{4} + {a}^{4}{c}^{2} + {b}^{2}{c}^{4} + {b}^{4}{c}^{2}. \] The solution \( t = - {2abc} \) easily l... | Yes |
For arbitrary real \( n \), let \( u = \cos {nA}, v = \cos {nB}, w = \cos {nC} \). A fixed point is \( X = \sin {nA} : \sin {nB} : \sin {nC} \) | \[ {\mathcal{C}}_{U}^{-1}\left( X\right) = \sin {nB}\cos {nC} + \sin {nC}\cos {nB} \] \[ \text{：}\sin {nC}\cos {nA} + \sin {nA}\cos {nC} \] \[ \text{：}\sin {nA}\cos {nB} + \sin {nB}\cos {nA} \] \[ = \sin \left( {{nB} + {nC}}\right) : \sin \left( {{nC} + {nA}}\right) : \sin \left( {{nA} + {nB}}\right) \] \[ = \sin {nA} ... | Yes |
Lemma 3. If \( X \) lies on a regular fixed line of \( {\mathcal{C}}_{U}^{-1} \) (or equivalently, a regular fixed line of \( {\mathcal{C}}_{U} \) ), then the sequence of points \( {\mathcal{C}}_{U}^{-n}\left( X\right) \) (or equivalently, the points \( \left. {{\mathcal{C}}_{U}^{n}\left( X\right) }\right) \) converges... | Next, suppose that \( P \) is an arbitrary point in the plane of \( {ABC} \) . We shall show that \( {\mathcal{C}}_{U}^{-n}\left( P\right) \) converges to a fixed point. Let \( {F}_{1},{F}_{2},{F}_{3} \) be distinct fixed points. Define\n\n\[ \n{P}_{2} = P{F}_{2} \cap {F}_{1}{F}_{3},\;{P}_{3} = P{F}_{3} \cap {F}_{1}{F}... | Yes |
Because \( {X}_{1113} \) and \( {X}_{1114} \) are antipodal points on the circumcircle, the lines \( {X}_{6}{X}_{2574} \) are \( {X}_{6}{X}_{2575} \) are perpendicular. | (x) Line \( {X}_{6}{X}_{2574} \) is parallel to the Simson line of \( {X}_{1114} \), and line \( {X}_{6}{X}_{2575} \) is parallel to the Simson line of \( {X}_{1113} \) . The two Simson lines are perpendicular ([1, p. 207]), so that the \( {\mathcal{C}}_{U} \) -fixed lines \( {X}_{6}{X}_{2574} \) and \( {X}_{6}{X}_{257... | Yes |
Proposition 1. Triangle \( {X}^{\prime }{Y}^{\prime }{Z}^{\prime } \) is perspective to (i) triangle \( {X}_{0}{Y}_{0}{Z}_{0} \) at the point\n\n\[ \n{P}_{0}/\left( {T\left( {{P}_{0},{P}_{1}}\right) }\right) = \left( {{u}_{0}{\left( \frac{{v}_{0}}{{v}_{1}} - \frac{{w}_{0}}{{w}_{1}}\right) }^{2} : {v}_{0}{\left( \frac{{... | Proof. Since \( {X}^{\prime }{Y}^{\prime }{Z}^{\prime } \) is an anticevian triangle, the perspectivity is clear in each case by the cevian nest theorem (see [10, §8.3] and [4, p.165, Supp. Exercise 7]). The perspectors are the cevian quotients \( {P}_{0}/\left( {T\left( {{P}_{0},{P}_{1}}\right) }\right) \) and \( {P}_... | Yes |
Proposition 3. The triangle \( {X}^{\prime }{Y}^{\prime }{Z}^{\prime } \) is self polar with respect to each of the inscribed conics with perspectors \( {P}_{0} \) and \( {P}_{1} \) . | Proof. Since \( {X}_{1}{Y}_{1}{Z}_{1} \) is a cevian triangle and \( {X}^{\prime }{Y}^{\prime }{Z}^{\prime } \) is an anticevian triangle with respect to \( {ABC} \), we have\n\n\[ \left( {{Y}^{\prime }{Z}_{0},{Y}^{\prime }A,{Y}^{\prime }{Y}_{0},{Y}^{\prime }C}\right) = \left( {{Y}^{\prime }{Z}_{0},{Y}^{\prime }A,{Y}^{... | Yes |
Proposition 5. Let \( P \) be a given point with anticevian triangle \( {X}^{\prime }{Y}^{\prime }{Z}^{\prime } \) . If \( {XYZ} \) is an inscribed triangle of \( {ABC} \) (with \( X, Y, Z \) on the sidelines \( {BC},{CA},{AB} \) respectively) such that \( {X}^{\prime },{Y}^{\prime },{Z}^{\prime } \) lie on the lines \... | Proof. Let \( P = \left( {u : v : w}\right) \) so that \[ {X}^{\prime } = \left( {-u : v : w}\right) ,\;{Y}^{\prime } = \left( {u : - v : w}\right) ,\;{Z}^{\prime } = \left( {u : v : - w}\right) . \] Since \( {XYZ} \) is an inscribed triangle of \( {ABC} \) , \[ X = \left( {0 : {t}_{1} : 1}\right) ,\;Y = \left( {1 : 0 ... | Yes |
Proposition 6. The locus of the perspector of the anticevian triangle of \( P \) and the cevian triangle of a point \( Q \) on the circumconic with perspector \( P \) is the trilinear polar of \( P \) . | Proof. Let \( Q = \left( {u : v : w}\right) \) be a point on the circumconic. The perspector is the cevian quotient\n\n\[ \left( {p\left( {-\frac{p}{u} + \frac{q}{v} + \frac{r}{w}}\right) : q\left( {\frac{p}{u} - \frac{q}{v} + \frac{r}{w}}\right) : r\left( {\frac{p}{u} + \frac{q}{v} - \frac{r}{w}}\right) }\right) \]\n\... | Yes |
Proposition 9. The cevian circumcircle of a point on a nondegenerate circumconic contains the center of the conic if and only if the conic is a rectangular hyperbola. | Proof. (a) The sufficiency part follows from Theorem 7.\n\n(b) For the converse, consider a nondegenerate conic through \( A, B, C, P \) whose center \( W \) lies on the cevian circumcircle of \( P \) . The locus of centers of conics through \( A, B, C, P \) is, by Theorem 8, a conic \( \mathcal{C} \) through the trace... | Yes |
Proposition 12. The lines \( {Y}_{0}{Z}_{0},{Y}_{1}{Z}_{1},{Y}_{2}{Z}_{2} \) are concurrent; similarly for the triples \( {Z}_{0}{X}_{0},{Z}_{1}{X}_{1},{Z}_{2}{X}_{2} \) and \( {X}_{0}{Y}_{0},{X}_{1}{Y}_{1},{X}_{2}{Y}_{2} \) . | Proof. The line \( {Y}_{2}{Z}_{2} \) has equation\n\n\( {u}_{1}\left( {\frac{x}{{u}_{0}}\left( {-\frac{{u}_{1}}{{u}_{0}} + \frac{{v}_{1}}{{v}_{0}} + \frac{{w}_{1}}{{w}_{0}}}\right) + \frac{y}{{v}_{0}}\left( {\frac{{u}_{1}}{{u}_{0}} - \frac{{v}_{1}}{{v}_{0}} + \frac{{w}_{1}}{{w}_{0}}}\right) + \frac{z}{{w}_{0}}\left( {\... | Yes |
Lemma 2. For a point \( X \) lying on a side of an acute triangle, the area at the opposite vertex is greater than one of the remaining two areas. | Proof. Without loss of generality, we may assume that \( X \in \overline{AB} \) and \( \left| {AX}\right| \leq \) \( \left| {BX}\right| \), see Figure 2. Straight line \( X{X}^{\prime } \) parallel to straight line \( {BC} \) cuts the triangle \( {AX}{X}^{\prime } \) greater than \( P\left( {A, X}\right) \) (as the ang... | Yes |
Theorem 3. If a triangle \( {ABC} \) is acute and \( f \) attains a maximum at \( {X}_{0} \), then \( P\left( {A,{X}_{0}}\right) = P\left( {B,{X}_{0}}\right) = P\left( {C,{X}_{0}}\right) = \frac{\left| \bigtriangleup ABC\right| }{3} \) . | Proof. \( f\left( A\right) = f\left( B\right) = f\left( C\right) = 0 \), and 0 is not a maximum of \( f \) . Therefore \( {X}_{0} \) is not a vertex of the triangle \( {ABC} \) . Let us assume that \( f\left( {X}_{0}\right) = P\left( {A,{X}_{0}}\right) \) . By Lemma \( 2,{X}_{0} \notin \overline{BC} \) . Suppose, on th... | Yes |
Lemma 1. Consider a point \( P \) inside the angular sector bounded by the half-lines \( {AB} \) and \( {AC} \), with projections \( {P}_{b} \) and \( {P}_{c} \) on \( {AC} \) and \( {AB} \) respectively. For a positive real number \( k,\Delta \left( {A{P}_{b}P{P}_{c}}\right) = k \cdot \Delta \left( {ABC}\right) \) if ... | Proof. We take \( A \) for pole and the bisector \( {AI} \) for polar axis; let \( \left( {\rho ,\theta }\right) \) be the polar coordinates of \( P \) . As \( A{P}_{b} = \rho \cos \left( {\frac{A}{2} - \theta }\right) \) and \( P{P}_{b} = \rho \sin \left( {\frac{A}{2} - \theta }\right) \), we have \( \Delta \left( {{A... | Yes |
Theorem 2. Let \( U \) be the point with barycentric coordinates \( \left( {u : v : w}\right) \) and \( {M}_{1} \) , \( {M}_{2},{M}_{3} \) be the antipodes on the circumcircle \( \Gamma \) of \( {ABC} \) of the points whose Simson lines pass through \( U \) and \( P \) the incenter of the triangle \( {M}_{1}{M}_{2}{M}_... | ## 2. Proof of Theorem 2\n\nIf \( P \) has homogeneous barycentric coordinates \( \left( {x : y : z}\right) \) with reference to triangle \( {ABC} \), then\n\n\[ \n{\left( x + y + z\right) }^{2}\Delta \left( {{AP}{P}_{b}}\right) = y\left( {z + \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2{b}^{2}}y}\right) \Delta , \n\]\n\n\[ \n... | Yes |
Theorem 1. Given \( \mathcal{P} \), there exist infinitely many cyclic quadrilaterals \( {ABCD} \) such that \( \mathcal{P} = \mathcal{V}\left( {ABCD}\right) \) . Such quadrilaterals can be constructed from \( \mathcal{P} \) by ruler and compass. | Proof. Consider a Cartesian system with origin at \( O \) and \( x \) -axis parallel to \( {PQ} \) (see Figure 3). The affix of a point \( Z \) is denoted by \( z \) . For example, \( q - p \) is a real number.\n\n... | Yes |
Theorem 2. Consider the cyclic quadrilaterals \( {ABCD} \) such that \( \mathcal{P} = \mathcal{V}\left( {ABCD}\right) \). If \( \mathcal{P} \) is not a rhombus, the locus of their diagonal midpoints is the rectangular hyperbola \( \mathcal{H}\left( \mathcal{P}\right) \) with the same center \( O \) as \( \mathcal{P} \)... | Proof. We use the same system of axes as in the preceding section and continue to suppose that \( {OP} = 1 \). We denote by \( \theta \) the directed angle \( \angle \left( {\overrightarrow{SR},\overrightarrow{SP}}\right) \) that is, \( \theta = \arg \left( {p + q}\right) \). Note that \( \sin \theta \neq 0 \). Let \( ... | Yes |
Theorem 3. If \( \mathcal{P} \) is not a rhombus, \( \mathcal{H}\left( \mathcal{P}\right) \) is the locus of the circumcenter of a cyclic quadrilateral \( {ABCD} \) such that \( \mathcal{P} = \mathcal{V}\left( {ABCD}\right) \) . | Of course, if \( \mathcal{P} \) is a rhombus, the locus is the pair of diagonals of \( \mathcal{P} \) . As another consequence of Theorem 2, we give a construction of a pair \( M,{M}^{\prime } \) of diagonal midpoints simpler than the one in \( §3 \) : through a vertex of \( \mathcal{P} \), say \( Q \) , draw a line in... | No |
For any point \( P \), the isogonal cubic \( \mathcal{K} = \mathrm{p}\mathcal{K}\left( {{X}_{6}, P}\right) \) meets the circumcircle at \( A, B, C \) and three other points \( {Q}_{1},{Q}_{2},{Q}_{3} \) such that \( P \) is the orthocenter of the triangle \( {Q}_{1}{Q}_{2}{Q}_{3} \) . | The lines \( {Q}_{1}{Q}_{1}^{ * },{Q}_{2}{Q}_{2}^{ * },{Q}_{3}{Q}_{3}^{ * } \) pass through the pivot \( P \) and are parallel to the asymptotes of the cubic. Since they are the axes of three inscribed parabolas, they must be tangent to the deltoid \( {\mathcal{H}}_{3} \), the anticomplement of the Steiner deltoid. Thi... | Yes |
Corollary 2. The isocubic \( \mathcal{K} \) contains the projections \( {R}_{1},{R}_{2},{R}_{3} \) of \( {P}^{ * } \) on the sidelines of \( {Q}_{1}{Q}_{2}{Q}_{3} \) . These three points lie on the bicevian conic \( \mathcal{C}\left( {G, P}\right) . | Proof. Let \( {R}_{1} \) be the third point of \( \mathcal{K} \) on the line \( {Q}_{2}{Q}_{3} \) . The following table shows the collinearity relations of nine points on \( \mathcal{K} \) and proves that \( {P}^{ * },{R}_{1} \) and \( {Q}_{1}^{ * } \) are collinear.\n\n--- \n\n\( {}^{2} \) This is the conic through th... | Yes |
Corollary 3. The lines \( {Q}_{i}{R}_{i}^{ * }, i = 1,2,3 \), pass through the cevian quotient \( P/{P}^{ * } \) . | Proof. This is obvious from the following table.\n\n<table><tr><td>\( {P}^{ * } \)</td><td>\( {P}^{ * } \)</td><td>\( P \)</td><td>\( \leftarrow P/{P}^{ * } \) is the tangential of \( {P}^{ * } \)</td></tr><tr><td>\( P \)</td><td>\( {Q}_{1}^{ * } \)</td><td>\( {Q}_{1} \)</td><td>\( \leftarrow {Q}_{1}{Q}_{1}^{ * } \) mu... | Yes |
Corollary 4. Let \( {S}_{1},{S}_{2},{S}_{3} \) be the reflections of \( P \) in \( {Q}_{1},{Q}_{2},{Q}_{3} \) respectively. The asymptotes of \( \mathcal{K} \) are the parallel at \( {S}_{i}^{ * } \) to the lines \( P{Q}_{i} \) or \( {P}^{ * }{R}_{i} \) . | Proof. These points \( {S}_{i} \) lie on the polar conic of the pivot \( P \) since they are the harmonic conjugate of \( P \) with respect to \( {Q}_{i} \) and \( {Q}_{i}^{ * } \) . The construction of the asymptotes derives from \( \left\lbrack {2,§{1.4.4}}\right\rbrack \) . | No |
Theorem 5. The inconic \( \mathcal{I}\left( P\right) \) concentric with \( \mathcal{C}{\left( G, P\right) }^{3} \) is also inscribed in the triangle \( {Q}_{1}{Q}_{2}{Q}_{3} \) and in the triangle formed by the Simson lines of \( {Q}_{1},{Q}_{2},{Q}_{3} \) . | Proof. Since the triangles \( {ABC} \) and \( {Q}_{1}{Q}_{2}{Q}_{3} \) are inscribed in the circumcircle, there must be a conic inscribed in both triangles. The rest is mere calculation. | No |
Theorem 9. \( \mathcal{K} \) meets the circumcircle at \( A, B, C \) and three other points \( {Q}_{1},{Q}_{2} \) , \( {Q}_{3} \) (one at least is real) lying on a same conic passing through \( \pi ,{\pi }^{ * } \) and \( \pi /{\pi }^{ * } \) . | Note that this conic meets the circumcircle again at the isogonal conjugate of the infinite point of the trilinear polar of the isoconjugate of \( \omega \) under the isoconjugation with fixed point \( \pi \) .\n\nWith \( \omega = p : q : r \) and \( \pi = u : v : w \), this conic has equation\n\n\[\n\mathop{\sum }\lim... | Yes |
Theorem 11. \( \mathcal{K} \) meets the circumcircle at the same points as the isogonal pivotal cubic with pivot \( P = u : v : w \) if and only if its pole \( \omega \) lie on the cubic \( {\mathcal{K}}_{\text{pole }} \) with equation | \[ \mathop{\sum }\limits_{\text{cyclic }}\left( {v + w}\right) \left( {{c}^{4}y - {b}^{4}z}\right) \frac{{x}^{2}}{{a}^{2}} - \left( {\mathop{\sum }\limits_{\text{cyclic }}\left( {{b}^{2} - {c}^{2}}\right) u}\right) {xyz} = 0 \] \[ \Leftrightarrow \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}u\left( {{c}^{2}y - {b}^{2}z... | Yes |
Theorem 12. \( \mathcal{K} \) meets the circumcircle at the same points as the isogonal pivotal cubic with pivot \( P = u : v : w \) if and only if its pivot \( \pi \) lie on the cubic \( {\mathcal{K}}_{\text{pivot }} \) with equation\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\left( {v + w}\right) \left( {{c}^{4}y - ... | In other words, for any point \( \pi \) on \( {\mathcal{K}}_{\text{pivot }} \), there is a pivotal cubic with pivot \( \pi \) meeting the circumcircle at the same points as the isogonal pivotal cubic with pivot \( P = u : v : w. \) | Yes |
Corollary 13. When \( P \) lies on the line \( {GK},{\mathcal{K}}_{\text{pole }} \) is a pivotal cubic and contains \( K,{X}_{25},{X}_{32} \) . Its pivot is \( {gcP} \) (on the circum-conic through \( G \) and \( K \) ) and its isopivot is \( {X}_{32} \) . Its pole is the barycentric product of \( {X}_{32} \) and \( {g... | All these cubics belong to a same pencil of pivotal cubics. Furthermore, \( {\mathcal{K}}_{\text{pole }} \) contains the cevian quotients of the pivot \( \operatorname{gc}P \) and \( K,{X}_{25},{X}_{32} \) . Each of these points is the third point of the cubic on the corresponding sideline of the triangle with vertices... | No |
Theorem 17. For a given pole \( \Omega \) or a given pivot \( P \), there is one and only one pivotal cubic \( \mathcal{K} = \mathrm{p}\mathcal{K}\left( {\Omega, P}\right) \) meeting the circumcircle at the vertices of an equilateral triangle. | With \( \Omega = K \) (or \( P = O \) ) we obviously obtain the McCay cubic and the equilateral triangle is the circumnormal triangle. More generally, a p \( \mathcal{K} \) meets the circumcircle at the vertices of circumnormal triangle if and only if its pole \( \Omega \) lies on the circum-cubic K378 passing through ... | Yes |
Proposition 1. Given a point \( P \) in the plane of triangle \( {ABC} \), the center \( h\left( P\right) \) of the Hagge circle associated with \( P \) is the point such the nine-point center \( N \) is the midpoint of \( h\left( P\right) {P}^{ * } \) where \( {P}^{ * } \) denotes the isogonal conjugate of \( P \) . | Proof. Let \( {AP} \) meet the circumcircle at \( D \), and reflect \( D \) in BC to the point \( U \) . The line \( {UH} \) is the doubled Simson line of \( D \), and the reflections of \( D \) in the other two sides are also on this line. The isogonal conjugate of \( D \) is well known to be the point at infinity in ... | Yes |
The triangle \( {ABC} \) has circumcircle \( \Gamma \), circumcenter \( O \) and orthocenter \( H \) . Choose a point \( P \) in the plane of \( {ABC} \) other than \( A, B, C \) . The cevian lines \( {AP},{BP},{CP} \) meet \( \Gamma \) again at \( D, E, F \) respectively. Reflect \( D \) in \( {BC} \) to a point \( U,... | Proof of Proposition 2. Let \( \angle {DAC} = {a}_{1} \) and \( \angle {BAD} = {a}_{2} \) . Similarly we define \( {b}_{1},{b}_{2},{c}_{1} \) and \( {c}_{2} \) . We deduce that the angles subtended by \( A, F, B, D, D \) and \( E \) at \( O \) as shown in Figure 2.\n\n carries a point \( {H}^{ + } \) on \( \Gamma \) to \( H \) . The same result applied to \( {XYZ}, P \) carries \( H \) to the orthocenter \( {H}^{ - } \) of \( {XYZ} \) . We may construct \( {H}^{ + } \) by drawing the ray \( P{H}^{ - } \) to meet... | Proof. The similarity associated with \( {ABC} \) and \( P \) is expressible as: reflect in \( {PA} \) , scale by a factor of \( \lambda \) from \( P \), and rotate about \( P \) through a certain angle. Note that if we repeat the process, constructing a similarity using the \( {XYZ} \) as the reference triangle, but s... | Yes |
Corollary 5. In a Brocard porism, as the poristic triangles vary, the locus of their orthocenters is contained in a circle with their common center \( h\left( F\right) \) on the Brocard axis, where \( F \) is the (areal) fourth power point of the triangles. The radius of this circle is \( R\left( {1 - 4{\sin }^{2}\omeg... | In fact there is a direct method to show that the locus of \( H \) in the Brocard porism is a subset of a circle, but this approach reveals neither center nor radius. We have already observed that \( \frac{J{K}^{2}}{J{G}^{2}} = 1 - 3{\tan }^{2}\omega \) so for triangles in a Brocard porism (with common \( O \) and \( K... | Yes |
Proposition 6. Let \( P \) be the center of the Brocard ellipse (the midpoint of the segment joining the Brocard points of \( {ABC} \) ). When the Hagge construction is made for the medial triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) using this point \( P \), then for each \( {ABC} \) in the porism, the Hagge... | Proof. If the areal coordinates of a point are \( \left( {l, m, n}\right) \) with respect to \( {ABC} \), then the areal coordinates of this point with respect to the medial triangle are \( (m + n - \) \( l, n + l - m, l + n - m) \) . The reference areals of \( P \) are \( \left( {{a}^{2}\left( {{b}^{2} + {c}^{2}}\righ... | Yes |
Proposition 8. Let three points \( X, Y \) and \( Z \) lie on a conic \( \sum \) and let \( {l}_{1},{l}_{2},{l}_{3} \) be three chords \( {XH},{YH},{ZH} \) all passing through a point \( H \) on \( \sum \). Suppose further that \( P \) is any point in the plane of \( \sum \), and let \( {XP},{YP},{ZP} \) meet \( \sum \... | Proof. Consider the hexagon HYVUWZ inscribed in \( \sum \). Apply Pascal’s hexagon theorem. It follows that \( M, P, N \) are collinear. By taking another hexagon \( N, P \), \( L \) are collinear. | No |
Proposition 10. The points \( {X}_{1},{Y}_{1},{Z}_{1},{U}_{1},{V}_{1},{W}_{1} \) lie on a conic (the Hagge midpoint conic). | Their coordinates are easily obtained and are\n\n\[ \n{X}_{1}\left( {{2u}\left( {{b}^{2}w + {c}^{2}v}\right) ,{vw}\left( {{a}^{2} + {b}^{2} - {c}^{2}}\right) ,{vw}\left( {{c}^{2} + {a}^{2} - {b}^{2}}\right) }\right) ,\n\]\n\n\[ \n{U}_{1}\left( {0, v\left( {2{c}^{2}v + w\left( {{b}^{2} + {c}^{2} - {a}^{2}}\right) }\righ... | Yes |
Proposition 11. \( {U}_{1},{X}_{1}, P \) are collinear. | This is proved by checking that when the coordinates of \( {X}_{1},{U}_{1}, P \) are placed as entries in the rows of a \( 3 \times 3 \) determinant, then this determinant vanishes. This shows that \( {X}_{1},{U}_{1}, P \) are collinear as are \( {Y}_{1},{V}_{1}, P \) and \( {Z}_{1},{W}_{1}, P \) . | Yes |
Proposition 12. The center of the Hagge midpoint conic is the midpoint of \( {Oh}\left( P\right) \) . It divides \( {P}^{ * }G \) in the ratio \( 3 : - 1 \) . | The proof is straightforward and is left to the reader. | No |
Proposition 13. The six points \( {X}_{k},{Y}_{k},{Z}_{k},{U}_{k},{V}_{k},{W}_{k} \) lie on a conic and the centers of these conics, for all values of \( k \), lie on the line \( {Oh}\left( P\right) \) and divide it in the ratio \( k : 1 \) . | This proposition was originally conjectured by us on the basis of drawings by the geometry software package CABRI and we are grateful to the Editor for confirming the conjecture to be correct. We have rechecked his calculation and for the record the coordinates of \( {X}_{k} \) and \( {U}_{k} \) are\n\n\[ \left( {\left... | Yes |
Proposition 14. \( {U}_{k},{X}_{k}, P \) are collinear. | The proof is by the same method as for Proposition 11. | No |
Proposition 15. The locus of centers of those Hagge circles which are tangent to the circumcircle is the Macbeath conic. | Proof. We address the elliptical case (see Figure 7) when \( {ABC} \) is acute and \( H \) is inside the circumcircle of radius \( R \) . The major axis of the Macbeath ellipse \( \sum \) is well known to have length \( R \) . Suppose that \( P \) is a point of the plane. Now \( h\left( P\right) \) is on \( \sum \) if ... | Yes |
Proposition 16. The locus of centers of those Hagge circles which cut the circumcircle at diametrically opposite points is a straight line perpendicular to the Euler line. | Proof. Let \( {ABC} \) have circumcenter \( O \) and orthocenter \( H \) . Choose \( {H}^{\prime } \) on \( {HO} \) produced so that \( {HO} \cdot O{H}^{\prime } = {R}^{2} \) where \( R \) is the circumradius of \( {ABC} \) . Now if \( X, Y \) are diametrically opposite points on \( S \) (but not on the Euler line), th... | Yes |
Lemma 1. If \( P \) has homogeneous barycentric coordinates \( x : y : z \) with reference to triangle \( {ABC} \), then\n\n\[ f\left( P\right) = \frac{\left( {y + z}\right) \left( {{2x} + y + z}\right) }{x} : \frac{\left( {z + x}\right) \left( {{2y} + z + x}\right) }{y} : \frac{\left( {x + y}\right) \left( {x + y + {2... | Proof. If \( P = x : y : z \), we have\n\n\[ \overrightarrow{A{P}_{c}} = \frac{y\overrightarrow{AB}}{x + y},\;\overrightarrow{AP} = \frac{y\overrightarrow{AB} + z\overrightarrow{AC}}{x + y + z},\;\overrightarrow{A{P}_{b}} = \frac{z\overrightarrow{AC}}{x + z}, \]\n\nso that\n\n\[ {\Delta }_{a}\left( P\right) = \Delta \l... | Yes |
Proposition 2. Let \( P = x : y : z \) and \( U = u : v : w \) . The lines \( {G}_{a}P \) and \( {P}_{a}^{ * }U \) are parallel if and only if \( P \) lies on the hyperbola \( {\mathcal{H}}_{a, U} \) through \( A,{G}_{a},{U}_{a}^{ * } \), the reflection of \( {U}_{b}^{ * } \) in \( C \) and the reflection of \( {U}_{c}... | Proof. As \( {P}_{a}^{ * } = 0 : z : y \) and \( {\left\lbrack {G}_{a}P\right\rbrack }_{\infty } = - \left( {{2x} + y + z}\right) : z + x : x + y \), the lines \( {G}_{a}P \) and \( {P}_{a}^{ * }U \) are parallel if and only if\n\n\[ \n{h}_{a, U}\left( P\right) \mathrel{\text{:=}} \det \left( {{\left\lbrack {G}_{a}P\ri... | Yes |
Corollary 3. The point \( f\left( P\right) \) is the intersection of the lines through \( {P}_{a}^{ * },{P}_{b}^{ * } \) and \( {P}_{c}^{ * } \) parallel to \( {G}_{a}P,{G}_{b}P,{G}_{c}P \) respectively. See Figure 1. | Proof. The lines \( {G}_{a}P,{G}_{b}P,{G}_{c}P \) are parallel to \( {P}_{a}^{ * }f\left( P\right) ,{P}_{b}^{ * }f\left( P\right) ,{P}_{c}^{ * }f\left( P\right) \) respectively. | Yes |
Lemma 4. If \( U \in \mathcal{T},{\mathcal{H}}_{a, U} \) and \( {\mathcal{H}}_{b, U} \) have a real common point in \( \mathcal{T} \) and a real common point in \( {\mathcal{T}}_{A} \), reflection in \( A \) of the open angular sector bounded by the half lines \( {AB} \) and \( {AC} \) . | Proof. Using the fact that \( {\mathcal{H}}_{a, U} \) passes through \( {\left\lbrack BC\right\rbrack }_{\infty } \), we can cut \( {\mathcal{H}}_{a, U} \) by lines parallel to \( {BC} \) to get a rational parametrization of \( {\mathcal{H}}_{a, U} \) . More precisely, let \( {B}_{t} \) and \( {C}_{t} \) be the images ... | Yes |
Theorem 5. If \( U \in \mathcal{T} \), the three hyperbolas \( {\mathcal{H}}_{a, U},{\mathcal{H}}_{b, U},{\mathcal{H}}_{c, U} \) have four distinct real common points, exactly one of which lies in T. This point is the only point \( P \in \mathcal{T} \) satisfying \( f\left( P\right) = U \) . | Proof. In a similar way as in Lemma 4, we can see that \( {\mathcal{H}}_{b, U} \) and \( {\mathcal{H}}_{c, U} \) have a common point in \( \mathcal{T} \) and a real common point in \( {\mathcal{T}}_{B} \) and that \( {\mathcal{H}}_{c, U} \) and \( {\mathcal{H}}_{a, U} \) have a real common point in \( \mathcal{T} \) an... | Yes |
Proposition 3. The lines \( {O}_{a}^{\prime }I,{O}_{b}^{\prime }I \) and \( {O}_{c}^{\prime }I \) are perpendicular to \( {BC},{CA} \) and \( {AB} \) respectively. | Proof. It is enough to prove that for the line \( {O}_{a}^{\prime }I \) . The other two cases are similar.\n\nLet \( M \) be the intersection (other than \( A \) ) of the circle \( \left( {O}_{a}^{\prime }\right) \) with the circumcircle of triangle \( {ABC} \) . Since \( I{O}_{a}^{\prime } = {OM} \) (circumradius) and... | Yes |
Lemma 1. Let \( x \) be the base of an isosceles triangle with given area \( A \) and perimeter \( P \) . Then\n\n\[ \n{2P}{x}^{3} - {P}^{2}{x}^{2} + {16}{A}^{2} = 0.\n\] | Proof. Working as in the above special case, we obtain \( y = \frac{P - x}{2} \) and \( \frac{x}{2}\sqrt{{y}^{2} - \frac{{x}^{2}}{4}} = \) \( A \) ; substituting the former condition into the latter, we arrive at (1). | No |
Theorem 2. There are exactly two distinct isosceles triangles of area \( A \) and perimeter \( P \) if and only if \( {P}^{2} > {12}\sqrt{3}A \) . There is exactly one if and only if \( {P}^{2} = {12}\sqrt{3}A \) and the triangle is equilateral. The vertex angles \( {\phi }_{1} < {\phi }_{2} \) of these two isosceles t... | Proof. Let \( f\left( x\right) \) be the cubic in (1). We first show that it has at most two distinct positive roots. Indeed the existence of three distinct positive roots would yield, by Rolle’s theorem, two distinct positive roots for \( {f}^{\prime }\left( x\right) = {6P}{x}^{2} - 2{P}^{2}x \) ; but the roots of \( ... | Yes |
Theorem 4. In every non-equilateral triangle of area \( A \) and perimeter \( P \) every angle \( \phi \) must satisfy the inequality \( {\phi }_{1} \leq \phi \leq {\phi }_{2} \), where \( {\phi }_{1} < \frac{\pi }{3} < {\phi }_{2} \) are the vertex angles of the two isosceles triangles of area \( A \) and perimeter \(... | Proof. As we have seen in Lemma 1, the cubic (1) yields the base \( x \) of each of the two isosceles triangles of area \( A \) and perimeter \( P \) ; and the formula above for the vertex angle \( \phi \) of an isosceles triangle follows from \( {x}^{2} = 2{y}^{2} - 2{y}^{2}\cos \phi \) (law of cosines) and \( y = \fr... | Yes |
Lemma 5. For some \( \psi \) in \( \left( {0,{\phi }_{1}}\right), s\left( \psi \right) = 0 \) . | Proof. Notice that \( \mathop{\lim }\limits_{{\phi \rightarrow {0}^{ + }}}s\left( \phi \right) = - A < 0 \) . On the other hand, the existence of an isosceles triangle with vertex angle \( {\phi }_{1} \) guarantees that \( s\left( {\phi }_{1}\right) > 0 \) (Theorem 3). By the continuity of \( s \) on \( \left( {0,\pi }... | Yes |
Lemma 6. The function \( s \) is strictly increasing on \( \left( {0,\pi }\right) \) and, for \( \phi \geq {\phi }_{1}, s\left( \phi \right) > \) 0. | Proof. Since the derivative \( {s}^{\prime }\left( \phi \right) = \frac{{P}^{2}}{4\left( {1 + \cos \phi }\right) } \) is positive on \( \left( {0,\pi }\right), s \) is strictly increasing; it follows that \( s\left( \phi \right) \geq s\left( {\phi }_{1}\right) > 0 \) for \( \phi \geq {\phi }_{1} \). | Yes |
Lemma 7. For \( \phi > {\phi }_{2}, h\left( \phi \right) < 0 \) . | Proof. Recall that \( h\left( \phi \right) = {\left( \frac{P}{2} + \frac{2A}{P}\left( \frac{1 + \cos \phi }{\sin \phi }\right) \right) }^{2} - \frac{8A}{\sin \phi } \) . By L’Hospital’s rule, we have \( \mathop{\lim }\limits_{{\phi \rightarrow \pi }}\frac{1 + \cos \phi }{\sin \phi } = \mathop{\lim }\limits_{{\phi \righ... | Yes |
Lemma 8. There is no \( \phi \) in \( \left( {0,\pi }\right) \) for which \( h\left( \phi \right) = {h}^{\prime }\left( \phi \right) = 0 \) . | Proof. Suppose \( h\left( \phi \right) = {h}^{\prime }\left( \phi \right) = 0 \) for some \( \phi \) in \( \left( {0,\pi }\right) \) . It follows that\n\n\[ \n{\left( \frac{P}{2} + \frac{2A}{P}\left( \frac{1 + \cos \phi }{\sin \phi }\right) \right) }^{2} = \frac{8A}{\sin \phi }\text{ and }\frac{P}{2} + \frac{2A}{P}\lef... | Yes |
Theorem 2. Let \( {A}_{0},{B}_{0},{C}_{0} \) be complex numbers which define an initial nondegenerate and non-equilateral triangle on the complex plane such that its centroid is at the origin (i.e., \( {A}_{0} + {B}_{0} + {C}_{0} = 0 \) ). Suppose we apply the FDRS with \( t = \frac{1}{2} + \frac{1}{2\sqrt{3}}\tan \lef... | Proof: Let \( r < 1 \), we have seen that there is \( {\varphi }_{k} \) which is arbitrarily close to any given angle on the unit circle. Since function \( \Phi \left( {\varphi }_{n}\right) \) is continuous with respect to \( {\varphi }_{n} \), it is apparent that \( {\widehat{A}}_{k} = \Phi \left( {\varphi }_{k}\right... | Yes |
Theorem 1 (Parry). Suppose triangle \( {ABC} \) has circumcenter \( O \) and orthocenter H. Parallel lines \( \alpha ,\beta ,\gamma \) are drawn through the vertices \( A, B, C \), respectively. Let \( {\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime } \) be the reflections of \( \alpha ,\beta ,\gamma \) in the... | We give a synthetic proof of this beautiful theorem below. C. F. Parry proposed this as a problem in the AMERICAN MATHEMATICAL MONTHLY, which was subsequently solved by R. L. Young using complex coordinates [6]. The point \( P \) in question is called the Parry reflection point. It appears as the triangle center \( {X}... | Yes |
Theorem 2 (Collings). Let \( \ell \) be a line in the plane of a triangle \( {ABC} \) . Its reflections in the sidelines \( {BC},{CA},{AB} \) are concurrent if and only if \( \ell \) passes through the orthocenter \( H \) of \( {ABC} \) . In this case, their point of concurrency lies on the circumcircle. | Synthetic proofs of Theorem 2 can be found in [1] and [3]. | No |
Lemma 3. The pedal triangle of the symmedian point is similar to the triangle of medians, the similarity factor being \( \tan \omega \) . | Proof. Since \( S = {bc}\sin A \), the distance from the centroid \( G \) to \( {AC} \) is clearly \( \frac{S}{3b} \). That from the symmedian point \( K \) to \( {AB} \) is \[ \frac{{c}^{2}}{{a}^{2} + {b}^{2} + {c}^{2}} \cdot \frac{S}{c} = \frac{S}{2{S}_{\omega }} \cdot c. \] Since \( K \) and \( G \) are isogonal con... | Yes |
Proposition 6. \( {G}^{\prime } \) is the intersection of \( {GK} \) with \( {HP} \), where \( H \) is the orthocenter of triangle \( {ABC} \) . | Proof. Apply Menelaus’ theorem to triangle \( {OGK} \) with transversal \( {HP} \), noting that \( {OH} : {HG} = 3 : - 2 \) . See Figure 7. | No |
Proposition 7. At most one of the vertices the Malfatti triangle and at most one of the vertices of the Malfatti squares on the sidelines can be outside the triangle. | Proof. If \( {Y}_{a} \) lies outside triangle \( {ABC} \), then \( \angle A{Z}_{a}{C}^{\prime } < \frac{\pi }{2} \), and \( \angle {Z}_{b}{Z}_{a}{C}^{\prime } > \frac{\pi }{2} \) . Since \( {Z}_{a}{C}^{\prime } \) is parallel to \( {AG},\angle {BAG} = \angle {Z}_{b}{Z}_{a}{C}^{\prime } \) is obtuse. Under the same hypo... | Yes |
Lemma 1 (Haruki). Given two nonintersecting chords \( {AB} \) and \( {CD} \) in a circle and a variable point \( P \) on the arc \( {AB} \) remote from points \( C \) and \( D \), let \( E \) and \( F \) be the intersections of chords \( {PC},{AB} \), and of \( {PD},{AB} \) respectively. The value of \( \frac{{AE} \cdo... | ## 2. Proof of Haruki's lemma\n\nA good interactive visualisation and proof of Haruki's lemma can be found in [1]. Here we present the proof essentially as it appeared in [3]. The proof is quite ingenious and relies on the fact that the angle \( \angle {CPD} \) is constant.\n\nWe begin by constructing a circumcirlce of... | Yes |
Lemma 2. Given two nonintersecting chords \( {AB} \) and \( {CD} \) in a circle and a variable point \( P \) on the arc \( {AB} \) remote from points \( C \) and \( D \), let \( E \) and \( F \) be the intersections of chords \( {PC},{AB} \), and of \( {PD},{AB} \) respectively. The following equalities hold:\n\n\[ \n\... | Proof. (1) Following the notation and proof of Lemma 1, we have \( \frac{{AE} \cdot {BF}}{EF} = {BG} \) . It remains to show that \( {BG} = \frac{{AC} \cdot {BD}}{CD} \), or, equivalently,\n\n\[ \n\frac{BG}{BD} = \frac{AC}{CD} \]\n\n(3)\n\n satisfying both (1) and (2), then \( A, B, C, D \) are concyclic. | Proof. First of all, points \( A, B, E, F \) are collinear, hence, they satisfy Euler’s distribution theorem (See [4, p.3] and [5]), i.e., if \( A, B, E, F \) are in this order, then, \( {AF} \cdot {BE} + {AB} \cdot {EF} = {AE} \cdot {BF} \) . Dividing through by \( {EF} \), we obtain\n\n\[ \frac{{AF} \cdot {BE}}{EF} +... | Yes |
Theorem 6. Given the points \( A, B, C, D \) and a point \( P \), define points \( E \) and \( F \) as the intersections of lines \( {PC} \) and \( {AB},{PD} \) and \( {AB} \) respectively. (a) The locus \( {\mathcal{L}}_{1} \) of points \( P \) satisfying (1) is the union of two circumconics of ABCD given by the equat... | Proof. In terms of signed lengths, (1) and (2) should be interpreted as \( {AE} \cdot {BF} \) . \( {CD} = \varepsilon \cdot {AC} \cdot {BD} \cdot {EF} \) and \( {AF} \cdot {BE} \cdot {CD} = \varepsilon \cdot {AD} \cdot {BC} \cdot {EF} \) for \( \varepsilon = \pm 1 \) . The results follow from direct substitutions. It i... | Yes |
Proposition 7. The four intersections of the bisectors of angles \( {ABD},{ACD} \), and the four intersections of the bisectors of angles \( {CAB} \) and \( {CDB} \) are points on \( {\mathcal{L}}_{1} \) . | Proof. First of all, it is routine to verify that for \( P = \left( {x : y : z}\right) \), we have\n\n\[ \left\lbrack {AEP}\right\rbrack \cdot \left\lbrack {BFP}\right\rbrack \cdot \left\lbrack {CDP}\right\rbrack = \left\lbrack {ACP}\right\rbrack \cdot \left\lbrack {BDP}\right\rbrack \cdot \left\lbrack {EFP}\right\rbra... | Yes |
Corollary 8. (a) When points \( A, B, C, D \) all belong to the same circle \( \mathcal{C} \), then one of the conics from \( {\mathcal{L}}_{1} \) and one from \( {\mathcal{L}}_{2} \) coincide with \( \mathcal{C} \). | Proof. (a) Assume \( {Q}_{1} \) not on the circle \( \mathcal{C} \). Suppose we have the situation as in Figure 8. In other cases the reasoning is similar. It is easy to see that \( \angle {AB}{Q}_{2} = \) \( \angle {AC}{Q}_{2} \) as \( {Q}_{2} \) belongs to the external bisector of the angle \( {ABD} \). This means th... | Yes |
Lemma 2. For arbitrary nonzero real numbers \( x, y, z \) ,\n\n\[ \n{x}^{2}{\sin }^{2}A + {y}^{2}{\sin }^{2}B + {z}^{2}{\sin }^{2}C \leq \frac{1}{4}{\left( \frac{yz}{x} + \frac{zx}{y} + \frac{xy}{z}\right) }^{2}.\n\]\n\nEquality holds if and only if \( {x}^{2} : {y}^{2} : {z}^{2} = \frac{1}{{a}^{2}\left( {{b}^{2} + {c}... | Proof. We make use of Kooi’s inequality [1, Inequality 14.1]: for real numbers \( {\lambda }_{1} \) , \( {\lambda }_{2},{\lambda }_{3} \) with \( {\lambda }_{1} + {\lambda }_{2} + {\lambda }_{3} \neq 0 \) ,\n\n\[ \n{\left( {\lambda }_{1} + {\lambda }_{2} + {\lambda }_{3}\right) }^{2}{R}^{2} \geq {\lambda }_{2}{\lambda ... | Yes |
For an interior point \( P \) and positive real numbers \( x, y, z \), we have\n\n\[ \n{x}^{2}{R}_{a} + {y}^{2}{R}_{b} + {z}^{2}{R}_{c} \geq \frac{2}{3}\left( {{yz}{w}_{a} + {zx}{w}_{b} + {xy}{w}_{c}}\right) .\n\]\n\nEquality holds if and only if the triangle \( {ABC} \) is equilateral, \( P \) its center, and \( x = y... | Proof. Replace in (9) \( x, y, z \) respectively by \( {yz}\sqrt{{R}_{b}{R}_{c}},{zx}\sqrt{{R}_{c}{R}_{a}},{xy}\sqrt{{R}_{a}{R}_{b}} \) . | No |
Corollary 4. For an interior point \( P \) in a triangle \( {ABC},{R}_{a}{R}_{b}{R}_{c} \geq \frac{64}{27}{w}_{a}{w}_{b}{w}_{c} \) . Equality holds if and only if \( {ABC} \) is equilateral and \( P \) its center. | Proof. This follows from (9) by putting \( x = y = z \) and applying the AM-GM inequality. | No |
Lemma 1. The radical center of the circles \( {\mathcal{C}}_{A},{\mathcal{C}}_{B},{\mathcal{C}}_{C} \) is the point \( Q \) . | Proof. The radical center of the circumcircle \( \mathcal{C} \) of triangle \( {ABC} \) and \( {\mathcal{C}}_{B},{\mathcal{C}}_{C} \) must be \( Q \) . Indeed, it must be the intersection of \( B{B}^{\prime } \) (the radical axis of \( \mathcal{C} \) and \( {\mathcal{C}}_{B} \) ) and \( C{C}^{\prime } \) (the radical a... | Yes |
Lemma 2. Let \( H \) be the orthocenter of triangle \( {ABC} \). For any point \( Q \neq H \) and \( P = H \), the circles \( {\mathcal{C}}_{A},{\mathcal{C}}_{B},{\mathcal{C}}_{C} \) are coaxial with radical axis \( {HQ} \). | Proof. When \( P = H \), the cevian triangle of \( P \) is the orthic triangle \( {H}_{a}{H}_{b}{H}_{c} \). The inversion with respect to the polar circle swaps \( A, B, C \) and \( {H}_{a},{H}_{b},{H}_{c} \) respectively. Hence the products of signed distances \( {HA} \cdot H{H}_{a},{HB} \cdot H{H}_{b} \), \( {HC} \cd... | Yes |
Theorem 3. In general, the locus of \( P \) for which the circles \( {\mathcal{C}}_{A},{\mathcal{C}}_{B},{\mathcal{C}}_{C} \) are coaxial is a circumcubic \( \mathcal{K}\left( Q\right) \) passing through \( H \) , \( Q \) and several other remarkable points. This cubic is tangent at \( A, B, C \) to the symmedians of t... | This is obtained through direct and easy calculation. It is sufficient to write that the radical circle of \( {\mathcal{C}}_{A},{\mathcal{C}}_{B},{\mathcal{C}}_{C} \) degenerates into the line at infinity and another line which is obviously the common radical axis of the circles. This calculation gives several equivale... | Yes |
Proposition 4. \( \mathcal{K}\left( Q\right) \) intersects the circumcircle at the same points as the pivotal isogonal cubic \( \mathrm{p}{\mathcal{K}}_{\text{circ }}\left( Q\right) \) with pivot \( \mathbf{{ag}}Q \) . | Proof. The equation of \( \mathcal{K}\left( Q\right) \) can be written in the form\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\left( {-{a}^{2}{qr} + {b}^{2}{rp} + {c}^{2}{pq}}\right) x\left( {{c}^{2}{y}^{2} - {b}^{2}{z}^{2}}\right) \]\n\n(2)\n\n\[ + \left( {{a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy}}\right) \mathop{\sum ... | Yes |
Theorem 5. The points \( {Q}_{1},{Q}_{2},{Q}_{3} \) are the antipodes on the circumcircle of the three points \( {Q}_{1}^{\prime },{Q}_{2}^{\prime },{Q}_{3}^{\prime } \) whose Simson lines pass through \( \mathbf{g}Q \) . | It follows that \( {Q}_{1},{Q}_{2},{Q}_{3} \) are three real distinct points if and only if \( \mathbf{g}Q \) lies inside the Steiner deltoid \( {\mathcal{H}}_{3} \). | Yes |
Theorem 8. The inconic \( \mathcal{I}\left( Q\right) \) with perspector \( \operatorname{tg}Q \) is inscribed in the two triangles \( {ABC} \) and \( {Q}_{1}{Q}_{2}{Q}_{3} \) . It is also inscribed in the triangle formed by the Simson lines of \( {Q}_{1},{Q}_{2},{Q}_{3} \) . | \( \mathcal{K}\left( Q\right) \) meets \( \mathcal{I}\left( Q\right) \) at six points which are the contacts of \( \mathcal{I}\left( Q\right) \) with the sidelines of the two triangles. Three of them are the vertices \( {A}_{1},{B}_{1},{C}_{1} \) of the cevian triangle of \( \operatorname{tg}Q \) in \( {ABC} \) . The o... | Yes |
Proposition 9. \( \mathcal{K}\left( Q\right) \) meets the line at infinity at the same points as the pivotal isogonal cubic \( \mathrm{p}{\mathcal{K}}_{\text{inf }}\left( Q\right) \) with pivot \( \mathbf{g}Q \) . | Proof. This follows by writing the equation of \( \mathcal{K}\left( Q\right) \) in the form\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}\operatorname{qr}x\left( {{c}^{2}{y}^{2} - {b}^{2}{z}^{2}}\right) + \left( {x + y + z}\right) \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}p\left( {{c}^{2}q - {b}^{2}r}\right) {... | Yes |
Proposition 10. The cubic \( \mathcal{K}\left( Q\right) \) contains the four foci of the inconic with center \( \mathbf{{cg}}Q \) and perspector \( \mathbf{{tg}}Q \) . | Proof. This follows by writing the equation of \( \mathcal{K}\left( Q\right) \) in the form\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{px}\left( {{c}^{2}q - {b}^{2}r}\right) \left( {{c}^{2}{y}^{2} + {b}^{2}{z}^{2}}\right) - 2\left( {\mathop{\sum }\limits_{\text{cyclic }}{a}^{2}\left( {{b}^{2} - {c}^{2}}\right) {qr}}\... | Yes |
Theorem 11. \( \mathcal{K}\left( Q\right) \) contains the center \( \operatorname{cg}Q \) of \( \mathcal{I}\left( Q\right) \) if and only if \( Q \) lies on the cubic \( \mathbf{{K172}} = \mathrm{p}\mathcal{K}\left( {{X}_{32},{X}_{3}}\right) \) . | Since we know that \( \mathcal{K}\left( Q\right) \) contains the perspector \( \operatorname{tg}Q \) of this same inconic when it is a pivotal cubic, it follows that there are only two cubics \( \mathcal{K}\left( Q\right) \) passing through the foci, the center, the perspector of \( \mathcal{I}\left( Q\right) \) and it... | Yes |
Proposition 12. The cubic \( \mathcal{K}\left( Q\right) \) meets the Steiner ellipse at the same points as \( \mathrm{p}\mathcal{K}\left( {\mathbf{{tg}}Q, Q}\right) \) . | Proof. This follows by writing the equation of \( \mathcal{K}\left( Q\right) \) in the form\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}{px}\left( {{b}^{2}r{y}^{2} - {c}^{2}q{z}^{2}}\right) + \left( {{xy} + {yz} + {zx}}\right) \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}\left( {{b}^{2} - {c}^{2}}\right) {qrx} =... | Yes |
Corollary 13. \( \mathcal{K}\left( Q\right) \) meets \( \mathcal{H}\left( P\right) \) again at two points on the line \( {\mathcal{L}}^{\prime }\left( Q\right) \) and \( \mathcal{H}\left( {\mathbf{g}P}\right) \) again at two points on the line \( \mathcal{L}\left( Q\right) \) . | For example, with \( P = G,\mathcal{H}\left( P\right) \) is the Kiepert hyperbola and \( {\mathcal{L}}^{\prime }\left( Q\right) \) is the line \( Q\mathbf{{gt}}Q,\mathcal{H}\left( {\mathbf{g}P}\right) \) is the Jerabek hyperbola and \( \mathcal{L}\left( Q\right) \) is the line \( Q\mathbf{{tg}}Q \) . | No |
Proposition 14. For varying \( Q \), the cubics \( \mathcal{K}\left( Q\right) \) form a net of cubics. | Proof. This follows by writing the equation of \( \mathcal{K}\left( Q\right) \) in the form\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}\operatorname{qr}x\left( {{c}^{2}y\left( {x + z}\right) - {b}^{2}z\left( {x + y}\right) }\right) = 0 \]\n\n(12)\n\n\[ \Leftrightarrow \mathop{\sum }\limits_{\text{cyclic }}{a}^{... | Yes |
Proposition 15. \( \mathcal{K}\left( Q\right) \) is a pivotal cubic \( \mathrm{p}\mathcal{K}\left( Q\right) \) if and only if \( Q \) lies on the cir-cumhyperbola \( \mathcal{H} \) passing through \( G \) and \( K \) . | Proof. We write the equation of \( \mathcal{K}\left( Q\right) \) in the form\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{b}^{2}{c}^{2}{px}\left( {r{y}^{2} - q{z}^{2}}\right) + \left( {\mathop{\sum }\limits_{\text{cyclic }}{a}^{2}\left( {{b}^{2} - {c}^{2}}\right) {qr}}\right) {xyz} = 0. \]\n\n(13)\n\nRecall that \( \ma... | Yes |
Proposition 16. The cubic \( \mathcal{K}\left( Q\right) \) belongs to another pencil of similar cubics generated by another pivotal cubic and another isogonal non-pivotal cubic. | Proof.\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{px}\left( {{c}^{2}q - {b}^{2}r}\right) \left( {{c}^{2}{y}^{2} + {b}^{2}{z}^{2}}\right) - \left( {\mathop{\sum }\limits_{\text{cyclic }}{a}^{2}\left( {{b}^{2} - {c}^{2}}\right) {qr}}\right) {xyz} \]\n\n(14)\n\n\[ + \mathop{\sum }\limits_{\text{cyclic }}{a}^{4}{qr}\left... | Yes |
Lemma 1. If \( P \) is a point in plane of triangle \( {ABC} \), with pedal triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \), then the perpendiculars from \( A \) to \( {B}^{\prime }{C}^{\prime } \), from \( B \) to \( {C}^{\prime }{A}^{\prime } \), from \( C \) to \( {A}^{\prime }{B}^{\prime } \) are concurrent ... | This is quite well-known. See, for example, [5, Theorem 237]. Figure 1 shows that \( {AP} \) and the perpendicular from \( A \) to \( {B}^{\prime }{C}^{\prime } \) are isogonal with reference to \( A \) . From this Lemma 1 follows. | No |
Proposition 1. For the two segments \( {CA} \) and \( {CB} \) , (1) \( {CQ} \) is the quadratic mean, (2) \( {CM} \) is the arithmetic mean, (3) \( {CG} \) is the geometric mean, (4) \( {CH} \) is the harmonic mean. | Proof. Note that the circle has radius \( \frac{1}{2}\sqrt{{a}^{2} + {b}^{2}} \) . (1) Since \( {CA} = {A}^{\prime }B, C{A}^{\prime }{BA} \) is an isosceles trapezoid, with \( {AB} \) parallel to \( C{A}^{\prime } \) . Since \( {CQ} \) is the bisector of angle \( {ACB}, Q \) is the midpoint of the arc \( {CAB} \) , and... | Yes |
Proposition 2. The lines \( {AB},{CQ} \), and \( {A}^{\prime \prime }G \) are concurrent. | Proof. Let the bisector \( {CQ} \) of angle \( {ACB} \) intersect \( {AB} \) at \( K \) . See Figure 4. Clearly, the triangles \( {ACK} \) and \( {A}^{\prime \prime }{CK} \) are congruent. Now,\n\n\[ \angle C{A}^{\prime \prime }K = \angle {CAK} = \angle {CAB} \]\n\n\[ = {180}^{ \circ } - \angle C{A}^{\prime }B\;\left( ... | Yes |
Lemma 1. ABCD is cyclic if and only if \( {ac} = {bd} \) . | Proof. Let \( {ABCD} \) be any convex quadrilateral, not necessarily admitting an incircle, and let its vertex angles be \( {2A},{2B},{2C} \), and \( {2D} \) . Then \( A, B, C \), and \( D \) are acute, and \( A + B + C + D = {180}^{ \circ } \) . We shall show that\n\n\[ \n{ABCD}\text{is cyclic} \Leftrightarrow \tan A\... | No |
Lemma 2. The radius \( r \) of the incircle is given by\n\n\[ \n{r}^{2} = \frac{{bcd} + {acd} + {abd} + {abc}}{a + b + c + d}.\n\] | Proof. Again, let the vertex angles of \( {ABCD} \) be \( {2A},{2B},{2C} \), and \( {2D} \), and let\n\n\[ \n\alpha = \tan A,\beta = \tan B,\gamma = \tan C,\delta = \tan D.\n\]\n\nLet \( {\varepsilon }_{1} = \sum \alpha ,{\varepsilon }_{2} = \sum {\alpha \beta },{\varepsilon }_{3} = \sum {\alpha \beta \gamma } \), and ... | Yes |
Lemma 3.\n\n\\[ \n{u}^{2} = \frac{a + c}{b + d}\\left( {\\left( {a + c}\\right) \\left( {b + d}\\right) + {4bd}}\\right) ,\\;\\text{ and }\\;{v}^{2} = \frac{b + d}{a + c}\\left( {\\left( {a + c}\\right) \\left( {b + d}\\right) + {4ac}}\\right) .\n\\] | Proof. Again, let the vertex angles of \\( {ABCD} \\) be \\( {2A},{2B},{2C} \\), and \\( {2D} \\) . Then\n\n\\[ \n\\cos {2A} = \\frac{1 - {\\tan }^{2}A}{1 + {\\tan }^{2}A} = \\frac{{a}^{2} - {r}^{2}}{{a}^{2} + {r}^{2}}\n\\]\n\n\\[ \n= \\frac{{a}^{2}\\left( {a + b + c + d}\\right) - \\left( {{bcd} + {acd} + {abd} + {abc... | Yes |
Lemma 3 ([2]). We assume that the two rotations are not equal to the identity, or to a rotation of angle \( \pi \) . Let \( \theta \) and \( u \) be the angle and axis of the first rotation, and denote by \( t \) the vector \( \tan \frac{\theta }{2} \cdot u \) and \( {t}^{\prime } \) the associated vector for the secon... | \[ {t}^{\prime \prime } = \frac{1}{1 - t \cdot {t}^{\prime }}\left( {t + {t}^{\prime } + t \land {t}^{\prime }}\right) . \] | Yes |
Lemma 4. Let \( s \) be an isometry of \( {\mathbb{R}}^{3} \) not equal to a translation. Let \( S \) be the associated linear map and \( u \) the vector of translation. Assume \( s \) is either a screw motion or a glide reflection. Then the points \( n \) which satisfy \( \overrightarrow{{ns}\left( n\right) } \in \mat... | Proof. We call \( \theta \) the eigenspace of \( S \) related to the eigenvalue one. We have \( s\left( n\right) = s\left( o\right) + {So}\overrightarrow{n} \) where \( o \), the origin of the base will be chosen later. Elementary geometry yields \( \overrightarrow{{ns}\left( n\right) } = \left( {S - {Id}}\right) X + Y... | Yes |
Proposition 5. Let \( P \) a polyhedron, the following properties are equivalent.\n\n(1) A word \( v \) is the prefix of a periodic word with period \( \\left| v\\right| \) .\n\n(2) There exists \( m \\in {v}_{0} \) such that \( \\overrightarrow{{s}_{v}\\left( m\\right) m} \) is admissible with base point \( m \) for \... | Proof of Proposition 5. First we claim the following fact. The vector connecting \( {T}^{\\left| v\\right| }\\left( {m,\\theta }\\right) \) to \( {s}_{v}\\left( m\\right) \) is parallel to the direction of \( {T}^{\\left| v\\right| }\\left( {m,\\theta }\\right) \) . For \( \\left| v\\right| = 1 \) if the billiard traje... | Yes |
Theorem 6. Let \( v \) be a periodic word of even length. The set of periodic points in the face \( {v}_{0} \) with code \( v \) and length \( \left| v\right| \) can have two shapes. Either it is an open set or it is a point. | Proof. Let \( \Pi \) be a face of the polyhedron, and let \( m \in \Pi \) be the starting point for a periodic billiard path. The first return map to \( m \) is an isometry of \( {\mathbb{R}}^{3} \) that fixes both \( m \) and the direction \( u \) of the periodic billiard path.\n\nAssume first \( \left| v\right| \) is... | No |
Theorem 7. Let \( P \) be a polyhedron and \( v \) the prefix of a periodic word of period \( \left| v\right| \) in \( P \) .\n\n(1) If the period is even, and \( {S}_{v} \) is different from the identity, then \( v \) is stable.\n\n(2) If the period is odd, then the word is stable if and only if \( {S}_{v} \) is const... | Proof of Theorem 7. First consider the case of period even. The matrix \( S = {S}_{v} \) is not the identity, and \( \theta = {\theta }_{v} \) is the eigenvector associated to the eigenvalue one. First note that by continuity \( v \) persists for sufficiently small perturbations of the polyhedron. Fix a perturbation an... | Yes |
Corollary 8. (1) All the words of odd length are stable in a polygon. | Proof. (1) was already mentioned in [4]. The proof is the same as that of Theorem 7. Indeed is \( \left| v\right| \) is odd then \( s \) has a real eigenvector, and we can apply the proof. | No |
Lemma 12. In the regular tetrahedron, consider the word \( v = {\left( abcd\right) }^{\infty } \). The first return map \( r \) on \( {\pi }_{a}\left( {\sigma }_{v}\right) \) is given by\n\n\[ r\left( \begin{array}{l} x \\ y \end{array}\right) = A\left( \begin{array}{l} x \\ y \end{array}\right) + B \]\n\nwhere\n\n\[ A... | Proof. If \( m \) is a point of the face \( a \), the calculation in Section 6 shows that\n\n\[ r\left( m\right) = \frac{1}{81}\left( \begin{matrix} - {79x} - {8y} + {16}\sqrt{2} \\ {66x} - {21y} + {42z} + \frac{3\sqrt{2}}{2} \\ {13x} + {29y} - {58z}\frac{{33}\sqrt{2}}{12} \end{matrix}\right) \]\n\nNow we compute \( m ... | No |
Theorem 1. The medians of \( \mathcal{S} \) are concurrent in \( C \) . Moreover, \( C \) divides the median \( {C}^{\prime }{C}^{\prime \prime } \) relative to a \( k \) -subsystem \( {\mathcal{S}}^{\prime } \) of \( \mathcal{S} \) into two parts such that:\n\n\[ \frac{{C}^{\prime }C}{C{C}^{\prime \prime }} = \frac{n ... | Proof. In fact, let \( \mathbf{v},{\mathbf{v}}^{\prime },{\mathbf{v}}^{\prime \prime } \) the position vectors of \( C,{C}^{\prime },{C}^{\prime \prime } \) respectively. It is easy to prove that\n\n\[ \mathbf{v} - {\mathbf{v}}^{\prime } = \frac{n - k}{k}\left( {{\mathbf{v}}^{\prime \prime } - \mathbf{v}}\right) \]\n\n... | Yes |
Corollary 3. The segment \( {C}_{1}^{\prime }{C}_{2}^{\prime } \) that joins the centroids of two \( k \) -subsystems \( {\mathcal{S}}_{1}^{\prime } \) , \( {\mathcal{S}}_{2}^{\prime } \) of \( \mathcal{S} \) is parallel to the segment \( {C}_{1}^{\prime \prime }{C}_{2}^{\prime \prime } \) that joins the centroids of t... | \[ \frac{{C}_{1}^{\prime }{C}_{2}^{\prime }}{{C}_{1}^{\prime \prime }{C}_{2}^{\prime \prime }} = \frac{n - k}{k} \] | Yes |
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