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By choosing \( {t}_{1} = {t}_{2} = \frac{n}{m} \) and putting \( t = \frac{v}{u} \), we obtain a generic Brahmagupta trapezoid: | \[ a = \left( {{m}^{2}u - {n}^{2}u + {2mnv}}\right) \left( {{2mnu} - {m}^{2}v + {n}^{2}v}\right) , \] \[ b = d = \left( {{m}^{2} + {n}^{2}}\right) \left( {{nu} - {mv}}\right) \left( {{mu} + {nv}}\right) , \] \[ c = {\left( {m}^{2} + {n}^{2}\right) }^{2}{uv} \] \[ e = f = {mn}\left( {{m}^{2} + {n}^{2}}\right) \left( {{u... | Yes |
Let \( {ECD} \) be the rational Heron triangle with \( c : \alpha : \beta = {14} : {15} : {13} \). Here, \( {t}_{1} = \frac{2}{3},{t}_{2} = \frac{1}{2} \) (and \( {t}_{3} = \frac{4}{7} \)). By putting \( t = \frac{v}{u} \) and clearing denominators, we obtain Brahmagupta quadrilaterals with sides | \[ a = \left( {{7u} - {4v}}\right) \left( {{4u} + {7v}}\right), b = {13}\left( {u - {2v}}\right) \left( {{2u} + v}\right), c = {65uv}, d = 5\left( {{2u} - {3v}}\right) \left( {{3u} + {2v}}\right) ,\] \[ diagonals e = {30}\left( {{u}^{2} + {v}^{2}}\right) ,\;f = {26}\left( {{u}^{2} + {v}^{2}}\right) ,\] \[ and area \big... | Yes |
Example 3. If we take \( {ECD} \) to be a right triangle with sides \( {CD} : {EC} : {ED} = \) \( {m}^{2} + {n}^{2} : {2mn} : {m}^{2} - {n}^{2} \), we obtain | \[ a = \left( {{m}^{2} + {n}^{2}}\right) \left( {{u}^{2} - {v}^{2}}\right) \] \[ b = \left( {\left( {m - n}\right) u - \left( {m + n}\right) v}\right) \left( {\left( {m + n}\right) u + \left( {m - n}\right) v}\right) , \] \[ c = 2\left( {{m}^{2} + {n}^{2}}\right) {uv}, \] \[ d = 2\left( {{nu} - {mv}}\right) \left( {{mu... | Yes |
If the angle \( \theta \) is chosen such that \( A + B - \theta = \frac{\pi }{2} \), then the side \( {BC} \) is a diameter of the circumcircle of \( {ABCD} \). | \[ t = \tan \frac{\theta }{2} = \frac{1 - {t}_{3}}{1 + {t}_{3}} = \frac{{t}_{1} + {t}_{2} - 1 + {t}_{1}{t}_{2}}{{t}_{1} + {t}_{2} + 1 - {t}_{1}{t}_{2}}. \] Putting \( {t}_{1} = \frac{n}{m},{t}_{2} = \frac{q}{p} \), and \( t = \frac{\left( {m + n}\right) q - \left( {m - n}\right) p}{\left( {m + n}\right) p - \left( {m -... | Yes |
Theorem 1. Let \( {B}_{a} \) and \( {C}_{a} \) be points respectively on the extensions of \( {CA} \) and \( {BA} \) beyond \( A \) such that \( {B}_{a}{C}_{a} \) is antiparallel to \( {BC} \) and has length \( s \), the semiperimeter of triangle \( {ABC} \) . Likewise, let \( {C}_{b},{A}_{b} \) be on the extensions of... | The vertices of the Tucker hexagon can be constructed as follows. Let \( {X}_{b} \) and \( {X}_{c} \) be the points of tangency of \( {BC} \) with excircles \( \left( {I}_{b}\right) \) and \( \left( {I}_{c}\right) \) respectively. Since \( B{X}_{b} \) and \( C{X}_{c} \) each has length \( s \), the parallel of \( {AB} ... | Yes |
Lemma 2. (1) \( {abc} = 4\operatorname{Rrs} \); | Proof. (1) follows from the formulae \( \bigtriangleup = {rs} \) and \( R = \frac{abc}{4\bigtriangleup } \) . | Yes |
Theorem 4. The radical circle of the excircles has center at the Spieker point \( S = \left( {b + c : c + a : a + b}\right) \), and radius \( \frac{1}{2}\sqrt{{r}^{2} + {s}^{2}} \) . | Proof. We compute the power of \( \left( {b + c : c + a : a + b}\right) \) with respect to the \( A \) - excircle. The powers of \( A, B, C \) with respect to the \( A \) -excircle are clearly\n\n\[ \n{p}_{1} = {s}^{2},\;{p}_{2} = {\left( s - c\right) }^{2},\;{p}_{3} = {\left( s - b\right) }^{2}.\n\]\n\nWith \( x = b +... | Yes |
Proposition 5. \( Q = \frac{1}{4Rr}\left( {\left( {{s}^{2} - {r}^{2}}\right) O - \frac{1}{2}\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) K}\right) \) . | Proof.\n\n\[ Q = \frac{1}{4Rr}\left( {\left( {{r}^{2} + {s}^{2}}\right) O + {6Rr} \cdot G - \left( {{r}^{2} + {s}^{2} + {2Rr}}\right) I}\right) \]\n\n\[ = \frac{1}{4Rr}\left( {\left( {{s}^{2} - {r}^{2}}\right) O + 2{r}^{2} \cdot O + {6Rr} \cdot G - \left( {{r}^{2} + {s}^{2} + {2Rr}}\right) I}\right) \]\n\n\[ = \frac{1}... | Yes |
Theorem 2. Let \( {ABCD} \) be a convex quadrilateral inscribed in a circle. Denote by \( {I}_{a}\left( {\rho }_{a}\right) ,{I}_{b}\left( {\rho }_{b}\right) ,{I}_{c}\left( {\rho }_{c}\right) ,{I}_{d}\left( {\rho }_{d}\right) \) the incircles of the triangles \( {BCD},{CDA},{DAB} \) , and \( {ABC} \) .\n\n(2.1) The ince... | Proof. In \( {ABCD} \) we have the following circles: \( {O}_{cd}\left( {r}_{cd}\right) ,{O}_{da}\left( {r}_{da}\right) ,{O}_{ab}\left( {r}_{ab}\right) \) , and \( {O}_{bc}\left( {r}_{bc}\right) \) inscribed respectively in angles \( {AEB},{BEC},{CED} \), and \( {DEA} \), each tangent internally to the circumcircle. Le... | Yes |
Theorem 1. The orthocorrespondent \( {P}^{ \bot } \) is a point at infinity if and only if \( P \) lies on the Monge (orthoptic) circle of the inscribed Steiner ellipse. | Proof. From (1), \( {P}^{ \bot } \) is a point at infinity if and only if\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{S}_{A}{x}^{2} - 2{a}^{2}{yz} = 0. \]\n\nThis is a circle in the pencil generated by the circumcircle and the nine-point circle, and is readily identified as the Monge circle of the inscribed Steiner el... | Yes |
Proposition 2. The orthocorrespondent \( {P}^{ \bot } \) lies on the sideline \( {BC} \) if and only if \( P \) lies on the circle \( {\Gamma }_{BC} \) with diameter \( {BC} \) . The perpendicular at \( P \) to \( {AP} \) intersects \( {BC} \) at the harmonic conjugate of \( {P}^{ \bot } \) with respect to \( B \) and ... | Proof. \( {P}^{ \bot } \) lies on \( {BC} \) if and only if its first barycentric coordinate is 0, i.e., if and only if \( u\left( {-u{S}_{A} + v{S}_{B} + w{S}_{C}}\right) + {a}^{2}{vw} = 0 \) which shows that \( P \) must lie on \( {\Gamma }_{BC} \) . | Yes |
Theorem 3. Let \( Q \) be a finite point. There are exactly two points \( {P}_{1} \) and \( {P}_{2} \) (not necessarily real nor distinct) such that \( Q = {P}_{1}^{ \bot } = {P}_{2}^{ \bot } \) . | Proof. Let \( Q \) be a finite point. The trilinear polar \( {\ell }_{Q} \) of \( Q \) intersects the sidelines of triangle \( {ABC} \) at \( {Q}_{a},{Q}_{b},{Q}_{c} \) . The circles \( {\Gamma }_{a},{\Gamma }_{b},{\Gamma }_{c} \) with diameters \( A{Q}_{a}, B{Q}_{b} \) , \( C{Q}_{c} \) are in the same pencil of circle... | Yes |
An orthopivotal cubic \( \mathcal{O}\left( P\right) \) is a pivotal circumcubic \( p\mathcal{K} \) if and only if the triangles \( {ABC} \) and \( {UVW} \) are perspective, i.e., if and only if \( P \) lies on the Napoleon cubic (isogonal \( p\mathcal{K} \) with pivot \( {X}_{5} \) ). | (1) the pivot \( Q \) of \( \mathcal{O}\left( P\right) \) lies on the cubic \( {\mathcal{K}}_{n} : {}^{26} \) it is the perspector of \( {ABC} \) and the \( \left( {-2}\right) \) -pedal triangle of \( P,{}^{27} \) and lies on the line \( P{X}_{5} \) ;\n\n(2) the pole \( \Omega \) of the isoconjugation lies on the cubic... | Yes |
Proposition 11. An orthopivotal cubic \( \mathcal{O}\left( P\right) \) is a focal if and only if \( P \) lies on \( {\mathcal{B}}_{2} \) . | This is the case of \( {\mathcal{B}}_{2} \) itself, which is an isogonal focal cubic passing through the following points: \( A, B, C, G, K,{X}_{13},{X}_{14},{X}_{15},{X}_{16},{X}_{111} \) (the singular focus), \( {X}_{368},{X}_{524} \), the vertices of the second Brocard triangle and their isogonal conjugates. All tho... | No |
Theorem 12. The locus of point \( P \) whose orthotransversal \( {\mathcal{L}}_{P} \) and trilinear polar \( {\ell }_{P} \) are parallel is the circular quintic\n\n\( {\mathcal{Q}}_{1} :\)\n\[ \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}{y}^{2}{z}^{2}\left( {{S}_{B}y - {S}_{C}z}\right) = 0. \] | Equivalently, \( {\mathcal{Q}}_{1} \) is the locus of point \( P \) for which\n\n(1) the lines \( P{P}^{ * } \) and \( {\ell }_{P} \) (or \( {\mathcal{L}}_{P} \) ) are perpendicular;\n\n(2) \( P \) lies on the Euler line of the pedal triangle of \( {P}^{ * } \),\n\n(3) \( P,{P}^{ * }, H/P \) (and \( {P}^{ \bot } \) ) a... | No |
Proposition 15. The quartic \( {\mathcal{Q}}_{2} \) contains the 61 following points:\n\n(1) the vertices \( A, B, C,{}^{44} \)\n\n(2) the circular points at infinity, \( {}^{45} \)\n\n(3) the three points where the Thomson cubic meets the circumcircle again,\n\n(4) the in/excenters \( I,{I}_{a},{I}_{b},{I}_{c} \), wit... | We give a proof of (10). Let \( {k}_{1},{k}_{2},{k}_{3} = 0, \pm 1 \), and consider\n\n\[{\varphi }_{1} = \frac{A + 2{k}_{1}\pi }{3},\;{\varphi }_{2} = \frac{B + 2{k}_{2}\pi }{3},\;{\varphi }_{3} = \frac{C + 2{k}_{3}\pi }{3}.\n\]\n\nDenote by \( M \) one of the 27 points with barycentric coordinates\n\n\[ \left( {a\cos... | Yes |
Lemma 3. Let \( \bigtriangleup \) denote the area of triangle \( {ABC} \), and \( R \) its circumradius.\n\n(1) \( \bigtriangleup = 2{R}^{2}\sin A\sin B\sin C \) ; | Proof. (1) By the law of sines,\n\n\[ \bigtriangleup = \frac{1}{2}{bc}\sin A = \frac{1}{2}\left( {{2R}\sin B}\right) \left( {{2R}\sin C}\right) \sin A = 2{R}^{2}\sin A\sin B\sin C. \] | Yes |
Proposition 5 (Dergiades [1]). The ratio of similarity of \( \bar{O}{B}_{a}{C}_{a},{A}_{b}\bar{O}{C}_{b} \), and \( {A}_{c}{B}_{c}\bar{O} \) with \( {ABC} \) is\n\n\[ k = \left| \frac{{R}^{2}}{3{R}^{2} - O{H}^{2}}\right| . \] | Proof. Since \( 2\bigtriangleup = a \cdot {h}_{a} + b \cdot {h}_{b} + c \cdot {h}_{c} \), and \( {h}_{a} = {ka}\sin {2A},{h}_{b} = {kb}\sin {2B} \) , and \( {h}_{c} = {kc}\sin {2C} \), the ratio of similarity is the absolute value of\n\n\[ = \frac{4{R}^{2}\sin A\sin B\sin C}{4{R}^{2}\left( {{\sin }^{2}A\sin {2A} + {\si... | Yes |
Suppose that \( f\left( {a, b, c}\right) : g\left( {b, c, a}\right) : h\left( {c, a, b}\right) \) is a point, as defined in [2] We abbreviate this point as \( {f}_{ab} : {g}_{bc} : {h}_{ca} \) and recall from \( \left\lbrack {5,7}\right\rbrack \) that bicentric triangles are defined by the forms\n\n\[ \left( \begin{mat... | Examples of bicentric pairs thus obtained will now be presented. An inductive method [6] of generating the non-circle-dependent objects of triangle geometry enumerates such objects in sets formally of size six. When the actual size is six, which means that no two of the six objects are identical, the objects form a pai... | Yes |
Lemma 2. The circumcenters of triangles \( {AP}{B}^{\prime } \) and \( {AP}{C}^{\prime } \) coincide if and only if \( P \) lies on the reflection of the circumcircle in the line \( {BC} \) . | Proof. Triangles \( {AP}{B}^{\prime } \) and \( {AP}{C}^{\prime } \) have the same circumcenter if and only if the four points \( A,{B}^{\prime }, P,{C}^{\prime } \) are concyclic. In terms of directed angles, \( \angle {BPC} = \) \( \angle {B}^{\prime }P{C}^{\prime } = \angle {B}^{\prime }A{C}^{\prime } = \angle {CAB}... | Yes |
Proposition 3. The 6 circumcenters of the cevasix configuration of \( P \) lie on a conic. | Proof. We need only consider the case when these 6 circumcenters are all distinct. The circumcenters \( {B}_{ + } \) and \( {C}_{ - } \) lie on the perpendicular bisector of the segment \( {AP} \) ; similarly, \( {B}_{ - } \) and \( {C}_{ + } \) lie on the perpendicular bisector of \( P{A}^{\prime } \) . These two perp... | Yes |
Proposition 4. The vertices of a hexagon \( {A}_{ + }{C}_{ - }{B}_{ + }{A}_{ - }{C}_{ + }{B}_{ - } \) with parallel opposite sides \( {B}_{ + }{C}_{ - }//{C}_{ + }{B}_{ - },{C}_{ + }{A}_{ - }//{A}_{ + }{C}_{ - },{A}_{ + }{B}_{ - }//{B}_{ + }{A}_{ - } \) lie on a circle if and only if the main diagonals \( {A}_{ + }{A}_... | Proof. If the vertices are concyclic, then \( {A}_{ + }{C}_{ - }{A}_{ - }{C}_{ + } \) is an isosceles trapezoid, and \( {A}_{ + }{A}_{ - } = {C}_{ + }{C}_{ - } \) . Similarly, \( {C}_{ + }{B}_{ - }{C}_{ - }{B}_{ + } \) is also an isosceles trapezoid, and \( {C}_{ + }{C}_{ - } = {B}_{ + }{B}_{ - } \) . Conversely, consi... | Yes |
Proposition 5. The vector sum \( {\mathbf{{AA}}}^{\prime } + {\mathbf{{BB}}}^{\prime } + {\mathbf{{CC}}}^{\prime } = \mathbf{0} \) if and only if \( P \) is the centroid. | Proof. Suppose with reference to triangle \( {ABC} \), the point \( P \) has absolute barycentric coordinates \( {uA} + {vB} + {wC} \), where \( u + v + w = 1 \) . Then,\n\n\[ \n{A}^{\prime } = \frac{1}{v + w}\left( {{vB} + {wC}}\right) ,\;{B}^{\prime } = \frac{1}{w + u}\left( {{wC} + {uA}}\right) ,\;{C}^{\prime } = \f... | Yes |
Proposition 6.\n\n\[ \n{\pi }_{b}\left( {{\mathbf{A}}_{ + }{\mathbf{A}}_{ - }}\right) = - \frac{1}{2}{\mathbf{{BB}}}^{\prime },\;{\pi }_{c}\left( {{\mathbf{A}}_{ + }{\mathbf{A}}_{ - }}\right) = \frac{1}{2}{\mathbf{{CC}}}^{\prime }, \n\] \n\n\[ \n{\pi }_{c}\left( {{\mathbf{B}}_{ + }{\mathbf{B}}_{ - }}\right) = - \frac{1... | Proof. The orthogonal projections of \( {A}_{ + } \) and \( {A}_{ - } \) on the cevian \( B{B}^{\prime } \) are respectively the midpoints of the segments \( P{B}^{\prime } \) and \( {BP} \) . Therefore, \n\n\[ \n{\pi }_{b}\left( {{\mathbf{A}}_{ + }{\mathbf{A}}_{ - }}\right) = \frac{B + P}{2} - \frac{P + {B}^{\prime }}... | Yes |
Lemma 7. If the four points \( {X}_{ + },{Y}_{ - },{Z}_{ + },{Z}_{ - } \) are concyclic, then \( P \) lies on the median through \( C \) . | Proof. Let \( x = \frac{AP}{A{A}^{\prime }} \) and \( y = \frac{BP}{B{B}^{\prime }} \) . If the four points \( {X}_{ + },{Y}_{ - },{Z}_{ + },{Z}_{ - } \) are concyclic, then \( \angle {Z}_{ + }{Z}_{ - }{X}_{ + } = \alpha \) and \( \angle {Y}_{ - }{Z}_{ + }{Z}_{ - } = \beta \) . Now,\n\n\[ \frac{\left| B{B}^{\prime }\ri... | Yes |
Theorem 2. Given triangles \( {A}_{k}{B}_{k}{C}_{k} \) and points \( {Z}_{k} \) for \( k = 1,2,3 + ,3 - \) and \( {D}_{ \pm }{E}_{ \pm }{F}_{ \pm } \) as described above, triangles \( {D}_{ + }{E}_{ + }{F}_{ + } \) and \( {D}_{ - }{E}_{ - }{F}_{ - } \) are equilateral triangles of negative orientation, congruent and pa... | Proof. To prove this we find the following affixes\n\n\[ \n{d}_{ + } = \frac{1}{3}\left( {{b}_{1} + {c}_{2} + \zeta {a}_{2} + \bar{\zeta }{a}_{1}}\right) ,\;{d}_{ - } = \frac{1}{3}\left( {{b}_{2} + {c}_{1} + \zeta {a}_{1} + \bar{\zeta }{a}_{2}}\right) ,\n\]\n\n\[ \n{e}_{ + } = \frac{1}{3}\left( {{c}_{1} + {a}_{2} + \ze... | Yes |
\[ {r}_{a} - {r}_{1} = \frac{\frac{2}{r} - \frac{1}{\sqrt{{r}_{2}{r}_{3}}}}{\left( {\frac{2}{r} - \frac{1}{\sqrt{{r}_{3}{r}_{1}}}}\right) \left( {\frac{2}{r} - \frac{1}{\sqrt{{r}_{1}{r}_{2}}}}\right) }, \] | Proof. For convenience we write \[ {t}_{1} \mathrel{\text{:=}} \tan \frac{A}{4},\;{t}_{2} \mathrel{\text{:=}} \tan \frac{B}{4},\;{t}_{3} \mathrel{\text{:=}} \tan \frac{C}{4}. \] Note that from \( \tan \left( {\frac{A}{4} + \frac{B}{4} + \frac{C}{4}}\right) = 1 \), we have \[ 1 - {t}_{1} - {t}_{2} - {t}_{3} - {t}_{1}{t}... | Yes |
Proposition 3. The points \( {P}_{ + } \) and \( {P}_{ - } \) divide the segment IP harmonically. | Proof. This follows from their coordinates given in (12), (9), and (11).\n\nFrom the coordinates of \( P,{P}_{ + } \) and \( {P}_{ - } \), it is easy to see that \( {P}_{ + } \) and \( {P}_{ - } \) divide the segment \( {IP} \) harmonically. | Yes |
Proposition 4. The lines \( {AX},{BY},{CZ} \) are concurrent at a point \( Q \) with homogeneous barycentric coordinates\n\n\[ \n\\left( {\\tan \\frac{A}{4} : \\tan \\frac{B}{4} : \\tan \\frac{C}{4}}\\right) .\n\] | Proof. In Figure 5, we have\n\n\[ \n{BX} = \\frac{1}{2}\\left( {a + B{X}_{2} - {X}_{3}C}\\right)\n\]\n\n\[ \n= \\frac{1}{2}\\left( {a + \\frac{{r}_{2}}{r}\\left( {s - b}\\right) - \\frac{{r}_{3}}{r}\\left( {s - c}\\right) }\\right)\n\]\n\n\[ \n= \\frac{1}{2}\\left( {a + {IB} - {IC}}\\right)\n\]\n\n(from (1))\n\n\[ \n= ... | Yes |
Lemma 1. The pedal and cevian triangles of \( P \) are directly (or indirectly) similar if and only if \( P \) (or \( \bar{P} \) ) lies on the three circles \( A{B}^{\prime }{C}^{\prime }, B{C}^{\prime }{A}^{\prime }, C{A}^{\prime }{B}^{\prime } \) . | Proof. This is an immediate consequence of the properties of the Miquel point above. | No |
Lemma 2. \( A,{B}^{\prime },{C}^{\prime }, P \) are concyclic if and only if \( P \) lies on the circle \( {BCH} \) . | Proof. \( A,{B}^{\prime },{C}^{\prime } \) and \( P \) are concyclic \( \Leftrightarrow \measuredangle {B}^{\prime }P{C}^{\prime } = \measuredangle {B}^{\prime }A{C}^{\prime } \Leftrightarrow \measuredangle {BPC} = \) \( \measuredangle {BHC} \Leftrightarrow P \) lies on the circle \( {BCH} \) . | Yes |
Proposition 3. The pedal and cevian triangles of \( P \) are directly similar only in the trivial case of \( P = H \) . | Proof. By Lemma 1, the pedal and cevian triangles of \( P \) are directly similar if and only if \( P \) lies on the three circles \( A{B}^{\prime }{C}^{\prime }, B{C}^{\prime }{A}^{\prime }, C{A}^{\prime }{B}^{\prime } \) . By Lemma 2, \( P \) lies on the three circles \( {BCH},{CAH},{ABH} \) . Hence, \( P = H \) . | Yes |
Lemma 4. \( A,{B}^{\prime },{C}^{\prime },\bar{P} \) are concyclic if and only if \( {P}^{ * } \) lies on the circle \( {\Gamma }_{A} \) . | Proof. If \( P = p : q : r \), the circle \( {\Phi }_{A} \) passing through \( A,{B}^{\prime },{C}^{\prime } \) is given by\n\n\[ \n{a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy} - p\left( {x + y + z}\right) \left( {\frac{{c}^{2}}{p + q}y + \frac{{b}^{2}}{p + r}z}\right) = 0, \n\]\n\nand its inverse in the circumcircle is the... | Yes |
Proposition 5. The pedal and cevian triangles of \( P \) are indirectly similar if and only if \( P \) is the isogonal conjugate of the Parry reflection point. | Proof. By Lemma 1, the pedal and cevian triangles of \( P \) are indirectly similar if and only if \( \bar{P} \) lies on the three circles \( A{B}^{\prime }{C}^{\prime }, B{C}^{\prime }{A}^{\prime }, C{A}^{\prime }{B}^{\prime } \) . By Lemma \( 4,{P}^{ * } \) lies on each of the circles \( {\Gamma }_{A},{\Gamma }_{B},{... | Yes |
Theorem 7 (Gibert). The lines \( A{A}_{1}, B{B}_{1}, C{C}_{1} \) concur at the isogonal conjugate of \( Q \) . | This is the point \( {X}_{1263} \) in [3]. The points \( A, B, C,{A}^{\prime },{B}^{\prime },{C}^{\prime }, O, Q,{A}_{1},{B}_{1} \) , \( {C}_{1} \) all lie on the Neuberg cubic of triangle \( {ABC} \), which is the isogonal cubic with pivot the infinite point of the Euler line. This cubic is also the locus of all point... | Yes |
Lemma 3. The line \( O{I}_{1} \) is the Euler line of triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . | Proof. Triangle \( {ABC} \) is the tangential triangle of \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . It is known that the circumcenter of the tangential triangle lies on the Euler line. See, for example, [1, p.71]. It follows that \( O{I}_{1} \) is the Euler line of triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prim... | Yes |
Theorem 1 (Harcourt). If the distances from the vertices \( A, B, C \) to a tangent to the incircle of triangle \( {ABC} \) are \( {a}_{1},{b}_{1},{c}_{1} \) respectively, then the algebraic sum \( a{a}_{1} + b{b}_{1} + c{c}_{1} \) is twice of the area of triangle \( {ABC} \) . | The distances are signed. Distances to a line from points on opposite sides are opposite in sign, while those from points on the same side have the same sign. For the tangent lines to the incircle, we stipulate that the distance from the incenter is positive. For example, in Figure 1, when the tangent line \( \ell \) s... | Yes |
Proposition 2. Let \( \ell \) be a line passing through a point \( P \) with homogeneous barycentric coordinates \( \left( {x : y : z}\right) \) . If the signed distances from the vertices \( A, B, C \) to a line \( \ell \) are \( {d}_{1},{d}_{2},{d}_{3} \) respectively, then\n\n\[ \n{d}_{1}x + {d}_{2}y + {d}_{3}z = 0.... | Proof. It is enough to consider the case when \( \ell \) separates \( A \) from \( B \) and \( C \) . We take \( {d}_{1} \) as negative, and \( {d}_{2},{d}_{3} \) positive. See Figure 2. If \( {A}^{\prime } \) is the trace of \( P \) on the side line \( {BC} \), it is well known that\n\n\[ \n\frac{AP}{P{A}^{\prime }} =... | Yes |
Theorem 3. If the distances from the vertices \( A, B, C \) to a tangent to the \( A \) - excircle of triangle \( {ABC} \) are \( {a}_{1},{b}_{1},{c}_{1} \) respectively, then \( - a{a}_{1} + b{b}_{1} + c{c}_{1} = 2\bigtriangleup \) . Analogous statements hold for the \( B \) - and \( C \) -excircles. | Proof. Apply Proposition 2 to the line \( \ell \) through the excenter \( {I}_{a} = \left( {-a : b : c}\right) \) parallel to the tangent. If the distances from \( A, B, C \) to \( \ell \) are \( {d}_{1},{d}_{2},{d}_{3} \) respectively, then\n\n\[ - a{d}_{1} + b{d}_{2} + c{d}_{3} = 0. \]\n\nSince \( {a}_{1} = {d}_{1} +... | Yes |
Theorem 4. \( {d}_{a2}{d}_{b3}{d}_{c1} = {d}_{a3}{d}_{b1}{d}_{c2} \) . | Proof. Applying Theorem 3 to the tangent \( {\ell }_{a} \) of the \( B \) -excircle (respectively the \( C \) -excircle), we have\n\n\[ a{d}_{a1} - b{d}_{a2} + c{d}_{a3} = 2\bigtriangleup ,\]\n\n\[ a{d}_{a1} + b{d}_{a2} - c{d}_{a3} = 2\bigtriangleup \text{.} \]\n\nFrom these it is clear that \( b{d}_{a2} = c{d}_{a3} \)... | Yes |
Theorem 5. The extangent triangle bounded by \( {\ell }_{a},{\ell }_{b},{\ell }_{c} \)\n\n(1) has incenter \( {I}^{\prime } \) and inradius \( {2R} + r \) ;\n\n(2) is perspective with the excentral triangle at \( {I}^{\prime } \) ;\n\n(3) is homothetic to the tangential triangle at the internal center of similitude of ... | Proof. It is enough to locate the homothetic center in (3). This is the point which divides \( {I}^{\prime }O \) in the ratio \( {2R} + r : - R \), i.e.,\n\n\[ \frac{\left( {{2R} + r}\right) O - R\left( {{2O} - I}\right) }{R + r} = \frac{r \cdot O + R \cdot I}{R + r}, \]\n\nthe internal center of similitude of the circ... | Yes |
Proposition 1. Isotomic inscribed triangles have equal areas. | Proof. Let \( X, Y, Z \) be points on the sidelines \( {BC},{CA},{AB} \) dividing the sides in the ratios\n\n\[ \n{BX} : {XC} = x : 1 - x,\;{CY} : {YA} = y : 1 - y,\;{AZ} : {ZB} = z : 1 - z.\n\]\n\nIn terms of barycentric coordinates with respect to \( {ABC} \), we have\n\n\[ \nX = \left( {1 - x}\right) B + {xC},\;Y = ... | Yes |
Proposition 2. The centroids of isotomic inscribed triangles are symmetric with respect to the centroid of the reference triangle. | Proof. The expressions in (2) allow one to determine the centroid of triangle \( {XYZ} \) easily. This is the point\n\n\[ \n{G}_{XYZ} = \frac{1}{3}\left( {X + Y + Z}\right) = \frac{\left( {1 + y - z}\right) A + \left( {1 + z - x}\right) B + \left( {1 + x - y}\right) C}{3}.\n\]\n\n(6)\n\nOn the other hand, with the coor... | Yes |
Corollary 3. The intouch and extouch triangles have equal areas, and the midpoint of their centroids is the centroid of triangle \( {ABC} \) . | Proof. These follow from the fact that the intouch triangle \( {XYZ} \) and the extouch triangle \( {X}^{\prime }{Y}^{\prime }{Z}^{\prime } \) are isotomic, as is clear from the following data, where \( a, b, c \) denote the lengths of the sides \( {BC},{CA},{AB} \) of triangle \( {ABC} \), and \( s = \frac{1}{2}\left(... | Yes |
Corollary 4. The triangles \( {A}_{b}{B}_{c}{C}_{a} \) and \( {A}_{c}{B}_{a}{C}_{b} \) have equal areas. The centroids of these triangles are symmetric with respect to the centroid \( G \) of triangle \( {ABC} \) . | These follow because \( {A}_{b}{B}_{c}{C}_{a} \) and \( {A}_{c}{B}_{b}{C}_{a} \) are isotomic inscribed triangles. Indeed,\n\n\[ B{A}_{b} : {A}_{b}C = s : - \left( {s - a}\right) = 1 + \frac{s - a}{a} : - \frac{s - a}{a} = C{A}_{c} : {A}_{c}B, \]\n\n\[ C{B}_{c} : {B}_{c}A = s : - \left( {s - b}\right) = 1 + \frac{s - b... | Yes |
Proposition 5. The triangles of residual centroids of isotomic inscribed triangles have equal areas. | Proof. From the coordinates given above, we obtain the area of the triangle of residual centroids as\n\n\[ \n\frac{1}{27}\left| \begin{matrix} 2 + y - z & z & 1 - y \\ 1 - z & 2 + z - x & x \\ y & 1 - x & 2 + x - y \end{matrix}\right| \bigtriangleup \n\]\n\n\[ \n= \frac{1}{9}\left( {3 - x - y - z + {xy} + {yz} + {zx}}\... | Yes |
Proposition 6. Let \( {XYZ} \) and \( {X}^{\prime }{Y}^{\prime }{Z}^{\prime } \) be isotomic inscribed triangles of \( {ABC} \) . The centroids of the following five triangles are collinear:\n\n- \( G \) of triangle \( {ABC} \) ,\n\n- \( {G}_{XYZ} \) and \( {G}_{{X}^{\prime }{Y}^{\prime }{Z}^{\prime }} \) of the inscri... | Proof. The centroid \( \widetilde{G} \) is the point\n\n\[ \widetilde{G} = \frac{1}{9}\left( {\left( {3 + {2y} - {2z}}\right) A + \left( {3 + {2z} - {2x}}\right) B + \left( {3 + {2x} - {2y}}\right) C}\right) .\n\nWe obtain the centroid \( \widetilde{{G}^{\prime }} \) by interchanging \( \left( {x, y, z}\right) \leftrig... | Yes |
Proposition 7. The triangles of residual orthocenters of isotomic inscribed triangles have equal areas. | See Figure 4. This is an immediate corollary of the following proposition (see Figure 5), which in turn is a special case of a more general situation considered in Proposition 8 below. | No |
Proposition 9. Given a triangle \( {ABC} \), if pairs of parallel lines \( {\mathcal{L}}_{1B},{\mathcal{L}}_{1C} \) through \( B, C,{\mathcal{L}}_{2C},{\mathcal{L}}_{2A} \) through \( C, A \), and \( {\mathcal{L}}_{3A},{\mathcal{L}}_{3B} \) through \( A, B \) are constructed, and if\n\n\[ \n{P}_{a} = {\mathcal{L}}_{2C}... | Proof. We write \( Y = {\mathcal{L}}_{2C} \cap {\mathcal{L}}_{3A} \) and \( Z = {\mathcal{L}}_{2A} \cap {\mathcal{L}}_{3B} \) . Consider the parallelogram \( {AZ}{P}_{a}Y \) in Figure 6. If the points \( B \) and \( C \) divide the segments \( Z{P}_{a} \) and \( Y{P}_{a} \) in the ratios\n\n\[ \n{ZB} : B{P}_{a} = v : 1... | Yes |
Lemma 11. Let \( X, Y, Z \) be points on \( {BC},{CA},{AB} \) such that\n\n\[ \n{BX} : {XC} = w : v,\;{CY} : {YA} = {u}_{c} : w,\;{AZ} : {ZB} = v : {u}_{b}.\n\]\n\nThe distance between the circumcenters \( {O}_{b} \) and \( {O}_{c} \) is the hypotenuse of a right triangle with one side \( \frac{a}{2} \) and another sid... | Proof. The distance between \( {O}_{b} \) and \( {O}_{c} \) along the side \( {BC} \) is clearly \( \frac{a}{2} \) . We calculate their distance along the altitude on \( {BC} \) . The circumradius of \( B\widetilde{Z}X \) is clearly \( {R}_{b} = \frac{ZX}{2\sin B} \) . The distance of \( {O}_{b} \) above \( {BC} \) is\... | Yes |
Theorem 12. The triangle of residual circumcenters of \( P \) is homothetic to \( {ABC} \) if and only if \( P \) lies on the Lucas cubic. | It is well known that the Lucas cubic is the locus of point \( P \) whose cevian triangle is also the pedal triangle of a point \( Q \) . In this case, the circumcircles of \( {AYZ},{BZX} \) and \( {CXY} \) intersect at \( Q \), and the circumcenters \( {O}_{a},{O}_{b},{O}_{c} \) are the midpoints of the segments \( {A... | Yes |
Proposition 1. The lines \( A{A}_{a}, B{B}_{a}, C{C}_{a} \) concur at the point with homogeneous barycentric coordinates\n\n\[ \left( {\frac{1}{\cos A} : \frac{1}{\cos B} : \frac{1}{\cos C}}\right) . \] | Proof. Let \( {l}_{a} \) be the length of \( B{C}_{a} = {C}_{a}{A}_{a} = {A}_{a}{B}_{a} = {B}_{a}C \) . It is clear that the directed length \( B{A}_{a} = 2{l}_{a}\cos B \) and \( {A}_{a}C = 2{l}_{a}\cos C \), and \( B{A}_{a} : {A}_{a}C = \) \( \cos B : \cos C \) . For the same reason, \( C{B}_{b} : {B}_{b}A = \cos C :... | Yes |
Proposition 2. Let \( {A}^{\prime } \) be the intersection of the bisector of angle \( A \) with the circumcircle of triangle \( {ABC} \) . (a) \( {A}_{a} \) is the intersection of \( {BC} \) with the parallel to \( A{A}^{\prime } \) through the orthocenter H. | Proof. (a) The line joining \( {A}_{a} = \left( {0 : \cos C : \cos B}\right) \) to \( H = \left( {\frac{a}{\cos A} : \frac{b}{\cos B} : \frac{c}{\cos C}}\right) \) has equation \[ \left| \begin{matrix} 0 & \cos C & \cos B \\ \frac{a}{\cos A} & \frac{b}{\cos B} & \frac{c}{\cos C} \\ x & y & z \end{matrix}\right| = 0. \]... | Yes |
Proposition 3. The circumcircle of triangle \( A{B}_{a}{C}_{a} \) contains (i) the circumcenter \( O \) of triangle \( {ABC} \) ,(ii) the orthocenter \( {H}_{a} \) of triangle \( {A}_{a}{B}_{a}{C}_{a} \), and (iii) the midpoint of the arc \( {BAC} \) . | Proof. (i) is an immediate corollary of Proposition 2(b) above.\n\n(ii) Let \( {H}_{a} \) be the orthocenter of triangle \( {A}_{a}{B}_{a}{C}_{a} \) . It is clear that\n\n\[ \angle {B}_{a}{H}_{a}{C}_{a} = \pi - \angle {B}_{a}{A}_{a}{C}_{a} = \pi - \angle {BAC} = \pi - \angle {C}_{a}A{B}_{a}. \]\n\nIt follows that \( {H... | Yes |
Proposition 4. The circumcenter \( {O}_{a} \) of triangle \( {A}_{a}{B}_{a}{C}_{a} \) is equidistant from \( O \) and \( H \) . | Proof. Construct the circle through \( O \) and \( H \) with center \( Q \) on the line \( {BC} \) . We prove that the midpoint \( P \) of the arc \( {OH} \) on the opposite side of \( Q \) is the circumcenter \( {O}_{a} \) of triangle \( {A}_{a}{B}_{a}{C}_{a} \) . See Figure 6. It will follow that \( {O}_{a} \) is equ... | Yes |
Theorem 5. The circumcenters of triangles \( {A}_{a}{B}_{a}{C}_{a},{A}_{b}{B}_{b}{C}_{b} \), and \( {A}_{c}{B}_{c}{C}_{c} \) are collinear. The line containing them is the perpendicular bisector of the segment \( {OH} \) . | One can check without much effort that in homogeneous barycentric coordinates, the equation of this line is\n\n\[ \n\frac{\sin {3A}}{\sin A}x + \frac{\sin {3B}}{\sin B}y + \frac{\sin {3C}}{\sin C}z = 0.\n\] | Yes |
Lemma 6. \( {g}_{{P}^{ * }}\left( B\right) = \frac{4{\cos }^{2}B}{{g}_{P}\left( B\right) } \) . | Proof. If \( P = \left( {{g}_{P}\left( B\right) : 1 : 1}\right) \) for an isosceles triangle \( {ABC} \) with \( B = C \), then\n\n\[ \n{P}^{ * } = \left( {\frac{{\sin }^{2}A}{{g}_{P}\left( B\right) } : {\sin }^{2}B : {\sin }^{2}B}\right) = \left( {\frac{4{\cos }^{2}B}{{g}_{P}\left( B\right) } : 1 : 1}\right) \n\]\n\ns... | Yes |
Proposition 7. \( \Phi \left( {P}^{ * }\right) = \Phi {\left( P\right) }^{ * } \) . | Proof. We make use of Lemma 6.\n\n\[ \Phi \left( {P}^{ * }\right) = \left( {{g}_{{P}^{ * }}\left( A\right) \tan A : {g}_{{P}^{ * }}\left( B\right) \tan B : {g}_{{P}^{ * }}\left( C\right) \tan C}\right) \]\n\n\[ = \left( {\frac{4{\cos }^{2}A}{{g}_{P}\left( A\right) }\tan A : \frac{4{\cos }^{2}B}{{g}_{P}\left( B\right) }... | Yes |
Proposition 1. For \( i = 1,2,3 \), let \( {P}_{i} \) be finite points with homogeneous barycentric coordinates \( \left( {{x}_{i} : {y}_{i} : {z}_{i}}\right) \) with respect to triangle \( {ABC} \) . The oriented area of the triangle \( {P}_{1}{P}_{2}{P}_{3} \) is  . | No |
Proposition 2. For \( i = 1,2,3 \), let \( {\ell }_{i} \) be a finite line with equation \( {p}_{i}x + {q}_{i}y + {r}_{i}z = 0 \). The oriented area of the triangle bounded by the three lines \( {\ell }_{1},{\ell }_{2},{\ell }_{3} \) is \[ \frac{{\left| \begin{array}{lll} {p}_{1} & {q}_{1} & {r}_{1} \\ {p}_{2} & {q}_{2... | A proof of this proposition can be found in [5]. | No |
Theorem 3. \( \frac{{\Delta }_{1}{\Delta }_{2}}{{\Delta }^{2}} = \frac{{\left( U + V + W - UVW\right) }^{2}}{4{\left( UVW\right) }^{2}} \) . | Proof. The coordinates of \( {A}_{1},{B}_{1},{C}_{1} \) are\n\n\[ \n{A}_{1} = \left( {-{a}^{2} : {S}_{C} + {SU} : {S}_{B} + {SU}}\right) , \n\] \n\n\[ \n{B}_{1} = \left( {{S}_{C} + {SV} : - {b}^{2} : {S}_{A} + {SV}}\right) \n\] \n\n\[ \n{C}_{1} = \left( {{S}_{B} + {SW} : {S}_{A} + {SW} : - {c}^{2}}\right) . \n\] \n\nBy... | Yes |
Proposition 4. The triangles \( {ABC} \) and \( {A}_{2}{B}_{2}{C}_{2} \) are orthologic. The perpendiculars from the vertices of one triangle to the corresponding lines of the other triangle concur at the point \( J \) . | Proof. As \( {C}_{1}{B}_{1} \) bisects \( A{A}_{3} \), we see \( {A}_{3} \) lies on \( {B}_{c}{C}_{b} \) and \( {AJ} \bot {B}_{c}{C}_{b} \) . Similarly, we have \( {BJ} \bot {C}_{a}{A}_{c} \) and \( {CJ} \bot {A}_{b}{B}_{a} \) . The perpendiculars from \( A, B, C \) to the corresponding sides of \( {A}_{2}{B}_{2}{C}_{2... | Yes |
Proposition 5. The perpendiculars from \( {A}_{2},{B}_{2},{C}_{2} \) to the corresponding sides of \( {A}_{3}{B}_{3}{C}_{3} \) meet at the reflection of \( J \) in the circumcenter \( {O}_{3} \) of triangle \( {A}_{3}{B}_{3}{C}_{3} \) . | Proof. Since triangle \( {A}_{3}{B}_{3}{C}_{3} \) is the pedal triangle of \( J \) in \( {A}_{2}{B}_{2}{C}_{2} \), and \( {A}_{2}J \) passes through the circumcenter of triangle \( {A}_{2}{B}_{3}{C}_{3} \), the perpendicular from \( {A}_{2} \) to \( {B}_{3}{C}_{3} \) passes through the orthocenter of \( {A}_{2}{B}_{3}{... | Yes |
Theorem 6. The perpendicular bisectors of \( {B}_{c}{C}_{b},{C}_{a}{A}_{c},{A}_{b}{B}_{a} \) are concurrent at a point which is the reflection of \( J \) in the circumcenter \( {O}_{1} \) of triangle \( {A}_{1}{B}_{1}{C}_{1} \) . | Proof. Let \( {M}_{1} \) and \( {M}_{a} \) be the midpoints of \( {B}_{1}{C}_{1} \) and \( {B}_{c}{C}_{b} \) respectively. Note that \( {M}_{1} \) is also the midpoint of \( A{M}_{a} \) . Also, let \( {O}_{1} \) be the circumcenter of \( {A}_{1}{B}_{1}{C}_{1} \) , and the perpendicular bisector of \( {B}_{c}{C}_{b} \) ... | Yes |
Proposition 7. The ratios \( \frac{{d}_{i}}{{d}_{i}^{\prime }}, i = 1,2,3 \), are independent of triangle \( {ABC} \) . More precisely,\n\n\[ \frac{{d}_{1}}{{d}_{1}^{\prime }} = \frac{1}{V} + \frac{1}{W},\;\frac{{d}_{2}}{{d}_{2}^{\prime }} = \frac{1}{W} + \frac{1}{U},\;\frac{{d}_{3}}{{d}_{3}^{\prime }} = \frac{1}{U} + ... | Proof. Since \( A{A}_{4} \bot {C}_{b}{B}_{c} \), the circumcircle of the cyclic quadrilateral \( {A}_{3}B{A}_{4}C \) meets \( {C}_{b}{B}_{c} \) besides \( {A}_{3} \) at the antipode \( {A}_{5} \) of \( {A}_{4} \) . See Figure 10. Let \( f, g, h \) denote, for vectors, the compositions of a rotation by \( \frac{\pi }{2}... | Yes |
Proposition 1. 1. (1) Triangles \( {A}_{g}{B}_{g}{C}_{g} \) and \( {A}_{s}{B}_{s}{C}_{s} \) are symmetric about the midpoint of segment \( {PQ} \) . | Proof. It is clear from the coordinates given above that the segments \( {A}_{g}{A}_{s},{B}_{g}{B}_{s} \) , \( {C}_{g}{C}_{s},{PQ} \) have a common midpoint\n\n\[ \left( {f\left( {u + v + w}\right) + u\left( {f + g + h}\right) : \cdots : \cdots }\right) . \]\n\nThe six points therefore lie on a conic with this common m... | Yes |
Proposition 4. (a) Given \( P \), the locus of \( Q \) so that \( {A}_{s}{B}_{s}{C}_{s} \) is perspective to the pedal triangle of \( Q \) is the line \( {}^{2} \)\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{S}_{A}\left( {{S}_{B}v - {S}_{C}w}\right) \left( {-u{S}_{A} + v{S}_{B} + w{S}_{C}}\right) x = 0. \] | This line passes through the orthocenter \( H \) and the point\n\n\[ \left( {\frac{1}{{S}_{A}\left( {-u{S}_{A} + v{S}_{B} + w{S}_{C}}\right) } : \cdots : \cdots }\right) ,\n\nwhich can be constructed as the perspector of \( {ABC} \) and the cevian triangle of \( P \) in the orthic triangle. | No |
Proposition 7. Given \( P \), there are two (real) points \( Q \) for which triangles \( {A}_{g}{B}_{g}{C}_{g} \) and \( {A}_{s}{B}_{s}{C}_{s} \) are equilateral. These two points divide \( P{P}^{ \bot } \) harmonically. | The points \( {Q}_{1,2} \) from Proposition 7 can be constructed in the following way, using the fact that \( P,{G}_{s},{G}_{g} \) and \( {P}^{ \bot } \) are collinear.\n\nStart with a point \( {G}^{\prime } \) on \( P{P}^{ \bot } \) . We shall construct an equilateral triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime ... | Yes |
Proposition 8. Every line through \( P \) intersects the circumconic \( {\mathcal{C}}_{P} \) at two real points. | Proof. For the special case of the symmedian point \( K \) this is clear, since \( K \) is the interior of the circumcircle. Now, there is a homography \( \varphi \) fixing \( A, B, C \) and transforming \( P = \left( {u : v : w}\right) \) into \( K = \left( {{a}^{2} : {b}^{2} : {c}^{2}}\right) \) . It is given by\n\n\... | Yes |
Theorem 9. For \( i = 1,2 \), the two lines containing the degenerate triangles of the parasix configuration \( \operatorname{Parasix}\left( {P,{Q}_{i}}\right) \) are parallel to a tangent from \( P \) to the inscribed conic \( {\mathcal{C}}_{\ell } \) with perspector the trilinear pole of \( {\ell }_{P} \) . The two t... | For example, for \( P = K \), the symmedian point, the circumconic \( {\mathcal{C}}_{P} \) is the circumcircle. The orthocorrespondent is the point\n\n\[ \n{K}^{ \bot } = \left( {{a}^{2}\left( {{a}^{4} - {b}^{4} + 4{b}^{2}{c}^{2} - {c}^{4}}\right) : \cdots : \cdots }\right) \n\]\n\non the Euler line. The line \( \ell \... | Yes |
Lemma 1. Under inversion with respect to a circle, center \( P \), radius \( \rho \), the image of the circle center \( {P}^{\prime } \), radius \( {\rho }^{\prime } \), is the circle, radius \( \left| {\frac{{\rho }^{2}}{{d}^{2} - {\rho }^{\prime 2}} \cdot {\rho }^{\prime }}\right| \) and center \( Q \) which divides ... | \[ {PQ} : Q{P}^{\prime } = {\rho }^{2} : {d}^{2} - {\rho }^{2} - {\rho }^{\prime 2}, \] where \( d \) is the distance between \( P \) and \( {P}^{\prime } \) . Thus, \[ Q = \frac{\left( {{d}^{2} - {\rho }^{2} - {\rho }^{\prime 2}}\right) P + {\rho }^{2} \cdot {P}^{\prime }}{{d}^{2} - {\rho }^{\prime 2}}. \] | Yes |
Theorem 2. The Apollonius circle has center\n\n\\[ \nQ = \\frac{1}{4Rr}\\left( {\\left( {{r}^{2} + {4Rr} + {s}^{2}}\\right) O + {2Rr} \\cdot H - \\left( {{r}^{2} + {2Rr} + {s}^{2}}\\right) I}\\right) \n\\]\n\nand radius \\( \\frac{{r}^{2} + {s}^{2}}{4r} \\) . | Proof. It is well known that the distance between \\( O \\) and \\( I \\) is given by\n\n\\[ \nO{I}^{2} = {R}^{2} - {2Rr} \n\\]\n\nSince \\( S \\) and \\( N \\) divide the segments \\( {IG} \\) and \\( {OG} \\) in the ratio \\( 3 : - 1 \\) ,\n\n\\[ \nS{N}^{2} = \\frac{{R}^{2} - {2Rr}}{4}. \n\\]\n\nApplying Lemma 1 with... | Yes |
\[ \overrightarrow{OQ} = - \frac{{s}^{2} - {r}^{2} - {4Rr}}{4Rr} \cdot \overrightarrow{OK} \] | Proof. The oriented areas of the triangles \( {KHI},{OKI} \), and \( {OHK} \) are as follows.\n\n\[ \bigtriangleup \left( {KHI}\right) = \frac{\left( {a - b}\right) \left( {b - c}\right) \left( {c - a}\right) f}{{16}\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) \cdot \bigtriangleup }, \]\n\n\[ \bigtriangleup \left( {OKI}... | Yes |
Proposition 5. The homogeneous barycentric coordinates (with respect to triangle \( {ABC} \) ) of the centers of similitude of the Apollonius circle with the various circles are as follows. | Proof. The homogenous barycentric coordinates (with respect to triangle \( {OHI} \) ) of the centers of similitude of the Apollonius circle with the various circles are as follows. | No |
Proposition 7. The four lines \( {L}_{i} = 0, i = 2,3,4,5 \), are concurrent at the point\n\n\[ \n{X}_{650} = \left( {a\left( {b - c}\right) \left( {s - a}\right) : b\left( {c - a}\right) \left( {s - b}\right) : c\left( {a - b}\right) \left( {s - c}\right) }\right) .\n\] | It follows that this point is the radical center of the five circles above. From this we obtain a circle orthogonal to the five circles. | No |
Lemma 3. The barycentric coordinates of the excenters with respect to the medial triangle are\n\n\[ \n{\mathbf{I}}_{a} = \frac{s \cdot {\mathbf{A}}_{1} - \left( {s - c}\right) {\mathbf{B}}_{1} - \left( {s - b}\right) {\mathbf{C}}_{1}}{s - a}, \]\n\n\[ \n{\mathbf{I}}_{b} = \frac{-\left( {s - c}\right) {\mathbf{A}}_{1} +... | Proof. It is enough to compute the coordinates of the excenter \( {I}_{a} \) :\n\n\[ \n{\mathbf{I}}_{a} = \frac{-a \cdot \mathbf{A} + b \cdot \mathbf{B} + c \cdot \mathbf{C}}{b + c - a} \]\n\n\[ \n= \frac{-a\left( {{\mathbf{B}}_{1} + {\mathbf{C}}_{1} - {\mathbf{A}}_{1}}\right) + b\left( {{\mathbf{C}}_{1} + {\mathbf{A}}... | Yes |
Theorem 1. Consider \( \bigtriangleup {ABC} \) and \( \phi \in \left( {-\frac{\pi }{2},\frac{\pi }{2}}\right) \smallsetminus \{ 0\} \) . There are unique rhombi \( {\mathcal{R}}_{A}\left( \phi \right) = A{A}_{c}{A}_{a}{A}_{b},{\mathcal{R}}_{B}\left( \phi \right) = B{B}_{a}{B}_{b}{B}_{c} \) and \( {\mathcal{R}}_{C}\left... | Proof. It is enough to show the construction of \( {\mathcal{R}}_{A} = {\mathcal{R}}_{A}\left( \phi \right) \).\n\nLet \( {B}_{r} \) be the image of \( B \) after a rotation through \( - 2\bar{\phi } \) about \( A \), and \( {C}_{r} \) the image of \( C \) after a rotation through \( 2\bar{\phi } \) about \( A \) . The... | Yes |
Theorem 4. The circumcenters of triangles \( {ABC} \) and \( {A}_{a}{B}_{b}{C}_{c} \) are collinear with \( K\left( \phi \right) \) . | Proof. Since \( P = K\left( \phi \right) \) is the radical center of \( {A}^{\phi }\left( B\right) ,{B}^{\phi }\left( C\right) \) and \( {C}^{\phi }\left( A\right) \) we see that\n\n\[ \overline{PA} \cdot \overline{P{A}_{a}} = \overline{PB} \cdot \overline{P{B}_{b}} = \overline{PC} \cdot \overline{P{C}_{c}}, \]\n\nwhic... | Yes |
Corollary 6. ABC is the Kiepert triangle \( \mathcal{K}\left( {-\phi }\right) \) with respect to \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . | See [5, Proposition 4]. | No |
Lemma 7. The triangle \( {X}^{\prime }{Y}^{\prime }{Z}^{\prime } \) is perspective to \( {ABC} \) at \( \left( {x : y : z}\right) \) and to \( {XYZ} \) at \( \left( {u + x : v + y : w + z}\right) \) . | See, for example, \( \left\lbrack {1,§4}\right\rbrack \) . | No |
Proposition 8. Triangle \( {A}_{a}^{\prime }{B}_{b}^{\prime }{C}_{c}^{\prime } \) is perspective to \( {ABC} \) at \( K\left( {-{2\phi }}\right) \) . It is also perspective to \( {A}_{a}{B}_{b}{C}_{c} \) at | \[ {P}_{1}\left( \phi \right) = \left( {\frac{2{S}_{A} + S\csc {2\phi }}{\left( {{S}_{A} + {S}_{\phi }}\right) \left( {{S}_{A} + {S}_{2\phi }}\right) } : \frac{2{S}_{B} + S\csc {2\phi }}{\left( {{S}_{B} + {S}_{\phi }}\right) \left( {{S}_{B} + {S}_{2\phi }}\right) } : \frac{2{S}_{C} + S\csc {2\phi }}{\left( {{S}_{C} + {... | Yes |
Proposition 9. Triangle \( {A}^{\prime \prime }{B}^{\prime \prime }{C}^{\prime \prime } \) is perspective to (1) \( {ABC} \) at \( {P}_{2}\left( \phi \right) = \left( {\left( {{S}_{A} - {S}_{2\phi }}\right) \left( {{a}^{2} + S\csc {2\phi }}\right) : \cdots : \cdots }\right) \) | Proof. (1) is clear from the coordinates given in (4). Since \( \left( {{a}^{2} + S\csc {2\phi }}\right) \left( {{S}_{A} - S\cot {2\phi }}\right) - \left( {{S}_{B} - S\cot {2\phi }}\right) \left( {{S}_{C} - S\cot {2\phi }}\right) = \left( {{a}^{2}{S}_{A} - {S}_{BC}}\right) + {S}^{2}\csc {2\phi }\cot A - S\cot {2\phi }\... | Yes |
Lemma 1. Through the vertex \( A \) of a triangle \( {ABC} \), a straight line \( {AD} \) is drawn, cutting the side \( {BC} \) at \( D \) . Let \( P \) be the center of the circle \( {\mathcal{C}}_{1} \) which touches \( {DC} \) , \( {DA} \) at \( E, F \) and the circumcircle \( {\mathcal{C}}_{2} \) of \( {ABC} \) at ... | Proof. Let \( M, N \) be the points of intersection of \( {KE},{KF} \) with \( {\mathcal{C}}_{2} \), and \( J \) the point of intersection of \( {AM} \) and \( {EF} \) (see Figure 3). \( {KE} \) is the internal bisector of \( \angle {BKC} \) [8, Théorème 119]. The point \( M \) being the midpoint of the arc \( {BC} \) ... | Yes |
Theorem 2. Through the vertex \( A \) of a triangle \( {ABC} \), a straight line \( {AD} \) is drawn, cutting the side \( {BC} \) at \( D \) . I is the center of the incircle of triangle \( {ABC} \) . Let \( P \) be the center of the circle which touches \( {DC},{DA} \) at \( E, F \), and the circumcircle of \( {ABC} \... | Proof. According to the hypothesis, \( {QG} \bot {BC},{BC} \bot {PE} \) ; so \( {QG}//{PE} \) . By Lemma 1, \( {GH} \) and \( {EF} \) pass through \( I \) . Triangles \( {DHG} \) and \( {QGH} \) being isosceles\nin \( D \) and \( Q \) respectively, \( {DQ} \) is\n(1) the perpendicular bisector of \( {GH} \) ,\n(2) the ... | Yes |
Proposition 2. Let \( \ell \) be a line through \( G \) intersecting \( {\Gamma }_{P} \) at two points \( M \) and \( N \) . The antiorthocorrespondents of \( M \) and \( N \) are four collinear points on \( {\mathcal{Q}}_{P} \) . | Proof. Let \( {M}_{1},{M}_{2} \) be the antiorthocorrespondents of \( M \), and \( {N}_{1},{N}_{2} \) those of \( N \) . By Lemma 1, each of the lines \( {M}_{1}{M}_{2} \) and \( {N}_{1}{N}_{2} \) intersects \( \ell \) at the same point on the Kiepert hyperbola. Since they both contain \( H,{M}_{1}{M}_{2} \) and \( {N}... | Yes |
Let the medians of \( {ABC} \) meet \( {\Gamma }_{P} \) again at \( {A}_{g},{B}_{g},{C}_{g} \) . The an-tiorthocorrespondents of these points are the third and fourth intersections of \( {\mathcal{Q}}_{P} \) with the altitudes of \( {ABC} \) . | The antiorthocorrespondents of \( A \) are \( A \) and \( {H}_{a} \) . In this case, the third and fourth points on \( {AH} \) are symmetric about the second tangent to \( {\gamma }_{P} \) which is parallel to \( {BC} \) . The first tangent is the perpendicular bisector of \( A{H}_{a} \) with contact \( \left( {v + w :... | Yes |
Proposition 7. The points of tangency of the two bitangents to \( {\mathcal{Q}}_{P} \) passing through \( H \) are the antiorthocorrespondents of the points where the polar line of \( G \) in \( {\Gamma }_{P} \) meets \( {\Gamma }_{P} \) . | Proof. Consider a line \( {\ell }_{H} \) through \( H \) which is supposed to be tangent to \( {\mathcal{Q}}_{P} \) at two (orthoassociate) points \( M \) and \( N \) . The orthocorrespondents of \( M \) and \( N \) must lie on \( {\Gamma }_{P} \) and on the orthocorrespondent of \( {\ell }_{H} \) which is a line throu... | Yes |
Theorem 9. The circle \( {\mathcal{C}}_{M} \) centered at \( {T}_{M} \) passing through \( {M}_{1} \) and \( {M}_{2} \) is bitangent to \( {\mathcal{Q}}_{P} \) at those points and orthogonal to the polar circle. | This is a consequence of the following result from [1, tome 3, p.170]. A bicircular quartic is a special case of \ | No |
Corollary 10. \( {\mathcal{Q}}_{P} \) is the envelope of circles \( {\mathcal{C}}_{M}, M \in {\Gamma }_{P} \), centered on \( {\gamma }_{P} \) and orthogonal to the polar circle. | Construction. It is easy to draw \( {\gamma }_{P} \) since we know its center \( {\omega }_{P} \) . For \( m \) on \( {\gamma }_{P} \), draw the tangent \( {t}_{m} \) at \( m \) to \( {\gamma }_{P} \) . The perpendicular at \( m \) to \( {Hm} \) meets the perpendicular bisector of \( A{H}_{a} \) at a point which is the... | No |
Theorem 12. \( {\mathcal{Q}}_{P} \) is invariant under three other inversions whose poles are the vertices of the triangle which is self-polar in both the polar circle and \( {\gamma }_{P} \) . | Proof. This is a consequence of [1, tome 3, p.172]. | No |
Theorem 1 (Thébault). Let \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be the orthic triangle of \( {ABC} \) . The Euler lines of the triangles \( A{B}^{\prime }{C}^{\prime }, B{C}^{\prime }{A}^{\prime }, C{A}^{\prime }{B}^{\prime } \) are concurrent at the Jerabek center. | We shall make use of homogeneous barycentric coordinates. With reference to triangle \( {ABC} \), the vertices of the orthic triangle are the points\n\n\[ \n{A}^{\prime } = \left( {0 : {S}_{C} : {S}_{B}}\right) ,\;{B}^{\prime } = \left( {{S}_{C} : 0 : {S}_{A}}\right) ,\;{C}^{\prime } = \left( {{S}_{B} : {S}_{A} : 0}\ri... | Yes |
Proposition 2. With reference to triangle \( {ABC} \), the barycentric equations of the Euler lines of \( {\mathbf{T}}_{i}, i = 1,2,3 \), are\n\n\[ \left( {1 - {t}_{1}}\right) {S}_{AA}\left( {{S}_{B} - {S}_{C}}\right) \left( {x + y + z}\right) = {c}^{2}\left( {{S}_{AB} + {S}_{BC} - 2{S}_{CA}}\right) y - {b}^{2}\left( {... | Proof. It is enough to establish the equation of the Euler line \( {\mathcal{L}}_{1} \) of \( {\mathbf{T}}_{1} \) . This is the image of the Euler line \( {\mathcal{L}}_{1}^{\prime } \) of triangle \( A{B}^{\prime }{C}^{\prime } \) under the homothety \( \mathrm{h}\left( {A,1 - {t}_{1}}\right) \) . A point \( \left( {x... | Yes |
Theorem 3. The three Euler lines \( {\mathcal{L}}_{i}, i = 1,2,3 \), are concurrent if and only if\n\n\[ \n{t}_{1}{a}^{2}\left( {{S}_{B} - {S}_{C}}\right) {S}_{AA} + {t}_{2}{b}^{2}\left( {{S}_{C} - {S}_{A}}\right) {S}_{BB} + {t}_{3}{c}^{2}\left( {{S}_{A} - {S}_{B}}\right) {S}_{CC} = 0.\n\] | Proof. From the equations of \( {\mathcal{L}}_{i}, i = 1,2,3 \), given in Proposition 2, it is clear that the condition for concurency is\n\n\[ \n\left( {1 - {t}_{1}}\right) {a}^{2}\left( {{S}_{B} - {S}_{C}}\right) {S}_{AA} + \left( {1 - {t}_{2}}\right) {b}^{2}\left( {{S}_{C} - {S}_{A}}\right) {S}_{BB} + \left( {1 - {t... | Yes |
Theorem 4. Given a point \( Q \) in the plane of a scalene triangle \( {ABC} \), there is a unque triple of antiparallels \( {\ell }_{i}, i = 1,2,3 \), for which the Euler lines \( {\mathcal{L}}_{i}, i = 1,2,3 \) , are concurrent at \( Q \) . | Proof. Construct the parallel through \( Q \) to the Euler line of \( A{B}^{\prime }{C}^{\prime } \) to intersect the line \( {AH} \) at \( {O}_{a} \) . The circle through \( A \) with center \( {O}_{a} \) intersects \( {AC} \) and \( {AB} \) at \( {B}_{a} \) and \( {C}_{a} \) respectively. The line \( {B}_{a}{C}_{a} \... | Yes |
Proposition 6. For points \( {P}_{1},{P}_{2},{P}_{3} \) on \( {\mathcal{L}}_{\mathrm{p}}, Q\left( {P}_{1}\right), Q\left( {P}_{2}\right), Q\left( {P}_{3}\right) \) are points on \( {\mathcal{L}}_{\mathrm{q}} \) satisfying | \[ Q\left( {P}_{1}\right) Q\left( {P}_{2}\right) : Q\left( {P}_{2}\right) Q\left( {P}_{3}\right) = {P}_{1}{P}_{2} : {P}_{2}{P}_{3}. \] | No |
Proposition 7. If the Euler lines \( {\mathcal{L}}_{i}, i = 1,2,3 \), are concurrent, the homothetic center \( P\left( \mathbf{T}\right) \) of \( \mathbf{T} \) and the orthic triangle lies on the line \( {\mathcal{L}}_{\mathrm{p}} \) . | Proof. If we write \( P\left( \mathbf{T}\right) = \left( {x : y : z}\right) \) . From (3), we obtain\n\n\[ \n{t}_{1} = \frac{-x{S}_{A} + y{S}_{B} + z{S}_{C}}{2{S}_{A}},\;{t}_{2} = \frac{-y{S}_{B} + z{S}_{C} + x{S}_{A}}{2{S}_{B}},\;{t}_{3} = \frac{-z{S}_{C} + x{S}_{A} + y{S}_{B}}{2{S}_{C}}. \n\]\n\nSubstitution in (2) y... | Yes |
Proposition 9. The triangles \( \mathbf{T} \) and \( {ABC} \) are perspective if and only if\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\left( {{S}_{B} - {S}_{C}}\right) \left( {{t}_{1}{S}_{AA} - {t}_{2}{t}_{3}{S}_{BC}}\right) = 0. \] | Proof. From the coordinates of the vertices of \( \mathbf{T} \), it is straightforward to check that \( \mathbf{T} \) and \( {ABC} \) are perspective if and only if\n\n\[ {t}_{1}{a}^{2}{S}_{AA} + {t}_{2}{b}^{2}{S}_{BB} + {t}_{3}{c}^{2}{S}_{CC} + 2{S}_{ABC} = 0 \]\n\nor (4) holds. Since the area of triangle \( \mathbf{T... | Yes |
Theorem 10. If the triangle \( \mathbf{T} \) is nondegenerate and is perspective to \( {ABC} \), then the perspector lies on the Jerabek hyperbola of \( {ABC} \). | Proof. If triangles \( {A}_{1}{B}_{1}{C}_{1} \) and \( {ABC} \) are perspective at \( P = \left( {x : y : z}\right) \), then\n\n\[ \n{A}_{1} = \left( {u + x : y : z}\right) ,\;{B}_{1} = \left( {x : v + y : z}\right) ,\;{C}_{1} = \left( {x : y : w + z}\right) \n\]\n\nfor some \( u, v, w \) . Since the line \( {B}_{1}{C}... | Yes |
Theorem 12. The Euler lines of the four triangles \( \mathbf{T} \) and \( {\mathbf{T}}_{i}, i = 1,2,3 \), are concurrent if and only if\n\n\[ \n{t}_{1} = - \frac{{16}{S}^{2} \cdot {S}_{ABC} + t\left( {{a}^{2}{b}^{4}{c}^{4} - 4{S}_{ABC}\left( {3{S}^{2} - {S}_{AA}}\right) }\right) }{4{S}_{AA}\left( {{a}^{2}{b}^{2}{c}^{2}... | Proof. The equation of the Euler line \( {\mathcal{L}}_{i}, i = 1,2,3 \), can be rewritten as\n\n\[ \n{t}_{1}{S}_{A}\left( {{S}_{B} - {S}_{C}}\right) \left( {x + y + z}\right) + {S}_{AA}\left( {{S}_{B} - {S}_{C}}\right) x \]\n\n\[ \n\left. {+\left( {{S}_{AB}\left( {{S}_{B} - {S}_{C}}\right) - \left( {{S}_{AA} - {S}_{BB... | Yes |
Proposition 13. The triangle \( \mathbf{T} \) is perspective with \( {ABC} \) and its Euler line contains the common point of the Euler lines of \( {\mathbf{T}}_{i}, i = 1,2,3 \) precisely in the following three cases. | (1) \( t = 0 \), with perspector \( {X}_{74} \) and common point of Euler line \( {X}_{184} \) .\n\n(2) \( t = \frac{-{12}{a}^{2}{b}^{2}{c}^{2}{S}_{ABC}}{{a}^{4}{b}^{4}{c}^{4} - {12}{a}^{2}{b}^{2}{c}^{2}{S}_{ABC} - {16}{\left( {S}_{ABC}\right) }^{2}} \), with perspector \( K \) . | Yes |
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