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Proposition 5. The triangle of reflections \( {A}^{\left( a\right) }{B}^{\left( b\right) }{C}^{\left( c\right) } \) is the image of the reflection triangle of \( N \) under the homothety \( \mathrm{h}\left( {O,2}\right) \) . | \n\nFigure 6. Homothety of triangle of reflections and reflection triangle of \( N \)\n\nFrom this we conclude that\n\n(1) the center of the circle \( {A}^{\left( a\right) }{B}^{\left( b\right) }{C}^{\left( c\right... | Yes |
Proposition 6 ([10]). The reflection triangle of \( P \) is perspective with its anticevian triangle at the cevian quotient \( Q = H/P \), which is also the isogonal conjugate of \( P \) in the orthic triangle. | Proof. Let \( {P}_{a}{P}_{b}{P}_{c} \) be the cevian triangle of \( P \), and \( {P}^{a}{P}^{b}{P}^{c} \) the anticevian triangle. Since \( P \) and \( {P}^{a} \) divide \( A{P}_{a} \) harmonically, we have \( \frac{1}{A{P}^{a}} + \frac{1}{AP} = \frac{2}{A{P}_{a}} \) . If the perpendicular from \( P \) to \( {BC} \) in... | Yes |
Proposition 7. The reflection triangle of a point \( P \) is perspective with \( {ABC} \) if and only if \( P \) lies on the Neuberg cubic | \[ \mathop{\sum }\limits_{\text{cyclic }}\left( {{S}_{AB} + {S}_{AC} - 2{S}_{BC}}\right) u\left( {{c}^{2}{v}^{2} - {b}^{2}{w}^{2}}\right) = 0. \] | No |
Proposition 8. The reflection triangle of \( P \) is perspective with the triangle of reflections if and only if \( P \) lies on the cubic (2). The locus of the perspector \( Q \) is the Neuberg cubic (1). | Proof. Note that \( {A}^{\left( a\right) },{P}^{\left( a\right) } \) and \( {P}_{a} \) are collinear, since they are the reflections of \( A, P \) and \( {P}_{a} \) in \( {BC} \) . Similarly, \( {B}^{\left( b\right) },{P}^{\left( b\right) },{P}_{b} \) are collinear, so are \( {C}^{\left( c\right) },{P}^{\left( c\right)... | Yes |
Proposition 9. The triangle of reflections is perspective to the anticevian triangle of \( P \) if and only if \( P \) lies on the Napoleon cubic, i.e., the isogonal cubic \( {pK}\left( {K, N}\right) \) | \[ \mathop{\sum }\limits_{\text{cyclic }}\left( {{a}^{2}\left( {{b}^{2} + {c}^{2}}\right) - {\left( {b}^{2} - {c}^{2}\right) }^{2}}\right) u\left( {{c}^{2}{v}^{2} - {b}^{2}{w}^{2}}\right) = 0. \] | Yes |
Proposition 10. The three circles \( {P}^{\left( a\right) }{BC},{P}^{\left( b\right) }{CA} \), and \( {P}^{\left( c\right) }{AB} \) have a common point | \[ {r}_{1}\left( P\right) = \left( {\frac{u}{\left( {{b}^{2} + {c}^{2} - {a}^{2}}\right) u\left( {u + v + w}\right) - \left( {{a}^{2}{vw} + {b}^{2}{wu} + {c}^{2}{uv}}\right) } : \cdots : \cdots }\right) . \] It is easy to see that \( {r}_{1}\left( P\right) = H \) if and only if \( P \) lies on the circumcircle. If \( P... | Yes |
Proposition 11. Let \( {P}^{\left\lbrack a\right\rbrack }{P}^{\left\lbrack b\right\rbrack }{P}^{\left\lbrack c\right\rbrack } \) be the antipedal triangle of \( P = \left( {u : v : w}\right) \) . The reflections of the circles \( {P}^{\left\lbrack a\right\rbrack }{BC} \) in \( {BC},{P}^{\left\lbrack b\right\rbrack }{CA... | Proof. Since \( B, P, C \), and \( {P}^{\left\lbrack a\right\rbrack } \) are concyclic, so are their reflections in the line \( {BC} \) . The circle \( {P}^{\left\lbrack a\right\rbrack }{BC} \) is identical with the reflection of the circle \( {P}^{\left( a\right) }{BC} \) in \( {BC} \) ; similarly for the other two ci... | Yes |
Proposition 12. Let \( {P}_{\left\lbrack a\right\rbrack }{P}_{\left\lbrack b\right\rbrack }{P}_{\left\lbrack c\right\rbrack } \) be the pedal triangle of \( P = \left( {u : v : w}\right) \) . The reflections of the circles \( A{P}_{\left\lbrack b\right\rbrack }{P}_{\left\lbrack c\right\rbrack } \) in \( {P}_{\left\lbra... | \[ {r}_{2}\left( P\right) = \left( {{a}^{2}\left( {2{a}^{2}{b}^{2}{c}^{2}u + {c}^{2}\left( {{\left( {a}^{2} + {b}^{2} - {c}^{2}\right) }^{2} - 2{a}^{2}{b}^{2}}\right) v + {b}^{2}\left( {{\left( {c}^{2} + {a}^{2} - {b}^{2}\right) }^{2} - 2{c}^{2}{a}^{2}}\right) w}\right) }\right) \left. {\cdot \left( {{b}^{2}{c}^{2}{u}^... | Yes |
Theorem 13 (Bailey [1, Theorem 5]). The isogonal conjugates of \( P \) and \( {r}_{1}\left( P\right) \) are inverse in the circumcircle. | Proof. Let \( P = \left( {u : v : w}\right) \), so that \( {P}^{ * } = \left( {{a}^{2}{vw} : {b}^{2}{wu} : {c}^{2}{uv}}\right) \) . From the above formula,\n\n\[{\left( {P}^{ * }\right) }^{-1} = \left( {{a}^{2}{vw}\left( {{a}^{2}{vw} + \left( {{a}^{2} - {b}^{2}}\right) {uv} + \left( {{a}^{2} - {c}^{2}}\right) {wu} - \l... | Yes |
Proposition 14. The inversive images of \( {A}^{\left( a\right) },{B}^{\left( b\right) },{C}^{\left( c\right) } \) in the circumcircle are perspective with \( {ABC} \) at \( {N}^{ * } \) . | Proof. These inversive images are\n\n\[ \n{\left( {A}^{\left( a\right) }\right) }^{-1} = \left( {-{a}^{2}\left( {{S}^{2} - 3{S}_{AA}}\right) : {b}^{2}\left( {{S}^{2} + {S}_{AB}}\right) : {c}^{2}\left( {{S}^{2} + {S}_{CA}}\right) }\right) , \n\] \n\n\[ \n{\left( {B}^{\left( b\right) }\right) }^{-1} = \left( {{a}^{2}\lef... | Yes |
Corollary 15 (Musselman [32]). The circles \( {AO}{A}^{\left( a\right) } \) , \( {BO}{B}^{\left( b\right) } \) , \( {CO}{C}^{\left( c\right) } \) are coaxial with common points \( O \) and \( {\left( {N}^{ * }\right) }^{-1} \) . | Proof. Invert the configuration in Proposition 14 in the circumcircle. | No |
Proposition 17. The inversive images of \( {P}_{a},{P}_{b},{P}_{c} \) in the circumcircle form a triangle perspective with \( {ABC} \) if and only if \( P \) lies on the circumcircle or the Euler line. | (a) If \( P \) lies on the circumcircle, the perspector is the isogonal conjugate of the inferior of \( P \) . The locus is the isogonal conjugate of the nine-point circle (see Figure 18).\n\n\n\nFigure 18. Isogona... | Yes |
Proposition 19. Let \( \begin{array}{l} X, Y, Z \\ {X}^{\prime },{Y}^{\prime }{Z}^{\prime } \end{array} \) be two triads of points. The triad of circles \( X{Y}^{\prime }{Z}^{\prime } \) , \( Y{Z}^{\prime }{X}^{\prime } \) and \( Z{X}^{\prime }{Y}^{\prime } \) have a common point if and only if the triad of circles \( ... | Proof. Let \( Q \) be a common point of the triad of circles \( X{Y}^{\prime }{Z}^{\prime }, Y{Z}^{\prime }{X}^{\prime }, Z{X}^{\prime }{Y}^{\prime } \) . Inversion with respect to a circle, center \( Q \) transforms the six points \( X, Y, Z,{X}^{\prime } \) , \( {Y}^{\prime },{Z}^{\prime } \) into \( x, y, z,{x}^{\pr... | Yes |
Proposition 22. Given \( P = \left( {u : v : w}\right) \), let \( X, Y, Z \) be the reflections of \( A, B, C \) in \( P \) .\n\n(a) The circles \( {AYZ},{BZX},{CXY} \) have a common point | \[ {r}_{3}\left( P\right) = \left( {\frac{1}{{c}^{2}v\left( {w + u - v}\right) - {b}^{2}w\left( {u + v - w}\right) } : \cdots : \cdots }\right) \]\nwhich is also the fourth intersection of the circumcircle and the circumconic with center \( P \) (see Figure 25). | No |
Proposition 23. The reflection triangle of \( P \) in the medial triangle is perspective with \( {ABC} \) if and only if \( P \) lies on the Euler line or the nine-point circle of \( {ABC} \) . | (a) If \( P \) lies on the Euler line, the perspector traverses the Jerabek hyperbola.\n\n(b) If \( P \) lies on the nine-point circle, the perspector is the infinite point which is the isogonal conjugate of the superior of \( P \) . | No |
Proposition 25. The reflection triangle of \( P \) in the orthic triangle \( {H}_{a}{H}_{b}{H}_{c} \) is perspective with \( {ABC} \) if and only if \( P \) lies on the cubic\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\frac{u}{{b}^{2} + {c}^{2} - {a}^{2}}\left( {f\left( {c, a, b}\right) {v}^{2} - f\left( {b, c, a}\rig... | where\n\n\[ f\left( {a, b, c}\right) = {a}^{4}\left( {{b}^{2} + {c}^{2}}\right) - 2{a}^{2}\left( {{b}^{4} - {b}^{2}{c}^{2} + {c}^{4}}\right) + \left( {{b}^{2} + {c}^{2}}\right) {\left( {b}^{2} - {c}^{2}\right) }^{2}. \] | Yes |
Proposition 29. Let \( \ell \) be a line through a given point \( P \), and \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) the reflections of \( A, B, C \) in \( \ell \) . The lines \( {A}^{\prime }P,{B}^{\prime }P,{C}^{\prime }P \) intersect the sidelines \( {BC},{CA} \) , \( {AB} \) respectively at \( X, Y, Z \) . T... | Proof. Let \( \ell \) be the line joining \( P = \left( {u : v : w}\right) \) and \( Q = \left( {x : y : z}\right) \) . The line \( \mathcal{L} \) containing \( X, Y, Z \) is\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\frac{u\mathbb{X}}{\left( {{b}^{2}{u}^{2} + 2{S}_{C}{uv} + {a}^{2}{v}^{2}}\right) {\left( uz - wx\rig... | Yes |
Proposition 30 ([9]). The reflection triangle of \( P = \left( {u : v : w}\right) \) in the cevian triangle of \( P \) is perspective with \( {ABC} \) at\n\n\[ \n{r}_{5}\left( P\right) = \left( {u\left( {-\frac{{a}^{2}}{{u}^{2}} + \frac{{b}^{2}}{{v}^{2}} + \frac{{c}^{2}}{{w}^{2}} + \frac{{b}^{2} + {c}^{2} - {a}^{2}}{vw... | Proof. Relative to the triangle \( {P}_{a}{P}_{b}{P}_{c} \), the coordinates of \( P \) are \( (v + w : w + u : \) \( u + v) \) . Similarly, those of \( A, B, C \) are\n\n\( \left( {-\left( {v + w}\right) : w + u : u + v}\right) ,\;\left( {v + w : - \left( {w + u}\right) : u + v}\right) ,\;\left( {v + w : w + u : - \le... | Yes |
Proposition 32 (Musselman [33]). Given a point \( P \), let \( X, Y, Z \) be the reflections of the orthocenter \( H \) in the lines \( {AP},{BP},{CP} \) respectively. The circles \( {APX} \) , \( {BPY},{CPZ} \) have a second common point | \[ {r}_{9}\left( P\right) = \left( {\frac{1}{-2{S}^{2}{vw} + {S}_{A}\left( {{a}^{2}{vw} + {b}^{2}{wu} + {c}^{2}{uv}}\right) } : \cdots : \cdots }\right) . \] | Yes |
Proposition 33 ([8]). Given a point \( P \) with reflections \( X, Y, Z \) in the perpendicular bisectors of \( {BC},{CA},{AB} \) respectively, the triangle \( {XYZ} \) is perspective with \( {ABC} \) if and only if \( P \) lies on the circumcircle or the Euler line. | (a) If \( P \) is on the circumcircle, the lines \( {AX},{BY},{CZ} \) are parallel. The per-spector is the isogonal conjugate of \( P \) (see Figure 42).\n\n(b) If \( P = {E}_{t} \) on the Euler line, then the perspector is \( {E}_{{t}^{\prime }}^{ * } \) on the Jerabek hyperbola, where\n\n\[ \n{t}^{\prime } = \frac{{a... | Yes |
Proposition 34. The reflections of \( {r}_{1}\left( {E}_{t}\right) \) in the altitudes are perspective with \( {ABC} \) at \( {r}_{1}\left( {E}_{{t}^{\prime }}\right) \) if and only if | \[ t{t}^{\prime } = \frac{{a}^{2}{b}^{2}{c}^{2}}{{a}^{2}{b}^{2}{c}^{2} - \left( {{b}^{2} + {c}^{2} - {a}^{2}}\right) \left( {{c}^{2} + {a}^{2} - {b}^{2}}\right) \left( {{a}^{2} + {b}^{2} - {c}^{2}}\right) }.\] | Yes |
Proposition 35 ([20,44]). The reflections of \( {I}_{b}{I}_{c} \) in \( A{I}_{a},{I}_{c}{I}_{a} \) in \( B{I}_{b} \), and \( {I}_{a}{I}_{b} \) in \( C{I}_{c} \) bound a triangle perspective with \( {ABC} \) at\n\n\[ \n{X}_{81} = \left( {\frac{a}{b + c} : \frac{b}{c + a} : \frac{c}{a + b}}\right) \n\] \n\n(see Figure 45... | Proof. The equations of these reflection lines are\n\n\[ \n- {bcx} + c\left( {c + a - b}\right) y + b\left( {a + b - c}\right) z = 0, \n\] \n\n\[ \nc\left( {b + c - a}\right) x - {cay} + a\left( {a + b - c}\right) z = 0, \n\] \n\n\[ \nb\left( {b + c - a}\right) x + a\left( {c + a - b}\right) y - {abz} = 0. \n\] \n\nThe... | Yes |
Proposition 36. The reflections of \( {BC} \) in \( A{I}_{a},{CA} \) in \( B{I}_{b} \), and \( {AB} \) in \( C{I}_{c} \) bound a triangle perspective with \( {I}_{a}{I}_{b}{I}_{c} \) at | \[ {X}_{55} = \left( {{a}^{2}\left( {b + c - a}\right) : {b}^{2}\left( {c + a - b}\right) : {c}^{2}\left( {a + b - c}\right) }\right) . \] | Yes |
Proposition 38. The reflections of the lines \( {AP} \) in \( {D}_{b}{D}_{c},{BP} \) in \( {D}_{c}{D}_{a} \), and \( {CP} \) in \( {D}_{a}{D}_{b} \) are concurrent. | Proof. Denote by \( x, y, z \) the distances of \( P \) from \( A, B, C \) respectively. The point \( {D}_{a} \) divides \( {BC} \) in the ratio \( y : z \) and has homogeneous barycentric coordinates \( \left( {0 : z : y}\right) \) . Similarly, \( {D}_{b} = \left( {z : 0 : x}\right) \) and \( {D}_{c} = \left( {y : x :... | Yes |
Proposition 39. The locus of \( P \) for which the reflections of the cevians \( {AP},{BP} \) , \( {CP} \) in the respective sidelines of the intouch triangle is the union of the circumcircle and the line OI: | (a) If \( P \) is on the circumcircle, the cevians are parallel, with infinite point the isogonal conjugate of \( P \) (see Figure 49).\n\n\n\nFigure 49. Reflections of cevians of \( P \) in the sidelines of the in... | Yes |
Proposition 1. Triangle \( {A}_{P}{B}_{P}{C}_{P} \) is oppositely similar to \( {ABC} \) . | Proof. The square lengths of the sides are\n\n\[ \n{\left( {B}_{P}{C}_{P}\right) }^{2} = k \cdot {a}^{2},\;{\left( {C}_{P}{A}_{P}\right) }^{2} = k \cdot {b}^{2},\;{\left( {A}_{P}{B}_{P}\right) }^{2} = k \cdot {c}^{2}, \n\] \n\nwhere \n\n\[ \n k = \frac{{S}^{2}\left( {\mathop{\sum }\limits_{\text{cyclic }}{a}^{4}{S}_{AA... | Yes |
Proposition 2. Triangle \( {A}_{P}{B}_{P}{C}_{P} \) is perspective with the medial triangle at\n\n\[ \left( {\frac{x}{{S}_{A}}\left( {-\frac{x}{{S}_{A}} + \frac{y}{{S}_{B}} + \frac{z}{{S}_{C}}}\right) : \cdots : \cdots }\right) . \] | Proof. Let \( {DEF} \) be the medial triangle. The line joining \( D{A}_{P} \) has equation\n\n\[ {S}_{A}\left( {{S}_{C}y - {S}_{B}z}\right) \mathbb{X} + {S}_{BC}x\mathbb{Y} - {S}_{BC}x\mathbb{Z} = 0, \]\n\nor\n\[ \left( {\frac{y}{{S}_{B}} - \frac{z}{{S}_{C}}}\right) \mathbb{X} + \frac{x}{{S}_{A}}\left( {\mathbb{Y} - \... | Yes |
Proposition 3. Triangle \( {A}_{P}{B}_{P}{C}_{P} \) is perspective with \( {ABC} \) if and only if \( P \) lies on the Jerabek hyperbola of \( {ABC} \) . In this case, the perspector traverses the Euler line. | Proof. The lines \( A{A}_{P}, B{B}_{P}, C{C}_{P} \) have equations\n\n\[ \left( {{S}_{C}\left( {{S}_{A} + {S}_{B}}\right) y + {S}_{B}\left( {{S}_{C} - {S}_{A}}\right) z}\right) \mathbb{Y} + \left( {{S}_{C}\left( {{S}_{A} - {S}_{B}}\right) y - {S}_{B}\left( {{S}_{C} + {S}_{A}}\right) z}\right) \mathbb{Z} = 0, \]\n\n\[ \... | Yes |
Proposition 4. The triangles \( {ABC} \) and \( {A}_{P}{B}_{P}{C}_{P} \) are orthologic. | (a) The perpendiculars from \( A \) to \( {B}_{P}{C}_{P}, B \) to \( {C}_{P}{A}_{P} \), and \( C \) to \( {A}_{P}{B}_{P} \) intersect at the point\n\n\[ \nQ = \left( {\frac{{a}^{2}{S}_{A}}{-\frac{x}{{S}_{ABC}} + \frac{y}{{b}^{2}{S}_{BB}} + \frac{z}{{c}^{2}{S}_{CC}}} : \frac{{b}^{2}{S}_{B}}{\frac{x}{{a}^{2}{S}_{AA}} - \... | Yes |
Theorem 1. A circle (O) passing through \( B, C \), tangent to the circles (I) (internally) and \( \left( P\right) \) (externally), exists if and only if one of the intangents of \( \left( P\right) \) and (I) is parallel to \( {BC} \) . | Proof. Consider \( P \) in general position (see Figure 3). Since the two right triangles\n\nformed by \( {AI} \), the two intangents, and radii of \( \left( I\right) \) are congruent, we see that\n\n(1) \( {DE},{FG} \), and \( {AI} \) are concurrent in \( X \) ,\n\n(2) the triangles \( {AEX} \) and \( {AGX} \) are con... | Yes |
Theorem 1. The pencil of lines through the circumcenter \( O \) is transformed to a pencil of rectangular hyperbolas passing through the vertices and orthocenter \( H \) of triangle \( {ABC} \) . The centers of these rectangular hyperbolas lie on the Euler circle. | Proof. The isogonal transform of \( O \) is \( {O}^{\prime } = H \) . Thus all the conics in the pencil pass through \( A, B, C, H \), an orthocentric set, i.e., any one of the points is the orthocenter of the other three. Hence the pencil of conics is a pencil of rectangular hyperbolas.\n\nIt is a well-known theorem o... | Yes |
Corollary 3. Let \( \mathcal{C} \) be an oval with center \( O \), having a circle \( \Gamma \) centered at \( O \) as orthoptic curve. By continuity, there is a point \( W \in \mathcal{C} \) such that the associated maximal-perimeter inscribed parallelogram is a rhombus, WNES (see Figure 3). Then \( \mathcal{C} \) is ... | The proof of the Corollary is a simple consequence of all the facts considered in the proof of the Theorem. | No |
Lemma 2. If \( x, y \) and \( z \) are the distances from the incenter to the vertices of a triangle, then the inradius \( r \) is a root of the cubic equation\n\n\[ \n{2xyz}{r}^{3} + \left( {{x}^{2}{y}^{2} + {y}^{2}{z}^{2} + {z}^{2}{x}^{2}}\right) {r}^{2} - {x}^{2}{y}^{2}{z}^{2} = 0.\n\] | Proof. If \( \alpha ,\beta \) and \( \gamma \) are the angles between these distances and the inradius (see Figure 5), we have \( \alpha + \beta + \gamma = \pi \), so \( \cos \left( {\alpha + \beta }\right) = \cos \left( {\pi - \gamma }\right) \) and it follows that \( \cos \alpha \cos \beta - \sin \alpha \sin \beta = ... | Yes |
Lemma 3. If \( u, v \) and \( z \) are the distances from an excenter to the vertices of a triangle, then the corresponding exradius \( {r}_{c} \) is a root of the cubic equation\n\n\[ \n{2uvz}{r}_{c}^{3} - \left( {{u}^{2}{v}^{2} + {v}^{2}{z}^{2} + {z}^{2}{u}^{2}}\right) {r}_{c}^{2} + {u}^{2}{v}^{2}{z}^{2} = 0. \n\] | Proof. Define angles \( \alpha ,\beta \) and \( \gamma \) to be between \( u, v, z \) and the sides of the triangle \( {ABC} \) or their extensions (see Figure 6). Then \( {2\alpha } + A = \pi ,{2\beta } + B = \pi \) and \( {2\gamma } = C \) . From the sum of angles in a triangle, \( A + B + C = \pi \), this simplifies... | Yes |
The triangle \( {K}_{a, a}{K}_{b, c}{K}_{c, c} \) of contact points is perspective with \( \mathbf{T} \) at a point with homogeneous barycentric coordinates\n\n\[ \left( {\frac{s - a}{{a}^{2}} : \frac{s - b}{{b}^{2}} : \frac{s - c}{{c}^{2}}}\right) . \] | Proof. The coordinates of \( {K}_{a, a},{K}_{b, b},{K}_{c, c} \) can be rewritten as\n\n\[ {K}_{a, a} = \left( {-\frac{{\left( b + c\right) }^{2}\left( {s - b}\right) \left( {s - c}\right) }{{b}^{2}{c}^{2}s} : \frac{s - b}{{b}^{2}} : \frac{s - c}{{c}^{2}}}\right) ,\]\n\n\[ {K}_{b, b} = \left( {\frac{s - a}{{a}^{2}} : -... | Yes |
Lemma 1. Let \( {BC} \) be a chord of a conic with center \( O \), and \( M \) the midpoint of \( {BC} \) . If the tangents to the conic at \( B \) and \( C \) intersect at \( A \), then the points \( A, M \) , O are collinear. | Proof. The harmonic conjugate \( S \) of the point \( M \) relative to \( B, C \) is the point at infinity of the line \( {BC} \) . The center \( O \) is the midpoint of every chord passing through \( O \) . Hence, the polar of \( O \) is the line at infinity. The polar of \( A \) is the line \( {BC} \) (see Figure 1).... | Yes |
Theorem 2. The foci \( {F}_{a},{F}_{b},{F}_{c} \) of the three parabolas are the vertices of the second Brocard triangle of \( {ABC} \) and hence are lying on the Brocard circle. The triangles \( {ABC} \) and \( {F}_{a}{F}_{b}{F}_{c} \) are perspective at the Lemoine point \( K \) . | Proof. With reference to Figure 3, in triangle \( {ADE} \), the median \( A{A}_{1} \) is parallel to the axis of the \( A \) -parabola \( {\mathcal{P}}_{a} \) . Hence, \( A{A}_{1} \) is an altitude of triangle \( {ADE} \) . The line \( A{F}_{a} \) is a diameter of the circumcircle of \( {ADE} \) and is the isogonal con... | Yes |
For an arbitrary point \( Q \) the line \( {X}^{\prime }Q \) intersects the line \( {AP} \) at the point \( R \), and we define the mapping \( h\left( Q\right) = S \), where \( S \) is the harmonic conjugate of \( Q \) with respect to \( {X}^{\prime } \) and \( R \) . The mapping \( h \) swaps the conics \( \left( {\ma... | The mapping \( h \) is involutive because \( h\left( Q\right) = S \) if and only if \( h\left( S\right) = Q \) . Since the line \( {AP} \) is the polar of \( {X}^{\prime } \) relative to the pair of lines \( {AB},{AC} \) we have \( h\left( A\right) = A, h\left( P\right) = P, h\left( C\right) = B \), so that \( h\left( ... | Yes |
Theorem 4. The line \( {Z}_{b}{Y}_{c} \) is a common tangent of \( \left( {\mathcal{B}}_{P}\right) \) and \( \left( {\mathcal{C}}_{P}\right) \), so is \( {X}_{c}{Z}_{a} \) of \( \left( {\mathcal{C}}_{P}\right) \) and \( \left( {\mathcal{A}}_{P}\right) \), and \( {Y}_{a}{X}_{b} \) of \( \left( {\mathcal{A}}_{P}\right) \... | Proof. We need only prove the case \( {Y}_{c}{Z}_{b} \) . The line has equation\n\n\[ \n{vwx} + {4wuy} + {4uvz} = 0.\n\]\n\nThis is tangent to \( \left( {\mathcal{B}}_{P}\right) \) at \( {Z}_{b} \) and to \( \left( {\mathcal{C}}_{P}\right) \) at \( {Y}_{c} \) as the following calculation confirms.\n\n\[ \n\left( \begin... | Yes |
Theorem 5. The six points \( {X}_{b},{X}_{c},{Y}_{c},{Y}_{a},{Z}_{a},{Z}_{b} \) lie on a conic (see Figure 7). | Proof. It is easy to verify that the six points satisfy the equation of the conic\n\n\[ 2{v}^{2}{w}^{2}{x}^{2} + 2{w}^{2}{u}^{2}{y}^{2} + 2{u}^{2}{v}^{2}{z}^{2} + 7{u}^{2}{vwyz} + {7u}{v}^{2}{wzx} + {7uv}{w}^{2}{xy} = 0, \] | Yes |
Theorem 5. The mapping \( \mu \) interchanges the following pairs of central points\n\n<table><tr><td>circumcenter \( O = {X}_{3} \)</td><td>Focus of Kiepert parabola \( E = {X}_{110} \)</td></tr><tr><td>orthocenter \( H = {X}_{4} \)</td><td>Jerabek center \( J = {X}_{125} \)</td></tr><tr><td>Lemoine point \( L = {X}_{... | All the \( \mu \) -coupling of points can be proved through the argument of \( §8 \) . The \( \mu \) -coupling of lines and circles follows from properties of Möbius transformations. Some central points mentioned in Theorem 5 are not listed in [3] but can be rather naturally characterized:\n\n(i) \( {K}^{\mu } \) (whos... | Yes |
Theorem 6. The foci of the Steiner inellipse of a triangle are the intersections of the major axis and the circle through the Fermat points and with center on the minor axis. | A direct analytic proof of this statement is achieved by considering, as usual, separate cases as shown below:\n\nCase 1: \( a > b.W = \left\lbrack {\frac{s}{3},\frac{{s}^{2}}{9p}}\right\rbrack \) . The circle in (4) has equation\n\n\[ \n{x}^{2} + {y}^{2} - x\frac{2s}{3} - y\frac{2{s}^{2}}{9p} + \frac{{s}^{2} + 9{p}^{2... | Yes |
Lemma 1. If \( \mathbf{Q} \) is orthodiagonal, the valtitudes \( {V}_{i}{H}_{i} \) and \( {V}_{i + 1}{H}_{i + 1}\left( {i = 1,2,3,4}\right) \) with respect to a \( v \) -parallelogram \( \mathbf{V} \) of \( \mathbf{Q} \) meet on the diagonal \( {A}_{i + 1}{A}_{i + 3} \) of \( \mathbf{Q} \) . | Proof. Let \( \mathbf{Q} \) be orthodiagonal and \( \mathbf{V} \) a v-parallelogram of \( \mathbf{Q} \) . Let us prove that the valtitudes \( {V}_{1}{H}_{1} \) and \( {V}_{2}{H}_{2} \) meet on the line \( {A}_{2}{A}_{4} \) (see Figure 4). The altitudes \( {V}_{3}{K}_{3},{V}_{4}{K}_{4},{A}_{4}H \) of triangle \( {V}_{3}... | Yes |
Theorem 2. Let \( \mathbf{Q} \) be orthodiagonal. Let \( \mathbf{V} \) be a \( v \) -parallelogram of \( \mathbf{Q} \) and \( {\mathbf{Q}}_{0} \) be the orthic quadrilateral of \( \mathbf{Q} \) relative to \( \mathbf{V} \) . The vertices of \( \mathbf{V} \) and those of \( {\mathbf{Q}}_{o} \) lie on the same circle. | Proof. In fact, since \( \mathbf{Q} \) is orthodiagonal, \( \mathbf{V} \) is a rectangle and it is inscribed in the circle \( \mathcal{C} \) of diameter \( {V}_{1}{V}_{3} = {V}_{2}{V}_{4} \) . The vertices of \( {\mathbf{Q}}_{\mathrm{o}} \) lie on \( \mathcal{C} \), because, for example, \( \angle {V}_{1}{H}_{1}{V}_{3}... | Yes |
Theorem 4. If \( \mathbf{Q} \) is cyclic and orthodiagonal and \( {\mathbf{Q}}_{\mathrm{o}} \) is an orthic quadrilateral of \( \mathbf{Q} \) that is inscribed in \( Q \), the valtitudes that detect \( {\mathbf{Q}}_{o} \) are the internal angle bisectors of \( {\mathbf{Q}}_{\mathrm{o}} \). | Proof. We prove that the valtitude \( {V}_{1}{H}_{1} \) is the bisector of \( \angle {H}_{2}{H}_{1}{H}_{4} \) (see Figure 6).\n\nSince \( \mathbf{Q} \) is cyclic, we have\n\n\[ \angle {A}_{1}{A}_{4}{A}_{2} = \angle {A}_{1}{A}_{3}{A}_{2} \]\n\n(1)\n\nbecause they are subtended by the same arc \( {A}_{1}{A}_{2} \) . Let ... | Yes |
Theorem 6. If \( \mathbf{Q} \) is cyclic and orthodiagonal, the bimedians of \( \mathbf{Q} \) are the axes of the diagonals of \( {\mathbf{Q}}_{\mathrm{{po}}} \) . | Proof. It is enough to consider the eight-point circle of \( \mathbf{Q} \) and prove that the bi-median \( {M}_{2}{M}_{4} \) is the axis of the diagonal \( {H}_{1}{H}_{3} \) of \( {\mathbf{Q}}_{\text{po }} \) (see Figure 8). Note that \( \angle {H}_{3}{M}_{4}{M}_{2} = \angle {H}_{3}{H}_{2}{M}_{2} \), because they are s... | Yes |
Theorem 7. If \( \mathbf{Q} \) is cyclic and orthodiagonal, the orthic quadrilaterals of \( \mathbf{Q} \) inscribed in \( \mathbf{Q} \) have the same perimeter. Moreover, they have the minimum perimeter of any quadrilateral inscribed in \( \mathbf{Q} \) . | Proof. Let \( \mathbf{Q} \) be cyclic and orthodiagonal and let \( {\mathbf{Q}}_{\mathrm{o}} \) be any orthic quadrilateral of \( \mathbf{Q} \) inscribed in \( \mathbf{Q} \) (see Figure 9). Let \( \overline{\mathbf{Q}} \) be any quadrilateral inscribed in \( \mathbf{Q} \) , different from \( {\mathbf{Q}}_{\mathrm{o}} \... | Yes |
Lemma 8. Let \( \mathbf{Q} \) be cyclic and orthodiagonal and let \( {\mathbf{Q}}_{\mathrm{o}} \) be an orthic quadrilateral of \( \mathbf{Q} \) inscribed in \( \mathbf{Q} \) . The triangle \( {A}_{i}{H}_{i + 1}{H}_{i + 2}\left( {i = 1,2,3,4}\right) \) is similar to the triangle \( {A}_{i}{A}_{i + 1}{A}_{i + 3} \) . | Proof. Let us prove that the triangles \( {A}_{1}{H}_{2}{H}_{3} \) and \( {A}_{1}{A}_{2}{A}_{4} \) are similar. Then all we need to prove is that \( \angle {A}_{1}{H}_{2}{H}_{3} = \angle {A}_{1}{A}_{2}{A}_{4} \) (see Figure 10).\n\nLet \( B \) be the common point to the valtitudes \( {V}_{1}{H}_{1} \) and \( {V}_{2}{H}... | Yes |
Theorem 9. If \( \mathbf{Q} \) is cyclic and orthodiagonal, the quadrilaterals \( {\mathbf{Q}}^{\prime } \) and \( \mathbf{Q} \) and the quadrilaterals \( {\mathbf{Q}}_{\text{po }} \) and \( {\mathbf{Q}}_{\mathrm{t}} \) are correspondent in a homothetic transformation whose center lies on the Euler line of \( \mathbf{Q... | Proof. It suffices to prove that the quadrilaterals \( {\mathbf{Q}}^{\prime } \) and \( \mathbf{Q} \) are homothetic (see Figure 11).\n\nLet us start proving that the sides of \( \mathbf{Q} \) are parallel to the sides of \( {\mathbf{Q}}^{\prime } \), for example that \( {A}_{1}{A}_{4} \) is parallel to \( {A}_{1}^{\pr... | Yes |
Theorem 10. If \( \mathbf{Q} \) is cyclic and orthodiagonal and \( {\mathbf{Q}}_{\mathrm{o}} \) is an orthic quadrilateral of \( \mathbf{Q} \) inscribed in \( \mathbf{Q} \), the perimeter of \( {\mathbf{Q}}_{\mathrm{o}} \) is twice the ratio between the area of \( \mathbf{Q} \) and the radius of the circumcircle of \( ... | Proof. In fact, from Theorem 7 all orthic quadrilaterals inscribed in \( \mathbf{Q} \) have the same perimeter, then it suffices to prove the property for \( {\mathbf{Q}}_{\mathrm{{po}}} \) . The segments \( {H}_{1}{H}_{2} \) and \( {T}_{1}{T}_{2} \) are parallel, because \( \mathbf{Q} \) and \( {\mathbf{Q}}_{\mathrm{t... | Yes |
Theorem 11. If \( \mathbf{Q} \) is orthodiagonal and is not a square, for any orthic quadrilateral \( {\mathbf{Q}}_{\mathrm{o}} \) of \( \mathbf{Q} \) the points \( {S}_{1},{S}_{2},{S}_{3},{S}_{4} \) lie on a line \( \mathcal{R} \) . | Proof. Set up an orthogonal coordinate system whose axes are the diagonals of \( \mathbf{Q} \) ; then the vertices of \( \mathbf{Q} \) have coordinates \( {A}_{1} = \left( {{a}_{1},0}\right) ,{A}_{2} = \left( {0,{a}_{2}}\right) ,{A}_{3} = \left( {{a}_{3},0}\right) \) , \( {A}_{4} = \left( {0,{a}_{4}}\right) \) . The eq... | Yes |
Theorem 1. The correspondences \( D{S}_{4} \leftrightarrow {Q}_{90}, D{S}_{3} \leftrightarrow {Q}_{120}^{\prime } \) and \( D{S}_{6} \leftrightarrow \) \( {Q}_{60}^{\prime } \) given by\n\n\[ \left( {R, r, d}\right) \mapsto \left( {\frac{1}{2}\left( {R + r - d}\right) ,\frac{1}{2}\left( {R + r + d}\right), R - r}\right... | Proof. It is straightforward that the above maps are mutually inverse and that they map the solution \( \left( {R, r, d}\right) \) of the equation \( {R}^{2} - {6Rr} + {r}^{2} - {d}^{2} = 0 \) into the solution \( \left( {a, b, c}\right) \) of the equation \( {a}^{2} + {b}^{2} - {c}^{2} = 0 \), and vice versa. The same... | Yes |
Proposition 1. Let \( {ABCD} \) be a cyclic quadrangle, \( E \) be the intersection point of its diagonals and \( k\left( E\right) \) the corresponding hippopede. Let the positions of the diagonal \( {AC} \) and \( E \) be fixed and consider the other diagonal \( {BD} \) turning about E. Further, let \( G \) be the cor... | The proof follows trivially from the formula \( \left| {AC}\right| \cdot \left| {BD}\right| \sin \theta \) expressing the area of the quadrangle in terms of the diagonals and their angle. Since \( {AC} \) is fixed the problem reduces to maximizing the product \( \left| {BD}\right| \sin \theta \), which is the length of... | Yes |
Proposition 2. Let \( c\left( {O, R}\right) \) be a circle of radius \( R \) centered at the origin and \( E : \left| {OE}\right| = {sR}, s \in \left( {0,1}\right) \) be a fixed point inside the circle.\n\n(1) The ellipse \( e\left( s\right) \) centered at \( O \) having one focus at \( E \) and eccentricity equal to \... | Using these facts and the symmetry of the figure we can eliminate the difficulties of selecting the furthest to \( O \) tangent of the hippopede and show that, by restricting the locations of the diagonal \( {AC} \) to the absolutely necessary needed to cover all possible quadrangles, the other diagonal \( {BD} \) of t... | No |
Proposition 3. The correspondence \( p : \alpha \mapsto \phi \) mapping each \( \alpha \in \left\lbrack {-\frac{\pi }{2},0}\right\rbrack \) to the polar angle \( \phi \in \left\lbrack {0,\Phi }\right\rbrack \) such that the tangent \( {t}_{G} \) of the hippopede at its point \( {G\phi } \) is parallel to \( {\bar{e}}_{... | This is obvious from the geometry of the figure. The analytic proof follows from the angle relations of the triangle \( {OJG} \) in Figure 6. In fact, if \( \left( {x, y}\right) = \) \( \left( {a\cos \theta, b\sin \theta }\right) \) is a parametrization of the ellipse, then the direction of \( {BD} \) is that of the no... | Yes |
For each point \( E \) there is a unique quadrangle \( {P\alpha } \in \mathcal{P} \) having the diagonals symmetric with respect to the diameter \( {OE} \) hence equal, so that \( {P\alpha } \) in this case is an isosceles trapezium. This quadrangle is also the maximal one in area among all quadrangles whose diagonals ... | In fact, by the previous discussion we see that the diametral position of \( {AC} \) does not deliver equal diagonals, except in the trivial case for \( s = 0 \) defining the square inscribed in the circle. Thus, we can assume \( \phi \neq \frac{\pi }{2} \) and the proof follows by observing the role of \( a = \tan \al... | Yes |
Proposition 1. (a) The tripolar of the right (left) Brocardian point of \( P \) is the right (left) Brocardian line of the tripolar of \( P \) . | Proof. \( \star {P}_{ \rightarrow } = {p}_{ \rightarrow } \) and \( \star {p}_{ \leftarrow } = {P}_{ \leftarrow } \) . | No |
Theorem 1. The lengths of the tangency chords \( {WY} \) and \( {XZ} \) in a tangential quadrilateral are respectively\n\n\[ k = \frac{2\left( {{efg} + {fgh} + {ghe} + {hef}}\right) }{\sqrt{\left( {e + f}\right) \left( {g + h}\right) \left( {e + g}\right) \left( {f + h}\right) }}, \]\n\n\[ l = \frac{2\left( {{efg} + {f... | Proof. If \( I \) is the incenter and angles \( \beta \) and \( \gamma \) are defined as in Figure 2, by the law of cosines in triangle \( {WYI} \) we get\n\n\[ {k}^{2} = 2{r}^{2} - 2{r}^{2}\cos \left( {{2\beta } + {2\gamma }}\right) = 2{r}^{2}\left( {1 - \cos \left( {{2\beta } + {2\gamma }}\right) }\right) . \]\n\nHen... | Yes |
Corollary 2. In a tangential quadrilateral with sides \( a, b, c \) and \( d \), the quotient of the tangency chords satisfy\n\n\[{\left( \frac{k}{l}\right) }^{2} = \frac{bd}{ac}\] | Proof. Taking the quotient of \( k \) and \( l \) from Theorem 1, after simplification we get\n\n\[ \frac{k}{l} = \sqrt{\frac{\left( {e + h}\right) \left( {f + g}\right) }{\left( {e + f}\right) \left( {h + g}\right) }} = \sqrt{\frac{db}{ac}} \]\n\nand the result follows. | Yes |
Corollary 3. The tangency chords in a tangential quadrilateral are of equal length if and only if it is a kite. | Proof. \( \left( \Rightarrow \right) \) If the quadrilateral is a kite it directly follows that the tangency chords are of equal length because of the mirrow symmetry in the longest diagonal (see Figure 3).\n\n\( \left( \Leftarrow \right) \) Conversely, if the tangency chords are of equal length in a tangential quadril... | Yes |
Lemma 4. The alternate angles between a chord and two tangents to a circle are supplementary angles, i.e., \( w + y = \pi \) in Figure 4. | Proof. Extend the tangents at \( W \) and \( Y \) to intersect at \( T \), see Figure 4. Triangle \( {TWY} \) is isosceles according to the two tangent theorem, so the angles at the base are equal, \( w = v \) . Also, \( v + y = \pi \) since they are angles on a straight line. Hence \( w + y = \pi \) . | Yes |
Theorem 5. If \( e, f, g \) and \( h \) are the tangent lengths in a tangential quadrilateral, the angle \( \varphi \) between the tangency chords is given by\n\n\[ \sin \varphi = \sqrt{\frac{\left( {e + f + g + h}\right) \left( {{efg} + {fgh} + {ghe} + {hef}}\right) }{\left( {e + f}\right) \left( {f + g}\right) \left(... | Proof. We start by relating the angle \( \varphi \) to two opposite angles in the tangential quadrilateral (see Figure 5).\n\nFrom the sum of angles in quadrilaterals \( {BWPX} \) and \( {DYPZ} \) we have \( w + x + \) \( \varphi + B = {2\pi } \) and \( y + z + \varphi + D = {2\pi } \) . Adding these,\n\n\[ w + x + y +... | Yes |
Corollary 6. The tangency chords in a tangential quadrilateral are perpendicular if and only if it is a bicentric quadrilateral. | Proof. In any tangential quadrilateral, \( B + D = {2\pi } - {2\varphi } \) by (10). The tangency chords are perpendicular if and only if\n\n\[ \varphi = \frac{\pi }{2} \Leftrightarrow B + D = \pi \]\n\nwhich is a well known characterization for a quadrilateral to be cyclic. Hence this is a characterization for the qua... | Yes |
Theorem 7. Ife, \( f, g \) and \( h \) are the tangent lengths in a tangential quadrilateral, then the contact quadrilateral has area\n\n\[ \n{K}_{\mathrm{c}} = \frac{2\sqrt{\left( {e + f + g + h}\right) {\left( efg + fgh + ghe + hef\right) }^{5}}}{\left( {e + f}\right) \left( {e + g}\right) \left( {e + h}\right) \left... | Proof. The area of any convex quadrilateral is\n\n\[ \nK = \frac{1}{2}{pq}\sin \theta \n\]\n\n(12)\n\nwhere \( p \) and \( q \) are the length of the diagonals and \( \theta \) is the angle between them (see [21, p.213] and [22]). Hence for the area of the contact quadrilateral we have\n\n\[ \n{K}_{c} = \frac{1}{2}{kl}... | No |
Theorem 8. If \( e, f, g \) and \( h \) are the tangent lengths in a tangential quadrilateral \( {ABCD} \), then its angles satisfy\n\n\[ \sin \frac{A}{2} = \sqrt{\frac{{efg} + {fgh} + {ghe} + {hef}}{\left( {e + f}\right) \left( {e + g}\right) \left( {e + h}\right) }}, \]\n\n\[ \sin \frac{B}{2} = \sqrt{\frac{{efg} + {f... | Proof. If the incircle has center \( I \) and is tangent to sides \( {AB} \) and \( {AD} \) at \( W \) and \( Z \) (see Figure 7), then by the law of cosines in triangle WZI\n\n\[ W{Z}^{2} = 2{r}^{2}\left( {1 - \cos {2\alpha }}\right) = \frac{4{e}^{2}{r}^{2}}{{r}^{2} + {e}^{2}} \]\n\nwhere we used\n\n\[ \cos {2\alpha }... | Yes |
Theorem 9. A bicentric quadrilateral with sides \( a, b, c \) and \( d \) has area\n\n\[ K = \sqrt{abcd}. \] | Proof. From formula (2) we get\n\n\[ {K}^{2} = \left( {{efg} + {fgh} + {ghe} + {hef}}\right) \left( {e + f + g + h}\right) \]\n\n\[ = {ef}\left( {g + h}\right) \left( {e + f}\right) + {ef}{\left( g + h\right) }^{2} + {gh}{\left( e + f\right) }^{2} + {gh}\left( {e + f}\right) \left( {g + h}\right) \]\n\n\[ = \left( {e +... | Yes |
Theorem 10. A bicentric quadrilateral with tangent lengths \( e, f, g \) and \( h \) has area\n\n\[ K = \sqrt[4]{efgh}\left( {e + f + g + h}\right) . \] | Proof. The quadrilateral has an incircle, so (see Figure 8)\n\n\[ r = e\tan \frac{A}{2} = f\tan \frac{B}{2} = g\tan \frac{C}{2} = h\tan \frac{D}{2}, \]\n\nhence\n\n\[ {r}^{4} = {efgh}\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2}\tan \frac{D}{2}. \]\n\n(13)\n\nIt also has a circumcircle, so \( A + C = \pi \) . Hence ... | Yes |
Theorem 11. A bicentric quadrilateral with tangency chords \( k \) and \( l \), and diagonals \( p \) and \( q \) has area | Proof. Using (12), Theorem 9 and Ptolemy's theorem, we have\n\n\[ K = \frac{1}{2}{pq}\sin \theta \Leftrightarrow \sqrt{abcd} = \frac{1}{2}\left( {{ac} + {bd}}\right) \sin \theta .\n\]\n\nHence\n\n\[ \frac{2}{\sin \theta } = \frac{{ac} + {bd}}{\sqrt{abcd}} = \sqrt{\frac{ac}{bd}} + \sqrt{\frac{bd}{ac}} = \frac{l}{k} + \f... | Yes |
Proposition 1. If a triangle and an inscribed ellipse are the orthogonal projections of a triangle and its incircle, the sidelengths of the latter triangle are proportional to the barycentric coordinates of the center of the ellipse. | Proof. Note that\n\n\[ \frac{BC}{C{A}_{ - }} = \frac{BC}{CE} = \frac{B{C}_{0}}{{C}_{0}A} = \frac{u}{v} \]\n\n\[ \frac{BC}{{A}_{ - }B} = \frac{BC}{DB} = \frac{{B}_{0}C}{A{B}_{0}} = \frac{u}{w}. \]\n\nFrom these, \( {BC} : C{A}_{ - } : {A}_{ - }B = u : v : w \) ; similarly for \( {BC} : C{A}_{ + } : {A}_{ + }B \) . | Yes |
Lemma 3. The lengths of \( A{A}_{ + } \) and \( A{A}_{ - } \) are given by\n\n\[ A{A}_{ + } = \frac{\sqrt{\mathcal{Q} + {16\Delta }{\Delta }^{\prime }}}{\sqrt{2}u}\;\text{ and }\;A{A}_{ - } = \frac{\sqrt{\mathcal{Q} - {16\Delta }{\Delta }^{\prime }}}{\sqrt{2}u}, \]\n\nwhere\n\n\[ \mathcal{Q} = \left( {{b}^{2} + {c}^{2}... | Proof. Applying the law of cosines to triangle \( {ABA} \), we have\n\n\[ A{A}_{ - }^{2} = A{B}^{2} + B{A}_{ - }^{2} - {2AB} \cdot B{A}_{ - }\cos \left( {{ABC} - {A}_{ - }{BC}}\right) \]\n\n\[ = {c}^{2} + {\left( \frac{aw}{u}\right) }^{2} - {2c} \cdot \frac{aw}{u}\left( {\cos B\cos {A}_{ - }{BC} + \sin B\sin {A}_{ - }{... | Yes |
Problem 8. Construct a triangle given, in position, the traces of its medians on the circumcircle. | Solution. Let a given triangle \( {A}_{1}{B}_{1}{C}_{1} \) be the circumcevian triangle of the (unknown) centroid \( G \) of the required triangle \( {ABC} \) . We construct the equilateral triangles \( {A}_{ + }{B}_{1}{C}_{1} \) and \( {A}_{ - }{B}_{1}{C}_{1} \) on the segment \( {B}_{1}{C}_{1} \) . Let \( {G}_{1} \) ... | Yes |
Theorem 1 (S. N. Collings). Let \( \rho \) be a line in the plane of a triangle \( {ABC} \) . Its reflections in the sidelines \( {BC},{CA},{AB} \) are concurrent if and only if \( \rho \) passes through the orthocenter \( H \) of \( {ABC} \) . In this case, their point of concurrency lies on the circumcircle. | Synthetic proofs of Theorem 1 can be found in [1] and [2]. | No |
Lemma 3. The three lines joining the vertices of a given triangle \( {ABC} \) with the reflections of the circumcenter \( O \) into the opposite sidelines are concurrent at the nine-point center of triangle \( {ABC} \) . | This is a simple consequence of the fact that the reflection \( {O}_{A} \) of \( O \) into the sideline \( {BC} \) is the circumcenter of triangle \( {BHC} \), where \( H \) is the orthocenter of \( {ABC} \) . In this case, according to the definition of the nine-point circle, the circumcircle of \( {BHC} \) is the hom... | No |
Theorem 4. The circumcircles of triangles \( A{N}_{B}{N}_{C}, B{N}_{C}{N}_{A}, C{N}_{A}{N}_{B}, A{N}_{B}^{\prime }{N}_{C}^{\prime } \) , \( B{N}_{C}^{\prime }{N}_{A}^{\prime }, C{N}_{A}^{\prime }{N}_{B}^{\prime } \) are concurrent at the Euler reflection point \( E \) of triangle \( {ABC} \) . | Proof. We shall show that each of these circles contains \( E \) . It is enough to consider the circle \( A{B}_{B}{N}_{C} \) .\n\nDenote by \( {O}_{B},{O}_{C} \) the reflections of the circumcenter \( O \) into the sidelines \( {CA} \) and \( {AB} \), respectively. The lines \( E{O}_{B}, E{O}_{C} \) are the reflections... | Yes |
Proposition 5. Let \( {ABC} \) be a triangle with circumcenter \( O \) and orthocenter \( H \) . Consider the points \( Y \) and \( Z \) on the sides \( {CA} \) and \( {AB} \) respectively such that the directed angles \( \left( {{AC},{HY}}\right) = - \frac{\pi }{3} \) and \( \left( {{AB},{HZ}}\right) = \frac{\pi }{3} ... | Proof. (a) Let \( V \) be the orthocenter of triangle \( {HYZ} \), and denote by \( {Y}^{\prime },{Z}^{\prime } \) the intersections of the lines \( {HZ} \) with \( {CA} \) and \( {HY} \) with \( {AB} \) . Since the quadrilateral \( {YZ}{Y}^{\prime }{Z}^{\prime } \) is cyclic, the lines \( {Y}^{\prime }{Z}^{\prime },{Y... | Yes |
Corollary 6. Given a triangle \( {ABC} \) with circumcenter \( O \), let \( Y, Z \) be points on the sides \( {AC},{AB} \) satisfying \( \left( {{AC},{OY}}\right) = - \frac{\pi }{3} \) and \( \left( {{AB},{OZ}}\right) = \frac{\pi }{3} \). The circumcenters of triangles \( {OYZ} \) and \( {BOC} \), and the vertex \( A \... | Now we complete the proof of Theorem 4. By applying Corollary 6 to triangle \( O{O}_{B}{O}_{C} \) with the points \( {N}_{B},{N}_{C} \) lying on the sidelines \( O{O}_{B} \) and \( O{O}_{C} \) such that \( \left( {O{O}_{B}, A{N}_{B}}\right) = - \frac{\pi }{3} \) and \( \left( {O{O}_{C}, A{N}_{C}}\right) = \frac{\pi }{3... | No |
Theorem 1. In a non-isosceles triangle the median and altitude to one of the sides divide the opposite angle into three parts. This angle is a right one if and only if the angle between the median and the longer of the sides at the considered vertex is equal to the angle between the altitude and the shorter side at tha... | Proof. We use notations as in Figure 3. If \( C = \frac{\pi }{2} \), we shall prove that \( \alpha = \beta \) . Triangle \( {AMC} \) is isosceles \( {}^{2} \) with \( {AM} = {CM} \), so \( A = \alpha \) . Triangles \( {ACB} \) and \( {CHB} \) are similar, so \( A = \beta \) . Hence \( \alpha = \beta \) .\n\nConversely,... | Yes |
Corollary 2. Let \( {CM},{CD} \) and \( {CH} \) be a median, an angle bisector and an altitude respectively in triangle \( {ABC} \) . The angle \( C \) is a right angle if and only if \( {CD} \) bisects angle HCM. | Proof. Since \( {CD} \) is an angle bisector in triangle \( {ABC} \), we have (see Figure 4)\n\n\[ \alpha + \angle {MCD} = \angle {HCD} + \beta . \]\n\n(2)\n\nUsing Theorem 1 and (2), we get\n\n\[ C = \frac{\pi }{2}\; \Leftrightarrow \;\alpha = \beta \; \Leftrightarrow \;\angle {MCD} = \angle {HCD}. \] | Yes |
Corollary 3. If the extensions of opposite sides in a tangential quadrilateral intersect at \( J \) and \( K \), and the extensions of opposite sides in its contact quadrilateral intersect at \( L \) and \( M \), then the four points \( J, L, K \) and \( M \) are collinear. | Proof. Consider the degenerate cyclic hexagon \( {WWXYYZ} \), where \( W \) and \( Y \) are double vertices. The extensions of the sides at these vertices are the tangents at \( W \) and \( Y \), see Figure 6. According to Pascal’s theorem, the points \( J, L \) and \( M \) are collinear.\n\nNext consider the degenerat... | Yes |
Corollary 4. If the extensions of opposite sides in a tangential quadrilateral intersect at \( J \) and \( K \), and the diagonals intersect at \( P \), then \( {JK} \) is perpendicular to the extension of IP where \( I \) is the incenter. | Proof. The contact quadrilateral \( {WXYZ} \) is a cyclic quadrilateral with circumcenter \( I \), see Figure 7. It is well known that the point of intersection of \( {WY} \) and \( {XZ} \) is also the point of intersection of the diagonals in the tangential quadrilateral \( {ABCD} \), see \( \left\lbrack {{10},{15},{1... | Yes |
Theorem 5. Let the extensions of opposite sides in a tangential quadrilateral intersect at \( J \) and \( K \) . If \( I \) is the incenter, then the quadrilateral is also cyclic if and only if \( {JIK} \) is a right angle. | Proof. We use notations as in Figure 8, where \( G \) and \( H \) are the midpoints of the tangency chords \( {WY} \) and \( {XZ} \) respectively and \( P \) is the point of intersection of \( {WY} \) and \( {XZ} \) . In isosceles triangles \( {WJY} \) and \( {XKZ},{IJ} \bot {WY} \) and \( {IK} \bot {XZ} \) . Hence opp... | Yes |
Theorem 6. A tangential quadrilateral is cyclic if and only if its Newton line is perpendicular to the Newton line of its contact quadrilateral. | Proof. We use notations as in Figure 9, where \( P \) is the point where both the diagonals and the tangency chords intersect (see \( \left\lbrack {{10},{15},{17}}\right\rbrack \) ) and \( L \) is the midpoint of \( {JK} \) . If \( I \) is the incenter, then the points \( E, I, F \) and \( L \) are collinear on the New... | Yes |
Theorem 2 (Lester). The Fermat points, the circumcenter, and the nine-point center of a triangle are concyclic. | Our starting point is a simple observation that the line joining the Fermat points intersects the Euler line at the midpoint of the orthocenter \( H \) and the centroid \( G \) . Clearing denominators in the homogeneous barycentric coordinates of the Fermat point \[ {F}_{ + } = \left( {\frac{1}{\sqrt{3}{S}_{A} + S},\fr... | Yes |
Proposition 4. The following statements are equivalent.\n\n(A) \( M{F}_{ + } \cdot M{F}_{ - } = {MO} \cdot {MN} \) .\n\n(B) The circle \( {F}_{ + }{F}_{ - }G \) is tangent to the Euler line at \( G \), i.e., \( M{F}_{ + } \cdot M{F}_{ - } = M{G}^{2} \) .\n\n(C) The circle \( {F}_{ + }{F}_{ - }H \) is tangent to the Eul... | Proof. Since \( M \) is the midpoint of \( {HG} \), the statements (B),(C),(D) are clearly equivalent. On the other hand, putting \( {OH} = {6d} \), we have\n\n\[ \n{MO} \cdot {MN} = {\left( MH\right) }^{2} = {\left( MG\right) }^{2} = 4{d}^{2},\n\]\n\nsee Figure 3. This shows that (A), (B), (C) are equivalent.\n\nNote ... | Yes |
Theorem 5. The Fermat points are inverse in the orthocentroidal circle. | Proof. The equation of the orthocentroidal circle is\n\n\[ 3\left( {{a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy}}\right) - 2\left( {x + y + z}\right) \left( {{S}_{A}x + {S}_{B}y + {S}_{C}z}\right) = 0, \]\n\nequivalently, \( {}^{1} \)\n\n\[ - 2\left( {{S}_{A}{x}^{2} + {S}_{B}{y}^{2} + {S}_{C}{z}^{2}}\right) + \left( {\left(... | Yes |
Theorem 6 (Gibert). Every circle whose diameter is a chord of the Kiepert hyperbola perpendicular to the Euler line passes through the Fermat points. | Proof. Since \( {F}_{ \pm } \) and \( G \) are on the Kiepert hyperbola, and the center of the circle \( {F}_{ + }{F}_{ - }G \) is on the perpendicular to the Euler line at \( G \), this line intersects the Kiepert hyperbola at a fourth point \( {Y}_{0} \) (see Figure 6), and the circle is a member of the pencil of con... | Yes |
Proposition 8. (a) The equation of the line \( {F}_{ + }{F}_{ - } \) is\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\left( {{b}^{2} - {c}^{2}}\right) {f}_{4,4}\left( {a, b, c}\right) x = 0. \] | Proof. (a) The line \( {F}_{ + }{F}_{ - } \) contains the symmedian point \( K \) and the Kiepert center\n\n\[ {K}_{\mathrm{i}} = \left( {{\left( {b}^{2} - {c}^{2}\right) }^{2},{\left( {c}^{2} - {a}^{2}\right) }^{2},{\left( {a}^{2} - {b}^{2}\right) }^{2}}\right) . \] | No |
Proposition 9. The center of the first Lester circle has homogeneous barycentric coordinates\n\n\[ \n\\left( {\\left( {{b}^{2} - {c}^{2}}\\right) {f}_{8,3}\\left( {a, b, c}\\right) : \\left( {{c}^{2} - {a}^{2}}\\right) {f}_{8,3}\\left( {b, c, a}\\right) : \\left( {{a}^{2} - {b}^{2}}\\right) {f}_{8,3}\\left( {c, a, b}\\... | Proof. This is the intersection of the lines (4) and (6). | No |
Proposition 11. Every circle through the isodynamic points can be represented by an equation\n\n\\[ \n3\\left( {{b}^{2} - {c}^{2}}\\right) \\left( {{c}^{2} - {a}^{2}}\\right) \\left( {{a}^{2} - {b}^{2}}\\right) \\left( {{a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy}}\\right) \n\\]\n\n\\[ \n+ \\left( {x + y + z}\\right) \\left... | Proof. Combining the above equations for the three Apollonian circles, we obtain\n\n\\[ \n3\\left( {{b}^{2} - {c}^{2}}\\right) \\left( {{c}^{2} - {a}^{2}}\\right) \\left( {{a}^{2} - {b}^{2}}\\right) \\left( {{a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy}}\\right) \n\\]\n\n\\[ \n+ \\left( {x + y + z}\\right) \\mathop{\\sum }\\... | Yes |
Theorem 13. The excentral triangle and the triangle of reflections are perspective at a point which is the inverse image of the incenter in the circumcircle of the excentral triangle. | Proof. We show that the lines \( {I}_{a}{A}^{ * },{I}_{b}{B}^{ * } \), and \( {I}_{c}{C}^{ * } \) intersect the line \( {OI} \) at the same point. Let \( X \) be the intersection of the lines \( A{A}^{ * } \) and \( {OI} \) (see Figure 10). If \( {h}_{a} \) is the \( A \) -altitude of triangle \( {ABC} \), and the para... | Yes |
Lemma 14. Let \( {XBC} \) and \( {X}^{\prime }{I}_{b}{I}_{c} \) be oppositely oriented similar isosceles triangles with bases \( {BC} \) and \( {I}_{b}{I}_{c} \) respectively. The lines \( {I}_{a}X \) and \( {I}_{a}{X}^{\prime } \) are isogonal with respect to angle \( {I}_{a} \) the excentral triangle (see Figure 11). | Proof. The triangles \( {I}_{a}{BC} \) and \( {I}_{a}{I}_{b}{I}_{c} \) are oppositely similar since \( {BC} \) and \( {I}_{b}{I}_{c} \) are antiparallel. In this similarity \( X \) and \( {X}^{\prime } \) are homologous points. Hence, the lines \( {I}_{a}X \) and \( {I}_{a}{X}^{\prime } \) are isogonal in the excentral... | Yes |
Theorem 15. The excentral triangle and the Kiepert triangle \( \mathcal{K}\left( \theta \right) \) are perspective at the isogonal conjugate of \( {K}_{\mathrm{e}}\left( {-\theta }\right) \) in the excentral triangle. | Proof. Let \( {XYZ} \) be a Kiepert triangle \( \mathcal{K}\left( \theta \right) \) . Construct \( {X}^{\prime },{Y}^{\prime },{Z}^{\prime } \) as in Lemma 14 (see Figure 11).\n\n(i) \( {I}_{a}{X}^{\prime },{I}_{b}{Y}^{\prime },{I}_{c}{Z}^{\prime } \) concur at the Kiepert perspector \( {K}_{\mathrm{e}}\left( {-\theta ... | Yes |
Proposition 16. The line \( {V}_{ + }{V}_{ - } \) has equation\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\left( {b - c}\right) \left( {{b}^{2} + {c}^{2} - {a}^{2}}\right) x = 0. \] | Proof. The line \( {V}_{ + }{V}_{ - } \) is the Brocard axis of the excentral triangle, with equation\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\frac{{b}^{\prime 2} - {c}^{\prime 2}}{{a}^{\prime 2}} \cdot {x}^{\prime } = 0 \]\n\nwith reference to the excentral triangle (see §7.1). Replacing these by parameter with re... | No |
Proposition 17. The Kiepert triangle \( \mathcal{K}\left( \theta \right) \) is perspective with the triangle of reflections \( {A}^{ * }{B}^{ * }{C}^{ * } \) if and only if \( \theta = \pm \frac{\pi }{3} \) . The perspector is \( {K}^{ * }\left( {-\theta }\right) \), the isogonal conjugate of \( K\left( {-\theta }\righ... | This means that for \( \varepsilon = \pm 1 \), the Fermat triangle \( \mathcal{K}\left( {\varepsilon \cdot \frac{\pi }{3}}\right) \) and the triangle of reflections are perspective at the isodynamic point \( {J}_{-\varepsilon } \) (see Figure 13 for the case \( \varepsilon = - 1 \) ).\n\n![bb70723e-cbb9-45f6-82d9-918ae... | Yes |
Lemma 19. The perpendicular bisector of the segment IW is the line\n\n\[ \n{bc}\left( {b + c}\right) x + {ca}\left( {c + a}\right) y + {ab}\left( {a + b}\right) z = 0. \n\] | Proof. If \( {M}^{\prime } \) is the midpoint of \( {IW} \), then since \( O \) is the midpoint of \( I{I}^{\prime } \), from the degenerate triangle \( I{I}^{\prime }W \) we have \( O{M}^{\prime } = \frac{{I}^{\prime }W}{2} = \frac{{R}^{2}}{OI} \) . This shows that the\n\nmidpoint of \( {IW} \) is the inverse of \( I ... | Yes |
Lemma 20. The perpendicular bisector of the segment \( {V}_{ + }{V}_{ - } \) is the line\n\n\[ \left( {b + c}\right) x + \left( {c + a}\right) y + \left( {a + b}\right) z = 0. \] | Proof. Since \( {V}_{ + } \) and \( {V}_{ - } \) are the isodynamic points of the excentral triangle, the perpendicular bisector of \( {V}_{ + }{V}_{ - } \) is the polar of the symmedian point of the excentral triangle with respect to its own circumcircle. With reference to the excentral triangle, its Lemoine axis has ... | Yes |
Proposition 21. The center of the first Evans circle is the point\n\n\[ \n\\left( {\\frac{a\\left( {b - c}\\right) }{b + c} : \\frac{b\\left( {c - a}\\right) }{c + a} : \\frac{c\\left( {a - b}\\right) }{a + b}}\\right) .\n\] | Proof. This is the intersection of the lines (20) and (21). | No |
Lemma 23. The line \( {EG} \) is parallel to the Fermat line \( {F}_{ + }{F}_{ - } \) . | Proof. The line \( {F}_{ + }{F}_{ - } \) is the same as the line \( K{K}_{\mathrm{i}} \), with equation given by (5). The line \( {EG} \) has equation\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}\left( {{b}^{2} - {c}^{2}}\right) {f}_{4,2}\left( {a, b, c}\right) x = 0. \]\n\n(26)\n\nBoth of these lines have the same inf... | Yes |
Proposition 24. The Euler reflection point \( E \) and the centroid are inverse in the Brocard circle. | Proof. Note that the line \( {F}_{ + }{F}_{ - } \) intersects the Euler line at the midpoint \( M \) of \( {HG} \) , and \( G \) is the midpoint of \( {OM} \) . Since \( {EG} \) is parallel to \( {MK} \), it intersects \( {OK} \) at its midpoint, the center of the Brocard circle. Since the circle through \( E, G,{J}_{ ... | Yes |
Proposition 25. The perpendicular bisector of the segment \( {GE} \) is the line\n\n\[ \frac{x}{{b}^{2} + {c}^{2} - 2{a}^{2}} + \frac{y}{{c}^{2} + {a}^{2} - 2{b}^{2}} + \frac{z}{{a}^{2} + {b}^{2} - 2{c}^{2}} = 0. \] | Proof. The midpoint of \( {EG} \) is the point\n\n\[ {Z}_{4} \mathrel{\text{:=}} \left( {\left( {{b}^{2} + {c}^{2} - 2{a}^{2}}\right) {f}_{4,5}\left( {a, b, c}\right) : \left( {{c}^{2} + {a}^{2} - 2{b}^{2}}\right) {f}_{4,5}\left( {b, c, a}\right) : \left( {{a}^{2} + {b}^{2} - 2{c}^{2}}\right) {f}_{4,5}\left( {c, a, b}\... | Yes |
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