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Proposition 6 ([3, p.10], [4, p.8]). The three medians of a triangle all pass through one point which divides each median into two segments in the ratio \( 2 : 1 \) . | It follows that the centroid of \( \mathcal{T} \) coincides with the centroid \( C \) of \( \mathcal{S} \) . | No |
Theorem 17. If \( {\mathcal{S}}_{{H}_{1}} \) and \( {\mathcal{S}}_{{H}_{2}} \) are two h-extension of \( \mathcal{S} \), then the systems \( {\mathcal{C}}_{{H}_{1, t}} \) and \( {\mathcal{C}}_{{H}_{2, t}} \) are correspondent in a translation. | It is easy to see that the vector of the translation transforming \( {\mathcal{C}}_{{H}_{1, t}} \) into \( {\mathcal{C}}_{{H}_{2, t}} \) is \( \frac{n + h}{h + t}\overrightarrow{{C}_{1}{C}_{2}} \) . | Yes |
Theorem 18. If \( \mathcal{S} \) is an \( n \) -system, \( {\mathcal{S}}_{H} \) is a 1-extension of \( \mathcal{S},{\mathcal{S}}_{K} \) is a \( \left( {n - 1}\right) \) - extension of \( \mathcal{S} \), then the systems \( {\mathcal{C}}_{H, n - 1} \) and \( {\mathcal{C}}_{K,1} \) are correspondent is a half-turn. | Proof. Let \( C \) and \( {C}_{K} \) be the centroids of \( {\mathcal{S}}_{H} \) and \( K \) respectively. From Corollary 2 the system \( {\mathcal{C}}_{H, n - 1} \) is the image of the system \( {\mathcal{C}}_{1}^{\prime } = \mathcal{S} \) in the dilatation with ratio \( \frac{1}{n} \) and center \( C \) that is, \( \... | Yes |
Proposition 19. The triangles \( {\mathcal{T}}_{H} \) and \( {\mathcal{T}}_{K} \) are correspondent in a half-turn. See Figure 8. | Let \( \left\{ {\mathcal{T}}_{H}\right\} \) be the family of triangles \( {\mathcal{T}}_{H} \) obtained by varying the point \( P \) and \( \left\{ {\mathcal{T}}_{K}\right\} \) be the family of triangles \( {\mathcal{T}}_{K} \) obtained by varying the points \( {P}_{1} \) and \( {P}_{2} \) . From Theorem 17 the triangl... | No |
Theorem 1 (Euclid III.7). Let \( {BC} \) be a chord in a circle \( \Omega \), let \( M \) be the mid-point of \( {BC} \), and let the line perpendicular to \( {BC} \) through \( M \) meet \( \Omega \) at \( E \) and \( F \) . As a point \( P \) moves from \( E \) to \( F \) along the arc \( {ECF} \) of \( \Omega \), th... | Proof. Referring to Figure 1, we shall show that if \( {EM} > {MF} \), i.e., if the center \( O \) of \( \Omega \) is between \( E \) and \( M \), and if \( P \) and \( Q \) are any points on the arc \( {ECF} \) such that \( P \) is closer to \( E \) than \( Q \), then \( {MP} > {MQ} \) . Under these assumptions,\n\n\[... | Yes |
Lemma 2. Let \( {ABC} \) be a triangle and let \( D \) and \( E \) be points on the sides \( {AB} \) and \( {AC} \) respectively (see Figure 2). Then \( \frac{AD}{AB} \) is greater than, less than, or equal to \( \frac{AE}{AC} \) according as \( \angle {ABC} \) is greater than, less than, or equal to \( \angle {ADE} \)... | Proof. Let \( {E}^{\prime } \) be the point on \( {AC} \) such that \( \frac{A{E}^{\prime }}{AC} = \frac{AD}{AB} \) ; i.e., \( D{E}^{\prime } \) is parallel to \( {BC} \) . If \( \frac{AE}{AC} = \frac{AD}{AB} \), then \( {E}^{\prime } = E \) and \( \angle {ABC} = \angle {ADE} \) . If \( \frac{AE}{AC} > \frac{AD}{AB} \)... | Yes |
Lemma 3. Two circles \( \Omega \) and \( {\Omega }^{\prime } \) intersect at \( B \) and \( C \), and the line perpendicular to \( {BC} \) through the midpoint \( M \) of \( {BC} \) meets \( \Omega \) and \( {\Omega }^{\prime } \) at \( E \) and \( {E}^{\prime } \), respectively, such that \( {E}^{\prime } \) is inside... | Proof. Let \( S \) be the point of intersection of \( {FP} \) and \( {F}^{\prime }{P}^{\prime } \) . Since \( \angle {EPF} = \frac{\pi }{2} = \) \( \angle {E}^{\prime }{P}^{\prime }{F}^{\prime } \), it follows that \( \angle M{E}^{\prime }{P}^{\prime } + \angle M{F}^{\prime }{P}^{\prime } = \frac{\pi }{2} = \angle {MEP... | Yes |
Lemma 4. Let \( {EBC} \) be an isosceles triangle having \( {EB} = {EC} \) . Let \( M \) be the midpoint of \( {BC} \) and let \( {E}^{\prime } \) be the circumcenter of \( {EBC} \) (see Figure 4). Then \( \frac{M{E}^{\prime }}{ME} \) is greater than, equal to, or less than \( \frac{1}{3} \) according as \( \angle {BEC... | Proof. Let \( \theta = \angle {BEC}, x = M{E}^{\prime } \), and let \( R \) be the circumradius of \( {EBC} \) . Then \( \angle M{E}^{\prime }C = \theta \) and\n\n\[ \frac{M{E}^{\prime }}{ME} - \frac{1}{3} = \frac{x}{x + R} - \frac{1}{3} = \frac{R\cos \theta }{R\cos \theta + R} - \frac{1}{3} = \frac{2\cos \theta - 1}{3... | Yes |
Theorem 5. Two circles \( \Omega \) and \( {\Omega }^{\prime } \) intersect at \( B \) and \( C \) and the line perpendicular to \( {BC} \) through the midpoint \( M \) of \( {BC} \) meets \( \Omega \) at \( E \) and \( F \) and meets \( {\Omega }^{\prime } \) at \( {E}^{\prime } \) and \( {F}^{\prime } \) . For every ... | Proof. Referring to Figure 5, suppose that \( {E}^{\prime } \) lies inside \( \Omega \) and let \( P \) and \( Q \) be two points on the arc \( {ECF} \) of \( \Omega \) such that \( P \) is closer to \( E \) than \( Q \) . we are to show that \( \frac{M{P}^{\prime }}{MP} < \frac{M{Q}^{\prime }}{MQ} \) .\n\nExtend \( {Q... | Yes |
Lemma 6. Let \( P \) be a point inside triangle \( {ABC} \) and let the rays \( {BP} \) and \( {CP} \) meet the circumcircle of \( {ABC} \) at \( {B}^{ * } \) and \( {C}^{ * } \) respectively (see Figure 6). Then (a) \( B{B}^{ * } = C{C}^{ * } \) if and only if \( {PB} = {PC} \) or \( \angle {BPC} = 2\angle {BAC} \) ; | Proof. (a) It is clear that\n\n\[ B{B}^{ * } = C{C}^{ * } \Leftrightarrow \angle {BA}{B}^{ * } = \angle {CA}{C}^{ * }\text{ or }\angle {BA}{B}^{ * } + \angle {CA}{C}^{ * } = \pi \]\n\n\[ \Leftrightarrow \angle {CA}{B}^{ * } = \angle {BA}{C}^{ * }\text{or}\angle {CA}{B}^{ * } + \angle {BA}{C}^{ * } + 2\angle {BAC} = \pi... | Yes |
Theorem 7. (i) If \( {ABC} \) is a triangle whose centroid \( G \) has the property that \( \angle {BGC} = 2\angle {BAC} \), then \( \frac{\pi }{3} \leq \angle {BAC} < \frac{\pi }{2} \) with \( \angle {BAC} = \frac{\pi }{3} \) if and only if \( {ABC} \) is equilateral. | Proof. (i) Let \( \Omega \) be the circumcircle of \( {ABC} \) and let \( {E}^{\prime } \) be its circumcenter. Let \( {\Omega }^{\prime } \) be the circumcircle of \( {E}^{\prime }{BC} \) . Let \( M \) be the midpoint of \( {BC} \) and let the perpendicular bisector of \( {BC} \) meet \( \Omega \) at \( E \) and \( F ... | Yes |
Lemma 8. Let \( {ABC} \) be a triangle with side-lengths \( a, b \), and \( c \) (in the standard order) and with centroid \( G \) . Let the rays \( {BG} \) and \( {CG} \) meet the circumcircle of \( {ABC} \) at \( {B}^{ * } \) and \( {C}^{ * } \) respectively. Then\n\n\[ B{B}^{*2} = \frac{{\left( {a}^{2} + {c}^{2}\rig... | Proof. Let \( m = B{B}^{\prime }, x = B{B}^{ * } \) . By Apollonius’ theorem, \( {m}^{2} = \frac{2\left( {{a}^{2} + {c}^{2}}\right) - {b}^{2}}{4} \) . Since \( B{B}^{\prime }{B}^{ * } \) and \( A{B}^{\prime }C \) are diagonals of a cyclic quadrilateral, \( m\left( {x - m}\right) = \frac{{b}^{2}}{4} \) . It follows that... | Yes |
Theorem 9. Let \( {ABC} \) be a triangle with side-lengths \( a, b \), and \( c \) (in the standard order) and with centroid \( G \) . Let the rays \( {BG} \) and \( {CG} \) meet the circumcircle of \( {ABC} \) at \( {B}^{ * } \) and \( {C}^{ * } \), respectively. If \( b \neq c \), then the following are equivalent: (... | Proof. Since \( b \neq c \), it follows that \( {GB} \neq {GC} \) . By Lemma 6,(i) is equivalent to (ii). To see that (i) is equivalent to (iii), let \( x = B{B}^{ * }, y = C{C}^{ * } \), and let \( s = {a}^{2} + {b}^{2} + {c}^{2} \) . By Lemma 8,\n\n\[ {x}^{2} = \frac{{\left( s - {b}^{2}\right) }^{2}}{{2s} - 3{b}^{2}}... | Yes |
Theorem 1. Let \( {V}_{1},{V}_{2},{V}_{3} \) and \( I \) be four points in the plane in general positions. Let \( {q}_{1},{q}_{2},{q}_{3} \) be three different oriented lines through \( I\left( {{V}_{i} \notin {q}_{i}}\right) \) . There exist at most two values \( x \in \mathbb{R} \smallsetminus \{ 0\} \) such that for... | Proof. For a real number \( x \) and \( i = 1,2,3 \), let \( {Q}_{i}\left( x\right) \) be a point on \( {q}_{i} \) for which \( I{Q}_{i}\left( x\right) = x \) . The correspondences \( {Q}_{i}\left( x\right) \leftrightarrow {Q}_{j}\left( x\right) \) define perspectivities \( \left( {q}_{i}\right) \bar{ \land } \) \( \le... | Yes |
Theorem 1. For \( x, y, z \in \mathbb{R} \) ,\n\n\[ \n{x}^{2}\sqrt{{R}_{2} + {R}_{3}} + {y}^{2}\sqrt{{R}_{3} + {R}_{1}} + {z}^{2}\sqrt{{R}_{1} + {R}_{2}} \]\n\n\[ \n\geq \sqrt{2}\left( {{yz}\sqrt{{r}_{2} + {r}_{3}} + {zx}\sqrt{{r}_{3} + {r}_{1}} + {xy}\sqrt{{r}_{1} + {r}_{2}}}\right) . \]\n\n(2) | Liu's proof, however, is quite complicated. We give a simple proof of Theorem 1 as a corollary of a more general result, also conjectured by Liu in [3]. | No |
Lemma 3 ([4,5]). For \( x, y, z \in \mathbb{R},{p}_{i} \in \left( {-\infty ,0}\right) \bigcup \left( {0, + \infty }\right) \), and \( {q}_{i} \in \mathbb{R} \) for \( i = 1,2,3 \), the quadratic inequality of three variables\n\n\[ \n{p}_{1}{x}^{2} + {p}_{2}{y}^{2} + {p}_{3}{z}^{2} \geq {q}_{1}{yz} + {q}_{2}{zx} + {q}_{... | \[ \n\left\{ \begin{array}{l} {p}_{i} > 0, i = 1,2,3 \\ 4{p}_{2}{p}_{3} > {q}_{1}^{2},4{p}_{3}{p}_{1} > {q}_{2}^{2},4{p}_{1}{p}_{2} > {q}_{3}^{2} \\ 4{p}_{1}{p}_{2}{p}_{3} \geq {p}_{1}{q}_{1}^{2} + {p}_{2}{q}_{2}^{2} + {p}_{3}{q}_{3}^{2} + {q}_{1}{q}_{2}{q}_{3} \end{array}\right.\n\] | Yes |
Lemma 4. In \( \\bigtriangleup {ABC} \), we have\n\n\[ \n{\\sin }^{2}\\frac{A}{2} + {\\sin }^{2}\\frac{B}{2} + {\\sin }^{2}\\frac{C}{2} + 2\\sin \\frac{A}{2}\\sin \\frac{B}{2}\\sin \\frac{C}{2} = 1.\n\] | Proof. This follows from the formula \( {\\sin }^{2}\\alpha = \\frac{1}{2}\\left( {1 - \\cos {2\\alpha }}\\right) \) and the known identity\n\n\[ \n\\cos A + \\cos B + \\cos C = 1 + 4\\sin \\frac{A}{2}\\sin \\frac{B}{2}\\sin \\frac{C}{2}.\n\] | Yes |
Lemma 2. With respect the circle of center \( \left( {{x}_{0},{y}_{0}}\right) \) and radius \( {R}_{0} \), the inversive image of the circle with center \( \left( {{x}_{C},{y}_{C}}\right) \) and radius \( R \) is the circle with center \( \left( {{x}_{C}^{\mathrm{i}},{y}_{C}^{\mathrm{i}}}\right) \) and radius \( {R}^{\... | \[ {x}_{C}^{\mathrm{i}} = {x}_{0} + \frac{{R}_{0}^{2}}{{\left( {x}_{C} - {x}_{0}\right) }^{2} + {\left( {y}_{C} - {y}_{0}\right) }^{2} - {R}^{2}}\left( {{x}_{C} - {x}_{0}}\right) ,\] \[ {y}_{C}^{\mathrm{i}} = {y}_{0} + \frac{{R}_{0}^{2}}{{\left( {x}_{C} - {x}_{0}\right) }^{2} + {\left( {y}_{C} - {y}_{0}\right) }^{2} - ... | Yes |
Theorem 3. For the circles in the Pappus chains and their inversive images in the incircle, the following identities hold. For \( n \geq 2 \) , | \[ \frac{{\rho }_{\mathrm{{inc}}}}{{\rho }_{\mathrm{r}n}^{\mathrm{i}}} - \frac{{\rho }_{\mathrm{{inc}}}}{{\rho }_{\mathrm{r}n}} = \frac{{\rho }_{\mathrm{{inc}}}}{{\rho }_{\mathrm{a}n}^{\mathrm{i}}} - \frac{{\rho }_{\mathrm{{inc}}}}{{\rho }_{\mathrm{a}n}} = \frac{{\rho }_{\mathrm{{inc}}}}{{\rho }_{\mathrm{b}n}^{\mathrm{... | Yes |
Example 1. The SISS\n\n\\[ \n\\left( {a, b, c}\\right) \\rightarrow \\left( {{bc},{ca},{ab}}\\right) \n\\]\n\n(9) | gives\n\n\\[ \nu\\left( {u, v, w}\\right) = \\nu\\left( {{bc},{ca},{ab}}\\right) \n\\]\n\n\\[ \n= \\left( {bc}\\right) \\left( {ca}\\right) \n\\]\n\n\\[ \n= {ta} \n\\]\n\nwhere \\( t = {abc} \\) . | Yes |
Example 2. The SISS\n\n\\[ \n\\left( {a, b, c}\\right) \\rightarrow \\left( {{a}^{2} - {bc},{b}^{2} - {ca},{c}^{2} - {ab}}\\right) \n\\]\n\n(10) | gives\n\n\\[ \nu\\left( {u, v, w}\\right) = \\nu\\left( {{a}^{2} - {bc},{b}^{2} - {ca},{c}^{2} - {ab}}\\right) \n\\]\n\n\\[ \n= {\\left( {a}^{2} - bc\\right) }^{2} - \\left( {{b}^{2} - {ca}}\\right) \\left( {{c}^{2} - {ab}}\\right) \n\\]\n\n\\[ \n= {ta}\\text{,}\n\\]\n\nwhere\n\n\\[ \n t = \\left( {a + b + c}\\right) \... | Yes |
Example 3. The SISS\n\n\\[ \n\\left( {a, b, c}\\right) \\rightarrow \\left( {{b}^{2} + {c}^{2} - {ab} - {ac},{c}^{2} + {a}^{2} - {bc} - {ba},{a}^{2} + {b}^{2} - {ca} - {cb}}\\right) \n\\]\n\n(11) | gives\n\n\\[ \nu\\left( {u, v, w}\\right) = {ta}, \n\\]\n\nwhere\n\n\\[ \nt = 2\\left( {a + b + c}\\right) \\left( {{a}^{2} + {b}^{2} + {c}^{2} - {bc} - {ca} - {ab}}\\right) .\n\\] | Yes |
Lemma 1. The triangle \( {A}^{ * }{B}^{ * }{C}^{ * } \) is the polar triangle of the medial triangle DEF of triangle \( {ABC} \) with respect to \( \Gamma \) . | Proof. Because \( {\mathcal{C}}_{a} \) is orthogonal to \( \Gamma \), the line \( {\ell }_{a} \) is the polar of \( D \) with respect to \( \Gamma \) . Similarly, \( {\ell }_{b} \) and \( {\ell }_{c} \) are the polars of \( E \) and \( F \) with respect to the same circle. | Yes |
Lemma 2. The lines \( {XY},{BI},{EF} \), and \( A{C}^{ * } \) are concurrent at a point of \( {\mathcal{C}}_{b} \), as are the lines \( {YZ},{BI},{DE} \), and \( A{B}^{ * } \) (see Figure 2). | Proof. Let \( {A}_{b} \) as the point on \( {EF} \), on the same side of \( F \) as \( E \), so that \( F{A}_{b} = {FA} \). (i) Because \( {FA} = F{A}_{b} = {FB} \), the points \( A,{A}_{b} \) and \( B \) all lie on a circle with center \( F \). This implies that \( \angle {ABC} = \angle {AF}{A}_{b} = 2\angle {AB}{A}_{... | Yes |
Theorem 3. The points \( {A}^{ * },{B}^{ * } \), and \( {C}^{ * } \) are the respective orthocenters of triangles \( {BIC},{CIA} \), and \( {AIB} \) . | Proof. Because the point \( {A}_{b} \) lies on the polar lines of \( {A}^{ * } \) and \( C \) with respect to \( \Gamma \), we know that \( {A}^{ * }C \bot {BI} \) . Combining this with the fact that \( {A}^{ * }I \bot {BC} \) we conclude that \( {A}^{ * } \) is indeed the orthocenter of triangle \( {BIC} \) . | Yes |
Theorem 4. The medial triangle DEF is perspective with triangle \( {A}^{ * }{B}^{ * }{C}^{ * } \), at the Mittenpunkt \( {M}_{\mathrm{t}}{}^{1} \) of triangle \( {ABC} \) (see Figure 3). | Proof. Because \( {A}^{ * }C \) is perpendicular to \( {BI} \), it is parallel to the external bisector of angle \( B \) . A similar argument holds for \( B{A}^{ * } \), so we conclude that \( {A}^{ * }B{I}_{a}C \) is a parallelogram. It follows that \( {A}^{ * }, D \), and \( {I}_{a} \) are collinear. This shows that ... | Yes |
Theorem 5. The Nagel point \( {N}_{\mathrm{a}} \) of triangle \( {ABC} \) is the common orthocenter of triangles \( A{B}^{ * }{C}^{ * },{A}^{ * }B{C}^{ * },{A}^{ * }{B}^{ * }C \) . | Proof. Consider the homothety \( \zeta \mathrel{\text{:=}} \mathrm{h}\left( {D, - 1}\right) \cdot {}^{2} \) This carries \( A \) into the vertex \( {A}^{\prime } \) of the anticomplementary triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) of \( {ABC} \) . It follows from Theorem 4 that \( \zeta \left( {A}^{ * }\r... | Yes |
Theorem 6 (Ehrmann). The centroid \( {G}^{ * } \) of triangle \( {A}^{ * }{B}^{ * }{C}^{ * } \) is the point dividing \( {IH} \) in the ratio \( I{G}^{ * } : {G}^{ * }H = 2 : 1 \) . | Proof. The four points \( A,{A}_{b}, I,{A}_{c} \) all lie on a circle with diameter \( {IA} \), which we will call \( {\mathcal{C}}_{a}^{\prime } \) . Let \( H \) be the orthocenter of triangle \( {ABC} \), and \( S \) the (second) intersection of \( {\mathcal{C}}_{a}^{\prime } \) with the altitude \( {AH} \) . Constru... | Yes |
Lemma 7. The six points \( {A}^{ * },{A}_{b}^{ * },{A}_{c}^{ * },{Y}^{ * },{Z}^{ * } \), and \( {N}_{\mathrm{a}} \) all lie on the circle with diameter \( {A}^{ * }{N}_{\mathrm{a}} \) (see Figure 6). | Proof. The points \( {A}_{b}^{ * } \) and \( {A}_{c}^{ * } \) lie on the circle with diameter \( {A}^{ * }{N}_{\mathrm{a}} \) by definition.\n\nWe know that the Nagel point and the Gergonne point are isotomic conjugates, so if we call \( {Y}^{\prime } \) the intersection of \( B{N}_{\mathrm{a}} \) and \( {AC} \), it fo... | Yes |
Lemma 8. The circle \( {\mathcal{C}}_{a}^{ * } \) intersects \( {\mathcal{C}}_{b} \) in the points \( {Y}^{ * } \) and \( {A}_{b}^{ * } \) . The point \( {A}_{b}^{ * } \) lies on \( {EF} \) (see Figure 7). | Proof. The point \( {Y}^{ * } \) lies on \( {\mathcal{C}}_{b} \) by definition, and on \( {\mathcal{C}}_{a}^{ * } \) by Lemma 7.\n\nConsider the homothety \( \phi \mathrel{\text{:=}} \mathrm{h}\left( {E, - 1}\right) \) . We already know that \( \phi \left( {A{C}^{ * }}\right) = C{A}^{ * } \) and \( \phi \left( {BI}\rig... | Yes |
Theorem 9. The point \( X \) has equal powers with respect to the circles \( {\mathcal{C}}_{b},{\mathcal{C}}_{c},{\mathcal{C}}_{a}^{ * } \) , and \( {\mathcal{C}}_{a}^{\prime } \) (see Figure 7). | Proof. Let us call \( {S}_{a} \) the intersection of \( E{Y}^{ * } \) and \( {BC} \), and \( {S}_{b} \) the intersection of \( X{Y}^{ * } \) and \( {EF} \) . Because \( E{Y}^{ * } \) is tangent to \( \Gamma \), we have \( {S}_{a}{Y}^{ * } = {S}_{a}X \) . Because triangles \( X{S}_{a}{Y}^{ * } \) and \( {S}_{b}E{Y}^{ * ... | Yes |
Theorem 10. The point \( U \) is a center of similitude of circles \( {\mathcal{C}}_{a}^{\prime } \) and \( {\mathcal{C}}_{a}^{ * } \) . Likewise, \( V \) is a center of similitude of circles \( {\mathcal{C}}_{b}^{\prime } \) and \( {\mathcal{C}}_{b}^{ * } \), and \( W \) of \( {\mathcal{C}}_{c}^{\prime } \) and \( {\m... | Proof. We know from Lemma 2 and Theorem 5 that \( {A}^{ * }{A}_{b}^{ * }\parallel A{A}_{b} \), and \( {AI}\parallel \) \( {A}^{ * }{N}_{\mathrm{a}} \), as well as \( {A}_{b}^{ * }{N}_{\mathrm{a}}\parallel {A}_{b}I \) . Hence triangles triangle \( {A}^{ * }{N}_{\mathrm{a}}{A}_{b}^{ * } \) and triangle \( {AI}{A}_{b} \) ... | Yes |
Theorem 11. The point \( U \) is a center of similitude of circles \( {\mathcal{C}}_{b} \) and \( {\mathcal{C}}_{c} \). Likewise, \( V \) is a center of similitude of circles \( {\mathcal{C}}_{c} \) and \( {\mathcal{C}}_{a} \), and \( W \) of \( {\mathcal{C}}_{a} \) and \( {\mathcal{C}}_{b} \) . | Proof. By Theorem 10, we know that\n\n\[ \frac{{A}_{b}U}{{A}_{c}U} = \frac{{A}_{b}^{ * }U}{{A}_{c}^{ * }U} \]\n\n(1)\n\nBy Table 1 and Theorem 8, we know that \( {A}_{b},{A}_{c}^{ * } \) lie on \( {\mathcal{C}}_{c} \) and \( {A}_{b},{A}_{b}^{ * } \) lie on \( {\mathcal{C}}_{b} \). Knowing that \( U \) lies on \( {EF} \... | Yes |
Theorem 12. The lines \( {BV} \) and \( {CW} \) intersect at a point on \( {EF} \) . This point is the center of similitude different from \( U \) of \( {\mathcal{C}}_{b} \) and \( {\mathcal{C}}_{c} \) (see Figure 9). | Proof. Let us call \( {U}^{\prime } \) the point of intersection of \( {BV} \) and \( {EF} \) . We have that \( G = \) \( {BE} \cap {CF} \) and \( V = {DF} \cap B{U}^{\prime } \) . By the theorem of Pappus-Pascal applied to the collinear triples \( E,{U}^{\prime }, F \) and \( C, D, B \), the intersection of \( {U}^{\p... | Yes |
Theorem 13. The point \( {X}^{\prime \prime } \) is the center of similitude different from \( U \) of circles \( {\mathcal{C}}_{a}^{\prime } \) and \( {\mathcal{C}}_{a}^{ * } \) . Likewise, \( {Y}^{\prime \prime } \) is a center of similitude of \( {\mathcal{C}}_{b}^{\prime } \) and \( {\mathcal{C}}_{b}^{ * } \), and ... | Proof. We construct the line \( {l}_{{X}^{\prime \prime }} \) which passes through \( {X}^{\prime \prime } \) and is parallel to \( {BC} \) . The triangle bounded by \( {AC},{AB},{l}_{{X}^{\prime \prime }} \) has \( \Gamma \) as its excircle opposite \( A \) . This implies that its Nagel point lies on \( A{X}^{\prime \... | Yes |
Theorem 15. The line connecting the centers of \( {\mathcal{C}}_{a}^{\prime } \) and \( {\mathcal{C}}_{a}^{ * } \) passes through the Feuerbach point of triangle \( {ABC} \) ; so do the lines joining the centers of \( {\mathcal{C}}_{b}^{\prime },{\mathcal{C}}_{b}^{ * } \) and those of \( {\mathcal{C}}_{c}^{\prime },{\m... | Proof. Let us call \( {H}_{a} \) the orthocenter of triangle \( A{A}_{b}{A}_{c} \) . Since \( {AI} \) is the diameter of \( {\mathcal{C}}_{a}^{\prime } \) (as in the proof of Theorem 6), we have \( A{H}_{a} = {AI} \cdot \cos {A}_{b}A{A}_{c} = {AI} \cdot \sin \frac{A}{2} \) , where the last equality follows from \( \fra... | Yes |
Theorem 16. The radical axis of \( {\mathcal{C}}_{a}^{\prime } \) and \( {\mathcal{C}}_{a}^{ * } \) passes through the Feuerbach point of triangle \( {ABC} \) ; so do the radical axes of \( {\mathcal{C}}_{b}^{\prime },{\mathcal{C}}_{b}^{ * } \), and of \( {\mathcal{C}}_{c}^{\prime },{\mathcal{C}}_{c}^{ * } \) (see Figu... | Proof. Because the radical axis of two circles is perpendicular to the line joining the centers of the circles, the radical axis \( {\mathcal{R}}_{a} \) of \( {\mathcal{C}}_{a}^{\prime } \) and \( {\mathcal{C}}_{a}^{ * } \) is perpendicular to the Euler line of triangle \( A{A}_{b}{A}_{c} \) . Since this Euler line con... | Yes |
Lemma 1. Suppose triangles \( {ABC} \) and \( A{B}^{\prime }{C}^{\prime } \) have a common angle at \( A \), and that the incircle of \( A{B}^{\prime }{C}^{\prime } \) is not greater than the incircle of \( {ABC} \) . If \( {C}^{\prime } > C \), then the bisector of \( {C}^{\prime } \) is less than the bisector of \( C... | Proof. Let \( {CF} \) and \( {C}^{\prime }{F}^{\prime } \) be the bisectors of angles \( C,{C}^{\prime } \) of triangles \( {ABC} \) , \( A{B}^{\prime }{C}^{\prime } \) . Assuming \( {C}^{\prime } > C \), we shall prove that \( {C}^{\prime }{F}^{\prime } < {CF} \) .\n\n![bb70723e-cbb9-45f6-82d9-918ae4727256_1894_0.jpg]... | Yes |
Lemma 2. Suppose triangles \( {ABC} \) and \( A{B}^{\prime }{C}^{\prime } \) have a common angle at \( A \), and a common angle bisector \( {AD} \), the common angle not greater than any other angle of \( A{B}^{\prime }{C}^{\prime } \) . If \( {C}^{\prime } > C \), then the bisector of \( {C}^{\prime } \) is less than ... | Proof. If the incirle of triangle \( A{B}^{\prime }{C}^{\prime } \) is not greater than that of \( {ABC} \), then the result follows from Lemma 1.\n\nAssume the incircle of \( A{B}^{\prime }{C}^{\prime } \) greater than the incircle of \( {ABC} \) (see Figure 3). The line \( {BC} \) cuts the incircle of \( A{B}^{\prime... | Yes |
Theorem 3. If three internal angle bisectors of triangle \( {ABC} \) are respectively equal to three internal angle bisectors of triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \), then the triangles are congruent. | Proof. Denote the angle bisectors of \( {ABC} \) by \( {AD},{BE},{CF} \) and let \( {AD} = {A}^{\prime }{D}^{\prime } \) , \( {BE} = {B}^{\prime }{E}^{\prime },{CF} = {C}^{\prime }{F}^{\prime } \) .\n\nIf for the angles of the triangles we have \( A = {A}^{\prime }, B = {B}^{\prime }, C = {C}^{\prime } \), then from th... | Yes |
For any complete quadrilateral \( ▱{ABCD} \) (as defined above), let \( {F}_{BC} \) be the unique point on \( {AD} \) such that \( {F}_{BC}{E}_{FG} \) is parallel to \( {BC} \) and let \( {F}_{DA} \) , \( {G}_{AB} \) and \( {G}_{CD} \) be defined similarly. Finally, let \( {F}_{G} \) and \( {G}_{F} \) be the midpoints ... | Proof. Note that by the harmonic property of quadrilaterals, the sides \( {DA} \) and \( {BC} \) are harmonically separated by \( F{E}_{FG} \) and \( {FG} \) . Therefore, the points \( {F}_{DA} \) and \( {F}_{BC} \) are harmonically separated by the points of intersection \( F{E}_{FG} \cap {F}_{DA}{F}_{BC} \) and \( {F... | Yes |
Corollary 2. With the notation introduced above, the directed ratios \( \frac{{F}_{BC}D}{{F}_{BC}A} \) and \( \frac{{F}_{DA}C}{{F}_{DA}B} \) are equal, as are the ratios \( \frac{{G}_{CD}A}{{F}_{CD}B} \) and \( \frac{{G}_{AB}D}{{F}_{AB}C} \) . | Proof. It suffices to prove the first part of the statement. Note that by construction the ratio \( \frac{{F}_{BC}D}{{F}_{BC}A} \) is equal to the cross ratio \( \left\lbrack {{E}_{FG}D,{E}_{FG}A;{E}_{FG}{F}_{BC},{E}_{FG}{F}_{DA}}\right\rbrack \) of the lines \( {E}_{FG}D,{E}_{FG}A,{E}_{FG}{F}_{BC} \), and \( {E}_{FG}{... | Yes |
Lemma 3. For any quadrilateral \( {ABCD} \) (with its sides in general position), the Miquel points of \( ▱{AB}{F}_{DA}{F}_{BC} \) and \( ▱{CD}{F}_{BC}{F}_{DA} \) both coincide with the Miquel point \( M \) of \( ▱{ABCD} \) . | Proof. Let \( M \) be constructed as the second point of intersection (other than \( F \) ) of the circumcircles of \( \bigtriangleup {FAB} \) and \( \bigtriangleup {FCD} \) . By Corollary 2, the ratio of the power of \( {F}_{BC} \) with respect to the circumcircle of \( \bigtriangleup {FCD} \) and the power of \( {F}_... | Yes |
Corollary 4. For any quadrilateral ABCD (with sides in general position), the (orthogonal) projection of \( M \) on \( {F}_{BC}{F}_{DA} \) lies on the pedal line of \( ▱{ABCD} \) . | Proof. By Lemma 3 and the properties of Miquel points, the (orthogonal) projection of \( M \) on \( {F}_{BC}{F}_{DA} \) is collinear with the (orthogonal) projections of \( M \) on \( {AB} \) , \( B{F}_{DA} \) and \( {F}_{BC}A \), i.e. its projections on \( {AB},{BC} \), and \( {DA} \) . But for \( {ABCD} \) in general... | Yes |
Corollary 6 (Emelyanov). For any quadrilateral ABCD (with sides in general position), the Miquel point \( M \) of \( ▱{ABCD} \) lies on the nine-point circle of the diagonal triangle \( \bigtriangleup {E}_{AC}{E}_{BD}{E}_{FG} \) of \( ▱{ABCD} \) . | Proof. Since the (orthogonal) projections of \( M \) on the sides of \( \bigtriangleup {M}_{AC}{M}_{BD}{M}_{FG} \) are collinear, \( M \) has to lie on the circumcircle of \( \bigtriangleup {M}_{AC}{M}_{BD}{M}_{FG} \) . But this is the same as saying that \( M \) lies on the nine-point circle of \( \bigtriangleup {E}_{... | Yes |
For any three distinct points \( A, B \) and \( C \), the line \( {AC} \) is perpendicular to \( {BC} \) if and only if \( C \) lies on the unique circle with diameter \( {AB} \) . | Let \( O \) be the center of the circle with diameter \( {AB} \) . Since both \( \bigtriangleup {AOC} \) and \( \bigtriangleup {BOC} \) are isosceles, the two line of the symmetry section of \( \{ {AB},{OC}\} \) are each perpendicular to one of \( {AC} \) and \( {BC} \) . Consequently, by BP 4 (Rotation), \( \{ {AC},{B... | Yes |
Lemma 7. Let \( O \) be the center of a circular inversion. Then, under this inversion (i) any circle not passing through \( O \) is mapped onto a circle not passing through \( O \) ,(ii) any line not passing through \( O \) is mapped onto a circle passing through \( O \) and vice versa (with the point at infinity of t... | Proof. Starting with (i), draw the line connecting \( O \) with the center of the circle not passing through \( O \) and let the points of intersection of this line with the latter circle be \( A \) and \( B \) (see Figure 1). Then, by Theorem 6 (Thales), the lines \( {AC} \) and \( {BC} \) are perpendicular. Let \( {A... | Yes |
Theorem 8 (Equal Angle). For four points on either a circle or a straight line, let \( X, Y, Z, W \) be any permutation of \( A, B, C, D \) . Then, any 2-section \( X\{ Y, Z\} \) is directly congruent to the 2-section \( W\{ Y, Z\} \) and the sections are either trivial (in case the points are collinear) or non-trivial... | Proof. It suffices to prove both statements for one permutation of \( A, B, C, D \) . Assume that \( A, B, C, D \) are co-cyclic or collinear. Let \( {B}^{\prime },{C}^{\prime } \) and \( {D}^{\prime } \) denote the images of \( B, C \) and \( D \), respectively, under circular inversion with respect to \( A \) . Then,... | Yes |
Corollary 9. Let \( A, B, C, D \) be any four co-cyclic points with \( E = {AC} \cap {BD} \) . Then the product of directed lengths \( \overline{AE} \cdot \overline{CE} \) equals the product \( \overline{BE} \cdot \overline{DE} \) . | Proof. Let \( {A}^{\prime } \) and \( {B}^{\prime } \) be the images under inversion of \( A \) and \( B \) with respect to \( E \) (and a circle of radius \( r \) ). Then \( \bigtriangleup {A}^{\prime }{B}^{\prime }E \) and \( \bigtriangleup {CDE} \) are directly similar with two legs in common. Therefore \( \overline... | No |
Lemma 10 (Bow, String and Arrow). For any triangle \( \bigtriangleup {ABC} \), let \( {C}^{\prime } \) be the midpoint of \( {AB} \) and let \( O \) be the circumcenter of the triangle and let \( {T}_{{AB}, C} \) denote the tangent line to the circumcircle of \( \bigtriangleup {ABC} \) at \( C \) . Then \( C\{ B, A\} \... | Proof. It suffices to prove the first statements of (i) and (ii). Let \( {A}^{\prime } \) be the midpoint of \( {BC} \) . Since \( O{C}^{\prime } \) is perpendicular to \( {AB} \) and \( O{A}^{\prime } \) is perpendicular to \( {BC} \), it follows that \( ▱{C}^{\prime }O{A}^{\prime }B \) is cyclic and therefore that \(... | Yes |
Theorem 11. An arbitrary 6-section \( \left\{ {{l}_{1},{l}_{2},{m}_{1},{m}_{2},{n}_{1},{n}_{2}}\right\} \) can be formed from the sides of a complete quadrangle \( \boxtimes {ABCD} \) (such that \( {l}_{1},{l}_{2} \) and so on are pairs of opposite sides) if and only if the three pairs of opposite sides can be rearrang... | Proof. Since any two quadrilaterals determine a projective map, every complete quadrangle is projective to the configuration of a rectangle and its diagonals. Therefore the diagonal points of a complete quadrangle are never collinear and every quadrangle in the affine plane has at least one pair of opposite sides which... | Yes |
For any complete quadrangle \( \boxtimes {ABCD} \), there is an involution that pairs the points of intersection of its opposite sides with the line at infinity. | Proof. Without loss of generality, we may assume that \( \{ {AB},{CD}\} \) is non-trivial. Let \( {L}_{1} = {AB} \cap {\ell }_{\infty } \) and so on. Then \( \left\lbrack {{L}_{1},{L}_{2},{M}_{1},{N}_{2}}\right\rbrack = \left\lbrack {{L}_{1},{L}_{2},{N}_{1},{M}_{2}}\right\rbrack \) . Now let \( \varphi \) be the involu... | Yes |
Corollary 13. Any Desarguesian 6-section is associated with two similarity classes of quadrilaterals (which may coincide). | Proof. Let the 6-section be denoted by \( \left\{ {{l}_{1},{l}_{2},{m}_{1},{m}_{2},{n}_{1},{n}_{2}}\right\} \) . If the cross ratio \( \left\lbrack {{l}_{1},{l}_{2};{m}_{1},{n}_{2}}\right\rbrack \) equals \( \left\lbrack {{l}_{1},{l}_{2};{n}_{1},{m}_{2}}\right\rbrack \), then the cross ratio \( \left\lbrack {{l}_{2},{l... | Yes |
Corollary 14. A complete quadrangle is directly reciprocal to itself if and only if it is orthocentric. | Proof. Since for any complete quadrangle directly reciprocal to itself all three 2- sections of opposite sides have to be both directly and inversely congruent, it follows that all opposite sides are perpendicular to each other. In other words, every vertex is the orthocenter of the triangle formed by the other three v... | Yes |
Corollary 15. A complete quadrangle is inversely reciprocal to itself if and only if it is cyclic. | Proof. Let \( \boxtimes {ABCD} \) denote the complete quadrangle. Then, if \( \boxtimes {ABCD} \) is inversely reciprocal to itself, \( \{ {AB},{AC}\} \) has to be inversely congruent to \( \{ {CD},{BD}\} \) or \( A\{ B, C\} { \cong }_{D}D\{ B, C\} \) . But this means that \( \boxtimes {ABCD} \) is cyclic. The converse... | No |
Lemma 16 (Invariance of Ratios). Let \( \boxtimes {ABCD} \) and \( \boxtimes {A}^{ * }{B}^{ * }{C}^{ * }{D}^{ * } \) be a pair of (affine) reciprocal quadrangles and diagonal points \( E, F, G \) and \( {E}^{ * },{F}^{ * },{G}^{ * } \) , respectively. Moreover, let \( X, Y \), and \( Z \) be any collinear triple of two... | Proof. The statement is trivial for any diagonal point on \( {\ell }_{\infty } \) . Without loss of generality, let us assume that the diagonal point \( F \) is in the affine plane. It now suffices to prove the statement for \( B, C \) and \( F \) . Under inversion with respect to \( F \) and a circle of radius \( r \)... | No |
Lemma 17 (Maxwell). Let \( \boxtimes {ABCD} \) and \( \boxtimes {A}^{ * }{B}^{ * }{C}^{ * }{D}^{ * } \) be a pair of reciprocal quadrangles. Then\n\n\[ \frac{\left| {AB}\right| \left| {CD}\right| }{\left| {{A}^{ * }{B}^{ * }}\right| \left| {{C}^{ * }{D}^{ * }}\right| } = \frac{\left| {AC}\right| \left| {BD}\right| }{\l... | Proof. Assume again that the point \( F \) is in the affine plane. Under inversion with respect to \( F \) and a circle of radius \( r \), we find \( \left| {{A}^{ * }{D}^{ * }}\right| = \left| {{r}^{2}/}\right| {FD}\left| {-{r}^{2}/}\right| {FA}\left| \right| = \) \( {r}^{2}\left| {AD}\right| /\left( {\left| {FA}\righ... | Yes |
For a given triangle \( \bigtriangleup {ABC} \) and any non-trivial 3-section \( \{ l, m, n\} \) not inversely congruent to \( \{ {BC},{CA},{AB}\} \) there is exactly one point \( D \) in the plane of \( \bigtriangleup {ABC} \) (and not on the sides of \( \bigtriangleup {ABC} \) ) such that \( \{ {AD},{BD},{CD}\} \) is... | Proof. Without loss of generality, we may assume that \( l, m \), and \( n \) are concurrent at a point \( Q \) . Let a point \( \mathrm{L} \) be a fixed point on \( l \) and let \( M \) be a variable point on \( m \) . Now construct a triangle \( \bigtriangleup {LMN} \) directly similar to \( \bigtriangleup {ABC} \) .... | Yes |
Theorem 19. A projective map of the plane is affine if and only if it can be obtained by reciprocation of a complete quadrangle \( \boxtimes {ABCD} \) with no parallel sides. Any such map reverses orientation if \( \boxtimes {ABCD} \) is convex and retains orientation when not. The map is Euclidean if and only if \( \b... | Proof. We already proved the if-part above. For an affine map, consider a triangle \( \bigtriangleup {ABC} \) and its image \( \bigtriangleup {A}^{ * }{B}^{ * }{C}^{ * } \) . By the previous lemma there is at least one point \( D \) (not on the sides of \( \bigtriangleup {ABC} \) ) such that \( \{ {AD},{BD},{CD}\} \) i... | Yes |
Lemma 20 (Generalized Ptolemy). For any six points \( A, B, C, D, P \), and \( Q \) in the (affine) plane\n\n\[ \left| {\bigtriangleup {PAB}}\right| \left| {\bigtriangleup {QCD}}\right| + \left| {\bigtriangleup {PCD}}\right| \left| {\bigtriangleup {QAB}}\right| \]\n\n\[ + \left| {\bigtriangleup {PAD}}\right| \left| {\b... | Proof. We represent the points \( A, B, C, D, P \), and \( Q \) by vectors \( \overrightarrow{a} = \left( {{a}_{1},{a}_{2},1}\right) \) and so on. Now consider the vectors \( {\left( \overrightarrow{a} \oplus \overrightarrow{a}\right) }^{T},\ldots ,{\left( \overrightarrow{d} \oplus \overrightarrow{d}\right) }^{T} \), a... | Yes |
For any complete quadrangle \( \boxtimes {ABCD} \) and \( E = {AC} \cap {BD} \) and a point \( P \) both in the (affine) plane of the quadrangle,\n\n\[ \left| {\bigtriangleup {PDA}}\right| \cdot \left| {\bigtriangleup {EBC}}\right| + \left| {\bigtriangleup {PBC}}\right| \cdot \left| {\bigtriangleup {EDA}}\right| \]\n\n... | Proof. Let \( Q \) coincide with \( E \) . \n\nNow, let \( \boxtimes {ABCD} \) be convex. Then \( E = {AC} \cap {BD} \) is in the affine plane and we can obtain \( \boxtimes {A}^{ * }{B}^{ * }{C}^{ * }{D}^{ * } \) by circular inversion with respect to \( E \) . Also, \( ▱{A}^{ * }{B}^{ * }{C}^{ * }{D}^{ * } \) is conve... | Yes |
Theorem 22. For any convex quadrilateral \( ▱{ABCD} \) with \( E = {AC} \cap {BD} \), let \( {I}_{AB} \) be the incenter of \( \bigtriangleup {EAB} \) and so on. Then \( \boxtimes {I}_{AB}{I}_{BC}{I}_{CD}{I}_{DA} \) is cyclic if and only if \( ▱{ABCD} \) is inscribable. | Proof. See [1] and the references there. The convexity requirement might not be necessary. | No |
Lemma 1. The slope \( m \) of the lines that pass through the point \( \left( {p, q}\right) \) and are tangent to the circle \( {x}^{2} + {\left( y - t\right) }^{2} = {t}^{2} \) satisfy\n\n\[ \n{m}^{2} + \frac{{2p}\left( {t - q}\right) }{{p}^{2} - {t}^{2}}m + \frac{{q}^{2} - {2qt}}{{p}^{2} - {t}^{2}} = 0.\n\] | Proof. In order for the line \( y = m\left( {x - p}\right) + q \) to be tangent to the circle \( {x}^{2} + (y - \) \( t{)}^{2} = {t}^{2} \) the minimum distance between the line and \( \left( {0, t}\right) \) must be \( t \) . Since the minimum distance between \( \left( {0, t}\right) \) and the line \( y = {mx} + \lef... | Yes |
Theorem 3. For \( \beta \neq {90}^{ \circ } \) and \( \alpha \) given there are at most two triangles \( T \) which can produce \( S \) in standard position. These triangles \( T \) have vertices located at\n\n\[ \left( {-1,0}\right) ,\left( {\frac{-{2t}}{{2t} + \left( {1 - {t}^{2}}\right) \tan \beta },\frac{{2t}\tan \... | Proof. The only thing left to prove are the bounds. For the upper bound, we must have that the second vertex is in the top half plane and so we need\n\n\[ \frac{{2t}\tan \beta }{{2t} + \left( {1 - {t}^{2}}\right) \tan \beta } > 0 \]\n\nIf \( \tan \beta > 0 \) then we need\n\n\[ {2t} + \left( {1 - {t}^{2}}\right) \tan \... | Yes |
Theorem 2. The vertex triangle of distinct circum-anticevian triangles is perspective to \( {ABC} \) . | Proof. The method of \( §2 \) applies, starting with\n\n\[ \left( \begin{array}{l} {A}^{\prime } \\ {B}^{\prime } \\ {C}^{\prime } \end{array}\right) = \left( \begin{array}{lll} {x}_{1} & {y}_{1} & {z}_{1} \\ {x}_{2} & {y}_{2} & {z}_{2} \\ {x}_{3} & {y}_{3} & {z}_{3} \end{array}\right) = \left( \begin{matrix} {ayz} & \... | Yes |
Theorem 3. Suppose \( X \) is a point not on a sideline of triangle \( {ABC} \) . Then the vertex triangle of the circumcevian triangles of \( X \) and \( {X}^{-1} \) is homothetic to \( {ABC} \) , and likewise for the pairs of vertex triangles in \( §§3 - 5 \) . | Proof. In accord with the definition of isogonal conjugate, trilinears for \( U = {X}^{-1} \) are given by \( u = {yz}, v = {zx}, w = {xy} \), so that\n\n\[ u : v : w = {x}^{-1} : {y}^{-1} : {z}^{-1}. \]\n\nIn the notation of \( §2 \), the vertex triangle (1) is given by its \( A \) -vertex \( {x}_{7} : {y}_{7} : {z}_{... | Yes |
Theorem 1. The nine-point circles of triangles \( {AIB},{AIC},{BIC} \) are concurrent at the Feuerbach point \( {F}_{\mathrm{e}} \) of triangle \( {ABC} \). | Proof. The Poncelet point of \( A, B, C, I \) must lie on the pedal circle of \( I \) with respect to triangle \( {ABC} \), and on the nine-point circle of triangle \( {ABC} \) (see Figure 1). Since these two circles have only the Feuerbach point \( {F}_{\mathrm{e}} \) in common, it must be the Poncelet point of \( A, ... | Yes |
Theorem 3 (Hatzipolakis). The Euler lines of triangles \( A{A}_{b}{A}_{c}, B{B}_{a}{B}_{c}, C{C}_{a}{C}_{b} \) are concurrent at \( {F}_{\mathrm{e}} \) (see Figure 4). | Proof. If \( {X}^{\prime } \) is the antipode of \( X \) in the incircle, \( {O}_{a} \) the midpoint of \( A \) and \( I,{H}_{a} \) the orthocenter of triangle \( A{A}_{b}{A}_{c} \), then clearly \( {\mathrm{H}}_{a}{\mathrm{O}}_{a} \) is the Euler line of triangle \( A{A}_{b}{A}_{c} \). Also, \( \angle {A}_{b}A{A}_{c} ... | Yes |
Theorem 4. If \( {X}^{\prime \prime },{Y}^{\prime \prime },{Z}^{\prime \prime } \) are the reflections of \( X, Y, Z \) in \( {AI},{BI},{CI} \), then the lines \( D{X}^{\prime \prime }, E{Y}^{\prime \prime }, F{Z}^{\prime \prime } \) concur at the Feuerbach point \( {F}_{\mathrm{e}} \) (see Figure 6). | Proof. We show that the line \( D{X}^{\prime \prime } \) contains the Feuerbach point \( {F}_{\mathrm{e}} \) . The same reasoning will apply to \( E{Y}^{\prime \prime } \) and \( F{Z}^{\prime \prime } \) as well.\n\nClearly, \( {X}^{\prime \prime } \) lies on the incircle. If we call \( N \) the nine-point center of tr... | Yes |
Theorem 5. The reflections of \( {OI} \) in the sidelines of the intouch triangle DEF are concurrent at the Feuerbach point of triangle \( {ABC} \) (see Figure 7). | Proof. Let us call \( {I}_{1} \) the reflection of \( I \) in \( {YZ} \) . By Theorem 1, the nine-point circle of triangle \( {AIC} \), which clearly passes through \( Y,{O}_{a},{C}_{a} \), also passes through \( {F}_{\mathrm{e}} \) . If \( S \) is the intersection of \( {YZ} \) and \( {AI} \), then clearly \( A \) is ... | Yes |
Theorem 7. The reflections of \( {OI} \) in the sidelines of the medial triangle \( {DEF} \) are concurrent at the Feuerbach point of triangle \( {ABC} \) (see Figure 8). | Proof. Call \( {I}_{2} \) the reflection of \( I \) in \( {EF} \), and \( {A}^{ * } \) the orthocenter of triangle \( {BIC} \) . The midpoint of \( I \) and \( {A}^{ * } \) is called \( M \) . Using Lemma 6, we know that \( {EF} \) is the polar line of \( {A}^{ * } \) with respect to the incircle. A similar argument as... | Yes |
Theorem 8. The three reflections of \( {\mathrm{H}}_{a}{\mathrm{O}}_{a} \) in the sidelines of triangle \( A{A}_{b}{A}_{c} \) and the line OI are concurrent at the reflection \( {E}_{a} \) of \( {F}_{\mathrm{e}} \) in \( {A}_{b}{A}_{c} \) . Similar theorems hold for triangles \( B{B}_{a}{B}_{c}, C{C}_{a}{C}_{b} \) (see... | Proof. The 3 reflections of \( {\mathrm{H}}_{a}{\mathrm{O}}_{a} \) in the sidelines of triangle \( A{A}_{b}{A}_{c} \) are concurrent at the Euler reflection point of triangle \( A{A}_{b}{A}_{c} \) . We will first show that this point is the reflection of \( {F}_{\mathrm{e}} \) in \( {A}_{b}{A}_{c} \) .\n\nThe circle wi... | Yes |
Theorem 9. The lines \( {AU},{BV},{CW} \) are concurrent at \( {X}_{80} \), the reflection of \( I \) in \( {F}_{\mathrm{e}} \) (see Figure 10). | Proof. The previous theorem tells us that \( {E}_{a} \) lies on \( {OI} \) . It follows that \( \angle {E}_{a}I{O}_{a} = \) \( \angle {OIA} \) . In the proof of Theorem 5, we prove that \( \angle {AIO} = \angle A{H}_{a}{O}_{a} \) . Since \( {F}_{e}{E}_{a} \) and \( A{H}_{a} \) are parallel, we deduce that \( {E}_{a}, I... | Yes |
Theorem 1. The relation (1) holds for triangles \( {ABC} \) and \( {DEF} \) if and only if they are orthologic. | Proof. The proofs in this paper will all be analytic.\n\nIn the rectangular co-ordinate system in the plane, we shall assume throughout that \( A\left( {0,0}\right), B\left( {1,0}\right), C\left( {u, v}\right), D\left( {d,\delta }\right), E\left( {e,\varepsilon }\right) \) and \( F\left( {f,\varphi }\right) \) for real... | Yes |
Theorem 2. The triangles \( {S}_{1}{S}_{3}{S}_{5} \) and \( {S}_{2}{S}_{4}{S}_{6} \) have equal area if and only if the triangles \( {ABC} \) and \( {DEF} \) are orthologic. | Proof. The vertices \( V \) and \( U \) of the square \( {DEVU} \) have co-ordinates \( (e + \varepsilon - \delta ,\varepsilon + d - e) \) and \( \left( {d + \varepsilon - \delta ,\delta + d - e}\right) \). From this we infer easily co-ordinates of all points in Figure 2. With the notation \( {u}_{ + } = u + v,{u}_{ - ... | Yes |
Theorem 3. The triangle \( {S}_{2}^{2}{S}_{4}^{2}{S}_{6}^{2} \) is the image of the triangle \( {S}_{1}^{2}{S}_{3}^{2}{S}_{5}^{2} \) under the rotation \( \rho \left( {{G}_{\sigma },\frac{\pi }{2}}\right) \) . The radical axis of their circumcircles goes through the centroid \( {G}_{\sigma } \) . | Proof. Since the point that divides the segment \( {DE} \) in the ratio \( 2 : 1 \) has coordinates \( \left( {\frac{d + {2e}}{3},\frac{\delta + {2\varepsilon }}{3}}\right) \), it follows that\n\n\[ \n{S}_{1}^{2}\left( {\frac{1 + {d}_{ + }}{3},\frac{1 - {d}_{ - }}{3}}\right) \;\text{ and }\;{S}_{2}^{2}\left( {\frac{{d}... | Yes |
For any real number \( t \) different from -1,0 and 2, the triangles \( {S}_{1}^{t}{S}_{3}^{t}{S}_{5}^{t} \) and \( {S}_{2}^{t}{S}_{4}^{t}{S}_{6}^{t} \) have equal area if and only if the triangles \( {ABC} \) and \( {DEF} \) are or-thologic. | Since the point that divides the segment \( {DE} \) in the ratio \( t : 1 \) has co-ordinates \( \left( {\frac{d + {te}}{t + 1},\frac{\delta + {t\varepsilon }}{t + 1}}\right) \), it follows that the points \( {S}_{i}^{t} \) have the co-ordinates\n\n\[ \n{S}_{1}^{t}\left( {\frac{t\left( {1 + {d}_{ + }}\right) }{2\left( ... | Yes |
Theorem 5. For any real number \( s \) different from -1 and 0, the triangles \( {T}_{1}^{s}{T}_{3}^{s}{T}_{5}^{s} \) and \( {T}_{2}^{s}{T}_{4}^{s}{T}_{6}^{s} \) have equal area if and only if the triangles \( {ABC} \) and \( {DEF} \) are orthologic. | Proof. As in the proof of Theorem 4, we find that the difference of areas of the triangles \( {T}_{1}^{s}{T}_{3}^{s}{T}_{5}^{s} \) and \( {T}_{2}^{s}{T}_{4}^{s}{T}_{6}^{s} \) is \( \frac{s\Delta }{4\left( {s + 1}\right) } \) . Hence, for \( s \neq - 1,0 \), the triangles \( {T}_{1}^{t}{T}_{3}^{t}{T}_{5}^{t} \) and \( {... | Yes |
Theorem 6. For any point \( P \) and any real numbers \( s \neq - 1 \) and \( t \neq - 1,\frac{2s}{s + 1} \), the triangles \( {U}_{1}^{\left( s, t\right) }{U}_{3}^{\left( s, t\right) }{U}_{5}^{\left( s, t\right) } \) and \( {U}_{2}^{\left( s, t\right) }{U}_{4}^{\left( s, t\right) }{U}_{6}^{\left( s, t\right) } \) have... | The proof is routine. See that of Theorem 4. | No |
Theorem 7. (a) The triangles \( {ABC} \) and \( {A}_{0}{B}_{0}{C}_{0} \) are orthologic if and only if the triangles \( {ABC} \) and \( {DEF} \) are orthologic. | Proof. Let \( {D}_{1}\left( {{d}_{1},{\delta }_{1}}\right) ,{E}_{1}\left( {{e}_{1},{\varepsilon }_{1}}\right) \) and \( {F}_{1}\left( {{f}_{1},{\varphi }_{1}}\right) \) . Recall from [2] that the triangles \( {DEF} \) and \( {D}_{1}{E}_{1}{F}_{1} \) are orthologic if and only if \( {\Delta }_{0} = 0 \), where\n\n\[ \n{... | Yes |
Theorem 8. The points \( {G}_{1} \) and \( {G}_{2} \) are the points \( {G}_{3} \) and \( {G}_{4} \) respectively. The points \( {G}_{1} \) and \( {G}_{2} \) divide the segments \( {G}_{\sigma }{G}_{\tau } \) and \( {G}_{\tau }{G}_{\sigma } \) in the ratio \( 1 : 2 \) . | Proof. The centroids \( {G}_{12A},{G}_{34B} \) and \( {G}_{56C} \) have the co-ordinates\n\n\( \left( {\frac{{2d} + 1 + v + u}{6},\frac{{2\delta } + 1 + v - u}{6}}\right) ,\left( {\frac{2\left( {e + 1}\right) + u - v}{6},\frac{{2\varepsilon } + u + v}{6}}\right) \) and \( \left( {\frac{2\left( {f + u}\right) + 1}{6},\f... | Yes |
Theorem 9. The following statements are equivalent:\n\n(a) The triangles \( {ABC} \) and \( {G}_{12A}{G}_{34B}{G}_{56C} \) are orthologic.\n\n(b) The triangles \( {ABC} \) and \( {G}_{12D}{G}_{34E}{G}_{56F} \) are orthologic.\n\n(c) The triangles DEF and \( {G}_{45A}{G}_{61B}{G}_{23C} \) are orthologic.\n\n(d) The tria... | Proof. The equivalence of (a) and (g) follows from the relation\n\n\[ \n{\Delta }_{0}\left( {{ABC},{G}_{12A}{G}_{34B}{G}_{56C}}\right) = \frac{\Delta }{3}.\n\]\n\nThe equivalence of (g) with (b), (c), (d), (e) and (f) one can prove in the same way. | No |
Theorem 10. (a) The following equality for areas of triangles holds:\n\n\[ \n\\left| {{A}^{\prime }{B}^{\prime }{C}^{\prime }}\\right| + \\left| {{D}^{\prime }{E}^{\prime }{F}^{\prime }}\\right| = \\left| {{A}^{\prime \prime }{B}^{\prime \prime }{C}^{\prime \prime }}\\right| + \\left| {{D}^{\prime \prime }{E}^{\prime \... | The proofs of both parts can be accomplished by a routine calculation. | No |
Theorem 11. (a) The following equality for areas of triangles holds:\n\n\[ \n\\left| {{A}_{1}^{\prime }{B}_{1}^{\prime }{C}_{1}^{\prime }}\\right| + \\left| {{D}_{1}^{\prime }{E}_{1}^{\prime }{F}_{1}^{\prime }}\\right| = \\left| {{A}_{1}^{\prime \prime }{B}_{1}^{\prime \prime }{C}_{1}^{\prime \prime }}\\right| + \\left... | The proofs of both parts can be accomplished by a routine calculation. | No |
Theorem 13. (a) The triangles \( {K}_{a}{K}_{b}{K}_{c},{L}_{a}{L}_{b}{L}_{c},{M}_{a}{M}_{b}{M}_{c},{N}_{a}{N}_{b}{N}_{c},{P}_{a}{P}_{b}{P}_{c} \) and \( {Q}_{a}{Q}_{b}{Q}_{c} \) are each homothetic with the triangle \( {ABC} \) . | Proof of parts (a) and (c) are routine while the simplest method to prove the part (b) is to show that the midpoints of the segments \( {K}_{x}{M}_{x} \) and \( {L}_{x}{N}_{x} \) coincide for \( x = a, b, c \) . Let \( {J}_{0},{K}_{0},{L}_{0},{M}_{0},{N}_{0},{P}_{0} \) and \( {Q}_{0} \) be centers of the above homothet... | No |
Theorem 14. (a) The symmedian point \( K \) of the triangle \( {ABC} \) lies on the line \( {K}_{0}{L}_{0} \) . | Proof. (a) It is straightforward to verify that the symmedian point with co-ordinates \( \left( {\frac{{u}^{2} + u + {v}^{2}}{2\left( {{u}^{2} - u + {v}^{2} + 1}\right) },\frac{v}{2\left( {{u}^{2} - u + {v}^{2} + 1}\right) }}\right) \) lies on the line \( {K}_{0}{L}_{0} \) . | Yes |
Theorem 15. The triangles \( {K}_{a}{K}_{b}{K}_{c} \) and \( {L}_{a}{L}_{b}{L}_{c} \) are congruent if and only if the triangles \( {ABC} \) and \( {DEF} \) are orthologic. | Proof. Since the triangles \( {K}_{a}{K}_{b}{K}_{c} \) and \( {L}_{a}{L}_{b}{L}_{c} \) are both homothetic to the triangle \( {ABC} \), we conclude that they will be congruent if and only if \( \left| {{K}_{a}{K}_{b}}\right| = \left| {{L}_{a}{L}_{b}}\right| \) . Hence, the theorem follows from the equality\n\n\[ \n{\le... | Yes |
Theorem 17. There is a unique central point \( P \) with the property that the triangles \( {S}_{1}{S}_{3}{S}_{5} \) and \( {S}_{2}{S}_{4}{S}_{6} \) are congruent. The first trilinear co-ordinate of this point \( P \) is \( a\left( {\left( {{b}^{2} + {c}^{2} + {2S}}\right) {a}^{2} - {b}^{4} - {c}^{4} - {2S}\left( {{b}^... | Proof. Let \( P\left( {p, q}\right) \) . The orthogonal projections \( {P}_{a},{P}_{b} \) and \( {P}_{c} \) of the point \( P \) onto the sidelines \( {BC},{CA} \) and \( {AB} \) have the co-ordinates\n\n\[ \left( {\frac{{\left( u - 1\right) }^{2}p + v\left( {u - 1}\right) q + {v}^{2}}{\xi - u + 1},\frac{v\left( {\left... | Yes |
Theorem 18. The triangles \( {S}_{1}{S}_{3}{S}_{5} \) and \( {S}_{2}{S}_{4}{S}_{6} \) have the same centroid if and only if the point \( P \) is the circumcenter of the triangle \( {ABC} \) . | Proof. We get \( {\left| {G}_{o}{G}_{e}\right| }^{2} = \frac{{M}^{2} + {N}^{2}}{9\left( {\xi - u + 1}\right) \left( {\xi + u}\right) \left( {1 + {4\xi }}\right) } \), with\n\n\[ M = {3\xi }\left( {{2u} - 1}\right) p + v\left( {1 + {4\xi }}\right) q - \xi \left( {{2\xi } + {3u} - 1}\right) \]\n\nand \( N = v\left( {1 + ... | Yes |
Theorem 19. For the pedal triangle DEF of a point \( P \) with respect to the triangle \( {ABC} \) the following statements are equivalent:\n\n(a) The triangles \( {A}_{0}{B}_{0}{C}_{0} \) and \( {D}_{0}{E}_{0}{F}_{0} \) are orthologic.\n\n(b) The triangles \( {ABC} \) and \( {G}_{45A}{G}_{61B}{G}_{23C} \) are ortholog... | (a) The orthology criterion \( {\Delta }_{0}\left( {{A}_{0}{B}_{0}{C}_{0},{D}_{0}{E}_{0}{F}_{0}}\right) \) is equal to the quotient \( \frac{-{vM}}{8\left( {\xi + u}\right) \left( {\xi - u + 1}\right) } \), with \( M \) the following linear polynomial in \( p \) and \( q \) .\n\n\[ M = 2\left( {{\xi }^{2} + \xi - {v}^{... | Yes |
Theorem 20. The following statements are equivalent:\n\n(a) The points \( {S}_{1},{S}_{3} \) and \( {S}_{5} \) are collinear.\n\n(b) The points \( {S}_{2},{S}_{4} \) and \( {S}_{6} \) are collinear.\n\n(c) The point \( P \) is on the circle with the center \( {K}_{-\omega } \) and the radius equal to the circumradius \... | Proof. Let \( M \) be the following quadratic polynomial in \( p \) and \( q \) :\n\n\[ \n{v}^{2}\left( {{p}^{2} + {q}^{2}}\right) + v\left( {{2u} - w}\right) p - \left( {{\xi }^{2} + {w\xi } - {v}^{2}}\right) q - \left( {\xi + u}\right) \left( {\xi - u + w}\right) , \n\] \n\nwhere \( w = v + 1 \) . The points \( {S}_{... | Yes |
Theorem 21. The triangles \( {A}_{0}{B}_{0}{C}_{0} \) and \( {D}_{0}{E}_{0}{F}_{0} \) always have different sums of squares of lengths of sides. | Proof. The difference \( {s}_{2}\left( {{A}_{0}{B}_{0}{C}_{0}}\right) - {s}_{2}\left( {{D}_{0}{E}_{0}{F}_{0}}\right) \) is equal to \( \frac{3{v}^{3}N}{4\left( {\xi - u + 1}\right) \left( {u + \xi }\right) } \), where \( N \) denotes the following quadratic polynomial in variables \( p \) and \( q \) :\n\n\[ \n{\left( ... | Yes |
Theorem 22. The triangles \( {A}_{0}{B}_{0}{C}_{0} \) and \( {D}_{0}{E}_{0}{F}_{0} \) have the same areas if and only if the point \( P \) lies on the circle \( {\theta }_{0} \) with the center at the symmedian point \( K \) of the triangle \( {ABC} \) and the radius \( R\sqrt{4 - 3{\tan }^{2}\omega } \), where \( R \)... | Proof. The difference \( \left| {{D}_{0}{E}_{0}{F}_{0}}\right| - \left| {{A}_{0}{B}_{0}{C}_{0}}\right| \) is equal to the quotient \( \frac{{v}^{2}{\zeta }^{2}M}{{16\mu }\left( {\zeta - u}\right) } \) , where \( \zeta = \xi + 1,\mu = \xi + u \) and \( M \) denotes the following quadratic polynomial in variables \( p \)... | Yes |
Theorem 23. The triangles \( {A}^{ * }{B}^{ * }{C}^{ * } \) and \( {D}^{ * }{E}^{ * }{F}^{ * } \) have the same sums of squares of lengths of sides if and only if the point \( P \) lies on the circle \( {\theta }_{0} \) . | Proof. The proof is almost identical to the proof of the previous theorem since the difference \( {s}_{2}\left( {{D}^{ * }{E}^{ * }{F}^{ * }}\right) - {s}_{2}\left( {{A}^{ * }{B}^{ * }{C}^{ * }}\right) \) is equal to \( \frac{{v}^{2}{\left( \xi + 1\right) }^{2}M}{2\left( {\xi - u + 1}\right) \left( {\xi + u}\right) } \... | No |
Theorem 24. For any point \( P \) the triangles \( {A}^{ * }{B}^{ * }{C}^{ * } \) and \( {D}^{ * }{E}^{ * }{F}^{ * } \) always have different areas. | Proof. The proof is similar to the proof of Theorem 21 since the difference\n\n\( \left| {{D}^{ * }{E}^{ * }{F}^{ * }}\right| - \left| {{A}^{ * }{B}^{ * }{C}^{ * }}\right| \) is equal to \( \frac{{v}^{3}N}{8\left( {\xi - u + 1}\right) \left( {\xi + u}\right) } \) . | No |
Theorem 25. The following statements are equivalent:\n\n(a) The triangles \( {A}_{0}{B}_{0}{C}_{0} \) and \( {D}_{0}{E}_{0}{F}_{0} \) are orthologic.\n\n(b) The triangles \( {ABC} \) and \( {G}_{45A}{G}_{61B}{G}_{23C} \) are orthologic.\n\n(c) The triangles \( {ABC} \) and \( {G}_{45D}{G}_{61E}{G}_{23F} \) are ortholog... | Proof. (g) \( {s}_{2}\left( {{A}^{\prime \prime }{B}^{\prime \prime }{C}^{\prime \prime }}\right) - {s}_{2}\left( {{A}^{\prime }{B}^{\prime }{C}^{\prime }}\right) = \frac{2vMN}{q\left( {{vp} - {uq}}\right) \left( {v\left( {p - 1}\right) - \left( {u - 1}\right) q}\right) } \), with\n\n\[ M = {\left( p - \frac{1}{2}\righ... | No |
Proposition 2. Given a closed polygon \( a = {A}_{1}\cdots {A}_{n} \) with \( n \) odd and a point \( P \) not lying on its side-lines, let \( b = {B}_{1}\cdots {B}_{n} \) be the conjugate polygon of a with respect to \( P \) . Then the transformation \( {G}_{1} \) is an affine reflection the axis of which passes throu... | That point \( P \) remains fixed under \( {G}_{1} \) is obvious, since \( {G}_{1} \) is a composition of affine reflections all of whose axes pass through \( P \) . From this, using the preservation of proportions by affinities and the invariance of \( {A}_{1}{A}_{2} \) follows also the that the parallels to \( {A}_{1}... | Yes |
Lemma 6. Let \( \{ {ABC}, D, e\} \) be correspondingly a triangle, a point and a line. Consider a variable line through \( D \) intersecting sides \( {AB},{BC} \) correspondingly at points \( E, F \) . Let \( G \) be the middle of \( {EF} \) and \( P \) the intersection point of lines \( e \) and \( {BG} \) . Let furth... | To prove the lemma introduce affine coordinates with axes along lines \( \{ {BC}, e\} \) and origin at \( J \), where \( I = e \cap {CA}, J = e \cap {CB} \) (see Figure 11). The points on line \( e \) are: \( M = e \cap {AB}, N = e \cap \left( {\parallel {BC}, D}\right), H = e \cap {DE}, Q = e \cap \left( {\parallel {D... | Yes |
Lemma 7. Let \( \\left\\{ {{A}_{1}\\cdots {A}_{n},{C}_{1}, e}\\right\\} \) be correspondingly a closed polygon, a point on side \( {A}_{1}{A}_{2} \) and a line. Consider a point \( P \) varying on line e and the corresponding conjugate polygon \( b = {B}_{1}\\cdots {B}_{n} \) . Construct the parallel to \( b \) polygon... | The proof results by inductively applying the previous lemma to each side of \( c \) , starting with side \( {C}_{1}{C}_{2} \), which by assumption passes through \( {C}_{1} \) (see Figure 12). Next prove that side \( {C}_{2}{C}_{3} \) passes through a point \( {O}_{23} \) by applying previous lemma to the triangle wit... | Yes |
Lemma 8. Let \( \left\{ {{A}_{1}\cdots {A}_{n},{C}_{1}, e}\right\} \) be correspondingly a closed polygon, a point on side \( {A}_{1}{A}_{2} \) and a line. Consider a point \( P \) varying on line \( e \), the corresponding conjugate polygon \( b = {B}_{1}\cdots {B}_{n} \) and the corresponding parallel to \( b \) poly... | Assume that the correspondence is not a constant one. Proceed then by applying the previous lemma and using the fixed points \( {O}_{23},{O}_{34},\cdots \) through which pass the sides of the inscribed polygons \( c \) as \( P \) varies on line \( e \) . It is easily shown inductively that correspondences \( {f}_{1} : ... | Yes |
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