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Theorem 6. \( {K}_{H}\left( \varphi \right) = K\left( \theta \right) \) if and only if\n\n\[ \cot \varphi \left( {\cot \omega + \cot \theta }\right) + \cot \theta \cot \omega + 1 = 0, \]\n\nwhere \( \omega \) is the Brocard angle of triangle \( {ABC} \) . | Proof. From (7),\n\n\[ {K}_{H}\left( \varphi \right) = \left( {\frac{1}{{S}_{BC} - {S}_{AA} + {a}^{2}{S}_{\varphi }} : \cdots : \cdots }\right) .\n\nThis is the same as \( K\left( \theta \right) \) if and only if\n\n\[ \left( {\left( {{S}_{CA} - {S}_{BB} + {b}^{2}{S}_{\varphi }}\right) \left( {{S}_{AB} - {S}_{CC} + {c}... | Yes |
Theorem 1. If a line goes through a vertex of the reference triangle \( \mathbf{ABC} \), the conjugate of this line is a line through the same vertex. | Proof. Choose vertex B. A line through this vertex has the form \( {nz} - \ell x = 0 \) . The isotomic conjugate is \( \frac{n}{z} - \frac{\ell }{x} = 0 \), which is the same as \( {nx} - \ell z = 0 \), a line through the same vertex. The isogonal conjugate works analogously. | Yes |
Theorem 2. Conics through \( {\mathrm{{AGCB}}}^{\mathrm{G}} \) and \( {\mathrm{{ACC}}}^{\mathrm{G}}{\mathrm{A}}^{\mathrm{G}} \) are self-isotomic. | Proof. The general conic is \( \ell {x}^{2} + m{y}^{2} + n{z}^{2} + {Lyz} + {Mzx} + {Nxy} = 0 \) . Choosing the case \( {\mathbf{{AGCB}}}^{\mathbf{G}} \), since \( \mathbf{A} \) and \( \mathbf{C} \) are on the conic, we have that \( \ell = n = 0 \) . From \( \mathbf{G} \) and \( {\mathbf{B}}^{\mathbf{G}} \) we get the ... | Yes |
Theorem 3. The ellipse \( {y}^{2} - {zx} = 0 \) | Proof. (1) can be verified by substituting coordinates as done above.\n\n(2) is true by the general principle that if an equation has the form (line \( {2}^{2} = \) (line 1)-(line 3), then the curve has a double intersection at the intersection of line 1 and line 2 and at the intersection of line 3 and line 2 and is ta... | Yes |
Theorem 2 (Goormaghtigh [2]). Let \( {A}_{1},{B}_{1},{C}_{1} \) be points on \( {OA},{OB},{OC} \) such that \[ \frac{O{A}_{1}}{OA} = \frac{O{B}_{1}}{OB} = \frac{O{C}_{1}}{OC} = t. \] (1) The intersections of the perpendiculars to \( {OA} \) at \( {A}_{1},{OB} \) at \( {B}_{1} \), and \( {OC} \) at \( {C}_{1} \) with th... | is the isogonal conjugate of the point \( P \) on the Euler line dividing \( {OH} \) in the ratio \( {OP} : {PH} = 1 : {2t} \) . See Figure 1. | Yes |
Theorem 3. Let \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be the tangential triangle of \( {ABC} \) . Consider points \( X \) , \( Y, Z \) dividing \( O{A}^{\prime }, O{B}^{\prime }, O{C}^{\prime } \) respectively in the ratio\n\n\[ \n\frac{OX}{O{A}^{\prime }} = \frac{OY}{O{B}^{\prime }} = \frac{OZ}{O{C}^{\prime }} ... | Proof. Let the isogonal line of \( {AX} \) (with respect to angle \( A \) ) intersect \( {OA} \) at \( {X}^{\prime } \) . The triangles \( {OAX} \) and \( O{X}^{\prime }A \) are similar. It follows that \( {OX} \cdot O{X}^{\prime } = O{A}^{2} \), and \( X,{X}^{\prime } \) are inverse in the circumcircle. Note also that... | Yes |
Theorem 1. Given \( a,{\ell }_{a},{\ell }_{b} > 0 \), there is a unique triangle \( {ABC} \) with \( {BC} = a \) and lengths of bisectors of angles \( A \) and \( B \) equal to \( {\ell }_{a} \) and \( {\ell }_{b} \) respectively if and only if \( {\ell }_{b} \leq a \) or\n\n\[ a < {\ell }_{b} < {2a}\;\text{ and }\;{\e... | Proof. In a triangle \( {ABC} \) with \( {BC} = a \) and given \( {\ell }_{a},{\ell }_{b} \), let \( y = {CA} \) and \( z = {AB} \) . We have \( {\ell }_{b} = \frac{2az}{a + z}\cos \frac{B}{2} \) and\n\n\[ z = \frac{a{\ell }_{b}}{{2a}\cos \frac{B}{2} - {\ell }_{b}}.\]\n\n(1)\n\nIt follows that \( \cos \frac{B}{2} > \fr... | Yes |
Lemma 1. If the intersection set \( {q}_{1}^{\psi } \cap {q}_{2}^{\psi } \) contains a vertex \( P \) of \( {q}_{2}^{\psi } \), then \( P \in \) \( S\left( {{q}_{2},{C}_{1}}\right) \) . | Proof. Let \( D \) be the straight line containing a diagonal of \( {q}_{2} \) and which does not contain \( P \) . Then the disc with center \( P \) and radius \( \psi \) contains \( {C}_{1} \) and \( {C}_{2} \) since \( d\left( {{C}_{1}, P}\right) \leq d\left( {{C}_{2}, P}\right) = \psi \) . Hence there is only one h... | Yes |
Lemma 2. We have\n\n\[ \n{q}_{1}^{\psi } \cap {q}_{2}^{\psi } \subset S\left( {{q}_{1},{C}_{2}}\right) \cap S\left( {{q}_{2},{C}_{1}}\right) .\n\]\n\n(3) | Proof. The proof is divided in three exclusive and exhaustive situations.\n\n(i) First, we suppose that the intersection set \( {q}_{1}^{\psi } \cap {q}_{2}^{\psi } = \{ P\} \) where \( P \) is a common vertex of \( {q}_{1}^{\psi } \) and \( {q}_{2}^{\psi } \) . We readily obtain \( P \in S\left( {{q}_{2},{C}_{1}}\righ... | Yes |
Lemma 3. We have \( \left. {\gamma = \gamma \left( {{q}_{1},{q}_{2}}\right) \in }\right\rbrack 0,\frac{\pi }{2}\lbrack \) . | Proof. If \( \gamma = 0 \), the two axes \( {A}_{{i}_{1}}\left( {q}_{1}\right) \) and \( {A}_{{i}_{2}}\left( {q}_{2}\right) \) are equal to some straight line \( D \) . The centers \( {C}_{1} \) and \( {C}_{2} \) lie on \( D \) . But by construction \( {A}_{{i}_{1}}\left( {q}_{1}\right) \) and \( {A}_{{i}_{2}}\left( {q... | Yes |
Lemma 4. We have \( {q}_{1}^{\psi } \cap {q}_{2}^{\psi } \subset B\left( {{q}_{1},{q}_{2}}\right) \) . | Proof. Let \( 0 < k \leq \psi \) . The homothetic square \( {q}_{1}^{k} \) (resp. \( {q}_{2}^{k} \) ) has two vertices in \( S\left( {{q}_{1},{C}_{2}}\right) \) (resp. \( S\left( {{q}_{2},{C}_{1}}\right) \) ). The straight line passing through those vertices of \( {q}_{1} \) (resp. \( {q}_{2} \) ) is parallel at distan... | Yes |
Lemma 5. We have\n\n(a) \( {\ell }_{1} = {\ell }_{2} \Leftrightarrow {I}_{1} = {I}_{2} < {I}_{3} = {I}_{4} \) .\n\n(b) \( {\ell }_{1} < {\ell }_{2} \Leftrightarrow {I}_{1} < {I}_{2} < {I}_{3} < {I}_{4} \) .\n\n(c) \( {\ell }_{2} < {\ell }_{1} \Leftrightarrow {I}_{2} < {I}_{1} < {I}_{4} < {I}_{3} \) . | Proof. If \( {\ell }_{1} = {\ell }_{2} \) then \( {I}_{1} = {I}_{2} < {I}_{3} = {I}_{4} \) . Shifting \( {C}_{1} \) along \( W{C}_{1} \) towards \( W \) causes \( {C}_{1}{I}_{1} \) and \( {C}_{1}{I}_{3} \) to slide in a parallel fashion, so that \( {I}_{1} < {I}_{2} \) and \( {I}_{3} < {I}_{4} \) . Since \( {C}_{1} \in... | Yes |
Theorem 6. (i) Among the four points \( {I}_{1},\ldots ,{I}_{4} \), the second one is the percussion point: \( P = {q}_{1}^{\psi } \cap {q}_{2}^{\psi } = \max \left\{ {{I}_{1},{I}_{2}}\right\} \) . We have\n\n\[ \psi = \max \left\{ {{\ell }_{1},{\ell }_{2}}\right\} \frac{\sqrt{2}}{1 + \cot \omega }.\n\] | Proof. (i). We suppose first that \( {\ell }_{2} > {\ell }_{1} \) . By Lemma 5 we have\n\n\[ {d}_{1} = \begin{Vmatrix}\overrightarrow{{C}_{1}{I}_{1}}\end{Vmatrix} < {d}_{2} = \begin{Vmatrix}\overrightarrow{{C}_{2}{I}_{2}}\end{Vmatrix} < {d}_{3} = \begin{Vmatrix}\overrightarrow{{C}_{1}{I}_{3}}\end{Vmatrix} < {d}_{4} = \... | Yes |
Corollary 7. We have\n\n\[ \n{\ell }_{1} < {\ell }_{2} \Leftrightarrow {q}_{2}\text{strikes}{q}_{1}\text{and}{q}_{1}\text{does not strike}{q}_{2}\text{,}\n\]\n\n\[ \n{\ell }_{2} < {\ell }_{1} \Leftrightarrow {q}_{1}\text{ strikes }{q}_{2}\text{ and }{q}_{2}\text{ does not strike }{q}_{1},\n\]\n\n\[ \n{\ell }_{1} = {\el... | Proof. The three implications from left to right are direct consequences of Theorem 6 and its proof. Since the three cases are exclusive and exhaustive, the three converse implications readily follow. | Yes |
Theorem 8. The square \( {q}_{2} \) strikes \( {q}_{1} \) if and only if \( \alpha \left( {{q}_{2},{q}_{1}}\right) \leq \alpha \left( {{q}_{1},{q}_{2}}\right) \) . The percussion point is the vertex of \( {q}_{1} \) or \( {q}_{2} \) which realizes the minimum of the eight angles appearing in \( \alpha \left( {{q}_{1},{... | Proof. Suppose that \( {q}_{2} \) strikes \( {q}_{1} = {ABCD} \) at \( P \) in the interior of side \( {AB} \), see Figure 7. Let \( {AB} \) be the \( x \) -axis and \( P \) the origin. Then for the interiors of \( {q}_{1} \) and \( {q}_{2} \) to be disjoint, the center \( {C}_{2} \) of \( {q}_{2} \) must be in \( \{ \... | Yes |
Theorem 1. The triangle centers for which the intercepts \( x, y, z \) are linear forms in \( a, b, c \) are the centroid, the Gergonne and the Nagel points. | Proof. Note first that if \( \left( {x, y, z}\right) \) are the intercepts corresponding to a center \( V \), and if\n\n\[ x = {\alpha a} + {\beta b} + {\gamma c}, \]\n\nthen it follows from reflecting \( {ABC} \) about the perpendicular bisector of the segment \( {BC} \) that\n\n\[ a - x = {\alpha a} + {\beta c} + {\g... | Yes |
Lemma 1. If \( {\alpha }_{i} \) and \( {\alpha }_{j} \) are semicircles in \( \Phi \left( {\alpha ,\beta ,\gamma }\right) \) with \( {\alpha }_{i} < {\alpha }_{j} \), the radius of the inscribed circle in the curvilinear triangle bounded by \( {\alpha }_{i},{\alpha }_{j} \) and \( \gamma \) is\n\n\[ \frac{{ab}\left( {{... | Proof. Let \( \mathcal{C} \) be the inscribed circle with radius \( r \) . First we invert \( \left\{ {{\alpha }_{i},{\alpha }_{j},\gamma ,\mathcal{C}}\right\} \) in the circle with center \( O \) and radius \( k \) . Then \( {\alpha }_{i} \) and \( {\alpha }_{j} \) are inverted to the lines \( \overline{{\alpha }_{i}}... | Yes |
Lemma 2. If \( {\alpha }_{i} \) (resp. \( {\alpha }_{j} \) ) is the line, then the radius of the inscribed circle is \[ \frac{-{ab}}{{a}_{j} - a}\left( {\text{ resp. }\frac{ab}{{a}_{i} + b}}\right) . \] | Proof. Even in this case (2) in the proof of Lemma 1 holds with \( s = 0 \) (resp. \( t = 0 \) ), and we get the conclusion. | No |
Theorem 3. Assume \( a \neq b \), and let \( {\alpha }_{i},{\alpha }_{j} \in \Phi \left( {\alpha ,\beta ,\gamma }\right) \) with \( {\alpha }_{i} < {\alpha }_{j} \) . The radius of the circle inscribed in the curvilinear triangle bounded by \( {\alpha }_{i},{\alpha }_{j} \) and \( \gamma \) is\n\n\[ \n\frac{{ab}\left( ... | Proof. If \( {\alpha }_{i} \) and \( {\alpha }_{j} \) are semicircles, then\n\n\[ \n\frac{{ab}\left( {\mu \left( {\alpha }_{i}\right) - \mu \left( {\alpha }_{j}\right) }\right) }{{b\mu }\left( {\alpha }_{i}\right) - {a\mu }\left( {\alpha }_{j}\right) } = \frac{{ab}\left( {\frac{{a}_{i} - a + b}{{a}_{i}} - \frac{{a}_{j}... | Yes |
Theorem 4. Assume \( a = b \), and let \( {\alpha }_{i},{\alpha }_{j} \in \Phi \left( {\alpha ,\beta ,\gamma }\right) \) with \( {\alpha }_{i} < {\alpha }_{j} \). The radius of the circle inscribed in the curvilinear triangle bounded by \( {\alpha }_{i},{\alpha }_{j} \) and \( \gamma \) is \[ \frac{{a}^{2}\left( {\mu \... | The functions \( x \mapsto \frac{{ab}\left( {1 - x}\right) }{b - {ax}}, a \neq b \) and \( x \mapsto \frac{{a}^{2}x}{{ax} - 1}, a > 0 \) are injective. | No |
Theorem 6. Let \( \\left\\{ {{\\alpha }_{0} = \\alpha ,{\\alpha }_{1},\\ldots ,{\\alpha }_{n} = \\beta ,\\gamma }\\right\\} \) be an arbelos in \( n \) -aliquot parts. The common radius of the Archimedean circles in n-aliquot parts is\n\n\[ \n\\left\\{ \\begin{array}{ll} \\frac{{ab}\\left( {{b}^{\\frac{2}{n}} - {a}^{\\... | Proof. First we consider the case \( a \\neq b \) . We can assume \( {\\alpha }_{0} < {\\alpha }_{1} < \\cdots < {\\alpha }_{n} \) by renaming if necessary. The sequence \( \\frac{b}{a} = \\mu \\left( {\\alpha }_{0}\\right) ,\\mu \\left( {\\alpha }_{1}\\right) ,\\ldots ,\\mu \\left( {\\alpha }_{n}\\right) = \\frac{a}{b... | Yes |
Theorem 7. Let \( \left\{ {{\alpha }_{0} = \alpha ,{\alpha }_{1},\ldots ,{\alpha }_{n} = \beta ,\gamma }\right\} \) be an arbelos in \( n \) -aliquot parts with \( {\alpha }_{0} < {\alpha }_{1} < \cdots < {\alpha }_{n} \) . Then \( {\alpha }_{i} \) is the line in \( \Phi \left( {\alpha ,\beta ,\gamma }\right) \) if \( ... | Proof. Suppose \( a \neq b \) . Since \( \frac{b}{a} = \mu \left( {\alpha }_{0}\right) ,\mu \left( {\alpha }_{1}\right) ,\ldots ,\mu \left( {\alpha }_{n}\right) = \frac{a}{b} \) is a geometric sequence with common ratio \( {\left( \frac{a}{b}\right) }^{\frac{2}{n}} \), we have \( \mu \left( {\alpha }_{i}\right) = {\lef... | Yes |
Lemma 8. (a) If \( a \neq b,{\left( \frac{{a}^{\prime }}{{b}^{\prime }}\right) }^{n} = {\left( \frac{a}{b}\right) }^{n + 2} \) . | Proof. If \( a \neq b \) we have\n\n\[ \n{a}^{\prime } = a - \frac{{ab}\left( {{a}^{\frac{2}{n}} - {b}^{\frac{2}{n}}}\right) }{{a}^{\frac{2}{n} + 1} - {b}^{\frac{2}{n} + 1}} = \frac{{a}^{\frac{2}{n} + 1}\left( {a - b}\right) }{{a}^{\frac{2}{n} + 1} - {b}^{\frac{2}{n} + 1}}, \n\]\n\n\[ \n{b}^{\prime } = b - \frac{{ab}\l... | Yes |
Theorem 9. \( \\left\\{ {{\\alpha }^{\\prime },{\\alpha }_{0},{\\alpha }_{1},\\ldots ,{\\alpha }_{n},{\\beta }^{\\prime },{\\gamma }^{\\prime }}\\right\\} \) is an arbelos in \( \\left( {n + 2}\\right) \) -aliquot parts. | Proof. Let us assume \( a \\neq b \) . By Lemma 8 and the proof of Theorem \( 6,\\mu \\left( {\\alpha }_{0}\\right) \) , \( \\mu \\left( {\\alpha }_{1}\\right) ,\\ldots ,\\mu \\left( {\\alpha }_{n}\\right) \) is a geometric sequence with common ratio \( {\\left( \\frac{{a}^{\\prime }}{{b}^{\\prime }}\\right) }^{\\frac{... | Yes |
Theorem 10. Let \( {\delta }_{{2n} - 1} \) be one of the Archimedean circles in\n\n\[ \left\{ {{\alpha }_{-n},{\alpha }_{-\left( {n - 1}\right) },\ldots ,{\alpha }_{-1},{\alpha }_{1},\ldots ,{\alpha }_{n},{\gamma }_{{2n} - 1}}\right\} .\n\]\n\nThen the radii of \( {\alpha }_{-n} \) and \( {\alpha }_{n} \) are\n\n\[ \fr... | Proof. Let \( \overline{{a}_{-n}} \) and \( \overline{{a}_{n}} \) be the radii of \( {\alpha }_{-n} \) and \( {\alpha }_{n} \) respectively. By Lemma 8 we have\n\n\[ {\left( \frac{\overline{{a}_{-n}}}{\overline{{a}_{n}}}\right) }^{\frac{1}{{2n} - 1}} = {\left( \frac{\overline{{a}_{-\left( {n - 1}\right) }}}{\overline{{... | Yes |
Corollary 12. Let \( {c}_{n} \) and \( {d}_{n} \) be the radii of \( {\gamma }_{n} \) and \( {\delta }_{n} \) respectively. | \[ {a}_{n} = {b}_{{2n} - 1} \] \[ {a}_{-n} = {b}_{-\left( {{2n} - 1}\right) }, \] \[ {c}_{{2n} - 1} = {c}_{2\left( {{2n} - 1}\right) } \] \[ {d}_{{2n} - 1} = {d}_{{4n} - 2} + {d}_{4n} \] Figure 3 shows the even pattern together with the odd pattern reflected in the \( x \) -axis. The trivial case of these patterns can ... | No |
Theorem 13. The external common tangent of \( {\beta }_{n} \) and \( {\beta }_{-n} \) touches \( {\gamma }_{4n} \) for any positive integer \( n \) . | Proof. The distance between the external common tangents of \( {\beta }_{n} \) and \( {\beta }_{-n} \) and the center of \( {\gamma }_{2n} \) is \( \frac{{\overline{{b}_{n}}}^{2} + {\overline{{b}_{-n}}}^{2}}{\overline{{b}_{n}} + \overline{{b}_{-n}}} \) where \( \overline{{b}_{n}} \) and \( \overline{{b}_{-n}} \) are th... | Yes |
Theorem 14. Let \( {\mathrm{{BK}}}_{n} \) be the circle orthogonal to \( \alpha ,\beta \) and \( {\delta }_{{2n} - 1} \), and let \( {\mathrm{{AR}}}_{n} \) be the inscribed circle of the curvilinear triangle bounded by \( {\beta }_{h},{\beta }_{0} \) and \( {\gamma }_{2n} \) . The circles \( {\mathrm{{BK}}}_{n} \) and ... | Proof. Assume \( a \neq b \) . Since \( {\mathrm{{AR}}}_{n} \) is the Archimedean circle of the arbelos in 2-aliquot parts \( \left\{ {{\beta }_{-n},{\beta }_{0},{\beta }_{n},{\gamma }_{2n}}\right\} \), the radius of \( {\mathrm{{AR}}}_{n} \) is\n\n\[ \frac{\overline{{b}_{n}}\overline{{b}_{-n}}\left( {\overline{{b}_{n}... | Yes |
Lemma 1. The number \( \cos \frac{\pi }{n}, n \geq 2 \), is rational if and only if \( n = 2 \) or \( n = 3 \) . | Proof. Suppose that \( \cos \frac{\pi }{n} \) is rational. Put\n\n\[{\zeta }_{2n} = \cos \frac{2\pi }{2n} + i\sin \frac{2\pi }{2n}\]\n\nthen \( {\zeta }_{2n} \) is a zero of the polynomial \( {X}^{2} - \left( {2 \cdot \cos \frac{\pi }{n}}\right) \cdot X + 1 \in \mathbb{Q}\left\lbrack X\right\rbrack \) . So, the minimal... | Yes |
Lemma 2. For every \( n \in \mathbb{N} \smallsetminus \{ 0\} \), there exist polynomials \( {f}_{n}\left( x\right) ,{g}_{n - 1}\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) such that\n\n(i) \( \deg \left( {f}_{n}\right) = n,{f}_{n}\left( x\right) = {2}^{n - 1}{x}^{n} + \cdots \) and \( \cos \left( {nx}\r... | Proof. From \( \cos x = \cos x,\frac{\sin x}{\sin x} = 1 \) ,\n\n\[ \cos \left( {k + 1}\right) x = \cos \left( {kx}\right) \cos x - \frac{\sin \left( {kx}\right) }{\sin x}\left( {1 - {\cos }^{2}x}\right) ,\]\n\n\[ \frac{\sin \left( {k + 1}\right) x}{\sin x} = \frac{\sin \left( {kx}\right) }{\sin x}\cos x + \cos \left( ... | Yes |
Lemma 3. Let \( n \in \mathbb{N} \smallsetminus \{ 0\}, q \in {\mathbb{Q}}^{ + } \smallsetminus \{ 0\} \) and \( {x}_{1},\ldots ,{x}_{n} \in \mathbb{R} \) . If\n\n\[ \cos {x}_{1},\sqrt{q} \cdot \sin {x}_{1},\ldots ,\cos {x}_{n},\sqrt{q} \cdot \sin {x}_{n} \]\n\nare rational, then so are \( \cos \left( {{x}_{1} + \cdots... | Proof. This follows by induction from the following equations \( \left( {k \geq 1}\right) \) .\n\n\[ \cos \left( {{x}_{1} + \cdots + {x}_{k + 1}}\right) = \cos \left( {{x}_{1} + \cdots + {x}_{k}}\right) \cdot \cos \left( {x}_{k + 1}\right) \]\n\n\[ - \frac{1}{q}\left( {\sqrt{q} \cdot \sin \left( {{x}_{1} + \cdots + {x}... | Yes |
Lemma 4. Let \( \Delta \) be a triangle with vertices \( A, B \) and \( C \) . Put \( a = \left| {BC}\right|, b = \left| {AC}\right| \) , \( c = \left| {AB}\right| ,\alpha = \widehat{BAC},\beta = \widehat{ABC} \) and \( \gamma = \widehat{BCA} \) . Let \( n \in \mathbb{N} \smallsetminus \{ 0\} \) and suppose that \( \co... | Proof. We have\n\n\[ \frac{b}{a} = \frac{\sin \beta }{\sin \alpha } = \frac{\sin \beta }{\sin \frac{\beta }{n}} \cdot \frac{\sin \frac{\alpha }{n}}{\sin \alpha } \cdot \frac{\sin \frac{\beta }{n}}{\sin \frac{\alpha }{n}}.\]\n\nBy Lemma 2, \( \frac{\sin \beta }{\sin \frac{\beta }{n}} \cdot \frac{\sin \frac{\alpha }{n}}{... | Yes |
Proposition 1. Given triangle \( {ABC} \) with Nagel point \( N \), let \( D \) be the midpoint of \( {BC} \) . The lines \( {ID} \) and \( {AN} \) are parallel. | Proof. The centroid \( G \) divides each of the segments \( {AD} \) and \( {NI} \) in the ratio \( {AG} \) : \( {GD} = {NG} \) : \( {GI} = 2 : 1 \) . See Figure 1. | No |
Proposition 2. Let \( X \) be the point of tangency of the incircle with \( {BC} \) . The antipode of \( X \) on the circle with diameter \( {ID} \) is a point on \( {AN} \) . | Proof. This follows from the fact that the antipode of \( X \) on the incircle lies on the segment \( {AN} \) . See Figure 2. | No |
Proposition 1. Let \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be a variable triangle of which \( {B}^{\prime } \) and \( {C}^{\prime } \) lie on \( {CA} \) and \( {AB} \) respectively. If the sidelines of triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) are tangent to a conic \( \mathcal{C} \) , then the locus... | Proof. Let \( {XYZ} \) be the points on \( \mathcal{C} \) and where \( {C}^{\prime }{A}^{\prime },{A}^{\prime }{B}^{\prime } \), and \( {B}^{\prime }{C}^{\prime } \) respectively meet \( \mathcal{C}.{ZX} \) is the polar (with respect to \( \mathcal{C} \) ) of \( {B}^{\prime } \), which passes through a fixed point \( {... | Yes |
Proposition 2. If \( {\mathcal{C}}_{A} \) degenerates into a line, the triangle \( {ABC} \) is selfpolar with respect to each conic tangent to the sides of two cevian triangles. The cevian triangle of the trilinear pole of any tangent to such a conic is tritangent to this conic. | Proof. Let \( P \) be a point and \( {A}^{P}{B}^{P}{C}^{P} \) its anticevian triangle. \( {ABC} \) is a polar triangle with respect to each conic through \( {A}^{P}{B}^{P}{C}^{P} \), as \( {ABC} \) are the diagonal points of the complete quadrilateral \( P{A}^{P}{B}^{P}{C}^{P} \) . Now consider a second anticevian tria... | Yes |
For every triangle \( {ABC} \), there exists a unique interior point \( M \) the cevians \( A{A}^{\prime }, B{B}^{\prime } \), and \( C{C}^{\prime } \) through which have the property that\n\n\[ \angle A{C}^{\prime }{B}^{\prime } = \angle B{A}^{\prime }{C}^{\prime } = \angle C{B}^{\prime }{A}^{\prime }\left( { = \Omega... | Proof. It is obvious that (1) is equivalent to the condition \( \left( {{A}^{\prime },{B}^{\prime },{C}^{\prime }}\right) = \left( {C, A, B}\right) \) , where \( {A}^{\prime },{B}^{\prime } \), and \( {C}^{\prime } \) are the angles of the cevian triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . Similarly,(2) is... | Yes |
Theorem 2. Let \( \Omega \) be the cevian Brocard angle of triangle \( {ABC} \) . (i) \( \cot \Omega \) satisfies the polynomial \( f \) given in (6), where \( {c}_{0} = \sin A\sin B\sin C \) and \( {c}_{3} = \cos A\cos B\cos C \) . (ii) \( \Omega \leq \pi /3 \) for all triangles. (iii) \( \Omega \) takes all values in... | Proof. (i) follows from the alternative proof of Theorem 1 given in the preceding section. To prove (ii), it suffices to prove that \( f\left( {1/\sqrt{3}}\right) \leq 0 \) for all triangles \( {ABC} \) . Let \[ G = f\left( \frac{1}{\sqrt{3}}\right) = \frac{4\sqrt{3}}{9}\sin A\sin B\sin C + \frac{4}{3}\cos A\cos B\cos ... | Yes |
Theorem 3. If two triangles have equal Brocard angles and equal cevian Brocard angles, then they are similar. | Proof. Let \( \omega \) and \( \Omega \) be the Brocard and cevian Brocard angles of triangle \( {ABC} \) , and let \( t = \cot \omega \) and \( T = \cot \Omega \) . From (10) it follows that \( t = {c}_{1}/{c}_{0} \) and therefore \( {c}_{1} = t{c}_{0} \) . Substituting this in (6), we see that \( {c}_{0}\left( {T + t... | Yes |
Theorem 4. If the cevians \( A{A}^{\prime }, B{B}^{\prime } \), and \( C{C}^{\prime } \) through a point \( P \) inside triangle \( {ABC} \) have the property that two of the quadrilaterals \( {ACP}{B}^{\prime }, B{A}^{\prime }P{C}^{\prime } \) , \( C{B}^{\prime }P{A}^{\prime },{AB}{A}^{\prime }{B}^{\prime },{BC}{B}^{\... | Proof. The first part is nothing but [4, Theorem 4] and is easy to prove. The second part follows from \( \omega = \pi /2 - A = \pi /2 - B = \pi /2 - C \) . | No |
Theorem 5. If any of the Brocard points \( L \) and \( {L}^{\prime } \) of triangle \( {ABC} \) coincides with any of its cevian Brocard points \( M \) and \( {M}^{\prime } \), then \( {ABC} \) is equilateral. | Proof. Let \( A{A}^{\prime }, B{B}^{\prime } \), and \( C{C}^{\prime } \) be the cevians through \( L \), and let \( \omega \) and \( \Omega \) be the Brocard and cevian Brocard angles of \( {ABC} \) ; see Figure 4A. By the exterior angle theorem, \( \angle {AL}{B}^{\prime } = \omega + \left( {B - \omega }\right) = B \... | Yes |
Theorem 6. Let \( L \) be the first Brocard point of \( {ABC} \), and let \( A{A}^{\prime }, B{B}^{\prime } \), and \( C{C}^{\prime } \) be the cevians through \( L \) . Then \( L \) coincides with one of the two Brocard points \( N \) and \( {N}^{\prime } \) of \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) if and only... | Proof. Let the angles of \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be denoted by \( {A}^{\prime },{B}^{\prime } \), and \( {C}^{\prime } \) . The proof of Theorem 5 shows that the condition \( L = {N}^{\prime } \) is equivalent to \( L = {M}^{\prime } \), which in turn implies that \( {ABC} \) is equilateral. This ... | Yes |
Theorem 7. The triangle centers for which the angles \( \alpha ,\beta ,\gamma \) are linear forms in \( A, B, C \) are the centroid, the orthocenter, and the Gergonne point. | Proof. Arguing as in Theorems 1 and 2 of [3], we see that \( \alpha ,\beta ,\gamma \) are of the form\n\n\[ \alpha = \frac{\pi - A}{2} + t\left( {B - C}\right) ,\;\beta = \frac{\pi - B}{2} + t\left( {C - A}\right) ,\;\gamma = \frac{\pi - C}{2} + t\left( {A - B}\right) . \]\n\nIn particular, \( \alpha + \beta + \gamma =... | Yes |
Theorem 1 (Archimedes). (1) The two circles each tangent to \( {PQ} \), the large semicircle and one of the smaller semicircles have equal radii \( t = \frac{ab}{a + b} \) . See Figure 10A. | Here is a simple construction of the Archimedean \ | No |
Proposition 1. The pedals of \( {O}_{a} \) on \( {BC},{O}_{b} \) on \( {CA} \), and \( {O}_{c} \) on \( {AB} \) form the cevian triangle of the Kiepert perspector \( K\left( {\arctan 2}\right) .{}^{1} \) | Proof. These pedals are the points \( \left( {0 : 2{S}_{C} + S : 2{S}_{B} + S}\right) ,\left( {2{S}_{C} + S : 0 : }\right. \) \( \left. {2{S}_{A} + S}\right) \), and \( \left( {2{S}_{B} + S : 2{S}_{A} + S : 0}\right) \) . | No |
Proposition 2. The pedals of \( {T}_{a} \) on \( {BC},{T}_{b} \) on \( {CA} \), and \( {T}_{c} \) on \( {AB} \) form the cevian triangle of the point \( \left( {{a}^{2} + S : {b}^{2} + S : {c}^{2} + S}\right) \) . | Proof. These pedals are the points \( \left( {0 : {b}^{2} + S : {c}^{2} + S}\right) ,\left( {{a}^{2} + S : 0 : {c}^{2}}\right) \), and \( \left( {{a}^{2} + S : {b}^{2} + S : 0}\right) \) . | Yes |
Proposition 3. The circumcircle, the radical circle, the inner Soddy circle, and the Brocard circles are coaxal, with the Lemoine axis as radical axis. | The Brocard circle has equation\n\n\[ \n{a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy} - \frac{{a}^{2}{b}^{2}{c}^{2}\left( {x + y + z}\right) }{{a}^{2} + {b}^{2} + {c}^{2}}\left( {\frac{x}{{a}^{2}} + \frac{y}{{b}^{2}} + \frac{z}{{c}^{2}}}\right) = 0. \]\n\nThe radical trace of these circles, namely, the intersection of the ra... | Yes |
Proposition 4. The circles of the Schoute coaxal system have centers \( {K}^{ * }\left( \theta \right) \) where \( \left| \theta \right| \geq \frac{\pi }{3} \), and radius \( \left| \frac{\sqrt{{\tan }^{2}\theta - 3}S}{2\left( {{S}_{\omega } + S \cdot \tan \theta }\right) }\right| \cdot R \) . It has equation\n\n\[ \n{... | Therefore, a circle with center \( \left( {{a}^{2}\left( {p{S}_{A} + {qS}}\right) : {b}^{2}\left( {p{S}_{B} + {qS}}\right) : {c}^{2}\left( {p{S}_{C} + {qS}}\right) }\right) \) and square radius \( \frac{\left( {{p}^{2} - 3{q}^{2}}\right) {a}^{2}{b}^{2}{c}^{2}}{{\left( 2pS + q\left( {a}^{2} + {b}^{2} + {c}^{2}\right) \r... | Yes |
Proposition 8. (a) The insimilicenters of the Lucas radical circle and the individual Lucas circles form a triangle perspective with \( {ABC} \) at \( {K}^{ * }\left( {\arctan 3}\right) \) . | Proof. These insimilicenters are the points\n\n\[ \left( {3{a}^{2}\left( {{S}_{A} + S}\right) : {b}^{2}\left( {3{S}_{B} + S}\right) : {c}^{2}\left( {3{S}_{C} + S}\right) }\right) ,\] \n\n\[ \left( {{a}^{2}\left( {3{S}_{A} + S}\right) : 3{b}^{2}\left( {{S}_{B} + S}\right) : {c}^{2}\left( {3{S}_{C} + S}\right) }\right) ,... | Yes |
Proposition 10. The six points \( {X}_{3},{X}_{4},{Y}_{3},{Y}_{4},{Z}_{3},{Z}_{4} \) lie on the conic | \[ \mathop{\sum }\limits_{\text{cyclic }}\frac{{a}^{2}}{{a}^{2} + S}{yz} = \frac{{a}^{2}{b}^{2}{c}^{2}S\left( {x + y + z}\right) }{\left( {{a}^{2} + S}\right) \left( {{b}^{2} + S}\right) \left( {{c}^{2} + S}\right) }\left( {\frac{x}{{a}^{2}} + \frac{y}{{b}^{2}} + \frac{z}{{c}^{2}}}\right) . \] | Yes |
\[ {3R}\sqrt{3} \geq \sum {PA} \geq l\sqrt{3} \] | where the last inequality follows by (6). | No |
Corollary 5. \[ T \leq \frac{\sqrt{3}}{12}{l}^{2} \] with equality when \( d = 0 \), i.e., when \( P \equiv O \) . | Now, since in any triangle of area \( T \), and sides \( {PA},{PB},{PC} \) one has \[ 2\sum {PA} \cdot {PB} - \sum P{A}^{2} \geq 4\sqrt{3} \cdot T \] (see for example [14], relation (8)), by (2) and (25) one can write \[ 2\sum {PA} \cdot {PB} \geq 3{d}^{2} + {l}^{2} + {l}^{2} - 3{d}^{2} = 2{l}^{2}, \] giving a new proo... | No |
Proposition 7. For the radii \( r \) and \( R \) of the Pompeiu triangle one has\n\n\[ r \leq \frac{l}{6} \leq \frac{R}{2} \] | The last inequality holds true by (23). This gives an improvement of Euler's inequality \( r \leq \frac{R}{2} \) for the Pompeiu triangle. Since \( T = \frac{{PA} \cdot {PB} \cdot {PC}}{4R} \), and \( r = \frac{T}{s} \), we get\n\n\[ {PA} \cdot {PB} \cdot {PC} = 2\operatorname{Rr}\left( {{PA} + {PB} + {PC}}\right) ,\] | No |
Proposition 9.\n\n\[ \n{PA} \cdot {PB} \cdot {PC} \leq \frac{\sqrt{3}{l}^{2}R}{3}. \n\] | This follows by \( T = \frac{{PA} \cdot {PB} \cdot {PC}}{4R} \) and (26). | No |
Theorem 13.\n\n\[ \sum {PA} \leq \sqrt{3\left( {{l}^{2} + 6{d}^{2}}\right) }.\n\] | This is related to (6). In fact,(6) and (35) imply that \( \sum {PA} = l\sqrt{3} \) if and only if \( d = 0 \), i.e., \( P \equiv O \) . | Yes |
We take the primitive \( \left( {a, b, c}\right) = \left( {{17},{21},{10}}\right) \), i.e., \( {T}_{2} \) and adjoin with itself (see Figure 3). Since \( \angle {CAD} = \angle {CBD},{ABCD} \) is cyclic. Ptolemy’s theorem shows that \( {AB} = \frac{341}{17} \) is rational. By enlarging the sides and diagonals 17 times e... | Since \( \angle {BAC} = \angle {BDC},\angle {ABD} = \angle {ACD} \), and \( \angle {BAD} = \angle {BAC} + \) \( \angle {CAD},\angle {BAD} \) is also a Heron angle and that triangle \( {ABD} \) is Heron. (Note: If \( \alpha \) and \( \beta \) are Heron angles then \( \alpha \pm \beta \) are also Heron angles. To see thi... | Yes |
Example 2. We adjoin the primitive triangles \( {T}_{4},{T}_{5} \) from Table 1. This can be done in two ways. | (i) Figure 4A illustrates one way. As in Example \( 1,{AB} = \frac{1500}{17} \), so Figure 4A is enlarged 17 times. The area is integral (reasoned as above). Hence the resulting quadrilateral is Brahmagupta. \n\n(i... | No |
Lemma 1. Let \( P \) be a point not on the sidelines of triangle \( {ABC} \), with traces \( {B}^{\prime },{C}^{\prime } \) on \( {AC},{AB} \) respectively. The circumcenters of triangles \( {AP}{B}^{\prime } \) and \( {AP}{C}^{\prime } \) coincide if and only if \( P \) lies on the reflection of the circumcircle \( {A... | The Proof of Lemma 1 is simple and can be found in [4]. | No |
Lemma 3. Let \( P \) be a point not on the sidelines of triangle \( {ABC} \), with traces \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) on \( {BC},{AC},{AB} \) respectively, and \( K \) the second intersection of the circumcircles of triangles \( {PC}{B}^{\prime } \) and \( P{C}^{\prime }B \) . The line \( {PK} \) is... | Proof. Triangles \( K{B}^{\prime }B \) and \( {KC}{C}^{\prime } \) are directly similar (see Figure 1). Therefore,\n\n\[ \n\frac{S\left\lbrack {K{B}^{\prime }B}\right\rbrack }{S\left\lbrack {{KC}{C}^{\prime }}\right\rbrack } = {\left( \frac{\overline{{B}^{\prime }B}}{\overline{C{C}^{\prime }}}\right) }^{2}.\n\]\n\nOn t... | Yes |
Lemma 4. Given a triangle \( {XYZ} \) and pairs of points \( M, N \) on \( {YZ}, P, Q \) on \( {ZX} \), and \( R, S \) on \( {XY} \) respectively. If the points in each of the quadruples \( P, Q \) , \( R, S;R, S, M, N;M, N, P, Q \) are concyclic, then all six points \( M, N, P, Q, R \) , \( S \) are concyclic. | Proof. Suppose that \( \left( {O}_{1}\right) ,\left( {O}_{2}\right) ,\left( {O}_{3}\right) \) are the circles passing through the quadruples \( \left( {P, Q, R, S}\right) ,\left( {R, S, M, N}\right) \), and \( \left( {M, N, P, Q}\right) \) respectively. If \( {O}_{1},{O}_{2},{O}_{3} \) are disctinct points, then \( {YZ... | Yes |
Proposition 1. Let \( p : {\mathbb{R}}^{4} \rightarrow \mathbb{R} \) a homogeneous function with the property that \( p\left( {{a}^{2},{b}^{2},{c}^{2},\bigtriangleup }\right) \geq 0 \), for any triangle in the Euclidean plane. Then for any \( x > 0 \) we have:\n\n\[ p\left( {x{a}^{2},\frac{1}{x}{b}^{2},{c}^{2} + \left(... | Proof. Consider \( q\left( x\right) = \left( {1 - \frac{1}{x}}\right) \left( {x{a}^{2} - {b}^{2}}\right) \), for \( x > 0 \) . In the triangle \( {ABC} \) we consider \( {A}_{1} \) and \( {B}_{1} \) on the sides \( {BC} \) and \( {AC} \), respectively, such that \( C{A}_{1} = {\alpha a} \) , \( {BC} = a, C{B}_{1} = {\b... | Yes |
Proposition 2. Let \( p : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be a homogeneous function of degree \( m \), and consider \( n \) collinear points \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) lying on the line \( d \) . Let \( S \) be a point exterior to the line \( \mathcal{L} \) and a secant \( {\mathcal{L}}^{\prime } ... | Proof. Denote \( {a}_{i} = \frac{{A}_{i}{A}_{i}^{\prime }}{{A}_{i}^{\prime }S} \), for \( i = 1,\ldots, n \) . Applying Menelaus’ Theorem in each of the triangles \( S{A}_{1}{A}_{2}, S{A}_{2}{A}_{3},\ldots, S{A}_{n - 1}{A}_{n} \) we have, for all \( i = 1,\ldots, n - 1 \),\n\n\[ \frac{1}{{a}_{i}} \cdot \frac{{A}_{i}K}{... | Yes |
Lemma 4. Let \( \left( O\right) \) be a circle tangent externally to two circles \( \left( {O}_{a}\right) \) and \( \left( {O}_{b}\right) \) respectively at \( A \) and \( B \) . If \( {PQ} \) is a common external tangent of \( \left( {O}_{a}\right) \) and \( \left( {O}_{b}\right) \) , then the quadrilateral \( {APQB} ... | Proof. Let \( {PA} \) intersect \( \left( O\right) \) at \( K \) . Since \( \left( O\right) \) and \( \left( {O}_{a}\right) \) touch each other externally at \( A,{OK} \) is parallel to \( {O}_{a}P \) . On the other hand, \( {O}_{a}P \) is also parallel to \( {O}_{b}Q \) as they are both perpendicular to the common tan... | Yes |
Proposition 7. The circle with diameter \( {A}_{a}{M}_{a} \) contains the point \( {F}_{a} \) . | Proof. Denote by \( {M}_{b}^{\prime } \) and \( {M}_{c}^{\prime } \) the midpoints of \( {I}_{a}B \) and \( {I}_{a}C \) respectively. The point \( {F}_{a} \) is common to the nine-point circles of \( {I}_{a}{BC},{I}_{a}{CA} \) and \( {I}_{a}{AB} \) . See Figure 5. We show that \( \left( {{A}_{a}{F}_{a},{F}_{a}{M}_{a}}\... | Yes |
Proposition 8. The points \( {X}_{b},{X}_{c} \) lie on the line \( {YZ} \) . | Proof. The collinearity of \( {C}_{a},{X}_{b},{X}_{c} \) follows from\n\n\[ \left( {{C}_{a}{X}_{b},{X}_{b}B}\right) = \left( {{C}_{a}{I}_{a},{I}_{a}B}\right) \]\n\n\[ = \left( {{C}_{a}{I}_{a},{AB}}\right) + \left( {{AB},{I}_{a}B}\right) \]\n\n\[ = {90}^{ \circ } + \left( {{I}_{a}B,{BC}}\right) \]\n\n\[ = \left( {{X}_{c... | Yes |
Proposition 9. The line \( {Y}_{a}{Z}_{a} \) contains the midpoints \( {B}^{\prime },{C}^{\prime } \) of \( {CA},{AB} \), and is parallel to \( {BC} \) . | Proof. Since \( A,{Y}_{a},{I}_{a},{Z}_{a} \) are concyclic,\n\n\[ \left( {A{Y}_{a},{Y}_{a}{Z}_{a}}\right) = \left( {A{I}_{a},{I}_{a}{Z}_{a}}\right) = \frac{C}{2} = \left( {{CA}, A{Y}_{a}}\right) . \]\n\nTherefore, the intersection of \( {AC} \) and \( {Y}_{a}{Z}_{a} \) is the circumcenter of the right triangle \( {AC}{... | Yes |
Proposition 10. The line \( {I}_{a}X \) contains the midpoint \( {A}^{\prime } \) of \( {BC} \) . | Proof. Since the diagonals of the parallelogram \( {I}_{a}{Y}_{a}X{Z}_{a} \) bisect each other, the line \( {I}_{a}X \) passes through the midpoint of the segment \( {Y}_{a}{Z}_{a} \) . Since \( {Y}_{a}{Z}_{a} \) and \( {BC} \) are parallel, with \( B \) on \( {I}_{a}{Z}_{a} \) and \( C \) on \( {I}_{a}{Y}_{a} \), the ... | Yes |
Theorem 11. The triangles \( {XYZ} \) and \( {I}_{a}{I}_{b}{I}_{c} \) are homothetic at the Mittenpunkt \( M \) of triangle \( {ABC} \), the ratio of homothety being \( {2R} + r : - {2R} \) . | Proof. The lines \( {I}_{a}X,{I}_{b}Y,{I}_{c}Z \) contain respectively the midpoints of \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) of \( {BC},{CA},{AB} \) . They intersect at the common point of \( {I}_{a}{A}^{\prime },{I}_{b}{B}^{\prime },{I}_{c}{C}^{\prime } \), the Mittenpunkt \( M \) of triangle \( {ABC} \) . ... | Yes |
Theorem 12. The Taylor circle of the excentral triangle is the radical circle of the excircles. | Proof. The perpendicular bisector of \( {Y}_{c}{Z}_{b} \) is a line parallel to the bisector of angle \( A \) and passing through the midpoint \( {A}^{\prime } \) of \( {BC} \) . This is the \( {A}^{\prime } \)-bisector of the medial triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . Similarly, the perpendicular ... | Yes |
Proposition 13. \( X \) lies on the radical axis of the circles \( \left( {W}_{b}\right) \) and \( \left( {W}_{c}\right) \) . | Proof. By Theorem 12, \( X{Z}_{a} \cdot X{Z}_{b} = X{Y}_{a} \cdot X{Y}_{c} \) . Since \( {Y}_{c},{Y}_{a} \) are on the nine-point circle \( \left( {W}_{b}\right) \) and \( {Z}_{a},{Z}_{b} \) on the the circle \( \left( {W}_{c}\right), X \) lies on the radical axis of these two nine-point circles. | Yes |
Proposition 14. \( {M}_{a} \) lies on the radical axis of the circles \( \left( {W}_{b}\right) \) and \( \left( {W}_{c}\right) \) . | Proof. Let \( {M}_{b}^{\prime \prime } \) and \( {M}_{c}^{\prime \prime } \) be the midpoints of \( A{I}_{b} \) and \( A{I}_{c} \) respectively. See Figure 9. Note that these lie on the nine-point circles \( \left( {W}_{b}\right) \) and \( \left( {W}_{c}\right) \) respectively. Since \( C,{I}_{b},{I}_{c}, B \) are conc... | Yes |
Proposition 15. \( X,{F}_{a} \), and \( {M}_{a} \) are collinear. | Proof. We prove that the Euler line of triangle \( A{Y}_{a}{Z}_{a} \) contains the point \( {F}_{a} \) . The points \( X \) and \( {M}_{a} \) are respectively the orthocenter and circumcenter of the triangle.\n\nLet \( {A}_{a}^{\prime } \) be the antipode of \( {A}_{a} \) on the \( A \) -excircle. Since \( {AX} \) has ... | Yes |
Proposition 16. \( X{F}_{a} \) is also the Euler line of triangle \( A{Y}_{a}{Z}_{a} \) . | Proof. The circumcenter of \( A{Y}_{a}{Z}_{a} \) is clearly \( {M}_{a} \) . On the other hand, since \( A \) is the orthocenter of triangle \( X{Y}_{a}{Z}_{a}, X \) is the orthocenter of triangle \( A{Y}_{a}{Z}_{a} \) . Therefore the line \( X{M}_{a} \), which also contains \( {F}_{a} \), is the Euler line of triangle ... | Yes |
Proposition 17. \( {M}_{a} \) is the orthocenter of triangle \( {I}^{\prime }{B}^{\prime }{C}^{\prime } \) . | Proof. Let \( {H}_{a} \) be the orthocenter of \( {IBC} \) . Since \( B{H}_{a} \) is perpendicular to \( {IC} \), it is parallel to \( {I}_{a}C \) . Similarly, \( C{H}_{a} \) is parallel to \( {I}_{a}B \) . Thus, \( B{H}_{a}C{I}_{a} \) is a parallelogram, and \( {A}^{\prime } \) is the midpoint of \( {I}_{a}{H}_{a} \) ... | Yes |
Proposition 18. \( {K}_{a} \) is the circumcenter of \( {I}^{\prime }{B}^{\prime }{C}^{\prime } \) . | Proof. By Lemma 4, the points \( {F}_{b},{F}_{c},{A}_{b} \) and \( {A}_{c} \) are concyclic, and the lines \( {A}_{b}{F}_{b} \) and \( {A}_{c}{F}_{c} \) intersect at a point \( {K}_{a} \) on the nine-point circle, which is the midpoint of the arc \( {B}^{\prime }{C}^{\prime } \) not containing \( {A}^{\prime } \) . See... | Yes |
Proposition 19. \( {K}_{a} \) lies on the radical axis of \( \left( {W}_{b}\right) \) and \( \left( {W}_{c}\right) \) . | Proof. Let \( D \) and \( E \) be the second intersections of \( {K}_{a}{F}_{b} \) with \( \left( {W}_{b}\right) \) and \( {K}_{a}{F}_{c} \) with \( \left( {W}_{c}\right) \) respectively. We shall show that \( {K}_{a}{F}_{b} \cdot {K}_{a}D = {K}_{a}{F}_{c} \cdot {K}_{a}E \) . \n\nSince \( {A}_{c},{F}_{c},{F}_{b},{A}_{b... | Yes |
Corollary 22. The radical center of the circles \( {\mathcal{C}}_{a},{\mathcal{C}}_{b},{\mathcal{C}}_{c} \) is \( {S}^{\prime } \) . | Proof. The points \( X \) and \( {F}_{a} \) are common to the circles \( {\mathcal{C}}_{b} \) and \( {\mathcal{C}}_{c} \) . The line \( X{F}_{a} \) is the radical axis of the two circles. Similarly the radical axes of the two other two pairs of circles are \( Y{F}_{b} \) and \( Z{F}_{c} \) . The radical center is there... | Yes |
Proposition 23. The line \( X{A}_{a} \) is perpendicular to \( {YZ} \) . | Proof. With reference to Figure 8, note that\n\n\[ \n{A}_{b}{Y}_{a} : {A}_{b}X = {A}_{b}C \cdot \frac{\sin \left( {C + \frac{A}{2}}\right) }{\sin \frac{C}{2}} : {A}_{b}{A}_{c} \cdot \frac{\sin \frac{A + B}{2}}{\sin \frac{B + C}{2}} \n\]\n\n\[ \n= {A}_{b}C : \left( {b + c}\right) \cdot \frac{\sin \frac{C}{2}\sin \frac{A... | Yes |
Proposition 25. The complement of the Schiffler point is the point \( {S}^{\prime } \) which divides\n\n\( {HW} \) in the ratio\n\n\[ H{S}^{\prime } : {S}^{\prime }W = 2\left( {{2R} + r}\right) : - R. \] | Proof. We define \( {K}_{b} \) and \( {K}_{c} \) similarly as \( {K}_{a} \) . Since \( {K}_{b} \) and \( {K}_{c} \) are the midpoints of the arcs \( {C}^{\prime }{A}^{\prime } \) and \( {A}^{\prime }{B}^{\prime },{K}_{b}{K}_{c} \) is perpendicular to the \( {A}^{\prime } \) -bisector of \( {A}^{\prime }{B}^{\prime }{C}... | Yes |
Theorem 1. Let \( a \) be a given positive number. (1). If \( d \leq r \), i.e. \( R \leq \left( {\sqrt{2} + 1}\right) r \), then there is a unique poristic triangle \( {ABC} \) with \( {BC} = a \) if and only if | \[ {4r}\left( {{2R} - r - {2d}}\right) \leq {a}^{2} \leq {4r}\left( {{2R} - r + {2d}}\right) . \] | Yes |
Proposition 4. The Kenmotu points \( {K}_{\mathrm{e}} \) and \( {K}_{\mathrm{e}}^{\prime } \) divide the segment \( {OK} \) in the ratio\n\n\[ O{K}_{\mathrm{e}} : {K}_{\mathrm{e}}K = {a}^{2} + {b}^{2} + {c}^{2} : {2S}, \]\n\n\[ O{K}_{\mathrm{e}}^{\prime } : {K}_{\mathrm{e}}^{\prime }K = {a}^{2} + {b}^{2} + {c}^{2} : - ... | Proof. A typical point on the Brocard axis has coordinates\n\n\[ {K}^{ * }\left( \theta \right) = \left( {{a}^{2}\left( {{S}_{A} + {S}_{\theta }}\right) : {b}^{2}\left( {{S}_{B} + {S}_{\theta }}\right) : {c}^{2}\left( {{S}_{C} + {S}_{\theta }}\right) }\right) .\n\nIt divides the segment \( {OK} \) in the ratio\n\n\[ O{... | Yes |
Theorem 1 (GPAT).\n\n\\[ \n{d}^{2}\left( {X, m, Y, n}\right) = {\sigma }^{2}\left( {X, m}\right) + \parallel \bar{x} - \bar{y}{\parallel }^{2} + {\sigma }^{2}\left( {Y, m}\right) .\n\\] | Proof.\n\n\\[ \n\frac{1}{MN}\mathop{\sum }\limits_{{x \in X, y \in Y}}m\left( x\right) n\left( y\right) \parallel x - y{\parallel }^{2}\n\\]\n\n\\[ \n= \frac{1}{MN}\mathop{\sum }\limits_{{x \in X, y \in Y}}m\left( x\right) n\left( y\right) \parallel x - \bar{x} + \bar{x} - \bar{y} + \bar{y} - y{\parallel }^{2}\n\\]\n\n... | Yes |
Proposition 3. Let \( p \) denote the perimeter of the triangle \( A, B, C \) . The distance between the incenter \( I \) and centroid \( G \) satisfies the following equation:\n\n\[ I{G}^{2} = \frac{{p}^{2}}{6} - \frac{5}{18}\left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) - {4Rr}. \] | Proof. Let \( {\bigtriangleup }_{G} \) denote the triangle weighted 1 at each vertex and \( {\bigtriangleup }_{I} \) denote the same triangle with weights attached to the vertices which are the lengths of the opposite sides. We apply the GPAT and a direct calculation:\n\n\[ {d}^{2}\left( {{\bigtriangleup }_{G},{\bigtri... | Yes |
Theorem 4 (Feuerbach). The nine point circle of \( \bigtriangleup {ABC} \) is internally tangent to the incircle. | Proof. (outline) The radius of the nine point circle is \( R/2 \) . The result will established if we show that \( \left| {IN}\right| = R/2 - r \) . However, in \( \bigtriangleup {INO} \) the point \( G \) is on the side \( {NO} \) and \( {NG} : {GO} = 1 : 2 \) . We know \( \left| {IO}\right| ,\left| {IG}\right| ,\left... | Yes |
Proposition 5. The incenter of a non-equilateral triangle lies strictly in the interior of the circle on diameter \( {GH} \) . | This was presumably known to Euler [5], and a stronger version of the result was proved in [4]. Given Feuerbach's theorem, this result almost proves itself. Let \( N \) be the nine-point center, the midpoint of the segment \( {OH} \), Feuerbach’s tangency result yields \( {IN} = R/2 - r \) . However \( O{I}^{2} = {R}^{... | Yes |
Theorem 1 (Steiner-Lehmus). If two internal angle bisectors of a triangle are equal, then the triangle is isosceles. | Figure 1 shows the bisectors \( {BE} \) and \( {CF} \) of \( \angle {ABC} \) and \( \angle {ACB} \) . We assume \( {BE} = {CF} \) . If \( {AB} \neq {AC} \), let \( {AB} < {AC} \), i.e., \( \angle {ACB} < \angle {ABC} \) or \( \frac{C}{2} < \frac{B}{2} \) . A\n\n--- \n\ncomparison of triangles \( {BEC} \) with \( {BFC} ... | Yes |
Theorem 2. If two Gergonne cevians of a triangle are equal, then the triangle is isosceles. | 3.1. First proof. Figure 2 shows the equal Gergonne cevians \( {BE},{CF} \) of triangle\n\n\( {ABC} \) . We consider \( \bigtriangleup {ABE},\bigtriangleup {ACF} \) and apply the law of cosines:\n\n\[ \nB{E}^{2} = {c}^{2} + {\left( s - a\right) }^{2} - {2c}\left( {s - a}\right) \cos A, \n\] \n\n\[ \nC{F}^{2} = {b}^{2} ... | Yes |
Theorem 3. The internal angle bisectors of the angles \( {ABC} \) and \( {ACB} \) of triangle \( {ABC} \) meet the Gergonne cevian \( {AD} \) at \( E \) and \( F \) respectively. If \( {BE} = {CF} \), then triangle \( {ABC} \) is isosceles. | Proof. We refer to Figure 4. If \( {AB} \neq {AC} \), let \( {AB} < {AC} \) . Hence \( b > c, s - b < \) \( s - c \) and \( E \) lies below \( F \) on \( {AD} \) . A simple calculation with the help of the angle bisector theorem shows that the Gergonne cevian \( {AD} \) lies to the left of the cevian that bisects \( \a... | Yes |
Proposition 2. Suppose \( {m}^{2} > 3 \) . The Euler conic is a hyperbola with eccentricity | \[ \varepsilon = \sqrt{\frac{2\sqrt{{m}^{2} + 1}}{\sqrt{{m}^{2} + 1} - 2}}. \] The foci are the points \[ \pm \left( {\operatorname{sgn}\left( m\right) \cdot \sqrt{\frac{3\left( {\sqrt{{m}^{2} + 1} + 1}\right) }{{m}^{2} - 3}},\sqrt{\frac{3\left( {\sqrt{{m}^{2} + 1} - 1}\right) }{{m}^{2} - 3}}}\right) , \] where \( \ope... | Yes |
Lemma 4. The Euler line of triangle \( {ABC} \) is perpendicular to \( {AB} \) if and only if \( {AB} = {AC} \) . In this case, the Euler line is the perpendicular bisector of \( {AB} \) . | We shall henceforth assume that the Euler line is not perpendicular to \( {AB} \) . It therefore has an equation of the form\n\n\[ y = {mx} + k \]\n\nThe circumcenter is the intersection of the Euler line with the line \( x = 0 \), the perpendicular bisector of \( {AB} \) . It is the point \( O = \left( {0, k}\right) \... | No |
Proposition 5. The number of points \( C \) for which triangle \( {ABC} \) has Euler line \( y = {mx} + k \) is 0,1, or 2 according as \( \left( {{m}^{2} - 3}\right) \left( {{k}^{2} + 1}\right) < , = \), or \( > - 4 \) . | In the hyperbolic and degenerate cases \( {m}^{2} \geq 3 \), there are always two such triangles. In the elliptic case, \( {m}^{2} < 3 \) . There are two such triangles if and only if \( {k}^{2} < \frac{{m}^{2} + 1}{3 - {m}^{2}} \) . | Yes |
For \( {m}^{2} < 3 \) and \( k = \pm \sqrt{\frac{{m}^{2} + 1}{3 - {m}^{2}}} \), there is a unique triangle \( {ABC} \) whose Euler line is the line \( y = {mx} + \dot{k} \) . The lines \( y = {mx} + {3k} \) are tangent to the Euler ellipse (3) at the points | \[ \pm \left( {\frac{-{2m}}{\sqrt{\left( {{m}^{2} + 1}\right) \left( {3 - {m}^{2}}\right) }},\frac{3 + {m}^{2}}{\sqrt{\left( {{m}^{2} + 1}\right) \left( {3 - {m}^{2}}\right) }}}\right) . \] | Yes |
Lemma 1. Referring to Figure 3, points \( X, U \) are intersections of opposite sides of \( q.{U}^{ * },{X}^{ * } \) are intersections of opposite sides of \( q * .V, W, Y, Z \) are intersections of opposite sides of other flanks of the complex. Points \( X, Y, Z \) and \( U, V, W \) are aligned on two parallel lines. | The proof is a trivial consequence of the similarity of opposite located main flanks. Thus, \( l, r \) are similar and their similarity center is \( X \) . Analogously \( t, b \) are similar and their similarity center is \( U \) . Besides triangles \( {XDY},{VDU} \) are anti-homothetic with respect to \( D \) and tria... | No |
Proposition 2. The following lines are common radical axes of the circle pairs:\n\n(1) Line \( X{X}^{ * } \) coincides with \( \left( {{lt},{lb}}\right) = \left( {t, b}\right) = \left( {{rt},{rb}}\right) = \left( {{ltr},{lbr}}\right) \) .\n\n(2) Line \( U{U}^{ * } \) coincides with \( \left( {{lt},{rt}}\right) = \left(... | Referring to Figure 4, I show that line \( {XYZ} \) is identical with \( \left( {{lbr}, b}\right) \) . Indeed, from the intersection of the two circles \( \{ {lbr}\} \) and \( \{ b\} \) with circle \( \{ {rb}\} \) we see that \( Z \) is on their radical axis. Similarly, from the intersection of these two circles with \... | No |
Proposition 3. Referring to Figure 5, the centers of the flanks build equal parallelograms with parallel sides: \( \left( {lt}\right) \left( t\right) \left( {tqb}\right) \left( {tlb}\right) ,\left( {ltr}\right) \left( {rt}\right) \left( {trb}\right) \left( {q * }\right) ,\left( l\right) \left( q\right) \left( b\right) ... | The proof of the various parallelities is a consequence of the coincidences of radical axes. For example, in the first figure, sides \( \left( t\right) \left( {rt}\right) ,\left( q\right) \left( r\right) ,\left( b\right) \left( {rb}\right) \) are parallel because, all, are orthogonal to the corresponding radical axis, ... | Yes |
Proposition 4. Referring to Figure 6, the following properties are valid:\n\n(1) The circumcircles of adjacent main flanks are tangent at the vertices of \( q \) . (2) The intersection point of the diagonals US, TV of the quadrilateral TSUV, formed by the centers of the main flanks, coincides with the anticenter \( M \... | The properties are immediate consequences of the definitions. (1) follows from the fact that triangles \( {FTC} \) and \( {CSB} \) are similar isosceli. To see that property (2) is valid, consider the parallelograms \( p = {OYMX},{p}^{* * } = {UX}{M}^{* * }Y \) and \( {p}^{ * } = S{Y}^{ * }{M}^{ * }Y \), tightly relate... | Yes |
Proposition 5. Referring to the previous figure, the lines \( {QM} \) and \( {QO} \) are symmetric with respect to the bisector of angle \( {AQD} \) . The same is true for lines \( {QX} \) and \( {QY} \) . | This is again obvious, since the trapezia \( {DAHG} \) and \( {CBLK} \) are similar and inversely oriented with respect to the sides of the angle \( {AQD} \) . | No |
Proposition 6. The quadrilateral STUV, of the centers of the main flanks, is circumscriptible, its incenter coincides with the circumcenter \( O \) of \( q \) and its radius is \( r \cdot \sin \left( {\phi + \xi }\right) .r \) being the circumcradius of \( q \) and \( {2\phi },{2\xi } \) being the measures of two angle... | The proof follows immediately from the similarity of triangles \( {UAO} \) and \( {OBS} \) in Figure 8. The angle \( \omega = \phi + \xi \), gives for \( \left| {OZ}\right| = r \cdot \sin \left( \omega \right) .Z,{Z}^{ * }, W,{W}^{ * } \) being the projections of \( O \) on the sides of STUV. Analogous formulas hold fo... | Yes |
Proposition 7. Referring to Figure 9, the anticenters of the flanks build equal parallelograms with parallel sides: \( \left\lbrack {lt}\right\rbrack \left\lbrack t\right\rbrack \left\lbrack q\right\rbrack \left\lbrack l\right\rbrack ,\left\lbrack {tlb}\right\rbrack \left\lbrack {tqb}\right\rbrack \left\lbrack b\right\... | The proof is similar to the one of Proposition 2. For example, segments \( \left\lbrack {lt}\right\rbrack \left\lbrack t\right\rbrack \) , \( \left\lbrack l\right\rbrack \left\lbrack q\right\rbrack ,\left\lbrack {tlb}\right\rbrack \left\lbrack {tqb}\right\rbrack ,\left\lbrack {lb}\right\rbrack \left\lbrack b\right\rbra... | No |
Proposition 8. The barycenters of the flanks build the pattern of equal parallelograms of Figure 11. | Indeed, this is a consequence of the corresponding results for centers and anti-centers of the flanks and the fact that linear combinations of parallelograms \( {c}_{i} = \) \( \left( {1 - t}\right) {a}_{i} + t{b}_{i} \), where \( {a}_{i},{b}_{j} \) denote the vertices of parallelograms, are again parallelograms. Here ... | Yes |
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