Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Theorem 1. The only poristic triangles perspective with \( {ABC} \) are:\n\n(1) the reflection of \( {ABC} \) in the line \( {OI} \), the perspector being the infinite point on a line perpendicular to \( {OI} \),\n\n(2) the circumcevian triangles of the two limit points of the coaxial system generated by the circumcirc... | In (1), the lines \( A{A}^{\prime }, B{B}^{\prime }, C{C}^{\prime } \) are all perpendicular to the line \( {OI} \) . See Figure 1. The perspector is the infinite point on a line perpendicular to \( {OI} \) . One such line is the trilinear polar of the incenter \( I = \left( {a : b : c}\right) \), with equation\n\n\[ \... | No |
Proposition 2. (1) \( {P}^{* * } = P \) . | Proof. (1) is trivial. | No |
Lemma 3. Let \( A = \left( {{x}_{1},{y}_{1}}\right) \) and \( B = \left( {{x}_{2},{y}_{2}}\right) \) be two points on the same circle \( {\mathcal{C}}_{b} \) . The segment \( {AB} \) is tangent to a circle \( {\mathcal{C}}_{{b}^{\prime }} \) at the point whose \( y \) -coordinate is \( \sqrt{{y}_{1}{y}_{2}} \) . | Proof. This is clear if \( {y}_{1} = {y}_{2} \) . In the generic case, extend \( {AB} \) to intersect the \( x \) -axis at a point \( C \) . The segment \( {AB} \) is tangent to a circle \( {\mathcal{C}}_{{b}^{\prime }} \) at a point \( P \) such that \( {CP} = {CF} \) . It follows that \( C{P}^{2} = C{F}^{2} = {CA} \c... | Yes |
Theorem 4. If a chord \( {AB} \) of \( {\mathcal{C}}_{b} \) is tangent to \( {\mathcal{C}}_{{b}^{\prime }} \) at \( P \), then the chord \( {A}^{ * }{B}^{ * } \) is tangent to the same circle \( {\mathcal{C}}_{{b}^{\prime }} \) at \( {P}^{ * } \) . | Proof. That \( P \) and \( {P}^{ * } \) lie on the same circle is clear from Proposition 2(1). It remains to show that \( {P}^{ * } \) is the correct point of tangency. This follows from noting that the \( y \) -coordinate of \( {P}^{ * } \), being \( \frac{{a}^{2}}{\sqrt{{y}_{1}{y}_{2}}} \), is the geometric mean of t... | No |
Corollary 6. The reflection of the triangle \( {A}^{ * }{B}^{ * }{C}^{ * } \) in the line \( {OI} \) also has \( I\left( r\right) \) as incircle, and is perspective with \( {ABC} \) at the point \( {F}^{\prime } \) . | Proof. This follows from Proposition 2 (3). | No |
Proposition 7. Let \( {XY} \) be the diameter of the circumcircle through the incenter \( I \) . If the tangents to the incircle from these two points are \( {XP},{XQ},{YQ} \), and \( Y{P}^{\prime } \) such that \( P \) and \( Q \) are on the same side of \( {OI} \), then \( P{P}^{\prime } \) intersects \( {OI} \) at \... | Proof. This follows from Theorem 4 by observing that \( \mathrm{Y} = {\mathrm{X}}^{ * } \) . | No |
Theorem 1 (van Aubel [2, p.163]). Let the cevians \( {AD},{BE},{CF} \) of triangle \( {ABC} \) concur at the point \( S \) . Then\n\n\[ \frac{AS}{SD} = \frac{AE}{EC} + \frac{AF}{FB} \] | Proof. Let \( \left\lbrack T\right\rbrack \) denote the area of triangle \( T \) . We use the known result: if two triangles have a common altitude, then their areas are proportional to the corresponding bases. Hence, from Figure 1,\n\n\[ \frac{AS}{SD} = \frac{\left\lbrack ABS\right\rbrack }{\left\lbrack SBD\right\rbra... | Yes |
In Figure 2, let \( {AD} \) denote the median, and \( {BE} \) the Gergonne cevian. Then \( \frac{AS}{SD} = \frac{2\left( {s - a}\right) }{s - c} \) . | Proof. The present hypothesis implies \( {BD} = {DC} \), and \( E \) is the point where the incircle is tangent with \( {AC} \) . It is well - known that \( {AE} = s - a,{EC} = s - c \) . Now, Ceva’s theorem, \( \frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1 \), yields \( \frac{AF}{FB} = \frac{s - a}{s - c} ... | No |
Theorem 3. Let \( \lambda \) be a rational number such that \( 0 < \lambda \leq 2 \) . The \( \lambda \) -family of Heron triangles is described by\n\n\[ \left( {a, b, c}\right) = \left( {2\left( {{m}^{2} + {\lambda }^{2}{n}^{2}}\right) ,\left( {2 + \lambda }\right) \left( {{m}^{2} - {2\lambda }{n}^{2}}\right) ,\lambda... | Proof. From the definition we have\n\n\[ \frac{2\left( {s - a}\right) }{s - c} = \lambda \;\text{ or }\;b = \frac{2 + \lambda }{2 - \lambda }\left( {a - c}\right) .\n\nIf \( \lambda \neq 2 \), we assume \( a - c = \left( {2 - \lambda }\right) p \) . This gives \( b = \left( {2 + \lambda }\right) p \) . If \( \lambda = ... | Yes |
Corollary 5. Theorem 3 yields the isosceles Heron triangles \( \left( {a, b, c}\right) = \left( {{m}^{2} + }\right. \) \( \left. {{n}^{2},2\left( {{m}^{2} - {n}^{2}}\right) ,{m}^{2} + {n}^{2}}\right) \) for \( \lambda = 2 \) . | Actually, \( \lambda = 2 \) yields \( \left( {a, b, c}\right) = \left( {{m}^{2} + 4{n}^{2},2\left( {{m}^{2} - 4{n}^{2}}\right) ,{m}^{2} + 4{n}^{2}}\right) \) . However, the transformation \( m \mapsto {2m}, n \mapsto n \) results in the more familiar form displayed in Corollary 5. | Yes |
Corollary 6. Theorem 3 describes the complete set of Heron triangles. | This is because the Gergonne cevian \( {BE} \) must intersect the median \( {AD} \) at a unique point. Therefore for all Heron triangles \( 0 < \lambda \leq 2 \) . Suppose first we fix \( \lambda \) at such a rational number. Then Theorem 3 gives the entire \( \lambda \) -family of Heron triangles each member of which ... | No |
Given a point \( Q \), construct through the traces \( {A}_{Q},{B}_{Q},{C}_{Q} \) forward parallelians to \( {AB},{BC},{CA} \), intersecting \( {CA},{AB},{BC} \) at \( Y, Z \) and \( X \) respectively. The lines \( {AX},{BY},{CZ} \) intersect at \( Q \rightarrow \) . | Proof. Suppose \( Q = \left( {x : y : z}\right) \) in homogeneous barycentric coordinates. In Figure 3a, \( {BX} : {XC} = B{C}_{Q} : {C}_{Q}A = x : y \) since \( {C}_{Q} = \left( {x : y : 0}\right) \) . It follows that \( X = \left( {0 : y : x}\right) \sim \left( {0 : \frac{1}{x} : \frac{1}{y}}\right) \) . Similarly, \... | Yes |
Proposition 2. The points \( {P}^{ \pm } \) are the forward Brocardian points of the Fermat points 7 | \[ {F}^{ \pm } = \left( {\frac{a}{\sin \left( {A \pm \frac{\pi }{3}}\right) } : \frac{b}{\sin \left( {B \pm \frac{\pi }{3}}\right) } : \frac{c}{\sin \left( {C \pm \frac{\pi }{3}}\right) }}\right) . \] | Yes |
Lemma 4. If \( {UGV} \) is a chord of the circumcircle \( \Gamma \) through \( G \) meeting \( \Gamma \) at \( U, V \) , then the tripolar of \( U \) is the line \( K\widehat{V} \) passing through the symmedian point \( K \) and the isogonal conjugate of \( V \) . | ## 7. The axes of \( \Phi \)\n\nTo proceed with the location of the axes of \( \Phi \), we start with the conditions of Lemma 2 where \( X, Y \) are the common points of \( {OGH} \) and \( \Gamma \) .\n\nFrom Lemma 4, since \( {XGY} \) are collinear, the tripolars of \( X, Y \) are respectively \( K\widehat{Y}, K\wideh... | No |
Proposition 1. If a circle through the focus of a parabola has its center on the directrix, there exists an equilateral triangle inscribed in the circle, whose side lines are tangent to the parabola. | Proof. Denote by \( p \) the distance from the focus \( F \) of the parabola to its directrix. In polar coordinates with the pole at \( F \), let the center of the circle be the point \( \left( {\frac{p}{\cos \alpha },\alpha }\right) \) . The radius of the circle is \( R = \frac{p}{\cos \alpha } \) . See Figure 4. If t... | Yes |
Proposition 3. Let \( {PQ} \) be a chord of the circumcircle. The following statements are equivalent: \( {}^{7} \n\n(a) \( {PQ} \) and \( {P}^{\prime }{Q}^{\prime } \) are parallel.\n\n(b) The line \( {PQ} \) is tangent to the Kiepert parabola \( \mathcal{P} \).\n\n(c) The Simson lines \( s\left( P\right) \) and \( s\... | Proof. If the line \( {PQ} \) is \( {ux} + {vy} + {wz} = 0 \), then \( {P}^{\prime }{Q}^{\prime } \) is \( {a}^{2}{ux} + {b}^{2}{vy} + {c}^{2}{wz} = 0 \) . These two lines are parallel if and only if\n\n\[ \n\frac{{b}^{2} - {c}^{2}}{u} + \frac{{c}^{2} - {a}^{2}}{v} + \frac{{a}^{2} - {b}^{2}}{w} = 0 \n\]\n\n(1)\n\nwhich... | Yes |
Proposition 4. Let \( P, Q, R \) be points on the circumcircle. The following statements are equivalent.\n\n(a) The Simson lines \( s\left( P\right), s\left( Q\right), s\left( R\right) \) are concurrent.\n\n(b) \( \left( {{AB},{PQ}}\right) + \left( {{BC},{QR}}\right) + \left( {{CA},{RP}}\right) = 0\left( {\;\operatorna... | Proof. See [4, §§2.16-20]. | No |
Proposition 5. A line \( \ell \) is parallel to a side of the Morley triangle if and only if\n\n\[ \left( {{AB},\ell }\right) + \left( {{BC},\ell }\right) + \left( {{CA},\ell }\right) = 0\;\left( {\;\operatorname{mod}\;\pi }\right) . | Proof. Consider the Morley triangle \( {M}_{a}{M}_{b}{M}_{c} \) . The line \( B{M}_{c} \) and \( C{M}_{b} \) intersecting at \( \mathrm{P} \), the triangle \( P{M}_{b}{M}_{c} \) is isoceles and \( \left( {{M}_{c}{M}_{b},{M}_{c}P}\right) = \frac{1}{3}\left( {B + C}\right) \) . Thus, \( \left( {{BC},{M}_{b}{M}_{c}}\right... | Yes |
Proposition 2. An equation of the form \( \ell {\lambda }^{2} + m{\mu }^{2} + n{\nu }^{2} + q = 0 \) represents a circle or a line according as \( \ell + m + n \) is nonzero or otherwise. | The center of the circle has homogeneous barycentric coordinates \( \left( {\ell : m : n}\right) \) . If \( \ell + m + n = 0 \), the line is orthogonal to the direction \( \left( {\ell : m : n}\right) \) . Among the applications one finds the equation of the Euler line in tripolar coordinates [op. cit. §26]. | No |
Proposition 3. The tripolar equation of the Euler line is\n\n\\[ \n\\left( {{b}^{2} - {c}^{2}}\\right) {\\lambda }^{2} + \\left( {{c}^{2} - {a}^{2}}\\right) {\\mu }^{2} + \\left( {{a}^{2} - {b}^{2}}\\right) {\\nu }^{2} = 0.\n\\]\n\n(1) | ## 5. Tripolar equation of the Euler line\n\nConsider the centroid \( G \) of triangle \( {ABC} \). By the Apollonius theorem, and the fact that \( G \) divides each median in the ratio \( 2 : 1 \), it is easy to see that the tripolar coordinates of \( G \) satisfy\n\n\\[ \n{\\lambda }^{2} : {\\mu }^{2} : {\\nu }^{2} =... | Yes |
Proposition 7. A complete list of finite real points common to all \( {\mathcal{C}}_{n} \) curves is as follows:\n\n(1) the vertices \( A, B, C \) and their reflections on the respective opposite side,\n\n(2) the apexes of the six equilateral triangles erected on the sides of \( {ABC} \) ,\n\n(3) the circumcenter, and\... | Proof. It is easy to see that each of these points lies on \( {\mathcal{C}}_{n} \) for every positive integer \( n \) . For the isodynamic points, recall that \( \lambda : \mu : \nu = \frac{1}{a} : \frac{1}{b} : \frac{1}{c} \) . We show that \( {\mathcal{C}}_{1} \) and \( {\mathcal{C}}_{2} \) meet precisely in these 15... | Yes |
Theorem 1. If \( O \) is a point on the plane \( \pi \) such that the centroidal line of the trihedron \( O.{ABC} \) is perpendicular to \( \pi \), then \( {OA} + {OB} + {OC} \leq {MA} + {MB} + {MC} \) for every point \( M \) on \( \pi \) . | Proof. Let \( M \) be an arbitrary point on \( \pi \), and \( \mathbf{i},\mathbf{j},\mathbf{k} \) the unit vectors along OA, \( \mathrm{{OB}},\mathrm{{OC}} \) respectively. The vector \( \mathbf{i} + \mathbf{j} + \mathbf{k} \) is parallel to the centroidal line of the trihedron \( O.{ABC} \) . Since this line is perpen... | Yes |
Lemma 2. The solution of\n\n\\[ \n{f}_{1}x + {g}_{1}y + {h}_{1}z = {f}_{2}x + {g}_{2}y + {h}_{2}z = {f}_{3}x + {g}_{3}y + {h}_{3}z \n\\]\n\nis\n\n\\[ \nx : y : z = \\left| \\begin{array}{lll} 1 & {g}_{1} & {h}_{1} \\\\ 1 & {g}_{2} & {h}_{2} \\\\ 1 & {g}_{3} & {h}_{3} \\end{array}\\right| : \\left| \\begin{array}{lll} {... | Proof. since there are two linear equations in three indeterminates, solution is unique up to a proportionality constant. To verify that this is the correct solution, note that for \\( i = 1,2,3 \\), substitution into the \\( i \\) -th linear form gives\n\n\\[ \n- \\left| \\begin{array}{llll} 0 & {f}_{i} & {g}_{i} & {h... | Yes |
Proposition 3. The point the shadows of whose pedal triangle are in the ratio \( p : q : r \) is the perspector of the cevian triangle of the point with normal coordinates \( \left( {\frac{1}{p} : \frac{1}{q} : \frac{1}{r}}\right) \) and the tangential triangle of \( {ABC} \) . | Proof. If \( Q \) is the point with normal coordinates \( \left( {\frac{1}{p} : \frac{1}{q} : \frac{1}{r}}\right) \), then \( P \), with coordinates given by (3), is the \( Q \) -Ceva conjugate of the symmedian point \( K = (a : b \) : c). See [3, p.57]. | No |
Proposition 4. \( {P}_{a}{P}_{b}{P}_{c} \) is the anticevian triangle of \( P \) with respect to the tangential triangle of \( {ABC} \) . | Proof. The vertices of the tangential triangle are\n\n\[ \n{A}^{\prime } = \left( {-a : b : c}\right) ,\;{B}^{\prime } = \left( {a : - b : c}\right) ,\;{C}^{\prime } = \left( {a : b : - c}\right) .\n\]\n\nFrom\n\n\[ \n(a\left( {-{ap} + {bq} + {cr}}\right), b\left( {{ap} - {bq} + {cr}}\right), c\left( {{ap} + {bq} - {cr... | Yes |
Proposition 5. There are exactly four points for each of which the midpoints of the sides of the pedal triangle are equidistant from the corresponding sides of \( {ABC} \) . | Proof. The midpoints of the sides of the pedal triangle have signed distances\n\n\[ \nx + \frac{1}{2}\left( {y\cos C + z\cos B}\right) ,\;y + \frac{1}{2}\left( {z\cos A + x\cos C}\right) ,\;z + \frac{1}{2}\left( {x\cos B + y\cos A}\right) \]\n\nfrom the respective sides of \( {ABC} \) . The segments joining the midpoin... | Yes |
Lemma 1. Suppose \( Z \) and \( {Z}^{\prime } \) are a pair of isogonal conjugate points. Let \( O \) and \( {O}^{\prime } \) be the circumcircles of the pedal triangles of \( Z \) and \( {Z}^{\prime } \) . Then \( O = {O}^{\prime } \), and the center of \( O \) is the midpoint between \( Z \) and \( {Z}^{\prime } \). | A proof is given in Johnson [1, pp.155-156]. See Figure 4. | No |
Lemma 2. Suppose that \( {G}_{P} \) is the centroid of the pedal triangle of a point \( P \), and that \( Q \) is the reflection of \( P \) in \( {G}_{P} \). Then there exists a symmetric pair of lines, one passing through \( P \) and the other passing through \( Q \). | Proof. With respect to the principal axes, write \( P = \left( {{x}_{P},{y}_{P}}\right) \) and \( Q = \left( {{x}_{Q},{y}_{Q}}\right) \).\n\nThen \( {G}_{P} = \left( {\frac{1}{3}{\lambda }_{2}{x}_{P},\frac{1}{3}{\lambda }_{1}{y}_{P}}\right) \), and \( \frac{2}{3}{\lambda }_{2}{x}_{P} = {x}_{P} + {x}_{Q} \), so that\n\n... | Yes |
If \( \Gamma \) is a line \( \ell \alpha + {m\beta } + {n\gamma } = 0 \), then \( P \cdot \Gamma \) is the line \( \left( {\ell /p}\right) \alpha + \) \( \left( {m/q}\right) \beta + \left( {n/r}\right) \gamma = 0 \) and \( \Gamma /P \) is the line \( p\ell \alpha + {qm\beta } + {rn\gamma } = 0 \). | Given the line \( {QR} \) of points \( Q \) and \( R \), it is easy to check that \( P \cdot {QR} \) is the line of \( P \cdot Q \) and \( P \cdot R \) . In particular, \( P \cdot \bigtriangleup {ABC} = \bigtriangleup {ABC} \), and if \( T \) is a cevian triangle, then \( P \cdot T \) is a cevian triangle. | No |
We ask for the locus of a point \( P \) for which the circumconic \( P \cdot {\Gamma }_{0} \) is a parabola. | As such a conic is the isogonal transform of a line tangent to \( {\Gamma }_{0} \), we begin with this statement of the problem: find \( P = p : q : r \) (trilinears) for which the line \( L \) given by \( \frac{p}{\alpha } + \frac{q}{\beta } + \frac{r}{\gamma } = 0 \) meets \( {\Gamma }_{0} \), given by \( \frac{a}{\a... | Yes |
Suppose points \( P \) and \( Q \) are given in trilinears: \( P = p : q : r \), and \( U = u : v : w \). We shall find the locus of a point \( X = \alpha : \beta : \gamma \) such that \( P \cdot X \) lies on the line \( {UX} \). | This on-lying is equivalent to the determinant equation\n\n\[ \left| \begin{matrix} u & v & w \\ \alpha & \beta & \gamma \\ {p\alpha } & {q\beta } & {r\gamma } \end{matrix}\right| = 0 \]\n\nexpressible as a circumconic:\n\n\[ \frac{u\left( {q - r}\right) }{\alpha } + \frac{v\left( {r - p}\right) }{\beta } + \frac{w\lef... | Yes |
Again in trilinears, let \( \Gamma \) be the self-isogonal cubic \( Z\left( U\right) \) given in [1, p. 240] by\n\n\[ \n{u\alpha }\left( {{\beta }^{2} - {\gamma }^{2}}\right) + {v\beta }\left( {{\gamma }^{2} - {\alpha }^{2}}\right) + {w\gamma }\left( {{\alpha }^{2} - {\beta }^{2}}\right) = 0.\n\]\n\nThis is the locus o... | The quotient \( \Gamma /P \) is the cubic\n\n\[ \n{up\alpha }\left( {{q}^{2}{\beta }^{2} - {r}^{2}{\gamma }^{2}}\right) + {vq\beta }\left( {{r}^{2}{\gamma }^{2} - {p}^{2}{\alpha }^{2}}\right) + {wr\gamma }\left( {{p}^{2}{\alpha }^{2} - {q}^{2}{\beta }^{2}}\right) = 0.\n\]\n\nAlthough \( \Gamma /P \) is not generally se... | No |
Proposition 1. The pivotal conic \( \mathcal{C} \) is tritangent to the Simson cubic \( \mathcal{E} \) at the tripoles of the Simson lines of the isogonal conjugates of the infinite points of the Morley sides. | Proof. Since \( \mathcal{C} \) is the dual of the nine-point circle, the following statements are equivalent:\n\n(1) \( \mathrm{t}\left( P\right) \) lies on \( \mathcal{C} \cap \mathcal{E} \).\n\n(2) \( \mathbf{s}\left( {P}^{\prime }\right) \) is tangent to the nine-point circle.\n\n(3) \( \mathbf{s}\left( P\right) \) ... | Yes |
Proposition 3. Let \( \ell \) be the tripolar of the point \( \left( {u : v : w}\right) \), intersecting the sidelines of triangle \( {ABC} \) at \( U, V, W \) with coordinates given by (1), and \( F = \) \( \left( {f : g : h}\right) \) a point not on \( \ell \) nor the side lines of the reference triangle. The locus o... | Proof. These intersections are the points\n\n\[ \n{AM} \cap {FU} = \left( {f\left( {{wy} + {vz}}\right) : \left( {{hv} + {gw}}\right) y : \left( {{hv} + {gw}}\right) z}\right) , \n\]\n\n\[ \n\begin{matrix} {BM} \cap {FV} & = & \left( {\left( {{fw} + {hu}}\right) x : g\left( {{uz} + {wx}}\right) : \left( {{fw} + {hu}}\r... | Yes |
Proposition 1. Let \( P \) be a point and let \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be its pedal triangle. Let \( {A}^{\prime \prime } \) be the point such that \( {B}^{\prime }P{C}^{\prime }{A}^{\prime \prime } \) is a parallelogram. In the same way construct \( {B}^{\prime \prime } \) and \( {C}^{\prime \prim... | Proof. This is equivalent to the construction of the isogonal conjugate of \( P \) as the point of concurrency of the perpendiculars through the vertices of \( {ABC} \) to the corresponding sides of the pedal triangle. Here we justify it directly by noting that \( {AP} \) and \( A{A}^{\prime \prime } \) are isogonal li... | Yes |
Proposition 2. A triangle \( {A}_{1}{B}_{1}{C}_{1} \) perspective with \( {ABC} \) induces a reciprocal conjugation : for every point \( M \) not on the side lines of \( {ABC} \) and \( {A}_{1}{B}_{1}{C}_{1} \), construct\n\n\[ \n{A}^{\prime } = {A}_{1}M \cap {BC},\;{B}^{\prime } = {B}_{1}M \cap {CA},\;{C}^{\prime } = ... | Proof. Since \( {A}_{1}{B}_{1}{C}_{1} \) is perspective with \( {ABC} \), we may write the coordinates of its vertices in the form\n\n\[ \n{A}_{1} = \left( {U : v : w}\right) ,\;{B}_{1} = \left( {u : V : w}\right) ,\;{C}_{1} = \left( {u : v : W}\right) . \n\]\n\n(1)\n\nThe perspector is \( P = \left( {u : v : w}\right)... | Yes |
Theorem 3. Let \( P, Q, R \) be collinear points. Denote by \( X, Y, Z \) the traces of \( R \) on the side lines \( {BC},{CA},{AB} \) of triangle \( {ABC} \), and construct triangle \( {A}_{1}{B}_{1}{C}_{1} \) with vertices\n\n\[ \n{A}_{1} = {PA} \cap {QX},\;{B}_{1} = {PB} \cap {QY},\;{C}_{1} = {PC} \cap {QZ}. \n\]\n\... | Proof. If \( P = \left( {u : v : w}\right), Q = \left( {U : V : W}\right) \), and \( R = \left( {1 - t}\right) P + {tQ} \) for some \( t \neq 0,1 \), then\n\n\[ \n{A}_{1} = \left( {\frac{-{tU}}{1 - t} : v : w}\right) ,\;{B}_{1} = \left( {u : \frac{-{tV}}{1 - t} : w}\right) ,\;{C}_{1} = \left( {u : v : \frac{-{tW}}{1 - ... | Yes |
Lemma 1. In Figure 1, circle \( O\left( R\right) \) is tangent externally to each of circles \( {O}_{1}\left( {r}_{1}\right) \) and \( {O}_{2}\left( {r}_{2}\right) \), at \( A \) and \( B \) respectively. If \( {A}_{1}{B}_{1} \) is a segment of an external common tangent to the circles \( \left( {O}_{1}\right) \) and \... | Proof. In the isosceles triangle \( {AOB},\cos {AOB} = \frac{2{R}^{2} - A{B}^{2}}{2{R}^{2}} = 1 - \frac{A{B}^{2}}{2{R}^{2}} \) . Applying the law of cosines to triangle \( {\mathrm{O}}_{1}{\mathrm{{OO}}}_{2} \), we have\n\n---\n\nPublication Date: September 4, 2001. Communicating Editor: Paul Yiu.\n\n\( {}^{1} \) The c... | Yes |
Lemma 2. If \( {A}_{1} \) and \( {B}_{1} \) are the feet of the internal bisectors of angles \( A \) and \( B \) , then\n\n\[ \n{A}_{1}{B}_{1} = \frac{{abc}\sqrt{R\left( {R + 2{r}_{3}}\right) }}{\left( {c + a}\right) \left( {b + c}\right) R}.\n\] | Proof. In Figure 2, let \( K \) and \( L \) be points on \( {I}_{3}{A}_{2} \) and \( {I}_{3}{B}_{2} \) such that \( {OK}//{CB} \) , and \( {OL}//{CA} \) . Since \( C{A}_{2} = C{B}_{2} = \frac{a + b + c}{2} \),\n\n\[ \n{OL} = \frac{a + b + c}{2} - \frac{b}{2} = \frac{c + a}{2},\;{OK} = \frac{a + b + c}{2} - \frac{a}{2} ... | Yes |
Proposition 2 (Distance formula). The square distance between two points with absolute barycentric coordinates \( P = {x}_{1}A + {y}_{1}B + {z}_{1}C \) and \( Q = {x}_{2}A + {y}_{2}B + {z}_{2}C \) is given by\n\n\[ \n{\left| PQ\right| }^{2} = {S}_{A}{\left( {x}_{1} - {x}_{2}\right) }^{2} + {S}_{B}{\left( {y}_{1} - {y}_... | (1) | No |
Proposition 3 (Conway). Let \( P \) be a point such that the directed angles \( {PBC} \) and \( {PCB} \) are respectively \( \phi \) and \( \psi \) . The homogeneous barycentric coordinates of \( P \) are \[ \left( {-{a}^{2} : {S}_{C} + {S}_{\psi } : {S}_{B} + {S}_{\phi }}\right) . \] | Since the cotangent function has period \( \pi \), we may always choose \( \phi \) and \( \psi \) in the range \( - \frac{\pi }{2} < \phi ,\psi \leq \frac{\pi }{2} \) . See Figure 1. | No |
Proposition 5. If \( \phi \neq \frac{\pi }{2}, \pm \frac{\pi }{6}, O = {K}_{\phi }\left( {-\left( {\frac{\pi }{2} - \phi }\right) }\right) \) . | Proof. Let \( \psi = - \left( {\frac{\pi }{2} - \phi }\right) \), so that \( {S}_{\psi } = - \frac{{S}^{2}}{{S}_{\phi }} \) . Note that\n\n\[ \n{A}^{\phi ,\psi } = \left( {-{a}^{2} : \frac{2{S}^{2}{S}_{\phi }}{{S}^{2} + {S}_{\phi }^{2}} + {S}_{C} : \frac{2{S}^{2}{S}_{\phi }}{{S}^{2} + {S}_{\phi }^{2}} + {S}_{B}}\right)... | Yes |
Theorem 1 (Bankoff [2]). Let \( P \) and \( Q \) be the midpoints of the semicircles \( \left( {BC}\right) \) and \( \left( {AC}\right) \) respectively. If the incircle of the arbelos is tangent to the semicircles \( \left( {BC}\right) ,\left( {AC}\right) \), and \( \left( {AB}\right) \) at \( X, Y, Z \) respectively, ... | Proof. Let \( D \) be the intersection of the semicircle \( \left( {AB}\right) \) with the line perpendicular to \( {AB} \) at \( C \) . See Figure 3. Note that \( {AB} \cdot {AC} = A{D}^{2} \) by Euclid’s proof of the Pythagorean theorem. \( {}^{2} \) Consider the inversion with respect to the circle \( A\left( D\righ... | Yes |
Corollary 2. The lines \( {AX},{BY} \), and \( {CZ} \) intersect at a point \( S \) on the incircle \( {XYZ} \) of the arbelos. | Proof. We have already proved that \( A, X, P \) are collinear, as are \( B, Y, Q \) . In Figure 4, let \( S \) be the intersection of the line \( {AP} \) with the circle \( {XYZ} \) . The inversive image \( {S}^{\prime } \) (in the circle \( A\left( D\right) \) ) is the intersection of the same line with the circle \(... | Yes |
Corollary 3. Let \( M \) be the midpoint of the semicircle \( \left( {AB}\right) \) on the opposite side of the arbelos.\n\n(i) The points \( A, B, X, Y \) lie on a circle, center \( M \).\n\n(ii) The line \( {CZ} \) passes through \( M \). | Proof. Consider Figure 5 which is a modification of Figure 3. Since \( C, P,{Y}^{\prime } \) are on a line making a \( {45}^{ \circ } \) angle with \( {AB} \), its inversive image (in the circle \( A\left( D\right) \) ) is a circle through \( A, B, X, Y \), also making a \( {45}^{ \circ } \) angle with \( {AB} \) . The... | Yes |
Theorem 1. Intersecting circles \( \left( O\right) \) and \( \left( {O}_{1}\right) \) are the circumcircle and an excircle of a triangle if and only if the tangent to \( \left( {O}_{1}\right) \) at an intersection of the circles meets \( \left( O\right) \) again at the touch point of a common tangent. | Proof. (Sufficiency) Let \( O\left( R\right) \) and \( {O}_{1}\left( {R}_{1}\right) \) be intersecting circles. (These circles are not assumed to be related to a triangle as in Figure 1.) Of the two lines tangent to both circles, let \( {AK} \) be one of them, as in Figure 2. Let \( P = {AK} \cap O{O}_{1} \) . Of the t... | Yes |
Theorem 2. For each of the rotating triangles \( {ABC} \) with fixed circumcircle and excircle corresponding to vertex \( A \), the feet of bisectors \( B{B}_{1} \) and \( C{C}_{1} \) traverse line \( {DE} \), where \( E \) is the touch point of the second common tangent. | ## 3. Proof of Theorem 2\n\nWe begin with the pole-polar correspondence between points and lines for the excircle with center \( {I}_{1} \), as in Figure 4.\n\nThe polars of \( A, B, C \) are \( {LM},{MK},{KL} \), respectively, where \( {\Delta KLM} \) is the \( A \) -extouch triangle. As \( B{B}_{1} \) is the internal... | Yes |
Lemma 3. The orthocenter \( {H}_{1} \) of triangle \( {KLM} \) stays fixed as triangle \( {ABC} \) rotates. | Proof. Let \( {KLM} \) be the extouch triangle of triangle \( {ABC} \), let \( {RST} \) be the orthic triangle of triangle \( {KLM} \), and let \( {H}_{1} \) and \( {E}_{1} \) be the orthocenter and nine-point center, respectively, of triangle \( {KLM} \), as in Figure 5.\n\n(1) The circumcircle of triangle \( {RST} \)... | Yes |
Theorem 1. Suppose a point \( X = {x}_{1} : {x}_{2} : {x}_{3} \) is given parametrically by\n\n\[ \n{x}_{1} = {d}_{1}{t}^{2} + {e}_{1}t + {f}_{1} \]\n\n(4)\n\n\[ \n{x}_{2} = {d}_{2}{t}^{2} + {e}_{2}t + {f}_{2} \]\n\n(5)\n\n\[ \n{x}_{3} = {d}_{3}{t}^{2} + {e}_{3}t + {f}_{3} \]\n\n(6)\n\nwhere the matrix\n\n\[ \nM = \lef... | Proof. Since \( M \) is nonsingular, its determinant \( \delta \) is nonzero, and \( {M}^{-1} = \frac{1}{\delta }{M}^{\# } \) . Let\n\n\[ \nX = \left( \begin{array}{l} {x}_{1} \\ {x}_{2} \\ {x}_{3} \end{array}\right) \text{ and }T = \left( \begin{matrix} {t}^{2} \\ t \\ 1 \end{matrix}\right) \]\n\nso that \( X = {MT} \... | Yes |
Theorem 2. Suppose \( P, U, V,{Q}_{t} \) are points as in (1)-(3); that is, \( {Q}_{t} \) traverses line \( {UV} \) . The locus of \( P(Q{Q}_{t} \) is the conic\n\n\[ \frac{{\alpha }^{2}}{{p}_{1}{q}_{1}} + \frac{{\beta }^{2}}{{p}_{2}{q}_{2}} + \frac{{\gamma }^{2}}{{p}_{3}{q}_{3}} - \left( {\frac{1}{{p}_{2}{q}_{3}} + \f... | Proof. First, it is easy to verify that equation (8) holds for \( \alpha : \beta : \gamma \) equal to any of these six vertices:\n\n\[ 0 : {p}_{2} : {p}_{3},{p}_{1} : 0 : {p}_{3},{p}_{1} : {p}_{2} : 0,\;0 : {q}_{2} : {q}_{3},{q}_{1} : 0 : {q}_{3},{q}_{1} : {q}_{2} : 0 \]\n\n(9)\n\n--- | Yes |
Corollary 2.1. Suppose \( P = {p}_{1} : {p}_{2} : {p}_{3} \) is a point and \( L \) given by \( {\ell }_{1}\alpha + {\ell }_{2}\beta + \) \( {\ell }_{3}\gamma = 0 \) is a line. Suppose the point \( {Q}_{t} \) traverses \( L \) . The locus of \( P(Q{Q}_{t} \) is the conic | Proof. Let \( U, V \) be distinct points on \( L \), and apply Theorem 2. | No |
Corollary 2.2. The conic (14) is inscribed to \( \\bigtriangleup {ABC} \) if and only if the line \( L = {UV} \) is the trilinear pole of \( P \) . | Proof. In this case, \( {\\ell }_{1} : {\\ell }_{2} : {\\ell }_{3} = 1/{p}_{1} : 1/{p}_{2} : 1/{p}_{3} \), so that \( P = Q \) . The cevian triangles indicated by (9) are now identical, and the six pass-through points are three tangency points. | No |
Let \( P = \) centroid and \( Q = \) orthocenter. Then line \( {UV} \) is given by | \[ \left( {\cos A}\right) \alpha + \left( {\cos B}\right) \beta + \left( {\cos C}\right) \gamma = 0, \] | Yes |
Theorem 3. The locus of the \( {P}_{t} \otimes Q \) is the circumconic\n\n\[ \n\left( {\frac{{p}_{3}}{{q}_{2}} + \frac{{p}_{2}}{{q}_{3}}}\right) {\beta \gamma } + \left( {\frac{{p}_{1}}{{q}_{3}} + \frac{{p}_{3}}{{q}_{1}}}\right) {\gamma \alpha } + \left( {\frac{{p}_{2}}{{q}_{1}} + \frac{{p}_{1}}{{q}_{2}}}\right) {\alph... | Proof. Following the proof of Theorem 1, let\n\n\[ \n{u}_{1}^{\prime } = - \frac{{u}_{1}}{{q}_{1}} + \frac{{u}_{2}}{{q}_{2}} + \frac{{u}_{3}}{{q}_{3}},{v}_{1}^{\prime } = - \frac{{v}_{1}}{{q}_{1}} + \frac{{v}_{2}}{{q}_{2}} + \frac{{v}_{3}}{{q}_{3}}, \n\]\n\nand similarly for \( {u}_{2}^{\prime },{u}_{3}^{\prime },{v}_{... | Yes |
Example 3. Regarding the conic (15), suppose \( P = {p}_{1} : {p}_{2} : {p}_{3} \) is an arbitrary triangle center and \( \Gamma \) is an arbitrary circumconic \( \ell /\alpha + m/\beta + n/\gamma = 0 \) . Let\n\n\[ Q = {q}_{1} : {q}_{2} : {q}_{3} \]\n\n\[ = \frac{1}{{p}_{1}\left( {-{p}_{1}\ell + {p}_{2}m + {p}_{3}n}\r... | For \( {P}_{t} \) ranging through the line \( L \) given by \( {p}_{1}\alpha + {p}_{2}\beta + {p}_{3}\gamma = 0 \), the locus of \( {P}_{t} \otimes Q \) is then \( \Gamma \), since\n\n\[ \frac{{p}_{3}}{{q}_{2}} + \frac{{p}_{2}}{{q}_{3}} : \frac{{p}_{1}}{{q}_{3}} + \frac{{p}_{3}}{{q}_{1}} : \frac{{p}_{2}}{{q}_{1}} + \fr... | Yes |
Example 4. Let \( \Gamma \) denote the circumconic \( p/\alpha + q/\beta + r/\gamma = 0 \), that is, the circumconic having as pivot the point \( P = p : q : r \) . The \( \Gamma \) -based polar of \( X \) is the trilinear polar of the cevapoint of \( P \) and \( X \), given by | \[ \left( {{qz} + {ry}}\right) \alpha + \left( {{rx} + {pz}}\right) \beta + \left( {{py} + {qx}}\right) \gamma = 0. \] | Yes |
Let \( \Gamma \) denote conic determined as in Theorem 2 by points \( P \) and \( Q \) . The conic is inscribed in \( \bigtriangleup {ABC} \) if and only if \( P = Q \), and in this case, the \( \Gamma \) -based polar of \( X \) is given by | \[ \frac{1}{p}\left( {-\frac{x}{p} + \frac{y}{q} + \frac{z}{r}}\right) \alpha + \frac{1}{q}\left( {\frac{x}{p} - \frac{y}{q} + \frac{z}{r}}\right) \beta + \frac{1}{r}\left( {\frac{x}{p} + \frac{y}{q} - \frac{z}{r}}\right) \gamma = 0. \] | Yes |
For \( P = u : v : w \), let \( \Gamma \left( P\right) \) be the circumconic \( {u\beta \gamma } + {v\gamma \alpha } + {w\alpha \beta } = 0 \). Assume that at least one point of \( \Gamma \left( P\right) \) lies inside \( \bigtriangleup {ABC} \); in other words, assume that \( \Gamma \left( P\right) \) is not an ellips... | For each \( \alpha : \beta : \gamma \) on the line \( {u\alpha } + {v\beta } + {w\gamma } = 0 \) and inside or on a side of \( \bigtriangleup {ABC} \), let \( P = p : q : r \), with \( p \geq 0, q \geq 0, r \geq 0 \), satisfy\n\n\[ \n \alpha = {p}^{2},\;\beta = {q}^{2},\;\gamma = {r}^{2}, \n\]\n\nand define\n\n\[ \n... | Yes |
For \( P = \left( {f : g : h}\right) \) and \( \ell = \left\lbrack {l : m : n}\right\rbrack \), the \( P\ell \) -perpendicularity is given by\n\n\[{\tau }_{P\ell } : \left( {{f}_{L} : {g}_{L} : {h}_{L}}\right) \mapsto \left( {\frac{f\left( {g{h}_{L} - h{g}_{L}}\right) }{l} : \frac{g\left( {h{f}_{L} - f{h}_{L}}\right) }... | Proof. Let \( L \) be a line passing through \( C \) with \( {\left\lbrack L\right\rbrack }^{\ell } = \left( {{f}_{L} : {g}_{L} : {h}_{L}}\right) \), and let \( {B}_{L} = L \cap {AB} = \left( {{f}_{L} : {g}_{L} : 0}\right) \) . We will consider triangle \( A{B}_{L}C \) . We have noted above that the \( P\ell \) -altitu... | Yes |
Theorem 2. The conic \( {\mathcal{O}}_{P\ell } \) :\n\n\[ f\left( {{gm} + {hn}}\right) {yz} + g\left( {{fl} + {hn}}\right) {xz} + h\left( {{fl} + {gm}}\right) {xy} = 0 \] \n\n(2)\n\nis the \( P\ell \) -circumcircle. | Proof. Clearly \( A, B \) and \( C \) are on the conic given by the equation. Let \( J = \left( {f}_{1}\right. \) : \( \left. {{g}_{1} : {h}_{1}}\right) \), then with (1) the condition that \( J \) is a fixed point of \( {\tau }_{P\ell } \) gives\n\n\[ \left( {\frac{{fg}{h}_{1} - f{g}_{1}h}{l} : \frac{{f}_{1}{gh} - {fg... | No |
Theorem 4. For \( I \in \{ A, B, C\} \), consider lines \( {l}_{I} \) and \( {l}_{I}^{\prime } \) unequal to sidelines of \( {ABC} \) that are images under \( {\tau }_{PI} \). Let \( {A}_{1} = {l}_{B} \cap {l}_{C}^{\prime },{B}_{1} = {\dot{l}}_{C} \cap {l}_{A}^{\prime } \) and \( {C}_{1} = {l}_{A} \cap {l}_{B}^{\prime ... | Proof. For \( I \in \{ A, B, C\} \), let \( {P}_{I} = \left( {{x}_{I} : {y}_{I} : {z}_{I}}\right) \in {l}_{I} \) be a point different from \( I \). We find, for instance, \( {l}_{A} = \left\lbrack {0 : {z}_{A} : - {y}_{A}}\right\rbrack \) and \( {l}_{C}^{\prime } = \left\lbrack {\widetilde{g}/{y}_{C} : - \widetilde{f}/... | No |
Theorem 1. Suppose \( {P}_{1} \) and \( {P}_{2} \) are distinct points, collinear with but not equal to \( {X}_{31} \) . Then \( U\left( {P}_{2}\right) = U\left( {P}_{1}\right) \) . | Proof. Write \( {P}_{1} = {p}_{1} : {q}_{1} : {r}_{1} \) and \( {P}_{2} = {p}_{2} : {q}_{2} : {r}_{2} \) . Then for some \( s = \) \( s\left( {a, b, c}\right) \neq 0, \)\n\n\[ \n{a}^{2} = s{p}_{1} + {p}_{2},\;{b}^{2} = s{q}_{1} + {q}_{2},\;{c}^{2} = s{r}_{1} + {r}_{2}, \n\] \n\nso that for \( f\left( {a, b, c}\right) \... | Yes |
Theorem 2. The generalized Brocard points defined by\n\n\[ \n\\frac{qc}{b} : \\frac{ra}{c} : \\frac{pb}{a}\\;\\text{ and }\\;\\frac{rb}{c} : \\frac{pc}{a} : \\frac{qa}{b} \n\]\n\nlie on \( \\Gamma \\left( P\\right) \) . | Proof. Writing ordered triples for the two points, we have\n\n\[ \n\\left( {a\\left( {r{b}^{2} - q{c}^{2}}\\right), b\\left( {p{c}^{2} - r{a}^{2}}\\right), c\\left( {q{a}^{2} - p{b}^{2}}\\right) }\\right) \n\]\n\n\[ \n= {abc}\\left( {\\frac{qc}{b},\\frac{ra}{c},\\frac{pb}{a}}\\right) + {abc}\\left( {\\frac{rb}{c},\\fra... | Yes |
Proposition 1. The lines \( {A}_{3}{A}_{4},{B}_{3}{B}_{4} \) and \( {C}_{3}{C}_{4} \) are concurrent. | Proof. Let \( {B}_{a}{C}_{a} = {C}_{b}{A}_{b} = {A}_{c}{B}_{c} = {2t} \), where \( t \) is given positive sign when \( {C}_{a}{B}_{a} \) and \( {BC} \) have equal directions, and positive sign when these directions are opposite. Note that \( {K}_{A}{K}_{B}{K}_{C} \) is homothetic to \( {ABC} \) and that \( {K}^{ * }\le... | Yes |
Suppose\n\n\[ \nP = p : q : r,\;U = u : v : w,\;{U}^{\prime } = {u}^{\prime } : {v}^{\prime } : {w}^{\prime } \]\n\nare points, none lying on a sideline of \( \bigtriangleup {ABC} \), and \( {U}^{\prime } \) is not on a sideline of the cevian triangle of \( P \) (whose vertices are the rows of matrix \( {\mathbb{P}}^{\... | We have\n\n\[ \n{\mathbb{P}}^{\prime } = \left( \begin{matrix} 0 & q & r \\ p & 0 & r \\ p & q & 0 \end{matrix}\right) ,\;\text{ and }\;{\left( {\mathbb{P}}^{\prime }\right) }^{-1} = \frac{1}{\left| {\mathbb{P}}^{\prime }\right| }\left( \begin{matrix} - p & q & r \\ p & - q & r \\ p & q & - r \end{matrix}\right) ,\n\]\... | Yes |
For any point \( X = x : y : z \) not on a sideline of \( \bigtriangleup {ABC} \), the isogonal conjugate of \( X \) is given by\n\n\[ F\left( X\right) = \frac{1}{x} : \frac{1}{y} : \frac{1}{z} \] | Suppose \( P, U,\varphi \) are as in Example 1. The involution \( m \) given by \( m\left( X\right) = \varphi \left( {F\left( {{\varphi }^{-1}\left( X\right) }\right) }\right) \) will be formulated: equation (3) implies\n\n\[ F\left( {{\varphi }^{-1}\left( X\right) }\right) = \frac{d}{-\frac{x}{p} + \frac{y}{q} + \frac... | Yes |
Continuing with isogonal conjugacy for \( F \) and with \( \varphi \) as in Example 3 (with \( U = 1 : 1 : 1 \) and \( {U}^{\prime } = p : q : r \) ), here we use \( {\varphi }^{-1} \) in place of \( \varphi \), so that \( m\left( X\right) = {\varphi }^{-1}(F\left( {\varphi \left( X\right) }\right) \) . The result is (... | Let \[ n\left( X\right) = \frac{1}{y + z} : \frac{1}{z + x} : \frac{1}{x + y}. \] Then \( X = n\left( X\right) \) -aleph conjugate of \( X \) . Another easily checked property is that a necessary and sufficient condition that \[ X = X\text{-aleph conjugate of the incenter} \] is that \( X = \) incenter or else \( X \) ... | No |
Example 5. Here, \( F \) is reflection about the circumcenter:\n\n\[ F\left( {x : y : z}\right) = {2R}\cos A - {hx} : {2R}\cos B - {hy} : {2R}\cos C - {hz}, \]\n\nwhere \( R = \) circumradius, and \( h{\text{normalizes}}^{2}X \) . | Keeping \( \varphi \) as in Example 4, we find\n\n\[ {m}_{1}\left( {x, y, z}\right) = {2abc}\left( {\cos B + \cos C}\right) \left( {\frac{x\left( {b + c - a}\right) }{p} + \frac{y\left( {c + a - b}\right) }{q} + \frac{z\left( {a + b - c}\right) }{r}}\right) - {16}{\sigma }^{2}x, \]\n\nwhich, via (4), defines the \( P \... | No |
Continuing Example 5 with \( {\varphi }^{-1} \) in place of \( \varphi \) leads to the \( P \) -gimel conjugate of \( X \), defined via (4) by\n\n\[ \n{m}_{1}\left( {x, y, z}\right) = {2abc}\left( {-\frac{\cos A}{p} + \frac{\cos B}{q} + \frac{\cos C}{r}}\right) S - 8{\sigma }^{2}x, \n\]\n\nwhere \( S = x\left( {{bq} + ... | It is easy to check that if \( P \in {\mathcal{L}}^{\infty } \), then \( m\left( {X}_{1}\right) = {X}_{1} \) | No |
For distinct points \( {X}^{\prime } = {x}^{\prime } : {y}^{\prime } : {z}^{\prime } \) and \( X = x : y : z \), neither lying on a sideline of \( \bigtriangleup {ABC} \), the \( {X}^{\prime } \) -Hirst inverse of \( X \) is defined [4, Glossary] by | \[ {y}^{\prime }{z}^{\prime }{x}^{2} - {x}^{\prime 2}{yz} : {z}^{\prime }{x}^{\prime }{y}^{2} - {y}^{\prime 2}{zx} : {x}^{\prime }{y}^{\prime }{z}^{2} - {z}^{\prime 2}{xy}. \] | Yes |
Continuing Example 7, we use \( {\varphi }^{-1} \) in place of \( \varphi \) and define the resulting image \( m\left( X\right) \) as the \( P \) -he conjugate of \( X.{}^{3} \) We have \( m \) as in (4) with | \[ {m}_{1}\left( {x, y, z}\right) = - p{\left( y + z\right) }^{2} + q{\left( z + x\right) }^{2} + r{\left( x + y\right) }^{2} \] \[ + \frac{qr}{p}\left( {x + y}\right) \left( {x + z}\right) - \frac{rp}{q}\left( {y + z}\right) \left( {y + x}\right) - \frac{pq}{r}\left( {z + x}\right) \left( {z + y}\right) . \] | Yes |
The complement of a point \( X \) not on \( {\mathcal{L}}^{\infty } \) is the point \( {X}^{\prime } \) satisfying the vector equation\n\n\[ \overrightarrow{{X}^{\prime }{X}_{2}} = \frac{1}{2}\overrightarrow{{X}_{2}X} \]\n\nIf \( X = x : y : z \), then\n\n\[ {X}^{\prime } = \frac{{by} + {cz}}{a} : \frac{{cz} + {ax}}{b}... | If \( X \in {\mathcal{L}}^{\infty } \), then (6) defines the complement of \( X \) . The mapping \( \varphi \left( X\right) = {X}^{\prime } \) is a collineation. | No |
The anticomplement of a point \( X \) is the point \( {X}^{\prime \prime } \) given by\n\n\[ \n{X}^{\prime \prime } = \frac{-{ax} + {by} + {cz}}{a} : \frac{{ax} - {by} + {cz}}{b} : \frac{{ax} + {by} - {cz}}{c}.\n\] | Keeping \( F \) and \( \varphi \) as in Example 11, we have \( {\varphi }^{-1}\left( X\right) = {X}^{\prime \prime } \) and define \( m \) by \( m = {\varphi }^{-1} \circ F \circ \varphi \), Thus, \( m\left( X\right) \) is determined as in (4) from\n\n\[ \n{m}_{1}\left( {x, y, z}\right) = \frac{1}{a}\left( {\frac{{b}^{... | No |
Let\n\n\\[ \n\\varphi \\left( {\\alpha : \\beta : \\gamma }\\right) = p\\left( {\\beta + \\gamma }\\right) : q\\left( {\\gamma + \\alpha }\\right) : r\\left( {\\alpha + \\beta }\\right) ,\n\\]\n\nso that\n\n\\[ \n{\\varphi }^{-1}\\left( {\\alpha : \\beta : \\gamma }\\right) = - \\frac{\\alpha }{p} + \\frac{\\beta }{q} ... | In accord with (13), the cubic \\( \\varphi \\left( {Z\\left( U\\right) }\\right) \\) has equation\n\n\\[ \n\\frac{u\\alpha }{p}\\left( {-\\frac{\\alpha }{p} + \\frac{\\beta }{q} + \\frac{\\gamma }{r}}\\right) \\left( {\\frac{\\beta }{q} - \\frac{\\gamma }{r}}\\right) + \\frac{v\\beta }{q}\\left( {\\frac{\\alpha }{p} -... | Yes |
Continuing Example 13 with \( {\varphi }^{-1} \) in place of \( \varphi \), the cubic \( {\varphi }^{-1}\left( {Z\left( U\right) }\right) \) is given by\n\n\[ \ns\left( {u, v, w, p, q, r,\alpha ,\beta ,\gamma }\right) + s\left( {v, w, u, q, r, p,\beta ,\gamma ,\alpha }\right) + s\left( {w, u, v, r, p, q,\gamma ,\alpha ... | Collinear triples \( X, U,{X}^{-1} \) on \( Z\left( U\right) \) yield collinear triples on \( {\varphi }^{-1}\left( {Z\left( U\right) }\right) \), so that \( {\varphi }^{-1}\left( U\right) \) is a pivot for \( {\varphi }^{-1}\left( {Z\left( U\right) }\right) \). The point \( {\varphi }^{-1}\left( {X}^{-1}\right) \) is ... | No |
Proposition 1. The sides of the chordal triangle \( {T}^{\prime } \) satisfy\n\n\[ \left( {{a}^{\prime }, a}\right) + \left( {{b}^{\prime }, b}\right) + \left( {{c}^{\prime }, c}\right) = 0\;\left( {\;\operatorname{mod}\;\pi }\right) . \] | Proof. Note that \( \left( {{a}^{\prime }, c}\right) = \left( {{B}_{c}{C}_{b},{B}_{c}A}\right) \) and\n\n\[ \left( {{c}^{\prime }, b}\right) = - \left( {{A}_{b}A,{A}_{b}{B}_{a}}\right) = \left( {{B}_{c}{B}_{a},{B}_{c}{C}_{b}}\right) \;\left( {\;\operatorname{mod}\;\pi }\right) \]\n\nwhile also\n\n\[ \left( {{b}^{\prime... | Yes |
Proposition 2. The triangle \( {T}^{\prime } \) is perspective to \( {ABC} \) . | Proof. From Pascal’s hexagon theorem applied to \( {C}_{a}{B}_{a}{A}_{b}{C}_{b}{B}_{c}{A}_{c} \) we see that the points of intersection \( {C}_{a}{B}_{a} \cap {C}_{b}{B}_{c},{B}_{c}{A}_{c} \cap {B}_{a}{A}_{b} \) and \( {A}_{b}{C}_{b} \cap {A}_{c}{C}_{a} \) are collinear. Therefore, triangles \( {ABC} \) and \( {A}^{\pr... | Yes |
Proposition 3. The corresponding sides of \( {T}^{\prime } \) and \( {T}^{\prime \prime } \) are antiparallel with respect to triangle \( T \) . | Proof. From the fact that \( {B}_{c}{A}_{c}{A}_{b}{C}_{b} \) is a cyclic quadrilateral, immediately we see \( \angle A{B}_{c}{C}_{b} = \angle A{A}_{b}{A}_{c} \), so that \( {a}^{\prime } \) and \( {a}^{\prime \prime } \) are antiparallel. By symmetry this proves the proposition. | Yes |
Theorem 5. Let \( \mathcal{F} \) be the family of circumcircles of equilateral triangles bounded by cevians whose locus of centers is the Napoleon circle \( \mathcal{N} \) passing through the Fermat point \( F \) . The envelope of this family \( \mathcal{F} \) is the pedal with respect to \( F \) of the circle \( {\mat... | Let \( i \) be the inversion with respect to a circle \( \mathcal{C} \) whose center is \( F \) and such that \( {\mathcal{C}}_{0} \) is invariant under it. The curve \( i\left( {\mathcal{P}}_{F}\right) \) is the image of \( {\mathcal{C}}_{0} \) by the reciprocal polar transformation with respect to \( \mathcal{C} \), ... | Yes |
Theorem 1. The inner and outer Fermat, isodynamic, and Napoleon points lie on a conic section. | Proof. Let \( O \) denote the circumcenter of a triangle, \( H \) its orthocenter, and \( G \) its centroid. Denote the inner Fermat point by \( {F}_{ + } \), the inner isodynamic point by \( {J}_{ + } \) , and the inner Napoleon point by \( {N}_{ + } \) . Similarly denote the outer Fermat, isodynamic, and Napoleon poi... | Yes |
Theorem 2. The inner and outer Fermat (isogonic) and Napoleon points along with the isogonal conjugates of the Napoleon points all lie on a conic consisting of two lines intersecting at the center of the nine-point circle. | Proof. Denote the isogonal conjugates of the inner and outer Napoleon points by \( {N}_{ + }^{ * } \) and \( {N}_{ - }^{ * } \) respectively. Consider the hexagon with vertices \( {F}_{ + },{F}_{ - },{N}_{ + },{N}_{ - } \) , \( {N}_{ + }^{ * } \), and \( {N}_{ - }^{ * } \) . Kimberling lists these collinearities: \( \m... | Yes |
Theorem 2. Let \( H \) be the orthocenter, \( \mathcal{C} \) the circumcircle of a triangle \( \mathcal{T} = \) \( {A}_{1}{A}_{2}{A}_{3} \)\n\n(i) For any point \( P \), let \( {P}_{i} \) denote the reflection of \( P \) across the side \( {A}_{j}{A}_{h} \) of \( \mathcal{T} \) . (Here, \( i, j, h \) is a permutation o... | All these statements are easy consequences of well-known properties of the Simson line, which is obviously parallel to \( r\left( P\right) \) . See, for example,[1, Theorems 2.5.1, 2.7.1,2]. This theorem defines a bijective mapping \( P \mapsto r\left( P\right) \) . Thus, given any line \( e \) through \( H \), there e... | Yes |
Theorem 4. Let \( G, O, E \) be three distinct points. Then there exists a triangle \( \mathcal{T} \) whose centroid, circumcenter and Euler point are \( G, O, E \), respectively, if and only if \( E \) lies outside the GO-cardioid. In this case the triangle \( \mathcal{T} \) is unique and can be constructed by paper-f... | Proof. Let us first look at isosceles (nonequilateral) triangles, which can be treated within ruler-and-compass geometry. \( {}^{9} \) Here, by symmetry, the Euler point \( E \) lies on the Euler line; indeed, by definition, it must be one of the vertices, say \( {A}_{3} = \) \( E = \left( {e,0}\right) \) . Then being ... | No |
Theorem 2. The locus of the orthology center \( \left\lbrack {\tau ,{\sigma }_{G}^{\lambda }}\right\rbrack \) of \( \tau \) and \( {\sigma }_{G}^{\lambda } \) is the Kiepert hyperbola of \( {ABC} \) . | Proof. The perpendicular from \( A \) onto the line \( {G}^{B}\left( \lambda \right) {G}^{C}\left( \lambda \right) \) has equation\n\n\[ b\left( {{T\lambda } + 3{k}_{b}}\right) y - c\left( {{T\lambda } + 3{k}_{c}}\right) z = 0. \]\n\nIt follows that \( \left\lbrack {\tau ,{\sigma }_{G}^{\lambda }}\right\rbrack \) is \(... | Yes |
Theorem 3. For every \( \lambda \in \mathbb{R} \), the triangles \( \tau \) and \( {\sigma }_{O}^{\lambda } \) are homothetic, with center of homothety at the symmedian point \( K \) . Hence, they are homologic with homology center \( K \) and their homology axis is the line at infinity. | Proof. The point \( \frac{T + {z}_{2a}\lambda }{-{abc}} : \frac{\lambda }{c} : \frac{\lambda }{b} \) is the circumcenter \( {O}^{A}\left( \lambda \right) \) of the flank \( A{R}_{4}{R}_{5} \) . Since the determinant \[ \left| \begin{matrix} 1 & 0 & 0 \\ a & b & c \\ \frac{T + {z}_{2a}\lambda }{-{abc}} & \frac{\lambda }... | Yes |
For every \( \lambda \in \mathbb{R} \), the triangles \( \tau \) and \( {\sigma }_{O}^{\lambda } \) are orthologic. The orthology center \( \left\lbrack {\tau ,{\sigma }_{O}^{\lambda }}\right\rbrack \) is the orthocenter \( H \) while the orthology center \( \left\lbrack {{\sigma }_{O}^{\lambda },\tau }\right\rbrack \)... | Proof. Since the triangles \( \tau \) and \( {\sigma }_{O}^{\lambda } \) are homothetic and their center of similitude is the symmedian point \( K \), it follows that \( \tau \) and \( {\sigma }_{O}^{\lambda } \) are orthologic and that \( \left\lbrack {\tau ,{\sigma }_{O}^{\lambda }}\right\rbrack = H \) . On the other... | Yes |
Theorem 5. The homology axis of \( \tau \) and \( {\sigma }_{H}^{\lambda } \) envelopes the parabola with directrix the line \( {HK} \) and focus the central point \( {X}_{112} \) . | Proof. The orthocenter \( {H}^{A}\left( \lambda \right) \) of the flank \( A{R}_{4}{R}_{5} \) is \( \frac{T - {2\lambda }{k}_{a}}{a{k}_{a}} : \frac{\lambda }{b} : \frac{\lambda }{c} \) . The line \( {H}^{B}\left( \lambda \right) {H}^{C}\left( \lambda \right) \) has equation\n\n\[ a\left( {3{k}_{b}{k}_{c}{\lambda }^{2} ... | Yes |
For every real number \( \lambda \) the triangles \( \tau \) and \( {\sigma }_{H}^{\lambda } \) are orthologic. The locus of the orthology center \( \left\lbrack {\tau ,{\sigma }_{H}^{\lambda }}\right\rbrack \) is the Kiepert hyperbola of \( {ABC} \). The locus of the orthology center \( \left\lbrack {{\sigma }_{H}^{\l... | Proof. The perpendicular \( p\left( {A,{H}^{B}\left( \lambda \right) {H}^{C}\left( \lambda \right) }\right) \) from \( A \) onto the line \( {H}^{B}\left( \lambda \right) {H}^{C}\left( \lambda \right) \) has equation \( {M}_{ - }\left( {b, c}\right) y - {M}_{ + }\left( {c, b}\right) z = 0 \), where\n\n\[ \n{M}_{ \pm }\... | Yes |
Theorem 7. For every \( \lambda \in \mathbb{R} \smallsetminus \{ 0\} \), the triangles \( {ABC} \) and \( {F}^{A}\left( \lambda \right) {F}^{B}\left( \lambda \right) {F}^{C}\left( \lambda \right) \) are homologic if and only if the triangle \( {ABC} \) is isosceles. | Proof. The center \( {F}^{A}\left( \lambda \right) \) of the nine-point circle of the flank \( A{R}_{4}{R}_{5} \) is\n\n\[ \frac{\left( {{k}_{a} - {a}^{2}}\right) \lambda - {2T}}{a} : \frac{\lambda {d}_{2b}}{-b} : \frac{\lambda {d}_{2c}}{c}. \]\n\nThe line \( A{F}^{A}\left( \lambda \right) \) has equation \( b{d}_{2c}y... | Yes |
Theorem 10. For every \( \lambda \in \mathbb{R} \smallsetminus \{ 0\} \), the triangles \( {ABC} \) and \( {K}^{A}\left( \lambda \right) {K}^{B}\left( \lambda \right) {K}^{C}\left( \lambda \right) \) are orthologic if and only if the triangle \( {ABC} \) is isosceles. | Proof. The symmedian point \( {K}^{A}\left( \lambda \right) \) of the flank \( A{R}_{4}{R}_{5} \) is\n\n\[ \frac{\left( {{d}_{2a}^{2} - {a}^{2}{z}_{2a}}\right) \lambda - T\left( {3{k}_{a} + 2{a}^{2}}\right) }{a} : {\lambda b}{k}_{b} : {\lambda c}{k}_{c}. \]\n\nIt follows that the perpendicular \( p\left( {{K}^{A}\left(... | Yes |
Lemma 3. Let \( {ABC} \) be any triangle, and \( V = \left( {A, B, C}\right) \) the corresponding vector of points in the complex plane with the centroid of \( {ABC} \) at the origin. If \( S \) is defined as in (4), then \( {ABC} \) is positively (negatively) oriented when \( \left| {s}_{1}\right| > \left| {s}_{2}\rig... | Proof. According to (7), \( \operatorname{Im}\left( {z\left( {A, B, C}\right) }\right) \sim {s}_{1}\overline{{s}_{1}} - {s}_{2}\overline{{s}_{2}} = {\left| {s}_{1}\right| }^{2} - {\left| {s}_{2}\right| }^{2} \) . | Yes |
Theorem 4. Let \( {A}_{0}{B}_{0}{C}_{0} \) be the base triangle, and \( {XYZ} \) the transformation triangle. If the sequence of triangles \( {A}_{n}{B}_{n}{C}_{n} \) is generated as described in \( §3 \) with every triangle \( {B}_{n - 1}{C}_{n - 1}{A}_{n}^{\prime } \) etc. oppositely oriented to \( {A}_{n - 1}{B}_{n ... | Proof. Without loss of generality, we may assume \( {A}_{0}{B}_{0}{C}_{0} \) positively oriented. It is sufficient to show that triangle \( {A}_{n}{B}_{n}{C}_{n} \) is positively oriented for every \( n \) . Then, every triangle \( {B}_{n - 1}{C}_{n - 1}{A}_{n}^{\prime } \) etc. is negatively oriented, and the result f... | Yes |
Theorem 6. The sequence of \( k \) -gons \( {\Pi }_{n} \) converges to a regular \( k \) -gon of \( q \) -type, if and only if \( \left| {\lambda }_{q}\right| > \left| {\lambda }_{i}\right| \) for every \( i \neq q \) such that \( {s}_{i} \neq 0 \) . | (i) \( {s}_{q} \neq 0 \) and \( {s}_{i} = 0 \) for every \( i \neq q \) . This corresponds to \( {\Pi }_{0} \) - and the whole sequence - being regular of \( q \) -type.\n\n(ii) There are two integers \( q, r \) such that \( {\lambda }_{r} = 0,{s}_{q},{s}_{r} \neq 0 \), and \( {s}_{i} = 0 \) for every \( i \neq q, r \)... | Yes |
Proposition 1. Consider a triangle \( {ABC} \) with \( a \leq b \leq c \) . Denote by \( X, Y, Z \) the midpoints of the sides \( {BC},{CA} \), and \( {AB} \) respectively. Let \( D, E, F \) be points on the sides \( {BC},{CA},{AB} \) satisfying the following two conditions:\n\n(1.1) \( D \) is between \( X \) and \( C... | Proof. Denote by \( \mathbf{i},\mathbf{j},\mathbf{k} \) the unit vectors along EF, FD, DE. See Figure 1. Since \( \angle {BFD} \leq \angle {AFE} \), we have \( \mathbf{i} \cdot \mathbf{{ZF}} \leq \mathbf{j} \cdot \mathbf{{ZF}} \) . Similarly, since \( \angle {CDE} \leq \angle {BDF} \) and \( \angle {CED} \leq \angle {A... | Yes |
Proposition 2. Suppose the side lengths of triangle \( {ABC} \) satisfy \( a \leq b \leq c \) . Let \( P \) be an interior point with (positive) homogeneous barycentric coordinates \( \left( {x : y : z}\right) \) satisfying\n\n(2.1) \( x \leq y \leq z \) ,\n\n(2.2) \( x\cot A \geq y\cot B \geq z\cot C \) .\n\nThen the ... | Proof. In Figure 1, \( {BD} = \frac{az}{y + z},{DC} = \frac{ay}{y + z} \), and \( {BF} = \frac{cx}{x + y} \) . Since \( y \leq z \), it is clear that \( {BD} \geq {DC} \) . Similarly, \( {AE} \geq {EC} \), and \( {AF} \geq {FB} \) . Condition (1.1) is satisfied. Applying the law of sines to triangle \( {BDF} \), we hav... | Yes |
Theorem 4. Consider the group structure on a pivotal isocubic with the pivot \( F \) as neutral element. The constant point is \( N = {F}_{t} \). | (1) \( F \) is the pivot, so \( P,{P}^{ * } \) and \( F \) are collinear.\n\n(2) Put \( P = F \) in (1).\n\n(3) \( P + {P}^{ * } = \left( {P \cdot {P}^{ * }}\right) \cdot F = F \cdot F = {F}_{t} \).\n\n(4) \( P + Q = \left( {P \cdot Q}\right) \cdot F = {\left( P \cdot Q\right) }^{ * } \) . (use (1))\n\n(5) This is Coro... | Yes |
Theorem 1. Consider a triangle \( {ABC} \) and a line \( l \), not through \( A, B \) or \( C \), in \( \mathcal{P} \). Put \( {AB} \cap l = {C}^{\prime \prime },{BC} \cap l = {A}^{\prime \prime },{CA} \cap l = {B}^{\prime \prime } \) and construct the points \( {A}^{\prime },{B}^{\prime } \) and \( {C}^{\prime } \) fo... | Proof. The line \( {A}^{\prime }{A}^{\prime \prime \prime } \) is clearly the polar line of \( {A}^{\prime \prime } \) with respect to the conic \( \mathcal{C}\left( {A, B, C, I,{I}^{\prime }}\right) \) and likewise for the line \( {B}^{\prime }{B}^{\prime \prime \prime } \) and \( {B}^{\prime \prime } \), and for the ... | Yes |
Theorem 2. If \( I,{I}^{\prime } \) are variable conjugate points in an involution \( \Omega \) on the line \( l \) with double (or fixed) points \( D \) and \( {D}^{\prime } \), then the locus of the point \( L \) is the conic \( \mathcal{L} = \mathcal{C}\left( {{A}^{\prime },{B}^{\prime },{C}^{\prime }, D,{D}^{\prime... | Proof. Since the conics \( \mathcal{C}\left( {A, B, C, I,{I}^{\prime }}\right) \) intersect the line \( l \) in the variable conjugate points \( I,{I}^{\prime } \) of an involution on \( l \), these conics must belong to a pencil with basis points \( A, B, C \) and a fourth point \( K \) : this follows from the Theorem... | Yes |
Theorem 2. Given positive real numbers \( u, v, w \), there are four \( \left( {u, v, w}\right) \) -cutting circles, at least two of which are real. When there are four distinct real circles, their centers form an orthocentric system, of which the circumcircle is the nine point circle. When two of these centers coincid... | 2.3. Let \( \left( {O}_{1}\right) \) and \( \left( {O}_{2}\right) \) be two \( \left( {u, v, w}\right) \) -cutting circles with centers \( {O}_{1} \) and \( {O}_{2} \) . Consider the midpoint \( M \) of \( {\mathrm{O}}_{1}{\mathrm{O}}_{2} \) . The orthogonal projection of \( M \) on \( {BC} \) clearly is the midpoint o... | No |
Corollary 4. The centers of the Stammler circles form an equilateral triangle circumscribing the circumcircle of \( {ABC} \), and tangent to the circumcircle at the vertices \( {A}_{0}{B}_{0}{C}_{0} \) of the circumtangential triangle. The radical axes of the Stammler circles among themselves are the Simson-Wallace lin... | \( {}^{5} \) These are the three Simson-Wallace lines passing through \( N \), i.e., the cevian lines of \( N \) in the triangle which is the translation of \( {A}_{0}{B}_{0}{C}_{0} \) by \( \mathbf{{ON}} \) . They are also the tangents to the Steiner deltoid at the cusps. | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.