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Theorem 15. \( {\left. \rho \right| }_{{\mathcal{T}}^{ * } \smallsetminus \Gamma } \) has \( n \) -periodic points for all integers \( n \geq 1 \) . | Proof. Consider the bottom half \( {\mathcal{S}}_{n}^{{lt} \downarrow } \) of \( {\mathcal{S}}_{n}^{lt} \) delimited by \( \Phi \) and by the upper parent curve of \( \Phi .{r}^{n} \) is a bijective continuous mapping from \( {\mathcal{S}}_{n}^{{lt} \downarrow } \) to the top \( {r}^{n}\left( {\mathcal{S}}_{n}^{{lt} \d... | Yes |
Theorem 16. \( \mathcal{T} \smallsetminus \left( {\mathcal{A} \cup \mathcal{D}}\right) \) is totally path-disconnected if \( \mathcal{T} = \left\{ {\left( {\alpha ,\beta }\right) \mid {0}^{ \circ } \leq }\right. \) \( \left. {\beta \leq \alpha \leq {90}^{ \circ } - \frac{\beta }{2}}\right\} \) . | Proof. Otherwise some fixed continuous curve between two different points of \( \mathcal{T} \smallsetminus \left( {\mathcal{A} \cup \mathcal{D}}\right) \) would be included in each member of an infinite nested family of shrinking fractal ancestor copies of \( {\mathcal{P}}_{1} \) or of \( {\mathcal{P}}_{1} \smallsetmin... | No |
Theorem 17. (1) The following situations are equivalent:\n\n(a) \( n \geq 1,{r}^{n}\left( z\right) = z \) and \( {r}^{n}\left( z\right) \) causes a left shift of \( {2m} \) digits.\n\n(b) \( m \geq 1 \),\n\n\[ z = \left( {{w}_{1} - \nu }\right) {y}_{1}{w}_{2}{y}_{2}\ldots {w}_{m}{y}_{m}\overline{{\sigma }_{y}\left( {{w... | Proof. (1) By setting \( {r}^{n}\left( z\right) = z \) in the paragraph preceding the theorem. | No |
Theorem 18. Under the reflection map \( r \), there are in \( {\mathcal{T}}_{1} \) uncountably many disjoint infinite forward orbits of classes of both finite and infinite triangles. | Proof. The infinite sequence \( z = \underset{1 \times }{\underbrace{1i}}{1ii1i}\underset{2 \times }{\underbrace{1i1i}}{1ii1i}\underset{3 \times }{\underbrace{1i1i1i}}\ldots \) codes a class of finite triangles with unique representation and generates an infinite forward orbit in \( {\mathcal{P}}_{1} : z \) begins inde... | Yes |
Theorem 19. \( \mathcal{A} \) is a dense open subset of \( \mathcal{T} = \left\{ {\left( {\alpha ,\beta }\right) \mid {0}^{ \circ } \leq \beta \leq \alpha \leq {90}^{ \circ } - \frac{\beta }{2}}\right\} \) . Any neighborhood of a point of \( \mathcal{T} \smallsetminus \mathcal{A} \) intersects countably many periodic o... | Proof. A point of \( {\mathcal{A}}_{n}, n \in \mathbf{N} \), has some neighborhood in \( {\mathcal{A}}_{n - 1} \cup {\mathcal{A}}_{n} \) if one sets \( {\mathcal{A}}_{-1} = {\mathcal{A}}_{0} \) . The rest follows from the fact that every neighborhood of a point of \( \mathcal{T} \smallsetminus \mathcal{A} \) contains (... | No |
Theorem 20. The backward orbit of a class of proper triangles of \( \mathcal{T} \smallsetminus \mathcal{A} \) is dense in \( \mathcal{T} \smallsetminus \mathcal{A} \) . | Proof. Consider a class of proper triangles \( {\Delta }_{0} \in \mathcal{T} \smallsetminus \mathcal{A} \) and suppose that \( {\Delta }_{0} \in \) \( {\mathcal{P}}_{N} \smallsetminus \alpha \) -axis. Fix a neighborhood of \( \Delta \in \mathcal{T} \smallsetminus \mathcal{A} \) and choose a fractal copy \( \mathcal{C} ... | Yes |
In terms of these expressions, (a) the line \( {OP} \) can be expressed as \[ \mathop{\sum }\limits_{\text{cyclic }}\left( {{b}^{2}{\lambda }_{c} + {c}^{2}{\lambda }_{b}}\right) x = 0 \] | Equations (2), (3) and (4) follow easily from (1) and the definitions. | No |
Theorem 5. The lines \( F{H}^{\prime }, E{P}^{ * } \) and \( {QR} \) concur at the point \( K \) on \( \mathcal{H} \) . | Proof. The equations of the lines \( F{H}^{\prime } \) and \( E{P}^{ * } \) are given by\n\n\[ F{H}^{\prime } : \;\mathop{\sum }\limits_{\text{cyclic }}\frac{{\lambda }_{a}}{{a}^{2}}\left( {{\lambda }_{b}{S}_{B} - {\lambda }_{c}{S}_{C}}\right) \left( {{\lambda }_{c}v - {\lambda }_{b}w}\right) x = 0 \]\n\nand\n\n\( E{P}... | Yes |
Theorem 6. The point \( D \) is on the line \( {EQ} \) . | Proof. The line \( {EQ} \) can be written as\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{\lambda }_{a}\left( {\left( {{\lambda }_{a}w - {\lambda }_{c}u}\right) {b}^{2} + \left( {{\lambda }_{a}v - {\lambda }_{b}u}\right) {c}^{2}}\right) \left( {{\lambda }_{c}v - {\lambda }_{b}w}\right) x = 0 \]\n\nA direct calculation ... | Yes |
Theorem 7. The following pairs of (perpendicular) lines are parallel to the asymptotes of \( \mathcal{H} \) :\n\n(a) the axes of \( \mathcal{E} \),\n\n(b) the tangents from \( K \) to the parabola \( \mathcal{P} \). | Proof. Let us denote with \( {L}_{1} \) and \( {L}_{2} \) the points of intersection of the line \( {OP} \) with the circumcircle of the triangle\n\n\[ \n{L}_{1} = \left( {{abc}\left( {{\lambda }_{b}{S}_{B} - {\lambda }_{c}{S}_{C}}\right) + {a}^{2}{S}_{A}\mu : \cdots : \cdots }\right) , \n\]\n\n\[ \n{L}_{2} = \left( {{... | Yes |
Theorem 8. The pole \( {P}^{\prime } \) of the line \( \mathcal{L} \) is on the line \( {FS} \) . | Proof. The line \( {FS} \) is given by\n\n\[ \frac{{\lambda }_{a}{\left( {\lambda }_{c}v - {\lambda }_{b}w\right) }^{2}u}{{a}^{2}}x + \frac{{\lambda }_{b}{\left( {\lambda }_{a}w - {\lambda }_{c}u\right) }^{2}v}{{b}^{2}}y + \frac{{\lambda }_{c}{\left( {\lambda }_{b}u - {\lambda }_{a}v\right) }^{2}w}{{c}^{2}}z = 0, \]\n\... | Yes |
Theorem 9. Points on \( \mathcal{C} \) are\n\n(a) the point \( K \) ,\n\n(b) the intersections of the line \( \mathcal{L} \) with the tangents from the point \( K \) to the parabola P. | Proof. (a) A long calculation allows one to show that indeed, the point \( T \) is equidistant to the points \( F \) and \( K \) . \( {}^{1} \) The common distance of the point \( T \) to the points \( F \) and \( S \) can be expressed as \( {d}_{1}/\left( {{d}_{2}{d}_{3}}\right) \) where\n\n\[ \n{d}_{1} = \mathop{\sum... | Yes |
Theorem 2 (Collings). If \( \mathcal{L} \) is a line passing through the orthocenter \( H \) of a triangle \( {ABC} \), then the reflections of \( \mathcal{L} \) in the sides \( {BC},{CA},{AB} \) are concurrent on the circumcircle of \( {ABC} \) at a point called the anti-Steiner point of \( \mathcal{L} \). | The proof for this is quite straightforward and it consists of a simple angle chasing (see [1] or [4]). | No |
Lemma 5. Let \( P \) be a point in the plane of a given triangle \( {ABC} \) with orthocenter H. Let \( {A}_{1},{B}_{1},{C}_{1} \) be the points where the lines \( {AP},{BP} \), and \( {CP} \), intersect again the circumcircle. Furthermore, let \( {A}_{2},{B}_{2},{C}_{2} \) be the reflections of \( P \) across the side... | Proof. The line \( {AH} \) cuts the circumcircle of triangle \( {ABC} \) again at the reflection \( D \) of \( H \) across \( {BC} \) . Thus, the line \( D{A}_{2} \) is the reflection of \( {PH} \) with respect to \( {BC} \) and intersects the circumcircle of triangle \( {ABC} \) again at the anti-Steiner point \( T \)... | Yes |
Lemma 6. Let \( P \) be a point in the plane of triangle \( {ABC} \) and \( {P}_{A}{P}_{B}{P}_{C} \) its pedal triangle with respect to \( {ABC} \) . Let \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) be the midpoints of the segments \( {PA} \) , \( {PB} \), and \( {PC} \), respectively, and let \( {P}_{1},{P}_{2},{P}... | Proof. Let \( U \) be the Poncelet point of the quadrilateral \( {ABCP} \) . By definition, this point lies on the pedal circle of \( P \) with respect to triangle \( {ABC} \) . Now, let \( D \) be the second intersection of \( {BC} \) with the pedal circle \( {P}_{A}{P}_{B}{P}_{C} \) and let \( R \) be the orthogonal ... | Yes |
Lemma 7. Given a triangle \( {ABC} \) with circumcenter \( O \) and medial triangle \( {DEF} \) , let \( P \) be a point with orthogonal projections \( {P}_{1},{P}_{2},{P}_{3} \) on these sides. Let \( {A}^{\prime } \) be the intersection of the lines \( {EF} \) and \( {P}_{2}{P}_{3} \), and define \( {B}^{\prime },{C}... | Proof. The orthogonal projection \( V \) of \( A \) on \( {OP} \) is clearly the second intersection of the circumcircles of the cyclic quadrilaterals \( P{P}_{2}A{P}_{3} \) and \( {OEAF} \) with diameters \( {AP} \) and \( {AO} \), respectively. Also, note that \( V \) is the Miquel point of the complete quadrilateral... | Yes |
If \( P \) is the midpoint of the segment \( {BF} \), the Newton - Gauss line of the complete quadrilateral EAFDBC determines with the line PM an angle equal to \( \angle {EFD} \) . | We show that triangles \( {NPM} \) and \( {EDF} \) are similar.\n\nSince \( {BE}\parallel {PN} \) and \( {FC}\parallel {PM},\angle {EAC} = \angle {NPM} \) and \( \frac{BE}{PN} = \frac{FC}{PM} = 2 \) .\n\nIn the cyclic quadrilateral \( {ABCD} \), we have\n\n\[ \angle {EDF} = \angle {EDA} + \angle {ADF} = \angle {ABC} + ... | Yes |
Theorem 2. The parallel from \( E \) to the Newton - Gauss line of the complete quadrilateral EAF DBC and the line EF are isogonal lines of angle BEC. | Proof. Since triangles \( {EDF} \) and \( {NPM} \) are similar, we have \( \angle {DEF} = \angle {PNM} \) . Let \( {E}^{\prime } \) be the intersection of the side \( {BC} \) with the parallel of \( {NM} \) through \( E \) . Because \( {PN}\parallel {BE} \) and \( {NM}\parallel E{E}^{\prime },\angle {BEF} = \angle {PNF... | Yes |
Theorem 3. The quadrilaterals MPGN and MQHN are cyclic. | Proof. By Theorem 1, \( \angle {EFD} = \angle {PMN} \) . The points \( P \) and \( N \) are the cir-cumcenters of the right triangles \( {BFG} \) and \( {EFG} \), respectively. It follows that \( \angle {PGF} = \angle {PFG} \) and \( \angle {FGN} = \angle {GFN} \) . Thus,\n\n\[ \angle {PGN} + \angle {PMN} = \left( {\an... | Yes |
Theorem 4. The complete quadrilaterals EGFHJI and EAFDBC have the same Newton-Gauss line. | Proof. The two complete quadrilaterals have a common diagonal \( {EF} \) . Its midpoint \( N \) lies on the Newton-Gauss lines of both quadrilaterals. Note that \( N \) is equidistant from \( G \) and \( H \) since it is the circumcenter of the cyclic quadrilateral \( {EGFH} \) . We show that triangles \( {MPG} \) and ... | Yes |
Theorem 1. The harmonic conjugate conic of the circle\n\n\\[ \n{a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy} - \\left( {x + y + z}\\right) \\left( {{Px} + {Qy} + {Rz}}\\right) = 0 \n\\]\n\n(3)\n\nis a circle if and only if \\( \\left( {P, Q, R}\\right) = m\\left( {{S}_{A},{S}_{B},{S}_{C}}\\right) \\) for some \\( m \\) . | Proof. The matrix of the circle (3) being\n\n\\[ \n\\left( \\begin{matrix} - {2P} & {c}^{2} - P - Q & {b}^{2} - R - P \\\\ {c}^{2} - P - Q & - {2Q} & {a}^{2} - Q - R \\\\ {b}^{2} - R - P & {a}^{2} - Q - R & - {2R} \\end{matrix}\\right)\n\\]\n\nits harmonic conjugate conic has matrix\n\n\\[ \n\\left( \\begin{matrix} - {... | Yes |
Proposition 2. If \( m \neq \frac{1}{2} \), the harmonic conjugate circle of \( {\mathcal{C}}_{m} \) is the circle \( {\mathcal{C}}_{{m}^{\prime }} \) , where \( {m}^{\prime } = \frac{m}{{2m} - 1} \) . | Proof. By the proof of Theorem 1, the harmonic conjugate circle of \( {\mathcal{C}}_{m} \) is the circle\n\n\[ \left( {{2m}\left( {{S}_{B} + {S}_{C}}\right) - {a}^{2}}\right) {yz} + \left( {{2m}\left( {{S}_{C} + {S}_{A}}\right) - {b}^{2}}\right) {zx} + \left( {{2m}\left( {{S}_{A} + {S}_{B}}\right) - {c}^{2}}\right) {xy... | Yes |
Proposition 3. The centers of a pair of harmonic conjugate circles divide the segment \( {OH} \) harmonically. | Proof. Let the harmonic conjugate circles be \( {\mathcal{C}}_{m} \) and \( {\mathcal{C}}_{{m}^{\prime }} \), with \( {m}^{\prime } = \frac{m}{{2m} - 1} \). Their centers are points \( {O}_{m} \) and \( {O}_{{m}^{\prime }} \) satisfying\n\n\[ O{O}_{{m}^{\prime }} : {O}_{{m}^{\prime }}H = {m}^{\prime } : 1 - {m}^{\prime... | Yes |
Theorem 2. The original quadrilateral \( {A}_{1}{B}_{1}{C}_{1}{D}_{1} \) can be reconstructed from the second generation quadrilateral \( {A}_{2}{B}_{2}{C}_{2}{D}_{2} \) using isogonal conjugation: | \[ {A}_{1} = {\operatorname{Iso}}_{{D}_{2}{A}_{2}{B}_{2}}\left( {C}_{2}\right) \] \[ {B}_{1} = {\operatorname{Iso}}_{{A}_{2}{B}_{2}{C}_{2}}\left( {D}_{2}\right) \] \[ {C}_{1} = {\operatorname{Iso}}_{{B}_{2}{C}_{2}{D}_{2}}\left( {A}_{2}\right) \] \[ {D}_{1} = {\operatorname{Iso}}_{{C}_{2}{D}_{2}{A}_{2}}\left( {B}_{2}\ri... | Yes |
Theorem 3. Let \( {Q}^{\left( 1\right) } \) be a quadrilateral. Then\n\n(1) \( {Q}^{\left( 2\right) } \) degenerates to a point if and only if \( {Q}^{\left( 1\right) } \) is cyclic. | Proof. The first and second statements follow immediately from the definition of the iterative process. | No |
Corollary 5. The angles between the sides and the diagonals of a quadrilateral satisfy the following identities:\n\n\[ \n\\left( {\\cot \\alpha + \\cot \\gamma }\\right) \\cdot \\left( {\\cot \\beta + \\cot \\delta }\\right) = \\left( {\\cot {\\alpha }_{1} - \\cot {\\beta }_{2}}\\right) \\cdot \\left( {\\cot {\\delta }... | Proof. Since the (directed) angles of \( {ACBD} \) are \( - {\\alpha }_{1},{\\beta }_{2},{\\gamma }_{1}, - {\\delta }_{2} \) and the directed angles of \( {ACDB} \) are \( {\\alpha }_{2},{\\beta }_{1}, - {\\gamma }_{2}, - {\\delta }_{1} \), the identities follow from formula (1) for the ratio of similarity. | Yes |
Lemma 6. Let \( {o}_{1} \) and \( {o}_{2} \) be two circles centered at \( {O}_{1} \) and \( {O}_{2} \) respectively and intersecting at points \( A \) and \( B \) . Let \( W, R, S \in {CS}\left( {{o}_{1},{o}_{2}}\right) \) be points on the circle of similitude such that \( R \) and \( S \) are symmetric to each other ... | Proof. Perform inversion in the mid-circle. The image of \( {CS}\left( {{o}_{1},{o}_{2}}\right) \) is the radical axis \( {RA}\left( {{o}_{1},{o}_{2}}\right) \), i.e., the line through \( A \) and \( B \) . The images of \( R \) and \( S \) lie on the line \( {AB} \) and are symmetric with respect to \( I \mathrel{\tex... | Yes |
Lemma 7. Spiral similarities \( {H}_{k, l}^{W} \) have the following properties:\n\n(1) \( {H}_{1,2}^{W}\left( {B}_{1}\right) = {A}_{1} \Leftrightarrow {H}_{2,4}^{W}\left( {A}_{1}\right) = {C}_{1} \) .\n\n(2) \( {H}_{1,2}^{W}\left( {B}_{1}\right) = {A}_{1} \Leftrightarrow {H}_{1,4}^{W}\left( {B}_{1}\right) = {C}_{1} \)... | Proof. Assume that \( {H}_{1,2}^{W}\left( {B}_{1}\right) = {A}_{1} \) . Let \( {P}_{1,2} \mathrel{\text{:=}} {A}_{1}{B}_{1} \cap {A}_{2}{B}_{2} \) be the joint point of the spiral similarity (centered at \( W \) ) taking \( {B}_{1} \) into \( {A}_{1} \) and \( {A}_{2} \) into \( {B}_{2} \) . Since points \( {B}_{1},{P}... | Yes |
Lemma 8. \( {H}_{1,4}^{W}\left( {B}_{1}\right) = {C}_{1},{H}_{1,2}^{W}\left( {D}_{1}\right) = {C}_{1},{H}_{4,2}^{W}\left( {D}_{1}\right) = {B}_{1} \) . | Proof. Lemma 7 shows that \( {H}_{1,2}^{W}\left( {B}_{1}\right) = {A}_{1} \) implies \( {H}_{1,4}^{W}\left( {B}_{1}\right) = {C}_{1} \) . Assume that \( {H}_{1,2}^{W}\left( {B}_{1}\right) \neq {A}_{1} \) . To find the image of \( {B}_{1} \) under \( {H}_{1,4}^{W} \), represent the latter as the composition \( {H}_{2,4}... | Yes |
Theorem 9. \( W \in {CS}\left( {{o}_{i},{o}_{j}}\right) \) for all \( i, j \in \{ 1,2,3,4\} \) . | Proof. By definition, \( W \in {CS}\left( {{o}_{1},{o}_{2}}\right) \cap {CS}\left( {{o}_{1},{o}_{4}}\right) \cap {CS}\left( {{o}_{2},{o}_{4}}\right) \) . We will show that \( W \in {CS}\left( {{o}_{3},{o}_{i}}\right) \) for any \( i \in \{ 1,2,4\} \) . Recall that \( {B}_{1} \in {CS}\left( {{o}_{1},{o}_{2}}\right) \cap... | Yes |
Theorem 11. (Isoptic property) All the triad circles \( {o}_{i} \) subtend equal angles at \( W \) . | In particular, \( W \) is inside of all of the triad circles in the case of a convex quadrilateral and outside of all of the triad circles in the case of a concave quadrilateral. (This was pointed out by Scimemi in [17]). If \( W \) is inside of a triad circle, the isoptic angle equals to \( \angle {TO}{T}^{\prime } \)... | Yes |
Corollary 12. The powers of \( W \) with respect to triad circles are proportional to the squares of the radii of the triad circles. | This property of the isoptic point was shown by Neville in [13] using tetracyclic coordinates and the Darboux-Frobenius identity. | No |
Corollary 14. (Isodynamic property of \( W \) ) The distances from \( W \) to the vertices of the quadrilateral are inversely proportional to the radii of the triad-circles going through the remaining three vertices:\n\n\[ \left| {W{A}_{1}}\right| : \left| {W{B}_{1}}\right| : \left| {W{C}_{1}}\right| : \left| {W{D}_{1}... | From analysis of similar triangles in the iterative process, it is easy to see that the limit point of the process satisfies the above distance relations. Therefore, \( W \) (defined at the beginning of this section as the second point of intersection of \( \left. {{CS}\left( {{o}_{1},{o}_{2}}\right) \text{and}{CS}\lef... | Yes |
Theorem 15 (Inversive property of W).\n\n\[ W = {\\operatorname{Inv}}_{{o}_{1}^{\\left( 2\\right) }}\\left( {A}_{1}\\right) = {\\operatorname{Inv}}_{{o}_{2}^{\\left( 2\\right) }}\\left( {B}_{1}\\right) = {\\operatorname{Inv}}_{{o}_{3}^{\\left( 2\\right) }}\\left( {C}_{1}\\right) = {\\operatorname{Inv}}_{{o}_{4}^{\\left... | Proof. To prove the first equality, perform inversion in a circle centered at \( {A}_{1} \) . The image of a point under the inversion will be denoted by the same letter with a prime. The images of the circles of similitude \( {CS}\\left( {{o}_{1},{o}_{2}}\\right) ,{CS}\\left( {{o}_{4},{o}_{1}}\\right) \) and \( {CS}\\... | Yes |
Lemma 19. Let \( M, N \in o \) and \( W \notin o \) . For every point \( L \in o \), define\n\n\[ J \mathrel{\text{:=}} \left( {MWL}\right) \cap {NL}. \]\n\nThe locus of points \( J \) is a straight line going through \( W \) . | Proof. For each point \( L \in o \), let \( K \) be the center of the circle \( k \mathrel{\text{:=}} \left( {MWL}\right) \) . The locus of centers of the circles \( k \) is the perpendicular bisector of the segment \( {MW} \) . Since \( M \in o \cap k \), there is a spiral similarity centered at \( M \) with joint poi... | Yes |
Lemma 19. Let \( {J}_{1} \) be the joint point corresponding to \( L = B \) . Then \( {J}_{1}, W \in l \) . | Let \( {J}_{2} \) be the joint point corresponding to \( M = C, N = Z \) and \( L = B \) in Lemma 19.\n\nLet \( Y = {J}_{2}C \cap {J}_{1}Z \) . By properties of spiral similarity, \( Y = {H}_{B, C}^{W}\left( Z\right) \) .\n\nNotice that by definition of \( {J}_{1} \), points \( {J}_{1}, B, C \) are on a line. Similarly... | Yes |
Lemma 21. Let \( {PQ} \) be a chord on a circle o centered at \( O \) . If \( W \notin \left( {POQ}\right) \), there is a spiral similarity centered at \( W \) that takes \( {PQ} \) into another chord of the circle \( O \) . | Proof. Let \( {H}_{P,{P}^{\prime }}^{W} \) be the spiral similarity centered at \( W \) that takes \( P \) into another point \( {P}^{\prime } \) on circle \( o \) . As \( {P}^{\prime } \) traces out \( o \), the images \( {H}_{P,{P}^{\prime }}^{W}\left( Q\right) \) of \( Q \) trace out another circle, \( {o}_{Q} \) . ... | Yes |
Corollary 23. \( W \in {CS}\left( {{o}_{i}^{\left( 1\right) },{o}_{j}^{\left( k\right) }}\right) \) for any \( i, j, k \) . | Proof. Since there is a spiral similarity centered at \( W \) that takes \( {A}_{1}{B}_{1} \) into \( {C}_{2}{D}_{2} \) , Theorem 22 implies that \( W \in {CS}\left( {{o}_{1},{o}_{4}^{\left( 2\right) }}\right) \) . Since \( W \in {CS}\left( {{o}_{1},{o}_{2}}\right) \), it follows that \( W \in {CS}\left( {{o}_{4}^{\lef... | Yes |
Theorem 25. (Inversion in a circle centered at \( W \) ) Consider the complete quadrangle determined by a nondegenerate quadrilateral. Inversion in \( W \) transforms\n\n- 6 lines of the complete quadrilateral into the 6 circles of similitude of the triad circles of the image quadrilateral;\n\n- 6 circles of similitude... | Proof. Observe that the 6 lines of the quadrangle are the radical axes of the triad circles taken in pairs. Since \( W \) belongs to all the circles of similitude of triad circles, by property 5 in section 3.1, inversion in a circle centered in \( W \) takes radical axes into the circles of similitude. This implies the... | Yes |
Theorem 27. The pedal quadrilateral of \( W \) is a parallelogram whose angles equal to those of the Varignon parallelogram. | Proof. Since \( W \in {CS}\left( {{o}_{1},{o}_{2}}\right) \cap {CS}\left( {{o}_{3},{o}_{4}}\right) \), property (5) of the circle of similitude implies that\n\n\[ \angle {AWB} = \angle {ACB} + \angle {ADB} = {\gamma }_{1} + {\delta }_{2}, \]\n\n\[ \angle {CWD} = \angle {CAD} + \angle {CBD} = {\alpha }_{1} + {\beta }_{2... | Yes |
Theorem 30. The pedal quadrilateral of a point with respect to quadrilateral \( {ABCD} \) is a nondegenerate parallelogram if and only if this point is \( W \) . | Proof. By Lemma 26, if \( P \in {CS}\left( {{o}_{1},{o}_{3}}\right) \cap {CS}\left( {{o}_{2},{o}_{4}}\right) \), then both pairs of opposite sides of the pedal quadrilateral \( {P}_{a}{P}_{b}{P}_{c}{P}_{d} \) are parallel.\n\nAssume that the pedal quadrilateral \( {P}_{a}{P}_{b}{P}_{c}{P}_{d} \) of \( P \) is a nondege... | Yes |
Lemma 32. The sides of the isogonal conjugate quadrilateral and the pedal quadrilateral of a given point are perpendicular to each other. | Proof. Let \( {b}_{A} \) be the bisector of the \( \angle {DAB} \) . Let \( I = {l}_{A} \cap {P}_{a}{P}_{d} \) and \( J = {b}_{A} \cap {P}_{a}{P}_{d} \) . Since \( A{P}_{a}P{P}_{d} \) is cyclic, it follows that \( \angle {P}_{d}{AP} = \angle {P}_{d}{P}_{a}P \) . Since \( P{P}_{a} \bot {P}_{a}A \) , it follows that \( {... | Yes |
Theorem 1 (Hyperbolic Euler's inequality). Suppose a hyperbolic triangle has a circumcircle and let \( R \) be its radius. Let \( r \) be the radius of the triangle’s incircle. Then\n\n\[ \tanh \frac{R}{k} \geq 2\tanh \frac{r}{k} \]\n\n(1)\n\nThe equality is achieved for an equilateral triangle for any fixed defect. | Proof. Recall that the radius \( R \) of the circumcircle of a hyperbolic triangle (if it exists) is given by\n\n\[ \tanh \frac{R}{k} = \sqrt{\frac{\sin \frac{\delta }{2}}{\prod \sin \left( {A + \frac{\delta }{2}}\right) }} = \frac{2\prod \sinh \frac{a}{2k}}{\sqrt{\sinh \frac{s}{k}\prod \sinh \frac{s - a}{k}}} \]\n\n(2... | Yes |
Lemma 3 (Cagnolli’s first formula). The area \( S = {k}^{2}\delta \) of a hyperbolic triangle \( {ABC} \) is given by\n\n\[ \sin \frac{S}{2{k}^{2}} = \frac{\sinh \frac{a}{2k}\sinh \frac{b}{2k}\sin C}{\cosh \frac{c}{2k}} \] | Proof. From the well known second (or \ | No |
Theorem 4 (Hyperbolic Finsler-Hadwiger's inequality). For a hyperbolic triangle \( {ABC} \) we have:\n\n\[ \sum \cosh \frac{a}{k} \geq \sum \cosh \frac{b - c}{k} + {12}\sin \frac{S}{2{k}^{2}}\prod \cosh \frac{a}{2k}\tan \frac{\pi - \delta }{6} \] | Proof. The idea is to try to mimic (as much as possible) the first proof of (8). Start with the hyperbolic law of cosines\n\n\[ \cosh \frac{a}{k} = \cosh \frac{b}{k}\cosh \frac{c}{k} - \sinh \frac{b}{k}\sinh \frac{c}{k}\cos A. \]\n\nBy adding and subtracting \( \sinh \frac{b}{k}\sinh \frac{c}{k} \), we obtain\n\n\[ \co... | Yes |
Theorem 5 (Spherical Finsler-Hadwiger's inequality). For a spherical triangle \( {ABC} \) on a sphere of radius \( \rho \) we have\n\n\[ \sum \cos \frac{a}{\rho } \geq \sum \cos \frac{b - c}{\rho } + {12}\sin \frac{S}{2{\rho }^{2}}\cos \frac{a}{2\rho }\cos \frac{b}{2\rho }\cos \frac{c}{2\rho }\tan \frac{\varepsilon - \... | The equality in (14) holds if and only if for any fixed \( \varepsilon \), the triangle is equilateral. | No |
Theorem 1. The sphere \( \mathcal{Q} \) with radius \( R \), and center at \( \frac{\sqrt{{bc} + {ac} + {ab}}}{2} \) above the point \( P \), is tangent externally to the four tritangent spheres. | Proof. Consider triangle \( {I}_{a}P{S}_{\mathrm{p}} \) with median \( {I}_{a}N \) . Note that \( {I}_{a}N = \frac{R}{2} + {r}_{a} \) and \( N{S}_{\mathrm{p}} = \frac{1}{2}{OI} \), where \( O \) is the circumcenter. It follows that \( N{S}_{\mathrm{p}}^{2} = \frac{1}{4}R\left( {R - {2r}}\right) \) by Euler’s formula. S... | Yes |
Proposition 1. If \( \mathcal{C} \) is the inscribed conic tangent to the sidelines at the traces of the point \( \left( {\frac{1}{u} : \frac{1}{v} : \frac{1}{w}}\right) \), its dual conic \( {\mathcal{C}}^{ * } \) is the circumconic | Proof. Since the barycentric equation of \( \mathcal{C} \) is\n\n\[ {u}^{2}{x}^{2} + {v}^{2}{y}^{2} + {w}^{2}{z}^{2} - {2vwyz} - {2wuzx} - {2uvxy} = 0, \]\n\nthe conic is represented by the matrix\n\n\[ M = \left( \begin{matrix} {u}^{2} & - {uv} & - {uw} \\ - {uv} & {v}^{2} & - {vw} \\ - {uw} & - {vw} & {w}^{2} \end{ma... | Yes |
Proposition 3. A line \( {px} + {qy} + {rz} = 0 \) is bisected by the nine-point circle if and only if\n\n\[ \frac{{a}^{2}\left( {{b}^{2} + {c}^{2} - {a}^{2}}\right) }{p} + \frac{{b}^{2}\left( {{c}^{2} + {a}^{2} - {b}^{2}}\right) }{q} + \frac{{c}^{2}\left( {{a}^{2} + {b}^{2} - {c}^{2}}\right) }{r} = 0. \] | Proof. The pedal of \( O \) on the line \( {px} + {qy} + {rz} = 0 \) is the point\n\n\[ P = - {b}^{2}{q}^{2} - {c}^{2}{r}^{2} + \left( {{b}^{2} + {c}^{2} - 2{a}^{2}}\right) {qr} + {a}^{2}{rp} + {a}^{2}{pq} \]\n\n\[ : - {c}^{2}{r}^{2} - {a}^{2}{p}^{2} + \left( {{c}^{2} + {a}^{2} - 2{b}^{2}}\right) {rp} + {b}^{2}{pq} + {... | Yes |
Corollary 4. A line is bisected by the nine-point circle \( \left( N\right) \) if and only if it is tangent to the inscribed conic with center the nine-point center \( N \) . | Proof. Let \( {px} + {qy} + {rz} = 0 \) be a line bisected by the nine-point circle. By Proposition 3, it is tangent to the inscribed conic with perspector \( \left( {\frac{1}{u} : \frac{1}{v} : \frac{1}{w}}\right) \) , where\n\n\[ u : v : w = {a}^{2}\left( {{b}^{2} + {c}^{2} - {a}^{2}}\right) : {b}^{2}\left( {{c}^{2} ... | Yes |
Proposition 5. The Sherman line is the trilinear polar of the intersection of (i) the trilinear polar of the Gergonne point, (ii) the isotomic line of the trilinear polar of the circumcenter (see Figure 2). | Proof. The point \( S \) is the intersection of the two lines with equations \[ \left( {b + c - a}\right) x + \left( {c + a - b}\right) y + \left( {a + b - c}\right) z = 0, \] \[ {a}^{2}\left( {{b}^{2} + {c}^{2} - {a}^{2}}\right) x + {b}^{2}\left( {{c}^{2} + {a}^{2} - {b}^{2}}\right) y + {c}^{2}\left( {{a}^{2} + {b}^{2... | Yes |
Lemma 4. Let \( a, b, c \) be real numbers such that \( a + b + c = 2 \) with \( 0 < a, b, c < 1 \) . Then we have\n\n\[ 1 + {abc} < {ab} + {bc} + {ca} \leq 1 + \frac{9}{8}{abc}. \] | Proof. The proof of Lemma 3 shows the right inequality. The left one follows from expanding the obvious inequality \( \left( {1 - a}\right) \left( {1 - b}\right) \left( {1 - c}\right) > 0 \), and noting \( a + b + c = 2 \) . | No |
Lemma 5. Let \( D \) consist of the set of ordered pairs \( \left( {p, u}\right) \) such that there exists a triangle of perimeter 2 having side lengths \( a, b, c \) with \( p = {abc} \) and \( u = \) \( \frac{{ab} + {bc} + {ca} - 1}{abc} \) . If \( 1 < {u}^{\prime } \leq \frac{9}{8} \) is fixed, then \( p = g\left( {... | Proof. Note first that \( \left( {p, u}\right) \in D \) implies \( 0 < p \leq \frac{8}{27} \) and \( 1 < u \leq \frac{9}{8} \), the latter by Lemma 4. Given \( {p}_{o} \in \left( {0,\frac{8}{27}}\right\rbrack \), let \( {u}_{o} \) denote the solution of the equation \( g\left( u\right) = {p}_{o} \) , where \( u \in \le... | Yes |
Lemma 6. Let \( f\left( u\right) \) be given by\n\n\[ f\left( u\right) = \frac{{\left\lbrack \left( 3 - 6u\right) g\left( u\right) + 2\right\rbrack }^{-\frac{1}{2}} - 3{\left( u - 1\right) }^{\frac{1}{2}}}{3{\left( u - 1\right) }^{\frac{1}{2}} - {24}{\left( u - 1\right) }^{\frac{3}{2}}} \]\n\nwhere \( g\left( u\right) ... | Proof. From the definitions, we have\n\n\[ \frac{d}{du}f\left( u\right) = \frac{\frac{{6g}\left( u\right) - \left( {3 - {6u}}\right) \frac{d}{du}g\left( u\right) }{2{\left( \left( 3 - 6u\right) g\left( u\right) + 2\right) }^{\frac{3}{2}}} - \frac{3}{2{\left( u - 1\right) }^{\frac{1}{2}}}}{3{\left( u - 1\right) }^{\frac... | Yes |
Lemma 7. Let \( h\left( a\right) \) be given by\n\n\[ h\left( a\right) = \frac{{a}^{3}{\left( 1 - a\right) }^{3} + 2{\left\lbrack a\left( 1 - a\right) \left( -{a}^{2} + 4a - 2\right) \right\rbrack }^{\frac{3}{2}} - 6{a}^{2}{\left( 1 - a\right) }^{2}{\left( 2a - 1\right) }^{2}}{6{\left( 1 - a\right) }^{2}{\left( 2a - 1\... | Proof. Using mathematical programming such as Maple, one can show that the equation \( \frac{d}{da}h\left( a\right) = 0 \) has a unique real solution \( {a}^{ * } \approx {0.741049808} \) on the interval \( 2 - \sqrt{2} < a < 1 \) . Since \( h\left( {2 - \sqrt{2}}\right) = {2.178511254}, h\left( \frac{2}{3}\right) = \m... | Yes |
Theorem 8. For any triangle \( {ABC} \) and point \( P \) in its plane, we have\n\n\[{\left( {R}_{1}{R}_{2}\right) }^{2} + {\left( {R}_{2}{R}_{3}\right) }^{2} + {\left( {R}_{3}{R}_{1}\right) }^{2} \geq 6\left( {{7R} - {6r}}\right) {r}^{3}.\] | Proof. By Lemma 2 when \( j = 2 \), it suffices to show\n\n\[\frac{{\left( abc\right) }^{2}}{{a}^{2} + {b}^{2} + {c}^{2}} \geq 6\left( {{7R} - {6r}}\right) {r}^{3}\]\n\nfor all triangles \( {ABC} \) with sides \( a, b \), and \( c \) such that \( a + b + c = 2 \) . Note that \( {4Rr} = {abc},{r}^{2} = {L}^{2} = \left( ... | Yes |
If a bicentric quadrilateral has an incircle and a circumcircle with radii \( r \) and \( R \) respectively, then it has the area\n\n\[ K = r\left( {r + \sqrt{4{R}^{2} + {r}^{2}}}\right) \sin \theta \] | Proof. We give two different proofs. Both of them uses the formula\n\n\[ K = \frac{1}{2}{pq}\sin \theta \]\n\n(1)\n\nwhich gives the area of a convex quadrilateral with diagonals \( p, q \) and angle \( \theta \) between them.\n\nFirst proof. In a cyclic quadrilateral it is easy to see that the diagonals satisfy \( p =... | Yes |
If a bicentric quadrilateral has an incircle and a circumcircle with radii \( r \) and \( R \) respectively, then its area satisfies\n\n\[ K \leq r\left( {r + \sqrt{4{R}^{2} + {r}^{2}}}\right) \]\n\nwhere there is equality if and only if the quadrilateral is a right kite. | Proof. There is equality if and only if the angle between the diagonals is a right angle, since \( \sin \theta \leq 1 \) with equality if and only if \( \theta = \frac{\pi }{2} \) . A tangential quadrilateral has perpendicular diagonals if and only if it is a kite according to Theorem 2 (i) and (iii) in [5]. Finally, a... | No |
Theorem 1. Suppose \( {M}_{1},{M}_{2},\ldots ,{M}_{m},\overline{{M}_{1}},\overline{{M}_{2}},\ldots ,\overline{{M}_{n}} \) are points in the complex plane. Also, suppose \( P = \mathop{\sum }\limits_{{i = 1}}^{m}{k}_{i}{M}_{i} + \mathop{\sum }\limits_{{i = 1}}^{n}\overline{{k}_{i}}\overline{{M}_{i}} \) where \( {k}_{1},... | As we illustrate in Section 6, the values of \( {x}_{1},\ldots ,{x}_{m},{y}_{1},\ldots ,{y}_{n}, z \) as rational functions of \( {k}_{1},{k}_{2},\ldots ,{k}_{m},\overline{{k}_{1}},\overline{{k}_{2}},\ldots ,\overline{{k}_{n}} \) can be computed adhoc from any specific situation that we face in practice. We observe tha... | No |
Proposition 1. Let \( A\left( {-1,0}\right), B\left( {0,1}\right) \), and \( D\left( {0, - 1}\right) \) be points in the Cartesian plane endowed with the Euclidean distance \( d \) . Let \( M \subset {\mathbb{R}}^{2} \) be the set\n\n\[ M = \{ A, B, D\} \cup \{ C \mid C\left( {x,0}\right), x \geq 0\} . \]\n\nThen the m... | Proof. We check that there exists a constant \( \delta \geq 0 \) such that\n\n\[ d\left( {X, Z}\right) + d\left( {Y, W}\right) \leq \left\lbrack {d\left( {Z, W}\right) + d\left( {Y, Z}\right) }\right\rbrack \vee \left\lbrack {d\left( {X, Y}\right) + d\left( {Z, W}\right) }\right\rbrack + {2\delta }, \]\n\nfor all \( X,... | Yes |
Proposition 2. Let \( A\left( {0,1}\right), B\left( {-1,0}\right), C\left( {0, - 1}\right) D\left( {a,0}\right) \), with \( a \in \left( {0,2}\right) \) be points in the interior of the disk centered at the origin of radius 2, endowed with the Cayley distance (see [3])\n\n\[ d\left( {X, Y}\right) = \frac{1}{2}\ln \frac... | Proof. A direct computation shows that\n\n\[ d\left( {A, D}\right) = d\left( {C, D}\right) = \frac{1}{2}\ln \left\lbrack {\frac{\sqrt{3{a}^{2} + 4} + 1}{\sqrt{3{a}^{2} + 4} - 1} \cdot \frac{\sqrt{3{a}^{2} + 4} + {a}^{2}}{\sqrt{3{a}^{2} + 4} - {a}^{2}}}\right\rbrack \]\n\n\[ d\left( {A, B}\right) = d\left( {B, C}\right)... | Yes |
Lemma 4. Let \( A, B, C, D \) be a projective base and \( f \) the corresponding quadratic transform. For every line \( l \) not coinciding with a side-line or vertex of \( {ABC} \), the lines \( l,{l}^{\prime } = {f}^{ * }\left( l\right) \) satisfy the following property: either both do not intersect the interior of \... | The proof is again an easy calculation in coordinates, which I omit. | No |
Theorem 5. Let \( {ABC} \) be a triangle and \( l,{l}^{\prime } \) be a pair of lines having the property of the previous lemma. Then there is a point \( D \), such that \( A, B, C, D \) is a projective base with quadratic transformation \( f \) and such that \( {l}^{\prime } = {f}^{ * }\left( l\right) \) . | To prove the theorem consider first the intersection points \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) of \( l \), and \( {A}^{\prime \prime },{B}^{\prime \prime },{C}^{\prime \prime } \) of \( {l}^{\prime } \) correspondingly with the sides \( {BC},{CA},{AB} \) of the triangle. By the hypothesis follows that the ... | Yes |
For every point \( D \) in the interior of the triangle \( {ABC} \) the sides of its cevian triangle and its tripolar are parallel to the axes of the four parabolas circumscribing the triangle and tangent to its middle-tripolar \( {m}_{D} \) . The intersections of these four lines with \( {m}_{D} \) are the contact poi... | Figure 7 shows a complete configuration of three points \( A, B, C \), a line \( l = {m}_{D} \) and the four parabolas passing through the points and tangent to the line. By the analysis made in \( §2 \), line \( l \) contains the middles of segments \( {A}_{1}{A}_{2},{B}_{1}{B}_{2} \) and \( {C}_{1}{C}_{2} \) . | No |
Theorem 9. Let \( c \) be a parabola with axis \( e \) circumscribing the triangle \( {ABC} \) . The locus of tripoles \( {D}_{P} \) of lines \( {e}_{p} \), which are the parallels to the axis from the points \( P \) of the parabola, is a hyperbola circumscribing the triangle and has, among others, the properties: | In (1) the result follows from the general property of circumconics to be generated by the tripoles of lines rotating about a fixed point (the perspector of the conic). In our case the fixed point is the point at infinity \( \left\lbrack e\right\rbrack \), determined by the direction of the axis of the parabola, and th... | Yes |
Theorem 10. Let \( {A}_{1}{B}_{1}{C}_{1} \) be the cevian triangle of point \( D \) with respect to triangle \( {ABC} \). The parabola tangent to the sides of \( {A}_{1}{B}_{1}{C}_{1} \) and the tripolar of \( D \) has its axis parallel to line \( l = {m}_{D} \). In addition, the triangle \( {ABC} \) is self-polar with... | The proof of the first part is a consequence of the theorem of Newton ([3, p. 208]), according to which, the centers of the conics which are tangent to four given lines is the line through the middles of the segments joining the diagonal points of the quadrilateral defined by the four lines (the Newton line of the quad... | Yes |
Lemma 11. If a parabola is inscribed in a triangle, then the anticomplementary triangle is self-polar with respect to the parabola. | To prove the lemma consider a parabola \( c \) inscribed in a triangle \( {ABC} \) . Consider also its anticomplementary \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) and the point \( {C}^{\prime \prime } \) of tangency with side \( {AB} \) (See Figure 10). The parallel to \( {AB} \) through \( C \), which is a side of... | No |
Theorem 12. Let the parabola \( c \) be tangent to the sides of the triangle \( {ABC} \) and to the tripolar \( \operatorname{tr}\left( D\right) \) of a point \( D \) . Then its contact point with \( \operatorname{tr}\left( D\right) \) is the intersection point of this line with the middle-tripolar \( l = {m}_{D} \) . | This is proved by an argument similar to that, used in the preceding theorem. In fact, define the homography \( f \) mapping triangle \( {ABC} \) to an equilateral \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) and point \( D \) to the centroid of \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . Then, see, as in the prec... | Yes |
Theorem 1. If the chord \( B{B}^{\prime } \) in a circle intersects the chord \( A{A}^{\prime } \) at the point \( P \) , then the division ratio\n\n\[ \n\frac{AP}{P{A}^{\prime }} = \frac{{AB} \cdot A{B}^{\prime }}{{A}^{\prime }B \cdot {A}^{\prime }{B}^{\prime }}.\n\] | Proof. If \( R \) is the radius of the circle (see Figure 1) and \( h,{h}^{\prime } \) are the heights of triangles \( {AB}{B}^{\prime },{A}^{\prime }B{B}^{\prime } \) from \( A \) and \( {A}^{\prime } \) respectively, then\n\n\[ \n\frac{AP}{P{A}^{\prime }} = \frac{h}{{h}^{\prime }} = \frac{\frac{{AB} \cdot A{B}^{\prim... | Yes |
Theorem 2 (Butterfly theorem, original version). If three chords \( A{A}^{\prime }, B{B}^{\prime }, C{C}^{\prime } \) in a circle are concurrent at the midpoint \( M \) of \( A{A}^{\prime } \), then the lines \( {BC} \) and \( {B}^{\prime }{C}^{\prime } \) intersect the line \( A{A}^{\prime } \) at two points \( P,{P}^... | Proof. It is sufficient to prove that \( \frac{AP}{P{A}^{\prime }} = \frac{{A}^{\prime }{P}^{\prime }}{{P}^{\prime }{A}^{\prime }} \) (see Figure 2). From Theorem 1 we have\n\n\[ 1 = \frac{AM}{M{A}^{\prime }} = \frac{{AB} \cdot A{B}^{\prime }}{{A}^{\prime }B \cdot {A}^{\prime }{B}^{\prime }} \]\n\nimplying\n\n\[ \frac{... | Yes |
Theorem 3 (Butterfly theorem, Mackay's version). Given a complete quadrangle inscribed in a circle; if any line cuts two opposite sides at equal distances from the center of the circle, it cuts each pair at equal distances from the center. | For a proof, see [1, p.105, Theorem 105]. | No |
Lemma 4. Two points \( P \) and \( {P}^{\prime } \) are conjugate relative to a circumconic of triangle \( {ABC} \) if and only if the conic passes through the cevian product of \( P \) and \( {P}^{\prime } \) . | Proof. Let \( P = \left( {u : v : w}\right) \) and \( {P}^{\prime } = \left( {{u}^{\prime } : {v}^{\prime } : {w}^{\prime }}\right) \) in barycentric coordinates with respect to triangle \( {ABC} \) . Their cevian product is the point\n\n\[ S = \left( {\frac{1}{v{w}^{\prime } + {v}^{\prime }w} : \frac{1}{w{u}^{\prime }... | Yes |
Theorem 5. Let \( {ABCD} \) be a cyclic quadrilateral, and \( M \) be the orthogonal projection of circumcenter \( O \) on a line \( \mathcal{L} \) . If a conic passing through \( A, B, C, D \) intersects \( \mathcal{L} \) at two points \( P \) and \( Q \) equidistant from \( M \), then for every conic passing through ... | Proof. Let \( N \) be the infinite point of the line \( \mathcal{L} \), which intersects the conic at \( P \) and \( Q \) (Figure 3). Since \( M \) and \( N \) are harmonic conjugate with respect to \( P \) and \( Q \) , the points \( M \) and \( N \) are conjugate relative to the conic. The polar of \( N \) relative t... | Yes |
Theorem 1 (9, Theorem 6). A tangential quadrilateral ABCD - without axes of symmetry - is cyclic if and only if its Newton line is perpendicular to the Newton line of its contact quadrilateral. | Proof of Theorem 1. Suppose that the Newton line of \( {ABCD} \), i.e., the line \( {n}_{1} \) through \( {M}_{AC}, Z,{M}_{BD},{M}_{IJ} \), and the Newton line of \( {EFGH} \), i.e., the line \( {n}_{2} \) through \( {M}_{EG},{M}_{FH} \), are perpendicular. Apply the inversion with respect to the incircle \( \mathcal{C... | Yes |
Theorem 2. ABCD is a bicentric quadrilateral with circumcircle \( \mathcal{C} \) and incircle \( \mathcal{C} \). Then bicentric quadrilaterals with circumcircle \( \mathcal{C} \) and incircle \( \mathcal{C} \) can be constructed starting from any point of the circumcircle \( \mathcal{C} \) (cf. [4],[6],[12],[13]). | Proof. If \( {ABCD} \) is bicentric (see Figure 1), R, \( r \) and \( d \) obey Fuss’ formula. Using inversion with respect to \( \mathcal{C} \), the circumcircle \( \mathcal{C} \) of \( {ABCD} \) is mapped onto the circle \( {\mathcal{C}}^{\prime } \) with center \( M \) and \( 2{R}^{\prime 2} + 2{d}^{\prime 2} = {r}^... | Yes |
Theorem 1. A convex quadrilateral \( {ABCD} \) is a trapezoid with parallel sides \( {AB} \) and \( {CD} \) if and only if\n\n\[ \cos A + \cos D = \cos B + \cos C = 0. \] | Proof. \( \left( \Rightarrow \right) \) If the quadrilateral is a trapezoid, then \( A + D = \pi \) . Hence\n\n\[ \cos A + \cos D = \cos A + \cos \left( {\pi - A}\right) = \cos A - \cos A = 0. \]\n\nThe second equality is proved in the same way.\n\n\( \left( \Leftarrow \right) \) We do an indirect proof of the converse... | Yes |
Proposition 2. A convex quadrilateral ABCD is a trapezoid with parallel sides \( {AB} \) and \( {CD} \) if and only if\n\n\[ \cot A + \cot D = \cot B + \cot C = 0. \] | Proof. Since the cotangent function is decreasing on the interval \( 0 < x < \pi \) and \( \cot \left( {\pi - x}\right) = - \cot x \), the proof is identical to that of Theorem 1 . | No |
Theorem 3. A convex quadrilateral \( {ABCD} \) is a trapezoid with parallel sides \( {AB} \) and \( {CD} \) if and only if\n\n\[ \tan \frac{A}{2}\tan \frac{D}{2} = \tan \frac{B}{2}\tan \frac{C}{2} = 1. \] | Proof. \( \left( \Rightarrow \right) \) If the quadrilateral is a trapezoid, then \( A + D = \pi = B + C \) . Using these, the equalities in the theorem directly follows since \( \tan \frac{D}{2} = \cot \frac{A}{2} \) and \( \tan \frac{C}{2} = \cot \frac{B}{2} \) .\n\n\( \left( \Leftarrow \right) \) Assume the quadrila... | Yes |
Proposition 4. A convex quadrilateral is a trapezoid if and only if one bimedian divide it into two quadrilaterals with equal areas. | Proof. \( \left( \Rightarrow \right) \) In a trapezoid, the bimedian between the bases (see the left half of Figure 2) divide it into two quadrilaterals with equal altitudes and two pairs of equal bases. Hence these two quadrilaterals, which are also trapezoids, have equal areas according to the well known formula for ... | Yes |
Proposition 5. If the diagonals in a convex quadrilateral ABCD intersect at \( P \) , then it is a trapezoid with parallel sides \( {AB} \) and \( {CD} \) if and only if the areas of the triangles \( {APD} \) and \( {BPC} \) are equal. | Proof. We have that the sides \( {AB} \) and \( {CD} \) are parallel if and only if (see Figure 1)\n\n\[ \n{h}_{ACD} = {h}_{BCD}\; \Leftrightarrow \;{T}_{ACD} = {T}_{BCD}\; \Leftrightarrow \;{T}_{APD} = {T}_{BPC} \]\n\nwhere \( {h}_{XYZ} \) and \( {T}_{XYZ} \) stands for the altitude and area of triangle \( {XYZ} \) re... | Yes |
Theorem 6. A convex quadrilateral is a trapezoid if and only if the product of the areas of the triangles formed by one diagonal is equal to the product of the areas of the triangles formed by the other diagonal. | Proof. We use notations on the subtriangle areas as in Figure 3. Then we have\n\n\[ \left( {S + {U}_{1}}\right) \left( {T + {U}_{2}}\right) = \left( {S + {U}_{2}}\right) \left( {T + {U}_{1}}\right) \]\n\n\[ \Leftrightarrow S{U}_{2} + T{U}_{1} = S{U}_{1} + T{U}_{2} \]\n\n\[ \Leftrightarrow S\left( {{U}_{2} - {U}_{1}}\ri... | Yes |
Lemma 7. The diagonals of a convex quadrilateral divide it into four non-overlapping triangles. The product of the areas of two opposite triangles is equal to the product of the areas of the other two triangles. | Proof. We denote the diagonal parts by \( w, x, y, z \) and the consecutive subtriangle areas by \( S,{U}_{1}, T,{U}_{2} \), see Figure 3. These areas satisfy\n\n\[ \n{ST} = \frac{1}{4}{wxyz}{\sin }^{2}\theta = {U}_{1}{U}_{2} \n\]\n\nwhere \( \theta \) is the angle between the diagonals. \( {}^{3} \) | Yes |
Theorem 8. The diagonals of a convex quadrilateral divide it into four non-overlapping triangles. If two opposite of these have areas \( S \) and \( T \), then the quadrilateral has the area\n\n\[ K = {\left( \sqrt{S} + \sqrt{T}\right) }^{2} \]\n\nif and only if it is a trapezoid whose parallel sides are the two sides ... | Proof. A convex quadrilateral has the area (see Figure 3)\n\n\[ K = S + T + {U}_{1} + {U}_{2} \]\n\n\[ = S + T + 2\sqrt{ST} - 2\sqrt{{U}_{1}{U}_{2}} + {U}_{1} + {U}_{2} \]\n\n\[ = {\left( \sqrt{S} + \sqrt{T}\right) }^{2} + {\left( \sqrt{{U}_{1}} - \sqrt{{U}_{2}}\right) }^{2} \]\n\nwhere we in the second equality used t... | Yes |
Proposition 9. The convex quadrilateral ABCD is a trapezoid with parallel sides \( {AB} \) and \( {CD} \) if and only if\n\n\[ \frac{DA}{BC} = \frac{\sin C}{\sin D}. \] | Proof. The quadrilateral is a trapezoid if and only if the triangles \( {ACD} \) and \( {BCD} \) have equal altitudes to the side \( {CD} \), which is equivalent to that the areas of these two triangles are equal. This in turn is equivalent to\n\n\[ \frac{1}{2}{CD} \cdot {DA}\sin D = \frac{1}{2}{CD} \cdot {BC}\sin C, \... | Yes |
Theorem 10. If a convex quadrilateral has consecutive sides \( a, b, c, d \) and diagonals \( p, q \), then\n\n\[ \n{p}^{2} + {q}^{2} = {b}^{2} + {d}^{2} + {2ac}\cos \xi \]\n\nwhere \( \xi \) is the angle between the extensions of the sides \( a \) and \( c \) . | Proof. In a convex quadrilateral \( {ABCD} \), let the extensions of \( {AB} \) and \( {CD} \) intersect at \( J \) . Other notations are as in Figure 4, where \( {AC} = p,{BD} = q,{AB} = a \) , and \( {AE} = x \) . We construct \( {GC} \) parallel to \( {AB} \) . Then \( \angle {DCG} = \angle {BJC} \) . We also have \... | Yes |
Corollary 11. A convex quadrilateral with consecutive sides \( a, b, c, d \) and diagonals \( p, q \) is a trapezoid with parallel sides \( a \) and \( c \) if and only if\n\n\[{p}^{2} + {q}^{2} = {b}^{2} + {d}^{2} + {2ac}.\] | Proof. This characterization is a direct consequence of Theorem 10, since the quadrilateral is a trapezoid if and only if \( \xi = 0 \) . | No |
Theorem 12. A convex quadrilateral is a trapezoid with parallel sides \( a \) and \( c \) if and only if the distance \( v \) between the midpoints of the diagonals has the length\n\n\[ v = \frac{\left| a - c\right| }{2}. \] | Proof. Inserting the equation in Corollary 11 into (3), we get that a convex quadrilateral is a trapezoid if and only if\n\n\[ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} = {b}^{2} + {d}^{2} + {2ac} + 4{v}^{2}\; \Leftrightarrow \;{\left( a - c\right) }^{2} = 4{v}^{2}. \]\n\nHence we get the characterization \( v = \frac{1}{2... | Yes |
Theorem 13. A convex quadrilateral with consecutive sides \( a, b, c, d \) is a trapezoid with parallel sides \( a \) and \( c \) if and only if the bimedian \( n \) that connects the midpoints of the sides \( b \) and \( d \) has the length\n\n\[ n = \frac{a + c}{2}. \] | Proof. The length of the bimedian \( n \) that connects the midpoints of the sides \( b \) and \( d \) in a convex quadrilateral is given by\n\n\[ 4{n}^{2} = {p}^{2} + {q}^{2} + {a}^{2} - {b}^{2} + {c}^{2} - {d}^{2} \]\n\naccording to [3, p.231] and post no 2 at [5] (both with other notations). Substituting \( {p}^{2} ... | Yes |
Theorem 14. A convex quadrilateral ABCD with consecutive sides \( a, b, c, d \) is a trapezoid with \( a\parallel c \) and \( a \neq c \) if and only if the length of the diagonals \( {AC} \) and \( {BD} \) are respectively\n\n\[ p = \sqrt{\frac{{ac}\left( {a - c}\right) + a{d}^{2} - c{b}^{2}}{a - c}}, \]\n\n\[ q = \sq... | Proof. We prove the second formula first. Using the law of cosines in the two triangles formed by diagonal \( {BD} = q \) in a convex quadrilateral, we have \( {d}^{2} = \) \( {a}^{2} + {q}^{2} - {2aq}\cos u \) and \( {b}^{2} = {c}^{2} + {q}^{2} - {2cq}\cos v \) (see Figure 5). Thus\n\n\[ \cos u = \frac{{a}^{2} + {q}^{... | Yes |
Theorem 15. Two opposite sides in a convex quadrilateral are parallel if and only if the midpoints of those sides and the intersection of the diagonals are three collinear points. | Proof. \( \left( \Rightarrow \right) \) In a trapezoid, let \( E \) and \( G \) be the midpoints of the sides \( {AB} \) and \( {CD} \) respectively, and \( P \) the intersection of the diagonals. Triangles \( {CDP} \) and \( {ABP} \) are similar due to two pairs of equal angles (see Figure 6). Note that \( {PG} \) and... | Yes |
Lemma 1. \( {f}_{p} \) is lower-semicontinuous, hence it achieves its minimum. | Proof. Let us introduce some notation: for \( p \in D \) and \( v \in {S}^{n - 1} \), we denote \( {\gamma }_{p, v} \) the connected component of \( \left( {p + \mathbb{R}v}\right) \cap D \) containing \( p \) . (So \( \overline{{\gamma }_{p, v}} = \left\lbrack {P, Q}\right\rbrack \) for some \( P, Q \in \partial D \) ... | Yes |
Proposition 2. The function \( \mathbf{r} : \bar{D} \rightarrow {\mathbb{R}}_{ \geq 0} \) is lower-semicontinuous. | Proof. We first study the function \( \mathbf{r} : D \rightarrow {\mathbb{R}}_{ + } \) . As \( \mathbf{r}\left( p\right) = \mathop{\min }\limits_{v}H\left( {p, v}\right) \), the lower-semicontinuity of \( H \) gives the lower-semicontinuity of \( \mathbf{r} \) : If \( {p}_{n} \rightarrow p \), take \( {v}_{n} \) such t... | Yes |
Lemma 3. The convex set \( D \) is the intersection\n\n\[ \mathop{\bigcap }\limits_{{\left| v\right| = 1}}{H}_{v}^{ - }\n\]\n\nand conversely, any such intersection is a convex set. Moreover,\n\n\[ \bar{D} = \mathop{\bigcap }\limits_{{\left| v\right| = 1}}{\bar{H}}_{v}^{ - }\n\] | Proof. The second assertion is clear, since the intersection of convex sets is convex.\n\nFor the first assertion, we have the trivial inclusion \( D \subset \mathop{\bigcap }\limits_{{\left| v\right| = 1}}{H}_{v} \) . Now suppose \( p \notin D \) . We have two cases:\n\n- \( p \notin \bar{D} \) . Then take \( q \in \b... | Yes |
Proposition 4. If \( D \) is convex, then \( \mathbf{r} \) achieves its supremum \( R \) at \( D \) . Moreover, \( {I}_{D} \cap D = {E}_{R} \cap D = \{ p \mid \mathbf{r}\left( p\right) = R\} \) and \( {I}_{D} = {E}_{R} \) (which is the closure of \( {E}_{R} \cap D \) ). | Proof. Let \( p \in {I}_{D} \cap \partial D \), and take a supporting hyperplane \( {H}_{v} \) at \( p \) . We claim that the open semiball \( {B}_{R}\left( p\right) \cap {H}_{v}^{ - } \subset D \) . If not, then there is a point \( q \in \partial D \) , \( d\left( {p, q}\right) < R, q \in {H}_{v}^{ - } \) . Then all t... | Yes |
Lemma 5. Let \( p \in D \) and \( r = \mathbf{r}\left( p\right) \) . Let \( \left\lbrack {P, Q}\right\rbrack \) be a segment of length \( r \) with \( P, Q \in \partial D \) and \( p \in \left\lbrack {P, Q}\right\rbrack \) . Let \( {v}_{P},{v}_{Q} \) be supporting vectors at \( P, Q \) respectively. Then\n\n(1) If \( {... | Proof. (1) Suppose first that \( {v}_{P},{v}_{Q} \) are parallel. So \( D \) is inside the region between the parallel hyperplanes \( {H}_{{v}_{P}} \) and \( {H}_{{v}_{Q}} \) . Clearly \( {v}_{P} = - {v}_{Q} \) . Let \( x \in D \), and draw the segment parallel to \( \left\lbrack {P, Q}\right\rbrack \) through \( x \) ... | Yes |
Theorem 6. The sets \( {D}_{r},{E}_{r} \) are convex sets, for \( r \in \left\lbrack {0, R}\right\rbrack, R = \max \mathbf{r} \) . Moreover, \( \partial {D}_{r} \cap D \) is \( {\mathbf{r}}^{-1}\left( r\right) \), for \( r \in \left( {0, R}\right) \) . | Proof. The assertion for \( {E}_{r} \) follows from that of \( {D}_{r} \) : knowing that \( {D}_{r} \) is convex, then\n\n\[ \n{E}_{r} = \overline{\mathop{\bigcap }\limits_{{\epsilon > 0}}{D}_{r - \epsilon }} \n\]\n\nis convex since the intersection of convex sets is convex, and the closure of a convex set is convex.\n... | Yes |
Proposition 7. Suppose \( D \) is a convex planar set. Let \( r \in \left( {0, R}\right) \) . Then \( \partial {D}_{r} \) is the envelope of the segments of length \( r \) with endpoints at \( \partial D \) . | Proof. As we proved before, the boundary of \( {D}_{r} \) is \( {\mathbf{r}}^{-1}\left( r\right) \), so the points of \( \partial {D}_{r} \) are \( r \) -accessible, but not \( {r}^{\prime } \) -accessible for \( {r}^{\prime } < r \) . Let \( p \in \partial {D}_{r} \) be a smooth point. Then there is a segment of lengt... | Yes |
Lemma 8. If \( D \) is strictly convex, then \( g \) is continuous. | Proof. Let \( {v}_{n} \in {S}^{n - 1},{v}_{n} \rightarrow v \) . Consider \( {p}_{n} = g\left( {v}_{n}\right) \in \partial D \), and the supporting hyperplane \( \left\langle {x - {p}_{n},{v}_{n}}\right\rangle \leq 0 \) . Let \( p = g\left( v\right) \), with supporting hyperplane \( \langle x - \) \( p, v\rangle \leq 0... | Yes |
Lemma 9. Suppose \( D \) strictly convex. For all \( 0 < r < R,\partial {D}_{r} \cap \partial D = \varnothing \), so \( \partial {D}_{r} = {\mathbf{r}}^{-1}\left( r\right) \) . | Proof. Take a point \( p \in \partial D \), and let \( {H}_{v} \) be a supporting hyperplante. Consider a small ball \( B \) around \( p \) of radius \( \leq r/2 \) . By strict convexity, \( d\left( {\partial B \cap D, H}\right) = \) \( {\epsilon }_{0} > 0 \) . Now we claim that \( {B}_{{\epsilon }_{0}}\left( p\right) ... | Yes |
Corollary 10. For \( D \) strictly convex, \( \mathbf{r} : \bar{D} \rightarrow {\mathbb{R}}_{ \geq 0} \) is continuous. | Proof. By Remark 2, \( \mathbf{r} \) is continuous on \( D \) . The continuity at \( \partial D \) follows from the proof of Lemma 9. | Yes |
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