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Proposition 9. For every parallelogram \( a = {A}_{1}{A}_{2}{A}_{3}{A}_{4} \) and every point \( P \) not lying on its side-lines the corresponding conjugate quadrangle \( b = {B}_{1}{B}_{2}{B}_{3}{B}_{4} \) is strongly periodic. | The proposition (see Figure 13) is equivalent to the property of the corresponding first recycler \( {G}_{1} \) to be the identity. To prove this it suffices to show that \( {G}_{1} \) fixes a point not lying on the parallel to \( {A}_{1}{A}_{2} \) through \( P \) (see the remark after corollary 5 of previous paragraph... | Yes |
Proposition 10. If for every point \( P \) not lying on the side-lines of the quadrangle \( a = {A}_{1}{A}_{2}{A}_{3}{A}_{4} \) the corresponding conjugate quadrangle \( b = {B}_{1}{B}_{2}{B}_{3}{B}_{4} \) is strongly periodic, then a is a parallelogram. | This is seen by taking for \( P \) the intersection point of the diagonals of the quadrangle. Consider then the parallel to \( b \) polygon starting at \( {A}_{2} \) . By assumption this must be closed, thus defining a triangle \( {A}_{2}{C}_{3}{C}_{4} \) (see Figure 15). The middles \( {D}_{1},{D}_{2} \) of the sides ... | Yes |
Lemma 12. Let \( a = E{E}^{\prime }F{F}^{\prime } \) be a quadrangle with diagonals \( {EF},{E}^{\prime }{F}^{\prime } \) and corresponding middles on them \( D, A \) . Draw from \( A \) parallels \( {AB},{AC} \) correspondingly to sides \( F{F}^{\prime },{F}^{\prime }E \) intersecting the diagonal \( {EF} \) correspon... | The proof is carried out using barycentric coordinates with respect to triangle \( {ABC} \) . Then points \( D, E, F,\ldots \) on line \( {BC} \) are represented using the corresponding small letters for parameters \( D = B + {dC}, E = B + {eC}, F = B + {fC},\ldots \) (see Figure 16). In addition \( P \) is represented... | Yes |
Lemma 13. Let \( a = {ABCD} \) be a quadrangle with diagonals \( {AC},{BD} \) and corresponding middles on them \( M, N \) . Draw from \( M \) parallels \( {ME},{MF} \) correspondingly to sides \( {AB},{AD} \) intersecting the diagonal \( {BD} \) at points \( E, F \) . Let \( P \) be a point of the Newton-line \( {MN} ... | In fact, consider the transformation \( {G}_{1} = {F}_{4} \circ {F}_{3} \circ {F}_{2} \circ {F}_{1} \) composed by the affine reflections with corresponding axes \( {PC},{PD},{PA},{PB} \) . By the discussion in the previous paragraph, the periodicity of the conjugate quadrilateral to \( P \) is equivalent to \( {G}_{1}... | Yes |
Theorem 14. For a non-parallelogramic quadrangle \( a = {A}_{1}{A}_{2}{A}_{3}{A}_{4} \) only the points \( P \) on its Newton-line have the corresponding conjugate quadrangle \( b = {B}_{1}{B}_{2}{B}_{3}{B}_{4} \) strongly periodic. | The previous theorem guarantees that all points of the Newton line have a strongly periodic corresponding conjugate polygon \( b \) . Assume now that there is an additional point \( {P}_{0} \), not on the Newton line, which has also a strongly periodic corresponding conjugate polygon. In addition fix a point \( {C}_{1}... | Yes |
Lemma 15. Let \( a = {A}_{1}{A}_{2}{A}_{3}{A}_{4} \) be a quadrangle and with diagonals intersecting at \( E \) . Let also \( \{ F, G\} \) be the two other diagonal points of its associated complete quadrilateral. Let also \( b = {B}_{1}{B}_{2}{B}_{3}{B}_{4} \) be the dual quadrangle of \( a \) .\n\n(1) Line \( {EG} \)... | \n\n\( {MN}/M{A}_{4} = 1 \), since Menelaus theorem applied to triangle \( {A}_{1}N{A}_{4} \) with secant line \( {B}_{1}{B}_{3}G \) gives \( \left( {{B}_{1}N/{B}_{1}{A}_{1}}\right) \left( {M{A}_{4}/{MN}}\right) \left( {G{A}_{1}/G{A}_{4}}\right) = 1 \) . But \( {B}_{1}N/{B}_{1}{A}_{1} = \) \( {B}_{4}{A}_{4}/{B}_{4}{A}_... | Yes |
Proposition 16. Let \( a = {A}_{1}{A}_{2}{A}_{3}{A}_{4} \) be a quadrangle and with diagonals intersecting at \( E \) . Let also \( \{ F, G\} \) be the two other diagonal points of its corresponding complete quadrilateral and \( \{ P, Q, R\} \) the middles of the diagonals \( \left\{ {{A}_{2}{A}_{4},{A}_{1}{A}_{3},{FG}... | Start with the intersection point \( T \) of diagonal \( {B}_{2}{B}_{4} \) with line \( {A}_{1}R \) (see Figure 20). Draw from \( T \) line \( {TV} \) parallel to side \( {A}_{1}{A}_{2} \) intersecting side \( {A}_{3}{A}_{4} \) at \( \mathrm{U} \) . Since the bundle \( F\left( {V, T, U,{A}_{1}}\right) \) is harmonic an... | Yes |
Proposition 1. Two points \( M \) and \( N \) on the hypotenuse \( {BD} \) of the isosceles, right-angled triangle \( {ABD} \), with \( M \) between \( B \) and \( N \), define an angle \( \measuredangle {MAN} = {45}^{ \circ } \) if and only if \( B{M}^{2} + N{D}^{2} = M{N}^{2} \) (see Figure 1). | Proof. Let \( R \) be the midpoint of \( {BD} \) so that \( {AR} = {BR} = {BR} \), and \( {AR} \) is an altitude of triangle \( {ABD} \) . We assume \( {AR} = 1 \) and denote \( {RM} = x,{RN} = y \) (see Figure 3). Note that\n\n\[ \tan \left( {\measuredangle {MAN}}\right) = \tan \left( {\measuredangle {MAR} + \measured... | Yes |
Theorem 2. Let \( {ABCD} \) be a unit square. Two half-lines through \( A \) meet the diagonal \( {BD} \) at \( M \) and \( N \), and the sides \( {BC},{CD} \) at \( M \) and \( P \) and \( Q \) respectively (see Figure 2). Assume \( {AP} \neq {AQ} \) .\n\nThe following statements are equivalent:\n\n(i) \( \measuredang... | Proof of Theorem 2. With Cartesian coordinates \( A\left( {0,0}\right), B\left( {1,0}\right), C\left( {1,1}\right), D\left( {0,1}\right) \) and \( P\left( {1, a}\right), Q\left( {b,1}\right) \) for some distinct \( a, b \in \left( {0,1}\right) \), we have \( M\left( {\frac{1}{1 + a},\frac{a}{1 + a}}\right) \) and \( N\... | Yes |
Proposition 3. If \( M \) and \( N \) are points inside a square \( {ABCD} \) such that \( \measuredangle {MAN} = \) \( \measuredangle {MCN} = {45}^{ \circ } \), then \( M{N}^{2} = B{M}^{2} + N{D}^{2} \) (see Figure 8). | This situation can be viewed as a surprising extension from the case of triangle \( {ABD} \) in Figure 6 is distorted into the polygon \( {ABMND} \) . In fact, by considering the symmetric of triangle \( {ABD} \) with respect to hypotenuse \( {BD} \) in Figure 1 , a particular case of Proposition 3 is obtained. This an... | Yes |
Proposition 5. If \( M \) and \( N \) are points inside a square \( {ABCD} \) such that \( \measuredangle {MAN} = \) \( \measuredangle {MCN} = {45}^{ \circ } \), then\n\n\[ \n{S}_{MCN} + {S}_{MAB} + {S}_{NAD} = {S}_{MAN} + {S}_{MBC} + {S}_{NCD}.\n\] | Having at hand the previous construction from Figure 8A (where \( F \) is defined by the conditions \( {\Delta CND} \equiv {\Delta CNF} \) and \( {\Delta CMB} \equiv {\Delta CMF} \) ), we have\n\n\[ \n{S}_{MCN} + {S}_{MAB} + {S}_{NAD} = {S}_{MCN} + {S}_{AMEN} = {S}_{AMCN} + {S}_{MEN}.\n\]\n\nSimilarly, \( {S}_{MAN} + {... | Yes |
Proposition 7. Let \( {ABCD} \) be a rhombus. Two rays through \( A \) meet the diagonal \( {BD} \) at \( M, N \), and the sides \( {BC} \) and \( {CD} \) at \( P, Q \) respectively (see Figure 14). Then \( {AN} = {NP} \) if and only if \( {AM} = {MQ} \) . | Proof. The key idea is that the statements \( {AN} = {NP} \) and \( {AM} = {MQ} \) are equivalent to \( \measuredangle {PAQ} = \frac{1}{2}\measuredangle {ABC} \).\n\nFirst, if \( \measuredangle {PAQ} = \frac{1}{2}\measuredangle {ABC} \), then \( \measuredangle {NAP} = \measuredangle {NBP} \), and the quadrilateral \( {... | Yes |
Problem 50. Given points \( {E}_{a},{E}_{b}, O \) . | Solution. Let \( P \) and \( Q \) be the midpoints of \( {E}_{a}O \) and \( {E}_{b}O \), respectively. Let \( R \) be the reflection of \( P \) through \( Q \) . The line through \( {E}_{b} \), perpendicular to \( {E}_{a}{E}_{b} \) , intersects the circle with diameter \( {OR} \) at \( {M}_{a} \) . The circumcircle, wi... | Yes |
Problem 72. Given points \( {E}_{a},{H}_{a},{H}_{b} \) . | Solution. The line through Ha perpendicular to the line \( {E}_{a}{H}_{a} \) is the side \( {BC} \) . All three given points lie on the nine-point circle, so it can be found. The second intersection of the nine-point circle with BC gives \( {M}_{a} \) . The circle with \( {M}_{a} \) as center and passing through \( {H}... | No |
Problem 103. Given points \( {E}_{a},{M}_{b}, O \) . | Solution. The line through \( {M}_{b} \), perpendicular to \( {M}_{b}O \) is \( {AC} \) . Reflecting \( {AC} \) through \( {E}_{a} \), then dilating this line with \( O \) as center and ratio \( \frac{1}{2} \) and finally intersecting this new line with the perpendicular bisector of \( {E}_{a}{M}_{b} \) gives \( N \) .... | Yes |
Proposition 2. Let \( O \) be the point where the diagonals of the convex quadrilateral \( {ABCD} \) meet and \( {r}_{1},{r}_{2},{r}_{3} \), and \( {r}_{4} \) respectively the radii of the circles inscribed in the triangles \( {AOB} \) , \( {BOC} \) , \( {COD} \) and \( {DOA} \) respectively. The following statements a... | Proof. (a) \( \Leftrightarrow \) (b). The inradius of a triangle is related to the altitudes by the simple\nrelation\n\[ \frac{1}{r} = \frac{1}{{h}_{a}} + \frac{1}{{h}_{b}} + \frac{1}{{h}_{c}} \]\n\nApplying this to the four triangles \( {AOB},{BOC},{COD} \), and \( {DOA} \), we have\n\n\[ \frac{1}{{r}_{1}} = \frac{1}{... | Yes |
Theorem 1. For \( \varepsilon = \pm 1 \), the circumcircles of triangles \( {A}^{\prime }{B}_{\varepsilon }{C}_{\varepsilon },{B}^{\prime }{C}_{\varepsilon }{A}_{\varepsilon },{C}^{\prime }{A}_{\varepsilon }{B}_{\varepsilon } \) are concurrent at the anticomplement \( {P}_{-\varepsilon } \) of the Fermat point \( {F}_{... | ## 2. Proof of Theorem 1\n\nFor \( \varepsilon = \pm 1 \), let \( {O}_{a,\varepsilon } \) be the center of the equilateral triangle \( {A}_{\varepsilon }{BC} \) ; similarly for \( {O}_{b,\varepsilon } \) and \( {O}_{c,\varepsilon } \).\n\n(1) We first note that \( {O}_{a, - \varepsilon } \) is the center of the circle ... | No |
Theorem 2. For \( \varepsilon = \pm 1 \), the circumcircles of the triangles \( {A}_{\varepsilon }{B}^{\prime }{C}^{\prime },{B}_{\varepsilon }{C}^{\prime }{A}^{\prime } \) , \( {C}_{\varepsilon }{A}^{\prime }{B}^{\prime } \) are concurrent. | Proof. Consider the inversion \( \Psi \) with respect to the anticomplement of the second Fermat point. According to Theorem 1, the images of the circumcircles of triangles \( {A}^{\prime }{B}_{ + }{C}_{ + },{B}^{\prime }{C}_{ + }{A}_{ + },{C}^{\prime }{A}_{ + }{B}_{ + } \) are three lines which bound a triangle \( {A}... | Yes |
Lemma 1 (Gauss). \( 1 + 2\left( {\zeta + {\zeta }^{2} + {\zeta }^{4}}\right) = \sqrt{7}i \) . | Proof. Although this can be directly verified, it is actually a special case of Gauss' famous theorem that if \( \zeta = \cos \frac{2\pi }{n} + i\sin \frac{2\pi }{n} \) for an odd integer \( n \), then\n\n\[\n\mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{\zeta }^{{k}^{2}} = \left\{ \begin{array}{lll} \sqrt{n} & \text{ if }... | No |
Lemma 2. If \( \alpha ,\beta ,\gamma \) are unit complex numbers, the reflection of \( \gamma \) in the line joining \( \alpha \) and \( \beta \) is \( {\gamma }^{\prime } = \alpha + \beta - {\alpha \beta }\bar{\gamma } \) . | Proof. As points in the complex plane, \( {\gamma }^{\prime } \) has equal distances from \( \alpha \) and \( \beta \) as \( \gamma \) does. This is clear from\n\n\[ \n{\gamma }^{\prime } - \alpha = \beta \left( {1 - \alpha \bar{\gamma }}\right) = \beta \bar{\gamma }\left( {\gamma - \alpha }\right) \n\]\n\n\[ \n{\gamma... | Yes |
Theorem 4. The Simson lines of \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) with respect to the heptagonal triangle \( \mathbf{T} \) are concurrent. | Proof. The pedals of \( A \) on \( {BC} \) is the midpoint \( {A}^{\prime } \) of \( A{A}^{ * } \) ; similarly for those of \( B \) on \( {CA} \) and \( C \) on \( {AB} \) . We tabulate the coordinates of the pedals of \( {A}^{\prime },{B}^{\prime } \) , \( {C}^{\prime } \) on the sidelines \( {BC},{CA},{AB} \) respect... | Yes |
Theorem 5. The medial triangle and the orthic triangle of \( \mathbf{T} \) are companion heptagonal triangles on the nine-point circle of \( \mathbf{T} \) . The residual vertex is the Euler reflection point \( E \) (on the circumcircle of \( \mathbf{T} \) ). | Proof. (1) The companionship of the medial and orthic triangles on the nine-point circle is clear from the table below.\n\n\[ \text{Center:}\;N = \frac{1}{2}\left( {\zeta + {\zeta }^{2} + {\zeta }^{4}}\right) \]\n\n\[ \text{ Residual vertex: }E = \frac{f}{2}\left( {-1 + \zeta + {\zeta }^{2} + {\zeta }^{4}}\right) \]\n\... | Yes |
Lemma 6. The distance between the nine-point center \( N \) and the \( A \) -excenter \( {I}_{a} \) is equal to the circumradius of the heptagonal triangle \( \mathbf{T} \) . | Proof. Note that \( {I}_{a} - N = \frac{2 + \zeta + {\zeta }^{2} + {\zeta }^{4}}{2} = \frac{3 + 1 + 2\left( {\zeta + {\zeta }^{2} + {\zeta }^{4}}\right) }{4} = \frac{3 + \sqrt{7}i}{4} \) is a unit complex number. | No |
Proposition 7. (1) The midpoint \( {F}_{a} \) of \( N{I}_{a} \) is the point of tangency of the nine-point circle and the A-excircle. | Proof. (1) By the Feuerbach theorem, the nine-point circle is tangent externally to each of the excircles. Since \( N{I}_{a} = R \), the circumradius, and the nine-point circle has radius \( \frac{1}{2}R \), the point of tangency with the \( A \) -excircle is the midpoint of \( N{I}_{a} \) , i.e.,\n\n\[ \n{F}_{a} = \fr... | Yes |
Proposition 8. \( {F}_{\mathrm{e}},{F}_{a},{F}_{b},{F}_{c} \) are the points of tangency of the nine-point circle with the incircle and the \( A \) -, \( B \) -, \( C \) -excircles respectively (see Figure 7). | Proof. We have already seen that \( {F}_{a} = \frac{1}{2} \cdot N + \frac{1}{2} \cdot {I}_{a} \) . It is enough to show that the points \( {F}_{\mathrm{e}},{F}_{b},{F}_{c} \) lie on the lines \( {NI}, N{I}_{b}, N{I}_{c} \) respectively:\n\n\[ \n{F}_{\mathrm{e}} = - \left( {{c}_{1} - {c}_{3}}\right) \cdot N + \left( {-{... | Yes |
Proposition 9. The vertices \( {F}_{a}^{\prime },{F}_{b}^{\prime },{F}_{c}^{\prime } \) of the companion of \( {F}_{b}{F}_{\mathrm{e}}{F}_{c} \) are the second intersections of the nine-point circle with the lines joining \( {F}_{a} \) to \( A, B, C \) respectively. | Proof.\n\n\[ \n{F}_{a}^{\prime } = - 2{c}_{2} \cdot {F}_{a} - 2\left( {{c}_{1} + {c}_{3}}\right) A, \]\n\n\[ \n{F}_{b}^{\prime } = - 2{c}_{3} \cdot {F}_{a} - 2\left( {{c}_{1} + {c}_{2}}\right) B, \]\n\n\[ \n{F}_{c}^{\prime } = - 2{c}_{1} \cdot {F}_{a} - 2\left( {{c}_{2} + {c}_{3}}\right) C. \]\n | Yes |
Theorem 10. \( D \) is a Kiepert perspector of the heptagonal triangle \( {ABC} \) . | Proof. What this means is that there are similar isosceles triangles \( {A}^{\prime \prime }{BC},{B}^{\prime \prime }{CA} \) , \( {C}^{\prime \prime }{AB} \) with the same orientation such that the lines \( A{A}^{\prime \prime }, B{B}^{\prime \prime }, C{C}^{\prime \prime } \) all pass through the point \( D \) . Let \... | Yes |
Corollary 11. The center of the Kiepert hyperbola is the point\n\n\[ \n{K}_{\mathrm{i}} = - \frac{1}{2}\left( {{\zeta }^{3} + {\zeta }^{5} + {\zeta }^{6}}\right) \n\] | Proof. Since \( D \) is the intersection of the Kiepert hyperbola and the circumcircle, the center of the Kiepert hyperbola is the midpoint of \( {DH} \), where \( H \) is the orthocenter of triangle \( {ABC} \) (see Figure 9). This has coordinate as given in (6) above. | No |
Proposition 12 (Bankoff and Garfunkel). The nine-point center \( N \) is the first Brocard point. | Proof. The relations\n\n\[ \frac{1}{2}\left( {\zeta + {\zeta }^{2} + {\zeta }^{4}}\right) - {\zeta }^{4} = \frac{\left( {-2{c}_{1} - 3{c}_{2} - 2{c}_{3}}\right) \left( {4 + \zeta + {\zeta }^{2} + {\zeta }^{4}}\right) }{7} \cdot \left( {\zeta - {\zeta }^{4}}\right) ,\]\n\n\[ \frac{1}{2}\left( {\zeta + {\zeta }^{2} + {\z... | Yes |
Proposition 13. The symmedian point \( K \) has coordinate \( \frac{2\left( {1 + 2\left( {\zeta + {\zeta }^{2} + {\zeta }^{4}}\right) }\right) }{7} = \frac{2i}{\sqrt{7}} \) . | Proof. It is known that on the Brocard circle with diameter \( {OK},\angle {NOK} = - \omega \) . From this,\n\n\[ K = \frac{1}{\cos \omega }\left( {\cos \omega - i\sin \omega }\right) \cdot N \]\n\n\[ = \left( {1 - \frac{i}{\sqrt{7}}}\right) \cdot N \]\n\n\[ = \frac{2\left( {4 + {\zeta }^{3} + {\zeta }^{5} + {\zeta }^{... | Yes |
Corollary 14. The second Brocard point is the Kiepert center \( {K}_{\mathrm{i}} \) . | Proof. By Proposition 13, the Brocard axis \( {OK} \) is along the imaginary axis. Now, the second Brocard point, being the reflection of \( N \) in \( {OK} \), is simply \( - \frac{1}{2}\left( {{\zeta }^{3} + }\right. \) \( {\zeta }^{5} + {\zeta }^{6} \) ). This, according to Corollary 11, is the Kiepert center \( {K}... | Yes |
Proposition 15. The first Brocard triangle is perspective with \( {ABC} \) at the point \( - \frac{1}{2} \) (see Figure 12). | Proof.\n\n\[ - \frac{1}{2} = \left( {-3{c}_{1} - 2{c}_{2} - 2{c}_{3}}\right) \cdot {A}_{-\omega } + {c}_{1} \cdot {\zeta }^{4}, \]\n\n\[ = \left( {-2{c}_{1} - 3{c}_{2} - 2{c}_{3}}\right) \cdot {B}_{-\omega } + {c}_{2} \cdot \zeta \]\n\n\[ = \left( {-2{c}_{1} - 2{c}_{2} - 3{c}_{3}}\right) \cdot {C}_{-\omega } + {c}_{3} ... | Yes |
Proposition 17. The orthocenter of the heptagonal triangle BCI lies on the line \( {OC} \) and the perpendicular from \( {C}_{I} \) to \( {AC} \) . | Proof. Since \( {ABC} \) has orthocenter \( H = \zeta + {\zeta }^{2} + {\zeta }^{4} \), the orthocenter of triangle \( {BCI} \) is the point\n\n\[ \n{H}^{\prime } = {f}_{1}\left( H\right) = - \left( {1 + {\zeta }^{4}}\right) = - \left( {{\zeta }^{2} + {\zeta }^{5}}\right) {\zeta }^{2}.\n\]\n\nThis expression shows that... | Yes |
Proposition 18. The nine-point circles of \( {AC}{C}_{I} \) and (the isosceles triangle) \( {B}^{\prime }{A}^{\prime }C \) are tangent internally at the midpoint of \( {B}^{\prime }C \) . | Proof. The nine-point circle of the isosceles triangle \( {B}^{\prime }{A}^{\prime }C \) clearly contains the midpoint \( M \) of \( {B}^{\prime }C \) . Since triangle \( A{B}^{\prime }C \) is also isosceles, the perpendicular from \( A \) to \( {B}^{\prime }C \) passes through \( M \) . This means that \( M \) lies on... | Yes |
Theorem 19. The following circles have a common point.\n\n(i) the circumcircle of \( {AC}{C}_{I} \) ,\n\n(ii) the nine-point circle of \( {AC}{C}_{I} \) ,\n\n(iii) the A-excircle of \( {AC}{C}_{I} \) ,\n\n(iv) the nine-point circle of \( B{B}_{I}C \) . | Proof. By Proposition 7(3), the first three circles concur at the \( A \) -Feuerbach point of triangle \( {AC}{C}_{I} \), which is the point\n\n\[ \n{f}_{2}\left( {F}_{a}\right) = \frac{1}{4}\left( {\zeta + 2{\zeta }^{2} + {\zeta }^{4} - {\zeta }^{5} + {\zeta }^{6}}\right) .\n\]\n\nIt is enough to verify that this poin... | Yes |
Lemma 21. The perpendicular bisector of the segment \( {ON} \) is the line containing \( X = - 1 \) and \( Y = \frac{1}{2}\left( {1 - \left( {{\zeta }^{3} + {\zeta }^{5} + {\zeta }^{6}}\right) }\right) \) . | Proof. (1) Complete the parallelogram \( O{I}_{a}{HX} \), then\n\n\[X = O + H - {I}_{a} = \left( {\zeta + {\zeta }^{2} + {\zeta }^{4}}\right) + \left( {{\zeta }^{3} + {\zeta }^{5} + {\zeta }^{6}}\right) = - 1\]\n\nis a point on the circumcircle. Note that \( N \) is the midpoint of \( {I}_{a}X \) . Thus, \( {NX} = \) \... | Yes |
Theorem 22. The circumcenter and the Fermat points of the heptagonal triangle \( \mathbf{T} \) form an equilateral triangle. | Proof. (1) Consider the circle through \( O \), with center at the point\n\n\[ L \mathrel{\text{:=}} - \frac{1}{3}\left( {{\zeta }^{3} + {\zeta }^{5} + {\zeta }^{6}}\right) \]\n\nThis is the center of the equilateral triangle with \( O \) as a vertex and \( {K}_{\mathrm{i}} = - \frac{1}{2}\left( {{\zeta }^{3} + }\right... | Yes |
Theorem 1. Let \( P \) be a triangle center of \( {ABC} \), and \( {P}_{a},{P}_{b},{P}_{c} \) the corresponding centers in triangles \( A{B}_{a}{C}_{a}, B{C}_{b}{A}_{b}, C{A}_{c}{B}_{c} \), which have centroids \( {G}_{a},{G}_{b},{G}_{c} \) respectively. The lines \( {G}_{a}{P}_{a},{G}_{b}{P}_{b},{G}_{c}{P}_{c} \) inte... | (i) The isogonal conjugate of the infinite point of the line \( {GP} \) is the point\n\n\[ \n{Q}^{\prime } = \left( {\frac{{a}^{2}}{-{2u} + v + w} : \frac{{b}^{2}}{u - {2v} + w} : \frac{{c}^{2}}{u + v - {2w}}}\right) \n\]\n\non the circumcircle.\n\n(ii) The lines \( {G}_{a}{P}_{a},{G}_{b}{P}_{b},{G}_{c}{P}_{c} \) inter... | Yes |
Proposition 2. The three lines \( {G}_{a}{P}_{a},{G}_{b}{P}_{b},{G}_{c}{P}_{c} \) are concurrent if and only if\n\n\[ f\left( {u, v, w}\right) {\left( x + y + z\right) }^{2}\left( {{b}^{2}{c}^{2}\left( {v - w}\right) x + {c}^{2}{a}^{2}\left( {w - u}\right) y + {a}^{2}{b}^{2}\left( {u - v}\right) z}\right) = 0, \]\n\nwh... | Computing the distance between \( G \) and \( P \), we obtain\n\n\[ f\left( {u, v, w}\right) = 9{\left( u + v + w\right) }^{2} \cdot G{P}^{2}. \]\n\nThis is nonzero for \( P \neq G \) . From this we obtain the following theorem. | Yes |
For a fixed point \( P = \left( {u : v : w}\right) \), the locus of a point \( T \) for which the \( {GP} \) -lines of triangles \( A{B}_{a}{C}_{a},{A}_{b}B{C}_{b} \), and \( {A}_{c}{B}_{c}C \) are concurrent is the line | \[ {b}^{2}{c}^{2}\left( {v - w}\right) \mathbb{X} + {c}^{2}{a}^{2}\left( {w - u}\right) \mathbb{Y} + {a}^{2}{b}^{2}\left( {u - v}\right) \mathbb{Z} = 0. \] | Yes |
In any triangle \( {ABC} \) with \( {H}_{a} \) not coinciding with \( {M}_{a} \), the line \( {M}_{a}P \) is parallel to the angle bisector \( A{T}_{a} \) . | Proof. Let \( O \) be the circumcenter of \( {ABC} \) (see Figure 2). The perpendicular bisector \( {M}_{a}O \) and the angle bisector \( A{T}_{a} \) intersect the circumcircle \( \left( O\right) \) at \( S \) . Let the midpoint of \( {E}_{a}O \) be \( R \), and reflect the entire figure through \( R \) . Let the refle... | Yes |
Problem 99. Given \( {E}_{a},{M}_{a} \) and \( {T}_{a} \) construct triangle \( {ABC} \) . | Solution. Draw line \( {M}_{a}{T}_{a} \) containing the side \( {BC} \) and then the altitude \( {E}_{a}{H}_{a} \) to this side (see Figure 3). The circle with center \( {E}_{a} \) and passing through \( {M}_{a} \) intersects the altitude at \( P \) . Draw \( {M}_{a}P \) . By Theorem 1, the line through \( {T}_{a} \) p... | Yes |
Problem 137. Given \( {M}_{a}, N \) and \( {T}_{a} \) construct triangle \( {ABC} \) . | Solution. Since \( N \) is the midpoint of \( {E}_{a}{M}_{a} \), both of these problems reduce to Problem 99. | No |
Problem 130. Given \( {H}_{a}, N \) and \( {T}_{a} \) construct triangle \( {ABC} \) . | Solution. The nine-point circle, with center \( N \) and passing through \( {H}_{a} \), intersects line \( {H}_{a}{T}_{a} \) again at \( {M}_{a} \), also reducing this problem to Problem 99. | No |
Proposition 1. The centers of the circles lie on the parabola with axis along \( {AB} \), focus at \( C \), and vertex the midpoint of \( {OB} \). | Proof. Consider a circle of the chain with center \( {O}^{\prime }\left( {x, y}\right) \), radius \( r \), tangent to the arc \( {GBH} \) at \( Q \). Since the segment \( {CQ} \) contains \( {O}^{\prime } \) (see Figure 2), we have, by taking into account that \( C \) has coordinates \( \left( {b - a,0}\right) \) and\n... | Yes |
Proposition 2. The points of tangency between consecutive circles of the chain lie on the circle with center \( A \) and radius \( {AG} \) . | Proof. Consider two neighboring circles with centers \( \left( {{X}_{i},{Y}_{i}}\right) ,\left( {{X}_{i + 1},{Y}_{i + 1}}\right) \), radii \( {r}_{i},{r}_{i + 1} \) respectively, tangent to each other at \( {T}_{i} \) (see Figure 3). By using Proposition 1 and noting that \( A \) has coordinates \( \left( {-{2a},0}\rig... | Yes |
Proposition 3. If a circle of the chain touches the chord \( {GH} \) at \( P \) and the arc \( {GBH} \) at \( Q \), then the points \( A, P, Q \) are collinear. | Proof. Suppose the circle has center \( {O}^{\prime } \) . It touches \( {GH} \) at \( P \) and the arc \( {GBH} \) at \( Q \) (see Figure 4). Note that triangles \( {CAQ} \) and \( {O}^{\prime }{PQ} \) are isosceles triangles with \( \angle {ACQ} = \angle P{O}^{\prime }Q \) . It follows that \( \angle {CQA} = \angle {... | No |
Theorem 1. (Inversive Invariants). The parameter \( {\epsilon }_{ij} \) is invariant under inversion in any circle whose center does not lie on either of the two given circles. | Proof. The circle \( {\mathcal{C}}_{0}\left( {{x}_{0},{y}_{0};{r}_{0}}\right) \) inverts \( \mathcal{C}\left( {x, y;r}\right) \) to \( {\mathcal{C}}^{\prime }\left( {{x}^{\prime },{y}^{\prime };{r}^{\prime }}\right) \), where if \( d \) is the\n\nEuclidean distance between the centers of \( \mathcal{C} \) and \( {\math... | Yes |
Lemma 2 (Second tangency point). If \( P\left( {{x}_{0},{y}_{0}}\right) \) is a point on a circle \( {\mathcal{C}}_{i} \) while \( {\mathcal{C}}_{j} \) is a second circle, then there exists exactly one circle \( {\mathcal{C}}_{a}\left( {{x}_{a},{y}_{a};{r}_{a}}\right) \) that is homogeneously tangent to \( {\mathcal{C}... | Proof. (Outline) \( {}^{2} \) The two tangency points \( P \) and \( {P}^{\prime } \) are collinear with a center of similitude \( {S}_{ij} \), which will be external or internal according as the radii have the same or different signs [7, Art. 117, p. 108]. It is then sufficient to find the intersections of \( {S}_{ij}... | Yes |
Theorem 3 (Apollonius Closure). Let \( {\mathcal{C}}_{1},{\mathcal{C}}_{2} \), and \( {\mathcal{C}}_{3} \) be three circles in the Circle Plane, and choose a point \( {P}_{1} \) on \( {\mathcal{C}}_{1} \) . Define \( {\mathcal{C}}_{12} \) to be the unique circle homogeneously tangent to \( {\mathcal{C}}_{1} \) at \( {P... | ## Proof. \( {}^{3} \) We first show that four consecutive \( {P}_{i} \) ’s lie on a circle, taking \( {P}_{1},{P}_{2},{P}_{3} \) , \( {P}_{4} \) as a typical example. See Figure 4.\n\n--- \n\n\( {}^{3} \) Rigby provides two proofs of this theorem in [6]. Searby independently rediscovered the result around 1987; he sho... | Yes |
Theorem 6 (Six-point Pencil). The equations (2) represent the complete set of six-point circles, which is part of a pencil of circles whose radical axis is \( \sigma \) . When the pencil consists of intersecting circles, \( \sigma \) might itself be a six-point circle. \( {}^{4} \) | Proof. \( {S}_{12} \) has the same power \( {}^{5} \), namely \( \frac{{e}_{12}{r}_{1}{r}_{2}}{{r}_{1}^{2} - {r}_{2}^{2}} \), with respect to all circles tangent to \( {\mathcal{C}}_{1} \) and \( {\mathcal{C}}_{2} \) ; but, for any point \( {P}_{1} \in {\mathcal{C}}_{1} \), the quantity \( {S}_{12}{P}_{1} \times {S}_{1... | No |
Theorem 8 (Gergonne-Desargues). For any given triple of circles, the six tangency points of a pair of Apollonius Circles, the three centers of similitude \( {S}_{ij} \), and the radical center \( {C}_{R} \) are ten points of a Desargues Configuration. | Proof. We should mention for completeness that by Gergonne’s construction \( {}^{6} \), the poles \( \left( {{x}_{i}^{\prime },{y}_{i}^{\prime }}\right) \) of \( \sigma \) with respect to \( {\mathcal{C}}_{i} \) are collinear with the radical center \( {C}_{R} \) and the tangency points of the two Apollonius Circles wi... | Yes |
Theorem 9 (Brianchon). The inverse image of \( \sigma \cap \alpha \) in \( \mathcal{S} \) is the Brianchon Point of the hexagon circumscribing \( \mathcal{S} \) and tangent to it at the six points where it intersects the given circles \( {\mathcal{C}}_{i} \), taken in the order indicated by the labels. | There is no need for any calculation here: Theorem 9 is the projective dual of Theorem 7 - the polarity defined by \( \mathcal{S} \) takes each point \( {P}_{i} \) to the line tangent there to \( \mathcal{S} \), while (because \( \sigma \bot \alpha \) ) it interchanges the axis of similitude \( \sigma \) with the inver... | No |
Theorem 4 (Dissecting arbitrary polygons). Every convex n-gon can be partitioned into \( 3\left( {n - 2}\right) \) cyclic kites (see Figure 5). | Proof. Notice first that every triangle can be dissected into three cyclic kites by dropping the radii from the incenter to the tangency points (see Figure 5 a). Partition the given \( n \) -gon into triangles. For instance, one can do this by drawing all the diagonals from a certain vertex. We obtain a triangulation c... | Yes |
Theorem 5. Suppose a cyclic quadrilateral \( Q \) has an \( m \) -by- \( n \) grid partition into mn cyclic quadrilaterals. Then:\na) If \( m \) and \( n \) are both even, \( Q \) is necessarily a rectangle.\nb) If \( m \) is odd and \( n \) is even, \( Q \) is necessarily an isosceles trapezoid. | We leave the easy proof for the reader. | No |
Theorem 7. (A class of cyclic quadrilaterals which have 3-by-3 grid dissections) Let \( {ABCD} \) be a cyclic quadrilateral such that the measure of each of the arcs \( \overset{⏜}{AB},\overset{⏜}{BC},\overset{⏜}{CD} \) and \( \overset{⏜}{DA} \) determined by the vertices on the circumcircle is greater than \( {60}^{ \... | Proof. Notice that the condition regarding the arc measures is stronger than the requirement that all angles of \( {ABCD} \) exceed \( {60}^{ \circ } \) . We will use the same assumptions and notations as in Theorem 6. The idea is to overlay the two constructions in Theorem 6 (see Figure 10).\n\nIt is straightforward t... | Yes |
Theorem 1. With reference to a given a triangle \( {ABC} \), a point \( P \) specified by the oriented angles\n\n\[ x = \measuredangle {BPC},\;y = \measuredangle {CPA},\;z = \measuredangle {APB}, \]\n\nhas homogeneous barycentric coordinates\n\n\[ \left( {\frac{1}{\cot A - \cot x} : \frac{1}{\cot B - \cot y} : \frac{1}... | Proof. Construct the circle through \( B, P, C \), and let it intersect the line \( {AP} \) at \( {A}^{\prime } \) (see Figure 2). Clearly, \( \angle {A}^{\prime }{BC} = \angle {A}^{\prime }{PC} = \pi - \angle {CPA} = \pi - y \) and similarly, \( \angle {A}^{\prime }{CB} = \pi - z \) . It follows from Conway’s formula ... | Yes |
Corollary 2 (Schaal). If for three points \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) the directed angles \( x = \) \( \left( {{A}^{\prime }B,{A}^{\prime }C}\right), y = \left( {{B}^{\prime }C,{B}^{\prime }A}\right) \) and \( z = \left( {{C}^{\prime }A,{C}^{\prime }B}\right) \) satisfy \( x + y + z \equiv 0{\;\oper... | Proof. Referring to Figure 2, if the circumcircles of triangles \( {A}^{\prime }{BC} \) and \( {B}^{\prime }{CA} \) intersect at \( P \), then from concyclicity,\n\n\[ \left( {{PB},{PC}}\right) = \left( {{A}^{\prime }B,{A}^{\prime }C}\right) = x, \]\n\n\[ \left( {{PC},{PA}}\right) = \left( {{B}^{\prime }C,{B}^{\prime }... | Yes |
Given a reference triangle \( {ABC} \) and two points \( P \) and \( Q \), let \( {R}_{a} \) be the intersection of the reflections of the lines \( {BP},{CP} \) in the lines \( {BQ},{CQ} \) respectively (see Figure 3). Similarly define the points \( {R}_{b} \) and \( {R}_{c} \) . The circumcircles of triangles \( {R}_{... | Proof. Let \( {x}^{\prime \prime } = \left( {{R}_{a}B,{R}_{a}C}\right) \) . Note that\n\n\[ {x}^{\prime \prime } = \left( {{R}_{a}B,{QB}}\right) + \left( {{QB},{QC}}\right) + \left( {{QC},{R}_{a}C}\right) \]\n\n\[ = \left( {{QB},{QC}}\right) + \left( {{R}_{a}B,{QB}}\right) + \left( {{QC},{R}_{a}C}\right) \]\n\n\[ = \le... | Yes |
Corollary 4. The mapping \( f \) preserves isogonal conjugation, i.e., \[ {f}^{ * }\left( {P, Q}\right) = f\left( {{P}^{ * },{Q}^{ * }}\right) . \] | Proof. If the points \( P \) and \( Q \) are defined by the directed angles in (3), and \( R = \) \( f\left( {P, Q}\right), S = f\left( {{P}^{ * },{Q}^{ * }}\right) \), then by Corollary \( 3,\left( {{R}^{ * }B,{R}^{ * }C}\right) = A - \left( {2{x}^{\prime } - x}\right) \) and \[ \left( {{SB},{SC}}\right) \equiv 2\left... | Yes |
Proposition 3. Given a conic homography \( f \) of the conic \( c \) with homograpy axis \( e \) , the transformation \( f \) preserves every member \( {c}^{\prime } = {\alpha c} + \beta {e}^{2} \) of the pencil generated by \( c \) and the (double) line e. The pole \( E \) of the axis \( e \) with respect to \( c \) i... | To prove the claims show first that line \( e \) is preserved by \( f \) (see Figure 3). For this take on \( c \) points \( A, B \) collinear with the pole \( E \) and consider their images \( {A}^{\prime } = f\left( A\right) ,{B}^{\prime } = f\left( B\right) \) . Since \( {AB} \) contains the pole of \( e \), the pole... | Yes |
Proposition 6. For every non-involutive conic homography \( f \) of a conic \( c \) and every point \( P \in c \) and \( {P}^{\prime } = f\left( P\right) \) lines \( P{P}^{\prime } \) envelope another conic \( {c}^{\prime } \) . Conic \( {c}^{\prime } \) is a member of the associated to \( f \) bitangent pencil. Invers... | An elegant proof of these statements up to the last is implied by a proposition proved in \( \left\lbrack {8,\mathrm{p}{.253}}\right\rbrack \), see also \( \left\lbrack {4\text{, vol.II, p.214}}\right\rbrack \) and \( \left\lbrack {5\text{, p.245}}\right\rbrack \) . Last statement follows from the fact that \( Q \) is ... | No |
Proposition 7. Every bitangent pencil of conics is projectively equivalent to one generated by a fixed conic \( c \) and a fixed line \( e \) in one of the following three possible configurations.\n\n(I) The line e non-intersecting the conic \( c \) (elliptic).\n\n(II) The line \( e \) intersecting the conic \( c \) at... | The proof follows by reducing each case to a kind of normal form. For case (I) select a projective basis \( A, B, C \) making a self-polar triangle with respect to \( c \) . For this take \( A \) to be the pole of \( e \) with respect to \( c \), take then \( B \) arbitrary on line \( e \) and define \( C \) to be the ... | Yes |
Proposition 8. Every homography preserving both, a conic c, an intersecting the conic line \( e \), and fixing one (A say) of the two intersection points \( A, B \) of \( c \) and \( e \) belongs to a group \( {\mathcal{G}}_{AB}\left( {c, e}\right) \) of homographies, which is isomorphic to the multiplicative group \( ... | Figure 6 illustrates the proof. Assume that homography \( f \) preserves both, the conic \( c \), the line \( e \), and also fixes \( A \) . Then it fixes also the other point \( B \) and also the pole \( C \) of line \( {AB} \) . Consequently \( f \) is uniquely determined by prescribing its value \( f\left( D\right) ... | Yes |
Proposition 9. For every normal point \( D \) of the conic \( c \) the isotropy group \( {\mathcal{G}}_{D}\left( {c, e}\right) \) is isomorphic to \( {\mathbb{Z}}_{2} \) . The different from the identity element of this group is the involution \( {I}_{D} \) with axis \( {DE} \) . | For types (I) and (II) of pencils a proof is the following. Let the homography \( f \) preserve the conic \( c \), the line \( e \) and fix point \( D \) . Then it preserves also the tangent \( {t}_{D} \) at \( D \) and consequently fixes also the intersection point \( A \) of this line with the axis \( e \) (see Figur... | Yes |
Proposition 10. For every point \( F \) not lying on the conic \( c \) and not lying on the side-lines of the invariant triangle \( {ABC} \) the orbit \( {\mathcal{G}}_{AB}F \) is the member conic \( {c}_{F} \) of the hyperbolic bitangent pencil \( \left( {c, e}\right) \) which passes through \( F \) . | In fact, \( {\mathcal{G}}_{AB}F \subset {c}_{F} \) since all \( f \in {\mathcal{G}}_{AB} \) preserve the member-conics of the pencil (see Figure 6). By the continuity of the action the two sets must then be identical. The second result that comes as byproduct is the one suggested by Figure 8. In its formulation as well... | No |
Proposition 11. For every point \( F \) not lying on the conic \( c \) and not lying on the side-lines of the invariant triangle \( {ABC} \), the intersection point \( H \) of lines \( {DE} \) and \( {FG} \), where \( G = {f}_{DE}\left( F\right) \), as \( E \) varies on the conic \( c \), describes a conic passing thro... | To prove this consider the projective basis and the matrix representation of \( {f}_{DE} \) given above. It is easy to describe in this basis the map sending line \( {DE} \) to \( {FG} \) . Indeed let \( E\left( {x, y, z}\right) \) be a point on the conic. Line \( {DE} \) has coefficients \( \left( {y - z, z - x, x - y... | Yes |
Proposition 12. For every point \( F \) not lying on the conic \( c \) and not lying on the side-lines of the invariant triangle \( {ABC} \), lines \( {EG} \) with \( G = {f}_{DE}\left( F\right) \) as \( E \) varies on \( c \) envelope a conic which belongs to the bitangent pencil \( \left( {c, e = {AB}}\right) \) . | The proof can be based on the dual of the argument of Chasles-Steiner ([3, p.89]), according to which the lines joining homologous points of a homographic relation between two ranges of points envelope a conic. Here lines \( {EG} \) (see Figure 9) join points \( \left( {x, y, z}\right) \) on the conic \( c \) with poin... | Yes |
Proposition 13. The group \( {\mathcal{G}}_{A}^{0} \) including the identity and all homographies \( f \) , which preserve a conic \( c \) and have \( A \) as a unique fixed point, is isomorphic to the additive group \( \mathbb{R} \) . Every non-indetity homography in this group induces in the tangent \( e \) at \( A \... | In fact consider the induced Moebius transformation on line \( e \) with respect to coordinates with origin at \( A \) (see Figure 10). Since \( A \) is a fixed point this transformation will have the form \( {x}^{\prime } = {ax}/\left( {{bx} + c}\right) \) . Since this is the only root of the equation \( x\left( {{bx}... | Yes |
Proposition 14. The isotropy group \( {\mathcal{G}}_{A} \) of conic homographies fixing a point \( A \) of the conic \( c \) is the semi-direct product of its subgroups \( {\mathcal{G}}_{A}^{0} \) of all homographies \( f \) which preserve \( c \) and have the unique fixed point \( A \) on \( c \) and the subgroup \( {... | To prove this apply the criterion ([1, p.285]) by which such a decomposition of the group is a consequence of the following two properties: (i) Every element \( g \) of the group \( {\mathcal{G}}_{A} \) is expressible in a unique way as a product \( g = {g}_{B} \circ {g}_{A} \) with \( {g}_{A} \in {\mathcal{G}}_{A}^{0}... | Yes |
Proposition 15. Only the elliptic bitangent pencils have homographies periodic of period \( n > 2 \) . Inversely, if a conic homography is periodic, then it is elliptic. | In fact, in the case of elliptic pencils, selecting the homography on \( e \) to be of the kind\n\n\[ \n{x}^{\prime } = \frac{\cos \left( \phi \right) x - \sin \left( \phi \right) }{\sin \left( \phi \right) x + \cos \left( \phi \right) },\;\phi = \frac{2\pi }{n},\n\]\n\nwe define by the procedure described above an \( ... | Yes |
Proposition 16. The group \( \mathcal{G}\left( {c, e}\right) \) of all homographies preserving a non-parabolic bitangent pencil is a subgroup of the group of homographies of the plane preserving line \( e \), fixing the center \( E \) of the pencil and commuting with involution \( {I}_{0} \) . | For the rest of the section I omit the reference to \( \left( {c, e}\right) \) and write simply \( \mathcal{G} \) instead of \( \mathcal{G}\left( {c, e}\right) \) . Involution \( {I}_{0} \) is a singularium and should be excluded from the set of all other involutions. It can be represented in infinite many ways as a pr... | No |
Proposition 17. For non-parabolic pencils two involutions \( I,{I}^{\prime } \) commute, if and only if their product is \( {I}_{0} \) . Further if the product of two involutions is an involution, then this involution is \( {I}_{0} \) . For parabolic pencils \( I \circ {I}^{\prime } \) is never commutative. | For the first claim notice that \( {I}^{\prime } \circ I = {I}_{0} \) implies \( {I}^{\prime } = {I}_{0} \circ I = I \circ {I}_{0} \) . Last because every element of \( \mathcal{G} \) commutes with \( {I}_{0} \) . Last equation implies \( I \circ {I}^{\prime } = \) \( {I}^{\prime } \circ I \) . Inversely, if last equat... | Yes |
Proposition 18. The automorphism group \( \mathcal{G} \) of a pencil \( \left( {c, e}\right) \) is the union of two cosets \( \mathcal{G} = {\mathcal{G}}^{\prime } \cup {\mathcal{G}}^{\prime \prime } \) . \( {\mathcal{G}}^{\prime } \) consists of the non-involutive automorphisms (and \( {I}_{0} \) for non-parabolic pen... | In fact, given an involutive \( I \in {\mathcal{G}}^{\prime \prime } \) and a non-involutive \( f \in {\mathcal{G}}^{\prime } \), we can, according to Proposition 5, represent \( f \) as a product \( f = I \circ {I}^{\prime } \) using involution \( I \) and another involution \( {I}^{\prime } \) completely determined b... | Yes |
Proposition 19. The subgroup \( {\mathcal{G}}^{\prime } \subset \mathcal{G} \) of non-involutive automorphisms of the bitangent pencil \( \left( {c, e}\right) \) is commutative. | The proof can be given on the basis of Figure 13, illustrating the case of elliptic pencils, the arguments though being valid also for the other types of pencils. In this figure the two products \( f \circ g \) and \( g \circ f \) of two non-involutive automorphisms of the pencil \( f \in {\mathcal{G}}^{\prime } \) and... | Yes |
Proposition 22. For every non-parabolic pencil the correspondence \( \mathcal{J} : Q \mapsto F \) between the centers of the involutions \( I \) and \( I \circ {I}_{0} \) defines an involutive homography on line e. The fixed points of \( \mathcal{J} \) coincide with the intersection points \( \{ A, B\} = \) \( c \cap e... | In fact considering the pencil \( {E}^{ * } \) of lines through \( E \) it is easy to see that the correspondence \( \mathcal{J} : F \mapsto Q \) (see Figure 14) is projective and has period two. The identification of the fixed points of \( \mathcal{J} \) with \( \{ A, B\} = c \cap e \) is equally trivial. | Yes |
Proposition 23. The automorphism group \( \mathcal{G}\left( {c, e}\right) \) of a non-parabolic pencil is uniquely determined by the triple\n\n\[ \left( {e, E,\mathcal{J}}\right) \]\n\nconsisting of a line \( e \) a point \( E \notin e \) and an involutive homography on line \( e \) . | In fact \( \mathcal{J} \) completely determines the involutive automorphisms \( {I}_{Q} \) of the pencil, since for each point \( Q \) on \( e \) point \( F = \mathcal{J}\left( Q\right) \) defines the axis \( {FE} \) of the involution \( {I}_{Q} \) . The involutive automorphisms in turn, through their compositions, det... | Yes |
Proposition 24. Given a line \( e \) and a point \( E \) consider a projective map \( \mathcal{I} : e \mapsto \) \( {E}^{ * } \) of the line onto the pencil \( {E}^{ * } \) of lines through \( E \) . Let \( {e}^{\prime } \) denote the complement in \( e \) of the set \( {e}^{\prime \prime } = \{ Q \in e : Q \in \mathca... | In fact, by the Chasles-Steiner construction method of conics ([3, p.73]), lines \( {XQ} \) and \( \mathcal{I}\left( Q\right) \) intersect at a point \( P \) describing a conic \( {c}^{\prime } \), which passes through \( X \) and \( E \) . Every point \( Q \in {e}^{\prime \prime } \) i.e. satisfying \( Q \in \mathcal{... | Yes |
Proposition 25. The conics generated by the previous method belong to a bitangent pencil with axis \( e \) and center \( E \) if and only if they are invariant by all involutions \( {I}_{Q} \) for \( Q \in {e}^{\prime } \) . The points in \( {e}^{\prime \prime } \) are the fixed points of the pencil. | The necessity of the condition is a consequence of Proposition 21. To prove the sufficiency assume that \( c \) is invariant under all \( \left\{ {{I}_{Q} : Q \in {e}^{\prime }}\right\} \) . Then for every \( Q \in {e}^{\prime } \) line \( \mathcal{I}\left( Q\right) \) is the polar of \( Q \) with respect to \( c \) . ... | Yes |
Proposition 29. The set of homologies having in common the axis \( e \) and the center \( E \) builds a commutative group \( \mathcal{K} \) which is isomorphic to the multiplicative group of real numbers. | That the composition \( h = g \circ f \) of two homologies with the previous characteristics is a homology follows directly from their definition. The homology coefficients multiply homomorphically \( {\kappa }_{h} = {\kappa }_{g}{\kappa }_{f} \), this being a consequence of Proposition 27 and the well-known identity f... | No |
Proposition 31. For every bitangent pencil \( \left( {c, e}\right) \) the elements of \( \mathcal{K}\left( {c, e}\right) \) commute with those of \( \mathcal{G}\left( {c, e}\right) \) . | The proposition is easily proved first for involutive automorphisms of the pencil, characterized by having their centers \( Q \) on the perspectivity axis \( e \) and their axis \( q \) passing through the perspectivity center \( E \) . Figure 17 suggests the proof of the commutativity of such an involution \( {f}_{Q} ... | Yes |
Proposition 32. For non-hyperbolic pencils and every two member-conics \( \left( {c,{c}^{\prime }}\right) \) of the pencil there is a perspectivity with center at \( E \) and axis the line \( e \), which maps \( c \) to \( {c}^{\prime } \) . For hyperbolic pencils this is true if \( c \) and \( {c}^{\prime } \) belong ... | To prove the claim consider a line through \( E \) intersecting two conics of the pencil at points \( P \in c,{P}^{\prime } \in {c}^{\prime } \) (see Figure 18). By Proposition 27 there is a perspectivity \( f \) mapping \( P \) to \( {P}^{\prime } \) . By the previous proposition \( f \) commutes with all \( g \in \ma... | Yes |
Proposition 33. Let \( f \) be a non-involutive automorphism of the pencil \( \left( {c, e}\right) \) and \( {c}^{\prime } \) be the member-conic determined by its tangential generation with respect to \( c \) (proposition-6). Then, for non-hyperbolic pencils there is a perspectivity \( {p}_{f} \in \mathcal{K} \) mappi... | In fact, given \( f \in \mathcal{G} \), according to proposition-6, there is a conic \( {c}^{\prime } \) of the bitangent pencil such that lines \( P{P}^{\prime },{P}^{\prime } = f\left( P\right) \) are tangent to \( {c}^{\prime } \) . The exceptional case for pencils of type (II) occurs when \( f \) interchanges the t... | Yes |
Lemma 2. The point \( X \) has coordinates\n\n\[ \left( {0 : s\sqrt{s - c} - \left( {s - a}\right) \sqrt{s - b} : s\sqrt{s - b} - \left( {s - a}\right) \sqrt{s - c}}\right) . \] | Proof. If \( \left( {O}_{1}\right) \) is the circle tangent to \( \left( {I}_{b}\right) ,\left( {I}_{c}\right) \), and to \( {BC} \) at \( X \) between \( {A}_{c} \) and \( {A}_{b} \), then \( {A}_{c}X : X{A}_{b} = \sqrt{{r}_{c}} : \sqrt{{r}_{b}} = \sqrt{s - b} : \sqrt{s - c} \) . Note that \( {A}_{c}{A}_{b} = b + c \)... | Yes |
Lemma 3. The polars of the vertices of \( {ABC} \) with respect to the corresponding excircles bound a triangle with vertices\n\n\[ U = \left( {-a\left( {b + c}\right) : {S}_{C} : {S}_{B}}\right) \]\n\n\[ V = \left( {{S}_{C} : - b\left( {c + a}\right) : {S}_{A}}\right) ,\]\n\n\[ W = \left( {{S}_{B} : {S}_{A} : - c\left... | Proof. The polar of \( A \) with respect to the excircle \( \left( {I}_{a}\right) \) is the line \( {B}_{a}{C}_{a} \), whose barycentric equation is\n\n\[ \left| \begin{matrix} x & y & z \\ - \left( {s - b}\right) & 0 & s \\ - \left( {s - c}\right) & s & 0 \end{matrix}\right| = 0 \]\n\nor\n\n\[ {sx} + \left( {s - c}\ri... | Yes |
Proposition 5. The triangle \( {UVW} \) has circumcenter \( H \) and circumradius \( {2R} + r \) . | Proof. Since \( H, B, V \) are collinear, \( {HV} \) is perpendicular to \( {CA} \) . Similarly, \( {HW} \) is perpendicular to \( {AB} \) . Since \( {VW} \) makes equal angles with \( {CA} \) and \( {AB} \), it makes equal angles with \( {HV} \) and \( {HW} \) . This means \( {HV} = {HW} \) . For the same reason, \( {... | Yes |
Proposition 6. The triangle UVW and the intouch triangle DEF are homothetic at the point\n\n\[ \nJ = \left( {\frac{b + c}{b + c - a} : \frac{c + a}{c + a - b} : \frac{a + b}{a + b - c}}\right) .\n\]\n\n(1)\n\nThe ratio of homothety is \( - \frac{{2R} + r}{r} \) . | Proof. The homothety follows easily from the parallelism of \( {VW} \) and \( {EF} \), and of \( {WU},{FD} \), and \( {UV},{DE} \) . The homothetic center is the common point \( J \) of the lines \( {DU},{EV} \), and \( {FW} \) (see Figure 5). These lines have equations\n\n\[ \n\left( {b - c}\right) \left( {b + c - a}\... | Yes |
Proposition 9. The radical center of \( \left( I\right) ,\left( {I}_{b}\right) ,\left( {I}_{c}\right) \) is the point\n\n\[ \n{J}_{a} = \left( {b + c : c - a : b - a}\right) .\n\]\n\nThis is also the midpoint of the segment DU. | Proof. The radical axis of \( \left( I\right) \) and \( \left( {I}_{b}\right) \) is the line joining the midpoints of the segments \( D{A}_{b} \) and \( F{C}_{b} \) . These midpoints have coordinates \( \left( {0 : a - c : a + c}\right) \) and \( \left( {c + a : c - a : 0}\right) \) . This line has equation\n\n\[ \n- \... | Yes |
Proposition 10. The triangle \( {J}_{a}{J}_{b}{J}_{c} \) is the image of the intouch triangle under the homothety \( \mathrm{h}\left( {J, - \frac{R}{r}}\right) \) . | Proof. Since \( {UVW} \) and \( {DEF} \) are homothetic at \( J \), and \( {J}_{a},{J}_{b},{J}_{c} \) are the midpoints of \( {DU},{EV},{FW} \) respectively, it is clear that \( {J}_{a}{J}_{b}{J}_{c} \) and \( {DEF} \) are also homothetic at the same \( J \) . Note that \( {J}_{b}{J}_{c} = \frac{1}{2}\left( {{VW} - {EF... | Yes |
Corollary 11. \( J \) is the radical center of the circles \( \left( {O}_{a}\right) ,\left( {O}_{b}\right) ,\left( {O}_{c}\right) \) . | Proof. Note that \( J{J}_{a} \cdot J{A}_{1} = \frac{R}{r} \cdot {DJ} \cdot J{A}_{1} \) . This is \( \frac{R}{r} \) times the power of \( J \) with respect to the incircle. The same is true for \( J{J}_{b} \cdot J{B}_{1} \) and \( J{J}_{c} \cdot J{C}_{1} \) . This shows that \( J \) is the radical center of the circles ... | Yes |
Proposition 12. \( {A}_{1}{B}_{1}{C}_{1} \) is perspective with \( {ABC} \) at the point\n\n\[ \nQ = \left( {\frac{1}{{a}^{2}\left( {s - a}\right) } : \frac{1}{{b}^{2}\left( {s - b}\right) } : \frac{1}{{c}^{2}\left( {s - c}\right) }}\right) ,\n\]\n\nwhich is the isotomic conjugate of the insimilicenter of the circumcir... | This is clear from the coordinates of \( {A}_{1},{B}_{1},{C}_{1} \) . The perspector \( Q \) is the iso-tomic conjugate of the insimilicenter of the circumcircle and the incircle. It is not in the current listing in [2]. | Yes |
Proposition 2. The perspector \( Q\left( P\right) \) coincides with \( P \) if and only if \( P \) is the orthocenter or the symmedian point. | Proof. The perspector \( R \) coincides with \( P \) if and only if the lines \( {AP},{\mathcal{L}}_{b},{\mathcal{L}}_{c} \) are concurrent, so are the triples \( {BP},{\mathcal{L}}_{c},{\mathcal{L}}_{a} \) and \( {CP},{\mathcal{L}}_{a},{\mathcal{L}}_{b} \) . Now, \( {AP},{\mathcal{L}}_{b},{\mathcal{L}}_{c} \) are conc... | Yes |
Proposition 3. Let \( P \) be a point distinct from the centroid \( G \), and \( \Gamma \) the circum-hyperbola containing \( G \) and \( P \) . If \( T \) traverses \( \Gamma \), the antiparallels through the intercepts of the triliner polar of \( T \) bound a triangle perspective with \( {ABC} \) with the same perspe... | Proof. The circum-hyperbola containing \( G \) and \( P \) is the isogonal transform of the line \( K{P}^{ * } \) . If we write \( {P}^{ * } = \left( {u : v : w}\right) \), then a point \( T \) on \( \Gamma \) has coordinates \( \left( {\frac{{a}^{2}}{u + t{a}^{2}} : \frac{{b}^{2}}{v + t{b}^{2}} : \frac{{c}^{2}}{w + t{... | Yes |
Lemma 1. Triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) has\n\n(a) angle measures \( {A}^{\prime } = \frac{\pi }{2} - \frac{A}{2},{B}^{\prime } = \frac{\pi }{2} - \frac{B}{2},{C}^{\prime } = \frac{\pi }{2} - \frac{C}{2} \), \n\n(b) sidelengths \( {a}^{\prime } = \sqrt{a\left( {b + c - a}\right) },{b}^{\prime } ... | Proof. (a) \( \angle {B}^{\prime }{A}^{\prime }{C}^{\prime } = \frac{1}{2}\angle {B}^{\prime }H{C}^{\prime } = \frac{1}{2}\angle {BHC} = \frac{\pi - A}{2} \) ; similarly for \( {B}^{\prime } \) and \( {C}^{\prime } \).\n\n(b) By the law of sines,\n\n\[ \n{a}^{\prime } = 2{R}^{\prime }\sin {A}^{\prime } = 2\sqrt{2Rr}\co... | Yes |
Proposition 2. (a) \( {a}^{\prime 2} + {b}^{\prime 2} + {c}^{\prime 2} = {a}^{2} + {b}^{2} + {c}^{2} - {\left( b - c\right) }^{2} - {\left( c - a\right) }^{2} - {\left( a - b\right) }^{2} \) . | Proof. (a) follows from Lemma 1(b); | No |
Lemma 3. The sequences \( {\left( {A}_{n}\right) }_{n \in \mathbb{N}},{\left( {B}_{n}\right) }_{n \in \mathbb{N}},{\left( {C}_{n}\right) }_{n \in \mathbb{N}} \) are convergent and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{A}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{B}_{n} = \mathop{\lim }\limit... | Proof. It is enough to consider the sequence \( {\left( {A}_{n}\right) }_{n \in \mathbb{N}} \) . Rewrite the relation \( {A}_{n + 1} = \) \( \frac{\pi }{2} - \frac{{A}_{n}}{2} \) as\n\n\[ {A}_{n + 1} - \frac{\pi }{3} = - \frac{1}{2}\left( {{A}_{n} - \frac{\pi }{3}}\right) . \]\n\nIt follows that the sequence \( {\left(... | Yes |
Proposition 4. The sequence \( {\left( {R}_{n}\right) }_{n \in \mathbb{N}} \) is convergent and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{R}_{n} = \frac{2}{3}\sqrt{\sqrt{3}\Delta } \) . | Proof. Since \( {R}_{n} = \frac{{a}_{n}{b}_{n}{c}_{n}}{4{\bigtriangleup }_{n}} = \frac{8{R}_{n}^{3}\sin {A}_{n}\sin {B}_{n}\sin {C}_{n}}{4{\bigtriangleup }_{n}} \), we have\n\n\[ \n{R}_{n}^{2} = \frac{\bigtriangleup }{2\sin {A}_{n}\sin {B}_{n}\sin {C}_{n}}.\n\]\n\nThe result follows from Lemma 3. | Yes |
Proposition 5. The sequences \( {\left( {a}_{n}\right) }_{n \in \mathbb{N}},{\left( {b}_{n}\right) }_{n \in \mathbb{N}},{\left( {c}_{n}\right) }_{n \in \mathbb{N}} \) are convergent and\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = \mathop{\lim }... | Proof. This follows from \( {a}_{n} = 2{R}_{n}\sin {A}_{n} \), Lemma 3 and Proposition 4. | No |
Proposition 2. The circle \( {P}^{\left( a\right) }{P}^{\left( b\right) }{P}^{\left( c\right) } \) has center \( {P}^{ * } \) . | Proof. Let \( Q \) be a point on the line isogonal to \( {AP} \) with respect to angle \( A \), i.e., the lines \( {AQ} \) and \( {AP} \) are symmetric with respect to the bisector of angle \( {BAC} \) (see Figure 2(a)). Clearly, the triangles \( {AQ}{P}^{\left( b\right) } \) and \( {AQ}{P}^{\left( c\right) } \) are co... | Yes |
Theorem 4 (Blanc [3]). Let \( \ell \) be a line through the circumcenter \( O \) of triangle \( {ABC} \), intersecting the sidelines at \( X, Y, Z \) respectively. The circles with diameters \( {AX},{BY},{CZ} \) are coaxial with two common points and radical axis \( \mathcal{L} \) containing the orthocenter \( H \) . | (a) One of the common points \( P \) lies on the nine-point circle, and is the center of the rectangular circum-hyperbola which is the isogonal conjugate of the line \( \ell \ | No |
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