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Theorem 1. Suppose \( {ABC} \) and \( {DEF} \) are triangles such that \( \{ A, B, C, D, E, F\} \) consists of at least five distinct points. Suppose \( L \) is a line and \( U = L \cap {L}^{\infty } \) . As \( {D}_{t} \) traverses the line \( {DU} \), the triangle \( {D}_{t}{E}_{t}{F}_{t} \) of translation of \( {DEF}... | Proof. The lines \( A{D}_{t}, B{E}_{t}, C{F}_{t} \) are given by the equations\n\n\[ \n- {z}_{1}\beta + {y}_{1}\gamma = 0,\;{z}_{2}\alpha - {x}_{2}\gamma = 0,\; - {y}_{3}\alpha + {x}_{3}\beta = 0, \]\n\nrespectively. Thus, the concurrence determinant,\n\n\[ \n\left| \begin{matrix} 0 & - {z}_{1} & {y}_{1} \\ {z}_{2} & 0... | Yes |
Theorem 2. Suppose \( {ABC} \) and \( {DEF} \) in Theorem 1 are homothetic. Then \( {D}_{t}{E}_{t}{F}_{t} \) is perspective to \( {ABC} \) for all \( t \) . | Proof. \( {ABC} \) and \( {DEF} \) are homothetic, and \( {DEF} \) and \( {D}_{t}{E}_{t}{F}_{t} \) are homothetic. Therefore \( {D}_{t}{E}_{t}{F}_{t} \) is homothetic to \( {ABC} \), which implies that \( {D}_{t}{E}_{t}{F}_{t} \) is perspective to \( {ABC} \) . | Yes |
Theorem 4. If, in Theorem 1, the line \( L \) is not parallel to a sideline of triangle \( {ABC} \), then the locus of each of the points \( {A}_{t},{B}_{t},{C}_{t} \) is a conic. | Proof. The point \( {A}_{t} = {a}_{t} : {b}_{t} : {c}_{t} \) is given by\n\n\[ \n{a}_{t} = {x}_{2}{x}_{3},\;{b}_{t} = {x}_{2}{y}_{3},\;{c}_{t} = {z}_{2}{x}_{3}, \n\]\n\nso that\n\n\[ \n{a}_{t} = \left( {{d}_{2} + {\epsilon tu}}\right) \left( {{d}_{3} + {\varphi tu}}\right) = {\epsilon \varphi }{u}^{2}{t}^{2} + \left( {... | Yes |
Theorem 5. Suppose \( U \) is any point on \( {L}^{\infty } \) but not on a sideline \( {BC},{CA},{AB} \) . Let \( X \) be the isotomic conjugate of \( U \) . The locus of the perspector \( {P}_{t} \) of triangles \( {D}_{t}{E}_{t}{F}_{t} \) and \( {ABC} \) is a conic that passes through \( A, B, C \), and the point\n\... | Proof. The perspector is the point \( {P}_{t} = {x}_{2}{x}_{3} : {x}_{2}{y}_{3} : {z}_{2}{x}_{3} \) . Substituting and simplifying give\n\n\[ \n{P}_{t} = {b}^{3}{c}^{3}{vw}\left( {{bv} - {acuwt}}\right) \left( {{cw} - {abuvt}}\right) \n\]\n\n\[ \n: {c}^{3}{a}^{3}{wu}\left( {{cw} - {bavut}}\right) \left( {{au} - {bcvwt}... | Yes |
Theorem 6. Suppose \( X \) is the isotomic conjugate of a point \( {U}_{1} \) on \( {L}^{\infty } \) but not on a sideline \( {BC},{CA},{AB} \) . Then the perspector \( {X}^{\prime } \) in Case 5.4 is invariant of the point \( U \) . In fact, \( {X}^{\prime } = {CD}\left( {{X}_{6},{U}_{1}^{-1}}\right) \), and \( {X}^{\... | Proof. Write \( X = x : y : z = {b}^{2}{c}^{2}{v}_{1}{w}_{1} : {c}^{2}{a}^{2}{w}_{1}{u}_{1} : {a}^{2}{b}^{2}{u}_{1}{v}_{1} \) where the point \( {U}_{1} = {u}_{1} : {v}_{1} : {w}_{1} \) is on \( {L}^{\infty } \) and \( {u}_{1}{v}_{1}{w}_{1} \neq 0 \) . Represent \( {U}_{1} \) parametrically by\n\n\[ \n{u}_{1} = \left( ... | Yes |
Corollary 7. As \( X \) traverses the Steiner circumellipse, the perspector \( {X}^{\prime } \) traverses the line at infinity. | Proof. The Steiner circumellipse is given by\n\n\[ \n{bc\beta \gamma } + {ca\gamma \alpha } + {ab\alpha \beta } = 0.\n\]\n\nThe corollary follows from the easy-to-verify fact that the isotomic conjugacy mapping carries the Steiner circumellipse to \( {L}^{\infty } \), to which Theorem 6 applies. | Yes |
Theorem 8. Suppose \( X = {CP}\left( {{X}_{2},{U}_{1}}\right) \), where \( {U}_{1} \) is a point on \( {L}^{\infty } \) but not on a sideline \( {BC},{CA},{AB} \) . Then the perspector \( {X}^{\prime } \) in Case 6.4 is invariant of the point \( U \) . In fact, \( {X}^{\prime } = {CD}\left( {{X}_{6},{U}_{1}^{-1}}\right... | Proof. Let \( X = x : y : z = b{u}_{1}{v}_{1} + c{u}_{1}{w}_{1} : c{v}_{1}{w}_{1} + a{v}_{1}{u}_{1} : a{w}_{1}{u}_{1} + b{w}_{1}{v}_{1} \) . Following the steps of the proof of Theorem 6, we have\n\n\[ {bw}{y}^{2} - {cv}{z}^{2} + \left( {{bv} - {cw}}\right) {yz} + {ax}\left( {{vz} - {wy}}\right) \]\n\n\[ = \widehat{\La... | Yes |
Corollary 9. As \( X \) traverses the Steiner inellipse, the perspector \( {X}^{\prime } \) traverses the circumconic (20). | Proof. The Steiner inellipse is given by\n\n\[ \n{a}^{2}{\alpha }^{2} + {b}^{2}{\beta }^{2} + {c}^{2}{\gamma }^{2} - {2bc\beta \gamma } - {2ca\gamma \alpha } - {2ab\alpha \beta } = 0.\n\]\n\n(21)\n\nFirst, we note that, using (16), it is easy to show that if \( {U}_{1} \) is on \( {L}^{\infty } \), then the point \( X ... | Yes |
Theorem 10. If \( X \) is the isotomic conjugate of a point \( {X}^{\prime } \) on the Euler line other than \( {X}_{2} \), then the perspector; in the case of the cevian triangle of \( X \) as given by (12)-(14), is \( {X}^{\prime } \) . | Proof. An arbitrary point \( {X}^{\prime } \) on the Euler line is given parametrically by\n\n\[ \n{a}_{1} + {su} : {b}_{1} + {sv} : {c}_{1} + {sw}, \n\]\n\nand the isotomic conjugate \( X = x : y : z \) by\n\n\[ \n{a}^{-2}{\left( {a}_{1} + su\right) }^{-1} : {b}^{-2}{\left( {b}_{1} + sv\right) }^{-1} : {c}^{-2}{\left(... | Yes |
Theorem 11. Suppose \( P \) is on the Euler line and \( P \neq {X}_{2} \) . Let \( X = {CP}\left( {{X}_{2}, P}\right) \) . Then the perspector \( {X}^{\prime } \), in the case of the anticevian triangle of \( X \), as given by (17)-(19), is the point \( P \) . | Proof. Write\n\n\[ p = {a}_{1} + {su},\;q = {b}_{1} + {sv},\;r = {c}_{1} + {sw}, \]\n\nwhere \( \left( {u, v, w}\right) = \left( {{a}_{1} - 2{b}_{1}{c}_{1},{b}_{1} - 2{c}_{1}{a}_{1},{c}_{1} - 2{a}_{1}{b}_{1}}\right) \), so that the point \( X = \) \( {CP}\left( {{X}_{2}, P}\right) \) is given by\n\n\[ x = p\left( {{bq}... | Yes |
Theorem 12. If \( P \) is on the circumcircle and \( X = {CS}\left( {{X}_{6}, P}\right) \), then the perspector \( {X}^{\prime } \), in the case of the anticevian triangle of \( X \) as given by (17)-(19), is the point \( {CD}\left( {{X}_{6}, P}\right) \) . | Proof. Represent an arbitrary point \( P = p : q : r \) on the circumcircle parametrically by\n\n\[ \left( {p, q, r}\right) = \left( {\frac{1}{\left( {b - c}\right) \left( {{bc} + s}\right) },\frac{1}{\left( {c - a}\right) \left( {{ca} + s}\right) },\frac{1}{\left( {a - b}\right) \left( {{ab} + s}\right) }}\right) .\n\... | Yes |
Theorem 1. Let \( {\lambda }_{j} \in \mathbf{R} \smallsetminus \{ 0\}, j = 1,\ldots, k \), and let \( {z}_{j}, j = 1,\ldots, k \) be distinct complex numbers. Then the zeros of the rational function \( R\left( z\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{j = 1}}^{k}\frac{{\lambda }_{j}}{z - {z}_{j}} \) are the ... | Since \( {f}^{\prime } = f \cdot \mathop{\sum }\limits_{{j = 1}}^{k}\frac{1}{z - {z}_{j}} \), where the \( {z}_{j} \) are the zeros of \( f \), Linfield’s Theorem can be used to locate the zeros of \( {f}^{\prime } \) which are not zeros of \( f \) . | No |
Proposition 2. Let \( n \geq 3 \), and let \( G \) be an \( n \) -gon with vertices \( {v}_{0},\ldots ,{v}_{n - 1} \), no three of which are collinear. The following are equivalent.\n\n(1) There is an ellipse which is tangent to the edges of \( G \) at their midpoints.\n\n(2) \( G \) is affine-equivalent to \( G\left( ... | Proof. 1) \( \Rightarrow \) 2): Applying an affine transformation if necessary, we may assume that the ellipse is a circle centered at 0 and that \( {v}_{0} = 1 \) . Let \( {m}_{0} \) be the midpoint of the edge \( {v}_{0}{v}_{1}.{v}_{0}{v}_{1} \) is then perpendicular to \( 0{m}_{0} \), and \( {v}_{0},{v}_{1} \) are e... | Yes |
Proposition 3. The foci of the Steiner inellipse of \( G \) are located at \( {f}_{ \pm } = g \pm (\theta + \) \( \left. {\theta }^{-1}\right) {xy} \) . | Proof. Translating if necessary, we may assume that \( g = 0 \), i.e., \( G \in {S}_{\zeta } \) . Note first that \( {f}_{ \pm } \in {K}_{0} \) . (This is to be expected, since the Steiner inellipse and its foci do not depend on the choice of initial vertex.) For \( k = 0,\ldots, n - 1 \), let \( {m}_{k} = \left( {{v}_... | Yes |
Corollary 4. The Steiner inellipse of \( G \) is a circle iff \( G \) is similar to \( G\left( \zeta \right) \) . | Proof. \( {f}_{ + } = {f}_{ - } \) iff \( {xy} = 0 \), i.e., iff one of \( u \) and \( v \) is zero. (Note that \( \theta + {\theta }^{-1} \neq \) 0.) | No |
Proposition 5. The foci of the Steiner inellipse of \( G \) are critical points of \( {f}_{G} \) . | Proof. Again, we may assume \( G \in {S}_{\zeta } \) . Since \( {f}_{G}^{\prime }/{f}_{G} = \mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{\left( z - {v}_{k}\right) }^{-1} \), it suffices to show that this sum vanishes at \( {f}_{ \pm } \) . Now \( {f}_{ + } \) is invariant under \( \varphi \), and \( {v}_{k} = {\varphi }^{... | Yes |
Lemma 6. Let \( k = 1,\ldots ,\lfloor n/2\rfloor \) . Then:\n\n(1) If \( k \) is relatively prime to \( n \), let \( {G}^{k} \) be the \( n \) -gon with vertices \( {v}_{0}^{k},\ldots ,{v}_{n - 1}^{k} \) given by \( {v}_{j}^{k} = {v}_{jk} \) . Then \( {G}^{k} \) is a Steiner n-gon with root \( {\zeta }^{k} \), and its ... | In the three given cases, we will say \( k \) -holomorph of \( G \) to refer to \( {G}^{k} \), the union of the \( m \) -gons \( {G}^{k, l} \), or the union of the line segments \( {v}_{j}{v}_{j + k} \) . We extend the definition of Steiner inellipse to the \( k \) -holomorphs in Cases 2 and 3, meaning the common Stein... | Yes |
Theorem 7. If \( G \) is a Steiner n-gon, the critical points of \( {f}_{G} \) are the foci of the Steiner inellipses of the holomorphs of \( G \), counted with multiplicities if \( G \) is regular. They are collinear, lying at the points \( g + \left( {2\cos {k\pi }/n}\right) {xy} \), as \( k \) ranges from 0 to \( n ... | (For the last statement, note that \( \cos \left( {n - k}\right) \pi /n = - \cos {k\pi }/n \).) | No |
Theorem 1. In any arbitrary quadrilateral the quasiorthocenter \( H \), the centroid \( G \), and the quasicircumcenter \( O \) are collinear. Furthermore, \( {OH} : {HG} = 3 : - 2 \) . | Proof. Consider three affine maps \( {f}_{G},{f}_{O} \) and \( {f}_{H} \) transforming the triangle \( {ABC} \) onto triangle \( {G}_{a}{G}_{b}{G}_{c},{O}_{a}{O}_{b}{O}_{c} \), and \( {H}_{a}{H}_{b}{H}_{c} \) respectively.\n\nIn the affine plane, write \( D = {xA} + {yB} + {zC} \) with \( x + y + z = 1 \).\n\n(i) Note ... | Yes |
Lemma 4. \( I = \left( {\left( {q + t}\right) A,\left( {p + r}\right) B,\left( {q + t}\right) C,\left( {p + r}\right) D}\right) \) . | Proof. Suppose the circumscriptible quadrilateral \( {ABCD} \) has a pair of non-parallel sides \( {AD} \) and \( {BC} \), which intersect at \( E \) . (If not, then \( {ABCD} \) is a rhombus, \( p = q = r = s \), and \( I = G \) ; the result is trivial). Let \( a = {EB} \) and \( b = {EA} \) .\n\n(i) As the incenter o... | Yes |
Lemma 5. \( G = \left( {\left( {p + q + t}\right) A,\left( {p + q + r}\right) B,\left( {q + r + t}\right) C,\left( {p + r + t}\right) D}\right) \) . | Proof. Denote the point of intersection of the diagonals by \( P \) . Note that \( \frac{AP}{CP} = \frac{p}{r} \) and \( \frac{BP}{DP} = \frac{q}{t} \) . Actually, according to one corollary of Brianchon’s theorem, the lines \( {T}_{1}{T}_{3} \) and \( {T}_{2}{T}_{4} \) also pass through \( P \) . For another proof, se... | Yes |
Theorem 7. For a circumscriptible quadrilateral, the Spieker point is the midpoint of the incenter and the Nagel point. | Proof. With reference to Figure 6, each side of the circumscriptible quadrilateral is equivalent to a mass equal to its length located at each of its two vertices. Thus,\n\n\[ \nS = \left( {\left( {{2p} + q + t}\right) A,\left( {p + {2q} + r}\right) B,\left( {q + {2r} + t}\right) C,\left( {p + r + {2t}}\right) D}\right... | Yes |
Proposition 1. Triangles \( {H}_{1, n}{H}_{3, n}{H}_{5, n} \) and \( {H}_{1, n - 2}{H}_{3, n - 2}{H}_{5, n - 2} \) have the same centroid, so do triangles \( {H}_{2, n}{H}_{4, n}{H}_{6, n} \) and \( {H}_{2, n - 2}{H}_{4, n - 2}{H}_{6, n - 2} \) . | Proof. Applying the relations \( \left( {1,2,3,4}\right) \) twice, we have\n\n\[ \n{H}_{1, n} = - \left( {1 - i}\right) {H}_{6, n - 2} + 2{H}_{1, n - 2} + \left( {1 - i}\right) {H}_{2, n - 2} - {H}_{3, n - 2}, \]\n\n\[ \n{H}_{3, n} = - \left( {1 - i}\right) {H}_{2, n - 2} + 2{H}_{3, n - 2} + \left( {1 - i}\right) {H}_{... | Yes |
Theorem 2. For each \( m = 1,2,3,4,5,6 \), the sequence of vertices \( {H}_{m, n} \) satisfies the recurrence relation\n\n\[ \n{H}_{m, n} = 6{H}_{m, n - 2} - 6{H}_{m, n - 4} + {H}_{m, n - 6}. \n\]\n\n(5) | Proof. By using the recurrence relations \( \left( {1,2,3,4}\right) \), we have\n\n\[ \n{H}_{1,2} = \left( {1 + i}\right) {H}_{1,1} - i{H}_{2,1} \n\]\n\n\[ \n{H}_{1,3} = 2{H}_{1,1} - \left( {1 + i}\right) {H}_{2,1} - {H}_{5,1} + \left( {1 + i}\right) {H}_{6,1}, \n\]\n\n\[ \n{H}_{1,4} = 3\left( {1 + i}\right) {H}_{1,1} ... | Yes |
Proposition 3. For a permutation \( \left( {j, k,\ell }\right) \) of the integers \( 1,2,3 \), the segments \( {M}_{j}{M}_{k}^{\prime } \) and \( {M}_{k}{M}_{j}^{\prime } \) are perpendicular to each other and equal in length, while \( {M}_{\ell }{M}_{\ell }^{\prime } \) is parallel to an angle bisector of \( {M}_{j}{M... | Proof. From the above expressions for \( {M}_{j}^{\prime }, j = 1,2,3 \), we have\n\n\[ \n{M}_{2}^{\prime } - {M}_{3} = \left( {1 - i}\right) {M}_{1} - {M}_{2} + i \cdot {M}_{3}, \n\]\n\n(6)\n\n\[ \n{M}_{3}^{\prime } - {M}_{2} = \left( {1 + i}\right) {M}_{1} - i \cdot {M}_{2} - {M}_{3} \n\]\n\n\[ \n= i\left( {\left( {1... | Yes |
Corollary 4. The triangles \( {M}_{1}{M}_{2}{M}_{3} \) and \( {M}_{1}^{\prime }{M}_{2}^{\prime }{M}_{3}^{\prime } \) are perspective. | The perspector is the outer Vecten point of \( {M}_{1}{M}_{2}{M}_{3} \) and the inner Vecten point of \( {M}_{1}^{\prime }{M}_{2}^{\prime }{M}_{3}^{\prime } \) . \( {}^{1} \) | Yes |
Theorem 12. Every triangle of the form\n\n\[ \n- {2fS} - g{a}^{2}\; : \;\left( {f + 1}\right) S + g{S}_{C}\; : \;\left( {f + 1}\right) S + g{S}_{B} \]\n\n\[ \n\left( {f + 1}\right) S + g{S}_{C}\; : \; - {2fS} - g{b}^{2}\; : \;\left( {f + 1}\right) S + g{S}_{A} \]\n\n\[ \n\left( {f + 1}\right) S + g{S}_{B} : g\left( {f ... | Proof. Clearly the triangle given above is perspective with \( {ABC} \) at the point\n\n\[ \n\left( {\frac{1}{\left( {f + 1}\right) S + g{S}_{A}} : \frac{1}{\left( {f + 1}\right) S + g{S}_{B}} : \frac{1}{\left( {f + 1}\right) S + g{S}_{C}}}\right) , \]\n\nwhich is the Kiepert perspector \( K\left( \phi \right) \) for \... | Yes |
Lemma 13. (a) If \( n \geq 2 \) is even, then\n\n\[ \overrightarrow{{A}_{1, n}{A}_{1, n + 1}} = - \frac{1}{2}\overrightarrow{{H}_{6, n}{H}_{6, n + 1}} = - \frac{1}{2}\overrightarrow{{H}_{1, n}{H}_{1, n + 1}}, \]\n\n\[ \overrightarrow{{B}_{1, n}{B}_{1, n + 1}} = - \frac{1}{2}\overrightarrow{{H}_{2, n}{H}_{2, n + 1}} = -... | Proof. Consider the case of \( {A}_{1, n}{A}_{1, n + 1} \) for even \( n \) . Translate the trapezoid \( {H}_{5, n + 1}{H}_{6, n + 1}{H}_{6, n}{H}_{5, n} \) by the vector \( \overrightarrow{{H}_{5, n + 1}{H}_{4, n + 1}} \) and the trapezoid \( {H}_{2, n + 1}{H}_{2, n}{H}_{1, n}{H}_{1, n + 1} \) by the vector \( \overri... | Yes |
Proposition 14. (1) If \( n \geq 3 \) is odd, the following pairs of triangles are congruent.\n\n(i) \( {\mathrm{H}}_{2, n}{\mathrm{\;B}}_{2, n + 1}{\mathrm{C}}_{1, n - 1} \) and \( {\mathrm{H}}_{5, n}{\mathrm{\;B}}_{1, n - 1}{\mathrm{C}}_{2, n + 1} \) ,\n\n(ii) \( {H}_{3, n}{A}_{1, n - 1}{B}_{2, n + 1} \) and \( {H}_{... | Proof. We consider the first of these cases. Let \( n \geq 3 \) be an odd number. Consider the triangles \( {H}_{2, n}{B}_{2, n + 1}{C}_{1, n - 1} \) and \( {H}_{5, n}{B}_{1, n - 1}{C}_{2, n + 1} \) . We show that \( {H}_{2, n}{B}_{2, n + 1} \) and \( {H}_{5, n}{B}_{1, n - 1} \) are perpendicular to each other and equa... | Yes |
Theorem 15. The locus of point \( P \) in \( {ABC} \) whose counterparts in the triangles \( {A}_{1,3}{H}_{4,2}{H}_{3,2},{H}_{5,2}{B}_{1,3}{H}_{6,2} \) and \( {H}_{2,2}{H}_{1,2}{C}_{1,3} \) form a triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) perspective to \( {ABC} \) is the union of the line at infinity and ... | Proof. The counterparts of \( P = \left( {x : y : z}\right) \) form a triangle perspective with \( {ABC} \) if and only if the parallels through \( A, B, C \) to \( {A}_{1,3}P,{B}_{1,3}P \) and \( {C}_{1,3}P \) are concurrent. These parallels have equations\n\n\[ \left( {\left( {S + {S}_{B}}\right) \left( {x + y + z}\r... | Yes |
Proposition 1 (Lalesco). If the Simson lines of \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) concur at \( P \), then (a) \( P \) is the midpoint of \( H{H}^{\prime } \), where \( {H}^{\prime } \) is the orthocenter of \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \), (b) for any point \( M \in \Gamma \), the Simson lin... | Let \( h \) be the rectangular hyperbola through \( A, B, C, P \) . If the hyperbola \( h \) intersects \( \Gamma \) again at \( U \), the center \( W \) of \( h \) is the midpoint of \( {HU} \) . Let \( {h}^{\prime } \) be the rectangular hyperbola through \( {A}^{\prime },{B}^{\prime },{C}^{\prime }, U \) . The cente... | Yes |
Proposition 2. Let \( P \) be a point inside \( {ABC} \) and \( {Q}_{a} \) the incenter of \( {PBC} \) . The following statements are equivalent.\n\n(a) \( {PB} - {PC} = {AB} - {AC} \) .\n\n(b) \( P \) lies on the open arc \( A{D}_{a} \) of \( {h}_{a} \) .\n\n(c) The quadrilateral \( A{B}^{\prime }P{C}^{\prime } \) has... | Proof. (a) \( \Leftrightarrow \) (b). As \( {2B}{D}_{a} = {AB} + {BC} - {AC} \) and \( {2C}{D}_{a} = {AC} + {BC} - {AB} \) , we have \( B{D}_{a} - C{D}_{a} = {AB} - {AC} \) and \( {D}_{a} \) is the vertex of the branch of \( {h}_{a} \) through A.\n\n(b) \( \Rightarrow \) (c). \( {AI} \) and \( P{Q}_{a} \) are the lines... | Yes |
Proposition 3. When the conditions of Proposition 2 are satisfied, the following statements are equivalent.\n\n(a) The incircles of \( {PBC} \) and \( A{B}^{\prime }P{C}^{\prime } \) are congruent.\n\n(b) \( P \) is the midpoint of \( {W}_{a}{Q}_{a} \) .\n\n(c) \( {W}_{a}{Q}_{a} \) and \( A{D}_{a} \) are parallel.\n\n(... | Proof. (a) \( \Leftrightarrow \) (b) is obvious.\n\nLet’s notice that, as \( I \) is the pole of \( A{D}_{a} \) with respect to \( {h}_{a},{M}_{a}I \) is the conjugate diameter of the direction of \( A{D}_{a} \) with respect to \( {h}_{a} \) .\n\nSo (c) \( \Leftrightarrow \) (d) because \( {W}_{a}{Q}_{a} \) is the tang... | No |
Proposition 4. The inner Soddy center \( X\left( {176}\right) \) is the only point \( P \) inside \( {ABC} \) (a) for which the incircles of \( {PBC},{PCA},{PAB} \) touch each other; (b) with cevian triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) for which each of the three quadrilaterals \( A{B}^{\prime }P{C}^{... | Proof. Proposition 2 shows that the conditions in (a) and (b) are both equivalent to\n\n\[ \n{PB} - {PC} = c - b,{PC} - {PA} = a - c,{PA} - {PB} = b - a.\n\] \n\nAs \( {PA} = \rho + s - a,{PB} = \rho + s - b,{PC} = \rho + s - c \), these conditions are satisfied for \( P = X\left( {176}\right) \) . Moreover, a point \(... | Yes |
Proposition 3. The following statements for a triangle \( {ABC} \) are equivalent.\n\n(1) \( {r}_{c} = s \) .\n\n(2) \( {r}_{a} = s - b \) .\n\n(3) \( {r}_{b} = s - a \) .\n\n(4) \( r = s - c \) .\n\n(5) \( C \) is a right angle. | Proof. By the formulas for the exradii and the Heron formula, each of (1), (2), (3), (4) is equivalent to the condition\n\n\[ \n\left( {s - a}\right) \left( {s - b}\right) = s\left( {s - c}\right) .\n\]\n\n(1)\n\n\n\n . | Proof. It is enough to prove that \( {MB} = {MI} \) . See Figure 3. This follows by an easy calculation of angles.\n\n(i) \( \angle {IB}{I}_{a} = {90}^{ \circ } \) since the two bisectors of angle \( B \) are perpendicular to each other.\n\n(ii) The midpoint \( N \) of \( {I}_{a}I \) is the circumcenter of triangle \( ... | Yes |
Proposition 5. The circle through the three excenters has radius \( {2R} \) and center \( 1 \) , the reflection of \( I \) in \( O \) . | Remark. Proposition 5 also follows from the fact that the circumcircle is the nine point circle of triangle \( {I}_{a}{I}_{b}{I}_{c} \), and \( I \) is the orthocenter of this triangle. | No |
Proposition 6. \( {r}_{a} + {r}_{b} + {r}_{c} = {4R} + r \) . | Proof. The line \( {I}_{a}{I}^{\prime } \) intersects \( {BC} \) at the point \( {X}^{\prime } \) of tangency with the excircle. Note that \( {I}^{\prime }{X}^{\prime } = {2R} - {r}_{a} \) . Since \( O \) is the midpoint of \( I{I}^{\prime } \), we have \( {IX} + {I}^{\prime }{X}^{\prime } = \) \( 2 \cdot {OD} \) . Fro... | Yes |
Proposition 7. A triangle \( {ABC} \) satisfies\n\n\[ {r}_{a}^{2} + {r}_{b}^{2} + {r}_{c}^{2} + {r}^{2} = {a}^{2} + {b}^{2} + {c}^{2} \]\n\n(3)\n\nif and only if it contains a right angle. | Proof. Using \( {AH} = {2R}\cos A \) and \( a = {2R}\sin A \), and similar expressions for \( {BH} \) , \( {CH}, b \), and \( c \), we have\n\n\[ A{H}^{2} + B{H}^{2} + C{H}^{2} + {\left( 2R\right) }^{2} - \left( {{a}^{2} + {b}^{2} + {c}^{2}}\right) \]\n\n\[ = 4{R}^{2}\left( {{\cos }^{2}A + {\cos }^{2}B + {\cos }^{2}C +... | Yes |
Lemma 8. (1) \( {r}_{a}{r}_{b} + {r}_{b}{r}_{c} + {r}_{c}{r}_{a} = {s}^{2} \) . | Proof. (1) follows from the formulas for the exradii and the Heron formula.\n\n\[ \n{r}_{a}{r}_{b} + {r}_{b}{r}_{c} + {r}_{c}{r}_{a} = \frac{{\bigtriangleup }^{2}}{\left( {s - a}\right) \left( {s - b}\right) } + \frac{{\bigtriangleup }^{2}}{\left( {s - b}\right) \left( {s - c}\right) } + \frac{{\bigtriangleup }^{2}}{\l... | Yes |
Proposition 9. \( {r}_{a}^{2} + {r}_{b}^{2} + {r}_{c}^{2} + {r}^{2} = {a}^{2} + {b}^{2} + {c}^{2} \) if and only if \( {2R} + r = s \) . | Proof. By Lemma 8(2) and (4), \( {r}_{a}^{2} + {r}_{b}^{2} + {r}_{c}^{2} + {r}^{2} = {a}^{2} + {b}^{2} + {c}^{2} \) if and only if \( {\left( 4R + r\right) }^{2} - 2{s}^{2} + {r}^{2} = 2{s}^{2} - 2\left( {{4R} + r}\right) r;4{s}^{2} = {\left( 4R + r\right) }^{2} + 2\left( {{4R} + r}\right) r + {r}^{2} = \) \( {\left( 4... | Yes |
Theorem 10. The following statements for a triangle \( {ABC} \) are equivalent.\n\n(1) \( {r}_{a} + {r}_{b} + {r}_{c} + r = a + b + c \) .\n\n(2) \( {r}_{a}^{2} + {r}_{b}^{2} + {r}_{c}^{2} + {r}^{2} = {a}^{2} + {b}^{2} + {c}^{2} \) .\n\n(3) \( R + {2r} = s \) .\n\n(4) One of the angles is a right angle. | Proof. (1) \( \Rightarrow \) (3): This follows easily from Proposition 6.\n\n\( \left( 3\right) \Leftrightarrow \left( 2\right) \) : Proposition 9 above.\n\n\( \left( 2\right) \Leftrightarrow \left( 4\right) \) : Proposition 7 above.\n\n\( \left( 4\right) \Rightarrow \left( 1\right) \) : Theorem 1 (1). | Yes |
Theorem 5. The centroid of the Kiepert cevian triangle of \( P \) lies on the Kiepert hyperbola if and only if \( P \) is\n\n(i) an infinite point, or\n\n(ii) on the tangent at \( {K}_{\mathrm{i}} \) to the Steiner inellipse. | Proof. Let \( P = \left( {u : v : w}\right) \) in homogeneous barycentric coordinates. Applying Proposition 2, we find that the centroid of \( {A}_{P}{B}_{P}{C}_{P} \) lies on the Kiepert hyperbola if and only if\n\n\[ \left( {u + v + w}\right) K{\left( u, v, w\right) }^{2}L\left( {u, v, w}\right) P\left( {u, v, w}\rig... | Yes |
Proposition 6. Two distinct Kiepert perspectors have parameters satisfying (3) if and only if the line joining them is parallel to the orthic axis. | Proof. The orthic axis \( {S}_{A}x + {S}_{B}y + {S}_{C}z = 0 \) has infinite point\n\n\[ P\left( \infty \right) = \left( {{S}_{B} - {S}_{C} : {S}_{C} - {S}_{A} : {S}_{A} - {S}_{B}}\right) .\n\]\n\nThe line joining \( K\left( s\right) \) and \( K\left( t\right) \) is parallel to the orthic axis if and only if\n\n\[ \lef... | Yes |
Theorem 9. The Kiepert cevian triangle of \( Q\left( t\right) \) is triply perspective to \( {ABC} \) . The three perspectors are collinear on the tangent of the Steiner inellipse at \( {K}_{\mathrm{i}} \) . | Proof. The triangles \( {B}^{\prime }\left( t\right) {C}^{\prime }\left( t\right) {A}^{\prime }\left( t\right) \) and \( {C}^{\prime }\left( t\right) {A}^{\prime }\left( t\right) {B}^{\prime }\left( t\right) \) are each perspective to \( {ABC} \), at the points\n\n\[ \n{Q}^{\prime }\left( t\right) = \left( {\frac{{S}_{... | Yes |
Theorem 1. The following statement for a tetrahedron P are equivalent.\n\n(1) \( \mathcal{P} \) has an edge-tangent sphere.\n\n(2) \( {a}_{01} + {a}_{23} = {a}_{02} + {a}_{13} = {a}_{03} + {a}_{12} \) ;\n\n(3) There exist \( {x}_{i} > 0, i = 0,1,2,3 \), such that \( {a}_{ij} = {x}_{i} + {x}_{j} \) for \( 0 \leq i < j \... | For \( i = 0,1,2,3,{x}_{i} \) is the length of a tangent from \( {P}_{i} \) to the edge-tangent sphere of \( \mathcal{P} \) . Let \( \ell \) denote the radius of this sphere. | No |
Theorem 2. [1, §793] The radius of the edge-tangent sphere of a circumscriptible tetrahedron of volume \( V \) is given by\n\n\[ \ell = \frac{2{x}_{0}{x}_{1}{x}_{2}{x}_{3}}{3V} \] | Lin and Zhu [4] have given the formula (1) in the form\n\n\[ {\ell }^{2} = \frac{{\left( 2{x}_{0}{x}_{1}{x}_{2}{x}_{3}\right) }^{2}}{2{x}_{0}{x}_{1}{x}_{2}{x}_{3}\mathop{\sum }\limits_{{0 \leq i < j \leq 3}}{x}_{i}{x}_{j} - \left( {{x}_{1}^{2}{x}_{2}^{2}{x}_{3}^{2} + {x}_{2}^{2}{x}_{3}^{2}{x}_{0}^{2} + {x}_{3}^{2}{x}_{... | Yes |
Lemma 4. If \( {x}_{i} > 0 \) for \( 0 \leq i \leq 3 \), then\n\n\[ \left( {\frac{{x}_{1} + {x}_{2} + {x}_{3}}{{x}_{1}{x}_{2}{x}_{3}} + \frac{{x}_{2} + {x}_{3} + {x}_{0}}{{x}_{2}{x}_{3}{x}_{0}} + \frac{{x}_{3} + {x}_{0} + {x}_{1}}{{x}_{3}{x}_{0}{x}_{1}} + \frac{{x}_{0} + {x}_{1} + {x}_{2}}{{x}_{0}{x}_{1}{x}_{2}}}\right... | Proof. From\n\n\[ {x}_{0}^{2}{x}_{1}^{2}{\left( {x}_{2} - {x}_{3}\right) }^{2} + {x}_{0}^{2}{x}_{2}^{2}{\left( {x}_{1} - {x}_{3}\right) }^{2} + {x}_{0}^{2}{x}_{3}^{2}{\left( {x}_{1} - {x}_{2}\right) }^{2} \]\n\n\[ + {x}_{1}^{2}{x}_{2}^{2}{\left( {x}_{0} - {x}_{3}\right) }^{2} + {x}_{1}^{2}{x}_{3}^{2}{\left( {x}_{0} - {... | Yes |
Corollary 5. For a circumscriptible tetrahedron \( \mathcal{P} \) with an edge-tangent sphere of radius \( \ell \), and faces with inradii \( {r}_{0},{r}_{1},{r}_{2},{r}_{3} \) , \[ \left( {\frac{1}{{r}_{0}^{2}} + \frac{1}{{r}_{1}^{2}} + \frac{1}{{r}_{2}^{2}} + \frac{1}{{r}_{3}^{2}}}\right) {\ell }^{2} \geq 6. \] Equal... | Proof. From the famous Heron formula, the inradius of a triangle ABC of side-lengths \( a = y + z, b = z + x \) and \( c = x + y \) is given by \[ {r}^{2} = \frac{xyz}{x + y + z}. \] Applying this to the four faces of \( \mathcal{P} \), we see that the first factor on the left hand side of (3) is \( \left( {\frac{1}{{r... | No |
Theorem 7. Suppose the edge lengths of an \( n \) -simplex \( \mathcal{P} = {P}_{0}{P}_{1}\cdots {P}_{n} \) are \( {P}_{i}{P}_{j} = {a}_{ij} \) for \( 0 \leq i < j \leq n \) . The \( n \) -simplex has an edge-tangent sphere if and only if there exist \( {x}_{i}, i = 0,1,\ldots, n \), satisfying \( {a}_{ij} = {x}_{i} + ... | \[ {\ell }^{2} = - \frac{{D}_{1}}{2{D}_{2}} \]\n\n(16)\n\nwhere\n\n\[ {D}_{1} = \left| \begin{matrix} - 2{x}_{0}^{2} & 2{x}_{0}{x}_{1} & \cdots & 2{x}_{0}{x}_{n - 1} \\ 2{x}_{0}{x}_{1} & - 2{x}_{1}^{2} & \cdots & 2{x}_{1}{x}_{n - 1} \\ \cdots & \cdots & \cdots & \cdots \\ 2{x}_{0}{x}_{n - 1} & 2{x}_{1}{x}_{n - 1} & \cd... | Yes |
Theorem 1. If \( {ABC} \) is a Euclidean triangle having largest angle \( \gamma < \arctan \left( \frac{24}{7}\right) \approx \) \( {74}^{ \circ } \), then \( {ABC} \) satisfies the strong triangle inequality. | An elegant trigonometric proof of Theorem 1 can by found in [3]. It should be noted that the bound of \( \arctan \left( \frac{24}{7}\right) \) is the best possible since any isosceles Euclidean triangle with \( \gamma = \arctan \left( \frac{24}{7}\right) \) violates the strong triangle inequality. | Yes |
Theorem 2. In neutral geometry a triangle \( {ABC} \) having largest angle \( \gamma \) satisfies the strong triangle inequality if \( \gamma \leq r \approx {1.15} \) radians or \( {65.87}^{ \circ } \) . | The proof of Theorem 2 is based on the fact that a model of neutral geometry is isomorphic to either the Euclidean plane or a hyperbolic plane. Given Theorem 1, it is enough to establish our result for the case of hyperbolic geometry. Moreover, since the strong triangle inequality holds if and only if \( {ka} + {kb} > ... | Yes |
Theorem 1. Every uniform elliptic cylindrical wedge is reducible. | Now assemble a finite collection of nonoverlapping elliptic cylindrical wedges with their horizontal semi axes, possibly of different lengths, in the same horizontal plane, but having a common vertical semi axis, which meets the base at a point \( O \) called the center. We assume the density of each component wedge is... | No |
Theorem 2. The following solids are reducible:\n\n(a) Any uniform polygonal elliptic shell.\n\n(b) Any wedge of a uniform polygonal elliptic shell.\n\n(c) Any horizontal slice of a wedge of type (b).\n\n(d) Any nonuniform assemblage of shells of type (a).\n\n(e) Any nonuniform assemblage of wedges of type (b).\n\n(f) A... | By using as building blocks horizontal slices of wedges cut from a polygonal elliptic shell, we can see intuitively how one might construct, from such building blocks, very general polygonal elliptic domes that are reducible and have more or less arbitrary mass distribution. By considering limiting cases of polygonal b... | No |
Corollary 5. The volume of a general elliptic dome is equal to the volume of its circumscribing punctured cylindrical container, that is, two-thirds the volume of the circumscribing unpunctured cylindrical container which, in turn, is simply the area of the base times the height. | The same formulas show that for a fixed altitude \( z \), we have \( {rdrd\theta } = {r}^{\prime }d{r}^{\prime }d{\theta }^{\prime } \) . In other words, the mapping also preserves areas of horizontal cross sections cut from the elliptic dome and its punctured container. This also implies Corollary 5 because of the sli... | No |
Corollary 6. (Lambert) Mapping (4), from the surface of a sphere to the lateral surface of its tangent cylinder, preserves areas. | If the hemisphere in this limiting case has radius \( a \), it is easily verified that (4) reduces to Lambert’s mapping: \( {\theta }^{\prime } = \theta ,\;{z}^{\prime } = z,\;{r}^{\prime } = a \) . | No |
Theorem 5. Any portion of a general nonuniform elliptic dome is reducible. | By analogy with Theorem 3, we can say that mapping (4) \ | No |
Corollary 8. (Sphere with cavity) Consider a spherically symmetric distribution of mass inside a solid sphere with a concentric cavity. Any slice between parallel planes pierced by the cavity has volume and mass proportional to the thickness of the slice, and is independent of the location of the slice. | Corollary 8 implies that the one-dimensional vertical projection of the density is constant along the cavity. This simple result has profound consequences in tomography, which deals with the inverse problem of reconstructing spatial density distributions from a knowledge of their lower dimensional projections. Details ... | No |
Theorem 7. Any uniform elliptic wedge or dome of altitude \( h \) has volume two-thirds that of its unpunctured prismatic container. Its centroid is located at height \( c \) above the plane of the base, where\n\n\[ c = \frac{3}{8}h \] | Proof. It suffices to prove (9) for the prismatic counterpart. For any prism of altitude \( h \), the centroid is at a distance \( h/2 \) above the plane of the base. For a cone or pyramid with the same base and altitude it is known that the centroid is at a distance \( {3h}/4 \) from the vertex. To determine the heigh... | Yes |
Theorem 8. Any nonuniform assemblage of elliptic shell elements with common altitude \( h \) and scaling factor \( \mu \) has volume \( {v}_{\mu }\left( h\right) \) given by (8). The height \( {c}_{\mu }\left( h\right) \) of the centroid above the plane of its base is given by\n\n\[ \n{c}_{\mu }\left( h\right) = \frac{... | Proof. Consider first a single uniform elliptic shell element. Again it suffices to do the calculation for the prismatic counterpart. The inner wedge has altitude \( {\mu h} \), so by (9) its centroid is at height \( {3\mu h}/8 \) . The centroid of the outer wedge is at height \( {3h}/8 \) . If the outer wedge has volu... | Yes |
Theorem 9. Any slice of altitude \( z \) cut from a uniform elliptic wedge of altitude \( h \) has volume given by (11), where \( \lambda = z/h \) and \( V \) is the volume of the unpunctured prismatic container. The height \( c\left( z\right) \) of the centroid is given by | \[ c\left( z\right) = \frac{3}{4}z\frac{2 - {\lambda }^{2}}{3 - {\lambda }^{2}} \] | Yes |
Theorem 11. Corresponding horizontal cross sections of a general profile uniform dome and its punctured prismatic counterpart have equal areas if, and only if, each profile is elliptic. | As already remarked in Section 5, an elliptic dome can be deformed in such a way that areas of horizontal cross sections are preserved but the deformed dome no longer has elliptic profiles. At first glance, this may seem to contradict Theorem 11. However, such a deformation will distort the vertical walls; the dome wil... | No |
Proposition 2. The locus of the center of inversion mapping two given circles \( {O}_{i}\left( {a}_{i}\right), i = 1,2 \), into two congruent circles is the union of their midcircles \( {\mathcal{M}}_{ + } \) and \( {\mathcal{M}}_{ - } \) . | Proof. Let \( d\left( {P, Q}\right) \) denote the distance between two points \( P \) and \( Q \) . Suppose inversion in \( P \) with power \( \Phi \) transforms the given circles into congruent circles. By Lemma 1,\n\n\[ \frac{d{\left( P,{O}_{1}\right) }^{2} - {r}_{1}^{2}}{d{\left( P,{O}_{2}\right) }^{2} - {r}_{2}^{2}... | Yes |
Given three circles, the common points of their midcircles taken by pairs are the centers of inversion that map the three given circles into three congruent circles. | For \( i, j = 1,2,3 \), let \( {\mathcal{M}}_{ij} \) be a midcircle of the circles \( {\mathcal{C}}_{i} = {O}_{i}\left( {R}_{i}\right) \) and \( {\mathcal{C}}_{j} = \) \( {O}_{j}\left( {R}_{j}\right) \) . By Proposition 2 we have \( {\mathcal{M}}_{ij} = {R}_{j} \cdot {\mathcal{C}}_{i} + {\varepsilon }_{ij} \cdot {R}_{i... | Yes |
Corollary 4. The locus of the centers of the circles that intersect three given circles at equal angles are \( 0,1,2,3 \) or 4 lines through their radical center \( P \) perpendicular to a line joining three of their centers of similitude. | Proof. Let \( {\mathcal{C}}_{1} = A\left( {R}_{1}\right) ,{\mathcal{C}}_{2} = B\left( {R}_{2}\right) \), and \( {\mathcal{C}}_{3} = C\left( {R}_{3}\right) \) be the given circles. Consider three midcircles with collinear centers. If \( X \) is an intersection of these midcircles, reflection in the center line gives ano... | Yes |
Proposition 7. The points \( {F}_{1} \) and \( {F}_{2} \) are the Fermat-Torricelli points of degenerate triangles \( {ABC} \) and \( {M}_{0}{M}_{1}{M}_{2} \) . | Let the diameter of \( \left( {O}^{\prime }\right) \) parallel \( {AB} \) meet \( \left( {O}^{\prime }\right) \) in \( {G}_{1} \) and \( {G}_{2} \) and Let \( {G}_{1}^{\prime } \) and \( {G}_{2}^{\prime } \) be their feet of the perpendicular altitudes on \( {AB} \) . From Pappus’ theorem we know that \( {G}_{1}{G}_{2}... | Yes |
Lemma 1. Suppose \( L = l : m : n \) and \( P = p : q : r \) are points. Let \( \mathcal{P} \) denote the circumconic \( {p\beta \gamma } + {q\gamma \alpha } + {r\alpha \beta } = 0 \) and \( \mathcal{L} \) the line \( {l\alpha } + {m\beta } + {n\gamma } = 0 \) . There exists a unique point \( U \) such that if \( X \in... | Proof. We wish to solve the containment \( X \odot U \in \mathcal{L} \) for \( U \), given that \( X \in \mathcal{P} \) . That is, we seek \( u : v : w \) such that\n\n\[ \nu\left( {-{uyz} + {vzx} + {wxy}}\right) l + v\left( {{uyz} - {vzx} + {wxy}}\right) m + w\left( {{uyz} + {vzx} - {wxy}}\right) n = 0. \]\n\n(2)\n\ng... | Yes |
Theorem 2. Suppose \( {\mathcal{L}}_{1} \) is the line \( {l}_{1}\alpha + {m}_{1}\beta + {n}_{1}\gamma = 0 \) and \( {\mathcal{L}}_{2} \) is the line \( {l}_{2}\alpha + {m}_{2}\beta + {n}_{2}\gamma = 0 \) . There exists a unique point \( U \) such that if \( X \in {\mathcal{L}}_{1} \), then \( {X}^{-1} \odot U \in {\ma... | Proof. The hypothesis that \( X \in {\mathcal{L}}_{1} \) is equivalent to \( {X}^{-1} \in \mathcal{P} \), the circumconic having equation \( {l}_{1}{\beta \gamma } + {m}_{1}{\gamma \alpha } + {n}_{1}{\alpha \beta } = 0 \) . Therefore, the lemma applies to the circumconic \( \mathcal{P} \) and the line \( {\mathcal{L}}_... | Yes |
Corollary 3. Given a circumconic \( \mathcal{P} \) and a point \( U \), there exists a line \( \mathcal{L} \) such that if \( X \in \mathcal{P} \), then \( X{\bigodot U} \in \mathcal{L} \) . | Proof. Assuming there is such a \( \mathcal{L} \), we have the point \( U = {L}^{-1} \odot P \) as Theorem 2, so that \( {L}^{-1} = \operatorname{cevapoint}\left( {U, P}\right) \), and\n\n\[ L = {\left( \operatorname{cevapoint}\left( U, P\right) \right) }^{-1}, \]\n\nso that \( \mathcal{L} \) is the line \( \left( {{wq... | Yes |
Corollary 4. Given a line \( \mathcal{L} \) and a point \( U \), there exists a circumconic \( \mathcal{P} \) such that if \( X \in \mathcal{P} \), then \( X(\underline{ \odot }U \in \mathcal{L} \) . | Proof. Assuming there is such a \( \mathcal{L} \), we have the point \( U = {L}^{-1}( \odot P \), and \( P = \) \( {L}^{-1}(\mathcal{C}U \), so that \( \mathcal{P} \) is the circumconic\n\n\[ u\left( {-{ul} + {vm} + {wn}}\right) {\beta \gamma } + v\left( {{ul} - {vm} + {wn}}\right) {\gamma \alpha } + w\left( {{ul} + {v... | No |
Proposition 1. The quadrilateral \( q = {ABCD} \) is cyclic if and only if its orthocycle \( c \) is orthogonal to the circle \( f \) passing through the midpoints \( X, Y \) of its diagonals and their intersection point \( E \) . | Proof. If \( q \) is cyclic, then we have already seen that its orthocycle \( c \) belongs to the bundle \( {\mathcal{C}}^{\prime } \) which is orthogonal to the one generated by its circumcircle \( k \) and the circle \( f \) passing through the diagonal midpoints and their intersection point.\n\nConversely, if the or... | Yes |
Proposition 2. (1) There is a 1-1 correspondance between the members of the family \( \left( {k, E, c}\right) \) and the points \( J \) of the open arc \( \left( {OMP}\right) \) of circle \( c \) . | Proof. In fact, from the Introduction, it is plain that each member \( q \) of the family \( \left( {k, E, c}\right) \) defines a \( J \) as required. Conversely, a point \( J \) on arc \( \left( {OMP}\right) \) of \( c \) defines two intersection points \( X, Y \) of \( {HJ} \) with \( f \), which are inverse with res... | Yes |
Lemma 3. Consider a circle bundle of non intersecting type and two chords of a member circle passing through the limit point \( E \) of the bundle (see Figure 3). The chords define a quadrilateral \( q = {ABCD} \) having these as diagonals. Extend two opposite sides \( {AD},{BC} \) until they intersect a second circle ... | Indeed, \( N, M \) can be taken as the intersection points of opposite sides of \( q \) . Then \( N \) is on the polar of \( E \), hence the polar \( p\left( N\right) \) of \( N \) contains \( E \) . Consider the intersection points \( O, P \) of this polar with sides \( {HK},{IJ} \) respectively. Then, (a) these sides... | Yes |
Lemma 4. \( q\left( c\right) \) is bicentric. | Indeed, by Proposition 2 the bisectors of angles \( \angle {AGB},\angle {BFC} \) will intersect at \( M \) . It suffices to show that the bisectors of two opposite angles of \( q\left( c\right) \) intersect also at \( M \) . Let us show that the bisector of angle \( \angle {ABC} \) passes through \( M \) (Figure 4). We... | Yes |
Proposition 5. (1) There is a unique member \( q = q\left( c\right) = {ABCD} \), of the family \( \left( {k, E, c}\right) \) which is bicentric. The corresponding \( J \) is the limit point \( M \) of bundle \( I \) contained in the circle \( k \) . | Proof. In fact, by the previous lemmas we know that \( q\left( c\right) \) is bicentric. To prove the uniqueness we assume that \( q = {ABCD} \) is bicircular and consider the incircle \( g \) and the tangential quadrilateral \( q = {UVWZ} \) . From Brianchon’s theorem the diagonals of \( {q}^{\prime } \) intersect als... | Yes |
Proposition 6. Consider all tripples \( \left( {k, E, c}\right) \) with fixed \( k, E \) and \( c \) running through the members of the circle bundle \( {\mathcal{C}}^{\prime } \) . Denote by \( q\left( c\right) \) the bicentric member of the corresponding family \( \left( {k, E, c}\right) \) and by \( {q}^{\prime } = ... | Proof. (1) follows from the fact that \( {UV} \) is orthogonal to the bisector \( {FM} \) of angle \( {BFC} \) . Analogously \( {VZ} \) is orthogonal to \( {GM} \) and \( {FM},{GM} \) are orthogonal ([2, §674]). | Yes |
Proposition 7. (1) For fixed \( \left( {k, E}\right) \), the set of all bicentrics \( \left\{ {q\left( c\right) : c \in {\mathcal{C}}^{\prime }}\right\} \) contains exactly one member which is simultaneously bicentric and orthodiagonal. It corresponds to the minimum circle of bundle \( {\mathcal{C}}^{\prime } \), is ki... | Proof. In fact, by applying the previous results to the tangential quadrilateral \( \dot{q} \) of \( q\left( c\right) \), we know that the orthocycle of \( {q}^{\prime } \) is orthogonal to the fixed incircle \( g \) and passes through two fixed points, depending only on \( \left( {k, E}\right) \) (the limit points of ... | Yes |
Proposition 8. Consider the tangential quadrilateral \( q = \) QRST circumscribed on the circumcircle \( k \) of the cyclic quadrilateral \( q = {ABCD} \) (Figure 9). The following facts are true:\n\n(1) The diagonals of \( {q}^{\prime } \) and \( q \) intersect at the same point \( E \) .\n\n(2) The pairs of opposite ... | Proof. (1) is a consequence of Brianchon's theorem (see a simpler proof in [5, p.157]). Identify the polar of \( E \) with the diameter \( e = {FG} \) of the orthocycle of \( q \) . The polar of \( F \) is \( {PG} \) and the polar of \( G \) is \( {OF} \) . \( T \) is the pole of \( {AB} \) which contains \( F \) . Hen... | Yes |
Proposition 9. For each quadrangle of the family \( q \in \left( {k, E, c}\right) \) construct the tangential quadrangle \( {q}^{\prime } = {QRST} \) of \( q = {ABCD} \) (Figure 10). The following facts are true.\n\n(1) There is exactly one \( {q}_{0} \in \left( {k, E, c}\right) \) whose corresponding tangential \( q \... | Proof. \( {q}^{\prime } \) being cyclic and circumscriptible it is bicentric. Hence the lines joining opposite contact points must be orthogonal and the orthocycle of \( \dot{q} \) passes through \( K \) . This follows from Proposition 2. Thus \( p = {EXKY} \) is a rectangle, \( {XY} \) being a diameter of the circle \... | Yes |
Proposition 10. (1) For each cyclic quadrilateral \( q = {ABCD} \) of the family \( \left( {k, E, c}\right) \) there is an orthodiagonal \( p = \) QRST whose diagonals coincide with the sides of the right angled triangle \( t = {FGM} \), defined by the limit point \( M \) of bundle \( \mathcal{C} \) and the intersectio... | Proof. Consider the lines orthogonal to \( {MA},{MB},{MC},{MD} \) at the vertices of \( q \) (Figure 11). They build a quadrilateral. To show the statement on the diagonals consider the two resulting cyclic quadrilaterals \( {q}_{1} = {MATB} \) and \( {q}_{2} = {MCRD} \) . Point \( F \) lies on the radical axis of thei... | Yes |
Lemma 1. When \( T \) traverses the Brocard axis, the locus of \( {X}_{a} \) is a conic \( {\gamma }_{a} \) . | Proof. This can be obtained through easy calculation. Here is a synthetic proof. Consider the projections \( {\pi }_{1} \) from the line \( {AC} \) onto the line \( {BC} \) in the direction of \( {H}_{a}{H}_{b} \), and \( {\pi }_{2} \) from the line \( {BC} \) onto the line \( {AB} \) in the direction of \( {AC} \) . C... | Yes |
Lemma 2. \( {\gamma }_{a},{\gamma }_{b} \) and \( {\gamma }_{c} \) have three common points \( {X}_{i}, i = 1,2,3 \), and one of them is always real. | Proof. Indeed, \( {\gamma }_{b} \) and \( {\gamma }_{c} \) for example meet at \( A \) and three other points, one of them being necessarily real. On the other hand, it is clear that any point \( X \) lying on two conics must lie on the third one. | Yes |
Theorem 4. The three lines \( {\mathcal{L}}_{i} \) passing through \( {X}_{i},{Y}_{i} \) are parallel and perpendicular to the Brocard axis \( {OK} \) . | Proof. We know (see [3]) that, for any bicevian conic \( \mathcal{C}\left( {P, Q}\right) \), there is an inscribed conic bitangent to \( \mathcal{C}\left( {P, Q}\right) \) at two points lying on the line \( {PQ} \) . On the other hand, any Tucker circle is bitangent to the Brocard ellipse and the line through the conta... | Yes |
Proposition 7. Each net of conics ( \( \mathcal{N} \) and \( {\mathcal{N}}^{\prime } \) ) contains one and only one circle. These circles \( \Gamma \) and \( {\Gamma }^{\prime } \) contain \( {X}_{110} \), the focus of the Kiepert parabola. | These circles are\n\n\( \Gamma : \;\mathop{\sum }\limits_{\text{cyclic }}{b}^{2}{c}^{2}\left( {{b}^{2} - {c}^{2}}\right) \left( {{a}^{2} - {b}^{2}}\right) {x}^{2} + {a}^{2}\left( {{b}^{2} - {c}^{2}}\right) \left( {{c}^{4} + {a}^{2}{b}^{2} - 2{a}^{2}{c}^{2}}\right) {yz} = 0 \)\n\nand\n\n\[{\Gamma }^{\prime } : \mathop{\... | Yes |
Theorem 1. A circle through \( O \) (not tangent internally to \( \beta \) ) is Archimedean if and only if its external common tangents with \( \beta \) intersect at a point on \( \alpha \) . | Proof. Consider a circle \( \delta \) with radius \( r \neq b \) and center \( \left( {r\cos \theta, r\sin \theta }\right) \) for some real number \( \theta \) with \( \cos \theta \neq - 1 \) . The intersection of the common external tangents of \( \beta \) and \( \delta \) is the external center of similitude of the t... | Yes |
Corollary 2. Let \( \delta \) be an Archimedean circle with a diameter \( {OT} \), and \( {T}_{\alpha } \) the intersection of the external common tangents of the circles \( \delta \) and \( \beta \) ; similarly define \( {T}_{\beta } \). (i) The vectors \( \overrightarrow{OT} \) and \( \overrightarrow{{O}_{\alpha }{T}... | Proof. We describe the center of \( \delta \) by \( \left( {{r}_{A}\cos \theta ,{r}_{A}\sin \theta }\right) \) for some real number \( \theta \) (see Figure 2). Then the point \( {T}_{\alpha } \) is described by \[ \left( {\frac{b{r}_{A}\left( {1 + \cos \theta }\right) }{b - {r}_{A}},\frac{b{r}_{A}\sin \theta }{b - {r}... | Yes |
Theorem 1. Let \( s \) and \( t \) be nonzero real numbers such that \( {tb} \pm {sa} \neq 0 \) . If there is a circle of radius \( \rho \) touching the circles \( \alpha \left( {nsa}\right) \) and \( \beta \left( {ntb}\right) \) appropriately for a real number \( n \), then its center lies on the line\n\n\[ x = \frac{... | Proof. Consider the center \( \left( {x, y}\right) \) of the circle with radius \( \rho \) touching \( \alpha \left( {nsa}\right) \) and \( \beta \left( {ntb}\right) \) appropriately. The distance between \( \left( {x, y}\right) \) and the centers of \( \alpha \left( {nsa}\right) \) and \( \beta \left( {ntb}\right) \) ... | No |
Theorem 2. If a circle \( \mathcal{C} \) of radius \( \rho \) touches \( \alpha \left\lbrack n\right\rbrack \) and \( \beta \left\lbrack n\right\rbrack \) appropriately for a real number \( n \), then the inclination of \( \mathcal{C} \) and \( \gamma \) is \( \frac{2r}{\rho } - n \) . | Proof. The square of the distance between the centers of the circles \( \mathcal{C} \) and \( \gamma \) is \( {\left( \rho + n\left( a + b\right) \right) }^{2} - {\left( n\left( a + b\right) \right) }^{2} + {\left( a - b\right) }^{2} \) by the Pythagorean theorem. Therefore their inclination is\n\n\[ \frac{{\rho }^{2} ... | Yes |
Corollary 3. A circle touching \( \alpha \left\lbrack n\right\rbrack \) and \( \beta \left\lbrack n\right\rbrack \) appropriately for a real number \( n \) is Archimedean if and only if the inclination of this circle and \( \gamma \) is \( 2 - n \) . | This gives an infinite set of Archimedean circles \( {\delta }_{n} \) with centers on the positive \( y \) -axis. The circle \( {\delta }_{n} \) exists if \( n \geq \frac{-r}{2\left( {a + b}\right) } \), and the maximal value of the inclination of \( \gamma \) and \( {\delta }_{n} \) is \( 2 + \frac{r}{2\left( {a + b}\... | Yes |
Theorem 1. For a given \( m \), the right-triangle solutions \( \left( {a, b, c}\right) \) to the problem \( A = {mP} \) are determined from the relations\n\n\[ 8{m}^{2} = \left( {a - {4m}}\right) \left( {b - {4m}}\right) \]\n\n(1)\n\n\[ c = a + b - {4m}\text{.} \]\n\n(2) | Each factorization\n\n\[ 8{m}^{2} = {d}_{1} \cdot {d}_{2} \]\n\n(3)\n\nwhere\n\n\[ {d}_{1} \leq \lfloor {2m}\sqrt{2}\rfloor \]\n\n(4)\n\ngenerates a solution triangle with sides given by the formulas\n\n\[ \left\{ \begin{array}{l} a = {d}_{1} + {4m} \\ b = {d}_{2} + {4m} \\ c = {d}_{1} + {d}_{2} + {4m} \end{array}\righ... | Yes |
For a given \( m \), all solutions \( \left( {a, b, c}\right) \) to the problem \( A = {mP} \) are determined as follows: Find all divisors \( u \) of \( {2m} \) ; for each \( u \), find all numbers \( v \) relatively prime to \( u \) and such that \( 1 \leq v \leq \lfloor \sqrt{3}u\rfloor \) ; to each pair \( u \) and... | Each factorization\n\n\[ 4{m}^{2}\left( {{u}^{2} + {v}^{2}}\right) = {\delta }_{1} \cdot {\delta }_{2} \]\n\n(8)\n\nwhere\n\n\[ {\delta }_{1} \leq \left\lfloor {{2m}\sqrt{{u}^{2} + {v}^{2}}}\right\rfloor \]\n\n(9)\n\nand only those factors \( {\delta }_{1},{\delta }_{2} \) for which \( v\left| {{\delta }_{1} + {2mu}\te... | Yes |
For a fixed \( m \), solving the problem \( A = {mP} \) is equivalent to determining all integer \( a, b, c \) that satisfy the equation\n\n\[ \n{\left\lbrack {c}^{2} - \left( {a}^{2} + {b}^{2}\right) \right\rbrack }^{2} + {\left\lbrack 4m\left( a + b + c\right) \right\rbrack }^{2} = {\left( 2ab\right) }^{2}, \n\]\n\no... | It is easy to see that the first equation in (13) can be interpreted as follows. | No |
Theorem 1. A projective linear function \( f\left( {x, y, z}\right) \) satisfies (2),(3), and (4) if and only if\n\n\[ f\left( {x, y, z}\right) = \frac{\left( {1 - {2t}}\right) x + t\left( {y + z}\right) }{x + y + z} \]\n\n(5)\n\nfor some \( t \) . If \( {\mathcal{S}}_{t} \) is the center function defined by (5) (and(1... | Proof. Let \( f\left( {x, y, z}\right) = {L}_{0}/{M}_{0} \), where \( {L}_{0} \) and \( {M}_{0} \) are linear forms in \( x, y \), and \( z \) , and suppose that \( f \) satisfies (2),(3), and (4). Let \( \sigma \) be the cycle \( \left( {xyz}\right) \), and let \( {L}_{i} = {\sigma }^{i}\left( {L}_{0}\right) \) and \(... | Yes |
Theorem 2. Let \( {ABC} \) be a scalene triangle and let \( V \) be any point in its plane. Then there exist unique real numbers \( t \) and \( v \) such that \( V \) is the center of \( {ABC} \) with respect to the center function \( {\mathcal{Q}}_{t, v} \) defined by the projective quadratic function \( f \) given by... | Proof. Clearly \( f \) satisfies the conditions (2),(3), and (4). Since \( V \) is in the plane of \( {ABC} \), it follows that \( V = {\xi A} + {\eta B} + {\zeta C} \) for some \( \xi ,\eta \), and \( \zeta \) with \( \xi + \eta + \zeta = 1 \) . Let \( a, b \), and \( c \) be the side-lengths of \( {ABC} \) as usual. ... | Yes |
Theorem 3. Let \( \mathcal{Z} \) be a center function, and let \( {ABC} \) be any scalene triangle in the domain of \( \mathcal{Z} \) . Then there exists another center function \( {\mathcal{Z}}^{\prime } \) defined by a projective function \( f \) such that \( \mathcal{Z}\left( {A, B, C}\right) = {\mathcal{Z}}^{\prime... | Proof. Let \( \mathbf{F} \) and \( \mathbf{G} \) be the families of centers defined in Theorem 2 and in Remark 4. Clearly, the centroid is the only center function that these two families have in common.\n\nIf \( \mathcal{Z} \notin \mathbf{F} \), then we use Theorem 2 to produce the center \( {\mathcal{Z}}^{\prime } = ... | Yes |
Lemma 1. If \( a, b, c \) denote the lengths of the sides, \( R \) the circumradius, and \( K \) the area of a triangle, then \( K = \frac{abc}{4R} \) . | Proof. From Figure 2,\n\n\[ \frac{h}{b} = \frac{\frac{a}{2}}{R} \Rightarrow h = \frac{ab}{2R} \Rightarrow K = \frac{1}{2}{hc} = \frac{abc}{4R}. \]\n | Yes |
Lemma 2 ([2]). Under the hypotheses of the Theorem, \( \frac{p}{q} = \frac{{ad} + {bc}}{{ab} + {cd}} \) . | Proof. From Figures 3 and 4 respectively,\n\n\[ \nQ = {K}_{1} + {K}_{2} = \frac{pab}{4R} + \frac{pcd}{4R} = \frac{p\left( {{ab} + {cd}}\right) }{4R}, \n\] \n\n\[ \nQ = {K}_{3} + {K}_{4} = \frac{qad}{4R} + \frac{qbc}{4R} = \frac{q\left( {{ad} + {bc}}\right) }{4R}. \n\] \n\nTherefore, \n\n\[ \np\left( {{ab} + {cd}}\right... | Yes |
Proposition 3. (2) The point \( {P}_{2} \) lies on the line joining the incenter to \( Q \) . | Proof. We need only prove (2). This is clear from\n\n\[ \n{P}_{2} = \left( {1 - \cos \frac{A}{2} - \cos \frac{B}{2} - \cos \frac{C}{2}}\right) I + \left( {\cos \frac{A}{2} + \cos \frac{B}{2} + \cos \frac{C}{2}}\right) Q.\n\]\n\nIn fact,\n\n\[ \n{P}_{2} = I + \left( {\cos \frac{A}{2} + \cos \frac{B}{2} + \cos \frac{C}{2... | Yes |
Proposition 4. The barycentric coordinates of \( {P}_{1}^{ * } \) are\n\n\[ \left( {\cos \frac{A}{2}{\sin }^{2}\frac{A}{4} : \cos \frac{B}{2}{\sin }^{2}\frac{B}{4} : \cos \frac{C}{2}{\sin }^{2}\frac{C}{4}}\right) . \] | Proof. This follows from\n\n\[ {P}_{1}^{ * } = \left( {\frac{{\sin }^{2}A}{\cos \frac{A}{2}{\cos }^{2}\frac{A}{4}} : \frac{{\sin }^{2}B}{\cos \frac{B}{2}{\cos }^{2}\frac{B}{4}} : \frac{{\sin }^{2}C}{\cos \frac{C}{2}{\cos }^{2}\frac{C}{4}}}\right) \]\n\n\[ = \left( {\frac{{\sin }^{2}\frac{A}{2}\cos \frac{A}{2}}{{\cos }^... | Yes |
Proposition 6. The perpendiculars to \( {B}_{a}{C}_{a} \) at \( {X}^{\prime } \), to \( {C}_{b}{A}_{b} \) at \( {Y}^{\prime } \), and to \( {A}_{c}{B}_{c} \) at \( {Z}^{\prime } \) are concurrent at the reflection of \( {P}_{2} \) in \( I \), which is the point\n\n\[ \n{P}_{2}^{\prime } = a\left( {1 + \cos \frac{B}{2} ... | Proof. Let \( {P}_{2}^{\prime } \) be the reflection of \( {P}_{2} \) in \( I \) . Since \( X \) and \( {X}^{\prime } \) are symmetric in the midpoint of \( {B}_{a}{C}_{a} \), and \( {P}_{2}X \) is perpendicular to \( {B}_{a}{C}_{a} \), it follows that \( {P}_{2}^{\prime }{X}^{\prime } \) is also perpendicular to \( {B... | Yes |
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