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Proposition 9. Referring to Figure 13, the centers of the sides of the main flanks are aligned as shown and the corresponding lines intersect at the outer diagonal of \( q \) i.e. the line joining the intersection points of opposite sides of \( q \) . | This is due to the fact that the main flanks are antihomothetic with respect to the vertices of \( q \) . Thus, the parallelograms of the main flanks are homothetic to each other and their homothety centers are aligned by three on a line. Later assertion can be reduced to the well known one for similarity centers of th... | No |
Proposition 11. Referring to Figure 16, the quadrilateral \( \left\lbrack t\right\rbrack \left\lbrack r\right\rbrack \left\lbrack b\right\rbrack \left\lbrack l\right\rbrack \) of the anticen-ters of the main flanks is symmetric to the quadrilateral of the anticenters of the peripheral flanks [tlb][ltr][trb][lbr]. The s... | By the way, the symmetry of center and anticenter about the barycenter leads to a simple proof of the chararacteristic property of the anticenter. Indeed, consider the symmetric \( {A}^{ * }{B}^{ * }{C}^{ * }{D}^{ * } \) of \( q \) with respect to the barycenter of \( q \) . The orthogonal from the middle of one side o... | No |
Lemma 14. Referring to Figure 23, lines \( {XZY} \) and \( {UVW} \) are defined by the intersection points of the opposite sides of main flanks of \( q \) . Line \( {X}^{ * }{U}^{ * } \) is defined by the intersection points of the opposite sides of \( \dot{q} \) . The figure has the properties: (1) Lines \( {XZY} \) a... | (1) is obvious, since lines \( {UVW} \) and \( {XZY} \) are diagonals of the parallelograms \( {X}^{ * }{WXV} \) and \( {U}^{ * }{YUZ} \) . (2) is also trivial since these triangles result from the extension of sides of similar quadrilaterals, namely \( q \) and its main flanks. (3) and (4) is a consequence of (2). The... | No |
Proposition 15. Referring to Figure 24, consider a quadrilateral \( {\dot{q}}^{ * } = {ABCD} \) and trisect the segmet \( {QR} \), with end-points the intersections of opposite sides of \( {q}^{ * } \) . From trisecting points \( U, V \) draw two arbitrary parallels \( {UY},{VX} \) intersecting the sides of \( {q}^{ * ... | In fact, (1) is trivial and (2) follows from (1) and an easy calculation of the ratios of the sides of the flanks. To prove (3) consider point \( S \) varying on side \( {BC} \) of \( \overset{*}{q} \) . Define the two parallels and in particular \( {VX} \) by joining \( V \) to \( S \) . Thus, the two parallels and th... | No |
Proposition 16. Given a circular quadrilateral \( {q}^{ * } = {A}^{ * }{B}^{ * }{C}^{ * }{D}^{ * } \) there is another circular quadrilateral \( q = {ABCD} \), whose cyclic complex has corresponding big flank the given one. | The proof follows immediately by applying (3) of the previous proposition to the given cyclic quadrilateral \( {q}^{ * } \) . In that case, the condition of the equality of ratios implies that the constructed by the proposition central quadrilateral \( q \) is similar to the main flanks. Thus the given \( {q}^{ * } \) ... | Yes |
Theorem 1. The pivotal cubic \( \mathrm{p}\mathcal{K}\left( {\Omega, P}\right) \) has normals at \( A, B, C \) concurrent if and only if\n\n(1) \( P \) lies on \( {\mathcal{C}}_{\Omega } \), equivalently, \( {P}^{ * } \) lies on the line at infinity, or\n\n(2) \( P \) lies on \( {\mathcal{L}}_{\Omega } \), equivalently... | More precisely, in (1), the tangents at \( A, B, C \) are parallel since \( {P}^{ * } \) lies on the line at infinity. Hence the normals are also parallel and \ | No |
Lemma 1. Let \( {ABC} \) be a non-degenerate triangle, and let \( x, y \), and \( z \) be real numbers such that\n\n\[ x{\left( A - B\right) }^{2} + y{\left( B - C\right) }^{2} + z{\left( C - A\right) }^{2} = 0.\]\n\n(1)\n\nThen either \( x = y = z = 0 \), or \( {xy} + {yz} + {zx} > 0 \) . | Proof. Let \( S = \frac{A - B}{A - C} \) be the shape of \( {ABC} \) . Dividing (1) by \( {\left( A - C\right) }^{2} \), we obtain\n\n\[ \left( {x + y}\right) {S}^{2} - {2yS} + \left( {y + z}\right) = 0.\]\n\n(2)\n\nSince \( {ABC} \) is non-degenerate, \( S \) is not real. Thus if \( x + y = 0 \), then \( y = 0 \) and ... | Yes |
Lemma 2. Suppose that the cevians through an interior point \( P \) of a triangle divide the sides in the ratios \( u : 1 - u, v : 1 - v \), and \( w : 1 - w \) . Then\n\n(i) \( {uvw} \leq \frac{1}{8} \), with equality if and only if \( u = v = w = \frac{1}{2} \), i.e., if and only if \( P \) is the centroid.\n\n(ii) \... | Proof. Let \( {uvw} = p \) . Then using the cevian condition \( {uvw} = \left( {1 - u}\right) \left( {1 - v}\right) \left( {1 - w}\right) \), we see that\n\n\[ p = \sqrt{u\left( {1 - u}\right) }\sqrt{v\left( {1 - v}\right) }\sqrt{w\left( {1 - w}\right) } \]\n\n\[ \leq \frac{u + \left( {1 - u}\right) }{2}\frac{v + \left... | Yes |
Theorem 3. Let \( A{A}^{\prime }, B{B}^{\prime } \), and \( C{C}^{\prime } \) be the cevians through an interior point \( P \) of triangle \( {ABC} \) . Then triangles \( {ABC} \) and \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) are similar if and only if \( P \) is the centroid of \( {ABC} \) . | Proof. One direction being trivial, we assume that \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) and \( {ABC} \) are similar, and we prove that \( P \) is the centroid.\n\nSuppose that the cevians \( A{A}^{\prime }, B{B}^{\prime } \), and \( C{C}^{\prime } \) through \( P \) divide the sides \( {BC} \) , \( {CA} \), an... | Yes |
Theorem 1. The orthocentroidal circle forms part of a coaxal system of circles including the circumcircle, the nine-point circle and the polar circle of the triangle. | It is possible to prove the next result by calculating that \( {JK} < {OG} \) directly (recall that \( J \) is the midpoint of \( {GH} \) ), but it is easier to use the equation of the orthocentroidal circle. | No |
Theorem 2. The symmedian point lies in the disc \( {\mathcal{D}}_{GH} \) . | Proof. Substituting \( u = {a}^{2}, v = {b}^{2}, w = {c}^{2} \) in the left hand side of equation (4) we get \( {a}^{4}{b}^{2} + {b}^{4}{c}^{2} + {c}^{4}{a}^{2} + {b}^{4}{a}^{2} + {c}^{4}{b}^{2} + {a}^{4}{c}^{2} - 3{a}^{2}{b}^{2}{c}^{2} - {a}^{6} - {b}^{6} - {c}^{6} \) and this quantity is negative for all real \( a, b... | Yes |
Theorem 3. One Brocard point lies in \( {\mathcal{D}}_{GH} \) and the other lies outside \( {\mathcal{S}}_{GH} \), or they both lie simultaneously on \( {\mathcal{S}}_{GH} \) (which happens if and only if the reference triangle is isosceles). | Proof. Let \( f\left( {u, v, w}\right) \) denote the left hand side of equation (4). One Brocard point has unnormalised areal co-ordinates\n\n\[ \left( {u, v, w}\right) = \left( {{a}^{2}{b}^{2},{b}^{2}{c}^{2},{c}^{2}{a}^{2}}\right) \]\n\nand the other has unnormalised areal co-ordinates\n\n\[ \left( {p, q, r}\right) = ... | Yes |
Theorem 4. Gergonne’s point lies in the orthocentroidal disk \( {\mathcal{D}}_{GH} \) . | Proof. Put \( u = \left( {c + a - b}\right) \left( {a + b - c}\right), v = \left( {a + b - c}\right) \left( {b + c - a}\right), w = \) \( \left( {b + c - a}\right) \left( {c + a - b}\right) \) and the left hand side of (5) becomes\n\n\[ \n- {18}{a}^{2}{b}^{2}{c}^{2} + \mathop{\sum }\limits_{\text{cyclic }}\left( {-{a}^... | Yes |
Theorem 7. The point \( {F}_{\mathrm{e}} \) is always outside the orthocentroidal circle. | Proof. Let \( J \) be the center of the orthocentroidal circle, and \( N \) be its nine-point center. In [9] it was established that\n\n\[ I{J}^{2} = O{G}^{2} - \frac{2r}{3}\left( {R - {2r}}\right) . \]\n\nWe have \( {IN} = R/2 - r, I{F}_{\mathrm{e}} = r \) and \( {JN} = {OG}/2 \). We may apply Stewart’s theorem to tri... | Yes |
Lemma 1. Let \( \left( K\right) \) be a circle with center \( K \) and \( {\ell }_{1} \) and \( {\ell }_{2} \) be two tangents to \( \left( \left( K\right) \) meeting in a point \( P \) . Let \( L \) be a point travelling through the line \( {PK} \) . When \( L \) travels linearly, then the radical axis of \( \left( K\... | Proof. Let \( P \) be the origin for Cartesian coordinates. Without loss of generality let \( {\ell }_{1} \) and \( {\ell }_{2} \) be lines making angles \( \pm \phi \) with the \( x \) -axis and let \( K\left( {{x}_{1},0}\right), L\left( {{x}_{2},0}\right) \) . With \( v = \sin \phi \) the circles \( \left( K\right) \... | Yes |
Corollary 6. In Theorem 2, the points \( S, P \), and \( I \) are collinear (Figure 6). | Proof. Since \( P \) is a center of similitude of circles \( \left( I\right) \) and \( \left( S\right), P \) must be collinear with the centers of the two circles. | Yes |
Corollary 10 (Kimberling [3]). Let \( \left( S\right) \) be the circle circumscribing the three ex-circles of triangle \( {ABC} \), i.e., internally tangent to each of them. Let the points of tangency of \( \left( S\right) \) with the excircles be \( X, Y \), and \( Z \) . (Figure 9). Then \( {AX},{BY} \) , and \( {CZ}... | The point of concurrence is known as the Apollonius point of the triangle. It is \( {X}_{181} \) in [5]. See also [4, p.102]. | Yes |
Theorem 14. The six centers of similitude of three spheres taken in pairs lie by threes on four straight lines. In particular, the three external centers of similitude are collinear; and any two internal centers of similitude are collinear with the third external one. | Proof. Consider the plane through the centers of the three spheres. This plane passes through all 6 centers of similitude. The plane cuts each sphere in a circle. Thus, on this plane, Proposition 5 applies, thus proving that the result holds for the spheres as well. | No |
Theorem 15. Let \( T = {A}_{1}{A}_{2}{A}_{3}{A}_{4} \) be a tetrahedron. Let \( \left( S\right) \) be any sphere in the interior of \( T \) (or let \( \left( S\right) \) be any sphere surrounding \( T \) ). Suppose there are four spheres, \( \left( {S}_{1}\right) ,\left( {S}_{2}\right) ,\left( {S}_{3}\right) \), and \(... | The proof of this theorem is exactly the same as the proof of Theorem 2, replacing the reference to Proposition 5 by Theorem 14. | Yes |
Lemma 1. If \( {\Delta }^{\prime } = \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) has \( {A}^{\prime } = {a}_{1} : {a}_{2} : {a}_{3},{B}^{\prime } = {b}_{1} : {b}_{2} : {b}_{3} \) , \( {C}^{\prime } = {c}_{1} : {c}_{2} : {c}_{3} \), then,\n\n(a) \( {GP}\left( {\Delta }^{\prime }\right) \) has equation\n\n\... | Proof. It is easy to see that, if \( P = x : y : z \), then\n\n\[ {A}_{P} = {A}^{\prime }P \cap {BC} = 0 : {a}_{2}x - {a}_{1}y : {a}_{3}x - {a}_{1}z, \]\n\n\[ {B}_{P} = {B}^{\prime }P \cap {CA} = {b}_{1}y - {b}_{2}x : 0 : {b}_{3}y - {b}_{2}z, \]\n\n\[ {C}_{P} = {C}^{\prime }P \cap {AB} = {c}_{1}z - {c}_{3}x : {c}_{2}z ... | Yes |
Lemma 2. If \( \Delta = \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) has \( {A}^{\prime } = {a}_{1} : {a}_{2} : {a}_{3},{B}^{\prime } = {b}_{1} : {b}_{2} : {b}_{3},{C}^{\prime } = {c}_{1} \) : \( {c}_{2} : {c}_{3} \), then,\n\n(a) \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) is perspective ... | Proof. We observe that the perspectivity in (a) is equivalent to the concurrence of \( A{A}^{\prime }, B{B}^{\prime } \) and \( C{C}^{\prime } \) . The given equality expresses the condition for the intersection of \( A{A}^{\prime } \) and \( B{B}^{\prime } \) to lie on \( C{C}^{\prime } \) . Part (b) follows by noting... | Yes |
Theorem 3. For a triangle \( {\Delta }^{\prime } = \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) , (a) \( {GP}\left( \Delta \right) \) is of type \( \mathrm{p}\mathcal{K} \) if and only if \( {\Delta }^{\prime } \) is perspective with \( \bigtriangleup {ABC} \) . | Proof. Suppose that \( {A}^{\prime } = {a}_{1} : {a}_{2} : {a}_{3},{B}^{\prime } = {b}_{1} : {b}_{2} : {b}_{3},{C}^{\prime } = {c}_{1} : {c}_{2} : {c}_{3} \), with \( {a}_{1}{b}_{2}{c}_{3} \neq 0 \) . (a) We begin by observing that equation (1) gives a cubic of type \( \mathrm{p}\mathcal{K} \) if and only if \( {f}_{1}... | Yes |
Theorem 4. Suppose that \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) is perspective to \( \bigtriangleup {ABC} \), with \( {A}^{\prime } \) not on \( {BC} \) , \( {B}^{\prime } \) not on \( {CA} \), and \( {C}^{\prime } \) not on \( {AB} \) . Let the perspector be \( {P}_{1} = {p}_{11} : {p}_{12} : {p}_... | Proof. (a) Since we have the perspector, the coordinates of the vertices \( A,{B}^{\prime },{C}^{\prime } \) must be as described. | No |
Theorem 5. Suppose that \( P \) is a point on \( \mathrm{p}\mathcal{K}\left( {W, S}\right) \) which is not fixed by \( W \) -isoconjugation, and is not \( S \) or its \( W \) -isoconjugate. Then there is a unique desmic structure with vertices \( A, B, C \) and perspector \( {P}_{1} = P \) with locus \( {GP} = \mathrm{... | Proof. For brevity, we shall write \( {X}^{ * } \) for the \( W \) -isoconjugate of a point \( X \) . As \( P \) is on \( \mathrm{p}\mathcal{K}\left( {W, S}\right), S \) is on \( P{P}^{ * } \) . Then \( S = {mP} + n{P}^{ * } \), for some constants \( m, n \) , with \( {mn} \neq 0 \) . If \( P = p : q : r, W = u : v : w... | Yes |
(a) The vertices of the desmic structure on \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) are \( A, B, C \) and intersections of \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) with the cubics \( \mathrm{p}\mathcal{K}\left( {W,{R}_{A}}\right) ,\mathrm{p}\mathcal{K}\left( {W,{R}_{B}}\right) ,\mathrm{p}\mathcal{K... | Proof. We observe that the vertices of the desmic structure can be derived from the normalized versions of the perspectors \( {P}_{1} \) and \( {P}_{2} \) . The normalization is such that \( R = {P}_{1} - {P}_{2} \) and \( S = {P}_{1} + {P}_{2} \) . Now consider the loci derived from the perspectors \( {P}_{1} \) and \... | Yes |
Theorem 7. If a cubic \( \mathcal{C} \) is of type \( \mathrm{n}\mathcal{K} \), but not of type \( \mathrm{c}\mathcal{K} \), then there is a unique desmic structure which defines \( \mathcal{C} \) as a Grassmann cubic. | Proof. Suppose that \( \mathcal{C} = \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) with \( W = u : v : w \), and \( R = r : s : t \) . We require perspectors \( {P}_{1} = {p}_{11} : {p}_{12} : {p}_{13} \), and \( {P}_{2} = {p}_{21} : {p}_{22} : {p}_{23} \) such that \( r : s : t = {p}_{11} - {p}_{21} : {p}_{12} - {p}... | Yes |
Theorem 8. Suppose that \( \mathcal{C} = \mathrm{p}\mathcal{K}\left( {W, S}\right) \) with \( W = u : v : w \), and \( S = r : s : t \) . (a) Any child of \( \mathcal{C} \) is of the form \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \), with \( R \) on\n\n\[ \mathrm{p}\mathcal{C}\left( {W, S}\right) : \;\left( {v{t}... | Proof. (a) We know from Theorem 7 that a child \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) of \( \mathcal{C} \) arises from a desmic structure. Suppose the perspectors are \( {P}_{1} \) and \( {P}_{2} \) . From Theorem 4(f) \( \left( {{P}_{1},{P}_{2}, R, S}\right) \) is a harmonic range. It follows that there ar... | Yes |
Theorem 9. Suppose that \( W = u : v : w, R = r : s : t \) and \( k \) are such that \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) is defined by a desmic structure. Let \( \mathrm{p}\mathcal{K}\left( {W, S}\right) \) be the parent of \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) .\n\n(a) The line \( {RS} \) i... | Proof. Suppose that the desmic structure has normalized perspectors \( {P}_{1} = f : g \) : \( h \) and \( {P}_{2} = {f}^{\prime } : {g}^{\prime } : {h}^{\prime } \) . Then we have\n\n\[ r = f - {f}^{\prime },\;s = g - {g}^{\prime },\;t = h - {h}^{\prime }; \]\n\n\[ u = f{f}^{\prime },\;v = g{g}^{\prime },\;w = h{h}^{\... | Yes |
The second Brocard cubic \( \mathrm{n}{\mathcal{K}}_{0}\left( {K, X\left( {523}\right) }\right) \), Gibert’s \( {K018} \). We cannot identify the perspectors of the desmic structure. They are complex. | Theorem 9(b) gives the parent as \( \mathrm{p}\mathcal{K}\left( {K, X\left( 5\right) }\right) \) - the Napolean cubic, and Gibert’s \( {K005} \) | No |
Our next result identifies the children of a given \( \mathrm{p}\mathcal{K}\left( {W, R}\right) \) which are of the form \( \mathrm{n}{\mathcal{K}}_{0}\left( {W, R}\right) \). | It turns out that the perspectors must lie on another cubic \( \mathrm{n}{\mathcal{K}}_{0}\left( {W, T}\right) \) . The root \( T \) is most neatly defined using the generalization of Gibert’s \( {PK} \) -transform. | No |
Theorem 10. Suppose that \( \left\{ {P,{P}^{ * }}\right\} \neq \left\{ {S,{S}^{ * }}\right\} \) are a pair of \( W \) -isoconjugates on \( \mathrm{p}\mathcal{K}\left( {W, S}\right) \) . Then they define a desmic structure with associated cubic of the form \( \mathrm{n}{\mathcal{K}}_{0}\left( {W, R}\right) \) if and onl... | Proof. Suppose that \( P = x : y : z, S = r : s : t, W = u : v : w \) . Then the point \( {P}^{ * } \) is \( u/x : v/y : w/z \) . As \( P \) is on \( \mathrm{p}\mathcal{K}\left( {W, S}\right) \), there exist constants \( m, n \) with \( P + m{P}^{ * } = {nS} \) . Then the normalized forms for the perspectors of the des... | Yes |
Example 4. Applying Theorem 10 to the McCay cubic \( \mathrm{p}\mathcal{K}\left( {K, O}\right) \) and the Or-thocubic \( \mathrm{p}\mathcal{K}\left( {K, H}\right) \) we get the second Brocard cubic \( {K019} = \mathrm{n}{\mathcal{K}}_{0}\left( {K, X\left( {647}\right) }\right) \). | In the former case, we can identify the perspectors of the desmic structures. From \( \left\lbrack {2,\mathrm{\;K}{019}}\right\rbrack \), we know that the points of \( {K019} \) are the foci of inconics with centers on the Brocard axis \( {OK} \) . Also, if \( P \) is on the McCay cubic, then \( P{P}^{ * } \) passes th... | No |
Lemma 11. Suppose that \( P = p : q : r \) and \( Q = u : v : w \) are points with distinct cevians. (a) Let\n\n\[ \n{A}^{\prime } = {BP} \cap {CQ},\;{B}^{\prime } = {CP} \cap {AQ},\;{C}^{\prime } = {AP} \cap {BQ}, \n\]\n\n\[ \n{A}^{\prime \prime } = {BQ} \cap {CP},\;{B}^{\prime \prime } = {CQ} \cap {AP},\;{C}^{\prime ... | Proof. (a) We begin by looking at the desmic structure which is derived from the given \( {P}_{1} \) and \( {P}_{2} \) as in Theorem 4. Thus, the first three vertices are obtained by replacing a coordinate of \( {P}_{1} \) by the corresponding coordinate of \( {P}_{2} \) . This has the vertices named, for example the f... | Yes |
Theorem 12. If \( P \) and \( Q \) are triangle centers with functions \( p\left( {a, b, c}\right) \) and \( q\left( {a, b, c}\right) \) , then the desmic structure \( \mathcal{D}\left( {P, Q}\right) \) has\n\n(a) perspectors \( {P}_{1},{P}_{2} \) with functions \( h\left( {a, b, c}\right) = \frac{1}{p\left( {b, c, a}\... | The proof requires only the observation that, as \( P \) and \( Q \) are centers, \( p\left( {a, b, c}\right) = \) \( p\left( {a, c, b}\right) \) and \( q\left( {a, b, c}\right) = q\left( {a, c, b}\right) \. | No |
Theorem 13. Suppose that \( P \neq Q \) . The cubics associated with the desmic structure \( \mathcal{D}\left( {P, Q}\right) \) are \( \mathrm{p}\mathcal{K}\left( {W,{NK}\left( {W, P}\right) }\right) \) and \( \mathrm{n}{\mathcal{K}}_{0}\left( {W,{PK}\left( {W, P}\right) }\right) \), where \( W \) is the isoconjugation... | Proof. We have the perspectors \( {P}_{1} \) and \( {P}_{2} \) of the desmic structure from Lemma 11(a). These give isoconjugation as that which interchanges \( {P}_{1} \) and \( {P}_{2} \) - see Theorem 4. Now observe that this also interchanges \( P \) and \( Q \), so \( Q = {P}^{ * } \) . From Theorem 4, we have coo... | No |
Theorem 14. Suppose that \( \mathcal{C} = \mathrm{n}{\mathcal{K}}_{0}\left( {W, R}\right) \) is not of type \( \mathrm{c}\mathcal{K} \), and does not have \( W = {R}^{2} \) . (a) The cevian points for \( \mathcal{C} \) are the \( W \) -isoconjugate points on \( \mathcal{T}\left( {R}^{ * }\right) \) . These are the inte... | Proof. From Theorems 3 and 7, we know that \( \mathcal{C} \) is a Grassmann cubic associated with a desmic structure which has triply perspective triangles. If the perspectors of the structure coincide at \( X \), then the equation shows that \( R = X \) and \( W = {X}^{2} \) . But we assumed that \( W \neq {R}^{2} \),... | Yes |
Theorem 15. If \( \mathcal{C} = \mathrm{n}\mathcal{K}\left( {{R}^{2}, R,{k}^{\prime }{rst}}\right) \) is not of type \( \mathrm{c}\mathcal{K} \), then \( \mathcal{C} \) is the Grassmann cubic associated with the desmic structure having perspectors \( R \) and \( {aR} \), where \( a \) is a root of\n\n\[{x}^{2} + \left(... | Proof. When we use Maple to solve the equations to identify \( {A}^{\prime } \) and \( {A}^{\prime \prime } \), we discover them as \( r : {as} : {at} \), with \( a \) as above. This identifies the perspectors as \( R \) and \( {aR} \) . It is easy to verify that this choice leads to \( \mathcal{C} \) . Note that the t... | No |
The third Brocard cubic, \( \mathrm{n}{\mathcal{K}}_{0}\left( {K, X\left( {647}\right) }\right) \), is \( {K019} \) in Gibert’s list. As it is of type \( \mathrm{n}{\mathcal{K}}_{0} \), but not of type \( \mathrm{c}\mathcal{K} \), Theorem 12(b) applies. The conic \( \mathcal{C}\left( R\right) \) is the Jerabek hyperbol... | This gives some new points on the cubic:\n\n\[ \n{A}^{\prime } = {BO} \cap {CH},\;{B}^{\prime } = {CO} \cap {AH},\;{C}^{\prime } = {AO} \cap {BH}, \n\] \n\n\[ \n{A}^{\prime \prime } = {BH} \cap {CO},\;{B}^{\prime \prime } = {CH} \cap {AO},\;{C}^{\prime \prime } = {AH} \cap {BO}. \n\] \n\nAlso, the cubic can be describe... | Yes |
The first Brocard cubic, \( \mathrm{n}{\mathcal{K}}_{0}\left( {K, X\left( {385}\right) }\right) \), is \( {K017} \) in Gibert’s list. Here, the vertices \( {A}^{\prime },{B}^{\prime },{C}^{\prime },{A}^{\prime \prime },{B}^{\prime \prime },{C}^{\prime \prime } \) have already been identified. They are the vertices of t... | Again, Gibert’s website shows the points on \( {K020} \) , together with \( \left\{ {{P}_{1},{P}_{2}}\right\} = \{ X\left( {32}\right), X\left( {76}\right) \} \) . | No |
The cubics \( \mathrm{p}\mathcal{K}\left( {K, X\left( {39}\right) }\right) \) and \( \mathrm{n}{\mathcal{K}}_{0}\left( {K, X\left( {512}\right) }\right) \). These cubics do not appear in Gibert's list, but the associated desmic structure is well-known. Take \( = G, Q = K \) in Lemma 11(a). Again we get isogonal cubics.... | This configuration is discussed in, for example, [4], but the proof that the triangles obtained from the intersections are perspective with \( \bigtriangleup {ABC} \) uses special properties of \( G \) and \( K \). Here, our Lemma 8(a) gives a very simple (geometric) proof of the general case. Here, \( {PK}\left( {K, G... | Yes |
Theorem 16. Suppose that \( D \) is a desmic structure with normalized perspectors \( {P}_{1},{P}_{2} \), and cubics \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) ,\mathrm{p}\mathcal{K}\left( {W, S}\right) \) . Then the desmic structure \( {D}^{\prime } \) with normalized perspectors \( {P}_{1}, - {P}_{2} \) has\n\n(... | Proof. We refer the reader to the proof of Theorem 6(a), which in turn uses the notation of Theorem 4(a).\n\nWe refer to the cubic \( \mathrm{n}\mathcal{K}\left( {W, S,{k}^{\prime }}\right) \) obtained in this way as the harmonic associate of \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) . | No |
The second Brocard cubic \( \mathrm{n}\mathcal{K}\left( {K, X\left( {385}\right) }\right) = {K017} \) was discussed in Example 7. It is \( {GN}\left( \Delta \right) \), where \( \Delta \) is either the first or third Brocard triangle. | From Corollary 17, the cubic \( \mathrm{p}\mathcal{K}\left( {K, X\left( {385}\right) }\right) = {K128} \) is \( {GP}\left( {\Delta }^{\prime }\right) \), where \( {\Delta }^{\prime } \) is the harmonic associate of either of these triangles. This gives us six new points on \( {K128} \) . | No |
Example 11. Let \( \mathcal{C} = {K216} \) of [2]. This was mentioned in \( §3 \) . It is of the form \( \mathrm{n}\mathcal{K}\left( {K, X\left( 5\right), k}\right) \) with parent \( \mathrm{p}\mathcal{K}\left( {K, X\left( {30}\right) }\right) = {K001} \) . From Theorem 16, the harmonic associate is of the form \( \mat... | Using Theorem 9(b), a calculation shows that \( {K067} = \mathrm{n}\mathcal{K}\left( {K, X\left( {30}\right) ,{k}^{\prime \prime }}\right) \) has parent \( {K005} \) . From Theorem 8(b), \( {K005} \) has a unique child with root \( X\left( {30}\right) \) . Thus \( {K067} \) is the harmonic associate of \( {K216} \) . T... | Yes |
Theorem 18. Suppose that \( X = x : y : z \) . The cubics associated with \( \mathcal{D}\left( X\right) \) are \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) and \( \mathrm{p}\mathcal{K}\left( {W, S}\right) \), where\n\n(i) \( W = x\left( {y + z}\right) : y\left( {z + x}\right) : z\left( {x + y}\right) \), the cent... | Proof. The coordinates of \( W, R \) and \( S \) follow at once from those of the perspectors and Theorem 4. The final part needs an equation for the harmonic associate. This is given by Theorem 4. The fact that \( G \) and \( x\left( {y - z}\right) : y\left( {z - x}\right) : z\left( {x - y}\right) \) lie on the cubic ... | No |
Theorem 19. Suppose that \( X = x : y : z \) . Let \( \Delta \) be the triangle \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) of \( \mathcal{E}\left( X\right) \) . Then \( {GN}\left( \Delta \right) = \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) and \( {GP}\left( \Delta \right) = \mathrm{p}\mathcal{K}\... | The harmonic associates are \( {GN}\left( {\Delta }^{\prime }\right) = \mathrm{n}\mathcal{K}\left( {W, G,{k}^{\prime }}\right) \), which degenerates as \( \mathcal{C}\left( W\right) \) and the line at infinity, and \( {GP}\left( {\Delta }^{\prime }\right) = \mathrm{p}\mathcal{K}\left( {W, R}\right) \), which is a centr... | Yes |
Theorem 20. For a given \( W \), suppose that \( R \) and \( S \) are distinct points, neither fixed by \( W \) -isoconjugation.\n\n(a) There exist harmonic triangles \( \Delta \) and \( {\Delta }^{\prime } \) with \( \mathrm{p}\mathcal{K}\left( {W, R}\right) = {GP}\left( \Delta \right) \) and\n\n\( \mathrm{p}\mathcal{... | Proof. (a) If \( \mathrm{p}\mathcal{K}\left( {W, R}\right) \) and \( \mathrm{p}\mathcal{K}\left( {W, S}\right) \) are loci of the given type, then \( {GN}\left( {\Delta }^{\prime }\right) = \) \( \mathrm{n}\mathcal{K}\left( {W, R, k}\right) \) by Theorem 16. Thus this \( \mathrm{n}\mathcal{K} \) has parent \( \mathrm{p... | Yes |
The complement of \( X\left( 1\right) \) is \( X\left( {10}\right) \), the Spieker center, the anticomplement of \( X\left( 1\right) \) is \( X\left( 8\right) \), the Nagel point. The center of the inconic with perspector \( X\left( 1\right) \) is \( X\left( {37}\right) \). If we apply Theorem 19 with \( X = X\left( 1\... | The cubic \( \mathrm{p}\mathcal{K}\left( {X\left( {37}\right), G}\right) \) does not appear in the current [2], but contains \( X\left( 1\right), X\left( 2\right), X\left( {10}\right) \) and \( X\left( {37}\right) \). The harmonic associate of C degenerates as the line at infinity and the circumconic with perspector \(... | Yes |
The complement of \( X\left( 7\right) \), the Gergonne point, is \( X\left( 9\right) \), the Mit-tenpunkt, the anticomplement of \( X\left( 7\right) \) is \( X\left( {144}\right) \) . The center of the inconic with perspector \( X\left( 7\right) \) is \( X\left( 1\right) \) . If we apply Theorem 19 with \( X = X\left( ... | The Grassmann points are \( X\left( 7\right) \) and \( X\left( 9\right) \) . The cubic \( \mathrm{p}\mathcal{K}\left( {X\left( 1\right), G}\right) \) does not appear in the current [2], but contains \( X\left( 1\right), X\left( 2\right), X\left( 7\right) \) and \( X\left( 9\right) \) . The harmonic associate of \( \mat... | Yes |
The complement of \( X\left( 8\right) \), the Nagel point, is \( X\left( 1\right) \), the incenter, the anticomplement of \( X\left( 8\right) \) is \( X\left( {145}\right) \) . The center of the inconic with perspector \( X\left( 1\right) \) is \( X\left( 9\right) \) . If we apply Theorem 19 with \( X = X\left( 8\right... | The Grassmann points are \( X\left( 8\right) \) and \( X\left( 1\right) \) . The cubic \( \mathrm{p}\mathcal{K}\left( {X\left( 9\right), G}\right) \) does not appear in the current [2], but contains \( X\left( 1\right), X\left( 2\right), X\left( 8\right) \) and \( X\left( 9\right) \) . The harmonic associate of \( \mat... | Yes |
The complement of \( X\left( {66}\right) \) is \( X\left( {206}\right) \), the anticomplement is not in [5], but appears in [2] as \( {P161} \) - see \( {K161} \) . The center of the inconic with perspector \( X\left( {66}\right) \) is \( X\left( {32}\right) \). | From Theorem 19 with \( X = X\left( {66}\right) \), we get a cubic \( \mathcal{C} = \mathrm{n}\mathcal{K}\left( {X\left( {32}\right) ,{P161}, k}\right) \) with parent \( {K177} = \mathrm{p}\mathcal{K}\left( {X\left( {32}\right), G}\right) \) . The Grassmann points are \( X\left( {66}\right) \) and \( X\left( {206}\righ... | Yes |
Theorem 21 (Gibert). Suppose that \( P \), and \( Q \) are two \( W \) -isoconjugate points on the cubic \( \mathrm{p}\mathcal{K}\left( {W, S}\right) \) . For \( X \) on \( \mathrm{p}\mathcal{K}\left( {W, S}\right) \), let \( {X}^{t} \) be the tangential of \( X \), and \( \mathrm{p}\mathcal{C}\left( X\right) \) be the... | This can be verified computationally. | No |
Proposition 1. Any set of \( {2n} + 1 \) concurrent lines, no two parallel, in \( {\mathbb{R}}^{2} \) are the medians of some \( \left( {{2n} + 1}\right) \) -gon. | Consider choosing a family of \( {2n} + 1 \) concurrent lines. Each line can be chosen by choosing a unit vector, the choice of each being a single degree of freedom (for instance, the angle that vector makes with the vector \( {\left( 1,0\right) }^{T} \) ). Another degree of freedom is the choice of point \( {A}_{0} \... | No |
Proposition 2. For all \( j \geq 0,{\Delta }_{j + n, j} = {\Delta }_{n,0} \) . | Proof. The case \( j = 0 \) is obvious. Suppose the result is true for \( j = k \) . Then\n\n\[ \n{\Delta }_{k + 1 + n, k + 1} = {a}_{k + 1 + n}{b}_{k + 1} - {b}_{k + 1 + n}{a}_{k + 1} \]\n\n\[ \n= \frac{{\Delta }_{k + n, k}}{{\Delta }_{k, k + 1}}\left( {\left( {{a}_{k} + {a}_{k + 1}}\right) {b}_{k + 1} - \left( {{b}_{... | Yes |
Proposition 3. For all \( j, k,\ell \) ,\n\n\[ \n{\Delta }_{j, k}{x}_{\ell } + {\Delta }_{\ell, j}{x}_{k} = {\Delta }_{\ell, k}{x}_{j} \n\] | Proof. We work component-by-component:\n\n\[ \n{\Delta }_{j, k}{a}_{\ell } + {\Delta }_{\ell, j}{a}_{k} = \left( {{a}_{j}{b}_{k} - {b}_{j}{a}_{k}}\right) {a}_{\ell } + \left( {{a}_{\ell }{b}_{j} - {b}_{\ell }{a}_{j}}\right) {a}_{k} \n\]\n\n\[ \n= {a}_{j}{b}_{k}{a}_{\ell } - {b}_{j}{a}_{k}{a}_{\ell } + a\ell {b}_{j}{a}_... | Yes |
Proposition 5. Given any sequence of \( n + 1 \) points, \( {x}_{0},{x}_{1},\ldots ,{x}_{n} \) such that the origin is not on any line \( \overline{{x}_{i},{x}_{i + 1}} \) or \( \overline{{x}_{n},{x}_{0}} \), then these points are \( n + 1 \) vertices in sequence of a unique \( \left( {{2n} + 1}\right) \) -gon whose me... | Here, we can choose \( n + 1 \) points to determine a \( \left( {{2n} + 1}\right) \) -gon whose medians intersect at the origin. Each point contributes two degrees of freedom for a total of \( {2n} + 2 \) degrees of freedom. Two more degrees of freedom are obtained for the point of concurrency, for a total of \( \left(... | No |
Theorem 2. In the configuration of the points \( A, B, P,{A}_{1},{B}_{1},{P}_{1} \) and the line \( {r}_{1} \) as defined above, the line \( {r}_{1} \) separates the point \( P \) from segment \( {AB} \) . | Proof of Theorem 2. We first augment our configuration by the images of another rigid, direct motion \( {\mu }_{2} \) which carries the points \( A, B, P,{P}_{2}^{ - } \) and the line \( r \) into \( {A}_{2},{B}_{2},{P}_{2}, P \) and \( {r}_{2} \) respectively where \( {A}_{2} \) lies on \( \overrightarrow{PA} \) and \... | Yes |
Proposition 1. The vertices of the orthic-of-intouch triangle are\n\n\[ \nU = \left( {\left( {b + c}\right) \left( {s - b}\right) \left( {s - c}\right) : b\left( {s - c}\right) \left( {s - a}\right) : c\left( {s - a}\right) \left( {s - b}\right) }\right) = \left( {\frac{b + c}{s - a} : \frac{b}{s - b} : \frac{c}{s - c}... | Proof. The intouch triangle \( {DEF} \) has vertices\n\n\[ \nD = \left( {0 : s - c : s - b}\right) ,\;E = \left( {s - c : 0 : s - a}\right) ,\;F = \left( {s - b : s - a : 0}\right) . \n\]\n\nThe sidelines of the intouch triangle have equations\n\n\[ \n{EF} : \; - \left( {s - a}\right) x + \left( {s - b}\right) y + \lef... | Yes |
Proposition 3. If \( {ABC} \) is acute angled, the vertices of the intouch-of-orthic triangle are\n\n\[ X = \left( {\left( {{b}^{2} + {c}^{2}}\right) {S}_{BC} : {b}^{2}{S}_{CA} : {c}^{2}{S}_{AB}}\right) = \left( {\frac{{b}^{2} + {c}^{2}}{{S}_{A}} : \frac{{b}^{2}}{{S}_{B}} : \frac{{c}^{2}}{{S}_{C}}}\right) ,\]\n\n\[ Y =... | Proof. The orthic triangle \( {D}^{\prime }{E}^{\prime }{F}^{\prime } \) has vertices\n\n\[ {D}^{\prime } = \left( {0 : {S}_{C} : {S}_{B}}\right) ,\;{E}^{\prime } = \left( {{S}_{C} : 0 : {S}_{A}}\right) ,\;{F}^{\prime } = \left( {{S}_{B} : {S}_{A} : 0}\right) . \]\n\nThe sidelines of the orthic triangle have equations\... | Yes |
Proposition 5. If triangle \( {ABC} \) is acute angled, then its intouch-of-orthic and orthic-of-intouch triangles are homothetic at the point\n\n\[ \nQ = \left( {\frac{a\left( {a\left( {b + c}\right) - \left( {{b}^{2} + {c}^{2}}\right) }\right) }{\left( {s - a}\right) {S}_{A}} : \frac{b\left( {b\left( {c + a}\right) -... | Proof. The homothetic center is the intersection of the lines \( {UX},{VY} \), and \( {WZ} \) . See Figure 5. Making use of the coordinates given in Propositions 1 and 3, we obtain the equations of these lines as follows.\n\n\( {UX} : \;{bc}\left( {s - a}\right) {S}_{A}\left( {c\left( {s - c}\right) {S}_{B} - b\left( {... | Yes |
Lemma 2. A moving angle with vertex in the fixed point \( A \) in the plane intersects a fixed line \( \mathcal{L}, A \) not on \( \mathcal{L} \), in a pair of points related by a projectivity. | Proof. As mentioned in the statement, let \( A \) be a fixed point and \( \mathcal{L} \) a fixed line such that \( A \) is not on \( \mathcal{L} \). Consider the rays \( h \) and \( k \) with origin in \( A \), the moving angle \( \angle {hk} \) with the vertex in \( A \) and of constant measure \( \alpha \). Denote by... | Yes |
Theorem 3. Every homothetic image of \( {ABC} \) in \( I \) is perspective with the intouch triangle DEF. | Proof. We work with homogeneous barycentric coordinates.\n\nThe image of \( {ABC} \) under the homothety \( \mathrm{h}\left( {I, t}\right) \) has vertices\n\n\[ \n{A}_{t} = \left( {a + t\left( {b + c}\right) : \left( {1 - t}\right) b : \left( {1 - t}\right) c}\right) , \n\]\n\n\[ \n{B}_{t} = \left( {\left( {1 - t}\righ... | Yes |
Proposition 4. The perspector of \( {A}_{t}{B}_{t}{C}_{t} \) and \( {H}_{a}{H}_{b}{H}_{c} \) is the reflection of \( {P}_{-t} \) in the incenter. | It is clear that the perspector \( {P}_{t} \) traverses a conic \( \Gamma \) as \( t \) varies, since its coordinates are quadratic functions of \( t \) . The conic \( \Gamma \) clearly contains \( I \) and the Gergonne point, corresponding respectively to \( t = 0 \) and \( t = 1 \) . Note also that \( D = {P}_{t} \) ... | Yes |
Proposition 1. Triangle \( {A}^{ * }{B}^{ * }{C}^{ * } \) is the anticevian triangle of the infinite point \( Q = \left( {u\left( {v - w}\right) : v\left( {w - u}\right) : w\left( {u - v}\right) }\right) \) of the trilinear polar of \( P \) . | We shall prove Proposition 1 in \( §2 \) below. As an anticevian triangle, \( {A}^{ * }{B}^{ * }{C}^{ * } \) is perspective with every cevian triangle. In particular, it is perspective with \( {XYZ} \) at the cevian quotient \( P/Q \), which depends on \( P \) only. We write\n\n\[ \tau \left( P\right) \mathrel{\text{:=... | No |
Proposition 3. \( \tau \left( P\right) \) is\n\n(1) the center of the circumconic through \( P \) and its isotomic conjugate \( \mathbf{P} \) ,\n\n(2) the perspector of the circum-hyperbola with asymptotes the trilinear polars of \( P \) and \( {P}^{ \bullet } \) . | Proof. (1) The circumconic through \( P \) and \( {P}^{ \bullet } \) has equation\n\n\[ \n\frac{u\left( {{v}^{2} - {w}^{2}}\right) }{x} + \frac{v\left( {{w}^{2} - {u}^{2}}\right) }{y} + \frac{w\left( {{u}^{2} - {v}^{2}}\right) }{z} = 0, \n\]\n\nwith perspector\n\n\[ \n{P}^{\prime } = \left( {u\left( {{v}^{2} - {w}^{2}}... | Yes |
Proposition 4. The trilinear polar of \( \tau \left( P\right) \) with respect to the cevian triangle of \( P \) passes through \( P \) . | Proof. The trilinear polar of \( \tau \left( P\right) \) with respect to \( {XYZ} \) is the perspectrix of the triangles \( {XYZ} \) and \( {A}^{ * }{B}^{ * }{C}^{ * } \) . Now, the sidelines of these triangle intersect at the points\n\n\[ \n{B}^{ * }{C}^{ * } \cap {YZ} = \left( {u\left( {v + w - {2u}}\right) : v\left(... | Yes |
Theorem 3. The lines \( {A}_{b}{A}_{c},{B}_{c}{B}_{a} \) and \( {C}_{a}{C}_{b} \) bound a triangle \( {\mathbf{T}}_{2} \) whose circumcircle is tangent to the nine-point circle at the Jerabek center. | Hence, the circumcircles of \( {\mathbf{T}}_{1} \) and \( {\mathbf{T}}_{2} \) are also tangent to each other at \( J \) . In this paper we work with homogeneous barycentric coordinates and adopt standard notations of triangle geometry. Basic results can be found in [2]. The Jerabek center \( J \), for example, has coor... | Yes |
Proposition 4. The homogeneous barycentric coordinates of the pedals of the vertices of triangle \( {ABC} \) on the circumradii are as follows. | Proof. We verify that the point\n\n\[ P = \left( {{S}_{A}\left( {{S}_{B} + {S}_{C}}\right) : {S}_{C}\left( {{S}_{C} - {S}_{A}}\right) : {S}_{C}\left( {{S}_{A} + {S}_{B}}\right) }\right) \]\n\nis the pedal \( {A}_{b} \) of \( A \) on the line \( {OB} \) . Since\n\n\[ \left( {{S}_{A}\left( {{S}_{B} + {S}_{C}}\right) ,{S}... | Yes |
Lemma 5. The quadrilateral \( {B}_{c}{C}_{b}{CB} \) is an isosceles trapezoid. | Proof. With reference to Figure 1, we have\n\n(i) \( \angle {B}_{c}{BC} = \frac{\pi }{2} - \angle {OCB} = \frac{\pi }{2} - \angle {OBC} = \angle {C}_{b}{CB} \) ,\n\n(ii) \( {B}_{c}B = {C}_{b}\widetilde{C} \) .\n\nIt follows that the quadrilateral \( {B}_{c}{C}_{b}{CB} \) is an isosceles trapezoid.\n\nTherefore, the lin... | Yes |
Proposition 6. Triangle \( {\mathbf{T}}_{1} \) is homothetic to\n\n(i) \( {ABC} \) at the procircumcenter \( \left( {{a}^{4}{S}_{A} : {b}^{4}{S}_{B} : {c}^{4}{S}_{C}}\right) ,{}^{1} \) | Proof. The lines \( {B}_{c}{C}_{b},{C}_{a}{A}_{c} \), and \( {A}_{b}{B}_{a} \) have equations\n\n\[ \n- \left( {{S}_{AA} + {S}_{BC}}\right) x + {S}_{A}\left( {{S}_{B} + {S}_{C}}\right) y + {S}_{A}\left( {{S}_{B} + {S}_{C}}\right) z = 0, \n\]\n\n\[ \n{S}_{B}\left( {{S}_{C} + {S}_{A}}\right) x - \left( {{S}_{BB} + {S}_{C... | Yes |
Proposition 7. Triangles \( {ABC} \) and \( {A}_{2}{B}_{2}{C}_{2} \) are perspective at\n\n\[ Q = \left( {\frac{1}{{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2} - {b}^{4} - {c}^{4}} : \frac{1}{{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2} - {c}^{4} - {a}^{4}} : \frac{1}{{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a... | Proof. From the coordinates of \( {A}_{2},{B}_{2},{C}_{2} \) given above, \( {\mathbf{T}}_{\mathbf{2}} \) is perspective with \( {ABC} \) at\n\n\[ Q = \left( {\frac{1}{3{S}_{BC} + {S}_{CA} + {S}_{AB} - {S}_{AA}} : \cdots : \cdots }\right) . \]\n\nThese are equivalent to those given above in terms of \( a, b, c \) . | Yes |
Theorem 1. There are two and only two non-degenerate inscribed conics whose perspector \( P \) is one focus : they are obtained when \( P \) is one of the isogonic centers. | Proof. If \( P \) is one focus of \( {\Gamma }_{i}\left( P\right) \), the other focus is the isogonal conjugate \( {P}^{ * } \) of \( P \) and the center is the midpoint of \( P{P}^{ * } \) . This center must be the isotomic conjugate of the anticomplement of \( P \) . A computation shows that \( P \) must lie on three... | Yes |
Theorem 2. The two (inscribed) Simmons conics generate a pencil of conics which contains the nine-point circle. | The four (not always real) base points of the pencil form a quadrilateral inscribed in the nine point circle and whose diagonal triangle is the anticevian triangle of \( {X}_{523} \), the infinite point of the perpendiculars to the Euler line. In Figure 1 we have four real base points on the nine point circle and on tw... | Yes |
Theorem 3. The two (inscribed) Simmons conics generate a tangential pencil of conics which contains the Steiner inellipse. | Indeed, their centers \( {X}_{396} \) and \( {X}_{395} \) lie on the line \( {GK} \) . The locus of foci of all inconics with center on this line is the (second) Brocard cubic K018 which is \( \mathrm{n}{\mathcal{K}}_{0}\left( {K,{X}_{523}}\right) \) (See [3]). These conics must be tangent to the trilinear polar of the... | No |
Theorem 4. There are two and only two non-degenerate circumconics whose per-spector \( P \) is one focus : they are obtained when \( P \) is one of the isogonic centers. | They will be called the Simmons circumconics denoted by \( {\sum }_{13} = {\Gamma }_{c}\left( {X}_{13}\right) \) and \( {\sum }_{14} = {\Gamma }_{c}\left( {X}_{14}\right) \) . See Figure 4.\n\nThe fourth common point of these conics is \( {X}_{476} \) (Tixier point) on the circumcircle. The centers and other real foci ... | No |
Theorem 5. The locus of the focus \( F \) of the inconic such that the corresponding directrix is parallel to the trilinear polar of \( F \) is the Euler-Morley quintic Q003. | Q003 is a very remarkable curve with equation\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}\left( {{S}_{B}y - {S}_{C}z}\right) {y}^{2}{z}^{2} = 0 \]\n\nwhich (at the time this paper is written) contains 70 points of the triangle plane. See [3] and [4]. | No |
Theorem 6. The locus of the perspector \( P \) of the inconic (or circumconic) such that \( P \) lies on one of its axis is the Stothers quintic Q012. | The Stothers quintic Q012 has equation\n\n\[ \mathop{\sum }\limits_{\text{cyclic }}{a}^{2}\left( {y - z}\right) \left( {{x}^{2} - {yz}}\right) {yz} = 0. \] | Yes |
Theorem 1 (Bellavitis, 1854). If the side lengths of a convex quadrilateral ABCD satisfy \( {AB} \cdot {CD} = {BC} \cdot {DA} \), then \( {w}_{1} + {w}_{2} + {w}_{3} + {w}_{4} = {180}^{ \circ } \) | Proof. If \( {AB} = {AD} \) then \( {BC} = {CD} \) and \( {AC} \) is the perpendicular bisector of \( {BD} \) . Hence \( {ABCD} \) is a kite, and it is obvious that \( {w}_{1} + {w}_{2} + {w}_{3} + {w}_{4} = {180}^{ \circ } \) .\n\nIf \( {ABCD} \) is not a kite, then from \( {AB} \cdot {CD} = {BC} \cdot {DA} \), we hav... | Yes |
In an arbitrary noncyclic quadrilateral ABCD, the side lengths satisfy the equality \( {AB} \cdot {CD} = {BC} \cdot {DA} \) if and only if | Proof. Since \( {ABCD} \) is not a cyclic quadrilateral the lines \( {DA},{DB},{DC} \) meet the circumcircle of triangle \( {ABC} \) at the distinct points \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) . The triangle \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) is the circumcevian triangle of \( D \) relative to \( ... | Yes |
Theorem 1. Let the diagonal points of a complete quadrilateral ABCD be \( P, Q \) and \( R \) . Let the intersections of \( {PQ} \) with \( {AD} \) and \( {BC} \) be \( H \) and \( F \) respectively and those of PR with CD and AB be G and E respectively (Figure 1). Then\n\n\[ \frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \fr... | Proof. We apply the Pappus-Pascal theorem to the triples \( \left( {Q, A, E}\right) \) and \( \left( {R, C, F}\right) \) and find that in triangle \( {ABC} \) the lines \( {AF},{BP} \) and \( {CE} \) are concurrent so that by\n\nCeva's theorem\n\[ \frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CP}{PA} = 1 \]\n\n(4)\n\nS... | Yes |
Theorem 2. If in the quadrilateral \( {ABCD} \) the points \( E,{F}_{1}, G \) and \( {H}_{1} \) lie on \( {AB} \) , \( {BC},{CD} \) and \( {DA} \) respectively such that \( S = {AG} \cap C{H}_{1} \) and \( T = A{F}_{1} \cap {CE} \) lie on \( {BD} \), then the points \( A, E,{F}_{1}, C, G \) and \( {H}_{1} \) lie on a c... | Proof. Here we have to switch to the field of Projective Geometry. We will use the cross ratio of pencils in relation to the cross-ratio of ranges. These concepts are extensively described by Eves [3]. Now consider the two pencils \( \left( {A{H}_{1},{AG}, A{F}_{1},{AE}}\right) \) and \( \left( {C{H}_{1},{CG}, C{F}_{1}... | Yes |
Theorem 3. If in the quadrilateral \( {ABCD} \) the points \( E, F, G \) and \( H \) lie on \( {AB} \) , \( {BC},{CD} \) and \( {DA} \) respectively and \( {EG} \) and \( {FH} \) concur in \( P = {AC} \cap {BD} \), then (1) is satisfied if and only if there is a conic inscribed in quadrilateral ABCD, which touches its ... | Proof. Assume that relation (1) holds. By Theorem 2 we know that BFGDHE and \( {AEFCGH} \) lie on two conics. Let \( V = {DE} \cap {BH} \) and \( W = {DF} \cap {BG} \) . First we apply Desargues’ theorem to triangle \( {GFC} \) and triangle \( {EHV} \) (Figure 5).\n\nThe lines \( {GE},{FH} \) and \( {CV} \) concur in \... | Yes |
Lemma 1. The locus \( \mathcal{K}\left( {P,\Delta, S}\right) \) of the points for which the distances \( {d}_{1},{d}_{2},{d}_{3} \) to the sides \( {a}_{1},{a}_{2},{a}_{3} \) of \( \Delta = {A}_{1}{A}_{2}{A}_{3} \) are connected by\n\n\[ \frac{{l}_{1}}{{\delta }_{1}}{d}_{1}^{2} + \frac{{l}_{2}}{{\delta }_{2}}{d}_{2}^{2... | Proof. Substituting the coordinates \( \left( {-\frac{1}{{l}_{1}},\frac{1}{{l}_{2}},\frac{1}{{l}_{3}}}\right) \), or \( \left( {\frac{1}{{l}_{1}}, - \frac{1}{{l}_{2}},\frac{1}{{l}_{3}}}\right) \), or \( \left( {\frac{1}{{l}_{1}},\frac{1}{{l}_{2}}, - \frac{1}{{l}_{3}}}\right) \) of \( {A}_{1}^{-1},{A}_{2}^{-1} \), and \... | Yes |
Lemma 2. The center \( M \) of \( \mathcal{C}\left( {P,\Delta }\right) \) has trilinear coordinates\n\n\[ \left( {{l}_{2}{l}_{3}\left( {{l}_{2}{p}_{2} + {l}_{3}{p}_{3}}\right) ,{l}_{3}{l}_{1}\left( {{l}_{3}{p}_{3} + {l}_{1}{p}_{1}}\right) ,{l}_{1}{l}_{2}\left( {{l}_{1}{p}_{1} + {l}_{2}{p}_{2}}\right) }\right) . \]\n\nI... | Proof. An easy calculation shows that the polar point of this point \( M \) with regard to the conic (4) is indeed the line at infinity, with equation \( {l}_{1}{x}_{1} + {l}_{2}{x}_{2} + {l}_{3}{x}_{3} = 0 \) . Moreover, if \( {P}_{\infty } \) is the point at infinity of the line \( {PG} \), the equation \( {2GM} = - ... | Yes |
Theorem 3. (1) The conics \( \mathcal{K}\left( {P,\Delta, S}\right) \) and \( \mathcal{C}\left( {{f}^{-1}\left( P\right) ,{f}^{-1}\left( \Delta \right) }\right) \) coincide. (2) The common axes of the conics \( \mathcal{K}\left( {P,\Delta, s}\right), s \in R \), and of the conic \( \mathcal{C}\left( {{f}^{-1}\left( P\r... | Proof. (1) Because of Lemma 1 and 2, both conics have center \( P \) and are circumscribed about the complementary triangle \( {f}^{-1}\left( \Delta \right) \) of \( {A}_{1}{A}_{2}{A}_{3} \ | Yes |
Theorem 1. Let \( {A}^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime } \) be an inscribed quadrilateral of \( {ABCD} \), and \( E = \) \( {A}^{\prime }{C}^{\prime } \cap {B}^{\prime }{C}^{\prime }, I = {AC} \cap {BD}, U = {AC} \cap {D}^{\prime }{A}^{\prime }, V = {AC} \cap {B}^{\prime }{C}^{\prime } \) . If \( E \) lie... | Proof. We shall use analytic geometry of the plane. Without loss of generality we can assume that \( A\left( {0,0}\right), B\left( {f, g}\right), C\left( {1,0}\right) \) and \( D\left( {p, q}\right) \) for some real numbers \( f \) , \( g, p \) and \( q \) . The points \( {A}^{\prime }\left( {\frac{fu}{u + 1},\frac{gu}... | Yes |
Theorem 2. (a) If \( E \) lies on the line \( {AC} \), then\n\n\[ \frac{\operatorname{area}\left( {t}_{1}\right) }{\operatorname{area}\left( {t}_{2}\right) } \cdot \frac{\operatorname{area}\left( {t}_{3}\right) }{\operatorname{area}\left( {t}_{4}\right) } = \frac{\operatorname{area}\left( {t}_{A}\right) }{\operatorname... | Proof. If we keep the same assumptions and notation from the proof of Theorem 1, then\n\n\[ \frac{\operatorname{area}\left( {t}_{1}\right) }{\operatorname{area}\left( {t}_{2}\right) } \cdot \frac{\operatorname{area}\left( {t}_{3}\right) }{\operatorname{area}\left( {t}_{4}\right) } - \frac{\operatorname{area}\left( {t}_... | Yes |
Theorem 3. \( \frac{R\left( {t}_{1}\right) }{R\left( {t}_{2}\right) } \cdot \frac{R\left( {t}_{3}\right) }{R\left( {t}_{4}\right) } = \frac{R\left( {t}_{A}\right) }{R\left( {t}_{C}\right) } \) . | Proof. Let us keep again the same assumptions and notation from the proof of Theorem 1. Since \( R\left( t\right) = \frac{\text{ product of side lengths }}{4\mathrm{{area}}\left( t\right) } \), we see that the square of the circumradius of a triangle with vertices in the points \( \left( {x, a}\right) ,\left( {y, b}\ri... | Yes |
Theorem 4. \( \frac{\left| AH\left( {t}_{1}\right) \right| }{\left| EH\left( {t}_{2}\right) \right| } \cdot \frac{\left| EH\left( {t}_{3}\right) \right| }{\left| CH\left( {t}_{4}\right) \right| } = \frac{\left| AH\left( {t}_{A}\right) \right| }{\left| CH\left( {t}_{C}\right) \right| } \). | Proof. This time one can see that \( {\left| AH\left( t\right) \right| }^{2} \) for the triangle \( t = {ABC} \) with the vertices in the points \( \left( {x, a}\right) ,\left( {y, b}\right) \) and \( \left( {z, c}\right) \) is given as\n\n\[ \frac{\left\lbrack {{\left( y - z\right) }^{2} + {\left( b - c\right) }^{2}}\... | Yes |
An I-triangle is perspective with the medial triangle if and only if it is perspective with the intouch triangle. | The homogeneous barycentric coordinates of the vertices of an \( I \) -triangle can be taken as\n\n\[ U = \left( {u : b : c}\right) ,\;V = \left( {a : v : c}\right) ,\;W = \left( {a : b : w}\right) \]\n\nfor some \( u, v, w \) . In each case, the condition for perspectivity is\n\n\[ F\left( {u, v, w}\right) \mathrel{\t... | Yes |
Theorem 2. Let \( {UVW} \) be an I-triangle perspective with the medial and the in-touch triangles. If \( {U}_{1},{U}_{1},{W}_{1} \) are the inversive images of \( U, V, W \) in the incircle, then \( {U}_{1}{V}_{1}{W}_{1} \) is also an I-triangle perspective with the medial and intouch triangles. | Proof. If the coordinates of \( U, V, W \) are as given in (1), then\n\n\[ \n{U}_{1} = \left( {{u}_{1} : b : c}\right) ,\;{V}_{1} = \left( {a : {v}_{1} : c}\right) ,\;{W}_{1} = \left( {a : b : {w}_{1}}\right) , \n\]\n\nwhere\n\n\[ \n{u}_{1} = \frac{\left( {a\left( {b + c}\right) - {\left( b - c\right) }^{2}}\right) u -... | Yes |
Theorem 3. Let \( {UVW} \) be an I-triangle perspective with the medial and the in-touch triangles. If \( {U}_{2} \) (respectively \( {V}_{2} \) and \( {W}_{2} \) ) is the inversive image of \( U \) (respectively \( V \) and \( W \) ) in the \( A \) - (respectively \( B \) - and \( C \) -) excircle, then \( {U}_{2}{V}_... | If an \( I \) -triangle \( {UVW} \) is perspective with the medial triangle at \( \left( {x : y : z}\right) \), then \( {U}_{2}{V}_{2}{W}_{2} \) is perspective with (i) the medial triangle at\n\n\[ \left( {\left( {s - a}\right) \left( {y + z - x}\right) \left( {\left( {a\left( {b + c}\right) + {\left( b - c\right) }^{2... | Yes |
Proposition 1. For integers \( h \) and \( k \) ,\n\n\[ \n{r}_{h}{r}_{k} = \mathop{\sum }\limits_{{j = 0}}^{{k - 1}}{r}_{h - k + {2j} + 1}. \n\] | Proof. Letting \( h \) and \( k \) be integers, we have\n\n\[ \n{r}_{h}{r}_{k} = {\left( \sin \frac{\pi }{n}\right) }^{-2}\sin \frac{h\pi }{n}\sin \frac{k\pi }{n} \n\]\n\n\[ \n= {\left( \sin \frac{\pi }{n}\right) }^{-2}\frac{1}{2}\left( {\cos \frac{\left( {h - k}\right) \pi }{n} - \cos \frac{\left( {h + k}\right) \pi }... | Yes |
Proposition 2. Given an integer \( k \) relatively prime to \( n \) , \n\n\[ \n\frac{1}{{r}_{k}} = \mathop{\sum }\limits_{{j = 1}}^{s}{r}_{k\left( {{2j} - 1}\right) } \n\] \n\nwhere \( s \) is any (not necessarily the smallest) positive integer such that \( {ks} \equiv \) \( \pm 1{\;\operatorname{mod}\;n} \) . | Proof. Starting at the right-hand side, we have \n\n\[ \n\mathop{\sum }\limits_{{j = 1}}^{s}{r}_{k\left( {{2j} - 1}\right) } = {\left( \sin \frac{\pi }{n}\sin \frac{k\pi }{n}\right) }^{-1}\mathop{\sum }\limits_{{j = 1}}^{s}\sin \frac{k\left( {{2j} - 1}\right) \pi }{n}\sin \frac{k\pi }{n} \n\] \n\n\[ \n= {\left( \sin \f... | Yes |
Proposition 3. For nonnegative integers \( m \) , \[ {r}_{2}{}^{m} = \mathop{\sum }\limits_{{i = 0}}^{m}\left( \begin{matrix} m \\ i \end{matrix}\right) {r}_{1 - m + {2i}} \] | Proof. We proceed by induction on \( m \) . When we have \( m = 0 \), the formula reduces to \( {r}_{2}{}^{0} = {r}_{1} \) and both sides equal 1 . This establishes the basis step. For the inductive step, we assume the result for \( m = n \) and begin with the sum on the right-hand side for \( m = n + 1 \) . \[ \mathop... | Yes |
Proposition 5. For all natural numbers \( k \) ,\n\n\[ \n{Q}_{2k}\left( t\right) = \sqrt{t + 1}\mathop{\sum }\limits_{i}\left( \begin{matrix} k - 1 - i \\ i \end{matrix}\right) {\left( -1\right) }^{i}{\left( t - 1\right) }^{k - 1 - {2i}}\n\]\n\n\[ \n{Q}_{{2k} + 1}\left( t\right) = \mathop{\sum }\limits_{i}\left( \begin... | Proof. We proceed by induction on \( k \) . These are easily checked to match the functions \( {Q}_{0},{Q}_{1},{Q}_{2} \), and \( {Q}_{3} \) for \( k = 0,1 \) . For \( k \geq 2 \), we have\n\n\[ \n{Q}_{2k}\left( t\right) = \left( {t - 1}\right) {Q}_{2\left( {k - 1}\right) }\left( t\right) - {Q}_{2\left( {k - 2}\right) ... | No |
Proposition 3. The orthic axis is perpendicular to the Euler line. | This proposition is very well known. It follows easily, for example, from the fact that the orthic axis \( {\ell }_{H} \) is the radical axis of the circumcircle and the nine-point circle. See, for example, [2, \$\$\$,4,5]. | No |
Proposition 5. If the trilinear polar \( {\ell }_{P} \) intersects \( {BC} \) at \( {A}_{1} \), then\n\n\[ \n{A}_{1}{X}^{2} = {A}_{1}B \cdot {A}_{1}C.\n\] | Proof. Let \( M \) is the midpoint of \( {BC} \) . Since \( {A}_{1},{A}^{\prime } \) divide \( B, C \) harmonically, we have \( M{B}^{2} = M{C}^{2} = M{A}_{1} \cdot M{A}^{\prime } \) (Newton’s theorem). Thus, \( M{X}^{\prime 2} = M{A}_{1} \) . \( M{A}^{\prime } \) . It follows that triangles \( M{X}^{\prime }{A}_{1} \)... | Yes |
Corollary 6. If \( {X}_{1} \) is the intersection of \( {YZ} \) and \( {BC} \), then \( {A}_{1} \) is the midpoint of \( X{X}_{1} \) . | Proof. If \( {X}_{1} \) is the intersection of \( {YZ} \) and \( {BC} \), then \( X,{X}_{1} \) divide \( B, C \) harmonically. The circle through \( X,{X}_{1} \), and with center on \( {BC} \) is orthogonal to the circle \( {\mathcal{C}}_{A} \) . By Proposition 5, this has center \( {A}_{1} \), which is therefore the m... | Yes |
Proposition 7. If \( {H}^{\prime } \) is the orthocenter of \( {XYZ} \), then the line \( O{H}^{\prime } \) is perpendicular to the trilinear polar \( {\ell }_{P} \) . | Proof. Consider the orthic triangle \( {DEF} \) of \( {XYZ} \) . Since \( {DEXY},{EFYZ} \), and \( {FDZX} \) are cyclic, and the common chords \( {DX},{EY},{FZ} \) intersect at \( {H}^{\prime },{H}^{\prime } \) is the radical center of the three circles, and\n\n\[ \n{H}^{\prime }D \cdot {H}^{\prime }X = {H}^{\prime }E ... | Yes |
Theorem 8. The orthocenter of the cevian triangle of \( \sqrt{P} \) lies on the Brocard axis if and only if \( P \) is an interior point on the Jerabek hyperbola. | Proof. The Brocard axis \( {OK} \) is orthogonal to the Lemoine axis. The locus of points whose trilinear polars are parallel to the Brocard axis is the Jerabek hyperbola. | No |
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