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Proposition 26. The center of the Parry circle \( {\mathcal{C}}_{\mathrm{P}} \) is the point\n\n\[ \left( {{a}^{2}\left( {{b}^{2} - {c}^{2}}\right) \left( {{b}^{2} + {c}^{2} - 2{a}^{2}}\right) ,\cdots ,\cdots }\right) . \] | Proof. This is the intersection of the line (27) above and the the Lemoine axis (14). | No |
Proposition 28. The Clawson point \( {C}_{\mathrm{w}} \) is the perspector of the triangle bounded by the radical axes of the circumcircle with the three excircles (see Figure 24). | Proof. The equations of the excircles are given in [15, §6.1.1]. The radical axes with circumcircle are the lines\n\n\[ \n{L}_{a} \mathrel{\text{:=}} {s}^{2}x + {\left( s - c\right) }^{2}y + {\left( s - b\right) }^{2}z = 0, \n\]\n\n\[ \n{L}_{b} \mathrel{\text{:=}} {\left( s - c\right) }^{2}x + {s}^{2}y + {\left( s - a\... | Yes |
Theorem 29 (Lester). The symmedian point, the Feuerbach point, the Clawson point, and the homothetic center of the orthic and the intangent triangles are concyclic. | There are a number of ways of proving this theorem, all very tedious. For example, it is possible to work out explicitly the equation of the circle containing these four points. Alternatively, one may compute distances and invoke the intersecting chords theorem. These proofs all involve polynomials of large degrees. We... | No |
Lemma 30. The equation of the circle passing through three given points \( {P}_{1} = \) \( \left( {{u}_{1} : {v}_{1} : {w}_{1}}\right) ,{P}_{2}\left( {{u}_{2} : {v}_{2} : {w}_{2}}\right) \) and \( {P}_{3} = \left( {{u}_{3} : {v}_{3} : {w}_{3}}\right) \) is\n\n\[ \n{a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy} - \left( {x + y... | Proof. This follows from applying Cramer's rule to the system of linear equations\n\n\[ \n{a}^{2}{v}_{1}{w}_{1} + {b}^{2}{w}_{1}{u}_{1} + {c}^{2}{u}_{1}{v}_{1} - {s}_{1}\left( {p{u}_{1} + q{v}_{1} + r{w}_{1}}\right) = 0, \n\]\n\n\[ \n{a}^{2}{v}_{2}{w}_{2} + {b}^{2}{w}_{2}{u}_{2} + {c}^{2}{u}_{2}{v}_{2} - {s}_{2}\left( ... | Yes |
Lemma 31. Four points \( {P}_{i} = \left( {{u}_{i} : {v}_{i} : {w}_{i}}\right), i = 1,2,3,4 \), are concyclic if and only if\n\n\[ \frac{D\left( {{u}_{1},{u}_{2},{u}_{4}}\right) }{D\left( {{u}_{1},{u}_{2},{u}_{3}}\right) } = \frac{D\left( {{v}_{1},{v}_{2},{v}_{4}}\right) }{D\left( {{v}_{1},{v}_{2},{v}_{3}}\right) } = \... | Proof. The circumcircles of triangles \( {P}_{1}{P}_{2}{P}_{3} \) and \( {P}_{1}{P}_{2}{P}_{4} \) have equations\n\n\[ {a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy} - \left( {x + y + z}\right) \left( {{px} + {qy} + {rz}}\right) = 0, \]\n\n\[ {a}^{2}{yz} + {b}^{2}{zx} + {c}^{2}{xy} - \left( {x + y + z}\right) \left( {{p}^{\pr... | Yes |
Theorem 1. The Hervey point \( h \) is the center of the circle which bears the four orthocenters \( {h}_{n} \) of the triangles \( {\Theta }_{n} \) . | Proof. From the relation (1) we deduce that the quadrilateral \( {O}_{m}{O}_{n}{h}_{m}{h}_{n} \) is a parallelogram, and its two diagonals \( {O}_{m}{h}_{m} \) and \( {O}_{n}{h}_{n} \) intersect at their common midpoints. The four triangles \( {\Theta }_{n} \) are endowed with the same role. Therefore the four midpoint... | Yes |
Theorem 2. The Hervey point \( h \) is the point of concurrence of the four altitudes of the triangles \( M{O}_{n}{N}_{n} \) drawn through the vertices \( {O}_{n} \) . | Proof. Since the triangles \( {\Theta }_{n} \) and \( {T}_{n} \) are directly similar, the circumcenter \( O \) and the orthocenter \( {h}_{n} \) respectively correspond to the circumcenter \( {O}_{n} \) and orthocenter \( {H}_{n} \) . We have the following equality of directed angles\n\n\[ \left( {{MO}, M{O}_{n}}\righ... | Yes |
Theorem 3. The Hervey point \( h \) is the point of concurrence of the four perpendicular bisectors of the segments \( {O}_{n}{H}_{n} \) of the Euler lines of the triangles \( {T}_{n} \) . | Proof. Since the triangles \( {\Theta }_{n} \) and \( {T}_{n} \) are directly similar, the circumcenter \( O \) and the orthocenter \( {h}_{n} \) respectively correspond to the circumcenter \( {O}_{n} \) and orthocenter \( {H}_{n} \) and we have the following equality between oriented angles of oriented lines\n\n\[ \le... | Yes |
Lemma 1. Let \( c \) be a circle and \( G \) a point. Let further \( D \) be a point on the circle and \( L \) a line parallel to the tangent \( {p}_{D} \) at \( D \) and not intersecting the circle. Then there is exactly one point \( A \) on \( L \) such that the tangents \( \left\{ {t,{t}^{\prime }}\right\} \) to \( ... | Figure 1 suggests a proof. Let \( \{ E, A\} \) be the intersections of lines \( E = L \cap {p}_{G} \) and \( A = L \cap {p}_{E} \) . Draw from \( A \) the tangents \( \left\{ {t = {AI},{t}^{\prime } = {AJ}}\right\} \) to \( c \) intersecting line \( {p}_{D} \) respectively at \( B \) and \( C \) . Let also \( M \) be t... | Yes |
Proposition 3. All triangles \( {ABC} \) sharing the same incircle \( c \) and centroid \( G \) have their side-middles on a conic \( {c}^{ * } \) and their vertices on a conic \( {c}^{* * } = f\left( {c}^{ * }\right) \) homothetic to \( {c}^{ * } \) by the homothety \( f \) centered at the centroid with ratio \( k = -... | The proof follows from the previous remarks and the fact that in this setting the (anti-)homothety \( f \) has center at \( G \) and ratio \( k = - 2 \) . By Lemma 5 below, the median \( {AM} \) of side \( {BC} \) coincides with the polar \( {p}_{E} \) of point \( E = L \cap {p}_{G} \), where \( L \) is the parallel to... | Yes |
Proposition 4. The homography \( H \) defined in the previous section is uniquely characterized by its properties:\n\n(i) H fixes points \( \{ P, Q\} \) which are the diameter points of \( c \) on the \( x \) -axis,\n\n(ii) \( H \) maps points \( \{ S, T\} \) to \( \left\{ {{S}^{\prime },{T}^{\prime }}\right\} \) . Her... | The proof of the proposition (see Figure 7) follows by applying the matrix to the coordinate vectors of these points which are \( P\left( {u + r,0,1}\right), Q\left( {u - r,0,1}\right), S\left( {u, r,1}\right) \) , \( T\left( {u, - r,1}\right) \) ,\n\n\( {S}^{\prime }\left( {{ku}, r,1}\right) \) and \( {T}^{\prime }\le... | Yes |
Lemma 5. Consider a hyperbola \( {c}^{ * } \) and its auxiliary circle \( c \) . Draw two tangents \( \left\{ {t,{t}^{\prime }}\right\} \) parallel to the asymptotes intersecting at a point \( J \) . Then every tangent \( p \) to the auxiliary circle intersects lines \( \left\{ {t,{t}^{\prime }}\right\} \) correspondin... | The proof starts by defining the polar \( {FE} \) of \( J \) with respect to the conic \( {c}^{ * } \) . This is simultaneously the polar of \( J \) with respect to circle \( c \) since \( \left( {G, H, I, J}\right) = - 1 \) are harmonic with respect to either of the curves (see Figure 12). Take then a tangent of \( c ... | Yes |
Corollary 6. Under the assumptions of Lemma 5, in the case conic \( {c}^{ * } \) is a hyperbola, point \( M = {c}^{ * } \cap {p}_{D} \) is the common middle of segments \( \{ {NO},{KL},{BC}\} \) on the tangent \( {p}_{D} \) of circle \( c \) . Of these segments the first \( {NO} \) is intercepted by the asymptotes of \... | The proof for the first segment \( {NO} \) follows directly from Lemma 5 . The proof for the second segment \( {KL} \) results from the well known fact [4, p.267], according to which \( {KL} \) and \( {NO} \) have common middle for every secant of the hyperbola. The proof for the third segment \( {BC} \) follows from t... | No |
Lemma 9. Let circle \( c \), be tangent to line \( {L}_{0} \) at its point \( C \) and line \( {L}_{1} \) be parallel to \( {L}_{0} \) and not intersecting \( c \) . From a point \( H \) on \( {L}_{1} \) draw the tangents to \( c \) which intersect \( {L}_{0} \) at points \( \{ K, L\} \) . Then \( {CL} \cdot {CK} \) is... | (1) is seen by drawing first the medial lines of segments \( \{ {AL},{AK}\} \) which meet at \( P \) and define there the circumcenter of \( {ALK} \) . Their parallels from \( \{ L, K\} \) respectively meet at the diametral \( M \) of \( A \) on the circumcircle \( {c}^{\prime } \) of \( {ALK} \) . Since they are ortho... | No |
Proposition 10. Let \( c\left( {I, r}\right) \) be a circle with center at point \( I \) and radius \( r \) . Let also \( G \) be situated at distance \( \left| {IG}\right| = \frac{r}{3} \) from \( I \) . Then the following statements are valid:\n\n(1) The homothety \( f \) centered at \( G \) and with ratio \( k = - 2... | Statement (1) is obvious. | No |
(1) The pencil \( \mathcal{I} \) of circles generated by \( c \) and \( {c}_{0} \) has limit points \( \left\{ {{S}_{p}, T}\right\} \), where \( T \) is the inverse of \( {S}_{p} \) with respect to \( c \) . | To complete the proof of the two first claims it suffices to identify the limit points of the pencil of circles generated by \( c \) and \( {c}_{0} \) . For this use can be made of the fact that these two points are simultaneously harmonic conjugate with respect to the diameter points of the circles \( c \) and \( {c}_... | Yes |
Lemma 2. Given three points \( A, B, C \) of a parabola and the direction \( m \) of its axis, the tangents to the parabola at \( A, B, C \) are constructible with ruler and compass. | Lemma 2 follows by applying Pascal’s theorem to the six points \( m, B, A, A, m, C \) on the parabola. The intersections \( D = {BC} \cap {Am}, E = {mC} \cap {AA} \) and \( F = {mm} \cap {BA} \) are collinear. Note that \( F \) is the infinite point of \( {AB} \) . Therefore, by constructing (i) the parallel to \( m \)... | Yes |
Lemma 3. If \( \\left( {f : g : h}\\right) \) is the infinite point of the line \( \\ell \), then the infinite point \( \\left( {{f}^{\\prime } : {g}^{\\prime } : {h}^{\\prime }}\\right) \) of the perpendiculars to \( \\ell \) is given by\n\n\[ \n{f}^{\\prime } = g{S}_{B} - h{S}_{C},\\;{g}^{\\prime } = h{S}_{C} - f{S}_... | Proof. Note that \( f + g + h = 0 = {f}^{\\prime } + {g}^{\\prime } + {h}^{\\prime } \) and that \( {S}_{A},{S}_{B},{S}_{C} \) are just the dot products \( \\overrightarrow{AB} \\cdot \\overrightarrow{AC},\\overrightarrow{BC} \\cdot \\overrightarrow{BA},\\overrightarrow{CA} \\cdot \\overrightarrow{CB} \), respectively.... | Yes |
Theorem 5. With the notations above,\n\n(a) \( {\mathcal{P}}^{\prime }{}_{m} \) is the image of \( {\mathcal{P}}_{m} \) under \( \mathrm{h}\left( {P,4}\right) \) ,\n\n(b) the locus of \( P \) is the cubic \( \mathcal{C} \) with barycentric equation \( {\left( x + y + z\right) }^{3} = {27xyz} \) , the centroid \( G \) o... | Proof. (a) As in \( §3 \), it is readily checked that a barycentric equation of \( {\mathcal{P}}^{\prime }{}_{m} \) is\n\n\[ {u}^{2}{x}^{2} + {v}^{2}{y}^{2} + {w}^{2}{z}^{2} - {2vwyz} - {2wuzx} - {2uvxy} = 0. \]\n\nLet \( {A}^{\prime } = \mathrm{h}\left( A\right) \) so that \( {uvw}{A}^{\prime } = \left( {{4uvw} - {u}^... | Yes |
Theorem 1. A convex quadrilateral is tangential if and only if the incircles in the two triangles formed by a diagonal are tangent to each other. | Proof. In a convex quadrilateral \( {ABCD} \), let the incircles in triangles \( {ABC},{CDA} \) , \( {BCD} \) and \( {DAB} \) be tangent to the diagonals \( {AC} \) and \( {BD} \) at the points \( X, Y, Z \) and \( W \) respectively (see Figure 2). First we prove that\n\n\[ \n{ZW} = \frac{1}{2}\left| {a - b + c - d}\ri... | Yes |
Theorem 2. The incircles in the four overlapping triangles formed by the diagonals of a convex quadrilateral are tangent to the sides in eight points, two per side, making one distance between tangency points per side. It is a tangential quadrilateral if and only if the sums of those distances at opposite sides are equ... | Proof. According to the two tangent theorem, \( {AZ} = {AY},{BS} = {BT},{CU} = {CV} \) and \( {DW} = {DX} \), see Figure 3. Using the Pitot theorem, we get\n\n\[ {AB} + {CD} = {BC} + {DA} \]\n\n\[ \Leftrightarrow {AZ} + {ZS} + {BS} + {CV} + {VW} + {DW} = {BT} + {TU} + {CU} + {DX} + {XY} + {AY} \]\n\n\[ \Leftrightarrow ... | Yes |
Theorem 3. A convex quadrilateral is subdivided into four nonoverlapping triangles by its diagonals. Consider the four tangency points of the incircles in these triangles on one of the diagonals. It is a tangential quadrilateral if and only if the distance between two tangency points on one side of the second diagonal ... | Proof. Here we cite the Russian proof given in [18]. We use notations as in Figure 4 and shall prove that the quadrilateral has an incircle if and only if \( {T}_{1}^{\prime }{T}_{2}^{\prime } = \) \( {T}_{3}^{\prime }{T}_{4}^{\prime } \) .\n\nBy the two tangent theorem we have\n\n\[ \nA{T}_{1} = A{T}_{1}^{\prime \prim... | Yes |
Theorem 4. A convex quadrilateral ABCD is tangential if and only if\n\n\[ \n\\frac{1}{{R}_{1}} + \\frac{1}{{R}_{3}} = \\frac{1}{{R}_{2}} + \\frac{1}{{R}_{4}} \n\]\n\nwhere \( {R}_{1},{R}_{2},{R}_{3} \) and \( {R}_{4} \) are the exradii to triangles \( {ABP},{BCP},{CDP} \) and \( {DAP} \) opposite the vertex \( P \), th... | Proof. In a triangle, an exradius \( {R}_{a} \) is related to the altitudes by the well known relation\n\n\[ \n\\frac{1}{{R}_{a}} = - \\frac{1}{{h}_{a}} + \\frac{1}{{h}_{b}} + \\frac{1}{{h}_{c}} \n\]\n\n(3)\n\n---\n\n\( {}^{5} \) Although he used different notations.\n\n---\n\n\n\nIf we denote the altitudes from \( A \... | Yes |
Theorem 5 (Dergiades). A convex quadrilateral ABCD with diagonals intersecting at \( P \) is tangential if and only if the four excenters to triangles \( {ABP},{BCP} \) , \( {CDP} \) and \( {DAP} \) opposite the vertex \( P \) are concyclic. | Proof. In a triangle \( {ABC} \) with sides \( a, b, c \) and semiperimeter \( s \), where \( I \) and \( {J}_{1} \) are the incenter and excenter opposite \( A \) respectively, and where \( r \) and \( {R}_{a} \) are the radii in the incircle and excircle respectively, we have \( {AI} = \frac{r}{\sin \frac{A}{2}} \) a... | Yes |
Theorem 6. A convex quadrilateral ABCD with diagonals intersecting at \( P \) is tangential if and only if\n\n\[ \frac{\left( {{AP} + {BP} - {AB}}\right) \left( {{CP} + {DP} - {CD}}\right) }{\left( {{AP} + {BP} + {AB}}\right) \left( {{CP} + {DP} + {CD}}\right) } = \frac{\left( {{BP} + {CP} - {BC}}\right) \left( {{DP} +... | Proof. In a triangle \( {ABC} \) with sides \( a, b \) and \( c \), the distance from vertex \( A \) to the incenter \( I \) is given by\n\n\[ {AI} = \frac{r}{\sin \frac{A}{2}} = \frac{\sqrt{\frac{\left( {s - a}\right) \left( {s - b}\right) \left( {s - c}\right) }{s}}}{\sqrt{\frac{\left( {s - b}\right) \left( {s - c}\r... | Yes |
Theorem 8. A convex quadrilateral ABCD with diagonals intersecting at \( P \) is tangential if and only if\n\n\[ \frac{\left( {{AB} + {AP} - {BP}}\right) \left( {{CD} + {CP} - {DP}}\right) }{\left( {{AB} - {AP} + {BP}}\right) \left( {{CD} - {CP} + {DP}}\right) } = \frac{\left( {{BC} - {BP} + {CP}}\right) \left( {{DA} -... | Proof. It is well known that in a triangle \( {ABC} \) with sides \( a, b \) and \( c \) ,\n\n\[ \tan \frac{A}{2} = \sqrt{\frac{\left( {s - b}\right) \left( {s - c}\right) }{s\left( {s - a}\right) }} = \sqrt{\frac{\left( {a - b + c}\right) \left( {a + b - c}\right) }{\left( {a + b + c}\right) \left( {-a + b + c}\right)... | Yes |
Lemma 9. If \( {J}_{1} \) is the excenter opposite \( A \) in a triangle \( {ABC} \) with sides \( a, b \) and c, then\n\n\[ \frac{{\left( B{J}_{1}\right) }^{2}}{ac} = \frac{s - c}{s - a} \]\n\nwhere \( s \) is the semiperimeter. | Proof. If \( {R}_{a} \) is the radius in the excircle opposite \( A \), we have (see Figure 10)\n\n\[ \sin \frac{\pi - B}{2} = \frac{{R}_{a}}{B{J}_{1}} \]\n\n\[ B{J}_{1}\cos \frac{B}{2} = \frac{T}{s - a} \]\n\n\[ {\left( B{J}_{1}\right) }^{2} \cdot \frac{s\left( {s - b}\right) }{ac} = \frac{s\left( {s - a}\right) \left... | Yes |
Theorem 10. A convex quadrilateral ABCD with diagonals intersecting at \( P \) is tangential if and only if the four excenters to triangles \( {ABP},{BCP},{CDP} \) and DAP opposite the vertices \( B \) and \( D \) are concyclic. | Proof. We use the notation \( {J}_{{AP} \mid B} \) for the excenter in the excircle tangent to side \( {AP} \) opposite \( B \) in triangle \( {ABP} \) . Using the Lemma in triangles \( {ABP},{BCP} \) , \( {CDP} \) and \( {DAP} \) yields (see Figure 11)\n\n\[ \n\frac{{\left( P{J}_{{AP} \mid B}\right) }^{2}}{{AP} \cdot ... | Yes |
Theorem 11. A convex quadrilateral ABCD with diagonals intersecting at \( P \) is tangential if and only if\n\n\[ \frac{1}{{R}_{a}} + \frac{1}{{R}_{c}} = \frac{1}{{R}_{b}} + \frac{1}{{R}_{d}} \]\n\nwhere \( {R}_{a},{R}_{b},{R}_{c} \) and \( {R}_{d} \) are the radii in the excircles to triangles \( {ABP},{BCP} \) , \( {... | Proof. We use notations on the altitudes as in Figure 12, which are the same as in the proof of Theorem 4. From (3) we have\n\n\[ \frac{1}{{R}_{a}} = - \frac{1}{{h}_{B}} + \frac{1}{{h}_{A}} + \frac{1}{{h}_{1}} \]\n\n\[ \frac{1}{{R}_{b}} = - \frac{1}{{h}_{B}} + \frac{1}{{h}_{C}} + \frac{1}{{h}_{2}} \]\n\n\[ \frac{1}{{R}... | Yes |
Theorem 2. Suppose DEF is the pedal triangle of a point \( X \) on \( {L}^{\infty } \) . Then the perspectors \( {P}^{ * },{P}_{1},{P}_{2} \) are invariant of \( X \), and \( {P}^{ * } \) lies on \( {L}^{\infty } \) . | Proof. The three perspectors as in Theorem 1 are given as in (5)-(7) by\n\n\[ \n{P}^{ * } = a\left( {{b}^{2}r - {c}^{2}q}\right) : b\left( {{c}^{2}p - {a}^{2}r}\right) : c\left( {{a}^{2}q - {b}^{2}p}\right) ,\n\]\n\n(8)\n\n\[ \n{P}_{1} = {qa}{c}^{2} : {rb}{a}^{2} : {pc}{b}^{2},\n\]\n\n\[ \n{P}_{2} = {ra}{b}^{2} : {pb}{... | Yes |
Theorem 3. Suppose \( X \) is a point on the circumcircle of \( {ABC} \), and \( P = {a}^{2} : {b}^{2} \) : \( {c}^{2} \) . Then the perspector \( {P}^{ * } \), given by\n\n\[ \n{P}^{ * } = {bcx}\left( {{y}^{2} - {z}^{2}}\right) \left( {{ax}\left( {{bz} - {cy}}\right) + {yz}\left( {{b}^{2} - {c}^{2}}\right) }\right) \n... | Proof. Since \( X \) satisfies \( \frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 0 \), we can and do substitute \( z = - \frac{cxy}{{ay} + {bx}} \) in (8), obtaining\n\n\[ \n{P}^{ * } = \alpha : \beta : \gamma \n\]\n\n(10)\n\nwhere\n\n\[ \n\alpha = {ycb}\left( {{ay} + {bx} - {cx}}\right) \left( {{ay} + {bx} + {cx}}\right) \... | Yes |
Theorem 4. Suppose \( X \) is a point on the circumcircle of \( {ABC} \), and let \( {X}^{\prime } \) denote the antipode of \( X \) . Then \( {P}^{ * } = X \oplus {X}^{\prime } \) . | Proof. \( {}^{1} \) Since \( X \) is an arbitrary point on \( \Gamma \), there exists \( \theta \) such that\n\n\[X = \csc \theta : \csc \left( {C - \theta }\right) : - \csc \left( {B + \theta }\right) ,\]\n\nwhere \( \theta \), understood here a function of \( a, b, c \), is defined ([6],[3, p. 39]) by\n\n\[0 \leq {2\... | Yes |
In Theorems 3 and 4, let \( X = {X}_{1113} \), this being a point of intersection of the Euler line and the circumcircle. The antipode of \( X \) is \( {X}_{1114} \), and we have | \[ {X}_{1113} \oplus {X}_{1114} = {X}_{125} \] the center of the Jerabek hyperbola, on the nine-point circle. | No |
The antipode of the Tarry point, \( {X}_{98} \), is the Steiner point, \( {X}_{99} \), and | \[ {X}_{98} \oplus {X}_{99} = {X}_{2679} \] | Yes |
Example 5. The antipode of the point, \( {X}_{101} = \frac{a}{b - c} : \frac{b}{c - a} : \frac{c}{a - b} \) is \( {X}_{103} \), and | \[ {X}_{101} \oplus {X}_{103} = {X}_{1566} \] | Yes |
The Euler line meets the line at infinity in the point \( {X}_{30} \), of which the isogonal conjugate on the circumcircle is the point\n\n\[ \n{X}_{74} = \frac{1}{\cos A - 2\cos B\cos C} : \frac{1}{\cos B - 2\cos C\cos A} : \frac{1}{\cos C - 2\cos A\cos B}.\n\] | The antipode of \( {X}_{74} \) is the center of the Kiepert hyperbola, given by\n\n\[ \n{X}_{110} = \csc \left( {B - C}\right) : \csc \left( {C - A}\right) : \csc \left( {A - B}\right) ,\n\]\n\nand we have\n\n\[ \n{X}_{74} \oplus {X}_{110} = {X}_{3258} \n\] | No |
Theorem 5. Let \( {X}^{\prime } \) be the antipode of \( X \) . Then \( X \oplus {X}^{\prime } \) is a point of intersection of the nine-point circle and the line of the following two points: the isogonal conjugate of \( X \) and \( X \) (medial). | Proof. Trilinears for \( X \) (medial) are given ([3, p. 86]) by\n\n\[ \left( {{by} + {cz}}\right) /a : \left( {{cz} + {ax}}\right) /b : \left( {{ax} + {by}}\right) /c. \]\n\nWriting \( u : v : w \) for trilinears for \( X \oplus {X}^{\prime } \) and \( {yz} : {zx} : {xy} \) for the isogonal conjugate of \( X \), and p... | Yes |
If \( {DEF} \) is the circumtangential triangle, then \( {D}^{\prime \prime }{E}^{\prime \prime }{F}^{\prime \prime } \) is homothetic to each of the three Morley equilateral triangles, as well as the circumtangential triangle (perspector \( {X}_{2} \), homothetic ratio \( - 1/2 \) ) the circumnormal triangle (per-spec... | The triangle \( {D}^{\prime \prime }{E}^{\prime \prime }{F}^{\prime \prime } \) is the second of two equilateral triangles described in the article on the Steiner deltoid at [5]; its vertices are given as follows:\n\n\[ \n{D}^{\prime \prime } = \cos \left( {B - C}\right) - \cos \left( {\frac{B}{3} - \frac{C}{3}}\right)... | Yes |
Proposition 1. A line tangent to a conic at the origin cannot intersect the conic section again at another point. | Proof. Choose coordinates so that the tangent line at the origin is \( x = 0 \) . Restricting the conic to \( x = 0 \) gives us a quadratic polynomial in \( y \), which has at most two zeroes with multiplicity. But since \( x = 0 \) is tangent to the conic at the origin, there is a double zero at \( y = 0 \), and so th... | Yes |
Lemma 4. Neither \( {L}_{1} \) nor \( {L}_{2} \) can be identically zero. | Proof. If \( {L}_{1} \) is identically zero, then the equation of \( {P}_{1} \) is a homogeneous quadratic form. Thus if \( {P}_{1}\left( {a, b}\right) = 0 \), then either \( a = b = 0 \) or the line \( \{ \left( {{at},{bt}}\right), t \in \mathbb{F}\} \) is contained in \( {P}_{1} \) . Thus \( {P}_{1} \) is either a po... | Yes |
Lemma 5. \( {L}_{1} \) cannot divide \( {Q}_{1} \), and \( {L}_{2} \) cannot divide \( {Q}_{2} \) . | Proof. If \( {L}_{1} \) divides \( {Q}_{1} \), then \( {L}_{1} \) divides \( {P}_{1} \) as well. But according to our definition, a conic cannot contain a zero set of a linear equation. A completely analogous proof works for \( {P}_{2} \) . | Yes |
Proposition 6. If a quadratic equation in two variables over an infinite field has two distinct solutions, it has infinitely many. | Proof. Let \( P\left( {x, y}\right) \) be a quadratic polynomial with two distinct solutions. We may choose coordinates so that one of the solutions is the origin, and the other is \( \left( {a, b}\right) \) , where \( a \) is nonzero. Substitute \( y = {mx} \) into \( P\left( {x, y}\right) \) . We then get\n\n\[ P\lef... | Yes |
If we define conics \( {P}_{1},{P}_{2} \) intersecting at the origin as in (1) and (2), then the medilocus is either the zero set of the equation\n\n\[ 1 = k\left( {-\frac{{L}_{1}\left( {x, y}\right) }{{Q}_{1}\left( {x, y}\right) }}\right) + \left( {1 - k}\right) \left( {-\frac{{L}_{2}\left( {x, y}\right) }{{Q}_{2}\lef... | Proof. Firstly, we write a parametric equation for a line through the origin and a non-origin point \( \left( {a, b}\right) \) as\n\n\[ l = \{ \left( {{at},{bt}}\right), t \in \mathbb{F}\} .\n\nWe define points \( C, D \) as in Definition 3, such that the line \( l \) intersects \( {P}_{1} \) at the origin and at \( C ... | Yes |
Lemma 10. Let \( l, g, f \in \mathbb{F}\left\lbrack {x, y}\right\rbrack \) be polynomials with degree \( l = 1 \), degree \( g = 2 \) and \( V\left( g\right) \) infinite. (i) If \( V\left( l\right) \subset V\left( f\right) \), then \( l \mid f \) . | Proof. (i) First, we claim that if \( l \) is linear and \( V\left( l\right) \subset V\left( f\right) \), then \( l \mid f \) . Corollary 1 of Proposition 2 in Chapter 1 of [1] says that if \( l \) is an irreducible polynomial in a closed field \( \overline{\mathbb{F}}, V\left( l\right) \subset V\left( f\right) \) and ... | Yes |
Proposition 9 (where \( {u}_{1} \) is \( h \) in the notation of that proposition). If \( {u}_{1}{\left( x, y\right) }^{2} \) divides each term of the right hand side, we have\n\n\[ \n{u}_{1}{\left( x, y\right) }^{2} \mid k{L}_{1}\left( {x, y}\right) {Q}_{2}\left( {x, y}\right) \n\]\n\nand\n\n\[ \n{u}_{1}{\left( x, y\r... | The first equation implies that \( {u}_{1}{\left( x, y\right) }^{2} \) is a scalar multiple of \( {Q}_{2}\left( {x, y}\right) \) . If \( {u}_{1}\left( {x, y}\right) \) is a scalar multiple of \( {L}_{2}\left( {x, y}\right) ,{u}_{1}\left( {x, y}\right) \) divides \( {P}_{2}\left( {x, y}\right) \) violating Lemma 5 . If ... | Yes |
Theorem 1. The vertices of the Varignon parallelogram and those of the principal orthic quadrilateral of \( \mathbf{Q} \), that lie on the lines containing two opposite sides of \( \mathbf{Q} \), belong to a circle with center \( G \) . | Proof. The circle with diameter \( {M}_{1}{M}_{3} \) passes through \( {H}_{1} \) and \( {H}_{3} \), because \( \angle {M}_{1}{H}_{1}{M}_{3} \) and \( \angle {M}_{1}{H}_{3}{M}_{3} \) are right angles (see Figure 3). Analogously, the circle with diameter \( {M}_{2}{M}_{4} \) passes through \( {H}_{2} \) and \( {H}_{4} \... | Yes |
Corollary 2. The vertices of the Varignon parallelogram and those of the principal orthic quadrilateral of \( \mathbf{Q} \) all lie on a circle (with center \( G \) ) if and only if \( \mathbf{Q} \) is orthodiagonal. | Proof. The two circles containing the vertices of \( V \) and \( H \) coincide if and only if \( {M}_{1}{M}_{3} = {M}_{2}{M}_{4} \), i.e., if and only if \( V \) is a rectangle. This is the case if and only if \( \mathbf{Q} \) is orthodiagonal. | Yes |
Theorem 3. If \( \mathbf{Q} \) is cyclic, the points \( {X}_{i},{X}_{i}^{\prime } \) that belong to the lines containing two opposite sides of \( \mathbf{Q} \) lie on a circle with center \( H \) . | Proof. Let us prove that the points \( {X}_{1},{X}_{1}^{\prime },{X}_{3},{X}_{3}^{\prime } \) are on a circle with center \( H \) (see Figure 5). Since \( H \) is on the perpendicular bisector of the segment \( {X}_{1}{X}_{1}^{\prime } \) , we have \( H{X}_{1} = H{X}_{1}^{\prime } \) . Moreover, since \( {X}_{1} \) lie... | Yes |
Corollary 4. For a cyclic quadrilateral \( \mathbf{Q} \), the eight points \( {X}_{i},{X}_{i}^{\prime }, i = 1,2,3,4 \) , all lie on a circle (with center \( H \) ) if and only if \( \mathbf{Q} \) is orthodiagonal. | Proof. The two circles that contains the points \( {X}_{i},{X}_{i}^{\prime } \) and coincide if and only if \( {H}_{1}G = {H}_{2}G = {H}_{3}G = {H}_{4}G \), i.e., if and only if the principal orthic quadrilateral is inscribed in a circle with center \( G \) . From Corollary 2, this is the case if and only if \( \mathbf... | Yes |
Theorem 5. If \( \mathbf{Q} \) is cyclic and orthodiagonal, the radius of the first Droz-Farny circle of \( \mathbf{Q} \) is the circumradius of \( \mathbf{Q} \) . | Proof. From the proof of Theorem 3 we have\n\n\[ H{X}_{1}^{2} = 2{H}_{1}{G}^{2} + \frac{1}{2}O{H}^{2}. \]\n\n(1)\n\nMoreover, \( {M}_{3}{A}_{3} = {M}_{3}H \) since \( \mathbf{Q} \) is orthodiagonal and \( \angle {A}_{3}H{A}_{4} \) is a right angle (see Figure 6). By applying Pythagoras’ theorem to the triangle \( O{M}_... | Yes |
Theorem 6. If \( \mathbf{Q} \) is cyclic, the points \( {Y}_{i},{Y}_{i}^{\prime } \) that belong to the lines containing two opposite sides of \( \mathbf{Q} \) lie on a circle with center \( O \) . | Proof. Let us prove that the points \( {Y}_{1},{Y}_{1}^{\prime },{Y}_{3},{Y}_{3}^{\prime } \) are on a circle with center \( O \) (see Figure 7).\n\nSince \( O \) is on the perpendicular bisector of the segment \( {Y}_{1}{Y}_{1}^{\prime }, O{Y}_{1} = O{Y}_{1}^{\prime } \) . Moreover, since \( {Y}_{1} \) lies on the cir... | Yes |
For a cyclic quadrilateral \( \mathbf{Q} \), the eight points \( {Y}_{i},{Y}_{i}^{\prime }, i = 1,2,3,4 \) , all lie on a circle (with center \( O \) ) if and only if \( \mathbf{Q} \) is orthodiagonal. | Proof. The two circles that contain the points \( {Y}_{i}^{\prime },{Y}_{i}^{\prime }, i = 1,2,3,4 \), coincide if and only if \( {M}_{1}G = {M}_{2}G \), i.e., if and only if \( {M}_{1}{M}_{3} = {M}_{2}{M}_{4} \) . This is the case if and only if \( \mathbf{Q} \) is orthodiagonal. | Yes |
Theorem 9. The eight points \( {Z}_{i},{Z}_{i}^{\prime }, i = 1,2,3,4 \), all lie on a circle if and only if \( \mathbf{Q} \) is orthodiagonal. | Proof. Suppose first that the eight points \( {Z}_{i},{Z}_{i}^{\prime }, i = 1,2,3,4 \), all lie on a circle. If \( C \) is the center of the circle, then each \( {C}_{i} \) coincides with \( C \) . Since the lines \( {A}_{1}{C}_{1} \) and \( {A}_{3}{C}_{3} \) both are perpendicular to \( {A}_{2}{A}_{4} \), the point \... | Yes |
Theorem 11. If \( \mathbf{Q} \) is a convex quadrilateral, the eight points \( {Z}_{i},{Z}_{i}^{\prime }, i = 1,2,3,4 \) , all lie on an ellipse whose axes are the bisectors of the angles between the diagonals of \( \mathbf{Q} \) . Moreover, the area of the ellipse is equal to the area of a circle with radius \( r \) . | Proof. We set up a Cartesian coordinate system with axes the bisectors of the angles between the diagonals of \( \mathbf{Q} \) . The equations of the diagonals are of the form \( y = {mx} \) and \( y = - {mx} \), with \( m > 0 \) . The vertices of \( \mathbf{Q} \) have coordinates \( {A}_{1}\left( {{a}_{1}, m{a}_{1}}\r... | Yes |
Corollary 2. We can solve the triangle problem when given \( O, H \) and either the Nagel point \( {N}_{\mathrm{a}} \) or the Spieker center \( {S}_{\mathrm{p}} \) . | Proof. We use the fact that the four points \( I, G,{S}_{\mathrm{p}},{N}_{\mathrm{a}} \) lie on a line with ratio \( {IG} : G{S}_{\mathrm{p}} : {S}_{\mathrm{p}}{N}_{\mathrm{a}} = 2 : 1 : 3 \) . Given \( O, H \) we can construct \( G \), and then given either \( {S}_{\mathrm{p}} \) or \( {N}_{\mathrm{a}} \), we can dete... | Yes |
Proposition 3. Let \( \mathcal{P} \) be the hyperbola of the Poncelet pencil whose center is \( W \). The point \( Z \) so that \( {HW} = {WZ} \) on the line \( {HW} \) is on the circumcircle of \( \Delta \) and the hyperbola \( \mathcal{P} \). | Proof. A rectangular hyperbola is symmetric about its center. Hence, any line through the center meets the hyperbola in two points of equal distance from the center. Thus the line \( {HW} \) meets \( \mathcal{P} \) at another point \( Z \) so that \( {WZ} = {HW} \). Since the central similarity at \( H \) with scale fa... | Yes |
Proposition 4. A hyperbola \( \mathcal{P} \) of the Poncelet pencil with center \( W \) is tangent to the circumcircle if and only if \( {HW} \) is perpendicular to \( \mathcal{L} = {\mathcal{P}}^{ * } \) if and only if the circumcircle point \( Z \) is a vertex of the triangle. | Proof. If \( Z \) is a vertex then its isogonal transform is at infinity on \( {BC} \) and on the transform \( \mathcal{L} = {\mathcal{P}}^{ * } \) . Hence \( \mathcal{L} \) is parallel to \( {BC} \) . Thus from the remarks above, \( {HW} \) is an altitude if and only if the circumcircle point is a vertex. Then tangenc... | Yes |
Theorem 5. For \( S \) on the circumcircle of triangle \( {ABC}, S{S}^{ * } \) is perpendicular to the Wallace-Simson line of \( S \), i.e., \( {S}^{ * } \) lies on the Wallace-Simson line of the antipode \( {S}^{\prime } \) of \( S \) . | Proof. From [4] the Wallace-Simson line of \( S \) passes through the isogonal conjugate \( {T}^{ * } \) of its antipodal point \( T = {S}^{\prime } \) . Thus it is perpendicular to \( S{S}^{ * } \) since the angle between \( {T}^{ * } \) and \( {S}^{ * } \) is 90 degrees, the angle being halved by the isogonal transfo... | No |
Theorem 6. Consider the line \( \mathcal{L} \) though the circumcenter of triangle \( {ABC} \), meeting the circumcircle at \( U,{U}^{\prime } \). Let \( \mathcal{K} = {\mathcal{L}}^{ * } \). (i) The Wallace-Simson lines of \( U,{U}^{\prime } \) are asymptotes of \( \mathcal{K} \) and meet at the center \( W\left( \mat... | Proof. The asymptotes of \( \mathcal{K} \) are the Wallace-Simson lines of the isogonal conjugates of the points at infinity of \( \mathcal{K} \) ,[4, p. 196]. Hence the center of \( \mathcal{K} \) is the orthopole of \( {\mathcal{K}}^{ * } = \mathcal{L} \) (see \( \left\lbrack {6,§{406}}\right\rbrack \) ). A dilation ... | Yes |
Theorem 8. We can uniquely determine the triangle when given \( O, H \) and \( K \) . | Proof. It is known that the center \( W \) of Kiepert’s hyperbola \( \mathcal{K} \) is the inverse of \( K \) in the orthocentroidal circle with diameter \( {GH} \) and center \( J \) [7]. Thus the center \( W \) is constructible given \( O, H, K \) . Since this point is the intersection of the ray \( {JK} \) with the ... | Yes |
Problem 1 (Fermat’s problem). Let \( {ABC} \) be a triangle. Find a point \( P \) such that \( {PA} + {PB} + {PC} \) is minimum. | The first published solution came from Evangelist Torricelli, published posthumously in 1659. Numerous subsequent solutions are readily to be found in books and journals. | No |
Problem 3. Let \( {ABC} \) be a triangle and \( x, y, z \) be non-zero real numbers. Find a point \( P \) such that \( x \cdot {PA} + y \cdot {PB} + z \cdot {PC} \) is minimum. | The solution of problem 3 requires the solution of Problem 2 as well as solutions to Problems 4 and 5 below. | No |
Problem 5. Let \( {ABC} \) be a triangle and \( x, y, z \) be positive real numbers. Find a point \( P \) such that \( - x \cdot {PA} - y \cdot {PB} + z \cdot {PC} \) is minimum. | Here is an easy solution of Problem 5. If \( z < x + y \), then \( - x \cdot {PA} - y \cdot {PB} + z \cdot {PC} \) decreases without bound as \( {PC} \) goes to \( \infty \), whence there exists no point \( P \) for which the minimum is attained; otherwise, when \( z \geq x + y \) the minimum is attained when \( P \) c... | Yes |
Give an isosceles triangle \( {ABC} \) with \( {AB} = {AC},{Ax} \) is the opposite ray of the ray \( {AB} \) . For every \( P \) lying inside \( {xAC} \), we have \( {PB} \geq {PC} \), with equality when \( P \) coincides with \( A \) . | Lemma 4 holds because every point to the side of the perpendicular bisector of \( {BC} \) is closer to \( C \) than to \( B \) . | No |
Lemma 2. The circles \( {\delta }_{t}^{\alpha } \) and \( {\delta }_{t}^{\beta } \) are congruent with common radii \( \left| {1 - {t}^{2}}\right| {r}_{\mathrm{A}} \) . | Proof. Let \( s \) be the radius of the circle \( {\delta }_{t}^{\alpha } \) . If \( {\gamma }_{t} \) touches \( \alpha \) and \( \beta \) internally, then we get\n\n\[ \left( {\frac{2ab}{a{t}^{2} + b} - 2{r}_{\mathrm{A}}}\right) : s = \frac{2ab}{a{t}^{2} + b} : b. \]\n\nSolving the equation we get \( s = \left( {1 - {... | Yes |
Theorem 3. The circles \( {\delta }_{t}^{\alpha } \) and \( {\delta }_{t}^{\beta } \) are congruent with common radius \( \left| {1 - {t}^{2}}\right| {r}_{\mathrm{A}} \) . The points of tangency \( {A}_{t},{B}_{t},{T}_{\alpha } \), and \( {T}_{\beta } \) are collinear. | The circle \( {\delta }_{t}^{\alpha } \) touches the line \( {\mathcal{L}}_{\alpha } \) at the point\n\n\[ \n{V}_{\alpha } = \left( {2{r}_{\mathrm{A}},{2t}{r}_{\mathrm{A}}\sqrt{\frac{a}{b}}}\right) .\n\]\n\nFrom these coordinates it is clear that \( {A}_{t},{V}_{\alpha } \) and \( O \) are collinear (see Figures 4 and ... | Yes |
For real numbers \( t \) and \( w \), the circles \( {\delta }_{t}^{\alpha } \) and \( {\delta }_{t}^{\beta } \) and the circles \( {\delta }_{w}^{\alpha } \) and \( {\delta }_{w}^{\beta } \) are congruent if and only if \( {\gamma }_{t} \) and \( {\gamma }_{w} \) are congruent. | In the case \( {t}^{2} + {w}^{2} = 2 \), one of the circles \( {\gamma }_{t} \) and \( {\gamma }_{w} \) touches \( \alpha \) and \( \beta \) internally and the other externally. In particular, the circles \( {\delta }_{t}^{\alpha } \) and \( {\delta }_{t}^{\beta } \) are Archimedean circles of the arbelos formed by \( ... | No |
Proposition 1. Two circles \( {\Gamma }_{1}\left( {r}_{1}\right) \) and \( {\Gamma }_{2}\left( {r}_{2}\right) \) are tangent to a circle \( \Gamma \left( R\right) \) through \( A, B \), respectively. The length \( {\delta }_{12} \) of the common external tangent of \( {\Gamma }_{1},{\Gamma }_{2} \) is given by\n\n\[ \n... | Proof. Without loss of generality assume that \( {r}_{1} \geq {r}_{2} \) . Let \( {\varepsilon }_{1} \) (respectively \( {\varepsilon }_{2} \) be \( + 1 \) or -1 according as the tangency of \( \left( O\right) \) and \( \left( {O}_{1}\right) \) (respectively \( \left( {O}_{2}\right) \) is external or internal. Figure 3... | Yes |
Proposition 2 (Casey’s theorem [2,§172]). Given four circles \( {\Gamma }_{i}, i = 1,2,3,4 \), let \( {\delta }_{ij} \) denote the length of a common tangent (either internal or external) between \( {\Gamma }_{i} \) and \( {\Gamma }_{j} \). The four circles are tangent to a fifth circle \( \Gamma \) (or line) if and on... | \[ {\delta }_{12} \cdot {\delta }_{34} \pm {\delta }_{13} \cdot {\delta }_{42} \pm {\delta }_{14} \cdot {\delta }_{23} = 0 \] | Yes |
Lemma 3. The lines through the incenter \( I \) parallel to \( {AB} \) and \( {AC} \) are tangent to the circle \( {A}_{1}\left( D\right) \) . | Proof. Let \( O \) be the circumcenter, and \( R \) the circumradius. Since \( {OD} = R\cos A \) , it is enough to show that\n\n\[ R\left( {1 - \cos A}\right) + r = A{A}_{1}\sin \frac{A}{2}. \]\n\n(3)\n\nThis follows from \( r = {IA}\sin \frac{A}{2} \) and \( 1 - \cos A = 2{\sin }^{2}\frac{A}{2} \) . (3) is equivalent ... | Yes |
Proposition 4. The sides of a triangle \( {ABC} \) are extended to points \( {X}_{b},{X}_{c},{Y}_{c} \) , \( {Y}_{a},{Z}_{a},{Z}_{b} \) such that\n\n\[ A{Y}_{a} = A{Z}_{a} = a,\;B{Z}_{b} = B{X}_{b} = b,\;C{X}_{c} = C{Y}_{c} = c. \]\n\nThe six points \( {X}_{b},{X}_{c},{Y}_{c},{Y}_{a},{Z}_{a},{Z}_{b} \) lie on a circle ... | Proof. If the incircle touches \( {BC} \) at \( X \), then \( {BX} = s - b \) . It follows that \( {X}_{b}X = \) \( b + \left( {s - b}\right) = s \), and \( I{X}_{b} = \sqrt{{r}^{2} + {s}^{2}} \) . The same result holds for the other five points. | Yes |
Proposition 5. The radical center of the circles \( {\omega }_{a},{\omega }_{b},{\omega }_{c} \) is the midpoint between the incenter \( I \) and Mittenpunkt of \( \bigtriangleup {ABC} \) . | Proof. Let \( {A}_{1}{P}_{2} \) and \( {A}_{1}{P}_{3} \) be the tangent segments from \( {A}_{1} \) to \( {\omega }_{b} \) and \( {\omega }_{c}\left( {{P}_{2} \in {\omega }_{b}}\right. \) and \( {P}_{3} \in {\omega }_{c} \) ) (See Figure 7). By Casey’s theorem for \( \left( {A}_{1}\right) ,\left( A\right) ,\left( C\rig... | Yes |
Proposition 6. The Apollonius circle \( \omega \) externally tangent to \( {\omega }_{a},{\omega }_{b},{\omega }_{c} \) is also tangent to the incircle of triangle \( {ABC} \) through its Feuerbach point \( {F}_{\mathrm{e}} \) . | Proof. Without loss of generality we assume that \( b \geq a \geq c \) . Let the incircle \( \left( I\right) \) touch \( {BC},{CA},{AB} \) at \( {X}_{a},{X}_{b},{X}_{c} \) respectively. By Casey’s theorem there exists a circle \( \omega \) tangent to \( {\omega }_{a},{\omega }_{a},{\omega }_{c} \) externally and tangen... | Yes |
Proposition 7. The center \( {I}_{0} \) of the inner Apollonius circle \( \omega \) of \( {\omega }_{a},{\omega }_{b},{\omega }_{c} \) is the intersection of the lines \( {X}_{3}{X}_{1001},{X}_{1}{X}_{11} \) and its radius \( \rho \) equals a third of the sum of the inradius and circumradius of triangle \( {ABC} \) . | Proof. By Proposition 5, the radical center \( L \) of \( {\omega }_{a},{\omega }_{b},{\omega }_{c} \) is the midpoint \( {X}_{1001} \) between the incenter \( I \) and the Mittenpunkt \( {X}_{9} \) of triangle \( {ABC} \) . Hence, the inversion with center \( {X}_{1001} \) and power equal to the power of \( {X}_{1001}... | Yes |
Theorem 1. A bicentric quadrilateral with sides \( a, b, c, d \) has the area\n\n\[ K = \sqrt{abcd}. \] | Proof. The diagonal \( {AC} \) divide a convex quadrilateral \( {ABCD} \) into two triangles \( {ABC} \) and \( {ADC} \) . Using the law of cosines in these, we have\n\n\[ {a}^{2} + {b}^{2} - {2ab}\cos B = {c}^{2} + {d}^{2} - {2cd}\cos D. \]\n\n(1)\n\nThe quadrilateral has an incircle. By the Pitot theorem \( a + c = b... | Yes |
Corollary 2. A bicentric quadrilateral with sides \( a, b, c, d \) has the area\n\n\[ K = {ac}\tan \frac{\theta }{2} = {bd}\cot \frac{\theta }{2} \]\n\nwhere \( \theta \) is the angle between the diagonals. | Proof. The angle \( \theta \) between the diagonals in a bicentric quadrilateral is given by\n\n\[ {\tan }^{2}\frac{\theta }{2} = \frac{bd}{ac} \]\n\naccording to \( \left\lbrack {8,\mathrm{p}{.30}}\right\rbrack \) . Hence we get\n\n\[ {K}^{2} = \left( {ac}\right) \left( {bd}\right) = {\left( ac\right) }^{2}{\tan }^{2}... | Yes |
Corollary 3. In a bicentric quadrilateral ABCD with sides \( a, b, c, d \) we have\n\n\[ \tan \frac{A}{2} = \sqrt{\frac{bc}{ad}} = \cot \frac{C}{2} \]\n\n\[ \tan \frac{B}{2} = \sqrt{\frac{cd}{ab}} = \cot \frac{D}{2}. \] | Proof. A well known trigonometric formula and (3) yields\n\n\[ \tan \frac{B}{2} = \sqrt{\frac{1 - \cos B}{1 + \cos B}} = \sqrt{\frac{cd}{ab}} \]\n\n(6)\n\nwhere we also used \( \cos D = - \cos B \) in (3). The formula for \( D \) follows from \( B = \pi - D \) . By symmetry in a bicentric quadrilateral, we get the form... | Yes |
Theorem 4. A tangential quadrilateral with sides \( a, b, c, d \) is also cyclic if and only if it has the area \( K = \sqrt{abcd} \) . | Proof. The area of a tangential quadrilateral is according to \( \left\lbrack {8,\mathrm{p}{.28}}\right\rbrack \) given by\n\n\[ K = \sqrt{abcd}\sin \frac{B + D}{2}.\n\]\n\nIt’s also cyclic if and only if \( B + D = \pi \) ; hence a tangential quadrilateral is cyclic if and only if it’s area is \( K = \sqrt{abcd} \) . | Yes |
Theorem 5. A bicentric quadrilateral ABCD with incenter I has the area\n\n\[ K = {AI} \cdot {CI} + {BI} \cdot {DI}. \] | Proof. The quadrilateral has an incircle, so \( \tan \frac{A}{2} = \frac{r}{e} \) where \( r \) is the inradius (see Figure 2). It also has a circumcircle, so \( A + C = B + D = \pi \) . Thus \( \cot \frac{C}{2} = \tan \frac{A}{2} \) and \( \sin \frac{C}{2} = \cos \frac{A}{2} \) . A bicentric quadrilateral can be parti... | Yes |
Corollary 6. A bicentric quadrilateral ABCD has the area\n\n\[ K = 2{r}^{2}\left( {\frac{1}{\sin A} + \frac{1}{\sin B}}\right) \]\n\nwhere \( r \) is the inradius. | Proof. Using one of the equalities in the proof of Theorem 5, we get\n\n\[ K = {r}^{2}\left( {\frac{1}{\sin \frac{A}{2}\cos \frac{A}{2}} + \frac{1}{\sin \frac{B}{2}\cos \frac{B}{2}}}\right) = {r}^{2}\left( {\frac{1}{\frac{1}{2}\sin A} + \frac{1}{\frac{1}{2}\sin B}}\right) \]\n\nand the result follows. | Yes |
Theorem 7. A convex quadrilateral with diagonals \( p, q \) and bimedians \( m, n \) has the area\n\n\[ K = \frac{1}{2}\sqrt{{p}^{2}{q}^{2} - {\left( {m}^{2} - {n}^{2}\right) }^{2}}. \] | Proof. A convex quadrilateral with sides \( a, b, c, d \) and diagonals \( p, q \) has the area\n\n\[ K = \frac{1}{4}\sqrt{4{p}^{2}{q}^{2} - {\left( {a}^{2} - {b}^{2} + {c}^{2} - {d}^{2}\right) }^{2}} \]\n\n(8)\n\naccording to [6, p.243],[11] and [16]. The length of the bimedians \( {}^{4}m, n \) in a convex quadrilate... | Yes |
Theorem 8. A bicentric quadrilateral with bimedians \( m, n \) and tangency chords \( k, l \) has the area\n\n\[ K = \left| \frac{{m}^{2} - {n}^{2}}{{k}^{2} - {l}^{2}}\right| {kl} \]\n\nif it is not a kite. | Proof. From Theorem 7 we get that in a convex quadrilateral\n\n\[ {\left( {m}^{2} - {n}^{2}\right) }^{2} = {\left( pq\right) }^{2} - 4{K}^{2}. \]\n\n(11)\n\nRewriting (7), we have in a bicentric quadrilateral\n\n\[ {pq} = \frac{{k}^{2} + {l}^{2}}{kl}K \]\n\nInserting this into (11) yields\n\n\[ {\left( {m}^{2} - {n}^{2... | Yes |
Theorem 9. Let a tangential quadrilateral have bimedians \( m, n \) and tangency chords \( k, l \) . Then\n\n\[ m < n\; \Leftrightarrow \;k > l \]\n\nwhere \( m \) and \( k \) connect the same pair of opposite sides. | Proof. Eulers extension of the parallelogram law to a convex quadrilateral with sides \( a, b, c, d \) states that\n\n\[ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} = {p}^{2} + {q}^{2} + 4{v}^{2} \]\n\nwhere \( v \) is the distance between the midpoints of the diagonals \( p, q \) (this is proved in [7, p.107] and [3, p.126]... | Yes |
Theorem 1. A tangential quadrilateral with sides \( a, b, c, d \) and diagonals \( p, q \) has the area | Proof. A convex quadrilateral with sides \( a, b, c, d \) and diagonals \( p, q \) has the area\n\n\[ K = \frac{1}{4}\sqrt{4{p}^{2}{q}^{2} - {\left( {a}^{2} - {b}^{2} + {c}^{2} - {d}^{2}\right) }^{2}} \]\n\n(2)\n\naccording to [6] and [14]. Squaring the Pitot theorem (1) yields\n\n\[ {a}^{2} + {c}^{2} + {2ac} = {b}^{2}... | Yes |
Corollary 3. The sums of the altitudes to opposite sides of a tangential quadrilateral in the nonoverlapping triangles formed by the diagonals are equal if and only if the quadrilateral is a kite. | Proof. If \( {h}_{1},{h}_{2},{h}_{3},{h}_{4} \) are the altitudes from the diagonal intersection \( P \) to the sides \( {AB},{BC},{CD},{DA} \) in triangles \( {ABP},{BCP},{CDP},{DAP} \) respectively, then according to [12] and Theorem 1 in [11] (with other notations)\n\n\[ \frac{1}{{h}_{1}} + \frac{1}{{h}_{3}} = \frac... | Yes |
Lemma 4. The incircle in a triangle \( {ABC} \) with sides \( a, b, c \) has the radius\n\n\[ r = \frac{a + b - c}{2}\tan \frac{C}{2}. \] | Proof. We use notations as in Figure 4, where there is one pair of equal tangent lengths \( x, y \) and \( z \) at each vertex due to the two tangent theorem. For the sides of the triangle we have \( a = y + z, b = z + x \) and \( c = x + y \) ; hence \( a + b - c = {2z} \) . \( {CI} \) is an angle bisector, so\n\n\[ \... | Yes |
Theorem 5. Let the diagonals in a tangential quadrilateral ABCD intersect at \( P \) and let the inradii in triangles \( {ABP},{BCP},{CDP},{DAP} \) be \( {r}_{1},{r}_{2},{r}_{3},{r}_{4} \) respectively. Then the quadrilateral is a kite if and only if\n\n\[ \n{r}_{1} + {r}_{3} = {r}_{2} + {r}_{4} \n\] | Proof. We use the same notations as in Figure 2. The four incircles and their radii are marked in Figure 5. Since \( \tan \frac{\pi - \theta }{2} = \cot \frac{\theta }{2} \), where \( \theta \) is the angle between the diagonals, Lemma 4 yields\n\n\[ \n{r}_{1} + {r}_{3} = {r}_{2} + {r}_{4} \n\]\n\n\[ \n\Leftrightarrow ... | Yes |
Corollary 6. If \( {r}_{1},{r}_{2},{r}_{3},{r}_{4} \) are the same inradii as in Theorem 5, then the tangential quadrilateral is a kite if and only if\n\n\[ \n{r}_{1}{r}_{3} = {r}_{2}{r}_{4} \n\] | Proof. In a tangential quadrilateral we have according to [12] and [15]\n\n\[ \n\frac{1}{{r}_{1}} + \frac{1}{{r}_{3}} = \frac{1}{{r}_{2}} + \frac{1}{{r}_{4}} \n\]\n\n(7)\n\nWe rewrite this as\n\n\[ \n\frac{{r}_{1} + {r}_{3}}{{r}_{1}{r}_{3}} = \frac{{r}_{2} + {r}_{4}}{{r}_{2}{r}_{4}} \n\]\n\nHence\n\n\[ \n{r}_{1} + {r}_... | Yes |
Lemma 7. The excircle to side \( {AB} = c \) in a triangle \( {ABC} \) with sides \( a, b, c \) has\nthe radius\n\[{R}_{c} = \frac{a + b + c}{2}\tan \frac{C}{2}.\] | Proof. We use notations as in Figure 6, where \( u + v = c \) . Also, according to the two tangent theorem, \( b + u = a + v \) . Hence \( b + u = a + c - u \), so\n\[u = \frac{a - b + c}{2}\]\nand therefore\n\[b + u = \frac{a + b + c}{2}.\]\nFor the exradius we have\n\[\tan \frac{C}{2} = \frac{{R}_{c}}{b + u}\]\nsince... | Yes |
Theorem 8. Let the diagonals in a tangential quadrilateral ABCD intersect at \( P \) and let the exradii in triangles \( {ABP},{BCP},{CDP},{DAP} \) opposite the vertex \( P \) be \( {R}_{1},{R}_{2},{R}_{3},{R}_{4} \) respectively. Then the quadrilateral is a kite if and only if\n\n\[ \n{R}_{1} + {R}_{3} = {R}_{2} + {R}... | Proof. The four excircles and their radii are marked in Figure 7. Since \( \tan \frac{\pi - \theta }{2} = \) \( \cot \frac{\theta }{2} \), where \( \theta \) is the angle between the diagonals, Lemma 7 yields\n\n\[ \n{R}_{1} + {R}_{3} = {R}_{2} + {R}_{4} \n\]\n\n\[ \n\Leftrightarrow \;\frac{w + y + a}{2}\tan \frac{\the... | Yes |
Corollary 9. If \( {R}_{1},{R}_{2},{R}_{3},{R}_{4} \) are the same exradii as in Theorem 8, then the tangential quadrilateral is a kite if and only if\n\n\[ \n{R}_{1}{R}_{3} = {R}_{2}{R}_{4} \n\] | Proof. In a tangential quadrilateral we have according to Theorem 4 in [8]\n\n\[ \n\frac{1}{{R}_{1}} + \frac{1}{{R}_{3}} = \frac{1}{{R}_{2}} + \frac{1}{{R}_{4}} \n\]\n\nWe rewrite this as\n\n\[ \n\frac{{R}_{1} + {R}_{3}}{{R}_{1}{R}_{3}} = \frac{{R}_{2} + {R}_{4}}{{R}_{2}{R}_{4}}. \n\]\n\nHence\n\n\[ \n{R}_{1} + {R}_{3}... | Yes |
Theorem 1. If \( {ABC} \) is a triangle and \( P \) is a point, then\n\n\[ \n{R}_{1}^{3} + {R}_{2}^{3} + {R}_{3}^{3} + 6{R}_{1}{R}_{2}{R}_{3} \geq {72}{r}^{3}. \n\]\n\n(1) | This answers a conjecture raised by \( \mathrm{{Wu}} \), Zhang, and Chu at the end of [5] in the affirmative. Furthermore, there is equality in (1) if and only if triangle \( {ABC} \) is equilateral with \( P \) its center. By homogeneity, one may take \( r = 1 \) in (1), which we will assume. Modifying our proof, one ... | Yes |
Lemma 2. Suppose \( B \) and \( C \) are fixed points on the \( x \) -axis. Consider all possible triangles \( {ABC} \) having inradius 1. If \( A = \left( {x, y}\right) \), where \( y > 0 \), then \( y \) achieves its minimum value only when triangle \( {ABC} \) is isosceles. | Proof. Let \( P \) denote the point of tangency of the incircle of triangle \( {ABC} \) with side \( {BC} \) . Without loss of generality, we may let \( B = \left( {0,0}\right), P = \left( {a,0}\right) \) and \( C = \left( {a + b,0}\right) \), where \( a \) and \( b \) are positive numbers such that \( {ab} > 1 \) . It... | Yes |
Lemma 3. Fix base \( {BC} \) . Let triangle \( {A}^{\prime }{BC} \) have inradius 1, and \( {P}^{\prime } \) be a point on or within this triangle. Then\n\n\[ \n{f}_{\bigtriangleup {A}^{\prime }{BC}}^{{P}^{\prime }} \geq {f}_{\bigtriangleup {ABC}}^{P} \n\]\n\nfor an isosceles triangle \( {ABC} \) with inradius 1 and so... | Proof. Let \( {R}_{1}^{\prime } = {A}^{\prime }{P}^{\prime },{R}_{2}^{\prime } = B{P}^{\prime } \), and \( {R}_{3}^{\prime } = C{P}^{\prime } \) . Without loss of generality, we may assume \( {R}_{1}^{\prime } = \min \left\{ {{R}_{1}^{\prime },{R}_{2}^{\prime },{R}_{3}^{\prime }}\right\} \), for we may relabel the figu... | Yes |
Corollary 6. If \( \alpha ,\beta \), and \( \gamma \) denote the half-angles of a triangle \( {ABC} \), then\n\n\[{\csc }^{3}\alpha + {\csc }^{3}\beta + {\csc }^{3}\gamma + 6\csc \alpha \csc \beta \csc \gamma \geq {72}.\] | Inequality (4) may also be realized by observing\n\n\[{\csc }^{3}\alpha + {\csc }^{3}\beta + {\csc }^{3}\gamma + 6\csc \alpha \csc \beta \csc \gamma \geq 9\csc \alpha \csc \beta \csc \gamma = 9\left( \frac{4R}{r}\right) \geq {72},\]\n\nusing Euler’s inequality and the fact \( \frac{r}{4R} = \sin \alpha \sin \beta \sin ... | Yes |
Corollary 7. If triangle \( {ABC} \) has median lengths \( {m}_{a},{m}_{b} \), and \( {m}_{c} \), then\n\n\[ {m}_{a}^{3} + {m}_{b}^{3} + {m}_{c}^{3} + 6{m}_{a}{m}_{b}{m}_{c} \geq {243}{r}^{3}. \] | Using the proof above of Theorem 1, it is possible to find lower bounds for the general sum \( {R}_{1}^{\alpha } + {R}_{2}^{\alpha } + {R}_{3}^{\alpha } + k{\left( {R}_{1}{R}_{2}{R}_{3}\right) }^{\frac{\alpha }{3}} \), where \( \alpha \) and \( k \) are positive constants, in several particular instances. | No |
Theorem 8. If \( {ABC} \) is a triangle and \( P \) is a point, then\n\n\[ {R}_{1}^{3} + {R}_{2}^{3} + {R}_{3}^{3} + k{R}_{1}{R}_{2}{R}_{3} \geq 8\left( {k + 3}\right) {r}^{3} \]\n\nfor \( k = 0,1,\ldots ,6 \) . | Furthermore, it appears that inequality (6) would hold for all real numbers \( k \) in the interval \( \left\lbrack {0,6}\right\rbrack \), though we do not have a complete proof. However, for any given number in this interval, one could apply the steps in the proof of Lemma 4 to test whether or not (6) holds. | No |
Proposition 2. The second Lemoine circle is the orthogonal projection on the plane of \( {ABC} \) of a sphere \( T \) tritangent externally to \( {T}_{A},{T}_{B} \), and \( {T}_{C} \) . Furthermore,\n\n(1) the center \( {T}_{K} \) of \( T \) has a distance of \( 2{r}_{L} \) to the plane of \( {ABC} \) ;\n\n(2) the high... | Proof. Since \( {A}^{\prime } \) is given in barycentrics by \( \left( {-{a}^{2} : {b}^{2} : {c}^{2}}\right) \), we find with help of the distance formula (see for instance [3], where some helpful information on circles in barycentric coordinates is given as well):\n\n\[{\rho }_{a} = {d}_{{A}^{\prime }, B} = \frac{abc}... | Yes |
Lemma 1. Let \( X \) lie on the sideline \( {BC} \) of triangle \( {ABC} \) . Then, with respect to triangle \( {ABC}, X \) has barycentric coordinates \( \left( {0,\frac{XC}{a},\frac{BX}{a}}\right) \) . | Proof. Since \( X \) lies on line \( {BC} \) between \( B \) and \( C \), there is a unique scalar \( t \) such that \( X - B = t\left( {C - B}\right) \) . In fact, the length of the directed segment \( {BX} = t \cdot {BC} = \) ta, i.e., \( t = \frac{BX}{a} \) . Rearranging, \( X = {0A} + \left( {1 - t}\right) B + {tC}... | Yes |
Theorem 2. Let \( X \) have barycentric coordinates \( \left( {\alpha ,\beta ,\gamma }\right) \) with respect to triangle \( {ABC} \) . Then for any point \( Y \) , \[ X{Y}^{2} = {\alpha A}{Y}^{2} + {\beta B}{Y}^{2} + {\gamma C}{Y}^{2} - \left( {{\beta \gamma }{a}^{2} + {\gamma \alpha }{b}^{2} + {\alpha \beta }{c}^{2}}... | Proof. Using the well known identity \( {\left| V\right| }^{2} = V \cdot V \), we compute first that \[ X{Y}^{2} = {\left| Y - X\right| }^{2} \] \[ = {\left| Y - \alpha A - \beta B - \gamma C\right| }^{2} \] \[ = {\left| \alpha \left( Y - A\right) + \beta \left( Y - B\right) + \gamma \left( Y - C\right) \right| }^{2} \... | Yes |
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