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Lemma 3. The centroid \( G \) has barycentric coordinates \( \left( {\frac{1}{3},\frac{1}{3},\frac{1}{3}}\right) \) . | Proof. Let \( {G}^{\prime } \) be the point with barycentric coordinates \( \left( {\frac{1}{3},\frac{1}{3},\frac{1}{3}}\right) \), and we will prove \( G = {G}^{\prime } \) . By Lemma \( 1,{A}^{\prime } = \frac{1}{2}B + \frac{1}{2}C \) . We calculate\n\n\[ \frac{1}{3}A + \frac{2}{3}{A}^{\prime } = \frac{1}{3}A + \frac... | Yes |
Lemma 4 (Euler Line Theorem). \( H - O = 3\left( {G - O}\right) \) . | Proof. Let \( {H}^{\prime } = O + 3\left( {G - O}\right) \) and we will prove \( H = {H}^{\prime } \) . By Lemma 3,\n\n\[ \n{H}^{\prime } - O = 3\left( {G - O}\right) = A + B + C - {3O} = \left( {A - O}\right) + \left( {B - O}\right) + \left( {C - O}\right) .\n\]\n\nAnd so,\n\n\[ \n\left( {{H}^{\prime } - A}\right) \cd... | Yes |
Lemma 5. \( \left( {A - O}\right) \cdot \left( {B - O}\right) = {R}^{2} - \frac{1}{2}{c}^{2} \) . | Proof. One has\n\n\[ \n{c}^{2} = {\left| A - B\right| }^{2} \n\] \n\n\[ \n= {\left| \left( A - O\right) - \left( B - O\right) \right| }^{2} \n\] \n\n\[ \n= O{A}^{2} + O{B}^{2} - 2\left( {A - O}\right) \cdot \left( {B - O}\right) \n\] \n\n\[ \n= 2{R}^{2} - 2\left( {A - O}\right) \cdot \left( {B - O}\right) \text{.} \n\] | Yes |
Theorem 6. \( {4A}{N}^{2} = {R}^{2} - {a}^{2} + {b}^{2} + {c}^{2} \) . | Proof. Since \( N \) is the midpoint of \( {OH} \), we have \( H - O = 2\left( {N - O}\right) \) . Combining this observation with Theorem 2, and using Lemma 5, we obtain\n\n\[ \n{4A}{N}^{2} = {\left| 2\left( A - O\right) - 2\left( N - O\right) \right| }^{2} \n\]\n\n\[ \n= {\left| \left( A - O\right) - \left( B - O\rig... | Yes |
Theorem 7. The incenter I has barycentric coordinates \( \left( {\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}}\right) \) . | Proof. Let \( {I}^{\prime } \) be the point with barycentric coordinates \( \left( {a/{2s}, b/{2s}, c/{2s}}\right) \), and we will prove \( I = {I}^{\prime } \) . Let \( F \) be the foot of the bisector of angle \( A \) on side \( {BC} \) . Applying the law of sines to triangles \( {ABF} \) and \( {ACF} \), and using \... | Yes |
Theorem 8 (Euler). \( O{I}^{2} = {R}^{2} - {2Rr} \) . | Proof. We use \( X = I \) and \( Y = O \) in Theorem 2 to obtain\n\n\[ O{I}^{2} = \frac{a}{2s}{R}^{2} + \frac{b}{2s}{R}^{2} + \frac{c}{2s}{R}^{2} - \left( {\frac{bc}{{\left( 2s\right) }^{2}}{a}^{2} + \frac{ca}{{\left( 2s\right) }^{2}}{b}^{2} + \frac{ab}{{\left( 2s\right) }^{2}}{c}^{2}}\right) \]\n\n\[ = {R}^{2} - \frac... | Yes |
Theorem 9. \( {IN} = \frac{1}{2}R - r \) and \( {I}_{a}N = \frac{1}{2}R + {r}_{a} \) | Proof. To find the distance \( {IN} \), we set \( X = I \) and \( Y = N \) in Theorem 2, with Theorems 6 and 7 supplying the distances \( {AN},{BN},{CN} \), and the barycentric coordinates of \( I \) . For brevity in our computation, we use cyclic sums, in which the displayed term is transformed under the permutations ... | Yes |
Theorem 10 (Feuerbach, 1822). In a nonequilateral triangle, the nine-point circle is internally tangent to the incircle and externally tangent to the three excircles. | Proof. Suppose first that the nine-point circle and the incircle are nonconcentric. Two nonconcentric circles are internally tangent if and only if the distance between their centers is equal to the positive difference of their radii. Since the nine-point circle is the circumcircle of the medial triangle \( {A}^{\prime... | Yes |
Theorem 1. If \( a, b, c, d \) are the consecutive sides of a cyclic quadrilateral, then its diagonal point triangle has the area\n\n\[ T = \frac{2abcdK}{\left| {{a}^{2} - {c}^{2}}\right| \left| {{b}^{2} - {d}^{2}}\right| } \]\n\nif it's not an isosceles trapezoid, where\n\n\[ K = \sqrt{\left( {s - a}\right) \left( {s ... | Proof. In a cyclic quadrilateral \( {ABCD} \) opposite angles are supplementary, so \( \sin C = \sin A \) and \( \sin D = \sin B \) . From these we get that \( \sin A = \frac{2K}{{ad} + {bc}} \) and \( \sin B = \frac{2K}{{ab} + {cd}} \) by dividing the quadrilateral into two triangles by a diagonal in two different way... | Yes |
Theorem 2. A convex quadrilateral is a trapezoid if and only if the product of the areas of the triangles formed by one diagonal is equal to the product of the areas of the triangles formed by the other diagonal. | Proof. We use the same notations as in the proof of Theorem 1. The following statements are equivalent.\n\n\[ \n{T}_{1}{T}_{2} = {T}_{3}{T}_{4} \n\]\n\n\[ \n\frac{1}{2}{ad}\sin A \cdot \frac{1}{2}{bc}\sin C = \frac{1}{2}{ab}\sin B \cdot \frac{1}{2}{cd}\sin D, \n\]\n\n\[ \n\sin A\sin C = \sin B\sin D, \n\]\n\n\[ \n\frac... | Yes |
Lemma 3. Let \( \ell : {px} + {qy} + {rz} = 0 \) be a line through the centroid \( G \) . The conjugate diameter in the Steiner circum-ellipse is the line \( \left( {q - r}\right) x + \left( {r - p}\right) y + \left( {p - q}\right) z = 0 \) . | Proof. The parallel of \( \ell \) through \( A \) intersects the ellipse again at the point\n\n\[ N = \left( {\frac{1}{q - r} : \frac{1}{p - q} : \frac{1}{r - p}}\right) . \]\n\nThe midpoint of \( {AN} \) is the point\n\n\[ \left( {2\left( {r - p}\right) \left( {p - q}\right) - {\left( q - r\right) }^{2} : \left( {q - ... | Yes |
Proposition 5. Let \( P \) and \( Q \) be points on the Steiner circum-ellipse. The lines \( {\mathcal{L}}_{P} \) and \( {\mathcal{L}}_{Q} \) are conjugate diameters if and only if \( P \) and \( Q \) are antipodal. | Proof. Suppose \( {\mathcal{L}}_{P} \) has equation \( {px} + {qy} + {rz} = 0 \) with \( p + q + r = 0 \) . The line \( {\mathcal{L}}_{Q} \) is the conjugate diameter of \( {\mathcal{L}}_{P} \) if and only if its has equation \( \left( {q - r}\right) x + \left( {r - p}\right) y + \) \( \left( {p - q}\right) z = 0 \) . ... | Yes |
Theorem 1. Let \( P \) be a point in the plane of a quadrilateral \( {ABCD} \), and let \( E \) , \( F, G \), and \( H \) be the centroid of triangles \( {ABP},{BCP},{CDP},{DAP} \), respectively. Then EFGH is a parallelogram. In particular, EFGH is cyclic if and only if \( {AC} \) and \( {BD} \) are perpendicular. | Proof. The quadrilateral EFGH is the image of the Varignon parallelogram under the homothety \( \mathrm{h}\left( {P,\frac{2}{3}}\right) \), hence is a parallelogram. A parallelogram is cyclic exactly when it is rectangle. | Yes |
Theorem 2. For a point \( P \) in the plane of a rectangle \( {ABCD} \), denote by \( E, F \) , \( G \) , \( H \) the circumcenters of triangles \( {ABP},{BCP},{CDP},{DAP} \), respectively. The geometrical locus of points \( P \) for which EFGH is a cyclic quadrilateral consists of the circumcircle of the rectangle and... | Proof. We first look for \( P \) lying on a side of the rectangle. For the sake of definiteness, suppose that \( P \) belongs to the line \( {AB} \) and \( P \neq A, B \) . Then \( E \) is at infinity, \( F \) and \( H \) sit on the perpendicular bisector of side \( {BC} \), and \( G \) is on the perpendicular bisector... | Yes |
Theorem 3. In a plane endowed with a Cartesian coordinate system centered at \( O \) , consider a rectangle \( {ABCD} \) with sides parallel to the axes and of length \( \left| {AB}\right| = \) \( {2a},\left| {BC}\right| = {2b} \), and diagonals intersecting at \( O \) . For a point \( P \) in the plane, not on the sid... | Proof. To take advantage of computations already performed, we use the fact that the coordinates of the orthocenter are the sums of the coordinates of the vertices with respect to a coordinate system centered in the circumcenter. Having in view the previous proof and the restriction on \( P \), we readily find\n\n\[ E\... | Yes |
Theorem 4. For a point \( P \) in the interior of a square \( {ABCD} \), one denotes by \( E \) , \( F, G \), and \( H \) the incenter of triangle \( {ABP},{BCP},{CDP},{DAP} \), respectively. The geometrical locus of points \( P \) for which EFGH is a cyclic quadrilateral consists of the diagonals \( {AC} \) and \( {BD... | Full details of the proof for Theorem 4 are given in [1]. As the quest for elegance and simplicity is still highly regarded by many mathematicians, we would like to have a less computationally involved proof to Theorem 4. It is to be expected that a satisfactory solution to this problem will be not only more conceptual... | No |
Lemma 3. The distance from the incenter to the centroid is less than one third the length of the longest median of the triangle. | Proof. We have just shown that both the incenter and centroid lie inside the medial triangle. Therefore the distance from the incenter to the centroid is less than the largest distance from the centroid to a vertex of the medial triangle. (Consider the circle centered at \( O \) passing through the most distant vertex.... | Yes |
Lemma 4. The length of a median is less than \( \mu \) . Hence, \( \nu < \mu < s \) . | Proof. Consider the median from \( A \) . If we rotate the triangle through \( \pi \) about \( {M}_{A} \) , the mid-point of the side opposite \( A \), we obtain the parallelogram \( {ABDC} \) . The diagonal \( {AD} \) has twice the length \( A{M}_{A} \) . As \( A, B \) and \( D \) form a non-degenerate triangle we hav... | Yes |
Proposition 5. The distance, \( d \), from the incenter to the Euler line satisfies\n\n\[ \frac{d}{s} < \frac{d}{\mu } < \frac{d}{\nu } < \frac{1}{3} \]\n\nwhere \( \nu \) is the length of the longest median, \( \mu \) is the length of the longest side and \( s \) is the semi-perimeter of the triangle. | Proof. As the centroid lies on the Euler line, the distance from the incenter to the Euler line is at most the distance from the incenter to the centroid. By Lemma 3, this distance is one third the length of the longest median. But, by Lemma 4, the length of each median is less than \( \mu < s \), and the result follow... | Yes |
Proposition 1. If \( A\left( \rho \right) \) is a Steiner circle tangent to \( \left( O\right) \) at \( P \) and \( \left( I\right) \) at \( Q \), then the line \( {PQ} \) contains \( {T}_{ + } \), the internal center of similitude of \( \left( O\right) \) and \( \left( I\right) \) . | Proof. Note that \( A \) divides \( {OP} \) internally in the ratio \( {OA} : {AP} = R - \rho : \rho \), so that\n\n\[ A = \frac{\rho \cdot O + \left( {R - \rho }\right) P}{R}. \]\n\nSimilarly, the same point \( A \) divides \( {IQ} \) externally in the \( {IA} : {AQ} = r + \rho : - \rho \) , so that\n\n\[ A = \frac{-\... | Yes |
Proposition 2. If two neighboring Steiner circles are tangent to each other at \( T \) , then \( T \) lies on a circle with center \( {T}_{ + } \) . | Proof. Applying the law of cosines to triangles \( {PO}{T}_{ + } \) and \( {AOI} \), we have\n\n\[ \frac{{R}^{2} + {\left( \frac{Rd}{R + r}\right) }^{2} - {T}_{ + }{P}^{2}}{{2R} \cdot \frac{Rd}{R + r}} = \frac{{\left( R - \rho \right) }^{2} + {d}^{2} - {\left( r + \rho \right) }^{2}}{2\left( {R - \rho }\right) d}. \]\n... | Yes |
Proposition 5. Let \( A\\left( \\rho \\right) \) be a Steiner circle between \( \\left( O\\right) \) and \( \\left( I\\right) \) . The radii of its two neighbors are the roots of the quadratic polynomial \( a{\\sigma }^{2} + {b\\sigma } + c \), where\n\n\\[ \na = {\\left( \\left( q + 1\\right) Rr - \\left( R - r\\right... | Proof. Let \( B\\left( \\sigma \\right) \) be a neighbor of \( A\\left( \\rho \\right) \) . Apply Proposition 4 to triangle \( {IAB} \) with sides \( \\rho + \\sigma, r + \\sigma, r + \\rho \), and the point \( O \) whose distances from \( I, A, B \) are respectively \( d, R - \\rho, R - \\sigma \) .\n\n\\[ \n\\left| \... | Yes |
Lemma 6. \( {b}^{2} - {4ac} = {16}{\left( q + 1\right) }^{2}{R}^{3}{r}^{3}{\rho }^{2}\left( {{\rho }_{1} - \rho }\right) \left( {\rho - {\rho }_{0}}\right) \) . | Proof. \( {b}^{2} - {4ac} = 4{\left( q + 1\right) }^{2}{R}^{2}{r}^{2}{\rho }^{2} \cdot D \), where\n\n\[ D = {\left( \left( q - 1\right) Rr - \left( R - r\right) \rho \right) }^{2} - \left( {{\left( \left( q + 1\right) Rr - \left( R - r\right) \rho \right) }^{2} + {4Rr}{\rho }^{2}}\right) \]\n\n\[ = {4Rr}\left( {-{qRr}... | Yes |
Proposition 7. The radius \( \rho \) of a Steiner circle and those of its two neighbors are rational if and only if\n\n\[ \rho = \mathcal{R}\left( \tau \right) \mathrel{\text{:=}} \frac{{\tau }^{2}{\rho }_{0} + \operatorname{Rr}{\rho }_{1}}{{\tau }^{2} + \operatorname{Rr}} \]\n\n(3)\n\nfor some rational number \( \tau ... | Proof. The roots of the quadratic polynomial \( a{\sigma }^{2} + {b\sigma } + c \) are rational if and only if \( {b}^{2} - {4ac} \) is the square of a rational number. With \( a, b, c \) given in (2), this discriminant is given by Lemma 6. Writing \( \operatorname{Rr}\left( {{\rho }_{1} - \rho }\right) \left( {\rho - ... | Yes |
Proposition 9. The standard rational Steiner pairs are parametrized by\n\n\[ R = 1,\;r = \frac{t}{\left( {q + t}\right) \left( {q + 1 + t}\right) },\;d = \frac{q\left( {q + 1}\right) - {t}^{2}}{\left( {q + t}\right) \left( {q + 1 + t}\right) }.\n\] | Proof. Since \( \left( {R, r, d}\right) = \left( {1,0,1}\right) \) is a rational solution of\n\n\[ {d}^{2} = {\left( 1 - r\right) }^{2} - {4qr} \]\n\nevery rational solution is of the form \( d = 1 - \left( {{2q} + 1 + {2t}}\right) r \) for some rational number \( t \) . Direct substitution leads to\n\n\[ \left( {q + t... | Yes |
Proposition 10. In a standard rational Steiner pair \( {\left( R, r, d\right) }_{q} \), a Steiner circle \( A\left( \rho \right) \) and its neighbors have rational radii if and only if\n\n\[ \rho = \mathcal{R}\left( \tau \right) = \frac{t\left( {q + {\left( q + t\right) }^{2}{\tau }^{2}}\right) }{\left( {q + t}\right) ... | Proof. The neighbors of \( A\left( \rho \right) \) have rational radii if and only if \( a{\sigma }^{2} + {b\sigma } + c \) (with \( a \) , \( b, c \) given in (2)) has rational roots. Therefore, the two neighbors have rational radii if and only if \( \operatorname{Rr}\left( {{\rho }_{1} - \rho }\right) \left( {\rho - ... | Yes |
Proposition 11. The iterations of \( {\tau }_{ + } \) (respectively \( {\tau }_{ - } \) ) have periods 3,4,6 according as \( q = 3,1 \), or \( \frac{1}{3} \) . | Proof. The iterations of \( {\tau }_{ + } \) are as follows. \n\nFigure 8. 6-cycle for \( q = \frac{1}{3} \)\n\nThe iterations of \( {\tau }_{ - } \) simply reverse the orientations of these cycles. | Yes |
Rational Steiner 4-cycles with simple rational radii: | <table><thead><tr><th>\\( t \\)</th><th>\\( \\left( {R, r, d}\\right) \\)</th><th>\\( \\mathcal{T} \\)</th><th>Steiner cycle</th></tr></thead><tr><td>1</td><td>\\( \\left( {1,\\frac{1}{6},\\frac{1}{6}}\\right) \\)</td><td>1 2 \\( \\frac{1}{3} \\)</td><td>\\( \\begin{array}{l} \\left( {\\frac{10}{21},\\;\\frac{5}{11},\\... | Yes |
Example 3. Rational Steiner 6-cycles with simple rational radii: | <table><thead><tr><th>\\( t \\)</th><th>\\( \\left( {R, r, d}\\right) \\)</th><th>\\( \\mathcal{T} \\)</th><th>Steiner cycle</th></tr></thead><tr><td>\\( \\frac{1}{2} \\)</td><td>\\( \\left( {1,\\;\\frac{18}{55},\\;\\frac{7}{55}}\\right) \\)</td><td>2</td><td>\\( \\left( {\\frac{1}{3},\\;\\frac{24}{61},\\;\\frac{8}{21}... | Yes |
Proposition 12. Let \( \left( O\right) \) and \( \left( I\right) \) be the circles in \( {\mathcal{S}}_{q}\left( t\right) \). (a) \( \left( I\right) \) is in the interior of \( \left( O\right) \) if and only if \( t > 0 \). | Proof. (a) The circle \( \left( I\right) \) is contained in the interior of \( \left( O\right) \) if and only if \( d + r < R \) and \( d - r > - R \). This means \( t\left( {q + 1 + t}\right) > 0 \) and \( q + t > 0 \). Therefore, \( t > 0 \). | Yes |
Proposition 13. The inversive image of \( {\mathcal{S}}_{q}\left( t\right) \) in the circle \( \left( O\right) \) is \( {\mathcal{S}}_{q}\left( {-t}\right) \) . | Proof. Let \( {I}^{\prime }\left( {r}^{\prime }\right) \) be the inversive image of \( \left( I\right) \) in \( \left( O\right) \), with \( {d}^{\prime } = O{I}^{\prime } \) . \n\n\[ \n{r}^{\prime } = \frac{1}{2}\left( {\frac{{R}^{2}}{d + r} - \frac{{R}^{2}}{d - r}}\right) = \frac{-t}{\left( {q - t}\right) \left( {q + ... | Yes |
Proposition 1. The area of the regular hexagon ABCDEF is 3 times the area of triangle \( {AMN} \) if and only if \( M \) divides \( {BC} \) in the golden ratio. | Proof. Let \( P \) is midpoint of the minor \( {BC} \) . It is clear that \( {AOP} \) is an isosceles right triangle and \( {AP} = \sqrt{2} \cdot {AO} \) (see Figure 1). The area of the regular hexagon \( {ABCDEF} \) is three times that of triangle \( {AMN} \) if and only if\n\n\[
\frac{\Delta AMN}{\Delta ABO} = 2 \Lef... | Yes |
Proposition 2. The sides of the equilateral triangle \( {AMN} \) are divided in the golden ratio as follows. | <table><thead><tr><th>Directed segment</th><th>\( {MN} \)</th><th>NM</th><th>\( {AM} \)</th><th>\( {AN} \)</th><th>\( {NA} \)</th><th>MA</th></tr></thead><tr><td>divided by</td><td>\( {OD} \)</td><td>\( {OC} \)</td><td>\( {OB} \)</td><td>\( {OF} \)</td><td>\( {BF} \)</td><td>perp. from \( O \) to \( {AB} \)</td></tr><t... | No |
Proposition 1. The conic \( \mathcal{C}\left( P\right) \) through the six points is a circle if and only if\n\n\[ \frac{u}{v + w} : \frac{v}{w + u} : \frac{w}{u + v} = {a}^{2} : {b}^{2} : {c}^{2}. \] | Proof. Note that the lines \( {B}_{a}{C}_{a},{C}_{b}{A}_{b},{A}_{c}{B}_{c} \) are parallel to the sidelines of \( {ABC} \) . These three lines bound a triangle homothetic to \( {ABC} \) at the point\n\n\[ \left( {\frac{u}{v + w} : \frac{v}{w + u} : \frac{w}{u + v}}\right) . \]\n\nIt is known (see, for example, \( \left... | Yes |
Proposition 3. If \( {ABC} \) is a scalene triangle, there are three distinct real points \( P \) for which the conic \( \mathcal{C}\left( P\right) \) is a circle. | Proof. Writing\n\n\[ \n\frac{u}{v + w} = \frac{{a}^{2}}{t},\;\frac{v}{w + u} = \frac{{b}^{2}}{t},\;\frac{w}{u + v} = \frac{{c}^{2}}{t}, \n\]\n\n(3)\n\nwe have\n\n\[ \n- {tu} + {a}^{2}v + {a}^{2}w = 0, \n\]\n\n\[ \n{b}^{2}u - {tv} + {b}^{2}w = 0, \n\]\n\n\[ \n{c}^{2}u + {c}^{2}v - {tw} = 0. \n\]\n\nHence,\n\n\[ \n\left|... | Yes |
Theorem 4. For a scalene triangle \( {ABC} \) with \( \rho = \frac{2}{\sqrt{3}}\sqrt{{a}^{2}{b}^{2} + {b}^{2}{c}^{2} + {c}^{2}{a}^{2}} \) and \( {\theta }_{0} \mathrel{\text{:=}} \frac{1}{3}\arccos \frac{8{a}^{2}{b}^{2}{c}^{2}}{{\rho }^{3}} \), the three points for which the corresponding conics \( \mathcal{P} \) are c... | Proof. From (3) the coordinates of \( P \) are\n\n\[ \nu : v : w = \frac{{a}^{2}}{{a}^{2} + t} : \frac{{b}^{2}}{{b}^{2} + t} : \frac{{c}^{2}}{{c}^{2} + t}, \]\n\nwith \( t \) a real root of the cubic equation (4). Writing \( t = \rho \cos \theta \) we transform (4) into\n\n\[ \frac{1}{4}{\rho }^{3}\left( {4{\cos }^{3}\... | Yes |
Proposition 5. The conic \( \mathcal{C}\left( P\right) \) is a circle if and only if \( P \) is an intersection, apart from the centroid \( G \), of\n\n(i) the rectangular hyperbola through \( G \) and the incenter \( I \) and their anticevian triangles,\n\n(ii) the circum-hyperbola through \( G \) and the symmedian po... | Proof. From (2), we have\n\n\[ f \mathrel{\text{:=}} {c}^{2}v\left( {u + v}\right) - {b}^{2}w\left( {w + u}\right) = 0, \]\n\n(5)\n\n\[ g \mathrel{\text{:=}} {a}^{2}w\left( {v + w}\right) - {c}^{2}u\left( {u + v}\right) = 0, \]\n\n(6)\n\n\[ h \mathrel{\text{:=}} {b}^{2}u\left( {w + u}\right) - {a}^{2}v\left( {v + w}\ri... | Yes |
Proposition 6. The three real points \( P \) for which \( \mathcal{C}\left( P\right) \) is a Tucker circle lie on a circle containing the following triangle centers: (i) the Euler reflection point\n\n\[ \n{X}_{110} = \left( {\frac{{a}^{2}}{{b}^{2} - {c}^{2}} : \frac{{b}^{2}}{{c}^{2} - {a}^{2}} : \frac{{c}^{2}}{{a}^{2} ... | Proof. The combination\n\n\[ \n{a}^{2}\left( {{c}^{2} - {a}^{2}}\right) \left( {{a}^{2} - {b}^{2}}\right) f + {b}^{2}\left( {{a}^{2} - {b}^{2}}\right) \left( {{b}^{2} - {c}^{2}}\right) g + {c}^{2}\left( {{b}^{2} - {c}^{2}}\right) \left( {{c}^{2} - {a}^{2}}\right) h\n\]\n\n(8)\n\nof (5)-(7) (with \( x, y, z \) replacing... | Yes |
Theorem 1. The segments \( {B}_{c}{C}_{b},{C}_{a}{A}_{c},{A}_{b}{B}_{a} \), and \( {OP} \) share a common midpoint. | Proof. Consider the lines perpendicular to \( {AP},{BP},{CP} \) at \( A, B, C \) respectively. These lines bound the antipedal triangle \( {A}_{ * }{B}_{ * }{C}_{ * } \) of \( P \) . If we draw the corresponding lines at \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) perpendicular to \( {A}^{\prime }P,{B}^{\prime }P,{... | Yes |
Lemma 2. The radii of the circles \( {PB}{C}^{\prime },{PC}{B}^{\prime },{PC}{A}^{\prime },{PA}{C}^{\prime },{PA}{B}^{\prime },{PB}{A}^{\prime } \) are | \[ {r}_{bc} = \frac{R}{a} \cdot {BP},\;{r}_{cb} = \frac{R}{a} \cdot {CP} \] \[ {r}_{ca} = \frac{R}{b} \cdot {CP},\;{r}_{ac} = \frac{R}{b} \cdot {AP} \] \[ {r}_{ab} = \frac{R}{c} \cdot {AP},\;{r}_{ba} = \frac{R}{c} \cdot {BP}. \] Proof. It is enough to establish the expression for \( {r}_{bc} \) . The others follow simi... | Yes |
Theorem 3. Let \( {A}_{1},{B}_{1},{C}_{1} \) be the midpoints of \( {BC},{CA},{AB} \) respectively. The six circumcenters lie on a circle if and only if\n\n\[ \n{A}_{1}P : {B}_{1}P : {C}_{1}P = {B}_{1}{C}_{1} : {C}_{1}{A}_{1} : {A}_{1}{B}_{1}. \n\] | Proof. Let \( M \) be the common midpoint of \( {OP},{B}_{c}{C}_{b},{C}_{a}{A}_{c},{A}_{b}{B}_{a} \) . Clearly the conic through the six circumcenter is a circle if and only if \( {B}_{c}{C}_{b} = {C}_{a}{A}_{c} = {A}_{b}{B}_{a} \) . Applying Apollonius’ theorem to the triangles \( P{B}_{c}{C}_{b} \) and \( {PBC} \), m... | Yes |
Theorem 5. The two points satisfying (1) are\n\n\[ \n{P}_{\varepsilon } \mathrel{\text{:=}} \left( {\sqrt{{S}_{A} + {S}_{B} + {S}_{C}} \cdot {S}_{BC} + {\varepsilon S} \cdot \sqrt{{S}_{ABC}} : \cdots : \cdots }\right) ,\;\varepsilon = \pm 1 \]\n\n in homogeneous barycentric coordinates. | Proof. Let \( P = \left( {{S}_{BC} + t : {S}_{CA} + t : {S}_{AB} + t}\right) \) . We have\n\n\[ 0 = \left( \begin{array}{lll} {S}_{BC} + t & {S}_{CA} + t & {S}_{AB} + t \end{array}\right) {M}_{a}\left( \begin{array}{l} {S}_{BC} + t \\ {S}_{CA} + t \\ {S}_{AB} + t \end{array}\right) \]\n\n\[ = \left( \begin{array}{lll} ... | Yes |
Proposition 6. The midpoint of the segment \( {P}_{ + }{P}_{ - } \) is the point\n\n\[ Q = \left( {{S}_{BC}\left( {{S}_{B} + {S}_{C} - 2{S}_{A}}\right) : {S}_{CA}\left( {{S}_{C} + {S}_{A} - 2{S}_{B}}\right) : {S}_{AB}\left( {{S}_{A} + {S}_{B} - 2{S}_{C}}\right) }\right) . | Proof. The midpoint between the two points \( \left( {{S}_{BC} + t : {S}_{CA} + t : {S}_{AB} + t}\right) \) and\n\n\( \left( {{S}_{BC} - t : {S}_{CA} - t : {S}_{AB} - t}\right) \) has coordinates\n\n\[ \left( {{S}_{BC} + {S}_{CA} + {S}_{AB} - {3t}}\right) \left( {{S}_{BC} + t,{S}_{CA} + t,{S}_{AB} + t}\right) )\]\n\n\[... | Yes |
Theorem 2. The circumconics \( {\mathcal{C}}_{\alpha \rightarrow },{\mathcal{C}}_{\beta \rightarrow },{\mathcal{C}}_{\gamma \rightarrow } \) (red in Figure 2) have the common point\n\n\[ \n{W}_{ \rightarrow } = \left( {\frac{1}{{p}_{ \rightarrow }} : \frac{1}{{q}_{ \rightarrow }} : \frac{1}{{r}_{ \rightarrow }}}\right)... | Hence, their perspectors are collinear on the tripolars of \( {W}_{ \rightarrow } \) and of \( {W}_{ \leftarrow } \), respectively. These lines are parallel and they intersect the infinite line at the point\n\n\[ \n{W}_{\infty } = \left( {{q}_{ \rightarrow } - {r}_{ \rightarrow } : {r}_{ \rightarrow } - {p}_{ \rightarr... | Yes |
Theorem 4. (a) The circumconics \( {\mathcal{C}}_{{W}_{ \rightarrow }^{ \bullet }} \) and \( {\mathcal{C}}_{{W}_{ \leftarrow }^{ \bullet }} \) intersect at the point \( {S}_{\delta } \mathrel{\text{:=}} {W}_{\infty }^{ \bullet } \) on the Steiner circumellipse. | Proof. (a) The conjugates of the circumconics \( {\mathcal{C}}_{{W}^{ \bullet }} \), and \( {\mathcal{C}}_{{W}^{ \bullet }} \), that are the lines \( \left\lbrack {{p}_{ \rightarrow } : {q}_{ \rightarrow } : {r}_{ \rightarrow }}\right\rbrack \) and \( \left\lbrack {{p}_{ \leftarrow } : {q}_{ \leftarrow } : {r}_{ \lefta... | No |
Theorem 7. The vertices of \( {\Delta }_{ \rightarrow } \) and \( {\Delta }_{ \leftarrow } \) lie on the \( {W}_{ \rightarrow } \) - circumconic and on the \( {W}_{ \leftarrow } \) - circumconic, respectively. | Proof. Easy verification. | No |
Theorem 9. The triangles \( \Delta ,\Delta {S}_{ \rightarrow } \) and \( \Delta {S}_{ \leftarrow } \) are pairwise triply perspective. The 9 centers of perspective lie on the infinite line, and the 9 axes of perspective pass through \( G \) . | We omit the elementary but long computational proof. Figure 5 illustrates the triple perspectivity of \( \Delta \) and \( \Delta {S}_{ \rightarrow } . | No |
Theorem 1. A convex quadrilateral ABCD is orthodiagonal if and only if\n\n\[ A{B}^{2} + C{D}^{2} = B{C}^{2} + D{A}^{2}. \]\n\n\n\nFigure 1. Normals to diagonal \( {AC} \) | Proof. Let \( X \) and \( Y \) be the feet of the normals from \( D \) and \( B \) respectively to diagonal \( {AC} \) in a convex quadrilateral \( {ABCD} \), see Figure 1. By the Pythagorean theorem we have \( B{Y}^{2} + A{Y}^{2} = A{B}^{2}, B{Y}^{2} + C{Y}^{2} = B{C}^{2}, D{X}^{2} + C{X}^{2} = C{D}^{2} \n\n--- \n\nan... | Yes |
Theorem 2. A convex quadrilateral ABCD is orthodiagonal if and only if\n\n\[ \angle {PAB} + \angle {PBA} + \angle {PCD} + \angle {PDC} = \pi \]\n\nwhere \( P \) is the point where the diagonals intersect. | Proof. By the sum of angles in triangles \( {ABP} \) and \( {CDP} \) (see Figure 3) we have\n\n\[ \angle {PAB} + \angle {PBA} + \angle {PCD} + \angle {PDC} = {2\pi } - {2\theta },\]\n\nwhere \( \theta \) is the angle between the diagonals. Hence \( \theta = \frac{\pi }{2} \) if and only if the equation in the theorem i... | Yes |
Theorem 3. A convex quadrilateral is orthodiagonal if and only if the projections of the diagonal intersection onto the sides are the vertices of a cyclic quadrilateral. | Proof. If the diagonals intersect in \( P \), and the projection points on \( {AB},{BC},{CD} \) and \( {DA} \) are \( K, L, M \) and \( N \) respectively, then \( {AKPN},{BLPK},{CMPL} \) and \( {DNPM} \) are cyclic quadrilaterals since they all have two opposite right angles (see Figure 3). Then \( \angle {PAN} = \angl... | Yes |
Theorem 4. In a convex quadrilateral ABCD where the diagonals intersect at \( P \) , let \( K, L, M \) and \( N \) be the projections of \( P \) onto the sides, and let \( R, S, T \) and \( U \) be the points where the lines \( {KP} \) , \( {LP} \) , \( {MP} \) and \( {NP} \) intersect the opposite sides. Then the quad... | Proof. \( \left( \Rightarrow \right) \) If \( {ABCD} \) is orthodiagonal, then \( K, L, M \) and \( N \) are concyclic by Theorem 3. We start by proving that \( {KTMN} \) has the same circumcircle as \( {KLMN} \) . To do this, we will prove that \( \angle {MNK} + \angle {MTK} = \pi \), which is equivalent to proving th... | Yes |
Theorem 5. If the normals to the sides of a convex quadrilateral ABCD through the diagonal intersection intersect the opposite sides in \( R, S, T \) and \( U \), then \( {ABCD} \) is orthodiagonal if and only if \( {RSTU} \) is a rectangle whose sides are parallel to the diagonals of \( {ABCD} \) . | Proof. \( \left( \Rightarrow \right) \) If \( {ABCD} \) is orthodiagonal, then \( {UTMN} \) is a cyclic quadrilateral according to Theorem 4 (see Figure 5). Thus\n\n\[ \angle {MTU} = \angle {DNM} = \angle {MPD} = \angle {TCP}, \]\n\nso \( {UT}\parallel {AC} \) . In the same way it can be proved that \( {RS}\parallel {A... | Yes |
Theorem 6. The first and second eight point circle of an orthodiagonal quadrilateral coincide if and only if the quadrilateral is also cyclic. | Proof. Since the second eight point circle is constructed from line segments through the diagonal intersection, the two eight point circles coincide if and only if the four maltitudes are concurrent at the diagonal intersection. The maltitudes of a convex quadrilateral are concurrent if and only if the quadrilateral is... | Yes |
Theorem 7. In a convex quadrilateral we have the following conditions: (i) The bimedians are congruent if and only if the diagonals are perpendicular. (ii) The bimedians are perpendicular if and only if the diagonals are congruent. | Proof. (i) According to the proof of Theorem 7 in [9], the bimedians \( m \) and \( n \) in a convex quadrilateral satisfy\n\n\[4\left( {{m}^{2} - {n}^{2}}\right) = - 2\left( {{a}^{2} - {b}^{2} + {c}^{2} - {d}^{2}}\right)\]\n\nwhere \( a, b, c \) and \( d \) are the sides of the quadrilateral. Hence\n\n\[m = n\; \Leftr... | Yes |
Theorem 8. A convex quadrilateral ABCD is orthodiagonal if and only if\n\n\[ \n{m}_{1}^{2} + {m}_{3}^{2} = {m}_{2}^{2} + {m}_{4}^{2} \]\n\nwhere \( {m}_{1},{m}_{2},{m}_{3} \) and \( {m}_{4} \) are the medians in the triangles \( {ABP},{BCP},{CDP} \) and \( {DAP} \) from the intersection \( P \) of the diagonals to the ... | Proof. Let \( P \) divide the diagonals in parts \( w, x \) and \( y, z \) (see Figure 8). By applying Apollonius' theorem in triangles \( {ABP},{CDP},{BCP} \) and \( {DAP} \) we get\n\n\[ \n{m}_{1}^{2} + {m}_{3}^{2} = {m}_{2}^{2} + {m}_{4}^{2} \]\n\n\( \Leftrightarrow 4{m}_{1}^{2} + 4{m}_{3}^{2} = 4{m}_{2}^{2} + 4{m}_... | Yes |
Theorem 9. A convex quadrilateral ABCD is orthodiagonal if and only if\n\n\[ \n{R}_{1}^{2} + {R}_{3}^{2} = {R}_{2}^{2} + {R}_{4}^{2} \]\n\nwhere \( {R}_{1},{R}_{2},{R}_{3} \) and \( {R}_{4} \) are the circumradii in the triangles \( {ABP},{BCP},{CDP} \) and DAP respectively and \( P \) is the intersection of the diagon... | Proof. According to the extended law of sines applied in the four subtriangles, \( a = 2{R}_{1}\sin \theta, b = 2{R}_{2}\sin \left( {\pi - \theta }\right), c = 2{R}_{3}\sin \theta \) and \( d = 2{R}_{4}\sin \left( {\pi - \theta }\right) \), see Figure 9. We get\n\n\[ \n{a}^{2} + {c}^{2} - {b}^{2} - {d}^{2} = 4{\sin }^{... | Yes |
Theorem 10. A convex quadrilateral ABCD is orthodiagonal if and only if the circumcenters of the triangles \( {ABP},{BCP},{CDP} \) and \( {DAP} \) are the midpoints of the sides of the quadrilateral, where \( P \) is the intersection of its diagonals. | Proof. The quadrilateral \( {ABCD} \) is orthodiagonal if and only if one of the triangles \( {ABP},{BCP},{CDP} \) and \( {DAP} \) have a right angle at \( P \) ; then all of them have it. Hence we only need to prove that the circumcenter of one triangle is the midpoint of a side if and only if the opposite angle is a ... | Yes |
Theorem 11. A convex quadrilateral ABCD is orthodiagonal if and only if\n\n\[ \n\frac{1}{{h}_{1}^{2}} + \frac{1}{{h}_{3}^{2}} = \frac{1}{{h}_{2}^{2}} + \frac{1}{{h}_{4}^{2}} \n\]\n\nwhere \( {h}_{1},{h}_{2},{h}_{3} \) and \( {h}_{4} \) are the altitudes in the triangles \( {ABP},{BCP},{CDP} \) and \( {DAP} \) from the ... | Proof. Let \( P \) divide the diagonals in parts \( w, x \) and \( y, z \) . From expressing twice the area of triangle \( {ABP} \) in two different ways we get (see Figure 10)\n\n\[ \na{h}_{1} = {wy}\sin \theta \n\]\n\nwhere \( \theta \) is the angle between the diagonals. Thus\n\n\[ \n\frac{1}{{h}_{1}^{2}} = \frac{{a... | Yes |
Theorem 1. If the longer base of an isosceles trapezoid is greater than the sum of the two isosceles sides, then there exists a point on the longer base of the trapezoid which when joined to the endpoints of the shorter base divides the trapezoid into three similar triangles. | Proof. To begin our construction we consider isosceles trapezoid \( {ABCD} \) with longer base \( {AD} \) and congruent sides \( {AB} \) and \( {CD} \) as shown in Figure 1. Additionally we let \( x = {AB} = {CD}, b = {BC}, e = {AD}, y = {BE} \), and \( z = {CE} \) . We propose that the point \( E \) can be located on ... | Yes |
Theorem 2. Using the notation introduced above we have the following equalities: (i) \( {y}^{2} = {ab},{x}^{2} = {ac},{z}^{2} = {bc} \) ;\n\n(ii) \( a = \frac{xy}{z}, b = \frac{yz}{x}, c = \frac{xz}{y} \) ;\n\n(iii) \( {xyz} = {abc} \), and\n\n(iv) the area of \( {ABCD} = \frac{1}{2}h\left( {a + b + c}\right) \) . | Proof. The first three follow immediately from the similar dissecting triangles, and (iv) follows directly from the formula for the area of a trapezoid. | No |
Theorem 3. Using the notation introduced above, the length of a diagonal, d, is given by\n\n\[ d = \sqrt{{ac} + {ab} + {bc}} = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}. \] | Proof. By the law of cosines for triangles \( {ABC} \) and \( {CDA} \), respectively, in Figure 1 , we have\n\n\[ {d}^{2} = A{C}^{2} = {x}^{2} + {b}^{2} - {2xb}\cos {ABC} \]\n\n\[ = {x}^{2} + {\left( a + c\right) }^{2} - {2x}\left( {a + c}\right) \cos \left( {{180}^{ \circ } - {ABC}}\right) \]\n\n\[ = {x}^{2} + {\left(... | Yes |
Theorem 4 (Generalization of the Pythagorean Theorem). Using the notation introduced above, \( {y}^{2} + {z}^{2} = b\left( {a + c}\right) \) . | Proof. Since the triangles are similar, the angles BEC, BAE and CDE are congruent. By Theorem 2(i), \[ {y}^{2} + {z}^{2} = {ab} + {bc} = b\left( {a + c}\right) = {b}^{2}, \] where the last equality holds whenever \( \angle {BAE} = {90}^{ \circ } \) . | No |
Theorem 5. Using the notation already introduced,\n\n(i) Triangle \( {G}_{a}{G}_{b}{G}_{c} \) is isosceles with \( {G}_{a}{G}_{b} = {G}_{c}{G}_{b} = \frac{1}{3}\sqrt{{ab} + {bc} + {ca}} \) ,\n\n(ii) the base of \( {G}_{a}{G}_{c} \) of triangle \( {G}_{a}{G}_{b}{G}_{c} \) is parallel to \( {AD} \) and its length is\n\n\... | Proof. We consider triangle \( {G}_{a}{G}_{b}{G}_{c} \) whose vertices are the respective centroids \( {G}_{a},{G}_{b} \), and \( {G}_{c} \) of triangles \( {BAE},{CEB} \) and \( {DEC} \) . Let \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \), and \( {D}^{\prime } \) be the respective midpoints of \( {AE},{BE},{CE} \), ... | Yes |
Corollary 7. If the dissecting triangles are right triangles, then\n\n(i) \( c = a + b \), and\n\n(ii) the area of triangle \( {O}_{a}{O}_{b}{O}_{c} \) is one-eighth the area of trapezoid \( {ABCD} \) . | Proof. For a right triangle the circumcenter is the midpoint of the hypotenuse of the right triangle. Therefore, \( {c}^{2} = {x}^{2} + {z}^{2} \) in triangle \( {BEC} \) in Figure 5. Substituting \( {x}^{2} = {ac} \) and \( {z}^{2} = {bc} \) yields \( {c}^{2} = {ac} + {bc} \) . From this the first result follows. Note... | Yes |
Theorem 8. If the isosceles trapezoid is literally cut apart, then the similar triangles can be rearranged to form two additional isosceles trapezoids which meet the same dissection criteria, have the same area, and have the same diagonal lengths as the original trapezoid. | Proof. With the trapezoid cut apart and reassembled we get the three cases shown in Figure 6 below. The triangles are numbered #1, #2, and #3 for clarity.\n\n\n\nFigure 6(i) Original trapezoid with similar triangle... | Yes |
Lemma 3. Let \( {ABC} \) be a triangle inscribed in a circle \( \left( O\right) \), and \( \mathcal{L} \) an arbitrary line. Let the parallels of \( \mathcal{L} \) through \( A, B, C \) intersect the circle at \( D, E, F \) respectively. The lines \( {\mathcal{L}}_{a},{\mathcal{L}}_{b},{\mathcal{L}}_{c} \) are the perp... | Proof. Let \( S \) be the intersection of \( {\mathcal{L}}_{a} \) and \( \left( O\right) \) . Let \( \ell \) be the line through \( O \) perpendicular to \( \mathcal{L} \) (see Figure 1).\n\n(a) Because \( A, B \), and \( C \) are the images of \( D, E \), and \( F \) through the reflections with axis \( \mathcal{L} \)... | Yes |
Lemma 5. Let \( {ABC} \) be a triangle inscribed in \( \left( O\right) .{A}_{1},{B}_{1},{C}_{1} \) are the images of \( A, B, C \) respectively through the symmetry with center \( O.{A}_{2},{B}_{2},{C}_{2} \) are the images of \( O \) through the reflections with axes \( {BC},{CA},{AB} \) respectively. \( {A}_{3},{B}_{... | Proof. (a) Let \( H \) be the orthocenter of \( {ABC} \) . Take the points \( D, S \) belonging to \( \left( O\right) \) such that \( {AD}//{OH} \) and \( {DS} \bot {BC} \) (see Figure 3).\n\nAccording to Lemma 3, the Steiner line of \( S \) with respect to \( {ABC} \) is parallel to \( {AD} \) . On the other hand, the... | Yes |
Lemma 6. If any of the three points in \( A, B, C, D \) are not collinear, then the nine-point circles of triangles \( {BCD},{CDA},{DAB},{ABC} \) all pass through one point. | Lemma 6 is familiar and its simple proof can be found in [2, p.242]. | No |
Theorem 1. The locus described by the common point of the diagonals of a \( v \) - parallelogram \( \mathbf{V} \) of \( \mathbf{Q} \) by varying \( \mathbf{V} \) is the third bimedian of \( \mathbf{Q} \) . | Proof. Let \( \mathbf{V} \) be any v-parallelogram of \( \mathbf{Q} \) and let \( {N}_{1}{N}_{2}{N}_{3}{N}_{4} \) be the Varignon parallelogram of \( \mathbf{V} \), with midpoint \( {N}_{i} \) of \( {V}_{i}{V}_{i + 1} \) (see Figure 2).\n\n is a \( v \) -parallelogram of \( \\mathbf{Q} \) and \( {G}^{\\prime } \) is the common point of the diagonals of \( \\mathbf{V} \), in the symmetry with center \( {G}^{\\prime } \) the valtitudes relative to \( \\mathbf{V} \) correspond with the vaxes relative to \( \\mathbf{V} \) . | Proof. In fact, \( {V}_{i} \) and \( {V}_{i + 2} \) are symmetric with respect to \( {G}^{\\prime } \) (see Figure 6). Then the vaxis \( {k}_{i} \) and the line parallel to it passing through \( {V}_{i + 2} \), i.e., the valtitude \( {h}_{i + 2} \) , are correspondent in the symmetry with center \( {G}^{\\prime } \) . | Yes |
Theorem 3. If \( \mathbf{V} \) is a \( v \) -parallelogram of \( \mathbf{Q} \) and \( {G}^{\prime } \) is the common point of the diagonals of \( \mathbf{V} \), the quadrilateral of the vaxes and the quadrilateral of the valtitudes are symmetric with respect to \( {G}^{\prime } \). | Proof. In fact, the valtitude \( {h}_{i + 2} \) is the correspondent of the vaxis \( {k}_{i} \) in the symmetry with center \( {G}^{\prime } \), and the point \( {B}_{i + 2} \) is the correspondent of the point \( {C}_{i} \) . | Yes |
Lemma 5. If \( \mathbf{Q} \) is orthodiagonal, the triangles \( {A}_{i}{A}_{i + 1}K \) and \( {C}_{i}{C}_{i + 3}{K}^{\prime },(i = \) \( 1,2,3,4 \) ) are similar. | Proof. Since \( \mathbf{Q} \) is orthodiagonal, the vertices \( {B}_{i} \) of \( {\mathbf{Q}}_{h} \) lie on the diagonals of \( \mathbf{Q} \) [6]. The diagonals of \( {\mathbf{Q}}_{h} \) and those of \( \mathbf{Q} \) lie on the same lines (see Figure 9). It follows that \( {\mathbf{Q}}_{h} \) is orthodiagonal. Then, by... | Yes |
Theorem 6. If \( \mathbf{Q} \) is orthodiagonal and \( \mathbf{V} \) is a \( v \) -parallelogram of \( \mathbf{Q} \), the quadrilateral of the vaxes and the quadrilateral of the valtitudes are affine to \( \mathbf{Q} \) . | Proof. From Lemma 5, we have\n\n\[ \frac{{A}_{1}K}{{A}_{2}K} = \frac{{C}_{1}{K}^{\prime }}{{C}_{4}{K}^{\prime }} \]\n\n(4)\n\n\[ \frac{{A}_{2}K}{{A}_{3}K} = \frac{{C}_{2}{K}^{\prime }}{{C}_{1}{K}^{\prime }} \]\n\n(5)\n\n\[ \frac{{A}_{3}K}{{A}_{4}K} = \frac{{C}_{3}{K}^{\prime }}{{C}_{2}{K}^{\prime }} \]\n\n(6)\n\nBy mul... | Yes |
Lemma 7. If \( \mathbf{Q} \) is cyclic, the angles of \( {\mathbf{Q}}_{k} \) are equal to those of \( \mathbf{Q} \) . Precisely, \( \angle {C}_{i}{C}_{i + 1}{C}_{i + 2} = \angle {A}_{i - 1}{A}_{i}{A}_{i + 1}\left( {i = 1,2,3,4}\right) \) . | Proof. Let us prove that \( \angle {C}_{1}{C}_{2}{C}_{3} = \angle {A}_{4}{A}_{1}{A}_{2} \) (see Figure 10). The other cases can be established analogously. Since \( \mathbf{Q} \) is cyclic, \( \angle {A}_{4}{A}_{1}{A}_{2} \) and \( \angle {A}_{2}{A}_{3}{A}_{4} \) are supplementary angles. Moreover, the angles at \( {V}... | Yes |
Theorem 8. If \( \mathbf{Q} \) is cyclic, then the quadrilateral of the vaxes and the quadrilateral of the valtitudes are cyclic. | Proof. Since \( \mathbf{Q} \) is cyclic, \( \angle {A}_{4}{A}_{1}{A}_{2} \) and \( \angle {A}_{2}{A}_{3}{A}_{4} \) are supplementary angles. Therefore, from Lemma \( 7,\angle {C}_{1}{C}_{2}{C}_{3} \) and \( \angle {C}_{1}{C}_{4}{C}_{2} \) are supplementary angles. Then, \( {\mathbf{Q}}_{k} \) is cyclic and, from Theore... | Yes |
Theorem 9. If \( \mathbf{Q} \) is cyclic and orthodiagonal and \( \mathbf{V} \) is a \( v \) -parallelogram of \( \mathbf{Q} \) , the quadrilateral of the vaxes and the quadrilateral of the valtitudes are similar to Q. | Proof. From Lemma 7, \( \mathbf{Q} \) and \( {\mathbf{Q}}_{k} \) have equal angles. Let us prove now that the sides of \( \mathbf{Q} \) are proportional to those of \( {\mathbf{Q}}_{k} \) . Consider the triangles \( {A}_{1}{A}_{2}{A}_{3} \) and \( {C}_{2}{C}_{3}{C}_{4} \) (see Figure 11). From Lemma 5 the triangles \( ... | Yes |
Lemma 10. If \( \mathbf{V} \) is a v-parallelogram of \( \mathbf{Q} \) and \( {M}_{i} \) is the midpoint of the side \( {A}_{i}{A}_{i + 1} \) of \( \mathbf{Q}\left( {i = 1,2,3,4}\right) \), then\n\n\[ \frac{{A}_{1}{V}_{1}}{{A}_{1}{M}_{1}} = \frac{{A}_{1}{V}_{4}}{{A}_{1}{M}_{4}} = \frac{{A}_{3}{V}_{2}}{{A}_{3}{M}_{2}} =... | Proof. In fact, since the triangles \( {A}_{1}{V}_{1}{V}_{4} \) and \( {A}_{1}{M}_{1}{M}_{4} \) are similar, as are triangles \( {A}_{3}{V}_{2}{V}_{3} \) and \( {A}_{3}{M}_{2}{M}_{3} \) (see Figure 12), we have\n\n\[ \frac{{A}_{1}{V}_{1}}{{A}_{1}{M}_{1}} = \frac{{A}_{1}{V}_{4}}{{A}_{1}{M}_{4}} = \frac{{V}_{1}{V}_{4}}{{... | Yes |
Theorem 11. If \( \mathbf{Q} \) is cyclic, the diagonals of the quadrilateral of the vaxes and those of the quadrilateral of the valtitudes are parallel to the diagonals of \( \mathbf{Q} \) . | Proof. Let \( O \) be the circumcenter of \( \mathbf{Q} \) (see Figure 13). Let \( {C}_{4}^{\prime } \) and \( {C}_{4}^{\prime \prime } \) be the common points of the line \( {A}_{1}O \) with the vaxes \( {k}_{1} \) and \( {k}_{4} \) respectively. Since the triangles \( {A}_{1}{V}_{1}{C}_{4}^{\prime } \) and \( {A}_{1}... | Yes |
Theorem 12. If \( \mathbf{Q} \) is cyclic and \( \mathbf{V} \) is a \( v \) -parallelogram of \( \mathbf{Q} \), the quadrilateral of the vaxes relative to \( \mathbf{V} \) has the same circumcenter of \( \mathbf{Q} \) . | Proof. From Theorem \( 8,{\mathbf{Q}}_{k} \) is cyclic. The axes of segments \( {C}_{2}{C}_{4} \) and \( {C}_{1}{C}_{3} \) meet at the circumcenter of \( {\mathbf{Q}}_{k} \) . The triangles \( O{C}_{2}{C}_{4} \) and \( O{A}_{1}{A}_{3} \) are correspondent in a homothetic transformation with center the circumcenter \( O... | Yes |
Theorem 13. If \( \mathbf{Q} \) is cyclic, all the quadrilaterals of the vaxes of \( \mathbf{Q} \) have the same Euler line. | Proof. Consider two v-parallelograms \( \mathbf{V} \) and \( {\mathbf{V}}^{\prime } \) and their quadrilaterals of the vaxes \( {\mathbf{Q}}_{k} \) and \( {\mathbf{Q}}_{k}^{\prime } \) respectively (see Figure 15). The vertices \( {C}_{i} \) and \( {C}_{i}^{\prime } \) of \( {\mathbf{Q}}_{k} \) and \( {\mathbf{Q}}_{k}^... | Yes |
Theorem 14. If \( \mathbf{Q} \) is cyclic and \( \mathbf{V} \) is a \( v \) -parallelogram of \( \mathbf{Q} \), the quadrilateral of the valtitudes relative to \( \mathbf{V} \) has the same anticenter of \( \mathbf{Q} \) . | Proof. Let \( H \) be the anticenter of \( \mathbf{Q} \) . Let \( {B}_{4}^{\prime } \) and \( {B}_{4}^{\prime \prime } \) be the common points of the line \( {A}_{1}H \) with the valtitudes \( {h}_{1} \) and \( {h}_{4} \), respectively (see Figure 16).\n\nSince the triangles \( {A}_{1}{V}_{1}{B}_{4}^{\prime } \) and \(... | Yes |
Theorem 15. If \( \mathbf{Q} \) is cyclic, all the quadrilaterals of the valtitudes of \( \mathbf{Q} \) have the same Euler line. | Proof. Given a v-parallelogram \( \mathbf{V} \) and the quadrilaterals \( {\mathbf{Q}}_{k} \) and \( {\mathbf{Q}}_{h} \) relative to it, from Theorem 3, the Euler line of \( {\mathbf{Q}}_{h} \) is the symmetric of the Euler line of \( {\mathbf{Q}}_{k} \) with respect to the point \( {G}^{\prime } \), common point to th... | No |
Theorem 16. If \( \mathbf{Q} \) is cyclic, the \( h \) -line and the \( k \) -line of \( \mathbf{Q} \) are parallel and are symmetric with respect to the line containing the third bimedian of \( \mathbf{Q} \) . | Proof. From Theorems 3, 13 and 15 it follows that the h-line and the k-line of \( \mathbf{Q} \) are symmetric with respect to \( {G}^{\prime } \), common point of the diagonals of any v-parallelogram of \( \mathbf{Q} \) . Therefore, in particular, they are parallel. Moreover, from Theorem 1, the points \( {G}^{\prime }... | Yes |
Theorem 1. Let \( {R}_{1},{R}_{2},{R}_{3},{R}_{4} \) be the circumradii in the triangles \( {ABP},{BCP} \) , \( {CDP} \), DAP respectively in a convex quadrilateral \( {ABCD} \) where the diagonals intersect at \( P \) . It has an excircle outside one of the vertices \( A \) or \( C \) if and only if\n\n\[ \n{R}_{1} + ... | Proof. According to the extended law of sines, the sides satisfies \( a = 2{R}_{1}\sin \theta \) , \( b = 2{R}_{2}\sin \theta, c = 2{R}_{3}\sin \theta \) and \( d = 2{R}_{4}\sin \theta \), where \( \theta \) is the angle between the diagonals, \( {}^{6} \) see Figure 3. Thus\n\n\[ \na + b - c - d = 2\sin \theta \left( ... | Yes |
Theorem 2. Let \( e, f, g, h \) be the distances from the vertices \( A, B, C, D \) respectively to the diagonal intersection in a convex quadrilateral ABCD with sides \( a, b, c, d \) . It has an excircle outside one of the vertices \( A \) or \( C \) if and only if\n\n\[ \n{agh} + {beh} = {cef} + {dfg} \n\]\n\nand an... | Proof. In [7] Hoehn proved that in a convex quadrilateral,\n\n\[ \n\operatorname{efgh}\left( {{a}^{2} + {c}^{2} - {b}^{2} - {d}^{2}}\right) = {a}^{2}{g}^{2}{h}^{2} + {c}^{2}{e}^{2}{f}^{2} - {b}^{2}{e}^{2}{h}^{2} - {d}^{2}{f}^{2}{g}^{2}. \n\]\n\nNow adding \( \operatorname{efgh}\left( {-{2ac} + {2bd}}\right) \) to both ... | Yes |
Theorem 3. Let \( {h}_{1},{h}_{2},{h}_{3},{h}_{4} \) be the altitudes from the diagonal intersection \( P \) to the sides \( {AB},{BC},{CD},{DA} \) in the triangles \( {ABP},{BCP},{CDP},{DAP} \) respectively in a convex quadrilateral ABCD. It has an excircle outside one of the vertices \( A \) or \( C \) if and only if... | Proof. The four equations (9) yields\n\n\[ \left( {\frac{1}{{h}_{1}} + \frac{1}{{h}_{2}} - \frac{1}{{h}_{3}} - \frac{1}{{h}_{4}}}\right) \sin \theta = \frac{a}{ef} + \frac{b}{fg} - \frac{c}{gh} - \frac{d}{he} = \frac{{agh} + {beh} - {cef} - {dfg}}{efgh}. \]\n\nSince \( \sin \theta \neq 0 \), we have that\n\n\[ \frac{1}... | Yes |
Theorem 4. Let \( x = \angle {ABD}, y = \angle {ADB}, z = \angle {BDC} \) and \( w = \angle {DBC} \) in a convex quadrilateral \( {ABCD} \) . It has an excircle outside one of the vertices \( A \) or \( C \) if and only if\n\n\[ \tan \frac{x}{2}\tan \frac{w}{2} = \tan \frac{y}{2}\tan \frac{z}{2} \]\n\nand an excircle o... | Proof. In [8], Theorem 7, we proved by using the law of cosines that\n\n\[ 1 - \cos x = \frac{\left( {d + a - q}\right) \left( {d - a + q}\right) }{2aq},\;1 + \cos x = \frac{\left( {a + q + d}\right) \left( {a + q - d}\right) }{2aq}, \]\n\n\[ 1 - \cos y = \frac{\left( {a + d - q}\right) \left( {a - d + q}\right) }{2dq}... | Yes |
Theorem 5. A convex quadrilateral with consecutive escribed circles of radii \( {R}_{a} \) , \( {R}_{b},{R}_{c} \) and \( {R}_{d} \) is tangential if and only if\n\n\[ \n{R}_{a}{R}_{c} = {R}_{b}{R}_{d} \n\] | Proof. We consider a convex quadrilateral \( {ABCD} \) where the angle bisectors intersect at \( {I}_{a},{I}_{b},{I}_{c} \) and \( {I}_{d} \) . Let the distances from these four intersections to the sides of the quadrilateral be \( {r}_{a},{r}_{b},{r}_{c} \) and \( {r}_{d} \), see Figure 6 . Then we have\n\n\[ \n{r}_{a... | Yes |
Corollary 6. In a bicentric quadrilateral \( {}^{9} \) and a tangential trapezoid with consecutive escribed circles of radii \( {R}_{a},{R}_{b},{R}_{c} \) and \( {R}_{d} \), the incircle has the radius | Proof. In these quadrilaterals, \( A + C = \pi = B + D \) or \( A + D = \pi = B + C \) (if we assume that \( {AB}\parallel {DC}) \) . Thus\n\n\[ \tan \frac{A}{2}\tan \frac{C}{2} = \tan \frac{B}{2}\tan \frac{D}{2} = 1 \]\n\nor\n\[ \tan \frac{A}{2}\tan \frac{D}{2} = \tan \frac{B}{2}\tan \frac{C}{2} = 1. \]\n\nIn either c... | Yes |
Theorem 8. An extangential quadrilateral with sides \( a, b, c \) and \( d \) has the exradius\n\n\[ \n\rho = \frac{K}{\left| a - c\right| } = \frac{K}{\left| b - d\right| }\n\]\n\nwhere \( K \) is the area of the quadrilateral. | Proof. We prove the formulas in the case that is shown i Figure 8. The area of the extangential quadrilateral \( {ABCD} \) is equal to the areas of the triangles \( {ABE} \) and \( {ADE} \) subtracted by the areas of \( {BCE} \) and \( {CDE} \) . Thus\n\n\[ K = \frac{1}{2}{a\rho } + \frac{1}{2}{d\rho } - \frac{1}{2}{b\... | Yes |
Lemma 1. The polynomial \( {u}^{3} - s{u}^{2} + {du} - p \) has roots \( {u}_{1} = {\sin }^{2}\alpha ,{u}_{2} = {\sin }^{2}\beta \) , \( {u}_{3} = {\sin }^{2}\gamma \) for some \( \{ \alpha ,\beta ,\gamma \} \in \mathcal{T} \) if and only if \( s, p \in \mathbf{R}, p \geq 0, d = \frac{{s}^{2}}{4} + p \) and \( D\left( ... | Proof. If \( d = \frac{{s}^{2}}{4} + p,\frac{p}{16}D\left( {s, p}\right) \) is the polynomial’s discriminant: for \( s \in \mathbf{R} \) and \( p > 0 \) one has then \( D\left( {s, p}\right) \geq 0 \) if and only if the roots are real; for \( s \in \mathbf{R} \) and \( p = 0 \) the roots are then 0 and \( \frac{s}{2} \... | Yes |
Theorem 2. The van IJzeren map is a bijection from \( \mathcal{T} \) to \[ {\mathcal{T}}^{ * } = \left\{ {\left( {s, p}\right) \mid D\left( {s, p}\right) = {\left( 9 - 4s\right) }^{3} - {\left( 8p + 2{s}^{2} - {18}s + {27}\right) }^{2} \geq 0, s \geq 0, p \geq 0}\right\} \] with inverse \( {V}^{-1} : {\mathcal{T}}^{ * ... | For \( \left( {s, p}\right) \in {\mathcal{T}}^{ * } \) the discriminant \( \frac{p}{16}D\left( {s, p}\right) \) of the above polynomial in \( u \) is 0 if and only if there are multiple roots among \( {\sin }^{2}\alpha ,{\sin }^{2}\beta \) and \( {\sin }^{2}\gamma \), i.e., if and only if \( \left( {s, p}\right) = {\Pi... | No |
Theorem 3. If \( r\left( \Delta \right) \) denotes the reflection triangle (class) of \( \Delta \), the map\n\n\[ \n\rho : {\mathcal{T}}^{ * } \rightarrow {\mathcal{T}}^{ * },\left( {s, p}\right) = {\Delta }^{ * } \mapsto r{\left( \Delta \right) }^{ * } \n\]\n\ninduced by \( r \) is given by \( \rho \left( {I}_{\pi /6}... | Proof. Consider the proper triangle \( \Delta = {ABC} \) with angles \( \alpha ,\beta ,\gamma \) and opposite sides \( a, b, c \) and reflect \( \Delta \) in all its sides to get the reflection triangle \( {\Delta }_{1} = \) \( {A}_{1}{B}_{1}{C}_{1} \) . Let \( \left( {s, p}\right) \) and \( \left( {S, P}\right) \) be ... | Yes |
Theorem 4. The reflection triangle of a nondegenerate triangle \( \Delta \) is degenerate if and only if \( s\left( \Delta \right) = \frac{5}{4} \), i.e., if and only if the point \( \left( {\alpha ,\beta }\right) \) formed by the two smallest angles of \( \Delta \) lies on the oval \( {\sin }^{2}\alpha + {\sin }^{2}\b... | Proof. Let first \( {ABC} \) be a proper triangle with opposite sides \( a, b, c \), circumcenter \( O \), circumradius \( R \), nine-point center \( N \), centroid \( G \) and medians \( {m}_{a},{m}_{b},{m}_{c} \), and let \( X \) be a point (not necessarily coplanar with \( {ABC} \) ). [10, p. 174] proves\n\n\[ X{A}^... | Yes |
Theorem 6. Let \( \Delta \) be an proper triangle with degenerate reflection triangle \( {\Delta }_{1} \) . The following properties are equivalent.\n\n(1) \( {\Delta }_{1} \) has two equal sides and three different vertices, i.e., \( \Delta \) has angles \( {45}^{ \circ } \) and \( {\omega }_{12} \) .\n\n(2) The middl... | Proof. (2) \( \Rightarrow \left( 1\right) \) : (14) shows that the upper part of \( \mathcal{L} \) cuts \( N\left( \frac{R}{2}\right) \) at \( {C}_{1} \) if and only if the polar angle of \( C \) is arccos \( \frac{7}{8} \) (infinite triangle’s case) or arccos \( \frac{3}{4} \) : in this second case, \( {C}_{1} = \left... | Yes |
Theorem 7. The action of \( r \) on a class of infinite triangles is given by \( r\left( {\Pi }_{\alpha }\right) = \) \( {\Pi }_{\left( {{2\alpha } + \arctan \left( {3\tan \alpha }\right) }\right) {\;\operatorname{mod}\;\pi }} \) (Figure 8) and \( r{\left( {\Pi }_{\alpha }\right) }^{ * } = \left( {\frac{s{\left( 4s - 5... | Proof. The theorem is true for \( \alpha = \frac{\pi }{2} \) . Take an acute angle \( \alpha \), consider a triangle with an angle \( {2\alpha } \) between sides of length 1 and 2 and define \( \delta \) as the acute or right angle formed by the bisector of \( {2\alpha } \) and the opposite side. Using the angle bisect... | Yes |
Theorem 9. The iterated reflection class of an isosceles base triangle class \( {I}_{\alpha } \) converges to an equilateral limit unless \( {I}_{\alpha } \) belongs to the backward orbit of \( {I}_{\pi /6} \) and converges thus to a degenerate limit in a finite number of steps, i.e., unless \( {I}_{\alpha } = {I}_{\pi... | See [4] for another proof, which iterates the formula\n\n\[ \n{\cos }^{2}{\alpha }_{1} = \frac{{\cos }^{2}\alpha {\left( 4{\cos }^{2}\alpha - 3\right) }^{2}}{1 + {16}{\cos }^{2}\alpha - {16}{\cos }^{4}\alpha }\n\]\n\nfor a nondegenerate \( r\left( {I}_{\alpha }\right) = {I}_{{\alpha }_{1}} \) and shows that \( \mathop{... | No |
Theorem 10 (Parents). The parents \( {\rho }^{-1}\left( {\Delta }_{1}^{ * }\right) \) (in \( {\mathbf{R}}^{\mathbf{2}} \) ) of any \( {\Delta }_{1}^{ * } = \left( {S, P}\right) \in \) \( {\mathcal{T}}^{ * } \smallsetminus \{ \left( {0,0}\right) \} \) are the points \( \left( {s, p}\right) \in {\mathbf{R}}^{\mathbf{2}} ... | The denominators are never zero. All between three and seven parents of \( \left( {S, P}\right) \) \( \in {\mathcal{T}}^{ * } \smallsetminus \Gamma \) lie in \( {\mathcal{T}}^{ * } \smallsetminus \Gamma \) except the rightmost parent \( \left( {\frac{5}{4} + \sin \alpha ,\frac{1 + \sin \alpha }{{64}\left( {1 - \sin \al... | Yes |
Theorem 11. The parents in \( \mathcal{T} \) of \( {I}_{\alpha },\alpha \neq \frac{\pi }{3} \), are - up to the exceptions mentioned below - the two non-isosceles classes \( \left\{ {{\alpha }_{ \pm }^{\prime },{\beta }_{ \pm }^{\prime },{\gamma }_{ \pm }^{\prime }}\right\} \) given by the non-obtuse angles\n\n\[{\alph... | Proof. Parts of this theorem have been already demonstrated in the proof of Theorem 10. Theorem 2 for \( s = \frac{5}{4} - \sin \alpha, p = \frac{1 - \sin \alpha }{{64}\left( {1 + \sin \alpha }\right) } \) gives an obtuse parent \( \left\{ {{\alpha }^{\prime },{\beta }^{\prime },{\gamma }^{\prime }}\right\} \) of \( {I... | Yes |
Theorem 14. \( \left( {s, p}\right) \in {\mathcal{T}}^{ * } \smallsetminus \left( {\Gamma \cup \left\{ \left( {\frac{7}{4},\frac{7}{64}}\right) \right\} }\right) \) converges to \( {I}_{\pi /3}^{ * } \) under iteration of \( \rho \) (with strictly growing coordinates except possibly for the first reflection step) when ... | Proof. We only have to prove that the corner of zone I near the heptagonal fixed point \( \left( {\frac{7}{4},\frac{7}{64}}\right) \) is mapped by \( \rho \) to zone I and not to zone III, and this is true: the points with \( {\rho }_{1}\left( {s, p}\right) = s, s > \frac{7}{4}, p > 0 \), are mapped upwards by \( \rho ... | Yes |
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