Difference between revisions of "Euler line"
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[[Image:Euler Line.PNG||500px|frame|center]] | [[Image:Euler Line.PNG||500px|frame|center]] | ||
+ | |||
+ | ==The points of intersection of the Euler line with the sides of the triangle== | ||
+ | <i><b>Acute triangle</b></i> | ||
+ | [[File:Euler line crosspoints.png|450px|right]] | ||
+ | Let <math>\triangle ABC</math> be the acute triangle where <math>AC > BC > AB.</math> | ||
+ | Denote <math>\angle A = \alpha, \angle B = \beta, \angle C = \gamma</math> | ||
+ | <cmath>\implies \beta > \alpha > \gamma.</cmath> | ||
+ | |||
+ | Let Euler line cross lines <math>AB, AC,</math> and <math>BC</math> in points <math>D, E,</math> and <math>F,</math> respectively. | ||
+ | |||
+ | Then point <math>D</math> lyes on segment <math>AB, \frac {\vec BD}{\vec DA} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0.</math> | ||
+ | |||
+ | Point <math>E</math> lyes on segment <math>AC, \frac {\vec AE}{\vec EC} = \frac {\tan \beta – \tan \gamma}{\tan \beta – \tan \alpha} > 0.</math> | ||
+ | |||
+ | Point <math>F</math> lyes on ray <math>BC, \frac {\vec {BF}}{\vec {CF}} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} > 0.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>n = \frac {\vec BD}{\vec DA}, m = \frac {\vec CE}{\vec EA}, p = \frac {\vec CX}{\vec XB}, X \in BC, q = \frac {AY}{YX}, Y \in AX.</math> | ||
+ | |||
+ | We use the formulae <math>m + pn = \frac {p+1}{q}</math> (see Claim “Segments crossing inside triangle” in “Schiffler point” in “Euler line”). | ||
+ | |||
+ | Centroid <math>G</math> lyes on median <math>AA'' \implies X = A'' , Y = G, p = 1, q = 2 \implies m+n=1.</math> | ||
+ | |||
+ | Orthocenter <math>H</math> lyes on altitude <math>AA' \implies X = A', Y = H, p = \frac {\vec {CX}}{\vec {XB}} = \frac {\tan \beta}{\tan \gamma}, q = \frac {AY}{YX}.</math> | ||
+ | <cmath>q = \frac {\vec {AH}}{\vec {HA'}} = \frac {\cos \alpha}{\cos \beta \cdot \cos \gamma} = \tan \beta \cdot \tan \gamma – 1 \implies</cmath> | ||
+ | <cmath>m \tan \gamma + n \tan \beta = \frac {\tan \beta + \tan \gamma}{1 – \tan \beta \cdot \tan \gamma} = – \tan \alpha.</cmath> | ||
+ | Therefore | ||
+ | <cmath>\frac {\vec {BD}}{\vec {DA}} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0,</cmath> | ||
+ | <cmath>\frac {\vec {CE}}{\vec {EA}} = m =\frac {\tan \beta – \tan \alpha}{\tan \beta – \tan \gamma} > 0.</cmath> | ||
+ | We use the signed version of Menelaus's theorem and get | ||
+ | <cmath>\frac {\vec {BF}}{\vec {FC}} = \frac {\tan \alpha – \tan \gamma}{ \tan \alpha –\tan \beta} < 0.</cmath> | ||
+ | |||
+ | <i><b>Obtuse triangle</b></i> | ||
+ | [[File:Euler line obtuse triangle.png|400px|right]] | ||
+ | Let <math>\triangle ABC</math> be the obtuse triangle where <math>BC > AC > AB \implies \alpha > \beta > \gamma.</math> | ||
+ | |||
+ | Let Euler line cross lines <math>AB, AC,</math> and <math>BC</math> in points <math>D, E,</math> and <math>F,</math> respectively. | ||
+ | |||
+ | Similarly we get <math>F \in BC, \frac {\vec {BF}}{\vec {FC}} = \frac {\tan \gamma – \tan \alpha}{ \tan \beta –\tan \alpha} > 0.</math> | ||
+ | |||
+ | <cmath>E \in AC, \frac {\vec {CE}}{\vec {EA}} = \frac {\tan \beta – \tan \alpha}{\tan \beta – \tan \gamma} > 0.</cmath> | ||
+ | <math>D \in</math> ray <math>BA, \frac {\vec {BD}}{\vec {AD}} = \frac {\tan \gamma – \tan \alpha}{\tan \beta – \tan \gamma} > 0.</math> | ||
+ | |||
+ | <i><b>Right triangle</b></i> | ||
+ | |||
+ | Let <math>\triangle ABC</math> be the right triangle where <math>\angle BAC = 90^\circ.</math> Then Euler line contain median from vertex <math>A.</math> | ||
+ | |||
+ | <i><b>Isosceles triangle</b></i> | ||
+ | |||
+ | Let <math>\triangle ABC</math> be the isosceles triangle where <math>AC = AB.</math> Then Euler line contain median from vertex <math>A.</math> | ||
+ | |||
+ | <i><b>Corollary: Euler line is parallel to side</b></i> | ||
+ | |||
+ | Euler line <math>DE</math> is parallel to side <math>BC</math> iff <math>\tan \beta \cdot \tan \gamma = 3.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>DE||BC \implies \frac {BD}{DA} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma}= | ||
+ | \frac {CE}{EA} = \frac {\tan \beta – \tan \alpha}{\tan \beta – \tan \gamma}.</math> | ||
+ | |||
+ | After simplification in the case <math>\beta \ne \gamma</math> we get <math>2 \tan \alpha = \tan \beta + \tan \gamma.</math> | ||
+ | <cmath>180^\circ – \alpha = \beta + \gamma \implies \tan \alpha = \frac{\tan \beta + \tan \gamma}{\tan \beta \cdot \tan \gamma – 1} \implies \tan \beta \cdot \tan \gamma = 3.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Angles between Euler line and the sides of the triangle== | ||
+ | [[File:Euler line side angle.png|450px|right]] | ||
+ | Let Euler line of the <math>\triangle ABC</math> cross lines <math>AB, AC,</math> and <math>BC</math> in points <math>D, E,</math> and <math>F,</math> respectively. | ||
+ | Denote <math>\angle A = \alpha, \angle B = \beta, \angle C = \gamma,</math> smaller angles between the Euler line and lines <math>BC, AC,</math> and <math>AB</math> as <math>\theta_A, \theta_B,</math> and <math>\theta_C,</math> respectively. | ||
+ | |||
+ | Prove that <math>\tan \theta_A = \vert \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma} \vert,</math> | ||
+ | <cmath>\tan \theta_B = |\frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}\vert, | ||
+ | \tan \theta_C = |\frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}\vert.</cmath> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, <math>AC > BC > AB \implies \frac {\vec {BF}}{\vec {CF}} = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} > 0.</math> | ||
+ | |||
+ | Let <math>|BC| = 2a, M</math> be the midpoint <math>BC, O</math> be the circumcenter of <math>\triangle ABC \implies OM = \frac {a}{\tan \alpha}.</math> | ||
+ | |||
+ | <cmath>MF = MC + CF, \frac {\vec {BF}}{\vec {CF}} = \frac {\vec {BC + CF}}{\vec {CF}} = \frac {2a}{|CF|}+ 1 =\frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \alpha} \implies</cmath> | ||
+ | <cmath>\frac {|MF|}{a} = \frac {\tan \beta – \tan \gamma}{2 \tan \alpha – \tan \beta – \tan \gamma} \implies</cmath> | ||
+ | <cmath>\tan \theta_A = \frac {|OM|}{|MF|} = |\frac {3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma}|.</cmath> | ||
+ | |||
+ | Symilarly, for other angles. | ||
+ | |||
+ | *[[Gossard perspector]] | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Distances along Euler line== | ==Distances along Euler line== | ||
[[File:OH distance.png|400px|right]] | [[File:OH distance.png|400px|right]] | ||
Let <math>H, G, O,</math> and <math>R</math> be orthocenter, centroid, circumcenter, and circumradius of the <math>\triangle ABC,</math> respectively. | Let <math>H, G, O,</math> and <math>R</math> be orthocenter, centroid, circumcenter, and circumradius of the <math>\triangle ABC,</math> respectively. | ||
− | <cmath>a = BC, b = AC, c = AB, \alpha = \angle A,\beta = \angle B,\gamma = \angle C.</cmath> | + | <cmath>a = BC, b = AC, c = AB,</cmath> |
+ | <cmath>\alpha = \angle A,\beta = \angle B,\gamma = \angle C.</cmath> | ||
− | Prove that <math>HO^2 = R^2 (1 | + | Prove that <math>HO^2 = R^2 (1 - 8 \cos A \cos B \cos C),</math> |
+ | <cmath>GO^2 = R^2 - \frac {a^2 + b^2 + c^2}{9}.</cmath> | ||
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
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WLOG, <math>ABC</math> is an acute triangle, <math>\beta \ge \gamma.</math> | WLOG, <math>ABC</math> is an acute triangle, <math>\beta \ge \gamma.</math> | ||
− | <cmath>OA = R, AH = 2 R \cos \alpha, \angle BAD = \angle OAC = 90^\circ | + | <cmath>OA = R, AH = 2 R \cos \alpha, \angle BAD = \angle OAC = 90^\circ - \beta \implies</cmath> |
− | <cmath>\angle OAH = \alpha | + | <cmath>\angle OAH = \alpha - 2\cdot (90^\circ - \beta) = \alpha + \beta + \gamma - 180^\circ + \beta - \gamma = \beta - \gamma .</cmath> |
− | <cmath>HO^2 = AO^2 + AH^2 | + | <cmath>HO^2 = AO^2 + AH^2 - 2 AH \cdot AO \cos \angle OAC = R^2 + (2 R \cos \alpha)^2 - 2 R \cdot 2R \cos \alpha \cdot \cos (\beta - \gamma).</cmath> |
− | <cmath>\frac {HO^2}{R^2} = 1 + 4 \cos \alpha (\cos \alpha | + | <cmath>\frac {HO^2}{R^2} = 1 + 4 \cos \alpha (\cos \alpha - \cos (\beta - \gamma) = 1 - 4 \cos \alpha (\cos (\beta + \gamma) + \cos (\beta – \gamma)) = 1 - 8 \cos \alpha \cos \beta \cos \gamma.</cmath> |
− | <cmath>\frac {HO^2}{R^2} = 1 + 4 \cos^2 \alpha | + | <cmath>\frac {HO^2}{R^2} = 1 + 4 \cos^2 \alpha - 4 \cos \alpha \cos (\beta - \gamma) = 5 - 4 \sin^2 \alpha + 4 \cos (\beta + \gamma) \cos (\beta - \gamma)</cmath> |
− | <cmath>HO^2 = 5R^2 | + | <cmath>HO^2 = 5R^2 - 4R^2 \sin^2 \alpha + 2R^2 \cos 2\beta + 2R^2 \cos 2 \gamma = 9R^2 - 4R^2 \sin^2 \alpha - 4R^2 \sin^2 \beta - 4 R^2\sin^2 \gamma</cmath> |
− | <cmath>HO^2 = 9R^2 | + | <cmath>HO^2 = 9R^2 - a^2 - b^2 - c^2, GO^2 = \frac {HO^2}{9} = R^2 - \frac {a^2 + b^2 + c^2}{9}.</cmath> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
==Position of Kimberling centers on the Euler line== | ==Position of Kimberling centers on the Euler line== | ||
[[File:Kimberling points on Euler line.png|450px|right]] | [[File:Kimberling points on Euler line.png|450px|right]] | ||
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Centroid <math>X(2)</math> | Centroid <math>X(2)</math> | ||
− | <cmath>X(2) = X(3) + \frac {1}{3} (X(4) | + | <cmath>X(2) = X(3) + \frac {1}{3} (X(4) - X(3)) \implies k_2 = \frac {1}{3}.</cmath> |
Nine-point center <math>X(5)</math> | Nine-point center <math>X(5)</math> | ||
− | <cmath>X(5) = X(3) + \frac {1}{2} (X(4) | + | <cmath>X(5) = X(3) + \frac {1}{2} (X(4) - X(3)) \implies k_5 = \frac {1}{2}.</cmath> |
de Longchamps point <math>X(20)</math> | de Longchamps point <math>X(20)</math> | ||
− | <cmath>X(20) = X(3) | + | <cmath>X(20) = X(3) - (X(4) - X(3)) \implies k_{20} = - 1.</cmath> |
Schiffler point <math>X(21)</math> | Schiffler point <math>X(21)</math> | ||
− | <cmath>X(21) = X(3) + \frac {1}{3+2r/R} (X(4) | + | <cmath>X(21) = X(3) + \frac {1}{3 + 2r/R} (X(4) + X(3)) \implies k_{21} = \frac {1}{3 + 2t}.</cmath> |
Exeter point <math>X(22)</math> | Exeter point <math>X(22)</math> | ||
− | <cmath>X(22) = X(3) + \frac {2}{J^ | + | <cmath>X(22) = X(3) + \frac {2}{J^2 - 3} (X(4) - X(3)) \implies k_{22} = \frac {2}{J^2 - 3}.</cmath> |
Far-out point <math>X(23)</math> | Far-out point <math>X(23)</math> | ||
− | <cmath>X(23) = X(3) + \frac {3}{J^2} (X(4) | + | <cmath>X(23) = X(3) + \frac {3}{J^2} (X(4) - X(3)) \implies k_{23} = \frac {3}{J^2}.</cmath> |
Perspector of ABC and orthic-of-orthic triangle <math>X(24)</math> | Perspector of ABC and orthic-of-orthic triangle <math>X(24)</math> | ||
− | <cmath>X(24) = X(3) + \frac {2}{J^2+1} (X(4) | + | <cmath>X(24) = X(3) + \frac {2}{J^2+1} (X(4) - X(3)) \implies k_{24} = \frac {2}{J^2 + 1}.</cmath> |
Homothetic center of orthic and tangential triangles <math>X(25)</math> | Homothetic center of orthic and tangential triangles <math>X(25)</math> | ||
− | <cmath>X(25) = X(3) + \frac {4}{J^2+3} (X(4) | + | <cmath>X(25) = X(3) + \frac {4}{J^2+3} (X(4) - X(3)) \implies k_{25} = \frac {4}{J^2 + 3}.</cmath> |
Circumcenter of the tangential triangle <math>X(26)</math> | Circumcenter of the tangential triangle <math>X(26)</math> | ||
− | <cmath>X(26) = X(3) + \frac {2}{J^ | + | <cmath>X(26) = X(3) + \frac {2}{J^2 - 1}(X(4) - X(3)) \implies k_{26} = \frac {2}{J^2 - 1}.</cmath> |
Midpoint of X(3) and <math>X(5)</math> | Midpoint of X(3) and <math>X(5)</math> | ||
− | <cmath>X(140) = X(3) + \frac {1}{4} (X(4) | + | <cmath>X(140) = X(3) + \frac {1}{4} (X(4) - X(3)) \implies k_{140} = \frac {1}{4}.</cmath> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
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==Euler lines of cyclic quadrilateral (Vittas’s theorem)== | ==Euler lines of cyclic quadrilateral (Vittas’s theorem)== | ||
+ | <i><b>Claim 1</b></i> | ||
[[File:2 chords 4 triangles 1.png|400px|right]] | [[File:2 chords 4 triangles 1.png|400px|right]] | ||
− | Let <math>ABCD</math> be a cyclic quadrilateral with diagonals intersecting at <math>P (\angle APB \ne 60^\circ).</math> | + | Let <math>ABCD</math> be a cyclic quadrilateral with diagonals intersecting at <math>P (\angle APB \ne 60^\circ).</math> The Euler lines of triangles <math>\triangle APB, \triangle BPC, \triangle CPD, \triangle DPA</math> are concurrent. |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
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We use <i><b>Claim</b></i> and get that lines <math>O_1H_1, O_2H_2, O_3H_3, O_4H_4</math> are concurrent (or parallel if <math>\angle APD = 60^\circ</math> or <math>\angle APD = 120^\circ</math>). | We use <i><b>Claim</b></i> and get that lines <math>O_1H_1, O_2H_2, O_3H_3, O_4H_4</math> are concurrent (or parallel if <math>\angle APD = 60^\circ</math> or <math>\angle APD = 120^\circ</math>). | ||
− | <i><b>Claim (Property of vertex of two parallelograms)</b></i> | + | <i><b>Claim 2 (Property of vertex of two parallelograms)</b></i> |
[[File:2 parallelograms.png|400px|right]] | [[File:2 parallelograms.png|400px|right]] | ||
Let <math>ABCD</math> and <math>EFGH</math> be parallelograms, <math>AB||EF, AD||EH.</math> Let lines <math>AE, BH,</math> and <math>CG</math> be concurrent at point <math>O.</math> Then points <math>D, O,</math> and <math>F</math> are collinear and lines <math>AG, BF, CE,</math> and <math>DH</math> are concurrent. | Let <math>ABCD</math> and <math>EFGH</math> be parallelograms, <math>AB||EF, AD||EH.</math> Let lines <math>AE, BH,</math> and <math>CG</math> be concurrent at point <math>O.</math> Then points <math>D, O,</math> and <math>F</math> are collinear and lines <math>AG, BF, CE,</math> and <math>DH</math> are concurrent. | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ~minor edit by Yiyj1 | ||
==Concurrent Euler lines and Fermat points== | ==Concurrent Euler lines and Fermat points== | ||
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==Thebault point== | ==Thebault point== | ||
− | [[File:Orthic triangle.png| | + | [[File:Orthic triangle.png|400px|right]] |
Let <math>AD, BE,</math> and <math>CF</math> be the altitudes of the <math>\triangle ABC,</math> where <math>BC> AC > AB, \angle BAC \ne 90^\circ.</math> | Let <math>AD, BE,</math> and <math>CF</math> be the altitudes of the <math>\triangle ABC,</math> where <math>BC> AC > AB, \angle BAC \ne 90^\circ.</math> | ||
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<i><b>Case 2 Obtuse triangle</b></i> | <i><b>Case 2 Obtuse triangle</b></i> | ||
− | [[File:Orthic triangle obtuse.png| | + | [[File:Orthic triangle obtuse.png|400px|right]] |
a) It is known, that Euler line of obtuse <math>\triangle ABC</math> cross AC and BC (middle and longest sides) in inner points. | a) It is known, that Euler line of obtuse <math>\triangle ABC</math> cross AC and BC (middle and longest sides) in inner points. | ||
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<i><b>Claim (Segment crossing the median)</b></i> | <i><b>Claim (Segment crossing the median)</b></i> | ||
− | [[File:Median cross segment.png| | + | [[File:Median cross segment.png|400px|right]] |
Let <math>M</math> be the midpoint of side <math>AB</math> of the <math>\triangle ABC, D \in AC,</math> | Let <math>M</math> be the midpoint of side <math>AB</math> of the <math>\triangle ABC, D \in AC,</math> | ||
− | < | + | <cmath>E \in BC, G = DE \cap CM.</cmath> |
+ | <cmath>\frac {BE}{CE} = m, \frac {AD}{CD} = n.</cmath> | ||
Then <math>\frac {DG}{GE} = \frac{1+m}{1+n}, \frac {MG}{GC} = \frac{n+m}{2}.</math> | Then <math>\frac {DG}{GE} = \frac{1+m}{1+n}, \frac {MG}{GC} = \frac{n+m}{2}.</math> | ||
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x = \frac{1}{(1+n)(2+m+n)}.</cmath> | x = \frac{1}{(1+n)(2+m+n)}.</cmath> | ||
<cmath>z+x = \frac {[CDM]}{2[CAM]} = \frac {1}{2(1 + n)}.</cmath> | <cmath>z+x = \frac {[CDM]}{2[CAM]} = \frac {1}{2(1 + n)}.</cmath> | ||
− | <cmath>\frac {MG}{GC} = \frac {z}{x} = \frac {z + x}{x} | + | <cmath>\frac {MG}{GC} = \frac {z}{x} = \frac {z + x}{x} - 1 =\frac {1}{2(1 + n)} (1 + n)(2 + m + n) - 1 = \frac {n + m}{2}.</cmath> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Schiffler point== | ==Schiffler point== | ||
− | [[File:Shiffler point.png| | + | [[File:Shiffler point.png|400px|right]] |
Let <math>I, O, G, R, \alpha,</math> and <math>r</math> be the incenter, circumcenter, centroid, circumradius, <math>\angle A,</math> and inradius of <math>\triangle ABC,</math> respectively. Then the Euler lines of the four triangles <math>\triangle BCI, \triangle CAI, \triangle ABI,</math> and <math>\triangle ABC</math> are concurrent at Schiffler point <math>S = X(21), \frac {OS}{SG} = \frac {3R}{2r}</math>. | Let <math>I, O, G, R, \alpha,</math> and <math>r</math> be the incenter, circumcenter, centroid, circumradius, <math>\angle A,</math> and inradius of <math>\triangle ABC,</math> respectively. Then the Euler lines of the four triangles <math>\triangle BCI, \triangle CAI, \triangle ABI,</math> and <math>\triangle ABC</math> are concurrent at Schiffler point <math>S = X(21), \frac {OS}{SG} = \frac {3R}{2r}</math>. | ||
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<cmath>m = \frac {GX}{XE} = \frac {2n}{3}.</cmath> | <cmath>m = \frac {GX}{XE} = \frac {2n}{3}.</cmath> | ||
− | <cmath>p = \frac {OE}{EY} = \frac {\cos \alpha}{(1 | + | <cmath>p = \frac {OE}{EY} = \frac {\cos \alpha}{(1 - \cos \alpha)/3} = \frac {3 \cos \alpha}{1 - \cos \alpha}</cmath> |
Using <i><b>Claim</b></i> we get | Using <i><b>Claim</b></i> we get | ||
− | <cmath>\frac {OS}{SG} = \frac {p+1}{m} | + | <cmath>\frac {OS}{SG} = \frac {p + 1}{m} - \frac {p}{n} = \frac {3(p + 1)}{2n} - \frac {p}{n} = \frac {p + 3}{2n} = \frac {3R}{2r}.</cmath> |
Therefore each Euler line of triangles <math>\triangle BCI, \triangle CAI, \triangle ABI,</math> cross Euler line of <math>\triangle ABC</math> in the same point, as desired. | Therefore each Euler line of triangles <math>\triangle BCI, \triangle CAI, \triangle ABI,</math> cross Euler line of <math>\triangle ABC</math> in the same point, as desired. | ||
<i><b>Claim (Segments crossing inside triangle)</b></i> | <i><b>Claim (Segments crossing inside triangle)</b></i> | ||
− | [[File:Segments crossing inside triangle.png| | + | [[File:Segments crossing inside triangle.png|400px|right]] |
Given triangle GOY. Point <math>S</math> lies on <math>GO, k = \frac {OS}{SG}.</math> | Given triangle GOY. Point <math>S</math> lies on <math>GO, k = \frac {OS}{SG}.</math> | ||
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Point <math>G'</math> lies on <math>GY, n = \frac {GG'}{G'Y}.</math> | Point <math>G'</math> lies on <math>GY, n = \frac {GG'}{G'Y}.</math> | ||
− | Point <math>X</math> lies on <math>GE, m = \frac {GX}{XE}.</math> Then <math>k = \frac {p+1}{m} | + | Point <math>X</math> lies on <math>GE, m = \frac {GX}{XE}.</math> Then <math>k = \frac {p + 1}{m} - \frac {p}{n}.</math> |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
Let <math>[OGY]</math> be <math>1</math> (We use sigh <math>[t]</math> for area of <math>t).</math> | Let <math>[OGY]</math> be <math>1</math> (We use sigh <math>[t]</math> for area of <math>t).</math> | ||
− | <cmath>[GSG'] = \frac{n}{(n+1)(k+1)}, [YEG'] = \frac{1}{(n+1)(p+1)},</cmath> | + | <cmath>[GSG'] = \frac{n}{(n + 1)(k + 1)}, [YEG'] = \frac{1}{(n + 1)(p + 1)},</cmath> |
− | <cmath>[SOE] = \frac{kp}{(k+1)(p+1)}, [ESG'] = \frac {[GSG']}{m},</cmath> | + | <cmath>[SOE] = \frac{kp}{(k + 1)(p + 1)}, [ESG'] = \frac {[GSG']}{m},</cmath> |
− | <cmath>[OGY] = [GSG'] + [YEG'] + [SOE] + [ESG'] = 1 \implies \frac{n(p+1)}{m}=nk + p \implies k = \frac {p+1}{m} | + | <cmath>[OGY] = [GSG'] + [YEG'] + [SOE] + [ESG'] = 1 \implies</cmath> |
+ | <cmath>\frac{n(p + 1)}{m}=nk + p \implies k = \frac {p + 1}{m} - \frac {p}{n}.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Line 494: | Line 591: | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | ==De Longchamps point | + | ==De Longchamps point X(20)== |
− | [[File:Longchamps.png| | + | [[File:Longchamps.png|400px|right]] |
<i><b>Definition 1</b></i> | <i><b>Definition 1</b></i> | ||
Line 515: | Line 612: | ||
Point <math>E</math> is the crosspoint of the center line of the <math>B-</math>power and <math>C-</math>power circles and there radical axis. <math>B'C' = \frac {a}{2}.</math> We use claim and get: | Point <math>E</math> is the crosspoint of the center line of the <math>B-</math>power and <math>C-</math>power circles and there radical axis. <math>B'C' = \frac {a}{2}.</math> We use claim and get: | ||
− | <cmath>C'E = \frac {a}{4} + \frac {R_C^2 | + | <cmath>C'E = \frac {a}{4} + \frac {R_C^2 - R_B^2}{a}.</cmath> |
<math>R_B</math> and <math>R_C</math> are the medians, so | <math>R_B</math> and <math>R_C</math> are the medians, so | ||
− | <cmath>R_B^2 = \frac {a^2}{2}+ \frac {c^2}{2} | + | <cmath>R_B^2 = \frac {a^2}{2}+ \frac {c^2}{2} - \frac {b^2}{4}, R_C^2 = \frac {a^2}{2}+ \frac {b^2}{2} - \frac {c^2}{4} \implies C'E = \frac {a}{4} + \frac {3(b^2 - c^2)}{4a}.</cmath> |
We use Claim some times and get: | We use Claim some times and get: | ||
− | <cmath>C'A_t = \frac {a}{4} | + | <cmath>C'A_t = \frac {a}{4} - \frac {b^2 - c^2}{4a}, |
− | A_tO_t = \frac {a}{2} | + | A_tO_t = \frac {a}{2} - 2 C'A_t = \frac {b^2 - c^2}{2a} \implies</cmath> |
− | <cmath>O_t L_t = C'E | + | <cmath>O_t L_t = C'E - C'A_t - A_t O_t = \frac {b^2 - c^2}{2a} = A_t O_t = H_t O_t \implies</cmath> |
radical axes of <math>B-</math>power and <math>C-</math>power cicles is symmetric to altitude <math>AH</math> with respect <math>O.</math> | radical axes of <math>B-</math>power and <math>C-</math>power cicles is symmetric to altitude <math>AH</math> with respect <math>O.</math> | ||
Line 530: | Line 627: | ||
<i><b>Claim (Distance between projections)</b></i> | <i><b>Claim (Distance between projections)</b></i> | ||
− | <cmath>x + y = a, c^2 | + | <cmath>x + y = a, c^2 - x^2 = h^2 = b^2 - y^2,</cmath> |
− | <cmath>y^2 | + | <cmath>y^2 - x^2 = b^2 - c^2 \implies y - x = \frac {b^2 - c^2}{a},</cmath> |
− | <cmath>x = \frac {a}{2} | + | <cmath>x = \frac {a}{2} - \frac {b^2 - c^2}{2a}, y = \frac {a}{2} + \frac {b^2 - c^2}{2a}.</cmath> |
<i><b>Definition 2</b></i> | <i><b>Definition 2</b></i> | ||
− | [[File:Longchamps 1.png| | + | [[File:Longchamps 1.png|400px|right]] |
We call <math>\omega_A = A-</math>circle of a <math>\triangle ABC</math> the circle centered at <math>A</math> with radius <math>BC.</math> The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of <math>A-</math>circle, <math>B-</math>circle, and <math>C-</math>circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under <i><b>Definition 1.</b></i> | We call <math>\omega_A = A-</math>circle of a <math>\triangle ABC</math> the circle centered at <math>A</math> with radius <math>BC.</math> The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of <math>A-</math>circle, <math>B-</math>circle, and <math>C-</math>circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under <i><b>Definition 1.</b></i> | ||
Line 580: | Line 677: | ||
==CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)== | ==CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)== | ||
− | [[File:X26.png| | + | [[File:X26.png|400px|right]] |
Prove that the circumcenter of the tangential triangle <math>\triangle A'B'C'</math> of <math>\triangle ABC</math> (Kimberling’s point <math>X(26))</math> lies on the Euler line of <math>\triangle ABC.</math> | Prove that the circumcenter of the tangential triangle <math>\triangle A'B'C'</math> of <math>\triangle ABC</math> (Kimberling’s point <math>X(26))</math> lies on the Euler line of <math>\triangle ABC.</math> | ||
Line 600: | Line 697: | ||
==PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)== | ==PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)== | ||
− | [[File:X25.png| | + | [[File:X25.png|400px|right]] |
Let <math>\triangle A_1B_1C_1</math> be the orthic triangle of <math>\triangle ABC. </math> Let <math>N</math> be the circumcenter of <math>\triangle A_1B_1C_1.</math> | Let <math>\triangle A_1B_1C_1</math> be the orthic triangle of <math>\triangle ABC. </math> Let <math>N</math> be the circumcenter of <math>\triangle A_1B_1C_1.</math> | ||
Let <math>\triangle A'B'C'</math> be the tangencial triangle of <math>\triangle ABC.</math> Let <math>O'</math> be the circumcenter of <math>\triangle A'B'C'.</math> | Let <math>\triangle A'B'C'</math> be the tangencial triangle of <math>\triangle ABC.</math> Let <math>O'</math> be the circumcenter of <math>\triangle A'B'C'.</math> | ||
Line 621: | Line 718: | ||
==Exeter point X(22)== | ==Exeter point X(22)== | ||
− | [[File:Exeter X22.png| | + | [[File:Exeter X22.png|400px|right]] |
− | Exeter point is the perspector of the circummedial triangle <math>A_0B_0C_0</math> and the tangential triangle <math>A'B'C'.</math> By another words, let <math>\triangle ABC</math> be the reference triangle (other than a right triangle). Let the medians through the vertices <math>A, B, C</math> meet the circumcircle <math>\Omega</math> of triangle <math>ABC</math> at <math>A_0, B_0,</math> and <math>C_0</math> respectively. Let <math>A'B'C'</math> be the triangle formed by the tangents at <math>A, B,</math> and <math>C</math> to <math>\Omega.</math> (Let <math>A'</math> be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent, the point of concurrence lies on Euler line of triangle <math>ABC,</math> the point of concurrence <math>X_{22}</math> lies on Euler line of triangle <math>ABC, \vec {X_{22}} = \vec O + \frac {2}{J^ | + | Exeter point is the perspector of the circummedial triangle <math>A_0B_0C_0</math> and the tangential triangle <math>A'B'C'.</math> By another words, let <math>\triangle ABC</math> be the reference triangle (other than a right triangle). Let the medians through the vertices <math>A, B, C</math> meet the circumcircle <math>\Omega</math> of triangle <math>ABC</math> at <math>A_0, B_0,</math> and <math>C_0</math> respectively. Let <math>A'B'C'</math> be the triangle formed by the tangents at <math>A, B,</math> and <math>C</math> to <math>\Omega.</math> (Let <math>A'</math> be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent, the point of concurrence lies on Euler line of triangle <math>ABC,</math> the point of concurrence <math>X_{22}</math> lies on Euler line of triangle <math>ABC, \vec {X_{22}} = \vec O + \frac {2}{J^2 - 3} (\vec H - \vec O), J = \frac {|OH|}{R},</math> where <math>O</math> - circumcenter, <math>H</math> - orthocenter, <math>R</math> - circumradius. |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
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<math>G = BB_0 \cap AA_0 \implies AG \cdot GA_0 = BG \cdot GB_0 \implies G</math> lies on radical axis of <math>\omega_A</math> and <math>\omega_B.</math> | <math>G = BB_0 \cap AA_0 \implies AG \cdot GA_0 = BG \cdot GB_0 \implies G</math> lies on radical axis of <math>\omega_A</math> and <math>\omega_B.</math> | ||
− | Therefore second crosspoint of <math>\omega_A</math> and <math>\omega_B</math> point <math>D</math> lies on line <math>OG</math> which is the Euler line of <math>\triangle ABC</math> as desired. | + | Therefore second crosspoint of <math>\omega_A</math> and <math>\omega_B</math> point <math>D</math> lies on line <math>OG</math> which is the Euler line of <math>\triangle ABC.</math> |
+ | Point <math>X_{22} = I_{\Omega}(D)</math> lies on the same Euler line as desired. | ||
− | Last we will find the length of <math> | + | Last we will find the length of <math>OX_{22}.</math> |
<cmath>A_1 = BC \cap AA_0 \implies AA_1 \cdot A_1A_0 = BA_1 \cdot CA_1 = \frac {BC^2}{4}.</cmath> | <cmath>A_1 = BC \cap AA_0 \implies AA_1 \cdot A_1A_0 = BA_1 \cdot CA_1 = \frac {BC^2}{4}.</cmath> | ||
− | <cmath>GO \cdot GD =GO \cdot (GO+ OD) = GA_1 \cdot GA_0</cmath> | + | <cmath>GO \cdot GD =GO \cdot (GO + OD) = GA_1 \cdot GA_0</cmath> |
− | <cmath> GA_1 \cdot GA_0 = \frac {AA_1}{3} \cdot ( \frac {AA_1}{3} + A_1A_0) = \frac {AA_1^2}{9} + | + | <cmath>GA_1 \cdot GA_0 = \frac {AA_1}{3} \cdot ( \frac {AA_1}{3} + A_1A_0) = \frac {AA_1^2}{9} + |
− | \frac {BC^2}{3 \cdot 4} = \frac {AB^2 + BC^2 +AC^2}{18}= \frac {R^2 | + | \frac {BC^2}{3 \cdot 4} = \frac {AB^2 + BC^2 + AC^2}{18}= \frac {R^2 - GO^2} {2}.</cmath> |
− | <cmath>2GO^2 + 2 GO \cdot OD = R^2 | + | <cmath>2GO^2 + 2 GO \cdot OD = R^2 - GO^2 \implies 2 GO \cdot OD = R^2 - 3GO^2.</cmath> |
− | <cmath> I_{Omega}(D) = | + | <cmath> I_{\Omega}(D) = X_{22} \implies OX_{22} = \frac {R^2} {OD} |
− | = \frac {R^2 \cdot 2 GO}{R^2 | + | = \frac {R^2 \cdot 2 GO}{R^2 - 3 GO^2} = \frac {2 HO}{3 - \frac {HO^2}{R^2}} = \frac {2}{3 - J^2} HO</cmath> as desired. |
<i><b>Mapping theorem</b></i> | <i><b>Mapping theorem</b></i> | ||
− | [[File:Transformation.png| | + | [[File:Transformation.png|400px|right]] |
Let triangle <math>ABC</math> and incircle <math>\omega</math> be given. | Let triangle <math>ABC</math> and incircle <math>\omega</math> be given. | ||
<cmath>D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega.</cmath> | <cmath>D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega.</cmath> | ||
Line 674: | Line 772: | ||
<i><b>Claim (Point on incircle)</b></i> | <i><b>Claim (Point on incircle)</b></i> | ||
− | [[File:Point on incircle.png| | + | [[File:Point on incircle.png|400px|right]] |
Let triangle <math>ABC</math> and incircle <math>\omega</math> be given. | Let triangle <math>ABC</math> and incircle <math>\omega</math> be given. | ||
<cmath>D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega, P \in \omega, F' \in AB,</cmath> | <cmath>D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega, P \in \omega, F' \in AB,</cmath> | ||
Line 690: | Line 788: | ||
We multiply and divide these equations and get: | We multiply and divide these equations and get: | ||
<cmath>PA'^2 = PF' \cdot PE', \frac {PF'}{PE'} = \frac {PF^2}{PE^2}.</cmath> | <cmath>PA'^2 = PF' \cdot PE', \frac {PF'}{PE'} = \frac {PF^2}{PE^2}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Far-out point X(23)== | ||
+ | [[File:Far-out point X23.png|400px|right]] | ||
+ | Let <math>\triangle A'B'C'</math> be the tangential triangle of <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>G, \Omega, O, R,</math> and <math>H</math> be the centroid, circumcircle, circumcenter, circumradius and orthocenter of <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that the second crosspoint of circumcircles of <math>\triangle AA'O, \triangle BB'O,</math> and <math>\triangle CC'O</math> is point <math>X_{23}.</math> Point <math>X_{23}</math> lies on Euler line of <math>\triangle ABC, X_{23} = O + \frac {3}{J^2} (H – O), J = \frac {OH}{R}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>I_{\Omega}</math> the inversion with respect <math>\Omega, A_1, B_1, C_1</math> midpoints of <math>BC, AC, AB.</math> | ||
+ | |||
+ | It is evident that <math>I_{\Omega}(A') = A_1, I_{\Omega}(B') = B_1, I_{\Omega}(C') = C_1.</math> | ||
+ | |||
+ | The inversion of circles <math>AA'O, BB'O, CC'O</math> are lines <math>AA_1, BB_1,CC_1</math> which crosses at point <math>G \implies X_{23} = I_{\Omega}(G).</math> | ||
+ | |||
+ | Therefore point <math>X_{23}</math> lies on Euler line <math>OG</math> of <math>\triangle ABC, OG \cdot OX_{23} = R^2 \implies \frac {OX_{23}} {OH} = \frac {R^2}{OG \cdot OH} = \frac {3}{J^2},</math> as desired. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Symmetric lines== | ||
+ | [[File:H line.png|400px|right]] | ||
+ | Let triangle <math>ABC</math> having the circumcircle <math>\omega</math> be given. | ||
+ | |||
+ | Prove that the lines symmetric to the Euler line with respect <math>BC, AC,</math> and <math>AB</math> are concurrent and the point of concurrence lies on <math>\omega.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | The orthocenter <math>H</math> lies on the Euler line therefore the Euler line is <math>H-line.</math> We use <i><b>H-line Clime</b></i> and finish the proof. | ||
+ | |||
+ | ==H–line Claim== | ||
+ | |||
+ | Let triangle <math>ABC</math> having the orthocenter <math>H</math> and circumcircle <math>\omega</math> be given. Denote <math>H–line</math> any line containing point <math>H.</math> | ||
+ | |||
+ | Let <math>l_A, l_B,</math> and <math>l_C</math> be the lines symmetric to <math>H-line</math> with respect <math>BC, AC,</math> and <math>AB,</math> respectively. | ||
+ | |||
+ | Prove that <math>l_A, l_B,</math> and <math>l_C</math> are concurrent and the point of concurrence lies on <math>\omega.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>D, E,</math> and <math>F</math> be the crosspoints of <math>H–line</math> with <math>AB, AC,</math> and <math>BC,</math> respectively. | ||
+ | |||
+ | WLOG <math>D \in AB, E \in AC.</math> | ||
+ | Let <math>H_A, H_B,</math> and <math>H_C</math> be the points symmetric to <math>H</math> with respect <math>BC, AC,</math> and <math>AB,</math> respectively. | ||
+ | |||
+ | Therefore <math>H_A \in l_A, H_B \in l_B, H_C \in l_C, AH = AH_B = AH_C, BH = BH_A = BH_C, CH = CH_A = CH_B \implies</math> | ||
+ | <cmath>\angle HH_BE = \angle EHH_B = \angle BHD = \angle BH_CD.</cmath> | ||
+ | |||
+ | Let <math>P</math> be the crosspoint of <math>l_B</math> and <math>l_C \implies BH_CH_BP</math> is cyclic <math>\implies P \in \omega.</math> | ||
+ | |||
+ | Similarly <math>\angle CH_BE = \angle CHE = \angle CH_A \implies CH_BH_AP</math> is cyclic <math>\implies P \in \omega \implies</math> the crosspoint of <math>l_B</math> and <math>l_A</math> is point <math>P.</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Line 702: | Line 854: | ||
*[[Kimberling’s point X(25)]] | *[[Kimberling’s point X(25)]] | ||
*[[Kimberling’s point X(26)]] | *[[Kimberling’s point X(26)]] | ||
+ | *[[De Longchamps point]] | ||
+ | *[[Gossard perspector]] | ||
+ | *[[Evans point]] | ||
+ | *[[Double perspective triangles]] | ||
+ | *[[Steiner line]] | ||
+ | *[[Miquel's point]] | ||
+ | *[[Spieker center]] | ||
+ | *[[Simson line]] | ||
+ | *[[Complete Quadrilateral]] | ||
+ | *[[Gauss line]] | ||
+ | *[[Isogonal conjugate]] | ||
+ | *[[Barycentric coordinates]] | ||
+ | *[[Symmetry]] | ||
+ | *[[Symmedian and Antiparallel]] | ||
*[[Central line]] | *[[Central line]] | ||
− | |||
*[[Gergonne line]] | *[[Gergonne line]] | ||
*[[Gergonne point]] | *[[Gergonne point]] | ||
− | + | ||
{{stub}} | {{stub}} |
Latest revision as of 03:47, 30 August 2023
In any triangle , the Euler line is a line which passes through the orthocenter , centroid , circumcenter , nine-point center and de Longchamps point . It is named after Leonhard Euler. Its existence is a non-trivial fact of Euclidean geometry. Certain fixed orders and distance ratios hold among these points. In particular, and
Euler line is the central line .
Given the orthic triangle of , the Euler lines of ,, and concur at , the nine-point circle of .
Contents
[hide]- 1 Proof Centroid Lies on Euler Line
- 2 Another Proof
- 3 Proof Nine-Point Center Lies on Euler Line
- 4 Analytic Proof of Existence
- 5 The points of intersection of the Euler line with the sides of the triangle
- 6 Angles between Euler line and the sides of the triangle
- 7 Distances along Euler line
- 8 Position of Kimberling centers on the Euler line
- 9 Triangles with angles of or
- 10 Euler lines of cyclic quadrilateral (Vittas’s theorem)
- 11 Concurrent Euler lines and Fermat points
- 12 Euler line of Gergonne triangle
- 13 Thebault point
- 14 Schiffler point
- 15 Euler line as radical axis
- 16 De Longchamps point X(20)
- 17 De Longchamps line
- 18 CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)
- 19 PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)
- 20 Exeter point X(22)
- 21 Far-out point X(23)
- 22 Symmetric lines
- 23 H–line Claim
- 24 See also
Proof Centroid Lies on Euler Line
This proof utilizes the concept of spiral similarity, which in this case is a rotation followed homothety. Consider the medial triangle . It is similar to . Specifically, a rotation of about the midpoint of followed by a homothety with scale factor centered at brings . Let us examine what else this transformation, which we denote as , will do.
It turns out is the orthocenter, and is the centroid of . Thus, . As a homothety preserves angles, it follows that . Finally, as it follows that Thus, are collinear, and .
Another Proof
Let be the midpoint of . Extend past to point such that . We will show is the orthocenter. Consider triangles and . Since , and they both share a vertical angle, they are similar by SAS similarity. Thus, , so lies on the altitude of . We can analogously show that also lies on the and altitudes, so is the orthocenter.
Proof Nine-Point Center Lies on Euler Line
Assuming that the nine point circle exists and that is the center, note that a homothety centered at with factor brings the Euler points onto the circumcircle of . Thus, it brings the nine-point circle to the circumcircle. Additionally, should be sent to , thus and .
Analytic Proof of Existence
Let the circumcenter be represented by the vector , and let vectors correspond to the vertices of the triangle. It is well known the that the orthocenter is and the centroid is . Thus, are collinear and
The points of intersection of the Euler line with the sides of the triangle
Acute triangle
Let be the acute triangle where Denote
Let Euler line cross lines and in points and respectively.
Then point lyes on segment
Point lyes on segment
Point lyes on ray
Proof
Denote
We use the formulae (see Claim “Segments crossing inside triangle” in “Schiffler point” in “Euler line”).
Centroid lyes on median
Orthocenter lyes on altitude Therefore We use the signed version of Menelaus's theorem and get
Obtuse triangle
Let be the obtuse triangle where
Let Euler line cross lines and in points and respectively.
Similarly we get
ray
Right triangle
Let be the right triangle where Then Euler line contain median from vertex
Isosceles triangle
Let be the isosceles triangle where Then Euler line contain median from vertex
Corollary: Euler line is parallel to side
Euler line is parallel to side iff
Proof
After simplification in the case we get
vladimir.shelomovskii@gmail.com, vvsss
Angles between Euler line and the sides of the triangle
Let Euler line of the cross lines and in points and respectively. Denote smaller angles between the Euler line and lines and as and respectively.
Prove that
Proof
WLOG,
Let be the midpoint be the circumcenter of
Symilarly, for other angles.
vladimir.shelomovskii@gmail.com, vvsss
Distances along Euler line
Let and be orthocenter, centroid, circumcenter, and circumradius of the respectively.
Prove that
Proof
WLOG, is an acute triangle,
vladimir.shelomovskii@gmail.com, vvsss
Position of Kimberling centers on the Euler line
Let triangle ABC be given. Let and are orthocenter, circumcenter, circumradius and inradius, respectively.
We use point as origin and as a unit vector.
We find Kimberling center X(I) on Euler line in the form of For a lot of Kimberling centers the coefficient is a function of only two parameters and
Centroid Nine-point center de Longchamps point Schiffler point Exeter point Far-out point Perspector of ABC and orthic-of-orthic triangle Homothetic center of orthic and tangential triangles Circumcenter of the tangential triangle
Midpoint of X(3) and vladimir.shelomovskii@gmail.com, vvsss
Triangles with angles of or
Claim 1
Let the in triangle be Then the Euler line of the is parallel to the bisector of
Proof
Let be circumcircle of
Let be circumcenter of
Let be the circle symmetric to with respect to
Let be the point symmetric to with respect to
The lies on lies on
is the radius of and translation vector to is
Let be the point symmetric to with respect to Well known that lies on Therefore point lies on
Point lies on
Let be the bisector of are concurrent.
Euler line of the is parallel to the bisector of as desired.
Claim 2
Let the in triangle be Then the Euler line of the is perpendicular to the bisector of
Proof
Let be circumcircle, circumcenter, orthocenter and incenter of the points are concyclic.
The circle centered at midpoint of small arc
is rhomb.
Therefore the Euler line is perpendicular to as desired.
Claim 3
Let be a quadrilateral whose diagonals and intersect at and form an angle of If the triangles PAB, PBC, PCD, PDA are all not equilateral, then their Euler lines are pairwise parallel or coincident.
Proof
Let and be internal and external bisectors of the angle .
Then Euler lines of and are parallel to and Euler lines of and are perpendicular to as desired.
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Euler lines of cyclic quadrilateral (Vittas’s theorem)
Claim 1
Let be a cyclic quadrilateral with diagonals intersecting at The Euler lines of triangles are concurrent.
Proof
Let be the circumcenters (orthocenters) of triangles Let be the common bisector of and Therefore and are parallelograms with parallel sides.
bisect these angles. So points are collinear and lies on one straight line which is side of the pare vertical angles and Similarly, points are collinear and lies on another side of these angles. Similarly obtuse so points and are collinear and lies on one side and points and are collinear and lies on another side of the same vertical angles.
We use Claim and get that lines are concurrent (or parallel if or ).
Claim 2 (Property of vertex of two parallelograms)
Let and be parallelograms, Let lines and be concurrent at point Then points and are collinear and lines and are concurrent.
Proof
We consider only the case Shift transformation allows to generalize the obtained results.
We use the coordinate system with the origin at the point and axes
We use and get points and are colinear.
We calculate point of crossing and and and and get the same result: as desired (if then point moves to infinity and lines are parallel, angles or
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Concurrent Euler lines and Fermat points
Consider a triangle with Fermat–Torricelli points and The Euler lines of the triangles with vertices chosen from and are concurrent at the centroid of triangle We denote centroids by , circumcenters by We use red color for points and lines of triangles green color for triangles and blue color for triangles
Case 1
Let be the first Fermat point of maximum angle of which smaller then Then the centroid of triangle lies on Euler line of the The pairwise angles between these Euler lines are equal
Proof
Let and be centroid, circumcenter, and circumcircle of respectevely.
Let be external for equilateral triangle is cyclic.
Point is centroid of Points and are colinear, so point lies on Euler line of
Case 2
Let be the first Fermat point of
Then the centroid of triangle lies on Euler lines of the triangles and The pairwise angles between these Euler lines are equal
Proof
Let be external for equilateral triangle, be circumcircle of is cyclic.
Point is centroid of
Points and are colinear, so point lies on Euler line of as desired.
Case 3
Let be the second Fermat point of Then the centroid of triangle lies on Euler lines of the triangles and
The pairwise angles between these Euler lines are equal
Proof
Let be internal for equilateral triangle, be circumcircle of
Let and be circumcenters of the triangles and Point is centroid of the is the Euler line of the parallel to
is bisector of is bisector of is bisector of is regular triangle.
is the inner Napoleon triangle of the is centroid of this regular triangle.
points and are collinear as desired.
Similarly, points and are collinear.
Case 4
Let and be the Fermat points of Then the centroid of point lies on Euler line is circumcenter, is centroid) of the
Proof
Step 1 We will find line which is parallel to
Let be midpoint of Let be the midpoint of
Let be point symmetrical to with respect to
as midline of
Step 2 We will prove that line is parallel to
Let be the inner Napoleon triangle. Let be the outer Napoleon triangle. These triangles are regular centered at
Points and are collinear (they lies on bisector
Points and are collinear (they lies on bisector
Points and are collinear (they lies on bisector angle between and is
Points and are concyclic Points and are concyclic
points and are concyclic
Therefore and are collinear or point lies on Euler line
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Euler line of Gergonne triangle
Prove that the Euler line of Gergonne triangle of passes through the circumcenter of triangle
Gergonne triangle is also known as the contact triangle or intouch triangle. If the inscribed circle touches the sides of at points and then is Gergonne triangle of .
Other wording: Tangents to circumcircle of are drawn at the vertices of the triangle. Prove that the circumcenter of the triangle formed by these three tangents lies on the Euler line of the original triangle.
Proof
Let and be orthocenter and circumcenter of respectively. Let be Orthic Triangle of
Then is Euler line of is the incenter of is the incenter of
Similarly,
where is the perspector of triangles and
Under homothety with center P and coefficient the incenter of maps into incenter of , circumcenter of maps into circumcenter of are collinear as desired.
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Thebault point
Let and be the altitudes of the where
a) Prove that the Euler lines of triangles are concurrent on the nine-point circle at a point T (Thebault point of )
b) Prove that if then else
Proof
Case 1 Acute triangle
a) It is known, that Euler line of acute triangle cross AB and BC (shortest and longest sides) in inner points.
Let be circumcenters of
Let and be centroids of
Denote is the circle (the nine-points circle).
is the midpoint where is the orthocenter of
Similarly
is the midline of
Let cross at point different from
spiral similarity centered at maps onto
This similarity has the rotation angle acute angle between Euler lines of these triangles is
Let these lines crossed at point Therefore points and are concyclic
Similarly, as desired.
b) Point lies on median of and divide it in ratio 2 : 1.
Point lies on Euler line of
According the Claim,
Similarly
Case 2 Obtuse triangle
a) It is known, that Euler line of obtuse cross AC and BC (middle and longest sides) in inner points.
Let be circumcenters of
Let and be centroids of
Denote is the circle (the nine-points circle).
is the midpoint where is the orthocenter of
Similarly
is the midline of
Let cross at point different from
spiral similarity centered at maps onto
This similarity has the rotation angle acute angle between Euler lines of these triangles is
Let these lines crossed at point Therefore points and are concyclic
Similarly, as desired.
b)
Point lies on median of and divide it in ratio
Point lies on Euler line of According the Claim,
Similarly
Claim (Segment crossing the median)
Let be the midpoint of side of the
Then
Proof
Let be (We use sign to denote the area of
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Schiffler point
Let and be the incenter, circumcenter, centroid, circumradius, and inradius of respectively. Then the Euler lines of the four triangles and are concurrent at Schiffler point .
Proof
We will prove that the Euler line of cross the Euler line of at such point that .
Let and be the circumcenter and centroid of respectively.
It is known that lies on circumcircle of
Denote
It is known that is midpoint point lies on median points belong the bisector of
Easy to find that ,
We use sigh [t] for area of t. We get
Using Claim we get Therefore each Euler line of triangles cross Euler line of in the same point, as desired.
Claim (Segments crossing inside triangle)
Given triangle GOY. Point lies on
Point lies on
Point lies on
Point lies on Then
Proof
Let be (We use sigh for area of vladimir.shelomovskii@gmail.com, vvsss
Euler line as radical axis
Let with altitudes and be given.
Let and be circumcircle, circumcenter, orthocenter and circumradius of respectively.
Circle centered at passes through and is tangent to the radius AO. Similarly define circles and
Then Euler line of is the radical axis of these circles.
If is acute, then these three circles intersect at two points located on the Euler line of the
Proof
The power of point with respect to and is
The power of point with respect to is
The power of point with respect to is
The power of point with respect to is
It is known that
Therefore points and lies on radical axis of these three circles as desired.
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De Longchamps point X(20)
Definition 1
The De Longchamps’ point of a triangle is the radical center of the power circles of the triangle. Prove that De Longchamps point lies on Euler line.
We call A-power circle of a the circle centered at the midpoint point with radius The other two circles are defined symmetrically.
Proof
Let and be orthocenter, circumcenter, and De Longchamps point, respectively.
Denote power circle by power circle by WLOG,
Denote the projection of point on
We will prove that radical axes of power and power cicles is symmetric to altitude with respect Further, we will conclude that the point of intersection of the radical axes, symmetrical to the heights with respect to O, is symmetrical to the point of intersection of the heights with respect to
Point is the crosspoint of the center line of the power and power circles and there radical axis. We use claim and get:
and are the medians, so
We use Claim some times and get: radical axes of power and power cicles is symmetric to altitude with respect
Similarly radical axes of power and power cicles is symmetric to altitude radical axes of power and power cicles is symmetric to altitude with respect Therefore the point of intersection of the radical axes, symmetrical to the heights with respect to is symmetrical to the point of intersection of the heights with respect to lies on Euler line of
Claim (Distance between projections)
Definition 2
We call circle of a the circle centered at with radius The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of circle, circle, and circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under Definition 1.
Proof
Let and be orthocenter, centroid, and De Longchamps point, respectively. Let cross at points and The other points are defined symmetrically. Similarly is diameter
Therefore is anticomplementary triangle of is orthic triangle of So is orthocenter of
as desired.
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De Longchamps line
The de Longchamps line of is defined as the radical axes of the de Longchamps circle and of the circumscribed circle of
Let be the circumcircle of (the anticomplementary triangle of
Let be the circle centered at (centroid of ) with radius where
Prove that the de Longchamps line is perpendicular to Euler line and is the radical axes of and
Proof
Center of is , center of is where is Euler line. The homothety with center and ratio maps into This homothety maps into and there is two inversion which swap and
First inversion centered at point Let be the point of crossing and
The radius of we can find using
Second inversion centered at point We can make the same calculations and get as desired.
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CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)
Prove that the circumcenter of the tangential triangle of (Kimberling’s point lies on the Euler line of
Proof
Let and be midpoints of and respectively.
Let be circumcircle of It is nine-points circle of the
Let be circumcircle of Let be circumcircle of
and are tangents to inversion with respect swap and Similarly, this inversion swap and and Therefore this inversion swap and
The center of and the center of lies on Euler line, so the center of lies on this line, as desired.
After some calculations one can find position of point on Euler line (see Kimberling's point
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PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)
Let be the orthic triangle of Let be the circumcenter of Let be the tangencial triangle of Let be the circumcenter of
Prove that lines and are concurrent at point, lies on Euler line of
Proof
and are antiparallel to BC with respect
Similarly,
Therefore homothetic center of and is the point of concurrence of lines and Denote this point as
The points and are the corresponding points (circumcenters) of and so point lies on line
Points and lies on Euler line, so lies on Euler line of
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Exeter point X(22)
Exeter point is the perspector of the circummedial triangle and the tangential triangle By another words, let be the reference triangle (other than a right triangle). Let the medians through the vertices meet the circumcircle of triangle at and respectively. Let be the triangle formed by the tangents at and to (Let be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through and are concurrent, the point of concurrence lies on Euler line of triangle the point of concurrence lies on Euler line of triangle where - circumcenter, - orthocenter, - circumradius.
Proof
At first we prove that lines and are concurrent. This follows from the fact that lines and are concurrent at point and Mapping theorem.
Let and be the midpoints of and respectively. The points and are collinear. Similarly the points and are collinear.
Denote the inversion with respect It is evident that
Denote
The power of point with respect is
Similarly the power of point with respect is
lies on radical axis of and
Therefore second crosspoint of and point lies on line which is the Euler line of Point lies on the same Euler line as desired.
Last we will find the length of as desired.
Mapping theorem
Let triangle and incircle be given. Let be the point in the plane Let lines and crossing second time at points and respectively.
Prove that lines and are concurrent.
Proof
We use Claim and get: Similarly,
We use the trigonometric form of Ceva's Theorem for point and triangle and get We use the trigonometric form of Ceva's Theorem for triangle and finish proof that lines and are concurrent.
Claim (Point on incircle)
Let triangle and incircle be given. Prove that
Proof
Similarly
We multiply and divide these equations and get:
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Far-out point X(23)
Let be the tangential triangle of
Let and be the centroid, circumcircle, circumcenter, circumradius and orthocenter of
Prove that the second crosspoint of circumcircles of and is point Point lies on Euler line of
Proof
Denote the inversion with respect midpoints of
It is evident that
The inversion of circles are lines which crosses at point
Therefore point lies on Euler line of as desired.
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Symmetric lines
Let triangle having the circumcircle be given.
Prove that the lines symmetric to the Euler line with respect and are concurrent and the point of concurrence lies on
Proof
The orthocenter lies on the Euler line therefore the Euler line is We use H-line Clime and finish the proof.
H–line Claim
Let triangle having the orthocenter and circumcircle be given. Denote any line containing point
Let and be the lines symmetric to with respect and respectively.
Prove that and are concurrent and the point of concurrence lies on
Proof
Let and be the crosspoints of with and respectively.
WLOG Let and be the points symmetric to with respect and respectively.
Therefore
Let be the crosspoint of and is cyclic
Similarly is cyclic the crosspoint of and is point
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See also
- Kimberling center
- Kimberling’s point X(20)
- Kimberling’s point X(21)
- Kimberling’s point X(22)
- Kimberling’s point X(23)
- Kimberling’s point X(24)
- Kimberling’s point X(25)
- Kimberling’s point X(26)
- De Longchamps point
- Gossard perspector
- Evans point
- Double perspective triangles
- Steiner line
- Miquel's point
- Spieker center
- Simson line
- Complete Quadrilateral
- Gauss line
- Isogonal conjugate
- Barycentric coordinates
- Symmetry
- Symmedian and Antiparallel
- Central line
- Gergonne line
- Gergonne point
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