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''' | ||
+ | |||
+ | ==Position of Kimberling centers on the Euler line== | ||
+ | [[File:Kimberling points on Euler line.png|450px|right]] | ||
+ | Let triangle ABC be given. Let <math>H = X(4), O = X(3), R</math> and <math>r</math> are orthocenter, circumcenter, circumradius and inradius, respectively. | ||
+ | |||
+ | We use point <math>\vec O = \vec X(3)</math> as origin and <math>\vec {HO}</math> as a unit vector. | ||
+ | |||
+ | We find Kimberling center X(I) on Euler line in the form of | ||
+ | <cmath>\vec X(I) = \vec O + k_i \cdot \vec {OH}.</cmath> | ||
+ | For a lot of Kimberling centers the coefficient <math>k_i</math> is a function of only two parameters <math>J = \frac {|OH|}{R}</math> and <math>t = \frac {r}{R}.</math> | ||
+ | |||
+ | Centroid <math>X(2)</math> | ||
+ | <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> | ||
+ | <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> | ||
+ | <cmath>X(20) = X(3) - (X(4) - X(3)) \implies k_{20} = - 1.</cmath> | ||
+ | Schiffler point <math>X(21)</math> | ||
+ | <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> | ||
+ | <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> | ||
+ | <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> | ||
+ | <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> | ||
+ | <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> | ||
+ | <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> | ||
+ | <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''' | ||
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'''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 483: | 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 498: | 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 548: | 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 568: | 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 585: | Line 714: | ||
Points <math>N</math> and <math>O' = X(26)</math> lies on Euler line, so <math>K</math> lies on Euler line of <math>\triangle ABC.</math> | Points <math>N</math> and <math>O' = X(26)</math> lies on Euler line, so <math>K</math> lies on Euler line of <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Exeter point X(22)== | ||
+ | [[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^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> | ||
+ | |||
+ | At first we prove that lines <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent. This follows from the fact that lines <math>AA_0, BB_0,</math> and <math>CC_0</math> are concurrent at point <math>G</math> and <i><b>Mapping theorem</b></i>. | ||
+ | |||
+ | Let <math>A_1, B_1,</math> and <math>C_1</math> be the midpoints of <math>BC, AC,</math> and <math>AB,</math> respectively. The points <math>A, G, A_1,</math> and <math>A_0</math> are collinear. Similarly the points <math>B, G, B_1,</math> and <math>B_0</math> are collinear. | ||
+ | |||
+ | Denote <math>I_{\Omega}</math> the inversion with respect <math>\Omega.</math> It is evident that <math>I_{\Omega}(A_0) = A_0, I_{\Omega}(A') = A_1, I_{\Omega}(B_0) = B_0, I_{\Omega}(B') = B_1.</math> | ||
+ | |||
+ | Denote <math>\omega_A = I_{\Omega}(A'A_0), \omega_B = I_{\Omega}(B'B_0) \implies</math> | ||
+ | <cmath>A_0 \in \omega_A, A_1 \in \omega_A, O \in \omega_A, B_0 \in \omega_B, B_1 \in \omega_B, O \in \omega_B \implies O = \omega_A \cap \omega_B.</cmath> | ||
+ | |||
+ | The power of point <math>G</math> with respect <math>\omega_A</math> is <math>GA_1 \cdot GA_0 = \frac {1}{2} AG \cdot GA_0.</math> | ||
+ | |||
+ | Similarly the power of point <math>G</math> with respect <math>\omega_B</math> is <math>GB_1 \cdot GB_0 = \frac {1}{2} BG \cdot GB_0.</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> | ||
+ | Point <math>X_{22} = I_{\Omega}(D)</math> lies on the same Euler line as desired. | ||
+ | |||
+ | 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>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} + | ||
+ | \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 - GO^2 \implies 2 GO \cdot OD = R^2 - 3GO^2.</cmath> | ||
+ | <cmath> I_{\Omega}(D) = X_{22} \implies OX_{22} = \frac {R^2} {OD} | ||
+ | = \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> | ||
+ | [[File:Transformation.png|400px|right]] | ||
+ | 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> | ||
+ | Let <math>P</math> be the point in the plane <math>ABC.</math> | ||
+ | Let lines <math>DP, EP,</math> and <math>FP</math> crossing <math>\omega</math> second time at points <math>D_0, E_0,</math> and <math>F_0,</math> respectively. | ||
+ | |||
+ | Prove that lines <math>AD_0, BE_0, </math> and <math>CF_0</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>k_A = \frac {\sin {D_0AE'}}{\sin {D_0AF'}} = \frac {D_0E'}{D_0A} \cdot \frac {D_0A}{D_0F'} = \frac {D_0E'}{D_0F'}.</cmath> | ||
+ | We use Claim and get: <math>k_A = \frac {D_0E^2}{D_0F^2}.</math> | ||
+ | <cmath>k_D = \frac {\sin {D_0DE}}{\sin {D_0DF}} = \frac {D_0E}{2R} \cdot \frac {2R}{D_0F} = \frac {D_0E}{D_0F} \implies k_A = k_D^2.</cmath> | ||
+ | Similarly, <math>k_B = k_E^2, k_C = k_F^2.</math> | ||
+ | |||
+ | We use the trigonometric form of Ceva's Theorem for point <math>P</math> and triangle <math>\triangle DEF</math> and get | ||
+ | <cmath>k_D \cdot k_E \cdot k_F = 1 \implies k_A \cdot k_B \cdot k_C = 1^2 = 1.</cmath> | ||
+ | We use the trigonometric form of Ceva's Theorem for triangle <math>\triangle ABC</math> and finish proof that lines <math>AD_0, BE_0, </math> and <math>CF_0</math> are concurrent. | ||
+ | |||
+ | |||
+ | <i><b>Claim (Point on incircle)</b></i> | ||
+ | [[File:Point on incircle.png|400px|right]] | ||
+ | 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>PF' \perp AB, E' \in AC, PE' \perp AC, A' \in EF, PA' \perp EF.</cmath> | ||
+ | Prove that <math>\frac {PF'}{PE'} = \frac {PF^2}{PE^2}, PA'^2 = PF' \cdot PE'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>AF = AE \implies \angle AFE = \angle AEF = \angle A'EE'.</cmath> | ||
+ | <cmath>\angle EFP = \angle PEE' \implies \angle PFF' = \angle PEE' \implies</cmath> | ||
+ | <cmath>\triangle PFF' \sim \triangle PEA' \implies \frac {PF}{PF'} = \frac {PE}{PA'}.</cmath> | ||
+ | |||
+ | Similarly <math>\triangle PEE' \sim \triangle PFA' \implies \frac {PE}{PE'} = \frac {PF}{PA'}.</math> | ||
+ | |||
+ | We multiply and divide these equations and get: | ||
+ | <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 590: | Line 847: | ||
==See also== | ==See also== | ||
*[[Kimberling center]] | *[[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(24)]] | ||
*[[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 04:47, 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
- 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
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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.
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Distances along Euler line
Let and
be orthocenter, centroid, circumcenter, and circumradius of the
respectively.
Prove that
Proof
WLOG, is an acute triangle,
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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
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Triangles with angles of
or ![$120^\circ$](//latex.artofproblemsolving.com/f/3/f/f3fd424e953ac76b50bf6103875acc049d21d3c2.png)
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
Denote
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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
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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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