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== | ||
+ | [[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. | ||
+ | <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 - 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> | ||
+ | |||
+ | 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 - \beta \implies</cmath> | ||
+ | <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 - 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 - \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 - 4 \cos \alpha \cos (\beta - \gamma) = 5 - 4 \sin^2 \alpha + 4 \cos (\beta + \gamma) \cos (\beta - \gamma)</cmath> | ||
+ | <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 - 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''' | ||
==Triangles with angles of <math>60^\circ</math> or <math>120^\circ</math>== | ==Triangles with angles of <math>60^\circ</math> or <math>120^\circ</math>== | ||
<i><b>Claim 1</b></i> | <i><b>Claim 1</b></i> | ||
− | [[File:120 orthocenter.png| | + | [[File:120 orthocenter.png|260px|right]] |
Let the <math>\angle C</math> in triangle <math>ABC</math> be <math>120^\circ.</math> Then the Euler line of the <math>\triangle ABC</math> is parallel to the bisector of <math>\angle C.</math> | Let the <math>\angle C</math> in triangle <math>ABC</math> be <math>120^\circ.</math> Then the Euler line of the <math>\triangle ABC</math> is parallel to the bisector of <math>\angle C.</math> | ||
Line 57: | Line 202: | ||
<i><b>Claim 2</b></i> | <i><b>Claim 2</b></i> | ||
− | [[File:Euler line 60.png| | + | [[File:Euler line 60.png|260px|right]] |
+ | [[File:4 Euler lines 60.png|260px|right]] | ||
Let the <math>\angle C</math> in triangle <math>ABC</math> be <math>60^\circ.</math> Then the Euler line of the <math>\triangle ABC</math> is perpendicular to the bisector of <math>\angle C.</math> | Let the <math>\angle C</math> in triangle <math>ABC</math> be <math>60^\circ.</math> Then the Euler line of the <math>\triangle ABC</math> is perpendicular to the bisector of <math>\angle C.</math> | ||
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<i><b>Claim 3</b></i> | <i><b>Claim 3</b></i> | ||
− | + | ||
Let <math>ABCD</math> be a quadrilateral whose diagonals <math>AC</math> and <math>BD</math> intersect at <math>P</math> and form an angle of <math>60^\circ.</math> If the triangles PAB, PBC, PCD, PDA are all not equilateral, then their Euler lines are pairwise parallel or coincident. | Let <math>ABCD</math> be a quadrilateral whose diagonals <math>AC</math> and <math>BD</math> intersect at <math>P</math> and form an angle of <math>60^\circ.</math> If the triangles PAB, PBC, PCD, PDA are all not equilateral, then their Euler lines are pairwise parallel or coincident. | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Euler lines of cyclic quadrilateral (Vittas’s theorem)== | ||
+ | <i><b>Claim 1</b></i> | ||
+ | [[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> The Euler lines of triangles <math>\triangle APB, \triangle BPC, \triangle CPD, \triangle DPA</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>O_1, O_2, O_3, O_4 (H_1, H_2, H_3, H_4)</math> be the circumcenters (orthocenters) of triangles <math>\triangle APD, \triangle CPD, \triangle BPC, \triangle APB.</math> | ||
+ | Let <math>I_2I_4</math> be the common bisector of <math>\angle APB</math> and <math>\angle CPD.</math> | ||
+ | <cmath>O_1O_4 \perp AC, H_3H_4 \perp AC, O_2O_3 \perp AC, H_1H_2 \perp AC \implies</cmath> | ||
+ | <cmath>O_1O_4 || H_3H_4 || O_2O_3 || H_1H_2.</cmath> | ||
+ | <cmath>O_1O_2 \perp BD, H_2H_3 \perp BD, O_3O_4 \perp BD, H_1H_4 \perp BD \implies</cmath> | ||
+ | <cmath>O_1O_2 || H_2H_3 || O_3O_4 || H_1H_4 .</cmath> | ||
+ | Therefore <math>O_1O_2O_3O_4</math> and <math>H_1H_2H_3H_4</math> are parallelograms with parallel sides. | ||
+ | |||
+ | <math>\triangle APB \sim \triangle DPC \implies \angle O_2PH_2 = \angle O_4PH_4 \implies I_2I_4</math> bisect these angles. | ||
+ | So points <math>O_2, P, H_4</math> are collinear and lies on one straight line which is side of the pare vertical angles <math> \angle O_2PH_2</math> and <math>\angle O_4PH_4.</math> Similarly, points <math>O_4, P, H_2</math> are collinear and lies on another side of these angles. | ||
+ | Similarly obtuse <math>\triangle APD \sim \triangle BPC</math> so points <math>H_1, P</math> and <math>O_3</math> are collinear and lies on one side and points <math>H_3, P</math> and <math>O_1</math> are collinear and lies on another side of the same vertical angles. | ||
+ | |||
+ | 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 2 (Property of vertex of two parallelograms)</b></i> | ||
+ | [[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. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | We consider only the case <math>AB \perp AD.</math> Shift transformation allows to generalize the obtained results. | ||
+ | |||
+ | We use the coordinate system with the origin at the point <math>O,</math> and axes <math>Ox||AD, Oy||BA.</math> | ||
+ | |||
+ | We use <math>x_A, y_A, y_B, x_E, y_F</math> and get | ||
+ | <math>x_B = x_A, y_E = \frac {x_E \cdot y_A}{x_A}, x_F = x_E, y_H = y_E, x_H=\frac {x_E \cdot y_A}{y_B}, y_D = y_A,</math> | ||
+ | <cmath>x_D=\frac {x_E \cdot y_A}{y_F}, x_C=x_D, y_C = y_B, x_G=\frac {x_E \cdot y_A}{y_B}, y_G = y_F \implies</cmath> | ||
+ | <math>\frac {y_C}{x_C}=\frac {y_G}{x_G} \implies</math> points <math>C, O,</math> and <math>G</math> are colinear. | ||
+ | |||
+ | We calculate point of crossing <math>AG</math> and <math>BF, AG</math> and <math>DH, AG</math> and <math>CE</math> and get the same result: | ||
+ | <cmath>x_I = \frac {y_A+ y_B – y_F – {\frac {x_E \cdot y_A}{x_A}}} {\frac{y_B}{x_E}– \frac {y_F}{x_A}}, | ||
+ | y_I = \frac {\frac{y_A \cdot y_B}{x_A}+ \frac{y_B \cdot y_F}{x_E}– \frac{y_A \cdot y_F}{x_A} – \frac {y_B \cdot y_F}{x_A}} {\frac{y_B}{x_E} – \frac {y_F}{x_A}}</cmath> as desired (if <math>\frac{y_B}{x_E}= \frac {y_F}{x_A}</math> then point <math>I</math> moves to infinity and lines are parallel, angles <math>\angle APD = 60^\circ</math> or <math>\angle APD = 120^\circ).</math> | ||
+ | |||
+ | '''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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Let <math>G', O,</math> and <math>\omega</math> be centroid, circumcenter, and circumcircle of <math>\triangle ABF,</math> respectevely. | Let <math>G', O,</math> and <math>\omega</math> be centroid, circumcenter, and circumcircle of <math>\triangle ABF,</math> respectevely. | ||
+ | |||
Let <math>\triangle ABD</math> be external for <math>\triangle ABC</math> equilateral triangle <math>\implies F = CD \cap \omega.</math> | Let <math>\triangle ABD</math> be external for <math>\triangle ABC</math> equilateral triangle <math>\implies F = CD \cap \omega.</math> | ||
<math>\angle AFB = 120^\circ \implies AFBD</math> is cyclic. | <math>\angle AFB = 120^\circ \implies AFBD</math> is cyclic. | ||
Line 108: | Line 298: | ||
<i><b>Case 2</b></i> | <i><b>Case 2</b></i> | ||
[[File:Fermat 1 150 Euler lines 3.png|500px|right]] | [[File:Fermat 1 150 Euler lines 3.png|500px|right]] | ||
− | Let <math>F</math> be the first Fermat point of <math>\triangle ABC, \angle BAC > 120^\circ.</math> Then the centroid <math>G</math> of triangle <math>ABC</math> lies on Euler lines of the triangles <math>\triangle ABF,\triangle ACF,</math> and <math>\triangle BCF.</math> The pairwise angles between these Euler lines are equal <math>60^\circ.</math> | + | Let <math>F</math> be the first Fermat point of <math>\triangle ABC, \angle BAC > 120^\circ.</math> |
+ | |||
+ | Then the centroid <math>G</math> of triangle <math>ABC</math> lies on Euler lines of the triangles <math>\triangle ABF,\triangle ACF,</math> and <math>\triangle BCF.</math> The pairwise angles between these Euler lines are equal <math>60^\circ.</math> | ||
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
Line 122: | Line 314: | ||
<i><b>Case 3</b></i> | <i><b>Case 3</b></i> | ||
[[File:Euler line used F2.png|400px|right]] | [[File:Euler line used F2.png|400px|right]] | ||
− | Let <math>F'</math> be the second Fermat point of <math>\triangle ABC.</math> Then the centroid <math>G</math> of triangle <math>ABC</math> lies on Euler lines of the triangles <math>\triangle ABF',\triangle ACF',</math> and <math>\triangle BCF'.</math> The pairwise angles between these Euler lines are equal <math>60^\circ.</math> | + | Let <math>F'</math> be the second Fermat point of <math>\triangle ABC.</math> Then the centroid <math>G</math> of triangle <math>ABC</math> lies on Euler lines of the triangles <math>\triangle ABF',\triangle ACF',</math> and <math>\triangle BCF'.</math> |
+ | |||
+ | The pairwise angles between these Euler lines are equal <math>60^\circ.</math> | ||
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
Line 146: | Line 340: | ||
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
− | <i><b>Step 1 | + | <i><b>Step 1</b></i> We will find line <math>F'D</math> which is parallel to <math>GG'.</math> |
Let <math>M</math> be midpoint of <math>BC.</math> Let <math>M'</math> be the midpoint of <math>FF'.</math> | Let <math>M</math> be midpoint of <math>BC.</math> Let <math>M'</math> be the midpoint of <math>FF'.</math> | ||
Line 152: | Line 346: | ||
Let <math>D</math> be point symmetrical to <math>F</math> with respect to <math>M.</math> | Let <math>D</math> be point symmetrical to <math>F</math> with respect to <math>M.</math> | ||
− | <math> | + | <math>M'M||F'D</math> as midline of <math>\triangle FF'D.</math> |
<cmath>\vec {G'G} = \frac {\vec A + \vec B + \vec C}{3} – \frac {\vec A + \vec F + \vec F'}{3} = \frac {2}{3} \cdot (\frac {\vec B + \vec C}{2} – \frac {\vec F + \vec F'}{3})</cmath> | <cmath>\vec {G'G} = \frac {\vec A + \vec B + \vec C}{3} – \frac {\vec A + \vec F + \vec F'}{3} = \frac {2}{3} \cdot (\frac {\vec B + \vec C}{2} – \frac {\vec F + \vec F'}{3})</cmath> | ||
<cmath>\vec {G'G} = \frac {2}{3} (\vec M – \vec M') = \frac {2}{3} \vec {M'M} \implies M'M||F'D||G'G.</cmath> | <cmath>\vec {G'G} = \frac {2}{3} (\vec M – \vec M') = \frac {2}{3} \vec {M'M} \implies M'M||F'D||G'G.</cmath> | ||
− | <i><b>Step 2 | + | <i><b>Step 2</b></i> We will prove that line <math>F'D</math> is parallel to <math>OG.</math> |
[[File:F1F2 Euler OG.png|450px|right]] | [[File:F1F2 Euler OG.png|450px|right]] | ||
Let <math>\triangle xyz</math> be the inner Napoleon triangle. Let <math>\triangle XYZ</math> be the outer Napoleon triangle. These triangles are regular centered at <math>G.</math> | Let <math>\triangle xyz</math> be the inner Napoleon triangle. Let <math>\triangle XYZ</math> be the outer Napoleon triangle. These triangles are regular centered at <math>G.</math> | ||
Line 169: | Line 363: | ||
<math>BC \perp Xy,</math> angle between <math>Zx</math> and <math>Xy</math> is <math>60^\circ \implies \angle BE'Z = 30^\circ.</math> | <math>BC \perp Xy,</math> angle between <math>Zx</math> and <math>Xy</math> is <math>60^\circ \implies \angle BE'Z = 30^\circ.</math> | ||
− | <cmath>\angle AF'B = 120^\circ, \overset{\Large\frown} { | + | <cmath>\angle AF'B = \angle AZB = 120^\circ, AZ = BZ \implies \overset{\Large\frown} {BZ} = 60^\circ \implies</cmath> |
Points <math>A, Z, F', B, E',</math> and <math>E</math> are concyclic <math>\implies \angle OZx = \angle CBF.</math> | Points <math>A, Z, F', B, E',</math> and <math>E</math> are concyclic <math>\implies \angle OZx = \angle CBF.</math> | ||
Line 177: | Line 371: | ||
<math>\angle GZO = \angle GxO = 30^\circ \implies</math> points <math>Z, O, G,</math> and <math>x</math> are concyclic | <math>\angle GZO = \angle GxO = 30^\circ \implies</math> points <math>Z, O, G,</math> and <math>x</math> are concyclic | ||
− | <cmath>\implies \angle GOx = \angle OZx – 30^\circ = \angle | + | <cmath>\implies \angle GOx = \angle OZx – 30^\circ = \angle DF'B – 30^\circ.</cmath> |
<cmath>\angle AF'B = 120^\circ, Ox \perp AF' \implies OG||F'D.</cmath> | <cmath>\angle AF'B = 120^\circ, Ox \perp AF' \implies OG||F'D.</cmath> | ||
Therefore <math>OG||G'G \implies O, G',</math> and <math>G</math> are collinear or point <math>G</math> lies on Euler line <math>OG'.</math> | Therefore <math>OG||G'G \implies O, G',</math> and <math>G</math> are collinear or point <math>G</math> lies on Euler line <math>OG'.</math> | ||
Line 210: | Line 404: | ||
==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> | ||
Line 259: | Line 453: | ||
<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. | ||
Line 298: | Line 492: | ||
<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> | ||
Line 317: | Line 512: | ||
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>. | ||
Line 343: | Line 538: | ||
<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> | ||
Line 357: | Line 552: | ||
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''' | ||
+ | |||
+ | ==Euler line as radical axis== | ||
+ | [[File:3 circles Euler line.png|500px|right]] | ||
+ | Let <math>\triangle ABC</math> with altitudes <math>AA_1, BB_1,</math> and <math>CC_1</math> be given. | ||
+ | |||
+ | Let <math>\Omega, O, H</math> and <math>R</math> be circumcircle, circumcenter, orthocenter and circumradius of <math>\triangle ABC,</math> respectively. | ||
+ | |||
+ | Circle <math>\omega_1</math> centered at <math>Q_1</math> passes through <math>A, A_1</math> and is tangent to the radius AO. Similarly define circles <math>\omega_2</math> and <math>\omega_3.</math> | ||
+ | |||
+ | Then Euler line of <math>\triangle ABC</math> is the radical axis of these circles. | ||
+ | |||
+ | If <math>\triangle ABC</math> is acute, then these three circles intersect at two points located on the Euler line of the <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | The power of point <math>O</math> with respect to <math>\omega_1, \omega_2,</math> and <math>\omega_3</math> is <math>R^2.</math> | ||
+ | |||
+ | The power of point <math>H</math> with respect to <math>\omega_1</math> is <math>AH \cdot HA_1.</math> | ||
+ | |||
+ | The power of point <math>H</math> with respect to <math>\omega_2</math> is <math>BH \cdot HB_1.</math> | ||
+ | |||
+ | The power of point <math>H</math> with respect to <math>\omega_3</math> is <math>CH \cdot HC_1.</math> | ||
+ | |||
+ | It is known that <math>AH \cdot HA_1 = BH \cdot HB_1 = CH \cdot HC_1.</math> | ||
+ | |||
+ | Therefore points <math>H</math> and <math>O</math> lies on radical axis of these three circles as desired. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==De Longchamps point X(20)== | ||
+ | [[File:Longchamps.png|400px|right]] | ||
+ | <i><b>Definition 1</b></i> | ||
+ | |||
+ | 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 <math>\triangle ABC</math> the circle centered at the midpoint <math>BC</math> point <math>A'</math> with radius <math>R_A = AA'.</math> The other two circles are defined symmetrically. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>H, O,</math> and <math>L</math> be orthocenter, circumcenter, and De Longchamps point, respectively. | ||
+ | |||
+ | Denote <math>B-</math>power circle by <math>\omega_B, C-</math>power circle by <math>\omega_C, D = \omega_B \cap \omega_C,</math> | ||
+ | <math>a = BC, b = AC, c = AB.</math> WLOG, <math>a \ge b \ge c.</math> | ||
+ | |||
+ | Denote <math>X_t</math> the projection of point <math>X</math> on <math>B'C', E = D_t.</math> | ||
+ | |||
+ | We will prove that 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> 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 <math>H</math> with respect to <math>O.</math> | ||
+ | |||
+ | 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 - R_B^2}{a}.</cmath> | ||
+ | <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} - \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: | ||
+ | <cmath>C'A_t = \frac {a}{4} - \frac {b^2 - c^2}{4a}, | ||
+ | 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 - 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> | ||
+ | |||
+ | Similarly radical axes of <math>A-</math>power and <math>B-</math>power cicles is symmetric to altitude <math>CH,</math> radical axes of <math>A-</math>power and <math>C-</math>power cicles is symmetric to altitude <math>BH</math> with respect <math>O.</math> | ||
+ | Therefore the point <math>L</math> of intersection of the radical axes, symmetrical to the heights with respect to <math>O,</math> is symmetrical to the point <math>H</math> of intersection of the heights with respect to <math>O \implies \vec {HO} = \vec {OL} \implies L</math> lies on Euler line of <math>\triangle ABC.</math> | ||
+ | [[File:Distances.png|350px|right]] | ||
+ | <i><b>Claim (Distance between projections)</b></i> | ||
+ | |||
+ | <cmath>x + y = a, c^2 - x^2 = h^2 = b^2 - y^2,</cmath> | ||
+ | <cmath>y^2 - x^2 = b^2 - c^2 \implies y - x = \frac {b^2 - c^2}{a},</cmath> | ||
+ | <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> | ||
+ | [[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> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>H, G,</math> and <math>L_o</math> be orthocenter, centroid, and De Longchamps point, respectively. Let <math>\omega_B</math> cross <math>\omega_C</math> at points <math>A'</math> and <math>E.</math> The other points <math>(D, F, B', C')</math> are defined symmetrically. | ||
+ | <cmath>AB' = BC, B'C = AB \implies \triangle ABC = \triangle CB'A \implies</cmath> | ||
+ | <cmath>AB||B'C \implies CH \perp B'C.</cmath> | ||
+ | Similarly <math>CH \perp A'C \implies A'B'</math> is diameter <math>\omega_C \implies</math> | ||
+ | <cmath>\angle A'EB' = 90^\circ, 2\vec {BG} = \vec {GB'}.</cmath> | ||
+ | |||
+ | Therefore <math>\triangle A'B'C'</math> is anticomplementary triangle of <math>\triangle ABC, \triangle DEF</math> is orthic triangle of <math>\triangle A'B'C'.</math> So <math>L_o</math> is orthocenter of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <math>2\vec {HG} = \vec GL_o, 2\vec {GO} = \vec HG \implies L_o = L</math> as desired. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==De Longchamps line== | ||
+ | [[File:Longchamps lime.png|450px|right]] | ||
+ | The de Longchamps line <math>l</math> of <math>\triangle ABC</math> is defined as the radical axes of the de Longchamps circle <math>\omega</math> and of the circumscribed circle <math>\Omega</math> of <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>\Omega'</math> be the circumcircle of <math>\triangle DEF</math> (the anticomplementary triangle of <math>\triangle ABC).</math> | ||
+ | |||
+ | Let <math>\omega'</math> be the circle centered at <math>G</math> (centroid of <math>\triangle ABC</math>) with radius <math>\rho = \frac {\sqrt{2}}{3} \sqrt {a^2 + b^2 + c^2},</math> where <math>a = BC, b = AC, c = AB.</math> | ||
+ | |||
+ | Prove that the de Longchamps line is perpendicular to Euler line and is the radical axes of <math>\Omega, \Omega', \omega,</math> and <math>\omega'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Center of <math>\Omega</math> is <math>O</math>, center of <math>\omega</math> is <math>L \implies OL \perp l,</math> where <math>OL</math> is Euler line. | ||
+ | The homothety with center <math>G</math> and ratio <math>-2</math> maps <math>\triangle ABC</math> into <math>\triangle DEF.</math> This homothety maps <math>\Omega</math> into <math>\Omega'.</math> | ||
+ | <math>R_{\Omega} \ne R_{\Omega'}</math> and <math>\Omega \cap \Omega' = K \implies </math> there is two inversion which swap <math>\Omega</math> and <math>\Omega'.</math> | ||
+ | |||
+ | First inversion <math>I_{\omega'}</math> centered at point <math>G = \frac {\vec O \cdot 2R + \vec H \cdot R}{2R + R} = \frac {2\vec O + \vec H}{3}.</math> Let <math>K</math> be the point of crossing <math>\Omega</math> and <math>\Omega'.</math> | ||
+ | |||
+ | The radius of <math>\omega'</math> we can find using <math>\triangle HKO:</math> | ||
+ | |||
+ | <cmath>OK = R, HK = 2R, HG = 2GO \implies GK^2 = 2(R^2 – GO^2), GO^2 = \frac {HO^2}{9} \implies</cmath> | ||
+ | <cmath>R_G = GK = \frac {\sqrt {2(a^2 + b^2 + c^2)}}{3}.</cmath> | ||
+ | |||
+ | Second inversion <math>I_{\omega}</math> centered at point <math>L = \frac {\vec O \cdot 2R – \vec H \cdot R}{2R – R} = 2 \vec O – \vec H.</math> We can make the same calculations and get <math>R_L = 4R \sqrt{– \cos A \cos B \cos C}</math> as desired. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)== | ||
+ | [[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> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>A_0, B_0,</math> and <math>C_0</math> be midpoints of <math>BC, AC,</math> and <math>AB,</math> respectively. | ||
+ | |||
+ | Let <math>\omega</math> be circumcircle of <math>\triangle A_0B_0C_0.</math> It is nine-points circle of the <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>\Omega</math> be circumcircle of <math>\triangle ABC.</math> Let <math>\Omega'</math> be circumcircle of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <math>A'B</math> and <math>A'C</math> are tangents to <math>\Omega \implies</math> inversion with respect <math>\Omega</math> swap <math>B'</math> and <math>B_0.</math> Similarly, this inversion swap <math>A'</math> and <math>A_0, C'</math> and <math>C_0.</math> Therefore this inversion swap <math>\omega</math> and <math>\Omega'.</math> | ||
+ | |||
+ | The center <math>N</math> of <math>\omega</math> and the center <math>O</math> of <math>\Omega</math> lies on Euler line, so the center <math>O'</math> of <math>\Omega'</math> lies on this line, as desired. | ||
+ | |||
+ | After some calculations one can find position of point <math>X(26)</math> on Euler line (see Kimberling's point <math>X(26)).</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)== | ||
+ | [[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'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> | ||
+ | |||
+ | Prove that lines <math>A_1A', B_1B',</math> and <math>C_1C'</math> are concurrent at point, lies on Euler line of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>B'C'</math> and <math>B_1C_1</math> are antiparallel to BC with respect <math>\angle BAC \implies B'C' ||B_1C_1.</math> | ||
+ | |||
+ | Similarly, <math>A'C' ||A_1C_1, A'B' ||A_1B_1.</math> | ||
+ | |||
+ | Therefore <math>\triangle A_1B_1C_1 \sim \triangle A'B'C' \implies</math> homothetic center of <math>\triangle A_1B_1C_1</math> and <math>\triangle A'B'C'</math> is the point of concurrence of lines <math>A_1A', B_1B',</math> and <math>C_1C'.</math> Denote this point as <math>K.</math> | ||
+ | |||
+ | The points <math>N</math> and <math>O'</math> are the corresponding points (circumcenters) of <math>\triangle A_1B_1C_1</math> and <math>\triangle A'B'C',</math> so point <math>K</math> lies on line <math>NO'.</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''' | ||
==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(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]] | *[[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.
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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,
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.
vladimir.shelomovskii@gmail.com, vvsss
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
vladimir.shelomovskii@gmail.com, vvsss ~minor edit by Yiyj1
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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