Difference between revisions of "Euler line"
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− | {{ | + | In any [[triangle]] <math>\triangle ABC</math>, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] <math>H</math>, [[centroid]] <math>G</math>, [[circumcenter]] <math>O</math>, [[nine-point center]] <math>N</math> and [[De Longchamps point | de Longchamps point]] <math>L</math>. It is named after [[Leonhard Euler]]. Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points. In particular, <math>\overline{OGNH}</math> and <math>OG:GN:NH = 2:1:3</math> |
+ | |||
+ | |||
+ | Euler line is the central line <math>L_{647}</math>. | ||
+ | |||
+ | |||
+ | Given the [[orthic triangle]] <math>\triangle H_AH_BH_C</math> of <math>\triangle ABC</math>, the Euler lines of <math>\triangle AH_BH_C</math>,<math>\triangle BH_CH_A</math>, and <math>\triangle CH_AH_B</math> [[concurrence | concur]] at <math>N</math>, the nine-point circle of <math>\triangle ABC</math>. | ||
+ | |||
+ | ==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]] <math>\triangle O_AO_BO_C</math>. It is similar to <math>\triangle ABC</math>. Specifically, a rotation of <math>180^\circ</math> about the midpoint of <math>O_BO_C</math> followed by a homothety with scale factor <math>2</math> centered at <math>A</math> brings <math>\triangle ABC \to \triangle O_AO_BO_C</math>. Let us examine what else this transformation, which we denote as <math>\mathcal{S}</math>, will do. | ||
+ | |||
+ | It turns out <math>O</math> is the orthocenter, and <math>G</math> is the centroid of <math>\triangle O_AO_BO_C</math>. Thus, <math>\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}</math>. As a homothety preserves angles, it follows that <math>\measuredangle O_AOG = \measuredangle AHG</math>. Finally, as <math>\overline{AH} || \overline{O_AO}</math> it follows that | ||
+ | <cmath>\triangle AHG = \triangle O_AOG</cmath> | ||
+ | Thus, <math>O, G, H</math> are collinear, and <math>\frac{OG}{HG} = \frac{1}{2}</math>. | ||
+ | |||
+ | ==Another Proof== | ||
+ | Let <math>M</math> be the midpoint of <math>BC</math>. | ||
+ | Extend <math>CG</math> past <math>G</math> to point <math>H'</math> such that <math>CG = \frac{1}{2} GH</math>. We will show <math>H'</math> is the orthocenter. | ||
+ | Consider triangles <math>MGO</math> and <math>AGH'</math>. Since <math>\frac{MG}{GA}=\frac{H'G}{GC} = \frac{1}{2}</math>, and they both share a vertical angle, they are similar by SAS similarity. Thus, <math>AH' \parallel OM \perp BC</math>, so <math>H'</math> lies on the <math>A</math> altitude of <math>\triangle ABC</math>. We can analogously show that <math>H'</math> also lies on the <math>B</math> and <math>C</math> altitudes, so <math>H'</math> is the orthocenter. | ||
+ | |||
+ | ==Proof Nine-Point Center Lies on Euler Line== | ||
+ | Assuming that the [[nine point circle]] exists and that <math>N</math> is the center, note that a homothety centered at <math>H</math> with factor <math>2</math> brings the [[Euler point]]s <math>\{E_A, E_B, E_C\}</math> onto the circumcircle of <math>\triangle ABC</math>. Thus, it brings the nine-point circle to the circumcircle. Additionally, <math>N</math> should be sent to <math>O</math>, thus <math>N \in \overline{HO}</math> and <math>\frac{HN}{ON} = 1</math>. | ||
+ | |||
+ | ==Analytic Proof of Existence== | ||
+ | Let the circumcenter be represented by the vector <math>O = (0, 0)</math>, and let vectors <math>A,B,C</math> correspond to the vertices of the triangle. It is well known the that the orthocenter is <math>H = A+B+C</math> and the centroid is <math>G = \frac{A+B+C}{3}</math>. Thus, <math>O, G, H</math> are collinear and <math>\frac{OG}{HG} = \frac{1}{2}</math> | ||
+ | |||
+ | [[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>== | ||
+ | <i><b>Claim 1</b></i> | ||
+ | [[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> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\omega</math> be circumcircle of <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>O</math> be circumcenter of <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>\omega'</math> be the circle symmetric to <math>\omega</math> with respect to <math>AB.</math> | ||
+ | |||
+ | Let <math>E</math> be the point symmetric to <math>O</math> with respect to <math>AB.</math> | ||
+ | |||
+ | The <math>\angle C = 120^\circ \implies O</math> lies on <math>\omega', E</math> lies on <math>\omega.</math> | ||
+ | |||
+ | <math>EO</math> is the radius of <math>\omega</math> and <math>\omega' \implies </math> translation vector <math>\omega'</math> to <math>\omega</math> is <math>\vec {EO}.</math> | ||
+ | |||
+ | Let <math>H'</math> be the point symmetric to <math>H</math> with respect to <math>AB.</math> Well known that <math>H'</math> lies on <math>\omega.</math> | ||
+ | Therefore point <math>H</math> lies on <math>\omega'.</math> | ||
+ | |||
+ | Point <math>C</math> lies on <math>\omega, CH || OE \implies CH = OE.</math> | ||
+ | |||
+ | Let <math>CD</math> be the bisector of <math>\angle C \implies E,O,D</math> are concurrent. | ||
+ | <math>OD = HC, OD||HC \implies CD || HO \implies </math> | ||
+ | |||
+ | Euler line <math>HO</math> of the <math>\triangle ABC</math> is parallel to the bisector <math>CD</math> of <math>\angle C</math> as desired. | ||
+ | |||
+ | <i><b>Claim 2</b></i> | ||
+ | [[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> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\omega, O, H, I</math> be circumcircle, circumcenter, orthocenter and incenter of the <math>\triangle ABC.</math> | ||
+ | <cmath>\angle AHB = 180^\circ – \angle ACB = 180^\circ – 60^\circ = 120^\circ.</cmath> | ||
+ | <cmath>\angle AOB = 2 \angle ACB = 120^\circ.</cmath> | ||
+ | <math>\angle AIB = 90^\circ + \frac {\angle ACB}{2} = 120^\circ \implies</math> points <math>A, H, I, O, B</math> are concyclic. | ||
+ | |||
+ | The circle <math>AIB</math> centered at midpoint of small arc <math>AB \implies</math> | ||
+ | |||
+ | <math>EH = EO = CO, EO||CH \implies COEH</math> is rhomb. | ||
+ | |||
+ | Therefore the Euler line <math>HO</math> is perpendicular to <math>CI</math> as desired. | ||
+ | |||
+ | <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. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>l</math> and <math>l'</math> be internal and external bisectors of the angle <math>\angle BPC, l \perp l'</math>. | ||
+ | |||
+ | Then Euler lines of <math>\triangle ABP</math> and <math>\triangle CDP</math> are parallel to <math>l'</math> and Euler lines of <math>\triangle BCP</math> and <math>\triangle ADP</math> are perpendicular to <math>l</math> as desired. | ||
+ | |||
+ | '''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== | ||
+ | [[File:Euler lines.png|350px|right]] | ||
+ | Consider a triangle <math>ABC</math> with Fermat–Torricelli points <math>F</math> and <math>F'.</math> The Euler lines of the <math>10</math> triangles with vertices chosen from <math>A, B, C, F,</math> and <math>F'</math> are concurrent at the centroid <math>G</math> of triangle <math>ABC.</math> We denote centroids by <math>g</math>, circumcenters by <math>o.</math> We use red color for points and lines of triangles <math>F**,</math> green color for triangles <math>F'**,</math> and blue color for triangles <math>FF'*.</math> | ||
+ | |||
+ | <i><b>Case 1</b></i> | ||
+ | [[File:Fermat 1 Euler lines 3.png|400px|right]] | ||
+ | Let <math>F</math> be the first Fermat point of <math>\triangle ABC</math> maximum angle of which smaller then <math>120^\circ.</math> Then the centroid of triangle <math>ABC</math> lies on Euler line of the <math>\triangle ABF.</math> The pairwise angles between these Euler lines are equal <math>60^\circ.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 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> | ||
+ | <math>\angle AFB = 120^\circ \implies AFBD</math> is cyclic. | ||
+ | |||
+ | Point <math>O</math> is centroid of <math>\triangle ABD \implies \vec O = \frac {\vec A + \vec B + \vec D}{3}.</math> | ||
+ | <cmath>\vec G' = \frac {A + B + F}{3}, G = \frac {A + B + C}{3} \implies </cmath> | ||
+ | <cmath>\vec {OG} = \frac {\vec C – \vec D}{3} = \frac {\vec DC}{3}, \vec {G'G} = \frac {\vec C – \vec F}{3} = \frac {\vec FC}{3} \implies</cmath> | ||
+ | <math>OG||G'G \implies</math> Points <math>O, G',</math> and <math>G</math> are colinear, so point <math>G</math> lies on Euler line <math>OG'</math> of <math>\triangle ABF.</math> | ||
+ | |||
+ | <math>\vec {GG_0} = \frac {A – F}{3} \implies GG_0||AF, \vec {GG_1} = \frac {B – F}{3} \implies GG_1||BF.</math> | ||
+ | |||
+ | <i><b>Case 2</b></i> | ||
+ | [[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> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\triangle ABD</math> be external for <math>\triangle ABC</math> equilateral triangle, <math>\omega</math> be circumcircle of <math>\triangle ABD \implies F = CD \cap \omega.</math> | ||
+ | <math>\angle ABD = 60^\circ, \angle AFD = 120^\circ \implies ABDF</math> is cyclic. | ||
+ | |||
+ | Point <math>O</math> is centroid of <math>\triangle ABD \implies</math> | ||
+ | <math>\vec {OG} = \frac {\vec DC}{3}, \vec {G'G} = \frac {\vec FC}{3} \implies OG||G'G \implies</math> | ||
+ | |||
+ | Points <math>O, G',</math> and <math>G</math> are colinear, so point <math>G</math> lies on Euler line <math>OG'</math> of <math>\triangle ABF</math> as desired. | ||
+ | |||
+ | <i><b>Case 3</b></i> | ||
+ | [[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> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\triangle ABD</math> be internal for <math>\triangle ABC</math> equilateral triangle, <math>\omega</math> be circumcircle of <math>\triangle ABD \implies F' = CD \cap \omega.</math> | ||
+ | |||
+ | Let <math>O_1, O_0,</math> and <math>O'</math> be circumcenters of the triangles <math>\triangle ABF',\triangle ACF',</math> and <math>\triangle BCF'.</math> | ||
+ | Point <math>O_1</math> is centroid of the <math>\triangle ABD \implies GO_1G_1</math> is the Euler line of the <math>\triangle ABF'</math> parallel to <math>CD.</math> | ||
+ | |||
+ | <math>O_1 O_0</math> is bisector of <math>BF', O'O_1</math> is bisector of <math>AF', O'O_0</math> is bisector of <math>CF' \implies \triangle O'O_1O_0</math> is regular triangle. | ||
+ | |||
+ | <math>\triangle O'O_0 O_1</math> is the inner Napoleon triangle of the <math>\triangle ABC \implies G</math> is centroid of this regular triangle. | ||
+ | <cmath>\angle GO_1O_0 = 30^\circ, O_1O_0 \perp F'B, \angle AF'B = 120^\circ \implies GO_0||F'A.</cmath> | ||
+ | |||
+ | <math>\vec {GG_0} = \frac {\vec {F'A}}{3} \implies GG_0||F'A \implies </math> points <math>O_0,G,</math> and <math>G_0</math> are collinear as desired. | ||
+ | |||
+ | Similarly, points <math>O',G,</math> and <math>G'</math> are collinear. | ||
+ | |||
+ | <i><b>Case 4</b></i> | ||
+ | [[File:F1F2 Euler.png|450px|right]] | ||
+ | Let <math>F</math> and <math>F'</math> be the Fermat points of <math>\triangle ABC.</math> Then the centroid of <math>\triangle ABC</math> point <math>G</math> lies on Euler line <math>OG' (O</math> is circumcenter, <math>G'</math> is centroid) of the <math>\triangle AFF'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <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>D</math> be point symmetrical to <math>F</math> with respect to <math>M.</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 {2}{3} (\vec M – \vec M') = \frac {2}{3} \vec {M'M} \implies M'M||F'D||G'G.</cmath> | ||
+ | |||
+ | <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]] | ||
+ | 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> | ||
+ | |||
+ | Points <math>O, z,</math> and <math>x</math> are collinear (they lies on bisector <math>AF').</math> | ||
+ | |||
+ | Points <math>O, Z,</math> and <math>Y</math> are collinear (they lies on bisector <math>AF).</math> | ||
+ | |||
+ | Points <math>M, X,</math> and <math>y</math> are collinear (they lies on bisector <math>BC).</math> | ||
+ | <cmath>E = YZ \cap BF, E' = Zx \cap BC.</cmath> | ||
+ | <cmath>BF \perp XZ \implies \angle BEZ = 30^\circ.</cmath> | ||
+ | <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 = \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> | ||
+ | <cmath>FM = MD, BM = MC \implies \angle CBF = \angle BCD.</cmath> | ||
+ | Points <math>C, D, X, B,</math> and <math>F'</math> are concyclic <math>\implies \angle BCD = \angle BF'D.</math> | ||
+ | |||
+ | <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 DF'B – 30^\circ.</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> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Euler line of Gergonne triangle== | ||
+ | [[File:Euler line of Gergonne triangle.png|500px|right]] | ||
+ | Prove that the Euler line of Gergonne triangle of <math>\triangle ABC</math> passes through the circumcenter of triangle <math>ABC.</math> | ||
+ | |||
+ | Gergonne triangle is also known as the contact triangle or intouch triangle. If the inscribed circle touches the sides of <math>\triangle ABC</math> at points <math>D, E,</math> and <math>F,</math> then <math>\triangle DEF</math> is Gergonne triangle of <math>\triangle ABC</math>. | ||
+ | |||
+ | Other wording: Tangents to circumcircle of <math>\triangle ABC</math> 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. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>H</math> and <math>I</math> be orthocenter and circumcenter of <math>\triangle DEF,</math> respectively. | ||
+ | Let <math>A'B'C'</math> be Orthic Triangle of <math>\triangle DEF.</math> | ||
+ | |||
+ | Then <math>IH</math> is Euler line of <math>\triangle DEF,</math> | ||
+ | <math>I</math> is the incenter of <math>\triangle ABC,</math> <math>H</math> is the incenter of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <math>\angle DEF = \angle DB'C' = \angle BDF = \frac { \overset{\Large\frown} {DF}}{2} \implies B'C' || BC.</math> | ||
+ | |||
+ | Similarly, <math>A'C' || AC, A'B' || AB \implies A'B'C'\sim ABC \implies</math> | ||
+ | |||
+ | <math>AA' \cap BB' \cap CC' = P,</math> where <math>P</math> is the perspector of triangles <math>ABC</math> and <math>A'B'C'.</math> | ||
+ | |||
+ | Under homothety with center P and coefficient <math>\frac {B'C'}{BC}</math> the incenter <math>I</math> of <math>\triangle ABC</math> maps into incenter <math>H</math> of <math>\triangle A'B'C'</math>, circumcenter <math>O</math> of <math>\triangle ABC</math> maps into circumcenter <math>I</math> of <math>\triangle A'B'C' \implies P,H,I,O </math> are collinear as desired. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Thebault point== | ||
+ | [[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> | ||
+ | |||
+ | a) Prove that the Euler lines of triangles <math>\triangle AEF, \triangle BFD, \triangle CDE</math> are concurrent on the nine-point circle at a point T (Thebault point of <math>\triangle ABC.</math>) | ||
+ | |||
+ | b) Prove that if <math>\angle BAC < 90^\circ</math> then <math>TE = TF + TD,</math> else <math>TF = TE + TD.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <i><b>Case 1 Acute triangle</b></i> | ||
+ | |||
+ | a) It is known, that Euler line of acute triangle <math>\triangle ABC</math> cross AB and BC (shortest and longest sides) in inner points. | ||
+ | |||
+ | Let <math>O_0, O, O'</math> be circumcenters of <math>\triangle AEF, \triangle BFD, \triangle CDE.</math> | ||
+ | |||
+ | Let <math>G_0, G,</math> and <math>G'</math> be centroids of <math>\triangle AEF, \triangle BFD, \triangle CDE.</math> | ||
+ | |||
+ | Denote <math>\angle ABC = \beta, K = DE \cap G'O', L = EF \cap G_0 O_0, M = EC \cap G'O', \omega </math> is the circle <math>DEF</math> (the nine-points circle). | ||
+ | |||
+ | <math>\angle CEH = \angle CDH = 90^\circ \implies O'</math> is the midpoint <math>CH,</math> where <math>H</math> is the orthocenter of <math>\triangle ABC \implies O' \in \omega.</math> | ||
+ | |||
+ | Similarly <math> O_0 \in \omega, O \in \omega.</math> | ||
+ | |||
+ | <math>CO' = HO', BO = OH \implies OO'</math> is the midline of <math>\triangle BHC \implies \triangle O_0OO' \sim \triangle ABC.</math> | ||
+ | |||
+ | Let <math>O'G'</math> cross <math>\omega</math> at point <math>T</math> different from <math>O'.</math> | ||
+ | |||
+ | <math>\triangle ABC \sim \triangle AEF \sim \triangle DBF \sim \triangle DEC \implies</math> spiral similarity centered at <math>E</math> maps <math>\triangle AEF</math> onto <math> \triangle DEC.</math> | ||
+ | |||
+ | This similarity has the rotation angle <math>180^\circ – \beta \implies</math> acute angle between Euler lines of these triangles is <math>\beta.</math> | ||
+ | |||
+ | Let these lines crossed at point <math>T'.</math> | ||
+ | Therefore <math>\angle O_0T'O' = \angle O_0OO' \implies </math> points <math>O, O_0, O',T</math> and <math>T'</math> are concyclic <math>\implies T = T'.</math> | ||
+ | |||
+ | Similarly, <math>OG \cap O'G' = T</math> as desired. | ||
+ | |||
+ | b) <math>\triangle AEF \sim \triangle DEC \implies \frac {FL}{LE} = \frac {CM}{ME}.</math> | ||
+ | Point <math>G'</math> lies on median of <math>\triangle DEC</math> and divide it in ratio 2 : 1. | ||
+ | |||
+ | Point <math>G'</math> lies on Euler line of <math>\triangle DEC.</math> | ||
+ | |||
+ | According the <i><b>Claim,</b></i> <math>\frac {DK}{KE}+ \frac {CM}{ME} = 1 \implies \frac {DK}{KE}+ \frac {FL}{LE} = 1.</math> | ||
+ | <math>FO_0 = EO_0 \implies \overset{\Large\frown} {FO_0} = \overset{\Large\frown} {EO_0} \implies \angle FTO_0 = \angle ETO_0 \implies \frac {TF}{TE} = \frac {FL}{LE}.</math> | ||
+ | |||
+ | Similarly <math>\frac {TD}{TE} = \frac {DK}{KE} \implies \frac {TF}{TE} + \frac {TD}{TE} = 1 \implies TE = TD + TF.</math> | ||
+ | |||
+ | |||
+ | |||
+ | <i><b>Case 2 Obtuse triangle</b></i> | ||
+ | [[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. | ||
+ | |||
+ | Let <math>O_0, O, O'</math> be circumcenters of <math>\triangle AEF, \triangle BFD, \triangle CDE.</math> | ||
+ | |||
+ | Let <math>G_0, G,</math> and <math>G'</math> be centroids of <math>\triangle AEF, \triangle BFD, \triangle CDE.</math> | ||
+ | |||
+ | Denote <math>\angle ABC = \beta, K = DF \cap GO, K' = DC \cap G'O',</math> | ||
+ | <math>L = EF \cap G_0 O_0, L' = EC \cap G'O', \omega </math> is the circle <math>DEF</math> (the nine-points circle). | ||
+ | |||
+ | <math>\angle AEH = \angle AFH = 90^\circ \implies O_0</math> is the midpoint <math>AH,</math> where <math>H</math> is the orthocenter of <math>\triangle ABC \implies O_0 \in \omega.</math> | ||
+ | |||
+ | Similarly <math> O' \in \omega, O \in \omega.</math> | ||
+ | |||
+ | <math>CO' = HO', BO = OH \implies OO'</math> is the midline of <math>\triangle BHC \implies \triangle O_0OO' \sim \triangle ABC.</math> | ||
+ | |||
+ | Let <math>O'G'</math> cross <math>\omega</math> at point <math>T</math> different from <math>O'.</math> | ||
+ | |||
+ | <math>\triangle ABC \sim \triangle AEF \sim \triangle DBF \sim \triangle DEC \implies</math> spiral similarity centered at <math>E</math> maps <math>\triangle AEF</math> onto <math> \triangle DEC.</math> | ||
+ | |||
+ | This similarity has the rotation angle <math>\beta \implies</math> acute angle between Euler lines of these triangles is <math>\beta.</math> | ||
+ | |||
+ | Let these lines crossed at point <math>T'.</math> | ||
+ | Therefore <math>\angle O_0T'O' = \angle O_0OO' \implies </math> points <math>O, O_0, O',T</math> and <math>T'</math> are concyclic <math>\implies T = T'.</math> | ||
+ | |||
+ | Similarly, <math>OG \cap O'G' = T</math> as desired. | ||
+ | |||
+ | b) <math>\triangle AEF \sim \triangle DEC \implies \frac {EL}{LF} = \frac {EL'}{L'C}.</math> | ||
+ | <math>\triangle AEF \sim \triangle DBF \implies \frac {DK}{KF} = \frac {DK'}{K'C}.</math> | ||
+ | |||
+ | Point <math>G'</math> lies on median of <math>\triangle DEC</math> and divide it in ratio <math>2 : 1.</math> | ||
+ | |||
+ | Point <math>G'</math> lies on Euler line of <math>\triangle DEC.</math> | ||
+ | According the <i><b>Claim,</b></i> <math>\frac {DK'}{K'C} + \frac {EL'}{L'C} = 1 \implies \frac {DK}{KF}+ \frac {EL}{LF} = 1.</math> | ||
+ | <math>FO_0 = EO_0 \implies \overset{\Large\frown} {FO_0} = \overset{\Large\frown} {EO_0} \implies \angle FTO_0 = \angle ETO_0 \implies \frac {TE}{TF} = \frac {EL}{LF}.</math> | ||
+ | |||
+ | Similarly <math>\frac {TD}{TF} = \frac {DK}{KF} \implies \frac {TE}{TF} + \frac {TD}{TF} = 1 \implies TF = TD + TE.</math> | ||
+ | |||
+ | <i><b>Claim (Segment crossing the median)</b></i> | ||
+ | [[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> | ||
+ | <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> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>[ABC]</math> be <math>1</math> (We use sign <math>[t]</math> to denote the area of <math>t).</math> | ||
+ | |||
+ | Denote <math>[CDG] = x, [CEG] = y, [DGM] = z.</math> | ||
+ | <cmath>[ACM] = [ BCM] = \frac {1}{2}.</cmath> | ||
+ | <cmath>x = \frac {CD \cdot CG}{2 \cdot CA \cdot CM}, | ||
+ | y = \frac {CE \cdot CG}{2 \cdot CB \cdot CM} \implies </cmath> | ||
+ | <cmath>\frac {DG}{GE} = \frac {x}{y} = \frac {CD \cdot CB}{CA \cdot CE} = \frac {1+m}{1+n}.</cmath> | ||
+ | <cmath>x + y = \frac {CD \cdot CE}{AC \cdot BC} = \frac {1}{(1+m)(1+n)} \implies</cmath> | ||
+ | <cmath>x + y = x(1+\frac {y}{x}) = x (1 + \frac {1+n}{1+m} )= \frac {1}{(1+n)(1+m)} \implies | ||
+ | x = \frac{1}{(1+n)(2+m+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} - 1 =\frac {1}{2(1 + n)} (1 + n)(2 + m + n) - 1 = \frac {n + m}{2}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Schiffler point== | ||
+ | [[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>. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | We will prove that the Euler line <math>O'G'</math> of <math>\triangle BCI</math> cross the Euler line <math>OG</math> of <math>\triangle ABC</math> at such point <math>S,</math> that <math>\frac {OS}{SG} = \frac {3R}{2r}</math>. | ||
+ | |||
+ | Let <math>O'</math> and <math>G'</math> be the circumcenter and centroid of <math>\triangle IBC,</math> respectively. | ||
+ | |||
+ | It is known that <math>O'</math> lies on circumcircle of <math>\triangle ABC, \overset{\Large\frown} {BO'} = \overset{\Large\frown} {CO'}.</math> | ||
+ | |||
+ | Denote <math>E = OO' \cap BC, X = AE \cap G'O', Y = GG' \cap OO'.</math> | ||
+ | |||
+ | It is known that <math>E</math> is midpoint <math>BC,</math> point <math>G</math> lies on median <math>AE,</math> points <math>A, I, O'</math> belong the bisector of <math>\angle A, \frac {AE}{GE} = \frac {IE}{IG'} = 3 \implies GY||AO', \frac {O'E}{YE} = 3, \frac {GG'}{G'Y} = \frac {AI}{IO'}.</math> | ||
+ | |||
+ | Easy to find that <math>AI = \frac {r} {\sin {\frac {\alpha}{2}}}</math>, | ||
+ | <math>IO' = BO' = 2 R {\sin {\frac {\alpha}{2}}} \implies \frac {AI}{IO'} = \frac {r}{R \cdot (1 – \cos \alpha)}.</math> | ||
+ | |||
+ | We use sigh [t] for area of t. We get | ||
+ | <cmath>n = \frac {GG'}{G'Y} = \frac {[GXO']}{[YXO'} = \frac {[GXO']}{[EXO']} \cdot \frac {[EXO']}{[YXO'} = \frac {GX}{XE} \cdot \frac {3}{2} \implies</cmath> | ||
+ | |||
+ | <cmath>m = \frac {GX}{XE} = \frac {2n}{3}.</cmath> | ||
+ | <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 | ||
+ | <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. | ||
+ | |||
+ | <i><b>Claim (Segments crossing inside triangle)</b></i> | ||
+ | |||
+ | [[File:Segments crossing inside triangle.png|400px|right]] | ||
+ | Given triangle GOY. Point <math>S</math> lies on <math>GO, k = \frac {OS}{SG}.</math> | ||
+ | |||
+ | Point <math>E</math> lies on <math>YO, p = \frac {OE}{EY}.</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} - \frac {p}{n}.</math> | ||
+ | |||
+ | <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> | ||
+ | <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>[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''' | ||
− | + | ==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]] | ||
− | |||
− | |||
− | + | {{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
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
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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