Difference between revisions of "Isogonal conjugate"
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If the abscissa axis coincides with the line <math>\ell</math> and the origin coincides with the point <math>O,</math> then the isogonals define the equations <math>y = \pm kx,</math> and the lines <math>(\frac{1}{x}, \pm k)</math> symmetrical with respect to the line <math>\ell</math> become their images. | If the abscissa axis coincides with the line <math>\ell</math> and the origin coincides with the point <math>O,</math> then the isogonals define the equations <math>y = \pm kx,</math> and the lines <math>(\frac{1}{x}, \pm k)</math> symmetrical with respect to the line <math>\ell</math> become their images. | ||
− | It is clear that, under the | + | It is clear that, under the converse transformation (also projective), such pairs of lines become isogonals, and the points equidistant from <math>\ell</math> lie on the isogonals. |
<i><b>The isogonal theorem</b></i> | <i><b>The isogonal theorem</b></i> | ||
+ | [[File:Isogonal.png|390px|right]] | ||
+ | Let two pairs of isogonals <math>OX - OX'</math> and <math>OY - OY'</math> with respect to the pair <math>(\ell,O)</math> be given. Denote <math>Z = XY \cap X'Y', Z' = X'Y \cap XY'.</math> | ||
− | + | Prove that <math>OZ</math> and <math>OZ'</math> are the isogonals with respect to the pair <math>(\ell,O).</math> | |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
+ | [[File:Transform isogonal.png|390px|right]] | ||
+ | Let us perform a projective transformation of the plane that maps the point <math>O</math> into a point at infinity and the line <math>\ell</math> maps to itself. In this case, the isogonals turn into a pair of straight lines parallel to <math>\ell</math> and equidistant from <math>\ell.</math> | ||
+ | |||
+ | The converse (also projective) transformation maps the points equidistant from <math>\ell</math> onto isogonals. We denote the image and the preimage with the same symbols. | ||
+ | |||
+ | Let the images of isogonals are vertical lines. Let coordinates of images of points be <cmath>X(-a, 0), X'(a,u), Y(-b,v), Y'(b,w).</cmath> | ||
+ | Equation of a straight line <math>XY</math> is <math>\frac{x + a}{a - b} = \frac {y}{v}.</math> | ||
+ | |||
+ | Equation of a straight line <math>X'Y'</math> is <math>\frac{x - a}{b - a} = \frac {y - u}{w - u}.</math> | ||
+ | |||
+ | The abscissa <math>Z_x</math> of the point <math>Z</math> is <math>Z_x = \frac {v a - a w + u b}{u - v - w}.</math> | ||
+ | |||
+ | Equation of a straight line <math>XY'</math> is <math>\frac{x + a}{b + a} = \frac {y}{w}.</math> | ||
+ | |||
+ | Equation of a straight line <math>X'Y</math> is <math>\frac{x - a}{- b - a} = \frac {y - u}{v - u}.</math> | ||
+ | |||
+ | The abscissa <math>Z'_x</math> of the point <math>Z'</math> is <math>Z'_x = \frac {v a - a w + u b}{- u + v + w} = - Z_x \implies</math> | ||
+ | |||
+ | Preimages of the points <math>Z</math> and <math>Z'</math> lie on the isogonals. <math>\blacksquare</math> | ||
+ | |||
+ | <i><b>The isogonal theorem in the case of parallel lines</b></i> | ||
+ | [[File:Parallels 1.png|330px|right]] | ||
+ | Let <math>OY</math> and <math>OY'</math> are isogonals with respect <math>\angle XOX'.</math> | ||
+ | |||
+ | Let lines <math>XY</math> and <math>X'Y'</math> intersect at point <math>Z, X'Y || XY'.</math> | ||
+ | |||
+ | Prove that <math>OZ</math> and line <math>l</math> through <math>O</math> parallel to <math>XY'</math> are the isogonals with respect <math>\angle XOX'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | The preimage of <math>Z'</math> is located at infinity on the line <math>l.</math> | ||
+ | |||
+ | The equality <math>Z'_x = -Z_x</math> implies the equality the slopes modulo of <math>OZ</math> and <math>l</math> to the bisector of <math>\angle XOX'. \blacksquare</math> | ||
+ | |||
+ | <i><b>Converse theorem</b></i> | ||
+ | [[File:Parallels 2.png|390px|right]] | ||
+ | Let lines <math>XY</math> and <math>X'Y'</math> intersect at point <math>Z, X'Y || XY'.</math> | ||
+ | |||
+ | Let <math>OZ</math> and <math>l</math> be the isogonals with respect <math>\angle XOX'.</math> | ||
+ | |||
+ | Prove that <math>OY</math> and <math>OY'</math> are isogonals with respect <math>\angle XOX' (\angle XOY' = \angle YOX').</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | The preimage of <math>Z'</math> is located at infinity on the line <math>l,</math> so the slope of <math>OZ</math> is known. | ||
+ | |||
+ | Suppose that <math>y' \in XY', y' \ne Y', \angle y'OX = \angle YOX'.</math> | ||
+ | |||
+ | The segment <math>XY</math> and the lines <math>XY', OZ</math> are fixed <math>\implies</math> | ||
+ | |||
+ | <math>y'X'</math> intersects <math>XY</math> at <math>z \ne Z,</math> | ||
+ | |||
+ | but there is the only point where line <math>OZ</math> intersect <math>XY.</math> Сontradiction. <math>\blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Parallel segments== | ||
+ | [[File:Parallels.png|350px|right]] | ||
+ | Let triangle <math>ABC</math> be given. Let <math>AD</math> and <math>AE</math> be the isogonals with respect <math>\angle BAC.</math> Let <math>BD ||CE, P = BE \cap CD.</math> | ||
+ | |||
+ | Prove that <math>P</math> lies on bisector of <math>\angle BAC</math> and <math>BD||AP.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Both assertions follow from <i><b>The isogonal theorem in the case of parallel lines</b></i> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Perpendicularity== | ||
+ | [[File:Right angles.png|400px|right]] | ||
+ | Let triangle <math>ABC</math> be given. Right triangles <math>ABD</math> and <math>ACE</math> with hypotenuses <math>AD</math> and <math>AE</math> are constructed on sides <math>AB</math> and <math>AC</math> to the outer (inner) side of <math>\triangle ABC.</math> Let <math>\angle BAD = \angle CAE, H = CD \cap BE.</math> | ||
+ | Prove that <math>AH \perp BC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\ell</math> be the bisector of <math>\angle BAC, F = BD \cap CE.</math> | ||
+ | |||
+ | <math>AB</math> and <math>AC</math> are isogonals with respect to the pair <math>(\ell,A).</math> | ||
+ | |||
+ | <math>AD</math> and <math>AE</math> are isogonals with respect to the pair <math>(\ell,A) \implies</math> | ||
+ | |||
+ | <math>AH</math> and <math>AF</math> are isogonals with respect to the pair <math>(\ell,A)</math> in accordance with <i><b>The isogonal theorem.</b></i> | ||
+ | |||
+ | <math>\angle ABD = \angle ACE = 90^\circ \implies</math> | ||
+ | |||
+ | <math>AF</math> is the diameter of circumcircle of <math>\triangle ABC.</math> | ||
+ | |||
+ | Circumradius and altitude are isogonals with respect bisector and vertex of triangle, so <math>AH \perp BC.</math> <math>\blacksquare</math> | ||
+ | |||
+ | *[[Circumradius]] | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Fixed point== | ||
+ | [[File:Fixed point.png|400px|right]] | ||
+ | Let fixed triangle <math>ABC</math> be given. Let points <math>D</math> and <math>E</math> on sidelines <math>BC</math> and <math>AB,</math> respectively be the arbitrary points. | ||
+ | |||
+ | Let <math>F</math> be the point on sideline <math>AC</math> such that <math>\angle BDE = \angle CDF.</math> | ||
+ | |||
+ | <math>G = BF \cap CE.</math> Prove that line <math>DG</math> pass through the fixed point. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | We will prove that point <math>A',</math> symmetric <math>A</math> with respect <math>\ell = BC,</math> lies on <math>DG</math>. | ||
+ | |||
+ | <math>\angle BDE = \angle CDF \implies DE</math> and <math>DF</math> are isogonals with respect to <math>(\ell, D).</math> | ||
+ | |||
+ | <math>A = BE \cap CF \implies </math> points <math>A</math> and <math>G</math> lie on isogonals with respect to <math>(\ell, D)</math> in accordance with <i><b>The isogonal theorem.</b></i> | ||
+ | |||
+ | Point <math>A'</math> symmetric <math>A</math> with respect <math>\ell</math> lies on isogonal <math>AD</math> with respect to <math>(\ell, D),</math> that is <math>DG.</math> <math>\blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Bisector== | ||
+ | [[File:Incircles.png|400px|right]] | ||
+ | Let a convex quadrilateral <math>ABCD</math> be given. Let <math>I</math> and <math>J</math> be the incenters of triangles <math>\triangle ABC</math> and <math>\triangle ADC,</math> respectively. | ||
+ | |||
+ | Let <math>I'</math> and <math>J'</math> be the A-excenters of triangles <math>\triangle ABC</math> and <math>\triangle ADC,</math> respectively. <math>E = IJ' \cap I'J.</math> | ||
+ | |||
+ | Prove that <math>CE</math> is the bisector of <math>\angle BCD.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\angle ICI' = \angle JCJ' = 90^\circ \implies</math> | ||
+ | |||
+ | <math>CI'</math> and <math>CJ'</math> are isogonals with respect to the angle <math>\angle ICJ.</math> | ||
+ | |||
+ | <math>A = II' \cap JJ' \implies AC</math> and <math>EC</math> are isogonals with respect to the angle <math>\angle ICJ</math> in accordance with <i><b>The isogonal theorem.</b></i> | ||
+ | |||
+ | Denote <math>\angle ACI = \angle BCI = \alpha, \angle ACJ = \angle DCJ = \beta.</math> | ||
+ | |||
+ | WLOG, <math>\beta \ge \alpha.</math> | ||
+ | <cmath>\angle ACJ = \angle ACE + \alpha = \beta,</cmath> | ||
+ | <cmath>\angle BCE = 2 \alpha + \beta - \alpha = \alpha + \beta = \angle DCE. \blacksquare</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Isogonal of the diagonal of a quadrilateral== | ||
+ | [[File:Quadrungle isogonals.png|250px|right]] | ||
+ | Given a quadrilateral <math>ABCD</math> and a point <math>P</math> on its diagonal such that <math>\angle APB = \angle APD.</math> | ||
+ | |||
+ | Let <math>E = AB \cap CD, F = AD \cap BC.</math> | ||
+ | |||
+ | Prove that <math>\angle BPE = \angle DPF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | [[File:Quadrungle transform.png|250px|right]] | ||
+ | Let us perform a projective transformation of the plane that maps the point <math>P</math> to a point at infinity and the line <math>\ell = AC</math> into itself. | ||
+ | |||
+ | In this case, the images of points <math>B</math> and <math>D</math> are equidistant from the image of <math>AC \implies</math> | ||
+ | |||
+ | the point <math>M</math> (midpoint of <math>BD)</math> lies on <math>\ell \implies</math> | ||
+ | |||
+ | <math>AC</math> contains the midpoints of <math>AC</math> and <math>BD \implies</math> | ||
+ | |||
+ | <math>\ell</math> is the Gauss line of the complete quadrilateral <math>ABCDEF \implies</math> | ||
+ | <math>\ell</math> bisects <math>EF \implies EE_0 = FF_0 \implies</math> | ||
+ | |||
+ | the preimages of the points <math>E</math> and <math>F</math> lie on the isogonals <math>PE</math> and <math>PF. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Isogonals in trapezium== | ||
+ | [[File:Trapezium ACFEE.png|370px|right]] | ||
+ | Let the trapezoid <math>AEFC, AC||EF,</math> be given. Denote <cmath>B = AE \cap CF, D = AF \cap CE.</cmath> | ||
+ | |||
+ | The point <math>M</math> on the smaller base <math>AC</math> is such that <math>EM = MF.</math> | ||
+ | |||
+ | Prove that <math>\angle AMB = \angle AMD.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>EM = MF \implies </cmath> | ||
+ | <cmath>\angle AME = \angle MEF = \angle MFE = \angle CMF.</cmath> | ||
+ | Therefore <math>EM</math> and <math>FM</math> are isogonals with respect <math>(AC,M).</math> | ||
+ | |||
+ | Let us perform a projective transformation of the plane that maps the point <math>M</math> to a point at infinity and the line <math>\ell = AC</math> into itself. | ||
+ | |||
+ | In this case, the images of points <math>E</math> and <math>F</math> are equidistant from the image of <math>\ell \implies AC</math> contains the midpoints of <math>AC</math> and <math>EF</math>, that is, <math>\ell</math> is the Gauss line of the complete quadrilateral <math>ABCDEF \implies</math> | ||
+ | |||
+ | <math>\ell</math> bisects <math>BD \implies BB_0 = DD_0 \implies</math> | ||
+ | |||
+ | The preimages of the points <math>B</math> and <math>D</math> lie on the isogonals <math>MB</math> and <math>MD. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Isogonals in complete quadrilateral== | ||
+ | [[File:Isogonals in complete quadrilateral.png|400px|right]] | ||
+ | Let complete quadrilateral <math>ABCDEF (E = AB \cap CD, F = AC \cap BD)</math> be given. Let <math>M</math> be the Miquel point of <math>ABCD.</math> | ||
+ | |||
+ | Prove that <math>AM</math> is isogonal to <math>DM</math> and <math>EM</math> is isogonal to <math>FM</math> with respect <math>\angle BMC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | <cmath>\angle BME = \angle BDE = \angle CDF = \angle CMF.</cmath> | ||
+ | <cmath>\angle BMD = \angle BED = \angle AEC = \angle AMC. \blacksquare</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Isogonal of the bisector of the triangle== | ||
+ | [[File:Bisector C.png|400px|right]] | ||
+ | The triangle <math>ABC</math> be given. The point <math>D</math> chosen on the bisector <math>AA'.</math> | ||
+ | |||
+ | Denote <cmath>B' = BD \cap AC, C' = CD \cap AB,</cmath> | ||
+ | <cmath>E = BB' \cap A'C', F = CC' \cap A'B'.</cmath> | ||
+ | Prove that <math>\angle BAE = \angle CAF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let us perform a projective transformation of the plane that maps the point <math>A</math> to a point at infinity and the line <math>\ell = AA'</math> into itself. | ||
+ | |||
+ | In this case, the images of segments <math>BC'</math> and <math>B'C</math> are equidistant from the image of <math>\ell \implies BC' || B'C || \ell.</math> | ||
+ | |||
+ | Image of point <math>D</math> is midpoint of image <math>BB'</math> and midpoint image <math>CC' \implies</math> | ||
+ | |||
+ | Image <math>BCB'C'</math> is parallelogramm <math>\implies</math> | ||
+ | |||
+ | <math>BC = B'C' \implies \frac {DE}{BD} = \frac {DF}{CD} \implies </math> distances from <math>E</math> and <math>F</math> to <math>\ell</math> are equal <math>\implies</math> | ||
+ | |||
+ | Preimages <math>AE</math> and <math>AF</math> are isogonals with respect <math>(\ell,A). \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Points on isogonals== | ||
+ | [[File:Points on isogonals.png|400px|right]] | ||
+ | The triangle <math>ABC</math> be given. The point <math>D</math> chosen on <math>BC.</math> | ||
+ | The point <math>E</math> chosen on <math>BC</math> such that <math>AD</math> and <math>AE</math> are isogonals with respect <math>\angle BAC.</math> | ||
+ | |||
+ | Prove that <math>\frac {AB^2}{BD \cdot BE} = \frac{ AC^2}{CD \cdot CE}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>\angle BAD = \angle CAE = \varphi,</math> | ||
+ | <math>\angle B = \beta, \angle C = \gamma \implies</math> | ||
+ | <math>\angle ADE = \beta + \varphi, \angle AED = \gamma + \varphi, \angle BAE = \angle CAD = \psi+\varphi.</math> | ||
+ | |||
+ | We use the Law of Sines and get: | ||
+ | <cmath>\frac {AB}{BD} = \frac {\sin (\beta + \varphi)}{\sin \varphi}, \frac {AB}{BE} = \frac {\sin (\gamma + \varphi)}{\sin (\psi +\varphi)},</cmath> | ||
+ | <cmath>\frac {AC}{CE} = \frac {\sin (\gamma + \varphi)}{\sin \varphi}, \frac {AC}{CD} = \frac {\sin (\beta + \varphi)}{\sin (\psi +\varphi)} \implies</cmath> | ||
+ | <cmath>\frac {AB \cdot AB}{BD \cdot BE} = \frac {AC \cdot AC}{CD \cdot CE} = \frac {\sin (\beta + \varphi) \cdot \sin (\gamma +\varphi)} {\sin \varphi \cdot \sin (\psi +\varphi)}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Trapezoid== | ||
+ | [[File:Trapezoid3.png|370px|right]] | ||
+ | The lateral side <math>CD</math> of the trapezoid <math>ABCD</math> is perpendicular to the bases, point <math>P</math> is the intersection point of the diagonals <math>ABCD</math>. | ||
+ | |||
+ | Point <math>Q</math> is taken on the circumcircle <math>\omega</math> of triangle <math>PCD</math> diametrically opposite to point <math>P.</math> | ||
+ | |||
+ | Prove that <math>\angle BQC = \angle AQD.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, <math>CD</math> is not the diameter of <math>\omega.</math> | ||
+ | Let sidelines <math>AD</math> and <math>BC</math> intersect <math>\omega</math> at points <math>D'</math> and <math>C',</math> respectively. | ||
+ | |||
+ | <math>DD' \perp CD, CC' \perp CD \implies CDD'C'</math> is rectangle <math>\implies</math> | ||
+ | <math>CC' = DD' \implies \angle CQC' = \angle DQD'.</math> | ||
+ | |||
+ | <math>QE||BC</math> is isogonal to <math>QO</math> with respect <math>\angle CQD \implies</math> | ||
+ | |||
+ | <math>QE||BC</math> is isogonal to <math>QP</math> with respect <math>\angle CQD \implies</math> | ||
+ | |||
+ | In accordance with <i><b>The isogonal theorem in case parallel lines</b></i> <math>\angle DQO = \angle CQE.</math> | ||
+ | |||
+ | <math>QE||BC</math> is isogonal to <math>QP</math> with respect <math>\angle CQD, P = AC \cap BD \implies</math> | ||
+ | |||
+ | <math>\angle AQD = \angle BQC</math> in accordance with <i><b>Converse theorem for The isogonal theorem in case parallel lines.</b></i> <math>\blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==IMO 2007 Short list/G3== | ||
+ | [[File:Trapezoid 17.png|400px|right]] | ||
+ | The diagonals of a trapezoid <math>ABCD</math> intersect at point <math>P.</math> | ||
+ | |||
+ | Point <math>Q</math> lies between the parallel lines <math>BC</math> and <math>AD</math> such that <math>\angle AQD = \angle CQB,</math> and line <math>CD</math> separates points <math>P</math> and <math>Q.</math> | ||
+ | |||
+ | Prove that <math>\angle BQP = \angle DAQ.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\angle AQD = \angle CQB \implies</math> | ||
+ | |||
+ | <math>BQ</math> and <math>AQ</math> are isogonals with respect <math>\angle CQD.</math> | ||
+ | |||
+ | <math>P =AC \cap BD, BC || AD \implies</math> | ||
+ | |||
+ | <math>QS || AD</math> is isogonal to <math>QP</math> with respect <math>\angle CQD.</math> | ||
− | + | From the converse of <i><b>The isogonal theorem</b></i> we get | |
− | + | <math>\angle BQP = \angle SQA = \angle DAQ \blacksquare</math> | |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
== Definition of isogonal conjugate of a point == | == Definition of isogonal conjugate of a point == | ||
− | [[File:Definitin 1.png| | + | [[File:Definitin 1.png|390px|right]] |
− | Let <math>P</math> be | + | Let triangle <math>\triangle ABC</math> be given. Let <math>\omega</math> be the circumcircle of <math>ABC.</math> Let point <math>P</math> be in the plane of <math>\triangle ABC, P \notin AB, P \notin BC, P \notin AC, P \notin \omega.</math> |
+ | Denote by <math>a,b,c</math> the lines <math>BC, CA, AB,</math> respectively. Denote by <math>p_a, p_b, p_c</math> the lines <math>PA</math>, <math>PB</math>, <math>PC</math>, respectively. | ||
+ | Denote by <math>q_a</math>, <math>q_b</math>, <math>q_c</math> the reflections of <math>p_a</math>, <math>p_b</math>, <math>p_c</math> over the angle bisectors of angles <math>A</math>, <math>B</math>, <math>C</math>, respectively. | ||
+ | |||
+ | Prove that lines <math>q_a</math>, <math>q_b</math>, <math>q_c</math> [[concurrence | concur]] at a point <math>Q.</math> | ||
+ | This point is called the isogonal conjugate of <math>P</math> with respect to triangle <math>ABC</math>. | ||
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
Line 37: | Line 331: | ||
<cmath> \frac{\sin \angle q_a b}{\sin \angle c q_a} \cdot \frac{\sin \angle q_b c}{\sin \angle a q_b} \cdot \frac{\sin \angle q_c a}{\sin \angle b q_c} = \frac{\sin \angle p_a c}{\sin \angle b p_a} \cdot \frac{\sin \angle p_b a}{\sin \angle c p_b} \cdot \frac{\sin \angle p_c b}{\sin \angle a p_c} = 1, </cmath> | <cmath> \frac{\sin \angle q_a b}{\sin \angle c q_a} \cdot \frac{\sin \angle q_b c}{\sin \angle a q_b} \cdot \frac{\sin \angle q_c a}{\sin \angle b q_c} = \frac{\sin \angle p_a c}{\sin \angle b p_a} \cdot \frac{\sin \angle p_b a}{\sin \angle c p_b} \cdot \frac{\sin \angle p_c b}{\sin \angle a p_c} = 1, </cmath> | ||
so again by the trigonometric form of Ceva, the lines <math>q_a, q_b, q_c</math> concur, as was to be proven. <math>\blacksquare</math> | so again by the trigonometric form of Ceva, the lines <math>q_a, q_b, q_c</math> concur, as was to be proven. <math>\blacksquare</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Let points P and Q lie on the isogonals with respect angles <math>\angle B</math> and <math>\angle C</math> of triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | Then these points lie on isogonals with respect angle <math>\angle A.</math> | ||
+ | |||
+ | <i><b>Corollary 2</b></i> | ||
+ | |||
+ | Let point <math>P</math> be in the sideline <math>BC</math> of <math>\triangle ABC, P \ne B, P \ne C.</math> | ||
+ | |||
+ | Then the isogonal conjugate of a point <math>P</math> is a point <math>A.</math> | ||
+ | |||
+ | Points <math>A,B,</math> and <math>C</math> do not have an isogonally conjugate point. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Three points== | ||
+ | [[File:3 points.png|400px|right]] | ||
+ | Let fixed triangle <math>ABC</math> be given. Let the arbitrary point <math>D</math> not be on sidelines of <math>\triangle ABC.</math> Let <math>E</math> be the point on isogonal of <math>CD</math> with respect angle <math>\angle ACB.</math> | ||
+ | Let <math>F</math> be the crosspoint of isogonal of <math>BD</math> with respect angle <math>\angle ABC</math> and isogonal of <math>AE</math> with respect angle <math>\angle BAC.</math> | ||
+ | |||
+ | Prove that lines <math>AD, BE,</math> and <math>CF</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>D' = BF \cap CE, S = CF \cap BE.</math> | ||
+ | |||
+ | <math>AE</math> and <math>AF</math> are isogonals with respect <math>\angle BAC \implies</math> | ||
+ | |||
+ | <math>D'</math> and S lie on isogonals of <math>\angle BAC.</math> | ||
+ | |||
+ | <math>\angle DBC = \angle D'BA, \angle DCB = \angle D'CA \implies</math> | ||
+ | |||
+ | <math>D'</math> is isogonal conjugated of <math>D</math> with respect <math>\triangle ABC \implies</math> | ||
+ | |||
+ | <math>D'</math> and <math>D</math> lie on isogonals of <math>\angle BAC.</math> | ||
+ | |||
+ | Therefore points <math>A, S</math> and <math>D</math> lie on the same line which is isogonal to <math>AD'</math> with respect <math>\angle BAC. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
== Second definition == | == Second definition == | ||
− | [[File:Definition 2.png| | + | [[File:Definition 2.png|370px|right]] |
Let triangle <math>\triangle ABC</math> be given. Let point <math>P</math> lies in the plane of <math>\triangle ABC,</math> | Let triangle <math>\triangle ABC</math> be given. Let point <math>P</math> lies in the plane of <math>\triangle ABC,</math> | ||
<cmath>P \notin AB, P \notin BC, P \notin AC.</cmath> | <cmath>P \notin AB, P \notin BC, P \notin AC.</cmath> | ||
Line 54: | Line 389: | ||
<cmath>\angle ACQ = \angle BCP_1 \implies \angle QCP_1 = \angle ACB.</cmath> | <cmath>\angle ACQ = \angle BCP_1 \implies \angle QCP_1 = \angle ACB.</cmath> | ||
<cmath>\angle BCQ = \angle ACP_2 \implies \angle QCP_2 = \angle ACB.</cmath> | <cmath>\angle BCQ = \angle ACP_2 \implies \angle QCP_2 = \angle ACB.</cmath> | ||
− | <math>\angle QCP_1 = \angle QCP_2, CP_1 = CP_2, QC</math> common | + | <math>\angle QCP_1 = \angle QCP_2, CP_1 = CP_2, QC</math> is common therefore |
<cmath>\triangle QCP_1 = \triangle QCP_2 \implies QP_1 = QP_2.</cmath> | <cmath>\triangle QCP_1 = \triangle QCP_2 \implies QP_1 = QP_2.</cmath> | ||
Similarly <math>QP_1 = QP_3 \implies Q</math> is the circumcenter of the <math>\triangle P_1P_2P_3.</math> <math>\blacksquare</math> | Similarly <math>QP_1 = QP_3 \implies Q</math> is the circumcenter of the <math>\triangle P_1P_2P_3.</math> <math>\blacksquare</math> | ||
Line 63: | Line 398: | ||
Let point <math>P</math> be the point with barycentric coordinates <math>(p : q : r),</math> | Let point <math>P</math> be the point with barycentric coordinates <math>(p : q : r),</math> | ||
− | <cmath>p = [(P-B),(P-C)], q = [(P-C),(P-A)], r = [(P-A),(P-B)].</cmath> | + | <cmath>p = [(P - B),(P - C)], q = [(P - C),(P - A)], r = [(P - A),(P - B)].</cmath> |
Then <math>Q</math> has barycentric coordinates <cmath>(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.</cmath> | Then <math>Q</math> has barycentric coordinates <cmath>(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.</cmath> | ||
Line 81: | Line 416: | ||
Let <math>\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.</math> | Let <math>\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.</math> | ||
− | <cmath>\frac {PE}{PD} = \frac {PC \sin \theta}{PC \sin \Theta} = \frac {QC \sin \theta}{QC \sin \Theta} = \frac {QD'}{QE'}.</cmath> | + | <cmath>\frac {PE}{PD} = \frac {PC \sin \theta}{PC \sin \Theta} = \frac {QC \sin \theta}{QC \sin \Theta} = \frac {QD'}{QE'}. \blacksquare</cmath> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Sign of isogonally conjugate points== | ==Sign of isogonally conjugate points== | ||
− | [[File:Isog dist.png| | + | [[File:Isog dist.png|350px|right]] |
− | [[File:Isog distance.png| | + | [[File:Isog distance.png|350px|right]] |
Let triangle <math>\triangle ABC</math> and points <math>P</math> and <math>Q</math> inside it be given. | Let triangle <math>\triangle ABC</math> and points <math>P</math> and <math>Q</math> inside it be given. | ||
Line 95: | Line 430: | ||
Let <math>\frac {PE}{PD} = \frac{QD'}{QE'}, \frac {PF}{PD} = \frac{QD'}{QF'}.</math> Prove that point <math>Q</math> is the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | Let <math>\frac {PE}{PD} = \frac{QD'}{QE'}, \frac {PF}{PD} = \frac{QD'}{QF'}.</math> Prove that point <math>Q</math> is the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | ||
− | One can prove similar theorem in the case <math>P</math> outside <math>\triangle ABC.</math> | + | One can prove a similar theorem in the case <math>P</math> outside <math>\triangle ABC.</math> |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
Line 105: | Line 440: | ||
<cmath>\sin \varphi \cdot \sin (\gamma - \psi) = \sin \psi \cdot \sin (\gamma - \varphi) \implies</cmath> | <cmath>\sin \varphi \cdot \sin (\gamma - \psi) = \sin \psi \cdot \sin (\gamma - \varphi) \implies</cmath> | ||
<cmath>\cos (\varphi - \gamma + \psi) - \cos(\varphi + \gamma - \psi) = \cos (\psi - \gamma + \varphi) - \cos(\psi + \gamma - \varphi)</cmath> | <cmath>\cos (\varphi - \gamma + \psi) - \cos(\varphi + \gamma - \psi) = \cos (\psi - \gamma + \varphi) - \cos(\psi + \gamma - \varphi)</cmath> | ||
− | <cmath>\cos (\gamma + \varphi -\psi) = \cos(\gamma - \psi + \varphi) \implies</cmath> | + | <cmath>\cos (\gamma + \varphi - \psi) = \cos(\gamma - \psi + \varphi) \implies</cmath> |
<cmath>\cos \gamma \cos (\varphi - \psi) - \sin \gamma \sin (\varphi - \psi) = \cos \gamma \cos (\varphi - \psi) + \sin \gamma \sin (\varphi - \psi)</cmath> | <cmath>\cos \gamma \cos (\varphi - \psi) - \sin \gamma \sin (\varphi - \psi) = \cos \gamma \cos (\varphi - \psi) + \sin \gamma \sin (\varphi - \psi)</cmath> | ||
<cmath>2 \sin \gamma \cdot \sin (\varphi - \psi) = 0, \varphi + \psi < 180^\circ \implies \varphi = \psi.</cmath> | <cmath>2 \sin \gamma \cdot \sin (\varphi - \psi) = 0, \varphi + \psi < 180^\circ \implies \varphi = \psi.</cmath> | ||
− | Similarly <math>\angle ABP = \angle CBQ | + | Similarly <math>\angle ABP = \angle CBQ.</math> |
− | point <math>Q</math> is the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | + | Hence point <math>Q</math> is the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC. \blacksquare</math> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Circumcircle of pedal triangles== | ==Circumcircle of pedal triangles== | ||
− | [[File:Common circle.png| | + | [[File:Common circle.png|330px|right]] |
Let <math>Q</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | Let <math>Q</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | ||
+ | |||
Let <math>E, D, F</math> be the projection <math>P</math> on sides <math>AC, BC, AB,</math> respectively. | Let <math>E, D, F</math> be the projection <math>P</math> on sides <math>AC, BC, AB,</math> respectively. | ||
Let <math>E', D', F'</math> be the projection <math>Q</math> on sides <math>AC, BC, AB,</math> respectively. | Let <math>E', D', F'</math> be the projection <math>Q</math> on sides <math>AC, BC, AB,</math> respectively. | ||
− | + | Prove that points <math>D, D', E, E', F, F'</math> are concyclic. | |
The midpoint <math>PQ</math> is circumcenter of <math>DD'EE'FF'.</math> | The midpoint <math>PQ</math> is circumcenter of <math>DD'EE'FF'.</math> | ||
Line 127: | Line 463: | ||
Let <math>\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.</math> | Let <math>\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.</math> | ||
− | <math>CE \cdot CE' = PC \cos \theta \cdot QC \cos \Theta = PC \cos \Theta \cdot QC \cos \theta = CD \cdot CD'.</math> | + | <math>CE \cdot CE' = PC \cos \theta \cdot QC \cos \Theta = PC \cos \Theta \cdot QC \cos \theta = CD \cdot CD'.</math> |
+ | |||
Hence points <math>D, D', E, E'</math> are concyclic. | Hence points <math>D, D', E, E'</math> are concyclic. | ||
Line 149: | Line 486: | ||
This follows from the uniqueness of the conjugate point and the fact that the line intersects the circle in at most two points. | This follows from the uniqueness of the conjugate point and the fact that the line intersects the circle in at most two points. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Two pares of isogonally conjugate points== | ||
+ | [[File:3 pare of points.png|400px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> and points <math>X</math> and <math>Y</math> be given. Let points <math>X'</math> and <math>Y'</math> be the isogonal conjugate of a points <math>X</math> and <math>Y</math> with respect to a triangle <math>\triangle ABC,</math> respectively. | ||
+ | |||
+ | Let <math>XY</math> cross <math>X'Y'</math> at <math>Z</math> and <math>XY'</math> cross <math>X'Y</math> at <math>Z'.</math> | ||
+ | |||
+ | Prove that point <math>Z'</math> is the isogonal conjugate of a point <math>Z</math> with respect to <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | There are two pairs of isogonals <math>CX - CX'</math> and <math>CY - CY'</math> with respect to the angle <math>\angle ACB \implies</math> | ||
+ | <math>CZ - CZ'</math> are isogonals with respect to the <math>\angle ACB</math> in accordance with <i><b>The isogonal theorem</b></i>. | ||
+ | |||
+ | Similarly <math>AZ - AZ'</math> are the isogonals with respect to the <math>\angle BAC.</math> | ||
+ | |||
+ | Therefore the point <math>Z'</math> is the isogonal conjugate of a point <math>Z</math> with respect to <math>\triangle ABC.</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Circles== | ==Circles== | ||
− | [[File:2 points isogon.png| | + | [[File:2 points isogon.png|300px|right]] |
− | Let <math>Q</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | + | Let <math>Q</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> |
+ | |||
Let <math>D</math> be the circumcenter of <math>\triangle BCP.</math> | Let <math>D</math> be the circumcenter of <math>\triangle BCP.</math> | ||
+ | |||
Let <math>E</math> be the circumcenter of <math>\triangle BCQ.</math> | Let <math>E</math> be the circumcenter of <math>\triangle BCQ.</math> | ||
+ | |||
Prove that points <math>D</math> and <math>E</math> are inverses with respect to the circumcircle of <math>\triangle ABC.</math> | Prove that points <math>D</math> and <math>E</math> are inverses with respect to the circumcircle of <math>\triangle ABC.</math> | ||
Line 163: | Line 521: | ||
The circumcenter of <math>\triangle ABC</math> point <math>O,</math> and points <math>D</math> and <math>E</math> lies on the perpendicular bisector of <math>BC.</math> | The circumcenter of <math>\triangle ABC</math> point <math>O,</math> and points <math>D</math> and <math>E</math> lies on the perpendicular bisector of <math>BC.</math> | ||
<cmath>\angle BOD = \angle COE = \angle BAC.</cmath> | <cmath>\angle BOD = \angle COE = \angle BAC.</cmath> | ||
− | <cmath>2 \angle BDO = \angle BDC = \overset{\Large\frown} {BC} = 360^\circ - \overset{\Large\frown} {CB} = 360^\circ - 2 \angle BPC.</cmath> | + | <cmath>2 \angle BDO = \angle BDC = \overset{\Large\frown} {BC} =</cmath> |
+ | <cmath>= 360^\circ - \overset{\Large\frown} {CB} = 360^\circ - 2 \angle BPC.</cmath> | ||
<cmath>\angle BDO = 180^\circ - \angle BPC = \angle PBC + \angle PCB.</cmath> | <cmath>\angle BDO = 180^\circ - \angle BPC = \angle PBC + \angle PCB.</cmath> | ||
Similarly <math>\angle CEO = 180^\circ - \angle BQC = \angle QBC + \angle QCB.</math> | Similarly <math>\angle CEO = 180^\circ - \angle BQC = \angle QBC + \angle QCB.</math> | ||
Line 175: | Line 534: | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | == Problems == | + | ==Equidistant isogonal conjugate points== |
+ | [[File:Equal distances.png|330px|right]] | ||
+ | [[File:Equidistant points.png|330px|right]] | ||
+ | Let triangle <math>ABC</math> with incenter <math>I</math> be given. | ||
+ | Denote <math>\omega = \odot BIC.</math> | ||
+ | |||
+ | Let point <math>P'</math> be the isogonal conjugate of the point <math>P</math> with respect to <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that <math>AP = AP'</math> iff <math>P \in \omega.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 1. Let <math>P \in \omega.</math> WLOG, <math>P \in \angle BAI.</math> | ||
+ | Point <math>P \in \omega \implies \angle PBI = \angle PCI.</math> | ||
+ | |||
+ | Point <math>P'</math> is the isogonal conjugate of the point <math>P</math> with respect to <math>\triangle ABC \implies</math> | ||
+ | <cmath>\angle PBI = \angle P'BI, \angle PCI = \angle P'CI \implies \angle P'BI = \angle P'CI.</cmath> | ||
+ | So points <math>B,C,I, P,</math> and <math>P'</math> are concyclic. | ||
+ | |||
+ | Let <math>E = AI \cap \odot ABC.</math> Then <math>E</math> is the center of <math>\omega \implies</math> | ||
+ | <cmath>EP = EP', \angle IEP = \angle IEP' = 2 \angle PBI.</cmath> | ||
+ | <cmath>\triangle AEP = \triangle AEP' \implies AP = AP'.</cmath> | ||
+ | |||
+ | |||
+ | 2. Let <math>AP = AP'.</math> | ||
+ | <math>\angle PAI = \angle PAE = \angle P'AI = \angle P'AE \implies</math> | ||
+ | |||
+ | Points <math>P</math> and <math>P'</math> are symmetric with respect <math>AI \implies PE = P'E.</math> | ||
+ | |||
+ | Suppose that <math>P \notin \odot BIC.</math> | ||
+ | |||
+ | Let <math>O</math> be the center of <math>\odot BPC, O'</math> be the center of <math>\odot BP'C.</math> | ||
+ | |||
+ | It is known that points <math>O</math> and <math>O'</math> are inverted with respect to the circumcircle of <math>\triangle ABC.</math> | ||
+ | |||
+ | Points <math>O, O',</math> and <math>E</math> belong to bisector <math>BC, E \in \odot ABC.</math> | ||
+ | |||
+ | Therefore <math>\overset{\Large\frown} {BIC}</math> divide <math>\overset{\Large\frown} {BPC}</math> and <math>\overset{\Large\frown} {BP'C}.</math> | ||
+ | |||
+ | WLOG (see diagram) <math>PE > IE > P'E,</math> contradiction. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==1995 USAMO Problems/Problem 3== | ||
+ | [[File:1995 USAMO 3.png|300px|right]] | ||
+ | Given a nonisosceles, nonright triangle <math>ABC,</math> let <math>O</math> denote the center of its circumscribed circle, and let <math>A_1, \, B_1,</math> and <math>C_1</math> be the midpoints of sides <math>BC, \, CA,</math> and <math>AB,</math> respectively. Point <math>A_2</math> is located on the ray <math>OA_1</math> so that <math>\triangle OAA_1</math> is similar to <math>\triangle OA_2A</math>. Points <math>B_2</math> and <math>C_2</math> on rays <math>OB_1</math> and <math>OC_1,</math> respectively, are defined similarly. Prove that lines <math>AA_2, \, BB_2,</math> and <math>CC_2</math> are concurrent. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | Let <math>AH</math> be the altitude of <math>\triangle ABC \implies</math> | ||
+ | <cmath>\angle BAH = 90^\circ - \angle ABC, \angle OAC = \angle OCA =</cmath> | ||
+ | <cmath>= \frac{180^\circ - \angle AOC}{2} = \frac{180^\circ - 2\angle ABC}{2} = \angle BAH.</cmath> | ||
+ | Hence <math>AH</math> and <math>AO</math> are isogonals with respect to the angle <math>\angle BAC.</math> | ||
+ | <cmath>\triangle OAA_1 \sim \triangle OA_2A, AH || A_1O \implies \angle AA_1O = \angle A_2AO = \angle A_1AH.</cmath> | ||
+ | <math>AA_2</math> and <math>AA_1</math> are isogonals with respect to the angle <math>\angle BAC.</math> | ||
+ | |||
+ | Similarly <math>BB_2</math> and <math>BB_1</math> are isogonals with respect to <math>\angle ABC.</math> | ||
+ | |||
+ | Similarly <math>CC_2</math> and <math>CC_1</math> are isogonals with respect to <math>\angle ACB.</math> | ||
+ | |||
+ | Let <math>G = AA_1 \cap BB_1</math> be the centroid of <math>\triangle ABC, L = AA_2 \cap BB_2.</math> | ||
+ | |||
+ | <math>L</math> is the isogonal conjugate of a point <math>G</math> with respect to a triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | <cmath>G \in CC_1 \implies L \in CC_2.</cmath> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | If median and symmedian start from any vertex of the triangle, then the angle formed by the symmedian and the angle side has the same measure as the angle between the median and the other side of the angle. | ||
+ | |||
+ | <math>AA_1, BB_1, CC_1</math> are medians, therefore <math>AA_2, BB_2, CC_2</math> are symmedians, so the three symmedians meet at a point which is triangle center called the Lemoine point. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==2011 USAMO Problems/Problem 5== | ||
+ | [[File:2011 USAMO 5.png|400px|right]] | ||
+ | Let <math>P</math> be a given point inside quadrilateral <math>ABCD</math>. Points <math>Q_1</math> and <math>Q_2</math> are located within <math>ABCD</math> such that <math>\angle Q_1 BC = \angle ABP</math>, <math>\angle Q_1 CB = \angle DCP</math>, <math>\angle Q_2 AD = \angle BAP</math>, <math>\angle Q_2 DA = \angle CDP</math>. | ||
+ | |||
+ | Prove that <math>\overline{Q_1 Q_2} \parallel \overline{AB}</math> if and only if <math>\overline{Q_1 Q_2} \parallel \overline{CD}</math>. | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | <i><b>Case 1</b></i> The lines <math>AB</math> and <math>CD</math> are not parallel. Denote <math>E = AB \cap CD.</math> | ||
+ | |||
+ | <cmath>\angle Q_1 BC = \angle ABP, \angle Q_1 CB = \angle DCP \implies</cmath> | ||
+ | Point <math>Q_1</math> isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle EBC \implies</math> | ||
+ | |||
+ | <math>EP</math> and <math>EQ_1</math> are isogonals with respect to <math>\angle BEC.</math> | ||
+ | |||
+ | Similarly point <math>Q_2</math> isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle EAD \implies</math> | ||
+ | |||
+ | <math>EP</math> and <math>EQ_2</math> are isogonals with respect to <math>\angle BEC.</math> | ||
+ | |||
+ | Therefore points <math>E, Q_1, Q_2</math> lies on the isogonal <math>EP</math> with respect to <math>\angle BEC \implies</math> | ||
+ | |||
+ | <math>Q_1Q_2</math> is not parallel to <math>AB</math> or <math>CD.</math> | ||
+ | |||
+ | <i><b>Case 2</b></i> <math>AB||CD.</math> We use <i><b>The isogonal theorem in case parallel lines</b></i> and get | ||
+ | <cmath>AB||Q_1Q_2 || CD. \blacksquare</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==2024 Sharygin olimpiad Problem 16== | ||
+ | [[File:2024 16.png|390px|right]] | ||
+ | Let <math>AA', BB',</math> and <math>CC'</math> be the bisectors of a triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | The segments <math>BB'</math> and <math>A'C'</math> meet at point <math>D.</math> Let <math>E</math> be the projection of <math>D</math> to <math>AC.</math> | ||
+ | |||
+ | Points <math>P</math> and <math>Q</math> on the sides <math>AB</math> and <math>BC,</math> respectively, are such that <math>EP = PD, EQ = QD.</math> | ||
+ | |||
+ | Prove that <math>\angle PDB' = \angle EDQ.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\triangle PDQ = \triangle PEQ (DQ = EQ, DP = PF, PQ</math> is the common side) <math>\implies</math> | ||
+ | |||
+ | <math>PQ \perp DE, F = PQ \cap DE</math> is the midpoint <math>DE.</math> | ||
+ | |||
+ | Denote <math>r</math> the inradius of <math>\triangle ABC, F'</math> is the base of perpendicular from <math>D</math> to <math>BC.</math> | ||
+ | <cmath>\frac {DF'}{r} = \frac {BD}{BI}, \frac {DE}{r} = \frac {B'D}{B'I},</cmath> | ||
+ | It is known that <cmath>\frac {B'D}{BD} = 2\frac {B'I}{BI} \implies \frac {DE}{r} = 2 \frac {DF'}{r}</cmath> | ||
+ | (see [[Bisector | Division of bisector]] for details.) | ||
+ | |||
+ | <math>DF = DF' \implies D</math> is incenter of <math>\triangle BPQ \implies</math> | ||
+ | |||
+ | <math>\angle BPD = \angle QPD = \angle QPE, \angle BQD = \angle PQD = \angle PQE \implies</math> | ||
+ | |||
+ | <math>B</math> is isogonal conjugate of E with respect <math>\triangle PDQ \implies \angle PDB' = \angle EDQ.</math> | ||
+ | |||
+ | Another solution see [[Sharygin_Olympiads,_the_best | 2024.2C_Problem_16]] | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Simplified distance formula for isogonal points== | ||
+ | [[File:1 pare and.png|370px|right]] | ||
+ | Let triangle <math>\triangle ABC,</math> points <math>P</math> and <math>P',</math> and <math>\odot ABC = \Omega</math> be given. Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | ||
+ | <cmath>D = AP \cap BC, E = AP' \cap BC, F = AP \cap \Omega, G = AP' \cap \Omega.</cmath> | ||
+ | Prove that <math>PF \cdot P'G= AF \cdot EG.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\angle AFC</math> and <math>\angle AGC</math> are both subtended by arc <math>\overset{\Large\frown} {AC} \implies \angle AFC = \angle AGC.</math> | ||
+ | <cmath>\angle PCF = \angle PCB + \angle BCF = \angle P'CA + \angle BAF = \angle P'CA + \angle P'AC=</cmath> | ||
+ | <cmath>= \angle GP'C \implies \triangle CPF \sim \triangle P'CG \implies PF \cdot P'G = FC \cdot CG.</cmath> | ||
+ | Similarly <cmath>\triangle CAF \sim \triangle ECG \implies AF \cdot EG = FC \cdot CG.</cmath> | ||
+ | [[Barycentric coordinates | Product of isogonal segments]] | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Point on circumcircle== | ||
+ | [[File:RADAX.png|350px|right]] | ||
+ | Let triangle <math>\triangle ABC,</math> points <math>D \in BC</math> and <math>E \in BC</math> be given. | ||
+ | |||
+ | Denote <math>\Omega = \odot ABC, \omega = \odot AED, G = \omega \cap \Omega \ne A, F = AG \cap BC,</math> | ||
+ | <cmath>K = AE \cap \Omega \ne A, L = GD \cap \Omega \ne G.</cmath> | ||
+ | Prove that <math>KL || BC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, the order of the points is <math>B,E,F,C,D,</math> as shown on diagram. | ||
+ | |||
+ | The spiral symilarity centered at <math>A</math> maps <math>\Omega</math> to <math>\omega</math> and point <math>L \in \Omega</math> to point <math>D \in \omega \implies \overset{\Large\frown} {AL} = \overset{\Large\frown} {AD} \implies \angle ABL = \angle AED.</math> | ||
+ | |||
+ | <math>\angle AED</math> is the external angle of <math>\triangle AEB \implies \angle AED = \angle ABC + \angle BAE \implies</math> | ||
+ | <cmath>\angle CBL = \angle BAK \implies \overset{\Large\frown} {BK} = \overset{\Large\frown} {CL} \implies BC ||KL. \blacksquare</cmath> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | <math>AL</math> is the isogonal conjugate to <math>AK</math> with respect <math>\angle BAC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | |||
+ | ==Fixed point on circumcircle== | ||
+ | [[File:Fixed point 2.png|280px|right]] | ||
+ | [[File:Fixed point 3.png|280px|right]] | ||
+ | [[File:Fixed point 4.png|280px|right]] | ||
+ | [[File:Fixed point 5.png|280px|right]] | ||
+ | Let triangle <math>\triangle ABC,</math> point <math>G \ne A</math> on circumcircle <math>\Omega = \odot ABC,</math> and point <math>D \in BC</math> be given. | ||
+ | |||
+ | Point <math>P</math> lies on <math>AG,</math> point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC, Q = DP' \cap AP, F = \odot DPQ \cap \Omega.</math> | ||
+ | |||
+ | Prove that <math>F</math> is fixed point and not depends from position of <math>P.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, the order of points on sideline is <math>B, C, D,</math> point <math>B</math> is closer to <math>AP</math> than to <math>AP'.</math> | ||
+ | |||
+ | Denote <math>Y = AP \cap BC, Z = AP' \cap BC,\omega = \odot ADY,</math> | ||
+ | <cmath>F' = \omega \cap \Omega \ne A, H = \Omega \cap F'D \ne F'.</cmath> | ||
+ | |||
+ | Spiral similarity centered at <math>A</math> which maps <math>\Omega</math> into <math>\odot AYD</math> transform point <math>H</math> into point <math>D \implies</math> | ||
+ | <cmath>\overset{\Large\frown} {ACH} = \overset{\Large\frown} {AD} \implies \angle AGH = \angle AYD \implies GH||BC \implies</cmath> <cmath>\overset{\Large\frown} {CH} = \overset{\Large\frown} {BG} \implies \angle BAG = \angle CAH \implies H \in AP'.</cmath> | ||
+ | Points <math>F', H,</math> and <math>D</math> are collinear. | ||
+ | |||
+ | It is known ([[Barycentric coordinates | Ratio of isogonal segments]]) that <math>\frac {AP'}{P'Z} \cdot \frac {AP}{PY} = \frac {AH}{HZ}.</math> | ||
+ | |||
+ | We use the ratio of the areas and get: | ||
+ | <cmath>\frac {[AQD]}{[QYD]} = \frac{AQ}{QY}, \frac {[ZQD]}{[YQD]} = \frac{ZD}{YD},</cmath> | ||
+ | <cmath>\frac {[AQD]}{[ZQD]} = \frac{AP'}{P'Z} \implies</cmath> | ||
+ | <cmath> \frac{AQ}{QY} = \frac {[AQD]}{[QYD]} = \frac {[AQD]}{[ZQD]} \cdot \frac {[ZQD]}{[YQD]} = \frac{AP'}{P'Z} \cdot \frac{ZD}{YD}.</cmath> | ||
+ | Denote <math>X = AP \cap DH.</math> | ||
+ | <cmath>\frac {[AXD]}{[YXD]} = \frac{AX}{XY}, \frac {[XZD]}{[XYD]} = \frac{ZD}{YD},</cmath> | ||
+ | <cmath>\frac {[AXD]}{[XZD]} = \frac{AH}{HZ} \implies</cmath> | ||
+ | <cmath> \frac{AX}{XY} = \frac {[AXD]}{[YXD]} = \frac {[AXD]}{[XZD]} \cdot \frac {[XZD]}{[YXD]} = \frac{AH}{HZ} \cdot \frac{ZD}{YD}.</cmath> | ||
+ | Therefore <math>\frac{AX}{XY} = \frac{AQ}{QY} \cdot \frac{AP}{PY}</math> which means ([[Radical axis | Problems | Simple]]) that <math>DX</math> is the radical axes of <math>\omega</math> and <math>\odot DPQ \implies</math> | ||
+ | |||
+ | <math>F = F'</math> and not depends from position of <math>P.</math> | ||
+ | |||
+ | [[Barycentric coordinates | Fixed point on circumcircle]] | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Distance formula for isogonal points== | ||
+ | [[File:Isogonal formulas.png|400px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> and point <math>P</math> be given. | ||
+ | |||
+ | Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | Let lines <math>AP</math> and <math>AP'</math> cross sideline <math>BC</math> at <math>D</math> and <math>E</math> and circumcircle of <math>\triangle ABC</math> at <math>F</math> and <math>G,</math> respectively. | ||
+ | |||
+ | We apply the Isogonal’s property and get <math>\frac {BD}{DC} \cdot \frac{BE}{EC} = \frac {AB^2}{AC^2}.</math> | ||
+ | |||
+ | <math>EF || BC.</math> We apply the Ptolemy's theorem to <math>BCGF</math> and get <cmath>BC \cdot FG = BG^2 – BF^2.</cmath> | ||
+ | |||
+ | We apply the barycentric coordinates and get | ||
+ | <cmath>\left| \frac{BE}{EC} - \frac {BD}{DC} \right| = \frac {AB \cdot BC \cdot FG}{AC \cdot PF \cdot P'G}.</cmath> | ||
+ | *[[Barycentric coordinates]] | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Miquel point for isogonal conjugate points== | ||
+ | [[File:Miquel of one pare.png|370px|right]] | ||
+ | [[File:Miquel 1 pare.png|370px|right]] | ||
+ | Let triangle <math>\triangle ABC,</math> points <math>Q \in BC</math> and <math>P</math> be given. Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math> | ||
+ | <cmath>D = QP \cap AP', E = AP \cap QP'.</cmath> | ||
+ | Let <math>M</math> be the Miquel point of a complete quadrilateral <math>PDP'E.</math> | ||
+ | |||
+ | Prove that <math>M</math> lies on the circumcircle of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Point <math>A</math> is the isogonal conjugate of a point <math>Q</math> with respect to a triangle <math>\triangle ABC,</math> so point <math>D</math> is the isogonal conjugate of a point <math>E</math> with respect to a triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | Points <math>P</math> and <math>E</math> lies on the same line, therefore <cmath>PF \cdot P'G = EF \cdot DG \implies</cmath> | ||
+ | |||
+ | <cmath>\frac {PF-EP}{PF} = \frac{DG-DP'}{DG} \implies \frac {EP}{PF} = \frac{DP'}{DG}.</cmath> | ||
+ | Point <math>M</math> lies on circles <math>APD</math> and <math>AEP' \implies </math> spiral similarity centered at <math>M</math> transform triangle <math>\triangle MPE</math> to <math>\triangle MDP' \implies</math> | ||
+ | <cmath>\angle MPE = \angle MDP', \frac {MP}{MD} = \frac {PE}{DP'} = \frac{PF}{DG} \implies</cmath> | ||
+ | <cmath>\triangle MPF \sim \triangle MDG \implies</cmath> <cmath>\angle AFM = \angle AGM \implies M \in \odot ABC.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Point on circumcircle== | ||
+ | [[File:Point on circumcircle 0.png|400px|right]] | ||
+ | Let triangle <math>\triangle ABC,</math> and points <math>D \in BC</math> and <math>E \in \odot ABC = \Omega</math> be given. | ||
+ | |||
+ | Let <math>\odot ADE = \omega, F = BC \cap \omega \ne D.</math> | ||
+ | |||
+ | Let lines <math>AF</math> and <math>AG (G \in BC)</math> be the isogonals with respect to the angle <math>\angle BAC, \odot AGD = \theta.</math> | ||
+ | |||
+ | Let <math>P</math> be an arbitrary point on <math>AF, Q = DP \cap AG, H = \theta \cap \Omega.</math> | ||
+ | |||
+ | Prove that <math>X = EP \cap HQ</math> lies on <math>\Omega.</math> | ||
+ | |||
+ | ===Simplified problem=== | ||
+ | Let <math>\triangle ABC,</math> and points <math>D \in BC</math> and <math>E \in \odot ABC = \Omega</math> be given, <math>\omega = \odot ADE, F = \omega \cap BC \ne D.</math> | ||
+ | |||
+ | Let lines <math>AF</math> and <math>AG (G \in BC)</math> be the isogonals with respect to <math>\angle BAC, \theta = \odot AGD, H = \theta \cap \Omega.</math> | ||
+ | |||
+ | Prove that <math>X = EF \cap HG \in \Omega.</math> | ||
+ | [[File:Point on circumcircle 1.png|400px|right]] | ||
+ | |||
+ | <i><b>Proof, Simplified problem</b></i> | ||
+ | <cmath>\angle CGH = \angle DAH = \frac{1}{2} \overset{\Large\frown} {HD} (\theta),</cmath> | ||
+ | <cmath>\angle DFE = \angle DAE = \frac{1}{2} \overset{\Large\frown} {ED} (\omega),</cmath> | ||
+ | <cmath>\angle EAH = \angle DAE - \angle DAH = \frac{1}{2} \overset{\Large\frown} {EH}(\Omega),</cmath> | ||
+ | <cmath>\angle EXH = \angle DFE - \angle DGH = \angle EAH \implies</cmath> | ||
+ | |||
+ | points <math>A, H, E, X</math> are concyclic on <math>\Omega.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | [[File:Point on circumcircle 2.png|400px|right]] | ||
+ | Let points <math>P'</math> and <math>Q'</math> be the isogonal conjugate of a points <math>P</math> and <math>Q</math> with respect to a triangle <math>\triangle ABC, \omega' = \odot Q'PD, \theta' = \odot P'QD.</math> | ||
+ | |||
+ | It is known that <math>E \in \omega', H \in \theta', \omega' \cap \theta' \cap \Omega = M.</math> | ||
+ | |||
+ | <cmath>\angle DQH = \angle DMH = \frac{1}{2} \overset{\Large\frown} {HD} (\theta'),</cmath> | ||
+ | <cmath>\angle DPE = \angle DME = \frac{1}{2} \overset{\Large\frown} {ED} (\omega'),</cmath> | ||
+ | <cmath>\angle EMH = \angle DME - \angle DMH = \frac{1}{2} \overset{\Large\frown} {EH}(\Omega),</cmath> | ||
+ | <cmath>\angle EXH = \angle DPE - \angle DQH = \angle EMH \implies</cmath> | ||
+ | points <math>M, H, E, X</math> are concyclic on <math>\Omega.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Isogonal of line BC with respect to angle BAC== | ||
+ | [[File:Isogonal of BC.png|350px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> be given, <math>\Omega = \odot ABC, AD || BC.</math> | ||
+ | |||
+ | Let lines <math>AE</math> and <math>AD</math> be the isogonals with respect to <math>\angle BAC.</math> | ||
+ | |||
+ | Prove that <math>AE</math> is tangent to <math>\Omega.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>O</math> and <math>H</math> be the circumcenter and the orthocenter of <math>\triangle ABC,</math> respectively. | ||
+ | <cmath>AH \perp BC, AD || BC \implies AH \perp AD.</cmath> | ||
+ | <math>AH</math> is isogonal to <math>AO</math> with respect to <math>\angle BAC \implies AE \perp AO \implies AE</math> is tangent to <math>\Omega.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Isogonal bijection lines and points== | ||
+ | [[File:Isogonal of l.png|350px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> and line <math>\ell, P \in \ell</math> be given, <math>\Omega = \odot ABC.</math> | ||
+ | |||
+ | Define <math>G \in \Omega</math> the point with property <math>G' \in \ell.</math> | ||
+ | |||
+ | Prove that <math>\angle ABG</math> is equal the angle <math>\theta</math> between <math>\ell</math> and <math>BC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, the configuration is the same as shown on diagram, <math>F = \ell \cap BC, AD' || \ell, \theta = \angle PFB, AD || BC, AE</math> is the tangent to <math>\Omega.</math> | ||
+ | |||
+ | <math>AD</math> is isogonal to <math>AE, AD'</math> is isogonal to <math>AG</math> with respect to <math>\angle BAC \implies</math> | ||
+ | <cmath>\theta = \angle PFB = \angle D'AD = \angle GAE = \angle GBA.</cmath> | ||
+ | A bijection has been established between the set of lines parallel to a given one and the set of points of the circumcircle. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | |||
+ | ==Miquel point for two pare isogonal points== | ||
+ | [[File:2 pare Miquel o.png|400px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> and points <math>P</math> and <math>Q</math> be given. | ||
+ | |||
+ | Let points <math>P'</math> and <math>Q'</math> be the isogonal conjugate of the points <math>P</math> and <math>Q</math> with respect to <math>\triangle ABC, \Omega = \odot ABC, M</math> is the Miquel point of quadrilateral <math>PQP'Q'.</math> | ||
+ | |||
+ | Prove that <math>M \in \Omega.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>R = PQ \cap P'Q', \Theta = \odot P'QR, \theta = \odot PQ'R.</math> | ||
+ | |||
+ | Then <math>M = \theta \cap \Theta</math> is the Miquel point of quadrilateral <math>PQP'Q'.</math> | ||
+ | |||
+ | Denote <math>E = \theta \cap \Omega \notin \Theta, F = \Theta \cap \Omega \notin \theta.</math> | ||
+ | |||
+ | Let <math>D \in \Omega</math> be the point with property <math>D' \in PQ.</math> | ||
+ | |||
+ | WLOG, configuration is similar as shown in diagram. | ||
+ | |||
+ | <math>P' \in DF, Q' \in DE</math> ([[Isogonal_conjugate | Isogonal_bijection_lines_and_points]]). | ||
+ | <cmath>\angle EMF = \angle RME - \angle RMF = \angle RQ'E - \angle RP'F = \angle P'Q'E - \angle DP'Q' = \angle P'DQ' = \angle EDF \implies M \in \odot DEF \blacksquare</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Isogonic center’s conjugate point== | ||
+ | [[File:Fermat 1.png|500px|right]] | ||
+ | [[File:Fermat 2.png|300px|right]] | ||
+ | Let triangle <math>ABC</math> with isogonic center <math>F (X(13)</math> or <math>X(14))</math> be given. Denote <math>\omega = \odot BIC.</math> | ||
+ | |||
+ | Let line <math>\ell_A</math> be the axial symmetry of line <math>AF</math> according to the sideline <math>BC.</math> | ||
+ | |||
+ | Define lines <math>\ell_B</math> and <math>\ell_C</math> similarly. | ||
+ | |||
+ | Prove that the lines <math>\ell_A, \ell_B,</math> and <math>\ell_C</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>I</math> be the incenter of <math>\triangle ABC, F =X(13).</math> | ||
+ | <cmath>A_1 = AF \cup BC, E = AI \cup BC, D \in BC, AD \perp AE, \omega = \odot AED.</cmath> | ||
+ | Let <math>F_1 = AF \cup \omega, F'</math> is simmetric to <math>F_1</math> with respect <math>BC \implies A_1F' = \ell_A.</math> | ||
+ | |||
+ | The diameter <math>DE</math> of <math>\omega</math> lies on <math>BC \implies F_1 \in \omega, \overset{\Large\frown} {EF_1} = \overset{\Large\frown} {EF'} \implies \angle EAF_1 = \angle EAF'.</math> | ||
+ | |||
+ | Therefore <math>AF'</math> is the isogonal conjugate of <math>AF</math> with respect to <math>\angle BAC.</math> | ||
+ | |||
+ | Similarly <math>\ell_B</math> and <math>\ell_C</math> are the isogonal conjugate of <math>BF</math> and <math>CF,</math> so point <math>F'</math> is the isogonal conjugate of point <math>F</math> with respect to <math>\triangle ABC.</math> | ||
+ | |||
+ | The second diagram show construction in the case <math>F =X(14).</math> The proof is similar. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Three pairs isogonal points== | ||
+ | [[File:Shar 2024 20.png|400px|right]] | ||
+ | Let a triangle <math>ABC,</math> points <math>D</math> and <math>E \in AD</math> be given, <math>F = CD \cap BE.</math> | ||
+ | Points <math>D', E'</math> and <math>F'</math> are the isogonal conjugate of the points <math>D, E,</math> and <math>F,</math> respectively, with respect to <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that <math>\frac {AD}{AE} \cdot \frac {BE}{BF} \cdot \frac {CF}{CD} = \frac {AE'}{AD'} \cdot \frac {BF'}{BE'} \cdot \frac {CD'}{CF'}.</math> | ||
− | === | + | <i><b>Proof</b></i> |
+ | |||
+ | Denote <math>\angle BAC = \alpha, \angle ABC = \beta, \angle ACB = \gamma,</math> | ||
+ | <cmath>\angle BAD = \varphi_A, \angle CBE = \varphi_B, \angle ACD = \varphi_C.</cmath> | ||
+ | We use isogonal properties and get | ||
+ | <cmath>\angle CAD' = \varphi_A, \angle ABE' = \varphi_B, \angle BCD' = \varphi_C.</cmath> | ||
+ | By applying the Law of Sines, we get <cmath>\frac {BE}{AE} = \frac {\sin \varphi_A}{\sin (\beta - \varphi_B)}, \frac {CF}{BF} = \frac {\sin \varphi_B}{\sin (\gamma - \varphi_C)}, \frac {AD}{CD} = \frac {\sin \varphi_C}{\sin (\alpha - \varphi_A)}.</cmath> | ||
+ | Symilarly, <cmath>\frac {AE'}{BE'} = \frac {\sin \varphi_B}{\sin (\alpha - \varphi_A)}, \frac {BF'}{CF'} = \frac {\sin \varphi_C}{\sin (\beta - \varphi_B)}, \frac {CD'}{AD'} = \frac {\sin \varphi_A}{\sin (\gamma - \varphi_C)}.</cmath> | ||
+ | We multiply these equations and get | ||
+ | <cmath>\frac {AE \cdot BF \cdot CD}{AD \cdot BE \cdot CF} = \frac{AD' \cdot BE' \cdot CF'}{AE' \cdot BF' \cdot CD'} = \frac {\sin \varphi_A \cdot \sin \varphi_B \cdot \sin \varphi_C}{\sin (\alpha - \varphi_A) \cdot \sin (\beta - \varphi_B) \cdot \sin (\gamma - \varphi_C)}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Ratio for three pairs of isogonal points== | ||
+ | [[File:Shar 2024 20 1.png|400px|right]] | ||
+ | [[File:Shar 2024 20 2.png|400px|right]] | ||
+ | Let a triangle <math>ABC,</math> points <math>D</math> and <math>E \in AD</math> be given, <math>F = CD \cap BE.</math> | ||
+ | |||
+ | Points <math>D', E'</math> and <math>F'</math> are the isogonal conjugate of the points <math>D, E,</math> and <math>F,</math> respectively, with respect to <math>\triangle ABC.</math> | ||
+ | |||
+ | Denote <math>R</math> and <math>R'</math> the circumradii of triangles <math>\triangle DEF</math> and <math>\triangle D'E'F',</math> respectively. | ||
+ | |||
+ | Prove that <math>\frac {AD \cdot BE \cdot CF}{R} = \frac {AE' \cdot BF' \cdot CD'}{R'}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
− | + | Denote <math>\frac {AD}{DE} = u, \frac {BE}{EF} = w, \frac {CF}{FD} = v.</math> | |
− | + | <math>[DEF] = 1,</math> where <math>[X]</math> is the area of the figure <math>X.</math> | |
+ | <cmath>\frac {[ADF]}{[DEF]} = \frac {AD}{DE} = u, \frac {[ACF]}{[DEF]} = \frac {[ACF]}{[ADF]} \cdot \frac {[ADF]}{[DEF]} = uv.</cmath> | ||
+ | Similarly, <cmath>\frac {[CFE]}{[DEF]} = v, \frac {[BCE]}{[DEF]} = vw, \frac {[BDE]}{[DEF]} = w, \frac {[ABD]}{[DEF]} = uw.</cmath> | ||
+ | <cmath>\frac {[ABC]}{[DEF]} = 1 + u + v + w + uv + uw + vw = (1 + u)(1 + v)(1 + w) - uvw,</cmath> | ||
+ | <cmath>\frac {[ABC]}{[DEF]} = \frac {AE \cdot BF \cdot CD}{DE \cdot DF \cdot EF} - \frac {AD \cdot BE \cdot CF}{DE \cdot DF \cdot EF}.</cmath> | ||
+ | <cmath>\frac {DE \cdot DF \cdot EF}{4R} = [DEF] \implies</cmath> | ||
+ | <cmath>[ABC] = \frac {AD \cdot BE \cdot CF}{4R} \cdot \left ( \frac {AE \cdot BF \cdot CD}{BE \cdot CF \cdot AD}-1 \right).</cmath> | ||
+ | Similarly, <cmath>[ABC] = \frac {AE' \cdot BF' \cdot CD'}{4R'} \cdot \left ( \frac {AD' \cdot BE' \cdot CD}{AE' \cdot BF' \cdot CD'}-1 \right).</cmath> | ||
+ | It is known that <math>\frac {AD}{AE} \cdot \frac {BE}{BF} \cdot \frac {CF}{CD} = \frac {AE'}{AD'} \cdot \frac {BF'}{BE'} \cdot \frac {CD'}{CF'}</math> ([[Isogonal_conjugate | Three pairs isogonal points]]), therefore | ||
+ | <cmath>\frac {AD \cdot BE \cdot CF}{4R} = \frac {AE' \cdot BF' \cdot CD'}{4R'}.</cmath> | ||
+ | Comment: The main idea of the proof was found by Leonid Shatunov. | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
[[Category:Geometry]] | [[Category:Geometry]] |
Latest revision as of 07:40, 18 April 2024
Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.
Contents
[hide]- 1 The isogonal theorem
- 2 Parallel segments
- 3 Perpendicularity
- 4 Fixed point
- 5 Bisector
- 6 Isogonal of the diagonal of a quadrilateral
- 7 Isogonals in trapezium
- 8 Isogonals in complete quadrilateral
- 9 Isogonal of the bisector of the triangle
- 10 Points on isogonals
- 11 Trapezoid
- 12 IMO 2007 Short list/G3
- 13 Definition of isogonal conjugate of a point
- 14 Three points
- 15 Second definition
- 16 Distance to the sides of the triangle
- 17 Sign of isogonally conjugate points
- 18 Circumcircle of pedal triangles
- 19 Common circumcircle of the pedal triangles as the sign of isogonally conjugate points
- 20 Two pares of isogonally conjugate points
- 21 Circles
- 22 Equidistant isogonal conjugate points
- 23 1995 USAMO Problems/Problem 3
- 24 2011 USAMO Problems/Problem 5
- 25 2024 Sharygin olimpiad Problem 16
- 26 Simplified distance formula for isogonal points
- 27 Point on circumcircle
- 28 Fixed point on circumcircle
- 29 Distance formula for isogonal points
- 30 Miquel point for isogonal conjugate points
- 31 Point on circumcircle
- 32 Isogonal of line BC with respect to angle BAC
- 33 Isogonal bijection lines and points
- 34 Miquel point for two pare isogonal points
- 35 Isogonic center’s conjugate point
- 36 Three pairs isogonal points
- 37 Ratio for three pairs of isogonal points
The isogonal theorem
Isogonal lines definition
Let a line and a point lying on be given. A pair of lines symmetric with respect to and containing the point be called isogonals with respect to the pair
Sometimes it is convenient to take one pair of isogonals as the base one, for example, and are the base pair. Then we call the remaining pairs as isogonals with respect to the angle
Projective transformation
It is known that the transformation that maps a point with coordinates into a point with coordinates is projective.
If the abscissa axis coincides with the line and the origin coincides with the point then the isogonals define the equations and the lines symmetrical with respect to the line become their images.
It is clear that, under the converse transformation (also projective), such pairs of lines become isogonals, and the points equidistant from lie on the isogonals.
The isogonal theorem
Let two pairs of isogonals and with respect to the pair be given. Denote
Prove that and are the isogonals with respect to the pair
Proof
Let us perform a projective transformation of the plane that maps the point into a point at infinity and the line maps to itself. In this case, the isogonals turn into a pair of straight lines parallel to and equidistant from
The converse (also projective) transformation maps the points equidistant from onto isogonals. We denote the image and the preimage with the same symbols.
Let the images of isogonals are vertical lines. Let coordinates of images of points be Equation of a straight line is
Equation of a straight line is
The abscissa of the point is
Equation of a straight line is
Equation of a straight line is
The abscissa of the point is
Preimages of the points and lie on the isogonals.
The isogonal theorem in the case of parallel lines
Let and are isogonals with respect
Let lines and intersect at point
Prove that and line through parallel to are the isogonals with respect
Proof
The preimage of is located at infinity on the line
The equality implies the equality the slopes modulo of and to the bisector of
Converse theorem
Let lines and intersect at point
Let and be the isogonals with respect
Prove that and are isogonals with respect
Proof
The preimage of is located at infinity on the line so the slope of is known.
Suppose that
The segment and the lines are fixed
intersects at
but there is the only point where line intersect Сontradiction.
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Parallel segments
Let triangle be given. Let and be the isogonals with respect Let
Prove that lies on bisector of and
Proof
Both assertions follow from The isogonal theorem in the case of parallel lines
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Perpendicularity
Let triangle be given. Right triangles and with hypotenuses and are constructed on sides and to the outer (inner) side of Let Prove that
Proof
Let be the bisector of
and are isogonals with respect to the pair
and are isogonals with respect to the pair
and are isogonals with respect to the pair in accordance with The isogonal theorem.
is the diameter of circumcircle of
Circumradius and altitude are isogonals with respect bisector and vertex of triangle, so
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Fixed point
Let fixed triangle be given. Let points and on sidelines and respectively be the arbitrary points.
Let be the point on sideline such that
Prove that line pass through the fixed point.
Proof
We will prove that point symmetric with respect lies on .
and are isogonals with respect to
points and lie on isogonals with respect to in accordance with The isogonal theorem.
Point symmetric with respect lies on isogonal with respect to that is
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Bisector
Let a convex quadrilateral be given. Let and be the incenters of triangles and respectively.
Let and be the A-excenters of triangles and respectively.
Prove that is the bisector of
Proof
and are isogonals with respect to the angle
and are isogonals with respect to the angle in accordance with The isogonal theorem.
Denote
WLOG,
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Isogonal of the diagonal of a quadrilateral
Given a quadrilateral and a point on its diagonal such that
Let
Prove that
Proof
Let us perform a projective transformation of the plane that maps the point to a point at infinity and the line into itself.
In this case, the images of points and are equidistant from the image of
the point (midpoint of lies on
contains the midpoints of and
is the Gauss line of the complete quadrilateral bisects
the preimages of the points and lie on the isogonals and
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Isogonals in trapezium
Let the trapezoid be given. Denote
The point on the smaller base is such that
Prove that
Proof
Therefore and are isogonals with respect
Let us perform a projective transformation of the plane that maps the point to a point at infinity and the line into itself.
In this case, the images of points and are equidistant from the image of contains the midpoints of and , that is, is the Gauss line of the complete quadrilateral
bisects
The preimages of the points and lie on the isogonals and
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Isogonals in complete quadrilateral
Let complete quadrilateral be given. Let be the Miquel point of
Prove that is isogonal to and is isogonal to with respect
Proof
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Isogonal of the bisector of the triangle
The triangle be given. The point chosen on the bisector
Denote Prove that
Proof
Let us perform a projective transformation of the plane that maps the point to a point at infinity and the line into itself.
In this case, the images of segments and are equidistant from the image of
Image of point is midpoint of image and midpoint image
Image is parallelogramm
distances from and to are equal
Preimages and are isogonals with respect
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Points on isogonals
The triangle be given. The point chosen on The point chosen on such that and are isogonals with respect
Prove that
Proof
Denote
We use the Law of Sines and get:
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Trapezoid
The lateral side of the trapezoid is perpendicular to the bases, point is the intersection point of the diagonals .
Point is taken on the circumcircle of triangle diametrically opposite to point
Prove that
Proof
WLOG, is not the diameter of Let sidelines and intersect at points and respectively.
is rectangle
is isogonal to with respect
is isogonal to with respect
In accordance with The isogonal theorem in case parallel lines
is isogonal to with respect
in accordance with Converse theorem for The isogonal theorem in case parallel lines.
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IMO 2007 Short list/G3
The diagonals of a trapezoid intersect at point
Point lies between the parallel lines and such that and line separates points and
Prove that
Proof
and are isogonals with respect
is isogonal to with respect
From the converse of The isogonal theorem we get
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Definition of isogonal conjugate of a point
Let triangle be given. Let be the circumcircle of Let point be in the plane of Denote by the lines respectively. Denote by the lines , , , respectively. Denote by , , the reflections of , , over the angle bisectors of angles , , , respectively.
Prove that lines , , concur at a point This point is called the isogonal conjugate of with respect to triangle .
Proof
By our constructions of the lines , , and this statement remains true after permuting . Therefore by the trigonometric form of Ceva's Theorem so again by the trigonometric form of Ceva, the lines concur, as was to be proven.
Corollary
Let points P and Q lie on the isogonals with respect angles and of triangle
Then these points lie on isogonals with respect angle
Corollary 2
Let point be in the sideline of
Then the isogonal conjugate of a point is a point
Points and do not have an isogonally conjugate point.
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Three points
Let fixed triangle be given. Let the arbitrary point not be on sidelines of Let be the point on isogonal of with respect angle Let be the crosspoint of isogonal of with respect angle and isogonal of with respect angle
Prove that lines and are concurrent.
Proof
Denote
and are isogonals with respect
and S lie on isogonals of
is isogonal conjugated of with respect
and lie on isogonals of
Therefore points and lie on the same line which is isogonal to with respect
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Second definition
Let triangle be given. Let point lies in the plane of Let the reflections of in the sidelines be
Then the circumcenter of the is the isogonal conjugate of
Points and have not isogonal conjugate points.
Another points of sidelines have points respectively as isogonal conjugate points.
Proof is common therefore Similarly is the circumcenter of the
From definition 1 we get that is the isogonal conjugate of
It is clear that each point has the unique isogonal conjugate point.
Let point be the point with barycentric coordinates Then has barycentric coordinates
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Distance to the sides of the triangle
Let be the isogonal conjugate of a point with respect to a triangle
Let and be the projection on sides and respectively.
Let and be the projection on sides and respectively.
Then
Proof
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Sign of isogonally conjugate points
Let triangle and points and inside it be given.
Let be the projections on sides respectively.
Let be the projections on sides respectively.
Let Prove that point is the isogonal conjugate of a point with respect to a triangle
One can prove a similar theorem in the case outside
Proof
Denote Similarly Hence point is the isogonal conjugate of a point with respect to a triangle
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Circumcircle of pedal triangles
Let be the isogonal conjugate of a point with respect to a triangle
Let be the projection on sides respectively.
Let be the projection on sides respectively.
Prove that points are concyclic.
The midpoint is circumcenter of
Proof
Let
Hence points are concyclic.
is trapezoid,
the midpoint is circumcenter of
Similarly points are concyclic and points are concyclic.
Therefore points are concyclic, so the midpoint is circumcenter of
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Common circumcircle of the pedal triangles as the sign of isogonally conjugate points
Let triangle and points and inside it be given. Let be the projections on sides respectively. Let be the projections on sides respectively.
Let points be concyclic and none of them lies on the sidelines of
Then point is the isogonal conjugate of a point with respect to a triangle
This follows from the uniqueness of the conjugate point and the fact that the line intersects the circle in at most two points.
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Two pares of isogonally conjugate points
Let triangle and points and be given. Let points and be the isogonal conjugate of a points and with respect to a triangle respectively.
Let cross at and cross at
Prove that point is the isogonal conjugate of a point with respect to
Proof
There are two pairs of isogonals and with respect to the angle are isogonals with respect to the in accordance with The isogonal theorem.
Similarly are the isogonals with respect to the
Therefore the point is the isogonal conjugate of a point with respect to
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Circles
Let be the isogonal conjugate of a point with respect to a triangle
Let be the circumcenter of
Let be the circumcenter of
Prove that points and are inverses with respect to the circumcircle of
Proof
The circumcenter of point and points and lies on the perpendicular bisector of Similarly
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Equidistant isogonal conjugate points
Let triangle with incenter be given. Denote
Let point be the isogonal conjugate of the point with respect to
Prove that iff
Proof
1. Let WLOG, Point
Point is the isogonal conjugate of the point with respect to So points and are concyclic.
Let Then is the center of
2. Let
Points and are symmetric with respect
Suppose that
Let be the center of be the center of
It is known that points and are inverted with respect to the circumcircle of
Points and belong to bisector
Therefore divide and
WLOG (see diagram) contradiction.
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1995 USAMO Problems/Problem 3
Given a nonisosceles, nonright triangle let denote the center of its circumscribed circle, and let and be the midpoints of sides and respectively. Point is located on the ray so that is similar to . Points and on rays and respectively, are defined similarly. Prove that lines and are concurrent.
Solution
Let be the altitude of Hence and are isogonals with respect to the angle and are isogonals with respect to the angle
Similarly and are isogonals with respect to
Similarly and are isogonals with respect to
Let be the centroid of
is the isogonal conjugate of a point with respect to a triangle
Corollary
If median and symmedian start from any vertex of the triangle, then the angle formed by the symmedian and the angle side has the same measure as the angle between the median and the other side of the angle.
are medians, therefore are symmedians, so the three symmedians meet at a point which is triangle center called the Lemoine point.
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2011 USAMO Problems/Problem 5
Let be a given point inside quadrilateral . Points and are located within such that , , , .
Prove that if and only if .
Solution
Case 1 The lines and are not parallel. Denote
Point isogonal conjugate of a point with respect to a triangle
and are isogonals with respect to
Similarly point isogonal conjugate of a point with respect to a triangle
and are isogonals with respect to
Therefore points lies on the isogonal with respect to
is not parallel to or
Case 2 We use The isogonal theorem in case parallel lines and get
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2024 Sharygin olimpiad Problem 16
Let and be the bisectors of a triangle
The segments and meet at point Let be the projection of to
Points and on the sides and respectively, are such that
Prove that
Proof
is the common side)
is the midpoint
Denote the inradius of is the base of perpendicular from to It is known that (see Division of bisector for details.)
is incenter of
is isogonal conjugate of E with respect
Another solution see 2024.2C_Problem_16
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Simplified distance formula for isogonal points
Let triangle points and and be given. Let point be the isogonal conjugate of a point with respect to a triangle Prove that
Proof
and are both subtended by arc Similarly Product of isogonal segments
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Point on circumcircle
Let triangle points and be given.
Denote Prove that
Proof
WLOG, the order of the points is as shown on diagram.
The spiral symilarity centered at maps to and point to point
is the external angle of
Corollary
is the isogonal conjugate to with respect
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Fixed point on circumcircle
Let triangle point on circumcircle and point be given.
Point lies on point be the isogonal conjugate of a point with respect to a triangle
Prove that is fixed point and not depends from position of
Proof
WLOG, the order of points on sideline is point is closer to than to
Denote
Spiral similarity centered at which maps into transform point into point Points and are collinear.
It is known ( Ratio of isogonal segments) that
We use the ratio of the areas and get: Denote Therefore which means ( Problems | Simple) that is the radical axes of and
and not depends from position of
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Distance formula for isogonal points
Let triangle and point be given.
Let point be the isogonal conjugate of a point with respect to a triangle
Let lines and cross sideline at and and circumcircle of at and respectively.
We apply the Isogonal’s property and get
We apply the Ptolemy's theorem to and get
We apply the barycentric coordinates and get
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Miquel point for isogonal conjugate points
Let triangle points and be given. Let point be the isogonal conjugate of a point with respect to a triangle Let be the Miquel point of a complete quadrilateral
Prove that lies on the circumcircle of
Proof
Point is the isogonal conjugate of a point with respect to a triangle so point is the isogonal conjugate of a point with respect to a triangle
Points and lies on the same line, therefore
Point lies on circles and spiral similarity centered at transform triangle to
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Point on circumcircle
Let triangle and points and be given.
Let
Let lines and be the isogonals with respect to the angle
Let be an arbitrary point on
Prove that lies on
Simplified problem
Let and points and be given,
Let lines and be the isogonals with respect to
Prove that
Proof, Simplified problem
points are concyclic on
Proof
Let points and be the isogonal conjugate of a points and with respect to a triangle
It is known that
points are concyclic on
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Isogonal of line BC with respect to angle BAC
Let triangle be given,
Let lines and be the isogonals with respect to
Prove that is tangent to
Proof
Let and be the circumcenter and the orthocenter of respectively. is isogonal to with respect to is tangent to
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Isogonal bijection lines and points
Let triangle and line be given,
Define the point with property
Prove that is equal the angle between and
Proof
WLOG, the configuration is the same as shown on diagram, is the tangent to
is isogonal to is isogonal to with respect to A bijection has been established between the set of lines parallel to a given one and the set of points of the circumcircle.
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Miquel point for two pare isogonal points
Let triangle and points and be given.
Let points and be the isogonal conjugate of the points and with respect to is the Miquel point of quadrilateral
Prove that
Proof
Denote
Then is the Miquel point of quadrilateral
Denote
Let be the point with property
WLOG, configuration is similar as shown in diagram.
( Isogonal_bijection_lines_and_points).
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Isogonic center’s conjugate point
Let triangle with isogonic center or be given. Denote
Let line be the axial symmetry of line according to the sideline
Define lines and similarly.
Prove that the lines and are concurrent.
Proof
Let be the incenter of Let is simmetric to with respect
The diameter of lies on
Therefore is the isogonal conjugate of with respect to
Similarly and are the isogonal conjugate of and so point is the isogonal conjugate of point with respect to
The second diagram show construction in the case The proof is similar.
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Three pairs isogonal points
Let a triangle points and be given, Points and are the isogonal conjugate of the points and respectively, with respect to
Prove that
Proof
Denote We use isogonal properties and get By applying the Law of Sines, we get Symilarly, We multiply these equations and get vladimir.shelomovskii@gmail.com, vvsss
Ratio for three pairs of isogonal points
Let a triangle points and be given,
Points and are the isogonal conjugate of the points and respectively, with respect to
Denote and the circumradii of triangles and respectively.
Prove that
Proof
Denote
where is the area of the figure Similarly, Similarly, It is known that ( Three pairs isogonal points), therefore Comment: The main idea of the proof was found by Leonid Shatunov.
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