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| | + | [[File:Isogonal.png|390px|right]] |
− | Let two pairs of isogonals <math>OX | + | 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> | 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| | + | [[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> | 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 | + | 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> | 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} | + | 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} | + | 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} | + | 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} | + | 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> | 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> | <i><b>The isogonal theorem in the case of parallel lines</b></i> | ||
− | [[File:Parallels 1.png| | + | [[File:Parallels 1.png|330px|right]] |
Let <math>OY</math> and <math>OY'</math> are isogonals with respect <math>\angle XOX'.</math> | Let <math>OY</math> and <math>OY'</math> are isogonals with respect <math>\angle XOX'.</math> | ||
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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> | 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> | + | <i><b>Converse theorem</b></i> |
− | [[File:Parallels 2.png| | + | [[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 lines <math>XY</math> and <math>X'Y'</math> intersect at point <math>Z, X'Y || XY'.</math> | ||
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Suppose that <math>y' \in XY', y' \ne Y', \angle y'OX = \angle YOX'.</math> | 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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Perpendicularity== | ==Perpendicularity== | ||
− | [[File:Right angles.png| | + | [[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> | 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> | Prove that <math>AH \perp BC.</math> | ||
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<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>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>\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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Fixed point== | ==Fixed point== | ||
− | [[File:Fixed point.png| | + | [[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 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. | |
− | |||
− | Prove that line <math>DG</math> pass through the fixed point. | ||
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
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We will prove that point <math>A',</math> symmetric <math>A</math> with respect <math>\ell = BC,</math> lies on <math>DG</math>. | 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 | + | <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> | <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> | ||
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==Bisector== | ==Bisector== | ||
− | [[File:Incircles.png| | + | [[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> | + | 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> | Prove that <math>CE</math> is the bisector of <math>\angle BCD.</math> | ||
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<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
− | <math>\angle ICI' = \angle JCJ' = 90^\circ \implies CI'</math> and <math>CJ'</math> are isogonals with respect to the angle <math>\angle ICJ.</math> | + | <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> | <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> | ||
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Let <math>E = AB \cap CD, F = AD \cap BC.</math> | Let <math>E = AB \cap CD, F = AD \cap BC.</math> | ||
− | Prove that <math>\angle | + | Prove that <math>\angle BPE = \angle DPF.</math> |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
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<math>\ell</math> is the Gauss line of the complete quadrilateral <math>ABCDEF \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> | |
− | <math>\ell</math> bisects <math>EF \implies | ||
− | |||
− | |||
the preimages of the points <math>E</math> and <math>F</math> lie on the isogonals <math>PE</math> and <math>PF. \blacksquare</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> | ||
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==Isogonals in trapezium== | ==Isogonals in trapezium== | ||
− | [[File:Trapezium ACFEE.png| | + | [[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> | Let the trapezoid <math>AEFC, AC||EF,</math> be given. Denote <cmath>B = AE \cap CF, D = AF \cap CE.</cmath> | ||
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<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
− | < | + | <cmath>EM = MF \implies </cmath> |
− | Therefore <math>EM</math> and <math>FM</math> are isogonals. | + | <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. | 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. | ||
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The preimages of the points <math>B</math> and <math>D</math> lie on the isogonals <math>MB</math> and <math>MD. \blacksquare</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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
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The triangle <math>ABC</math> be given. The point <math>D</math> chosen on the bisector <math>AA'.</math> | The triangle <math>ABC</math> be given. The point <math>D</math> chosen on the bisector <math>AA'.</math> | ||
− | Denote < | + | 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> | Prove that <math>\angle BAE = \angle CAF.</math> | ||
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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> | 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> | ||
− | <math>D</math> is midpoint of <math>BB'</math> and midpoint <math>CC' | + | Image of point <math>D</math> is midpoint of image <math>BB'</math> and midpoint image <math>CC' \implies</math> |
− | <math> | + | Image <math>BCB'C'</math> is parallelogramm <math>\implies</math> |
− | <math>\implies</math> Preimages <math>AE</math> and <math>AF</math> are isogonals with respect <math>(\ell,A). \blacksquare</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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | == | + | ==Points on isogonals== |
− | [[File: | + | [[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 | + | Prove that <math>\angle BQC = \angle AQD.</math> |
<i><b>Proof</b></i> | <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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | ==IMO 2007 Short list== | + | ==IMO 2007 Short list/G3== |
− | [[File:Trapezoid 17.png| | + | [[File:Trapezoid 17.png|400px|right]] |
The diagonals of a trapezoid <math>ABCD</math> intersect at point <math>P.</math> | 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> | 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> | ||
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<math>QS || AD</math> is isogonal to <math>QP</math> with respect <math>\angle CQD.</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 | + | <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 268: | Line 337: | ||
Then these points lie on isogonals with respect angle <math>\angle A.</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== | ==Three points== | ||
− | [[File:3 points.png| | + | [[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 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> | 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> | ||
Line 289: | Line 369: | ||
<math>D'</math> and <math>D</math> lie on isogonals of <math>\angle BAC.</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. | + | 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''' | '''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 309: | 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 318: | 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 340: | Line 420: | ||
==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 350: | 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 360: | 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. \blacksquare</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> | ||
Line 376: | Line 456: | ||
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 383: | 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 405: | 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 419: | 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 431: | 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 08:40, 18 April 2024
Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.
Contents
- 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
Let
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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
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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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