Difference between revisions of "Symmedians, Lemoine point"
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<cmath>\angle BDA = \angle ACB = \angle ACM \implies \triangle ABD \sim \triangle AMC \implies</cmath> | <cmath>\angle BDA = \angle ACB = \angle ACM \implies \triangle ABD \sim \triangle AMC \implies</cmath> | ||
<cmath>\frac {AM}{MC}= \frac {AB}{BD}.</cmath> | <cmath>\frac {AM}{MC}= \frac {AB}{BD}.</cmath> | ||
− | Similarly <math>\triangle | + | Similarly <math>\triangle ABM \sim \triangle ADC \implies \frac {AM}{MB}= \frac {AC}{CD}.</math> |
<cmath>BM = MC \implies \frac {BD}{CD} = \frac{AB}{AC}.</cmath> | <cmath>BM = MC \implies \frac {BD}{CD} = \frac{AB}{AC}.</cmath> | ||
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Then <math>BE</math> is the <math>B-</math> symmedian of <math>\triangle ABD, CE</math> is the <math>C-</math> symmedian of <math>\triangle ACD, DE</math> is the <math>D-</math> symmedian of <math>\triangle BCD.</math> | Then <math>BE</math> is the <math>B-</math> symmedian of <math>\triangle ABD, CE</math> is the <math>C-</math> symmedian of <math>\triangle ACD, DE</math> is the <math>D-</math> symmedian of <math>\triangle BCD.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Radical axis of circumcircle and Apollonius circle== | ||
+ | [[File:Circumcircle circle.png|390px|right]] | ||
+ | The bisectors of the external and internal angles at vertex <math>A</math> of <math>\triangle ABC</math> intersect line <math>BC</math> at points <math>D</math> and <math>E.</math> The circle <math>\omega = \odot ADE</math> intersects the circumcircle of <math>\triangle ABC</math> at points <math>A</math> and <math>Y.</math> Prove that line <math>AY</math> contains the <math>A-</math>symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | The circle <math>\omega</math> is the Apollonius circle for points <math>B</math> and <math>C \implies</math> | ||
+ | <cmath>\frac {BE}{CE} = \frac {AB}{AC} = \frac {BY}{CY} \implies</cmath> | ||
+ | <math>AY</math> is the <math>A-</math>symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Simson line== | ||
+ | [[File:Simson line symmedian.png|390px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> be given, <math>\Omega = \odot ABC.</math> Point <math>D</math> lies on arc <math>BC</math> of <math>\Omega.</math> | ||
+ | |||
+ | Let points <math>E, F,</math> and <math>G</math> be the the foots from <math>D</math> to <math>\overline{AB}, \overline{AC},</math> and to <math>\overline{BC},</math> respectively. | ||
+ | |||
+ | Prove that <math>EG = GF</math> iff <math>D</math> lies on <math>A-</math>symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Points <math>E, F,</math> and <math>G</math> lies on Simson line. | ||
+ | |||
+ | <math>DE \perp AE, DG \perp BC \implies BD</math> is diameter of circle <math>BEDG.</math> | ||
+ | |||
+ | Similarly, <math>CD</math> is diameter of circle <math>BEDG.</math> | ||
+ | |||
+ | 1. Let <math>D</math> lies on <math>A-</math>symmedian of <math>\triangle ABC \implies</math> | ||
+ | <cmath>\frac {BD}{CD} = \frac{AB}{AC} = \frac {\sin \angle ACB}{\sin \angle ABC}.</cmath> | ||
+ | <cmath>\frac {EG}{GF} = \frac {BD \sin \angle EBG}{CD \sin \angle GCF} = \frac {BD}{CD} \cdot \frac {\sin \angle ABC}{\sin \angle ACB} = 1.</cmath> | ||
+ | 2. Let <math>EG = GF \implies \frac {BD}{CD} = \frac{AB}{AC} \implies D</math> lies on <math>A-</math>symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | == Lemoine point of Gergonne triangle== | ||
+ | [[File:Lemoine and Gergone.png|430px|right]] | ||
+ | 1. Prove that the Lemoine point of the Gergonne triangle serves as the Gergonne point of the base triangle. | ||
+ | |||
+ | 2. The inscribed circle <math>\omega</math> touches the sides of given triangle <math>ABC</math> at points <math>A', B', C'.</math> | ||
+ | Prove that the line <math>AA'</math> is <math>A'-</math>symmedian of Gergonne triangle <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | 3. The inscribed circle touches the sides of given triangle <math>ABC</math> at points <math>A', B', C'.</math> Prove that <math>\angle AA'C' = \angle B'A'M,</math> where <math>M</math> is the midpoint <math>B'C'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 2. Denote <math>P = AA' \cap \omega, P \ne A'.</math> | ||
+ | <cmath>\triangle APC' \sim \triangle AC'A' \implies \frac {AC'}{AA'} = \frac {PC'}{A'C'}.</cmath> | ||
+ | |||
+ | Similarly, <math>\frac {AB'}{AA'} = \frac {PB'}{A'B'}.</math> | ||
+ | <cmath>AB' = AC' \implies \frac {PC'}{A'C'} = \frac {PB'}{A'B'} \implies \frac {PB'}{PC'} = \frac {A'B'}{A'C'}.</cmath> | ||
+ | Therefore <math>A'P(</math> and <math>A'A)</math> is <math>A'-</math>symmedian of Gergonne triangle <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | 1. Similarly, <math>B'B</math> is <math>B'-</math>symmedian and <math>C'C</math> is <math>C'-</math>symmedian of Gergonne triangle <math>\triangle A'B'C'.</math> So the Lemoine point of the Gergonne triangle <math>\triangle A'B'C'</math> serves as the Gergonne point of the base triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | 3. Let <math>M</math> be the midpoint <math>B'C'.</math> The median <math>A'M</math> is isogonal conjugate <math>A'A</math> with respect <math>\angle B'A'C' \implies \angle AA'C' = \angle B'A'M.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Symmedian and tangents== | ||
+ | [[File:Tangents and symmedian.png|220px|right]] | ||
+ | Let <math>\triangle ABC</math> and it’s circumcircle <math>\Omega</math> be given. | ||
+ | |||
+ | Tangents to <math>\Omega</math> at points <math>B</math> and <math>C</math> intersect at point <math>F.</math> | ||
+ | |||
+ | Prove that <math>AF</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>D = AF \cap \Omega \ne A.</math> WLOG, <math>\angle BAC < 180^\circ.</math> | ||
+ | <cmath>\triangle FDB \sim \triangle FBA \implies \frac {BD}{AB} = \frac{DF}{BF}.</cmath> | ||
+ | <cmath>\triangle FDC \sim \triangle FCA \implies \frac {CD}{AC} = \frac{DF}{CF}.</cmath> | ||
+ | <math>BF = CF \implies \frac {BD}{CD} = \frac{AB}{AC} \implies AD</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | [[File:Tangents to symmedian.png|220px|right]] | ||
+ | Let <math>\triangle ABC</math> and it’s circumcircle <math>\Omega</math> be given. | ||
+ | |||
+ | Let tangent to <math>\Omega</math> at points <math>A</math> intersect line <math>BC</math> at point <math>F.</math> | ||
+ | |||
+ | Let <math>FD</math> be the tangent to <math>\Omega</math> different from <math>FA.</math> | ||
+ | |||
+ | Then <math>AD</math> is <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Lemoine point properties== | ||
+ | [[File:L and G.png|390px|right]] | ||
+ | Let <math>\triangle ABC</math> be given. Let <math>L</math> be the Lemoine point of <math>\triangle ABC.</math> | ||
+ | |||
+ | <cmath>LD \perp BC, D \in BC, LE \perp AC, E \in AC, LF \perp AB, F \in AB.</cmath> | ||
+ | |||
+ | Prove that <math>\frac{LD}{BC} = \frac{LE}{AC} = \frac{LF}{AB}, L</math> is the centroid of <math>\triangle DEF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>G</math> be the centroid of <math>\triangle ABC, GD' \perp BC, D' \in BC,</math> | ||
+ | <cmath>LE' \perp AC, E' \in AC, LF' \perp AB, F' \in AB.</cmath> | ||
+ | |||
+ | The double area of <math>\triangle AGC</math> is <math>2[AGC] = GE' \cdot AC = 2[BGC] = GD' \cdot BC \implies \frac {GD' }{GE' } = \frac {AC}{BC}.</math> | ||
+ | |||
+ | Point <math>L</math>is the isogonal conjugate of point <math>G</math> with respect to <math>\triangle ABC \implies \frac {LE}{LD} =\frac {GD' }{GE' } =\frac {AC}{BC}.</math> | ||
+ | |||
+ | Similarly, one can get <math>\frac {LE}{AC} = \frac {LD}{BC} = \frac {LF}{AB} = k.</math> | ||
+ | |||
+ | The double area of <math>\triangle DLE</math> is <math>2[DLE] = LD \cdot LE \sin \angle DLE = k BC \cdot k AC \cdot \sin \angle ACB = k^2 \cdot 2[ABC].</math> | ||
+ | |||
+ | Similarly, one can get <math>[DLE] = [DLF] = [DEF] = k^2 [ABC] \implies L</math> is the centroid of <math>\triangle DEF.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Vector sum <math>\vec {LE} + \vec {LD} + \vec {LF} = \vec 0.</math> | ||
+ | |||
+ | Each of these vectors is obtained from the triangle side vectors by rotating by <math>90^\circ</math> and multiplying by a constant <math>k^2,</math> | ||
+ | <cmath>\vec {AC} + \vec {CB} + \vec {BA} = \vec 0.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Parallel lines== | ||
+ | [[File:Symmedians perp and par.png|390px|right]] | ||
+ | Let <math>\triangle ABC</math> and it’s Lemoine point <math>L</math> be given. | ||
+ | |||
+ | Let <math>D</math> be an arbitrary point. Let <math>D'</math> be the foot from <math>D</math> to line <math>\overline{AC}</math>. | ||
+ | |||
+ | Denote <math>\ell</math> the line through <math>D</math> and parallel to <math>AC.</math> | ||
+ | |||
+ | Denote <math>\ell'</math> the line parallel to <math>AB</math> such that distance <math>EE' = DD' \cdot \frac {AB}{AC}</math> and points <math>E</math> and <math>D</math> are both in the exterior (interior) of <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that points <math>F = \ell \cap \ell', A,</math> and <math>L</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>P(Q)</math> the foot from <math>L</math> to <math>\overline{AB}(\overline{AC})</math>. | ||
+ | |||
+ | <cmath>\frac {PL}{AB} = \frac {QL}{AC} \implies \frac {PL}{EE'} = \frac {QL}{DD'}.</cmath> | ||
+ | Denote <math>F = AL \cap \ell, F' = AL \cap \ell' \implies</math> | ||
+ | <cmath>\frac {FA}{AL} = \frac {DD'}{QL} = \frac {EE'}{PL} = \frac {F'A}{AL} \implies F = F'.</cmath> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | If squares <math>ABGF</math> and <math>ACDE</math> are constructed in the exterior of <math>\triangle ABC,</math> then <math>AO,</math> where <math>O</math> is the center of circle <math>\odot AEF,</math> is the symmedian in <math>\triangle ABC</math> through <math>A.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Radical axis== | ||
+ | [[File:Circles symmedians.png|390px|right]] | ||
+ | [[File:6 points line.png|390px|right]] | ||
+ | Circle <math>\omega</math> passes through points <math>A</math> and <math>B</math> and touches line <math>AC,</math> circle <math>\theta</math> passes through points <math>A</math> and <math>C</math> and touches line <math>AB.</math> Let <math>L</math> be the Lemoine point of <math>\triangle ABC.</math> | ||
+ | |||
+ | Prove that the radical axis of these circles contains the symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote centers of <math>\omega, \theta,</math> and <math>\odot ABC</math> throught <math>Q,P,</math> and <math>O,</math> respectively. | ||
+ | |||
+ | Denote <math>\ell</math> line throught <math>P</math> parallel to <math>AC, \ell'</math> line throught <math>Q</math> parallel to <math>AB.</math> | ||
+ | <math>\angle QAB = |90^\circ - \angle BAC| = \angle PAC \implies \triangle ABQ \sim \triangle ACP \implies</math> | ||
+ | |||
+ | The ratio of distance from <math>P</math> to <math>AC</math> to <math>AC</math> is equal to the ratio of distance from <math>Q</math> to <math>AB</math> to <math>AB \implies X = \ell \cap \ell' \in AL.</math> | ||
+ | <math>AP \perp AB \implies AP \perp \ell', QA \perp AC \implies QA \perp \ell \implies</math> | ||
+ | |||
+ | <math>A</math> is the orthocenter of <math>\triangle PQX \implies XA \perp QP \implies</math> | ||
+ | |||
+ | the radical axis of these circles contains the <math>A-</math> symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Corollary</b></i> | ||
+ | |||
+ | Circumcenter of <math>A-</math>Apollonius circle point <math>M,</math> circumcenter <math>O,</math> the points on <math>\odot ABT</math> and <math>\odot ACT</math> opposite <math>A</math> belong the line perpendicular <math>A-</math>symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Construction of symmedian’s point== | ||
+ | [[File:Symmedian from parallel.png|390px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> be given. | ||
+ | |||
+ | Let <math>D</math> be the arbitrary point on sideline <math>AB, DE||BC, E \in AC.</math> | ||
+ | |||
+ | The lines <math>BE</math> and <math>CD</math> meet at point <math>F.</math> The circumcircles of triangles <math>\triangle BFD</math> and <math>\triangle CEF</math> meet at two distinct points <math>F</math> and <math>Q.</math> | ||
+ | |||
+ | Prove that the line <math>AQ</math> is the <math>A-</math>symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | The spiral similarity centered at <math>Q</math> with the angle of rotation <math>\angle BQE</math> maps point <math>B</math> to point <math>E</math> and point <math>D</math> to point <math>C.</math> | ||
+ | |||
+ | So <math>\triangle BDQ \sim \triangle ECQ \implies </math> | ||
+ | |||
+ | <math>\angle DBQ = \angle CEQ \implies \angle ABQ + \angle AEQ = 180^\circ \implies ABQE</math> is concyclic. | ||
+ | |||
+ | Let <math>H</math> and <math>G</math> be the foot from <math>Q</math> to <math>\overline{AB}</math> and to <math>\overline{AC},</math> respectively. | ||
+ | |||
+ | <math>\frac {AB}{AC} = \frac {BD}{EC} = \frac {BQ}{QE} = \frac {HQ}{QG}</math> which means that <math>Q</math> lies on <math>A-</math>symmedian. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Common Lemoine point== | ||
+ | [[File:L to L.png|440px|right]] | ||
+ | [[File:2 46 Prasolov.png|440px|right]] | ||
+ | Let <math>\triangle ABC</math> be given, <math>\Omega = \odot ABC.</math> | ||
+ | |||
+ | Let <math>L</math> be the Lemoine point of <math>\triangle ABC.</math> | ||
+ | |||
+ | <math>A' = AL \cap \Omega \ne A, B' = BL \cap \Omega \ne B, C' = CL \cap \Omega \ne C.</math> | ||
+ | |||
+ | Prove that the point <math>L</math> is the Lemoine point of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote point <math>D</math> so that <math>LD \perp BC, D \in BC.</math> | ||
+ | |||
+ | Similarly denote <math>E \in AC</math> and <math>F \in AB.</math> | ||
+ | <math>L</math> is the centroid of <math>\triangle DEF.</math> | ||
+ | |||
+ | <math>\triangle DEF \sim \triangle A'B'C'</math> (see Claim). | ||
+ | |||
+ | Let point <math>G</math> be the centroid of <math>\triangle A'B'C' \implies</math> | ||
+ | <cmath>\angle LDE = \angle GA'B'.</cmath> | ||
+ | <math>CDLE</math> is cyclic so <math>\angle LDE = \angle LCE = \angle LCA = \angle C'CA = \angle C'A'A = \angle C'A'L</math> | ||
+ | therefore <math>A'L</math> and <math>A'G</math> are isogonals with respect <math>\angle C'A'B'.</math> | ||
+ | |||
+ | Similarly <math>B'L</math> and <math>B'G</math> are isogonals with respect <math>\angle A'B'C' \implies</math> | ||
+ | |||
+ | <math>L</math> is the isogonal conjugate of a point <math>G</math> with respect to a triangle <math>\triangle A'B'C'</math> | ||
+ | |||
+ | so <math>L</math> is the Lemoine point of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <i><b>Claim</b></i> | ||
+ | |||
+ | Lines AP, BP and CP intersect the circumcircle of <math>\triangle ABC</math> at points <math>A', B',</math> and <math>C'.</math> | ||
+ | |||
+ | Points <math>D, E,</math> and <math>F</math> are taken on the lines <math>BC, CA,</math> and <math>AB</math> so that <math>\angle PDB = \angle PFA = \angle PEC</math> (see diagram). | ||
+ | |||
+ | Prove that <math>\triangle A'B'C' \sim \triangle DEF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\angle PFA = \angle PDB \implies PDBF</math> is cyclic so <math>\angle PDF = \angle PBF = \angle ABB' = \angle AA'B'.</math> | ||
+ | |||
+ | Similarly, <math>\angle PDE = \angle AA'C' \implies</math> | ||
+ | <math>\angle FDE = \angle PDF + \angle PDE = \angle AA'B' + \angle AA'C' = \angle B'A'C'.</math> | ||
+ | |||
+ | Similarly, <math>\angle DEF = \angle A'B'C'. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Lemoine point extreme properties== | ||
+ | Lemoine point <math>L</math> minimizes the sum of the squares of the distances to the sides of the triangle (among all points internal to <math>\triangle ABC.)</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let us denote the desired point by <math>X.</math> Let us imagine that point <math>X</math> is connected to springs of equal stiffness attached to the sides at points <math>D, E,</math> and <math>F</math> and contacts sliding along them without friction. The segments modeling the springs will be perpendicular to the corresponding side. The energy of each spring is proportional to the square of its length. The minimum energy of the system corresponds to the minimum of the sum of the squares of the lengths of these segments, that is, the sum of the squares of the distances from <math>X</math> to the sides. | ||
+ | |||
+ | It is known that the minimum spring energy corresponds to the equilibrium position. The condition of equilibrium at a point <math>X</math> is the equality to zero of the vector sum of forces applied from the springs to the point <math>X.</math> The force developed by each spring is proportional to its length, that is, the equilibrium condition is that the sum of the vectors <math>\vec {XE} + \vec {XD} + \vec {XF} = \vec 0.</math> It is clear that the point <math>L</math> corresponds to this condition. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Lemoine point and perpendicularity== | ||
+ | [[File:Symmedians perp.png|430px|right]] | ||
+ | Let <math>\triangle ABC</math> be given. Let <math>L</math> be the Lemoine point of <math>\triangle ABC,</math> | ||
+ | <math>LD \perp BC, D \in BC, LE \perp AC, E \in AC, LF \perp AB, F \in AB.</math> | ||
+ | |||
+ | <math>M</math> is the midpoint <math>BC.</math> | ||
+ | |||
+ | Prove that <math>FE \perp AM.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>AL</math> is isogonal conjugated <math>AM</math> with respect <math>\angle A \implies \angle BAL = \angle CAM.</math> | ||
+ | |||
+ | <math>LE \perp AE, LF \perp AF \implies AELD</math> is cyclic. | ||
+ | |||
+ | <math>\angle CAM = \angle BAL = \angle FEL.</math> | ||
+ | |||
+ | <math>LE \perp AC \implies EF \perp AM.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Lemoine point line== | ||
+ | [[File:L M P line.png|430px|right]] | ||
+ | Let <math>\triangle ABC</math> be given. Let <math>L</math> be the Lemoine point of <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>AH</math> be the height, <math>AM</math> be the median, <math>LD \perp BC, D \in BC,</math> | ||
+ | |||
+ | <math>LE \perp AC, E \in AC, LF \perp AB, F \in AB, P</math> be the midpoint <math>AH</math>. | ||
+ | |||
+ | Prove that the points <math>L, P,</math> and <math>M</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>O</math> the circumcenter <math>\odot AELF, LO = AO.</math> | ||
+ | |||
+ | Denote <math>T</math> the midpoint <math>FE \implies OT \perp FE.</math> | ||
+ | |||
+ | <math>L</math> is centroid of <math>\triangle DEF \implies DLT</math> is <math>D-</math>median of <math>\triangle DEF.</math> | ||
+ | |||
+ | Denote <math>Q</math> the point symmetric <math>L</math> with respect <math>T \implies QT</math> is the midline of <math>\triangle LAQ \implies AQ \perp EF \implies Q \in AM \perp EF.</math> | ||
+ | |||
+ | <math>LD = 2 TL \implies DL = LQ \implies ML</math> is the median of <math>\triangle MDQ.</math> | ||
+ | |||
+ | <math>MP</math> is the median of <math>\triangle MHA, HA || DQ \implies</math> the points <math>L, P,</math> and <math>M</math> are collinear. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Antiparallel lines and segments== | ||
+ | Two lines <math>BC</math> and <math>B'C'</math> are said to be antiparallel with respect to the sides of an angle <math>A</math> if they make the same angle in the opposite senses with the bisector of that angle. | ||
+ | |||
+ | A segment <math>B'C',</math> where points <math>B'</math> and <math>C'</math> lie on rays <math>AC</math> and <math>AB,</math> is called antiparallel to side <math>BC</math> if <math>\angle AB'C' = \angle ABC</math> and <math>\angle AC'B' = \angle ACB.</math> The points <math>B', C', B,</math> and <math>C</math> are concyclic. | ||
+ | |||
+ | Prove that the symmedian <math>AS</math> bisects any segment <math>B'C'</math> iff it is antiparallel to side <math>BC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 1) Let segment <math>B'C'</math> be the antiparallel to side <math>BC.</math> Reflection through the bisector of angle <math>A</math> maps the segment <math>B'C'</math> into a segment parallel to side <math>BC,</math> and maps the symmedian <math>AS</math> into the median which bisects image of <math>B'C'.</math> | ||
+ | |||
+ | 2) Suppose the symmedian <math>AS</math> bisects the segment <math>DE,D \in AB, D \ne B', E \in AC, E \ne C'</math> in point <math>M.</math> There is a segment <math>B'C'</math> with ends on the sides of angle <math>A</math> which contain point <math>M</math> and is antiparallel to side <math>BC.</math> <math>M</math> is the midpoint <math>B'C' \implies</math> | ||
+ | |||
+ | <math>DB'EC'</math> is parallelogram, so <math>DB' || EC'. DB' \in AB, EC' \in AC -</math> contradiction. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Bisectors and antiparallel== | ||
+ | [[File:Bisectors and antiparallel.png|450px|right]] | ||
+ | Let <math>CD</math> be the antiparallel <math>AC</math> in the triangle <math>\triangle ABC.</math> | ||
+ | Prove that | ||
+ | |||
+ | 1) the bisectors of the angles <math>\angle ABC</math> and <math>\angle ACD</math> are perpendicular, | ||
+ | |||
+ | 2) the point <math>Q</math> of intersection of bisectors lies on the midline <math>A_1B_1, 2QA_1 = BC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 1) Lines <math>CA</math> and <math>CD</math> are symmetrical with respect to the bisector <math>CQ.</math> | ||
+ | |||
+ | Denote <math>H = AC \cap BQ.</math> Line <math>\ell</math> throught <math>H</math> parallel to <math>CD</math> is symmetrical to <math>AC</math> with respect to the bisector <math>BQ.</math> | ||
+ | |||
+ | The axes of symmetry of two lines are perpendicular. | ||
+ | |||
+ | 2) Angle <math>\angle BQC = 90^\circ, A_1</math> is the midpoint <math>BC \implies 2QA_1 = BC.</math> | ||
+ | |||
+ | Denote <math>K = CQ \cap AB.</math> Bisector <math>BQ</math> is height in <math>\triangle AKC \implies KQ = QC \implies Q \in A_1B_1.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Symmetry of angles== | ||
+ | [[File:Crcle median symm 1.png|400px|right]] | ||
+ | Let <math>CD</math> be the antiparallel <math>AC</math> of the triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>CD</math> crosses the median <math>BB_1</math> of <math>\triangle ABC</math> at the point <math>F</math> and crosses the simedian <math>BE</math> at the point <math>G.</math> | ||
+ | |||
+ | Prove that the points <math>E, G, F,</math> and <math>B_1</math> are concyclic, <math>FE || BC.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Under reflection along a bisector <math>BQ</math> the median <math>BB_1</math> maps into symmedian <math>BF.</math> | ||
+ | |||
+ | Under reflection along a bisector <math>BQ</math> antiparallel <math>CD</math> maps into the line parallel <math>AC.</math> | ||
+ | |||
+ | Under reflection along a bisector <math>BQ</math> the sides of <math>\angle BFC</math> maps into the lines parallel to the sides of the <math>\angle BEA,</math> so <math>\angle BFC = \angle BEA \implies</math> points <math>E, G, F,</math> and <math>B_1</math> are concyclic. | ||
+ | <cmath>\angle BAC = \angle A_1B_1C = \angle GB_1E = \angle GFE = \angle BCF \implies FE||BC.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Three intersecting antiparallel segments== | ||
+ | [[File:3 Symmedians.png|390px|right]] | ||
+ | Let triangle <math>ABC</math> and a point <math>L</math> lying inside it be given. Let <math>D'E, DF',</math> and <math>FE'</math> be three segments antiparallel to <math>AB, AC,</math> and <math>BC,</math> respectively. <math>L \in D'E, L \in DF', L \in FE'.</math> | ||
+ | |||
+ | Prove that <math>DF' = D'E = E'F</math> iff <math>L</math> is a Lemoine point. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>\angle BAC = \angle BDF' = \angle CD'E \implies LD = LD'.</cmath> | ||
+ | Similarly, <math>LE = LE', LF = LF'.</math> | ||
+ | |||
+ | 1. Let <math>L</math> be the Lemoine point. So <math>LD = LF', LF = LE', LE = LD' \implies DF' = D'E = E'F = 2 LD.</math> | ||
+ | |||
+ | 2. Let <math>DF' = D'E = E'F \implies</math> | ||
+ | <cmath>LD + LF' = LD' + LE = LE' + LF \implies</cmath> | ||
+ | <cmath>LD + LF = LD + LE = LE + LF \implies</cmath> | ||
+ | <math>LD = LE = LF = LD' = LE' = LF' \implies L</math> lies on each symmedian. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Three intersecting parallel to sides segments== | ||
+ | [[File:Parallel segments.png|390px|right]] | ||
+ | Let triangle <math>ABC</math> be given. <math>O</math> is it’s circumcenter, <math>L</math> is it’s Lemoine point. | ||
+ | |||
+ | Let <math>A''B', B''C',</math> and <math>C''A'</math> be three segments parallel to <math>AB, BC,</math> and <math>CA,</math> respectively. <math>L \in A''B', L \in B''C', L \in C''A'.</math> | ||
+ | |||
+ | Prove that points <math>A', B', C', A'', B'',</math> and <math>C''</math> lies on the circle centered at midpoint <math>LO</math> (the first Lemoine circle). | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>AB' || C''L, AC'' || B'L \implies AB'LC''</math> is the parallelogram. | ||
+ | |||
+ | Denote <math>P = AL \cap B'C'', B'P = PC'' \implies B'C''</math> is antiparallel to <math>BC.</math> | ||
+ | |||
+ | Similarly, <math>A''C'</math> is antiparallel to <math>AC, A'B''</math> is antiparallel to <math>AB.</math> | ||
+ | <math>\triangle AB'C'' \sim \triangle ABC.</math> Denote <math>c = AB, b = AC, a = BC.</math> | ||
+ | |||
+ | The coefficient of similarity is | ||
+ | <cmath>k = \frac {AC''}{AC} = \frac {AC''}{AB} \cdot \frac {AB}{AC} = \frac {b^2}{a^2 +b^2+c^2} \cdot \frac {c}{b} = \frac {bc}{a^2 +b^2+c^2}.</cmath> | ||
+ | |||
+ | Therefore <math>B'C'' = k \cdot BC = \frac {abc}{a^2 +b^2+c^2}.</math> Similarly, <math>A'B'' = B'C'' = C'A''.</math> | ||
+ | |||
+ | Denote <math>P' = BL \cap A''C', P'' = CL \cap A'B'' \implies AP = PL, BP' = P'L, CP'' = P''L.</math> | ||
+ | |||
+ | <math>\triangle PP'P'' \sim \triangle ABC.</math> The coefficient of similarity is <math>\frac{1}{2}.</math> | ||
+ | |||
+ | Denote <math>Q</math> the midpoint <math>OL.</math> The midline of <math>\triangle LAO</math> is <math>QP = \frac {OA}{2}</math> | ||
+ | |||
+ | Similarly, <math>P'Q = P''Q = PQ \implies Q</math> is circumcenter of <math>\triangle PP'P''.</math> | ||
+ | |||
+ | <math>B'C''</math> is antiparallel to <math>BC,</math> so <math>B'C''</math> is tangent to <math>\odot PP'P'', P</math> is the midpoint <math>B'C''.</math> | ||
+ | |||
+ | Similarly, <math>A'B'' = B'C'' = C'A''</math> are tangent to <math>\odot PP'P'', P'</math> is the midpoint <math>A''C', P''</math> is the midpoint <math>A'B''.</math> | ||
+ | |||
+ | Therefore <math>A'Q = B'Q = C'Q = A''Q = B''Q = C''Q. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Tucker circle== | ||
+ | [[File:Tucker circle.png|450px|right]] | ||
+ | [[File:Tucker circle A.png|450px|right]] | ||
+ | Let triangle <math>ABC</math> be given. <math>O</math> is it’s circumcenter, <math>L</math> is it’s Lemoine point. | ||
+ | |||
+ | Let homothety centered at <math>L</math> with factor <math>k</math> maps <math>\triangle ABC</math> into <math>\triangle DEF</math>. | ||
+ | |||
+ | Denote the crosspoints of sidelines these triangles as | ||
+ | <cmath>A' = BC \cap DF, B' = AC \cap DE, C' = AB \cap EF,</cmath> | ||
+ | <cmath> A'' = BC \cap DE, B'' = AC \cap EF, C'' = AB \cap DF.</cmath> | ||
+ | |||
+ | Prove that points <math>A', B', C', A'', B'',</math> and <math>C''</math> lies on the circle centered at <math>LO</math> (Tucker circle). | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>AB' || C''D, AC'' || B'D \implies AB'DC''</math> is the parallelogram. | ||
+ | |||
+ | Denote <math>K = AD \cap B'C'', AK = KD, B'K = KC'' \implies</math> | ||
+ | <math>B'C''</math> is antiparallel to <math>BC.</math> | ||
+ | |||
+ | Similarly, <math>A''C'</math> is antiparallel to <math>AC, A'B''</math> is antiparallel to <math>AB.</math> | ||
+ | |||
+ | <math>M = BE \cap A''C'</math> is midpoint <math>BE, N = A'B'' \cap CF</math> is the midpoint <math>CF.</math> | ||
+ | |||
+ | <math>\triangle AB'C'' \sim \triangle ABC.</math> | ||
+ | |||
+ | <math>\frac {LD}{AL} = k, AL = LD \implies \frac {AK}{AL} = \frac{2}{k+1}, \frac {KL}{AL} = \frac{1-k}{1+k}.</math> | ||
+ | |||
+ | Similarly, <math>\frac {BM}{BL} = \frac {CN}{CL} = \frac {AK}{AL} =\frac{2}{k+1}, \frac {KL}{AL} = \frac{ML}{BM}= \frac {NL}{CL} = \frac{1-k}{1+k}.</math> | ||
+ | |||
+ | Let <math>B'''C'''</math> be the symmedian <math>BC</math> through <math>L.</math> | ||
+ | <cmath>B'''C''' || B'C'' \implies B'C'' = B'''C''' \cdot \frac {AK}{AL} =\frac{2B'''C'''}{k+1}.</cmath> | ||
+ | |||
+ | It is known that three symmedians through <math>L</math> are equal, so <math>A''C' = C''B' = B''A' = \frac{2B'''C'''}{k+1}.</math> | ||
+ | |||
+ | <math>\triangle KMN</math> is homothetic to <math>\triangle ABC</math> with center <math>L</math> and factor <math>\frac{1-k}{1+k}.</math> | ||
+ | |||
+ | So segments <math>A''C' = C''B' = B''A'</math> are tangents to <math>\odot KMN</math> and points of contact are the midpoints of these segments. | ||
+ | |||
+ | Denote <math>Q</math> the circumcenter of <math>\triangle KMN, Q \in LO.</math> | ||
+ | |||
+ | Therefore <math>A'Q = B'Q = C'Q = A''Q = B''Q = C''Q. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Tucker circle 2== | ||
+ | [[File:Tucker circle B.png|440px|right]] | ||
+ | Let triangle <math>ABC</math> be given. Let <math>D</math> be the arbitrary point on sideline <math>AC.</math> | ||
+ | |||
+ | Let <math>DD'</math> be the antiparallel to side <math>AB, D' \in BC.</math> | ||
+ | |||
+ | Denote point <math>E \in AB, D'E || AC.</math> | ||
+ | |||
+ | Let <math>EE'</math> be the antiparallel to side <math>BC, E' \in AC.</math> | ||
+ | |||
+ | Denote point <math>F \in BC, E'F || AB.</math> | ||
+ | |||
+ | Let <math>FF'</math> be the antiparallel to side <math>AC, F' \in AB.</math> | ||
+ | |||
+ | Prove that points <math>D, D', E, E', F,</math> and <math>F'</math> lies on the circle centered at <math>LO</math> (Tucker circle). | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>\angle ABC = \angle CDD' = \angle AE'E, DE' || DE' \implies DD'EE'</math> is isosceles trapezoid. | ||
+ | |||
+ | So <math>DD' = EE'.</math> | ||
+ | |||
+ | <math>\angle ACB = \angle BF'F = \angle AEE', FE' || EF' \implies EFE'F'</math> is isosceles trapezoid. | ||
+ | |||
+ | So <math>FF' = EE' = DD'. \angle BFF' = \angle CD'D \implies F'D || BC.</math> | ||
+ | |||
+ | Denote <math>A'</math> the midpoint <math>EE', B'</math> the midpoint <math>FF', C'</math> the midpoint <math>DD'.</math> | ||
+ | <math>AB || FE' \implies A'B' ||AB.</math> Similarly, <math>A'C' ||AC, C'B' ||CB.</math> | ||
+ | |||
+ | <math>A'</math> is the midpoint of antiparallel of <math>BC \implies AA'</math> is the <math>A-</math>symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | Similarly, <math>BB'</math> is the <math>B-</math>symmedian, <math>CC'</math> is the <math>C-</math>symmedian of <math>\triangle ABC.</math> | ||
+ | |||
+ | Therefore Lemoine point <math>L = AA' \cap BB' \cap CC', \triangle A'B'C'</math> is homothetic to <math>\triangle ABC</math> with center <math>L.</math> | ||
+ | |||
+ | So segments <math>DD' = EE' = FF'</math> are tangents to <math>\odot A'B'C'</math> and points of contact are the midpoints of these segments. | ||
+ | |||
+ | Denote <math>Q</math> the circumcenter of <math>\triangle A'B'C', Q \in LO,</math> where <math>O</math> is the circumcenter of <math>\triangle ABC.</math> | ||
+ | |||
+ | Therefore <math>DQ = D'Q = EQ = E'Q = FQ = F'Q. \blacksquare</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 16:38, 6 August 2024
The reflecting of the median over the corresponding angle bisector is the symmedian. The angle formed by the symmedian and the angle bisector has the same measure as the angle between the median and the angle bisector, but it is on the other side of the angle bisector. The symmedian is isogonally conjugate to the median
There are three symmedians. They are meet at a triangle center called the Lemoine point.
Contents
[hide]- 1 Proportions
- 2 Radical axis of circumcircle and Apollonius circle
- 3 Simson line
- 4 Lemoine point of Gergonne triangle
- 5 Symmedian and tangents
- 6 Lemoine point properties
- 7 Parallel lines
- 8 Radical axis
- 9 Construction of symmedian’s point
- 10 Common Lemoine point
- 11 Lemoine point extreme properties
- 12 Lemoine point and perpendicularity
- 13 Lemoine point line
- 14 Antiparallel lines and segments
- 15 Bisectors and antiparallel
- 16 Symmetry of angles
- 17 Three intersecting antiparallel segments
- 18 Three intersecting parallel to sides segments
- 19 Tucker circle
- 20 Tucker circle 2
Proportions
Let be given.
Let be the median,
Prove that iff is the symmedian than
Proof
1. Let be the symmedian. So
Similarly
By applying the Law of Sines we get
Similarly,
2.
As point moves along the fixed arc
from
to
, the function
monotonically increases from zero to infinity. This means that there is exactly one point at which the condition is satisfied. In this case, point
lies on the symmedian.
Similarly for point
Corollary
Let be the
symmedian of
Then is the
symmedian of
is the
symmedian of
is the
symmedian of
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Radical axis of circumcircle and Apollonius circle
The bisectors of the external and internal angles at vertex of
intersect line
at points
and
The circle
intersects the circumcircle of
at points
and
Prove that line
contains the
symmedian of
Proof
The circle is the Apollonius circle for points
and
is the
symmedian of
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Simson line
Let triangle be given,
Point
lies on arc
of
Let points and
be the the foots from
to
and to
respectively.
Prove that iff
lies on
symmedian of
Proof
Points and
lies on Simson line.
is diameter of circle
Similarly, is diameter of circle
1. Let lies on
symmedian of
2. Let
lies on
symmedian of
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Lemoine point of Gergonne triangle
1. Prove that the Lemoine point of the Gergonne triangle serves as the Gergonne point of the base triangle.
2. The inscribed circle touches the sides of given triangle
at points
Prove that the line
is
symmedian of Gergonne triangle
3. The inscribed circle touches the sides of given triangle at points
Prove that
where
is the midpoint
Proof
2. Denote
Similarly,
Therefore
and
is
symmedian of Gergonne triangle
1. Similarly, is
symmedian and
is
symmedian of Gergonne triangle
So the Lemoine point of the Gergonne triangle
serves as the Gergonne point of the base triangle
3. Let be the midpoint
The median
is isogonal conjugate
with respect
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Symmedian and tangents
Let and it’s circumcircle
be given.
Tangents to at points
and
intersect at point
Prove that is
symmedian of
Proof
Denote WLOG,
is
symmedian of
Corollary
Let and it’s circumcircle
be given.
Let tangent to at points
intersect line
at point
Let be the tangent to
different from
Then is
symmedian of
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Lemoine point properties
Let be given. Let
be the Lemoine point of
Prove that is the centroid of
Proof
Let be the centroid of
The double area of is
Point is the isogonal conjugate of point
with respect to
Similarly, one can get
The double area of is
Similarly, one can get is the centroid of
Corollary
Vector sum
Each of these vectors is obtained from the triangle side vectors by rotating by and multiplying by a constant
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Parallel lines
Let and it’s Lemoine point
be given.
Let be an arbitrary point. Let
be the foot from
to line
.
Denote the line through
and parallel to
Denote the line parallel to
such that distance
and points
and
are both in the exterior (interior) of
Prove that points and
are collinear.
Proof
Denote the foot from
to
.
Denote
Corollary
If squares and
are constructed in the exterior of
then
where
is the center of circle
is the symmedian in
through
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Radical axis
Circle passes through points
and
and touches line
circle
passes through points
and
and touches line
Let
be the Lemoine point of
Prove that the radical axis of these circles contains the symmedian of
Proof
Denote centers of and
throught
and
respectively.
Denote line throught
parallel to
line throught
parallel to
The ratio of distance from to
to
is equal to the ratio of distance from
to
to
is the orthocenter of
the radical axis of these circles contains the symmedian of
Corollary
Circumcenter of Apollonius circle point
circumcenter
the points on
and
opposite
belong the line perpendicular
symmedian of
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Construction of symmedian’s point
Let triangle be given.
Let be the arbitrary point on sideline
The lines and
meet at point
The circumcircles of triangles
and
meet at two distinct points
and
Prove that the line is the
symmedian of
Proof
The spiral similarity centered at with the angle of rotation
maps point
to point
and point
to point
So
is concyclic.
Let and
be the foot from
to
and to
respectively.
which means that
lies on
symmedian.
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Common Lemoine point
Let be given,
Let be the Lemoine point of
Prove that the point is the Lemoine point of
Proof
Denote point so that
Similarly denote and
is the centroid of
(see Claim).
Let point be the centroid of
is cyclic so
therefore
and
are isogonals with respect
Similarly and
are isogonals with respect
is the isogonal conjugate of a point
with respect to a triangle
so is the Lemoine point of
Claim
Lines AP, BP and CP intersect the circumcircle of at points
and
Points and
are taken on the lines
and
so that
(see diagram).
Prove that
Proof
is cyclic so
Similarly,
Similarly,
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Lemoine point extreme properties
Lemoine point minimizes the sum of the squares of the distances to the sides of the triangle (among all points internal to
Proof
Let us denote the desired point by Let us imagine that point
is connected to springs of equal stiffness attached to the sides at points
and
and contacts sliding along them without friction. The segments modeling the springs will be perpendicular to the corresponding side. The energy of each spring is proportional to the square of its length. The minimum energy of the system corresponds to the minimum of the sum of the squares of the lengths of these segments, that is, the sum of the squares of the distances from
to the sides.
It is known that the minimum spring energy corresponds to the equilibrium position. The condition of equilibrium at a point is the equality to zero of the vector sum of forces applied from the springs to the point
The force developed by each spring is proportional to its length, that is, the equilibrium condition is that the sum of the vectors
It is clear that the point
corresponds to this condition.
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Lemoine point and perpendicularity
Let be given. Let
be the Lemoine point of
is the midpoint
Prove that
Proof
is isogonal conjugated
with respect
is cyclic.
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Lemoine point line
Let be given. Let
be the Lemoine point of
Let be the height,
be the median,
be the midpoint
.
Prove that the points and
are collinear.
Proof
Denote the circumcenter
Denote the midpoint
is centroid of
is
median of
Denote the point symmetric
with respect
is the midline of
is the median of
is the median of
the points
and
are collinear.
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Antiparallel lines and segments
Two lines and
are said to be antiparallel with respect to the sides of an angle
if they make the same angle in the opposite senses with the bisector of that angle.
A segment where points
and
lie on rays
and
is called antiparallel to side
if
and
The points
and
are concyclic.
Prove that the symmedian bisects any segment
iff it is antiparallel to side
Proof
1) Let segment be the antiparallel to side
Reflection through the bisector of angle
maps the segment
into a segment parallel to side
and maps the symmedian
into the median which bisects image of
2) Suppose the symmedian bisects the segment
in point
There is a segment
with ends on the sides of angle
which contain point
and is antiparallel to side
is the midpoint
is parallelogram, so
contradiction.
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Bisectors and antiparallel
Let be the antiparallel
in the triangle
Prove that
1) the bisectors of the angles and
are perpendicular,
2) the point of intersection of bisectors lies on the midline
Proof
1) Lines and
are symmetrical with respect to the bisector
Denote Line
throught
parallel to
is symmetrical to
with respect to the bisector
The axes of symmetry of two lines are perpendicular.
2) Angle is the midpoint
Denote Bisector
is height in
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Symmetry of angles
Let be the antiparallel
of the triangle
Let crosses the median
of
at the point
and crosses the simedian
at the point
Prove that the points and
are concyclic,
Proof
Under reflection along a bisector the median
maps into symmedian
Under reflection along a bisector antiparallel
maps into the line parallel
Under reflection along a bisector the sides of
maps into the lines parallel to the sides of the
so
points
and
are concyclic.
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Three intersecting antiparallel segments
Let triangle and a point
lying inside it be given. Let
and
be three segments antiparallel to
and
respectively.
Prove that iff
is a Lemoine point.
Proof
Similarly,
1. Let be the Lemoine point. So
2. Let
lies on each symmedian.
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Three intersecting parallel to sides segments
Let triangle be given.
is it’s circumcenter,
is it’s Lemoine point.
Let and
be three segments parallel to
and
respectively.
Prove that points and
lies on the circle centered at midpoint
(the first Lemoine circle).
Proof
is the parallelogram.
Denote is antiparallel to
Similarly, is antiparallel to
is antiparallel to
Denote
The coefficient of similarity is
Therefore Similarly,
Denote
The coefficient of similarity is
Denote the midpoint
The midline of
is
Similarly, is circumcenter of
is antiparallel to
so
is tangent to
is the midpoint
Similarly, are tangent to
is the midpoint
is the midpoint
Therefore
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Tucker circle
Let triangle be given.
is it’s circumcenter,
is it’s Lemoine point.
Let homothety centered at with factor
maps
into
.
Denote the crosspoints of sidelines these triangles as
Prove that points and
lies on the circle centered at
(Tucker circle).
Proof
is the parallelogram.
Denote
is antiparallel to
Similarly, is antiparallel to
is antiparallel to
is midpoint
is the midpoint
Similarly,
Let be the symmedian
through
It is known that three symmedians through are equal, so
is homothetic to
with center
and factor
So segments are tangents to
and points of contact are the midpoints of these segments.
Denote the circumcenter of
Therefore
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Tucker circle 2
Let triangle be given. Let
be the arbitrary point on sideline
Let be the antiparallel to side
Denote point
Let be the antiparallel to side
Denote point
Let be the antiparallel to side
Prove that points and
lies on the circle centered at
(Tucker circle).
Proof
is isosceles trapezoid.
So
is isosceles trapezoid.
So
Denote the midpoint
the midpoint
the midpoint
Similarly,
is the midpoint of antiparallel of
is the
symmedian of
Similarly, is the
symmedian,
is the
symmedian of
Therefore Lemoine point is homothetic to
with center
So segments are tangents to
and points of contact are the midpoints of these segments.
Denote the circumcenter of
where
is the circumcenter of
Therefore
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