Difference between revisions of "Symmedians, Lemoine point"
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''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''' | ||
+ | |||
==Symmedian and tangents== | ==Symmedian and tangents== | ||
[[File:Tangents and symmedian.png|220px|right]] | [[File:Tangents and symmedian.png|220px|right]] | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''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''' | ||
+ | |||
+ | ==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 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> | 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> | ||
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<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. | <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''' | ||
+ | ==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. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <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''' |
Revision as of 17:34, 3 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
- 1 Proportions
- 2 Radical axis of circumcircle and Apollonius circle
- 3 Symmedian and tangents
- 4 Lemoine point properties
- 5 Parallel lines
- 6 Radical axis
- 7 Common Lemoine point
- 8 Lemoine point extreme properties
- 9 Lemoine point and perpendicularity
- 10 Lemoine point line
- 11 Antiparallel lines and segments
- 12 Three intersecting antiparallel segments
- 13 Three intersecting parallel to sides segments
- 14 Tucker circle
- 15 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
vladimir.shelomovskii@gmail.com, vvsss
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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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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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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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
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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