Difference between revisions of "Symmedians, Lemoine point"
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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> | 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. | + | Prove that the points <math>E, G, F,</math> and <math>B_1</math> are concyclic, <math>EF || BC.</math> |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
Line 387: | Line 387: | ||
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. | 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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 16:36, 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
- 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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