Difference between revisions of "Symmedians, Lemoine point"
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<math>DB'EC'</math> is parallelogram, so <math>DB' || EC'. DB' \in AB, EC' \in AC -</math> contradiction. | <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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 05:56, 2 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 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
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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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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