Symmedians, Lemoine point
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 Three intersecting antiparallel segments
- 16 Three intersecting parallel to sides segments
- 17 Tucker circle
- 18 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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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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