Difference between revisions of "Symmedians, Lemoine point"
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<math>LE \perp AC \implies EF \perp AM.</math> | <math>LE \perp AC \implies EF \perp AM.</math> | ||
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+ | '''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> | ||
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+ | <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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 00:59, 23 July 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]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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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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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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