Difference between revisions of "Symmedians, Lemoine point"
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Common Lemoine point== | ||
+ | [[File:L to L.png|390px|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 \angle LDE = \angle GA'B'.</math> | ||
+ | |||
+ | <math>CDLE</math> is cyclic so <math>\angle LDE = \angle LCE = \angle LCA = \angle C'CA = \angle C'A'A = \angle C'A'L \implies 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 L</math> is the isogonal conjugate of a point <math>G</math> with respect to a triangle <math>\triangle A'B'C' \implies 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 \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> |
Revision as of 15:46, 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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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 and are isogonals with respect
Similarly and are isogonals with respect is the isogonal conjugate of a point with respect to a triangle 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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