Difference between revisions of "De Longchamps point"
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==De Longchamps circle== | ==De Longchamps circle== | ||
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De Longchamps circle <math>\Omega</math> of the obtuse triangle <math>ABC</math> is circle centered at de Longchamps point <math>L</math> which is orthogonal to the <math>\omega_C.</math> Prove that de Longchamps circle is orthogonal to the <math>\omega_A, \omega_B, A-</math>power, <math>B-</math>power and <math>C-</math>power cicles and the radius of de Longchamps circle is <math>R_{\Omega} = 4R \sqrt { – \cos A \cos B \cos C}.</math> | De Longchamps circle <math>\Omega</math> of the obtuse triangle <math>ABC</math> is circle centered at de Longchamps point <math>L</math> which is orthogonal to the <math>\omega_C.</math> Prove that de Longchamps circle is orthogonal to the <math>\omega_A, \omega_B, A-</math>power, <math>B-</math>power and <math>C-</math>power cicles and the radius of de Longchamps circle is <math>R_{\Omega} = 4R \sqrt { – \cos A \cos B \cos C}.</math> | ||
− | + | [[File:Longchamps circle.png|350px|right]] | |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
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<math>L</math> is the radical center of all six circles, therefore <math>\Omega</math> is perpendicular to each of these circles. | <math>L</math> is the radical center of all six circles, therefore <math>\Omega</math> is perpendicular to each of these circles. | ||
− | <math>L</math> is orthocenter of the anticomplementary triangle of <math>\triangle ABC</math> so radius of <math>\Omega</math> is twice radius of circle finded by Claim < | + | <math>L</math> is orthocenter of the anticomplementary triangle of <math>\triangle ABC</math> so radius of <math>\Omega</math> is twice radius of circle finded by Claim <cmath>\implies R_{\Omega} = 4R \sqrt { – \cos A \cos B \cos C}.</cmath> |
<i><b>Claim (Radius)</b></i> | <i><b>Claim (Radius)</b></i> | ||
− | + | [[File:Longchamps Claim.png|300px|right]] | |
Let <math>ABC</math> be obtuse triangle <math>(\angle A > 90^\circ)</math> with circumcircle <math>\Omega,</math> circumradius <math>R,</math> and orthocenter <math>H.</math> | Let <math>ABC</math> be obtuse triangle <math>(\angle A > 90^\circ)</math> with circumcircle <math>\Omega,</math> circumradius <math>R,</math> and orthocenter <math>H.</math> | ||
Let <math>\omega'</math> be the circle with diameter <math>AB.</math> | Let <math>\omega'</math> be the circle with diameter <math>AB.</math> | ||
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Let altitude <math>AH</math> cross <math>BC</math> at <math>D</math> and cross <math>\Omega</math> second time at <math>H'.</math> | Let altitude <math>AH</math> cross <math>BC</math> at <math>D</math> and cross <math>\Omega</math> second time at <math>H'.</math> | ||
− | <math>AD \perp BD \implies D \in \omega',</math> points <math>H,A, D</math> are collinear <math>\implies</math> | + | |
+ | <math>AD \perp BD \implies D \in \omega',</math> points <math>H,A, D</math> are collinear <math>\implies</math> | ||
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+ | Inversion with respect <math>\omega</math> swap <math>A</math> and <math>D \implies R_\omega^2 = HA \cdot HD.</math> | ||
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Well known that <math>HA = – 2R \cos A.</math> | Well known that <math>HA = – 2R \cos A.</math> | ||
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<cmath>BC \perp HD, AC \perp BH \implies \angle C = \angle BHD \implies HD = BH \cos C.</cmath> | <cmath>BC \perp HD, AC \perp BH \implies \angle C = \angle BHD \implies HD = BH \cos C.</cmath> | ||
Points <math>H</math> and <math>H'</math> are symmetric with respect <math>BC \implies BH' = BH.</math> | Points <math>H</math> and <math>H'</math> are symmetric with respect <math>BC \implies BH' = BH.</math> | ||
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<cmath>R_\omega^2 = – 2R \cos A \cdot 2R \cos B \cos C \implies R_\omega = 2R \sqrt{– \cos A \cos B \cos C}.</cmath> | <cmath>R_\omega^2 = – 2R \cos A \cdot 2R \cos B \cos C \implies R_\omega = 2R \sqrt{– \cos A \cos B \cos C}.</cmath> | ||
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+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==De Longchamps line== | ||
+ | [[File:Longchamps lime.png|450px|right]] | ||
+ | The de Longchamps line <math>l</math> of <math>\triangle ABC</math> is defined as the radical axes of the de Longchamps circle <math>\omega</math> and of the circumscribed circle <math>\Omega</math> of <math>\triangle ABC.</math> | ||
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+ | Let <math>\Omega'</math> be the circumcircle of <math>\triangle DEF</math> (the anticomplementary triangle of <math>\triangle ABC).</math> | ||
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+ | Let <math>\omega'</math> be the circle centered at <math>G</math> (centroid of <math>\triangle ABC</math>) with radius <math>\rho = \frac {\sqrt{2}}{3} \sqrt {a^2 + b^2 + c^2},</math> where <math>a = BC, b = AC, c = AB.</math> | ||
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+ | Prove that the de Longchamps line is perpendicular to Euler line and is the radical axes of <math>\Omega, \Omega', \omega,</math> and <math>\omega'.</math> | ||
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+ | <i><b>Proof</b></i> | ||
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+ | Center of <math>\Omega</math> is <math>O</math>, center of <math>\omega</math> is <math>L \implies OL \perp l,</math> where <math>OL</math> is Euler line. | ||
+ | The homothety with center <math>G</math> and ratio <math>-2</math> maps <math>\triangle ABC</math> into <math>\triangle DEF.</math> This homothety maps <math>\Omega</math> into <math>\Omega'.</math> | ||
+ | <math>R_{\Omega} \ne R_{\Omega'}</math> and <math>\Omega \cap \Omega' = K \implies </math> there is two inversion which swap <math>\Omega</math> and <math>\Omega'.</math> | ||
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+ | First inversion <math>I_{\omega'}</math> centered at point <math>G = \frac {\vec O \cdot 2R + \vec H \cdot R}{2R + R} = \frac {2\vec O + \vec H}{3}.</math> Let <math>K</math> be the point of crossing <math>\Omega</math> and <math>\Omega'.</math> | ||
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+ | The radius of <math>\omega'</math> we can find using <math>\triangle HKO:</math> | ||
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+ | <cmath>OK = R, HK = 2R, HG = 2GO \implies GK^2 = 2(R^2 – GO^2), GO^2 = \frac {HO^2}{9} \implies</cmath> | ||
+ | <cmath>R_G = GK = \frac {\sqrt {2(a^2 + b^2 + c^2)}}{3}.</cmath> | ||
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+ | Second inversion <math>I_{\omega}</math> centered at point <math>L = \frac {\vec O \cdot 2R – \vec H \cdot R}{2R – R} = 2 \vec O – \vec H.</math> We can make the same calculations and get <math>R_L = 4R \sqrt{– \cos A \cos B \cos C}</math> as desired. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
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==See Also== | ==See Also== | ||
*[[Circumcircle]] | *[[Circumcircle]] | ||
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*[[Euler line]] | *[[Euler line]] | ||
Latest revision as of 09:23, 17 November 2022
- The title of this article has been capitalized due to technical restrictions. The correct title should be de Longchamps point.
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The de Longchamps point ( ![]() orthocenter ( ![]() through the circumcenter ( ![]() |
The de Longchamps point of a triangle is the reflection of the triangle's orthocenter through its circumcenter.
The point is collinear with the orthocenter and circumcenter.
de Longchamps point
Definition 1
The de Longchamps point of a triangle is the radical center of the power circles of the triangle. Prove that De Longchamps point lies on Euler line.
We call A-power circle of a the circle centered at the midpoint
point
with radius
The other two circles are defined symmetrically.
Proof
Let and
be orthocenter, circumcenter, and De Longchamps point, respectively.
Denote power circle by
power circle by
WLOG,
Denote the projection of point
on
We will prove that radical axes of power and
power cicles is symmetric to altitude
with respect
Further, we will conclude that the point of intersection of the radical axes, symmetrical to the heights with respect to O, is symmetrical to the point of intersection of the heights
with respect to
Point is the crosspoint of the center line of the
power and
power circles and there radical axis.
We use claim and get:
and
are the medians, so
We use Claim some times and get:
radical axes of
power and
power cicles is symmetric to altitude
with respect
Similarly radical axes of power and
power cicles is symmetric to altitude
radical axes of
power and
power cicles is symmetric to altitude
with respect
Therefore the point
of intersection of the radical axes, symmetrical to the heights with respect to
is symmetrical to the point
of intersection of the heights with respect to
lies on Euler line of
Claim (Distance between projections)
Definition 2
We call circle of a
the circle centered at
with radius
The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of
circle,
circle, and
circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under Definition 1.
Proof
Let and
be orthocenter, centroid, and De Longchamps point, respectively. Let
cross
at points
and
The other points
are defined symmetrically.
Similarly
is diameter
Therefore is anticomplementary triangle of
is orthic triangle of
So
is orthocenter of
as desired.
vladimir.shelomovskii@gmail.com, vvsss
De Longchamps circle
De Longchamps circle of the obtuse triangle
is circle centered at de Longchamps point
which is orthogonal to the
Prove that de Longchamps circle is orthogonal to the
power,
power and
power cicles and the radius of de Longchamps circle is
Proof
– diameter
Let be the foot of perpendicular from
to
Let be crosspoint of
and
Points and
are simmetric with respect to
Points and
lies on
power circle and on
.
points
and
are collinear
power circle.
is the radical center of all six circles, therefore
is perpendicular to each of these circles.
is orthocenter of the anticomplementary triangle of
so radius of
is twice radius of circle finded by Claim
Claim (Radius)
Let be obtuse triangle
with circumcircle
circumradius
and orthocenter
Let
be the circle with diameter
Let
be the circle perpendicular to
centered at
Find
the radius of
Let altitude cross
at
and cross
second time at
points
are collinear
Inversion with respect swap
and
Well known that
Points
and
are symmetric with respect
vladimir.shelomovskii@gmail.com, vvsss
De Longchamps line
The de Longchamps line of
is defined as the radical axes of the de Longchamps circle
and of the circumscribed circle
of
Let be the circumcircle of
(the anticomplementary triangle of
Let be the circle centered at
(centroid of
) with radius
where
Prove that the de Longchamps line is perpendicular to Euler line and is the radical axes of and
Proof
Center of is
, center of
is
where
is Euler line.
The homothety with center
and ratio
maps
into
This homothety maps
into
and
there is two inversion which swap
and
First inversion centered at point
Let
be the point of crossing
and
The radius of we can find using
Second inversion centered at point
We can make the same calculations and get
as desired.
vladimir.shelomovskii@gmail.com, vvsss
See Also
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