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Difference between revisions of "De Longchamps point"

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The point is [[collinear]] with the orthocenter and circumcenter.
 
The point is [[collinear]] with the orthocenter and circumcenter.
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==de Longchamps point==
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[[File:Longchamps.png|450px|right]]
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<i><b>Definition 1</b></i>
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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.
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We call A-power circle of a <math>\triangle ABC</math> the circle centered at the midpoint <math>BC</math> point <math>A'</math> with radius <math>R_A = AA'.</math> The other two circles are defined symmetrically.
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<i><b>Proof</b></i>
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Let <math>H, O,</math> and <math>L</math> be orthocenter, circumcenter, and De Longchamps point, respectively.
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Denote <math>B-</math>power circle by <math>\omega_B, C-</math>power circle by <math>\omega_C, D = \omega_B \cap \omega_C,</math>
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<math>a = BC, b = AC, c = AB.</math> WLOG, <math>a \ge b \ge c.</math>
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Denote <math>X_t</math> the projection of point <math>X</math> on <math>B'C', E = D_t.</math>
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We will prove that radical axes of <math>B-</math>power and <math>C-</math>power cicles is symmetric to altitude <math>AH</math> with respect <math>O.</math> 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 <math>H</math> with respect to <math>O.</math>
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Point <math>E</math> is the crosspoint of the center line of the <math>B-</math>power and <math>C-</math>power circles and there radical axis. <math>B'C' = \frac {a}{2}.</math> We use claim and get:
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<cmath>C'E =  \frac {a}{4} + \frac {R_C^2 – R_B^2}{a}.</cmath>
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<math>R_B</math> and <math>R_C</math> are the medians, so
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<cmath>R_B^2 = \frac {a^2}{2}+ \frac {c^2}{2} – \frac {b^2}{4}, R_C^2 = \frac {a^2}{2}+ \frac {b^2}{2} – \frac {c^2}{4} \implies C'E =  \frac {a}{4} + \frac {3(b^2 – c^2)}{4a}.</cmath>
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We use Claim some times and get:
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<cmath>C'A_t =  \frac {a}{4} – \frac {b^2 – c^2}{4a},
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A_tO_t =  \frac {a}{2} – 2 C'A_t = \frac {b^2 – c^2}{2a} \implies</cmath>
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<cmath>O_t L_t = C'E – C'A_t - A_t O_t = \frac {b^2 – c^2}{2a}  = A_t O_t = H_t O_t \implies</cmath>
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radical axes of <math>B-</math>power and <math>C-</math>power cicles is symmetric to altitude <math>AH</math> with respect <math>O.</math>
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Similarly radical axes of <math>A-</math>power and <math>B-</math>power cicles is symmetric to altitude <math>CH,</math> radical axes of <math>A-</math>power and <math>C-</math>power cicles is symmetric to altitude <math>BH</math> with respect <math>O.</math>
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Therefore the point <math>L</math> of intersection of the radical axes, symmetrical to the heights with respect to <math>O,</math> is symmetrical to the point <math>H</math> of intersection of the heights with respect to <math>O \implies \vec {HO} = \vec {OL} \implies L</math> lies on Euler line of <math>\triangle ABC.</math>
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[[File:Distances.png|350px|right]]
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<i><b>Claim (Distance between projections)</b></i>
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<cmath>x + y = a, c^2 – x^2 = h^2 = b^2 – y^2,</cmath>
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<cmath>y^2 – x^2 = b^2 – c^2 \implies y – x = \frac {b^2 – c^2}{a},</cmath>
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<cmath>x = \frac {a}{2} –  \frac {b^2 – c^2}{2a}, y = \frac {a}{2} +  \frac {b^2 – c^2}{2a}.</cmath>
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<i><b>Definition 2</b></i>
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[[File:Longchamps 1.png|500px|right]]
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We call <math>\omega_A = A-</math>circle of a <math>\triangle ABC</math> the circle centered at <math>A</math> with radius <math>BC.</math> The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of <math>A-</math>circle, <math>B-</math>circle, and  <math>C-</math>circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under <i><b>Definition 1.</b></i>
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<i><b>Proof</b></i>
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Let <math>H, G,</math> and <math>L_o</math> be orthocenter, centroid, and De Longchamps point, respectively. Let <math>\omega_B</math> cross <math>\omega_C</math> at points <math>A'</math> and <math>E.</math> The other points <math>(D, F, B', C')</math> are defined symmetrically.
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<cmath>AB' = BC, B'C = AB \implies \triangle ABC = \triangle CB'A \implies</cmath>
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<cmath>AB||B'C \implies CH \perp B'C.</cmath>
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Similarly <math>CH \perp A'C \implies A'B'</math> is diameter <math>\omega_C \implies</math>
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<cmath>\angle A'EB' = 90^\circ, 2\vec {BG} = \vec {GB'}.</cmath>
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Therefore <math>\triangle A'B'C'</math> is anticomplementary triangle of <math>\triangle ABC, \triangle DEF</math> is orthic triangle of  <math>\triangle A'B'C'.</math> So <math>L_o</math> is orthocenter of <math>\triangle A'B'C'.</math>
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<math>2\vec {HG} = \vec GL_o, 2\vec {GO} = \vec HG \implies L_o = L</math> as desired.
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==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>
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[[File:Longchamps circle.png|350px|right]]
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<i><b>Proof</b></i>
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<math>\vec {A'B'} = 2\vec {BA}, A'B'</math> – diameter <math>\omega_C.</math>
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Let <math>E</math> be the foot of perpendicular from <math>L</math> to <math>AB'.</math>
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Let <math>E_0</math> be crosspoint of <math>\omega_C</math> and <math>AB' \implies A'E_0 \perp AB' \implies E = E_0.</math>
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Points <math>A'</math> and <math>E</math> are simmetric with respect to <math>BC, MA' = AM \implies</math>
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Points <math>A'</math> and <math>E</math> lies on <math>A-</math>power circle and on <math>\omega_C</math>.
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<math>\Omega \perp \omega_C,</math> points <math>L, A',</math> and <math>E</math> are collinear <math>\implies \Omega \perp A-</math> power circle.
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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.
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<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>
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<i><b>Claim (Radius)</b></i>
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[[File:Longchamps Claim.png|300px|right]]
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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>
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Let <math>\omega'</math> be the circle with diameter <math>AB.</math>
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Let <math>\omega</math> be the circle perpendicular to <math>\omega'</math> centered at <math>H.</math> Find <math>R_\omega,</math> the radius of <math>\omega.</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>
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<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>
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<cmath>BC \perp HD, AC \perp BH \implies \angle C = \angle BHD \implies HD = BH \cos C.</cmath>
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Points <math>H</math> and <math>H'</math> are symmetric with respect <math>BC \implies BH' = BH.</math>
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<cmath>\angle BAH' = 90^\circ – \angle B \implies BH' = 2R \sin \angle BAH' = 2R \cos B \implies</cmath>
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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>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==De Longchamps line==
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[[File:Longchamps lime.png|450px|right]]
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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.
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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>
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<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>
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<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.
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'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==See Also==
 
==See Also==
 
*[[Circumcircle]]
 
*[[Circumcircle]]
−
*[[Excircle]]
+
*[[Euler line]]
−
*[[Incircle]]
 
  
 
{{stub}}
 
{{stub}}
 
[[Category:Definition]]
 
[[Category:Definition]]
 
[[Category:Geometry]]
 
[[Category:Geometry]]

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.

[asy] draw((0,0)--(44,60)--(44,-10)--cycle); draw((0,0)--(44,0),blue+dashed); draw((44,60)--(22,-5),blue+dashed); draw((44,-10)--(6.5,10),blue+dashed); label("H",(24,0),(1,1)); dot((24,0)); draw((22,30)--(44,14),red); draw((22,-5)--(34,46),red); draw((44,25)--(18,25),red); dot((29,25)); label("C",(29,25),(1,1)); draw(Circle((29,25),25),dashed); dot((34,50)); label("L",(34,50),(1,1)); [/asy]

Enlarge.png
The de Longchamps
point ($L$) is the the
orthocenter ($H$) reflected
through the circumcenter
($C$).

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

Longchamps.png

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 $\triangle ABC$ the circle centered at the midpoint $BC$ point $A'$ with radius $R_A = AA'.$ The other two circles are defined symmetrically.

Proof

Let $H, O,$ and $L$ be orthocenter, circumcenter, and De Longchamps point, respectively.

Denote $B-$power circle by $\omega_B, C-$power circle by $\omega_C, D = \omega_B \cap \omega_C,$ $a = BC, b = AC, c = AB.$ WLOG, $a \ge b \ge c.$

Denote $X_t$ the projection of point $X$ on $B'C', E = D_t.$

We will prove that radical axes of $B-$power and $C-$power cicles is symmetric to altitude $AH$ with respect $O.$ 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 $H$ with respect to $O.$

Point $E$ is the crosspoint of the center line of the $B-$power and $C-$power circles and there radical axis. $B'C' = \frac {a}{2}.$ We use claim and get:

\[C'E = \frac {a}{4} + \frac {R_C^2 – R_B^2}{a}.\] $R_B$ and $R_C$ are the medians, so \[R_B^2 = \frac {a^2}{2}+ \frac {c^2}{2} – \frac {b^2}{4}, R_C^2 = \frac {a^2}{2}+ \frac {b^2}{2} – \frac {c^2}{4} \implies C'E = \frac {a}{4} + \frac {3(b^2 – c^2)}{4a}.\]

We use Claim some times and get: \[C'A_t = \frac {a}{4} – \frac {b^2 – c^2}{4a}, A_tO_t = \frac {a}{2} – 2 C'A_t = \frac {b^2 – c^2}{2a} \implies\] \[O_t L_t = C'E – C'A_t - A_t O_t = \frac {b^2 – c^2}{2a} = A_t O_t = H_t O_t \implies\] radical axes of $B-$power and $C-$power cicles is symmetric to altitude $AH$ with respect $O.$

Similarly radical axes of $A-$power and $B-$power cicles is symmetric to altitude $CH,$ radical axes of $A-$power and $C-$power cicles is symmetric to altitude $BH$ with respect $O.$ Therefore the point $L$ of intersection of the radical axes, symmetrical to the heights with respect to $O,$ is symmetrical to the point $H$ of intersection of the heights with respect to $O \implies \vec {HO} = \vec {OL} \implies L$ lies on Euler line of $\triangle ABC.$

Distances.png

Claim (Distance between projections)

\[x + y = a, c^2 – x^2 = h^2 = b^2 – y^2,\] \[y^2 – x^2 = b^2 – c^2 \implies y – x = \frac {b^2 – c^2}{a},\] \[x = \frac {a}{2} – \frac {b^2 – c^2}{2a}, y = \frac {a}{2} + \frac {b^2 – c^2}{2a}.\]

Definition 2

Longchamps 1.png

We call $\omega_A = A-$circle of a $\triangle ABC$ the circle centered at $A$ with radius $BC.$ The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of $A-$circle, $B-$circle, and $C-$circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under Definition 1.

Proof

Let $H, G,$ and $L_o$ be orthocenter, centroid, and De Longchamps point, respectively. Let $\omega_B$ cross $\omega_C$ at points $A'$ and $E.$ The other points $(D, F, B', C')$ are defined symmetrically. \[AB' = BC, B'C = AB \implies \triangle ABC = \triangle CB'A \implies\] \[AB||B'C \implies CH \perp B'C.\] Similarly $CH \perp A'C \implies A'B'$ is diameter $\omega_C \implies$ \[\angle A'EB' = 90^\circ, 2\vec {BG} = \vec {GB'}.\]

Therefore $\triangle A'B'C'$ is anticomplementary triangle of $\triangle ABC, \triangle DEF$ is orthic triangle of $\triangle A'B'C'.$ So $L_o$ is orthocenter of $\triangle A'B'C'.$

$2\vec {HG} = \vec GL_o, 2\vec {GO} = \vec HG \implies L_o = L$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

De Longchamps circle

De Longchamps circle $\Omega$ of the obtuse triangle $ABC$ is circle centered at de Longchamps point $L$ which is orthogonal to the $\omega_C.$ Prove that de Longchamps circle is orthogonal to the $\omega_A, \omega_B, A-$power, $B-$power and $C-$power cicles and the radius of de Longchamps circle is $R_{\Omega} = 4R \sqrt { – \cos A \cos B \cos C}.$

Longchamps circle.png

Proof

$\vec {A'B'} = 2\vec {BA}, A'B'$ – diameter $\omega_C.$

Let $E$ be the foot of perpendicular from $L$ to $AB'.$

Let $E_0$ be crosspoint of $\omega_C$ and $AB' \implies A'E_0 \perp AB' \implies E = E_0.$

Points $A'$ and $E$ are simmetric with respect to $BC, MA' = AM \implies$

Points $A'$ and $E$ lies on $A-$power circle and on $\omega_C$.

$\Omega \perp \omega_C,$ points $L, A',$ and $E$ are collinear $\implies \Omega \perp A-$ power circle.

$L$ is the radical center of all six circles, therefore $\Omega$ is perpendicular to each of these circles. $L$ is orthocenter of the anticomplementary triangle of $\triangle ABC$ so radius of $\Omega$ is twice radius of circle finded by Claim \[\implies R_{\Omega} = 4R \sqrt { – \cos A \cos B \cos C}.\]

Claim (Radius)

Longchamps Claim.png

Let $ABC$ be obtuse triangle $(\angle A > 90^\circ)$ with circumcircle $\Omega,$ circumradius $R,$ and orthocenter $H.$ Let $\omega'$ be the circle with diameter $AB.$ Let $\omega$ be the circle perpendicular to $\omega'$ centered at $H.$ Find $R_\omega,$ the radius of $\omega.$

Let altitude $AH$ cross $BC$ at $D$ and cross $\Omega$ second time at $H'.$

$AD \perp BD \implies D \in \omega',$ points $H,A, D$ are collinear $\implies$

Inversion with respect $\omega$ swap $A$ and $D \implies R_\omega^2 = HA \cdot HD.$

Well known that $HA = – 2R \cos A.$

\[BC \perp HD, AC \perp BH \implies \angle C = \angle BHD \implies HD = BH \cos C.\] Points $H$ and $H'$ are symmetric with respect $BC \implies BH' = BH.$ \[\angle BAH' = 90^\circ – \angle B \implies BH' = 2R \sin \angle BAH' = 2R \cos B \implies\] \[R_\omega^2 = – 2R \cos A \cdot 2R \cos B \cos C \implies R_\omega = 2R \sqrt{– \cos A \cos B \cos C}.\]


vladimir.shelomovskii@gmail.com, vvsss

De Longchamps line

Longchamps lime.png

The de Longchamps line $l$ of $\triangle ABC$ is defined as the radical axes of the de Longchamps circle $\omega$ and of the circumscribed circle $\Omega$ of $\triangle ABC.$

Let $\Omega'$ be the circumcircle of $\triangle DEF$ (the anticomplementary triangle of $\triangle ABC).$

Let $\omega'$ be the circle centered at $G$ (centroid of $\triangle ABC$) with radius $\rho = \frac {\sqrt{2}}{3} \sqrt {a^2 + b^2 + c^2},$ where $a = BC, b = AC, c = AB.$

Prove that the de Longchamps line is perpendicular to Euler line and is the radical axes of $\Omega, \Omega', \omega,$ and $\omega'.$

Proof

Center of $\Omega$ is $O$, center of $\omega$ is $L \implies OL \perp l,$ where $OL$ is Euler line. The homothety with center $G$ and ratio $-2$ maps $\triangle ABC$ into $\triangle DEF.$ This homothety maps $\Omega$ into $\Omega'.$ $R_{\Omega} \ne R_{\Omega'}$ and $\Omega \cap \Omega' = K \implies$ there is two inversion which swap $\Omega$ and $\Omega'.$

First inversion $I_{\omega'}$ centered at point $G = \frac {\vec O \cdot 2R + \vec H \cdot R}{2R + R} = \frac {2\vec O + \vec H}{3}.$ Let $K$ be the point of crossing $\Omega$ and $\Omega'.$

The radius of $\omega'$ we can find using $\triangle HKO:$

\[OK = R, HK = 2R, HG = 2GO \implies GK^2 = 2(R^2 – GO^2), GO^2 = \frac {HO^2}{9} \implies\] \[R_G = GK = \frac {\sqrt {2(a^2 + b^2 + c^2)}}{3}.\]

Second inversion $I_{\omega}$ centered at point $L = \frac {\vec O \cdot 2R – \vec H \cdot R}{2R – R} = 2 \vec O – \vec H.$ We can make the same calculations and get $R_L = 4R \sqrt{– \cos A \cos B \cos C}$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

See Also

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