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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 of Euler line==
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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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==See Also==
 
==See Also==

Revision as of 14:37, 15 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 of Euler line

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


See Also

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