View source for Kimberling’s point X(20)

==De Longchamps point X(20)==
[[File:Longchamps.png|450px|right]]
<i><b>Definition 1</b></i>

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 <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. 

<i><b>Proof</b></i>

Let <math>H, O,</math> and <math>L</math> be orthocenter, circumcenter, and De Longchamps point, respectively.

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>
<math>a = BC, b = AC, c = AB.</math> WLOG, <math>a \ge b \ge c.</math>

Denote <math>X_t</math> the projection of point <math>X</math> on <math>B'C', E = D_t.</math>

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>
 
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:

<cmath>C'E =  \frac {a}{4} + \frac {R_C^2 – R_B^2}{a}.</cmath>
<math>R_B</math> and <math>R_C</math> are the medians, so 
<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>

We use Claim some times and get:
<cmath>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</cmath>
<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>
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>

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>
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>
[[File:Distances.png|350px|right]]
<i><b>Claim (Distance between projections)</b></i>

<cmath>x + y = a, c^2 – x^2 = h^2 = b^2 – y^2,</cmath>
<cmath>y^2 – x^2 = b^2 – c^2 \implies y – x = \frac {b^2 – c^2}{a},</cmath>
<cmath>x = \frac {a}{2} –  \frac {b^2 – c^2}{2a}, y = \frac {a}{2} +  \frac {b^2 – c^2}{2a}.</cmath>

<i><b>Definition 2</b></i>
[[File:Longchamps 1.png|500px|right]]
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>

<i><b>Proof</b></i>

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.
<cmath>AB' = BC, B'C = AB \implies \triangle ABC = \triangle CB'A \implies</cmath>
<cmath>AB||B'C \implies CH \perp B'C.</cmath>
Similarly <math>CH \perp A'C \implies A'B'</math> is diameter <math>\omega_C \implies</math>
<cmath>\angle A'EB' = 90^\circ, 2\vec {BG} = \vec {GB'}.</cmath> 

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>

<math>2\vec {HG} = \vec GL_o, 2\vec {GO} = \vec HG \implies L_o = L</math> as desired.

'''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> 

Let <math>\Omega'</math> be the circumcircle of <math>\triangle DEF</math> (the anticomplementary triangle of <math>\triangle ABC).</math>

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>

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>

<i><b>Proof</b></i>

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> 

First inversion <math>I_{\omega'}</math> centered at point <math>\vec 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>

The radius of <math>\omega'</math> we can find using <math>\triangle HKO:</math>

<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>
 
Second inversion <math>I_{\omega}</math> centered at point <math>\vec 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'''