Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Kimberling’s point X(20)"

(Created page with "==De Longchamps point X(20)== 450px|right <i><b>Definition 1</b></i> The De Longchamps’ point of a triangle is the radical center of the power circl...")
 
(→‎De Longchamps line)
 
(2 intermediate revisions by the same user not shown)
Line 54: Line 54:
  
 
<math>2\vec {HG} = \vec GL_o, 2\vec {GO} = \vec HG \implies L_o = L</math> as desired.
 
<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'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Latest revision as of 10:37, 26 November 2022

De Longchamps point X(20)

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 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 $\vec G = \frac {\vec O \cdot 2R + \vec H \cdot R}{2R + R} = \frac {2\vec O + \vec H}{3}.$ 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 $\vec L = \frac {\vec O \cdot 2R – \vec H \cdot R}{2R – R} = 2 \vec O – \vec H.$ 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Kimberling’s_point_X(20)&oldid=183097"