# Kimberling’s point X(20)

## De Longchamps point X(20)

**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 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**