De Longchamps point
- The title of this article has been capitalized due to technical restrictions. The correct title should be de Longchamps point.
|
The de Longchamps point ( ![]() orthocenter ( ![]() through the circumcenter ( ![]() |
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
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
See Also
This article is a stub. Help us out by expanding it.