De Longchamps point
- The title of this article has been capitalized due to technical restrictions. The correct title should be de Longchamps point.
point () is the the
orthocenter () reflected
through the circumcenter
The point is collinear with the orthocenter and circumcenter.
de Longchamps point
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.
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)
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.
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
De Longchamps circle
De Longchamps circle of the obtuse triangle is circle centered at de Longchamps point which is orthogonal to the Prove that de Longchamps circle is orthogonal to the power, power and power cicles and the radius of de Longchamps circle is
Let be the foot of perpendicular from to
Let be crosspoint of and
Points and are simmetric with respect to
Points and lies on power circle and on .
points and are collinear power circle.
is the radical center of all six circles, therefore is perpendicular to each of these circles. is orthocenter of the anticomplementary triangle of so radius of is twice radius of circle finded by Claim
Let be obtuse triangle with circumcircle circumradius and orthocenter Let be the circle with diameter Let be the circle perpendicular to centered at Find the radius of
Let altitude cross at and cross second time at
points are collinear
Inversion with respect swap and
Well known that
Points and are symmetric with respect
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
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.
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