Kiepert conics explained

In triangle geometry, the Kiepert conics are two special conics associated with the reference triangle. One of them is a hyperbola, called the Kiepert hyperbola and the other is a parabola, called the Kiepert parabola. The Kiepert conics are defined as follows:

If the three triangles

A\primeBC

,

AB\primeC

and

ABC\prime

, constructed on the sides of a triangle

ABC

as bases, are similar, isosceles and similarly situated, then the triangles

ABC

and

A\primeB\primeC\prime

are in perspective. As the base angle of the isosceles triangles varies between

-\pi/2

and

\pi/2

, the locus of the center of perspectivity of the triangles

ABC

and

A\primeB\primeC\prime

is a hyperbola called the Kiepert hyperbola and the envelope of their axis of perspectivity is a parabola called the Kiepert parabola.

It has been proved that the Kiepert hyperbola is the hyperbola passing through the vertices, the centroid and the orthocenter of the reference triangle and the Kiepert parabola is the parabola inscribed in the reference triangle having the Euler line as directrix and the triangle center X110 as focus.[1] The following quote from a paper by R. H. Eddy and R. Fritsch is enough testimony to establish the importance of the Kiepert conics in the study of triangle geometry:[2]

"If a visitor from Mars desired to learn the geometry of the triangle but could stay in the earth's relatively dense atmosphere only long enough for a single lesson, earthling mathematicians would, no doubt, be hard-pressed to meet this request. In this paper, we believe that we have an optimum solution to the problem. The Kiepert conics ...."

Kiepert hyperbola

The Kiepert hyperbola was discovered by Ludvig Kiepert while investigating the solution of the following problem proposed by Emile Lemoine in 1868: "Construct a triangle, given the peaks of the equilateral triangles constructed on the sides." A solution to the problem was published by Ludvig Kiepert in 1869 and the solution contained a remark which effectively stated the locus definition of the Kiepert hyperbola alluded to earlier.[2]

Basic facts

Let

a,b,c

be the side lengths and

A,B,C

the vertex angles of the reference triangle

ABC

.

Equation

The equation of the Kiepert hyperbola in barycentric coordinates

x:y:z

is
b2-c2+
x
c2-a2+
y
a2-b2
z

=0.

Center, asymptotes

(b2-c2)2:(c2-a2)2:(a2-b2)2

.

\sqrt{2}

.

Properties

  1. The center of the Kiepert hyperbola lies on the nine-point circle. The center is the midpoint of the line segment joining the isogonic centers of triangle

ABC

which are the triangle centers X(13) and X(14) in the Encyclopedia of Triangle Centers.
  1. The image of the Kiepert hyperbola under the isogonal transformation is the Brocard axis of triangle

ABC

which is the line joining the symmedian point and the circumcenter.
  1. Let

P

be a point in the plane of a nonequilateral triangle

ABC

and let

p

be the trilinear polar of

P

with respect to

ABC

. The locus of the points

P

such that

p

is perpendicular to the Euler line of

ABC

is the Kiepert hyperbola.

Kiepert parabola

The Kiepert parabola was first studied in 1888 by a German mathematics teacher Augustus Artzt in a "school program".[2] [3]

Basic facts

x:y:z

is

f2x2+g2y2+h2z2-2fgxy-2ghyz-2hfzx=0


where

f=(b2-c2)/a,g=(c2-a2)/b,h=(a2-b2)/c

.

a2/(b2-c2):b2/(c2-a2):c2/(a2-b2)

ABC

.

See also

External links

Notes and References

  1. Web site: Kimberling, C. . X(110)=Focus of Kiepert Parabola . Encyclopedia of Triangle Centers . 4 February 2022.
  2. Eddy, R. H. and Fritsch, R. . The Conics of Ludwig Kiepert: A Comprehensive Lesson in the Geometry of the Triangle . Math. Mag. . 1994 . 67 . 3 . 188–205. 10.1080/0025570X.1994.11996212 .
  3. Sharp, J. . Artzt parabolas of a triangle . The Mathematical Gazette . 2015 . 99 . 546 . 444–463 . 10.1017/mag.2015.81. 123814409 .