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Forum GeometricorumVolume 1 (2001) 9197.

FORUM GEOM

ISSN 1534-1178

Some Properties of the Lemoine Point

Alexei Myakishev

Abstract. The Lemoine point, K, ofABC has special properties involving

centroids of pedal triangles. These properties motivate a definition of Lemoine

field, F, and a coordinatization of the plane ofABC using perpendicular axes

that pass through K. These principal axes are symmetrically related to two other

lines: one passing through the isodynamic centers, and the other, the isogonic

centers.

1. Introduction

Let ABC be the pedal triangle of an arbitrary point Z in the plane of a triangleABC, and consider the vector field F defined by F(Z) = ZA + ZB + ZC. Itis well known that F(Z) is the zero vector if and only if Z is the Lemoine point,K, also called the symmedian point. We call F the Lemoine field ofABC andK the balance point ofF.

B

C

A

Z

F(Z)

A

B

C

Figure 1

Z

Fx

Fy

F(Z)

A B x

y C

Figure 2

The Lemoine field may be regarded as a physical force field. Any point Z in this

field then has a natural motion along a certain curve, or trajectory. See Figure 1.

We shall determine parametric equations for these trajectories and find, as a result,

special properties of the lines that bisect the angles between the line of the isogoniccenters and the line of the isodynamic centers ofABC.

Publication Date: June 21, 2001. Communicating Editor: Clark Kimberling.

blagodar dorogu Lenu za okazannu mne moralnu podderku. Take oqen priznatelen professoru Kimberlingu za razrexenie mnogoqislennyh prob-lem, kasawihs angliiskogo zyka. The author dedicates his work to Helen and records hisappreciation to the Communicating Editor for assistance in translation.

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92 Alexei Myakishev

2. The Lemoine equation

In the standard cartesian coordinate system, place ABC so that A = (0, 0),B = (c, 0), C = (m, n), and write Z = (x, y). For any line P x + Qy + R = 0,the vector H from Z to the projection ofZ on the line has components

hx = P

P2 + Q2 (P x + Qy + R), hy = Q

P2 + Q2 (P x + Qy + R).

From these, one find the components of the three vectors whose sum defines F(Z):

vector x component y componentZA

n(nx+y(cm)cn)n2+(cm)2

(mc)(nx+y(cm)cn)n2+(cm)2

ZB n(nxmy)

m2+n2m(nxmy)m2+n2

ZC 0 y

The components of the Lemoine field F(Z) = ZA + ZB +ZC are given by

Fx = (x + y) + dx, Fy = (x + y) + dy,where

= n2

m2+n2+ n

2

n2+(cm)2 , =mnm2+n2

+ n(cm)n2+(cm)2 ,

= 1 + m2

m2+n2+ (cm)

2

n2+(cm)2 ;

dx =cn2

n2+(cm)2 , dy =cn(cm)

n2+(cm)2 .

See Figure 2. Assuming a unit mass at each point Z, Newtons Second Law now

gives a system of differential equations:

x = (x + y) + dx, y = (x + y) + dy,where the derivatives are with respect to time, t. We now translate the origin from

(0, 0) to the balance point (dx, dy), which is the Lemoine point K, thereby obtain-ing the system

x = (x + y), y = (x + y),which has the matrix form

x

y

= M

x

y

, (1)

where M =

. We shall refer to (1) as the Lemoine equation.

3. Eigenvalues of the matrix M

In order to solve equation (1), we first find eigenvalues 1 and 2 of M. Theseare the solutions of the equation |M I| = 0, i.e., ( )( ) 2 = 0, or

2 ( + ) + ( 2) = 0.

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Some properties of the Lemoine point 93

Thus

1 + 2 = + = 1 +m2 + n2

m2 + n2+

n2 + (c m)2n2 + (c m)2 = 3.

Writing a, b, c for the sidelengths |BC|, |CA|, |AB| respectively, we find thedeterminant

|M| = 2 = n2

a2b2(a2 + b2 + c2) > 0.

The discriminant of the characteristic equation 2 ( + ) + ( 2) = 0 isgiven by

D = ( + )2 4( 2) = ( )2 + 42 0. (2)Case 1: equal eigenvalues 1 = 2 =

32 . In this case, D = 0 and (2) yields = 0

and = . To reduce notation, write p = c m. Then since = 0, we havem

m2+n2= p

p2+n2, so that

(mp)(mp n2) = 0. (3)Also, since = , we find after mild simplifications

n4 (m2 +p2)n2 3m2p2 = 0. (4)Equation (3) imples that m = p or mp = n2. If m = p, then equation (4) has

solutions n =

3m =

3p. Consequently, C =12c,

32 c

, so that ABC isequilateral. However, ifmp = n2, then equation (4) leads to (m +p)2 = 0, so thatc = 0, a contradiction. Therefore from equation (3) we obtain this conclusion: ifthe eigenvalues are equal, then ABC is equilateral.

Case 2: distinct eigenvalues 1,2 =3

D

2 . Here D > 0, and 1,2 > 0 accordingto (2). We choose to consider the implications when

= 0, = . (5)We omit an easy proof that these conditions correspond to ABC being a right

triangle or an isosceles triangle. In the former case, write c2 = a2 + b2. Then thecharacteristic equation yields eigenvalues and , and

=n2

b2+

n2

a2=

n2(a2 + b2)

a2b2=

n2c2

a2b2= 1,

since ab = nc = twice the area of the right triangle. Since + = 3, = 2.

4. General solution of the Lemoine equation

According to a well known theorem of linear algebra, rotation of the coordinate

system about K gives the system x = 1x, y = 2y. Let us call the axes ofthis coordinate system the principal axes of the Lemoine field.Note that ifABC is a right triangle or an isosceles triangle (cf. conditions(5)), then the angle of rotation is zero, and K is on an altitude of the triangle. In

this case, one of the principal axes is that altitude, and the other is parallel to the

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94 Alexei Myakishev

corresponding side. Also ifABC is a right triangle, then K is the midpoint ofthat altitude.

In the general case, the solution of the Lemoine equation is given by

x = c1 cos 1t + c2 sin 2t, y = c3 cos 1t + c4 sin 2t, (6)

where 1 =

1, 2 =

2. Initial conditions x(0) = x0, y(0) = y0, x(0) = 0,

y(0) = 0 reduce (6) to

x = x0 cos 1t, y = y0 cos 2t, (7)

with 1 > 0, 2 > 0, 21 +

22 = 3. Equations (7) show that each trajectory is

bounded. If1 = 2, then the trajectory is a line segment; otherwise, (7) representsa Lissajous curve or an almost-everywhere rectangle-filling curve, according as1

2is rational or not.

5. Lemoine sequences and centroidal orbits

Returning to the Lemoine field, F, suppose Z0 is an arbitrary point, and GZ0 is

the centroid of the pedal triangle ofZ0. Let Z0 be the point to which F translates

Z0. It is well known that GZ0 lies on the line Z0Z0 at a distance 13 of that fromZ0 to Z

0. With this in mind, define inductively the Lemoine sequence of Z0 as

the sequence (Z0, Z1, Z2, . . .), where Zn, for n 1, is the centroid of the pedaltriangle ofZn1. Writing the centroid of the pedal triangle ofZ0 as Z1 = (x1, y1),we obtain 3(x1 x0) = 1x0 and

x1 =1

3(3 1)x0 = 1

32x0; y1 =

1

31y0.

Accordingly, the Lemoine sequence is given with respect to the principal axes by

Zn =

x0

2

3

n, y0

1

3

n. (8)

Since 131 and 132 are between 0 and 1, the points Zn approach (0, 0) as n .That is, the Lemoine sequence of every point converges to the Lemoine point.

Representation (8) shows that Zn lies on the curve (x, y) = (x0ut, y0v

t), whereu = 132 and v =

131. We call this curve the centroidal orbit of Z0. See Figure

3. Reversing the directions of axes if necessary, we may assume that x0 > 0 andy0 > 0, so that elimination oft gives

y

y0=

x

x0

k, k =

ln v

ln u. (9)

Equation (9) expresses the centroidal orbit of Z0 = (x0, y0). Note that if1 =2, then v = u, and the orbit is a line. Now let XZ and YZ be the points in which

line ZGZ meets the principal axes. By (8),

|ZGZ||GZXZ| =

2

1,

|ZGZ||GZYZ| =

1

2. (10)

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Some properties of the Lemoine point 95

K

C

A

B

Figure 3

These equations imply that ifABC is equilateral with center O, then the centroidGZ is the midpoint of segment OGZ.

As another consequence of (10), suppose ABC is a right triangle; let H bethe line parallel to the hypotenuse and passing through the midpoint of the altitude

H to the hypotenuse. Let X and Y be the points in which line ZGZ meets H andH, respectively. Then |ZGZ| : |XGZ| = |Y GZ| : |ZGZ| = 2 : 1.6. The principal axes of the Lemoine field

Physically, the principal axes may be described as the locus of points in the plane

ofABC along which the direction of the Lemoine sequence remains constant.That is, ifZ0 lies on one of the principal axes, then all the points Z1, Z2, . . . lie on

that axis also.

In this section, we turn to the geometry of the principal axes. Relative to the

coordinate system adopted in 5, the principal axes have equations x = 0 and y =0. Equation (8) therefore shows that if Z0 lies on one of these two perpendicularlines, then Zn lies on that line also, for all n 1.

Let A1, A2 denote the isodynamic points, and F1, F2 the isogonic centers, of

ABC. Call lines A1A2 and F1F2 the isodynamic axis and the isogonic axisrespectively. 1

Lemma 1. Suppose Z and Z are a pair of isogonal conjugate points. Let O andO be the circumcircles of the pedal triangles ofZ and Z. Then O = O, and thecenter ofO is the midpoint between Z andZ.

1The points F1, F2, A1, A2 are indexed as X13, X14, X15, X16 and discussed in [2].

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96 Alexei Myakishev

OF1

A1

C

A

B

Figure 4

A proof is given in Johnson [1, pp.155156]. See Figure 4.

Now suppose that Z = A1. Then Z = F1, and, according to Lemma 1, the

pedal triangles of Z and Z have the same circumcircle, whose center O is themidpoint between A1 and F1. Since the pedal triangle of A1 is equilateral, thepoint O is the centroid of the pedal triangle ofA1.

Next, suppose L is a line not identical to either of the principal axes. Let L bethe reflection ofL about one of the principal axes. Then L is also the reflection ofL about the other principal axis. We call L and L a symmetric pair of lines.

Lemma 2. Suppose thatGP is the centroid of the pedal triangle of a point P, and

thatQ is the reflection of P in GP. Then there exists a symmetric pair of lines, one

passing through P and the other passing through Q.

Proof. With respect to the principal axes, write P = (xP, yP) and Q = (xQ, yQ).Then GP = (

132xP,

131yP), and

232xP = xP + xQ, so that

xQ =

2

32 1

xP =

1

3(22 (1 + 2))xP = 1

3(2 1)xP.

Likewise, yQ =13yP(1 2). It follows that xPyP =

xQyQ

. This equation shows

that the line y = yPxP x passing through P and the line y = yQ

xQ x passing through

Q are symmetric about the principal axes y = 0 and x = 0. See Figure 5.

Theorem. The principal axes of the Lemoine field are the bisectors of the angles

formed at the intersection of the isodynamic and isogonic axes in the Lemoine

point.

Proof. In Lemma 2, take P = A1 and Q = F1. The symmetric pair of lines arethen the isodynamic and isogonic axes. Their symmetry about the principal axesis equivalent to the statement that these axes are the asserted bisectors.

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Some properties of the Lemoine point 97

F1

A1

F2

C

A

B

K

Z1

G1

Z2

G2

Figure 5

References

[1] R.A. Johnson, Advanced Euclidean Geometry, Dover reprint, 1960.

[2] C. Kimberling, Encyclopedia of Triangle Centers, 2000,

http://cedar.evansville.edu/ck6/encyclopedia/.

Alexei Myakishev: Smolnaia 61-2, 138, Moscow, Russia, 125445.

E-mail address: alex geom@mtu-net.ru