Forum Geometricorum
Volume 16 (2016) 257–267.
FORUM GEOM
ISSN 1534-1178
Locus of Centroids of Similar Inscribed Triangles
Francisco Javier Garcı́a Capitán
Abstract. We study the locus of the centroids of families of similar triangles
inscribed in a given triangle.
1. Miquel circles
Given a triangle ABC and three points X, Y , Z on the sidelines BC, CA, AB
respectively, the three Miquel circles are the circumcircles of the triangles AY Z,
BZX, and CXY . According to Miquel’s theorem, the three Miquel circles concur
at a point, the Miquel point of X, Y , Z.
A
Y
Z
M
B
C
X
Figure 1
It is well known that if XY Z remains similar to a given triangle, the point M
is fixed. The converse is also true: given a point M with homogeneous barycentric
coordinates (u, v, w) with reference to ABC, if for X, Y , Z on the lines BC, CA,
AB respectively, the circumcircles of AY Z, BZX, and CXY pass through M ,
then all such triangles XY Z are mutually similar. If X0 Y0 Z0 is the pedal triangle
of M , then every such triangle XY Z satisfies
∠X0 M X = ∠Y0 M Y = ∠Z0 M Z (= θ).
Publication Date: June 20, 2016. Communicating Editor: Paul Yiu.
258
F. J. Garcı́a Capitán
If t = tan θ, the vertices of the triangle are
Xt = (0, (a2 + b2 − c2 )u + 2a2 v − 2Stu, (c2 + a2 − b2 )u + 2a2 w + 2Stu),
Yt = (2b2 u + (a2 + b2 − c2 )v + 2Stv, 0, (b2 v + c2 − a2 )v + 2b2 w − 2Stv),
2
2
2
2
2
2
2
(1)
2
Zt = (2c u + (c + a − b )w − 2Stw, 2c v + (b + c − a )w + 2Stw, 0)
in homogeneous barycentric coordinates. For basic formulas in barycentric coordinates, see [2].
2. The locus of the centroid of triangles XY Z
Note that in (1) above, the coordinate sums of Xt , Yt , Zt are respectively 2a2 (u+
v + w), 2b2 (u + v + w), 2c2 (u + v + w). The centroid of Xt Yt Zt is the point
v
w w
u u
v
Gt = G0 + 2t · a2 b2 c2 S 2 − 2 , 2 − 2 , 2 − 2 ,
b
c c
a a
b
where
G0 = (a2 (4b2 c2 u + c2 (a2 + b2 − c2 )v + b2 (c2 + a2 − b2 )w),
b2 (c2 (a2 + b2 − c2 )u + 4a2 c2 v + a2 (b2 + c2 − a2 )w),
c2 (b2 (c2 + a2 − b2 )u + a2 (b2 + c2 − a2 )v + 4a2 b2 w))
is the centroid of the pedal triangle of M . Note that for M = K = (a2 , b2 , c2 ), the
symmedian point of triangle ABC, Gt = G0 = K.
For M = K, the coordinates of Gt are linear functions of t, the locus of Gt is a
straight line (M ). The line (M ) clearly contains G0 and the infinite point
v
w w
u u
v
J(M ) := 2 − 2 , 2 − 2 , 2 − 2 .
(2)
b
c c
a a
b
This is the infinite point of the line
v
w
u
x + 2 y + 2 z = 0,
2
a
b
c
2 2 2
the trilinear polar of M ∗ := au , bv , cw , the isogonal conjugate of M . We
summarize this in the following theorem.
Theorem 1. Let M = K be a point with homogeneous barycentric coordinates
(u, v, w) with reference to ABC. The locus of the centroids of triangles XY Z with
Miquel point M is the line (M ) through the centroid G0 of the pedal triangle of
M parallel to the trilinear polar of the isogonal conjugate of M .
The line (M ) has barycentric equation
b2 c2 u2 − 2c2 a2 v 2 − 2a2 b2 w2 − a2 (b2 + c2 − a2 )vw x = 0.
+b2 (−a2 + b2 + 2c2 )wu + c2 (−a2 + 2b2 + c2 )uv
cyclic
(3)
Locus of centroids of similar inscribed triangles
259
(M )
A
Z0
Y0
M
M*
G0
X0
B
C
Figure 2
Example 1.
M
(i)
G
(ii)
O
(iii) H
(iv) I
(v)
X(55)
(vi) X(56)
(vii) X(99)
(viii) X(110)
(M )
(a4 − 4a2 (b2 + c2 ) + b4 + 5b2 c2 + c4 )x = 0
cyclic 2
2
2
cyclic (b + c − 2a )x = 0
line
O and
joining
2
2
2
cyclic (a − 2a(b + c) + b + bc + c )x = 0
line joining X(7) and
line
and
joining X(8)
x
2
2
2
2 2 = 0
cyclic a (b x+c )−2b c
cyclic b2 +c2 −2a2 = 0
J(M )
X(512)
X(523)
X(520)
X(513)
X(514)
X(522)
X(888)
X(690)
3. Parallelism and orthogonality
Proposition 2. The lines (M ) and (M ) are parallel if and only if the line M M passes through the symmedian point of triangle ABC.
Proof. Let M = (u, v, w) and M = (u , v , w ) in homogeneous barycentric
coordinates. The lines (M ) and (M ) are parallel if and only if the trilinear polars
z = 0:
of M and M are parallel. Equivalently, J(M ) lies on the line au2 x+ bv2 y+ w
c2
w v w
u w u
v
u v
+
+
−
−
−
a2 b2 c2
b2 c2 a2
c2 a2 b2
1
= 2 2 2 a2 (v w − vw ) + b2 (w u − wu ) + c2 (u v − uv ) .
a b c
0=
260
F. J. Garcı́a Capitán
Since (v w − vw )x + (w u − wu )y + (u v − uv )z = 0 represents the line M M ,
the condition is equivalent to the line M M containing (a2 , b2 , c2 ), the symmedian
point of triangle ABC.
Corollary 3. If M lies on the Brocard axis, the line (M ) is perpendicular to Euler
line.
Proof. If M lies on the Brocard axis, by Theorem 1,
2
b SB
c2 SC c2 SC
a2 SA a2 SA b2 SB
J(M ) = J(O) =
− 2 , 2 − 2 , 2 − 2
b2
c
c
a
a
b
= (SB − SC , SC − SA , SA − SB ),
which is the triangle center X(523) in [1], the infinite point of the perpendicular to
the Euler line. Therefore, (M ) is perpendicular to the Euler line.
Corollary 4. The locus of the centroids of equilateral inscribed triangles is formed
by two lines perpendicular to Euler line.
Proof. This follows from the fact that equilateral inscribed triangles have Miquel
points the isodyanamic points X(15) or X(16) on the Brocard axis. The locus is
formed by the lines
√
( 3(a2 (b2 + c2 ) − (b4 + c4 )) ± 2(b2 + c2 − 2a2 )S)x = 0,
cyclic
S being twice the area of triangle ABC.
Proposition 5. Let M = (u, v, w). The locus of M for which (M ) ⊥ (M ) is
the line
(−2b2 c2 u + c2 (a2 + b2 − c2 )v + b2 (c2 + a2 − b2 )w)x = 0.
(4)
cyclic
Proof. For M = (x, y, z), (M ) ⊥ (M ) if and only if
SA
v
b2
−
w
u
w y
z
u z
x
v x
y
+S
+S
= 0.
−
−
−
−
−
B
C
c2
b2
c2
c2
a2
c2
a2
a2
b2
a2
b2
Rearrangement with the substitutions SA =
b2 +c2 −a2
2
etc leads to (4) above.
Example 2.
(i)
M
G
(ii)
O
(iii) I
(iv)
X(55)
(v)
X(56)
(vi)
X(110)
locus of M for which (M ) ⊥ (M )
2 2
2
4
4
cyclic (a (b + c ) − (b + c ))x = 0
2a4 −a2 (b2 +c2 )−(b2 −c2 )2
x=0
cyclic
a2
a2 (b+c)−2abc−(b+c)(b−c)2
x=0
cyclic
a
2a3 −a2 (b+c)−(b+c)(b−c)2
x=0
cyclic
a2
2a4 −a3 (b+c)−a2 (b−c)2 +a(b+c)(b−c)2 −(b2 −c2 )2
cyclic
a2
2a6 −2a4 (b2 +c2 )+a2 (b4 +c4 )−(b2 +c2 )(b2 −c2 )2
x
cyclic
a2
inf. point
X(523)
X(8675)
X(9001)
X(9000)
x=0
X(8999)
=0
X(526)
Locus of centroids of similar inscribed triangles
261
Proposition 6. Let Li : pi x + qi y + ri z = 0, i = 1, 2, be two lines through the
symmedian point K, and J1 , J2 the infinite points of (M1 , (M2 ) for Mi on Li
respectively. Then points J1 and J2 correspond to perpendicular lines if and only
if
(q1 + r1 )p2 + (r1 + p1 )q2 + (p1 + q1 )r2 = 0.
Remark. In other words, the point Q = (q1 + r1 : r1 + p1 : p1 + q1 ) lies on the
line L2 .
Construction 7. Given a line L1 containing the symmedian point K, construct
(i) Q = the complement of the isotomic conjugate of the trilinear pole of L1 ,
(ii) the line L2 = KQ.
For arbitrary points M on L1 and M on L2 , (M ) ⊥ (M ).
L1
A
M
L2
(M )
P
M'
K
G
Q
(M )
B
C
Figure 3
Proposition 8. The locus of M for which (M ) contains a given point P (u, v, w)
is the circle Γ(P )
(a2 + b2 + c2 )(u + v + w)(a2 yz + b2 zx + c2 xy)
⎞
⎛
b2 c2 (2v + 2w − u)x⎠ = 0.
− (x + y + z) ⎝
cyclic
with center
O(P ) = (a2 ((a4 − 2a2 (b2 + c2 ) + b4 − 8b2 c2 + c4 )u
+ (a4 − a2 (2b2 − c2 ) + (b2 − c2 )(b2 + 2c2 ))v
+ (a4 + a2 (b2 − 2c2 ) − (b2 − c2 )(2b2 + c2 )w),
· · · , · · · ).
and passing through the symmedian point K.
For M = G, the centroid, this is the circle
2
(a + b2 + c2 )(a2 yz + b2 zx + c2 xy) − (x + y + z)(b2 c2 x + c2 a2 + a2 b2 z) = 0
with center X(182), the midpoint of OK.
262
F. J. Garcı́a Capitán
Proposition 9. The tangent of Γ(P ) at K is parallel to (P ).
Proof. The tangent of Γ(P ) at K is the line1
(b2 + c2 )u − a2 v − a2 w
−b2 u + (c2 + a2 )v − b2 w
−c2 u − c2 v + (a2 + b2 )w
x+
y+
z = 0.
a2
b2
c2
This has infinite point
2
−b u + (c2 + a2 )v − b2 w −c2 u − c2 v + (a2 + b2 )w
−
,
b2
c2
−c2 u − c2 v + (a2 + b2 )w (b2 + c2 )u − a2 v − a2 w
−
,
c2
a2
(b2 + c2 )u − a2 v − a2 w −b2 u + (c2 + a2 )v − b2 w
−
a2
b2
2
(c + a2 + b2 )v (a2 + b2 + c2 )w (a2 + b2 + c2 )w (b2 + c2 + a2 )u
−
,
−
,
=
b2
c2
c2
a2
(b2 + c2 + a2 )u (c2 + a2 + b2 )v
−
a2
b2
v
w
w
u u
v
= (a2 + b2 + c2 ) 2 − 2 , 2 − 2 , 2 − 2
b
c c
a a
b
equal to J(P ). Therefore the tangent is parallel to (P ).
Here is a construction of the center O(P ) of the circle Γ(P ).
A
K
B
T
(P )
C
S
P
O(P)
U
(Q)
Q
Figure 4
1Given the homogeneous (quadratic) equation of a conic, the tangent at a point (u, v, w) can be
obtained by replacing x2 , y 2 , z 2 by ux, vy, wz, and yz, zx, xy by
1
(vx + uy) respectively.
2
1
(wy
2
+ vz),
1
(uz
2
+ wx),
Locus of centroids of similar inscribed triangles
263
Construction 10. Given a point P = K, construct
(1) the line (P ) to intersect KP at S;
(2) the orthogonal projections
T of K on (P ), and
U of P on KT ;
(3) the parallel of P T through U to intersect the line KP at Q, (the line (Q)
passes through P );
(4) the perpendicular bisector of KQ to intersect KT at O(P ).
O(P ) is the center of Γ(P ).
4. Envelopes
Proposition 11. If M traverses a line L , the lines (M ) envelope a parabola
whose axis is parallel to the trilinear polar of the isogonal conjugate of the infinite
point of L .
Focus
F = a4 p2 + b2 (c2 + a2 − b2 )q 2 + c2 (a2 + b2 − c2 )r2
− (c2 + a2 − b2 (a2 + b2 − c2 )qr − c2 a2 rp − a2 b2 pq,
··· ,···).
Directrix
(a2 ((b2 + c2 − a2 )2 + 8b2 c2 )p
cyclic
+ b2 (a4 + a2 (−2b2 + c2 ) + (b2 − c2 )(b2 + 2c2 ))q
+ c2 (a4 + a2 (b2 − 2c2 ) − (b2 − c2 )(2b2 + c2 ))r)x
= 0.
For example, if L is the Lemoine axis
barycentric equation
x
a2
+
⎛
(b2 + c2 − 2a2 )2 yz − (x + y + z) ⎝
cyclic
y
b2
+
z
c2
= 0, the parabola has
⎞
(a4 − 4b2 c2 )x⎠ = 0.
cyclic
2
It has focus the Parry point
a2
b2
c2
,
X(111) =
,
,
b2 + c2 − 2a2 c2 + a2 − 2b2 a2 + b2 − 2c2
and directrix the line
(a4 − a2 (b2 + c2 ) + 4b2 c2 )x = 0
cyclic
2The Parry point is the isogonal conjugate of the infinite point of the line GK.
264
F. J. Garcı́a Capitán
which is the perpendicular to the line GK from the intersection of the Euler line
and the Simson line of the Steiner point. 3
A
B
(M )
C
M
L
Figure 5
Proposition 12. If M is a point on the circumcircle, the line (M ) is tangent to
the Steiner inellipse.
2
2
2
Proof. If M = (b2 −c2a)(a2 +τ ) , (c2 −a2b)(b2 +τ ) , (a2 −b2c)(c2 +τ ) on the circumcircle,
the centroid of the (degenerate) pedal triangle of M is the point
G0 = (c2 a2 + a2 b2 − 2b2 c2 − (b2 + c2 − 2a2 )τ )
· (a2 (a2 (b2 + c2 ) − (b4 + c4 )) + (2a4 − a2 (b2 + c2 ) − (b2 − c2 )2 )τ ),
· · · , · · · ).
The trilinear polar of M ∗ is the line
x
y
z
+
+
=0
(b2 − c2 )(a2 + τ ) (c2 − a2 )(b2 + τ ) (a2 − b2 )(c2 + τ )
with infinite point
J(τ ) = ((b2 − c2 )(a2 + τ )(a2 (b2 + c2 ) − 2b2 c2 − (b2 + c2 − 2a2 )τ ),
(c2 − a2 )(b2 + τ )(b2 (c2 + a2 ) − 2c2 a2 − (c2 + a2 − 2b2 )τ ),
(a2 − b2 )(c2 + τ )(c2 (a2 + b2 ) − 2a2 b2 − (a2 + b2 − 2c2 )τ )).
The line (M ) contains G0 and J(τ ). It has barycentric equation
x
= 0.
2
2
2
2
2
a (b + c ) − 2b c − (b2 + c2 − 2a2 )τ
cyclic
This is the tangent to the Steiner inellipse
x2 + y 2 + z 2 − 2yz − 2zx − 2xy = 0
3This intersection is the triangle center X(1513) = ((a2 (b2 + c2 ) − (b4 + c4 ))(3a4 + (b2 −
c2 )2 ), · · · , · · · ).
Locus of centroids of similar inscribed triangles
265
at the point4
T (τ ) = ((a2 (b2 + c2 ) − 2b2 c2 − (b2 + c2 − 2a2 )τ )2 , · · · , · · · ).
A
M
(M )
G
O
C
B
Figure 6
Corollary 13. The Steiner inellipse is the envelope of (M ) for M on the circumcircle of triangle ABC.
Remark. If St is the Steiner point, the fourth intersection of the cirucumcircle
and the Steiner circum-ellipse, and the line M (τ )St intersects the Steiner circumellipse at T (τ ), then T (τ ) is the midpoint of G and T (τ ).
5. The inverse problem
We solve the inverse problem of finding the point M (u, v, w) so that (M ) is a
given line L : px + qy + rz = 0 not containing the symmedian point K. This
has to satisfy two conditions:
(i) J(M ) is the infinite point of L , and
(ii) the centroid of the pedal triangle of M lies on L .
r−p
p−q
q−r
u + 2 v + 2 w = 0,
a2
b
c
2
2
2
2
4a p + (a + b − c )q + (c2 + a2 − b2 )r
u
a2
(a2 + b2 − c2 )p + 4b2 q + (b2 + c2 − a2 )r
+
v
b2
(c2 + a2 − b2 )p + (b2 + c2 − a2 )q + 4c2 r
w = 0.
+
c2
4If (u, v, w) is an infinite point, then (u2 , v 2 , w 2 ) is a point on the Steiner inellipse, and the
z
= 0.
tangent at that point is ux + yv + w
266
F. J. Garcı́a Capitán
Solving these equations, we obtain
u:v:w
= a2 (−a2 p2 + 2b2 q 2 + 2c2 r2 + (b2 + c2 − a2 )qr − (b2 + 2c2 − a2 )rp − (2b2 + c2 − a2 )pq)
: b2 (2a2 p2 − b2 q 2 + 2c2 r2 − (2c2 + a2 − b2 )qr + (c2 + a2 − b2 )rp − (c2 + 2a2 − b2 )pq)
: c2 (2a2 p2 + 2b2 q 2 − c2 r2 − (a2 + 2b2 − c2 )qr − (2a2 + b2 − c2 )rp + (a2 + b2 − c2 )pq).
Example 3.
L
M
(i) orthic axis
X(187)
(ii) Lemoine axis X(352)
Remarks. (1) X(187) is the midpoint of the isodynamic points .
(2) X(352) is a point on the circle through the centroid and the isodynamic
points.
We conclude with a construction of M from (M ).
N
A
N'
C'' K
B
B''
M'
A'
C
M
L = (M )
Q
B'
C'
(M )
Figure 7
Construction 14. Given a line L , construct
(i) the isogonal conjugate Q of the infinite point of L ,
(ii) the cevian triangle A B C of Q and the points B = C A ∩ CA and C =
A B ∩ AB, (the line B C passes through the symmedian point K),
(iii) the line (M ) for any point M on the line B C , (this line is parallel to L ),
(iv) any line through K intersecting L and (M ) at N and N respectively,
(v) the parallel through N to M N to intersect B C at M .
The point M has (M ) equal to the given line L .
Locus of centroids of similar inscribed triangles
267
References
[1] C. Kimberling, Encyclopedia of Triangle Centers, available at
http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.
[2] P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes,
2001; with corrections, 2013, available at
http://math.fau.edu/Yiu/Geometry.html
Francisco Javier Garcı́a Capitán: Departamento de Matemáticas, I.E.S. Alvarez Cubero, Avda.
Presidente Alcalá-Zamora, s/n, 14800 Priego de Córdoba, Córdoba, Spain
E-mail address: [email protected]