b
Forum Geometricorum
Volume 3 (2003) 125–134. b b
FORUM GEOM
ISSN 1534-1178
Isotomic Inscribed Triangles and Their Residuals
Mario Dalcı́n
Abstract. We prove some interesting results on inscribed triangles which are
isotomic. For examples, we show that the triangles formed by the centroids
(respectively orthocenters) of their residuals have equal areas, and those formed
by the circumcenters are congruent.
1. Isotomic inscribed triangles
The starting point of this investigation was the interesting observation that if
we consider the points of tangency of the sides of a triangle with its incircle and
excircles, we have two triangles of equal areas.
Cb
Bc
A
Z
Y
Z Y
B
Ac X X C Ab
Ba
Ca
Figure 1
In Figure 1, X, Y , Z are the points of tangency of the incircle with the sides BC,
CA, AB of triangle ABC, and X , Y , Z those with the corresponding excircles.
In [2], XY Z and X Y Z are called the intouch and extouch triangles of ABC
respectively. That these two triangles have equal areas is best explained by the fact
that each pair of points X, X ; Y , Y ; Z, Z are isotomic on their respective sides,
i.e.,
BX = X C, CY = Y A, AZ = Z B. (1)
Publication Date: June 16, 2003. Communicating Editor: Paul Yiu.
126 M. Dalcı́n
We shall say that XY Z and X Y Z are isotomic inscribed triangles. The fol-
lowing basic proposition follows from simple calculations with barycentric coor-
dinates.
Proposition 1. Isotomic inscribed triangles have equal areas.
Proof. Let X, Y , Z be points on the sidelines BC, CA, AB dividing the sides in
the ratios
BX : XC = x : 1 − x, CY : Y A = y : 1 − y, AZ : ZB = z : 1 − z.
In terms of barycentric coordinates with respect to ABC, we have
X = (1 − x)B + xC, Y = (1 − y)C + yA, Z = (1 − z)A + zB. (2)
The area of triangle XY Z, in terms of the area of ABC, is
0 1−x x
XY Z = y 0 1 − y
1 − z z 0
=(1 − (x + y + z) + (xy + yz + zx))
=(xyz + (1 − x)(1 − y)(1 − z)). (3)
See, for example, [4, Proposition 1]. If X , Y , Z are points satisfying (1), then
BX : X C = 1−x : x, CY : Y A = 1−y : y, AZ : Z B = 1−z : z, (4)
and
X = xB + (1 − x)C, Y = yC + (1 − y)A, Z = zA + (1 − z)B. (5)
The area of triangle X Y Z
can be obtained from (3) by replacing x, y, z by 1 − x,
1 − y, 1 − z respectively. It is clear that this results in the same expression. This
completes the proof of the proposition.
Proposition 2. The centroids of isotomic inscribed triangles are symmetric with
respect to the centroid of the reference triangle.
Proof. The expressions in (2) allow one to determine the centroid of triangle XY Z
easily. This is the point
1 (1 + y − z)A + (1 + z − x)B + (1 + x − y)C
GXY Z = (X +Y +Z) = . (6)
3 3
On the other hand, with the coordinates given in (5), the centroid of triangle X Y Z
is
1 (1 − y + z)A + (1 − z + x)B + (1 − x + y)C
GX Y Z = (X + Y + Z ) = .
3 3
(7)
It follows easily that
1 1
(GXY Z + GX Y Z ) = (A + B + C) = G,
2 3
the centroid of triangle ABC.
Isotomic inscribed triangles and their residuals 127
Corollary 3. The intouch and extouch triangles have equal areas, and the midpoint
of their centroids is the centroid of triangle ABC.
Proof. These follow from the fact that the intouch triangle XY Z and the extouch
triangle X Y Z are isotomic, as is clear from the following data, where a, b, c
denote the lengths of the sides BC, CA, AB of triangle ABC, and s = 12 (a+b+c).
BX = X C = s − b, BX = XC = s − c,
CY = Y A = s − c, CY = Y A = s − a,
AZ = Z B = s − a, AZ = ZB = s − b.
In fact, we may take
s−b s−c s−a
x= , y= , z= ,
a b c
and use (3) to obtain
2(s − a)(s − b)(s − c)
XY Z = X Y Z = .
abc
Let R and r denote respectively the circumradius and inradius of triangle ABC.
Since = rs and
abc (s − a)(s − b)(s − c)
R= , r2 = ,
4 s
we have
r
XY Z = X Y Z = · .
2R
If we denote by Ab and Ac the points of tangency of the line BC with the B-
and C-excircles, it is easy to see that Ab and Ac are isotomic points on BC. In
fact,
BAb = Ac C = s, BAc = Ab C = −(s − a).
Similarly, the other points of tangency Bc , Ba , Ca , Cb form pairs of isotomic points
on the lines CA and AB respectively. See Figure 1.
Corollary 4. The triangles Ab Bc Ca and Ac Ba Cb have equal areas. The centroids
of these triangles are symmetric with respect to the centroid G of triangle ABC.
These follow because Ab Bc Ca and Ac Bb Ca are isotomic inscribed triangles.
Indeed,
s−a s−a
BAb : Ab C =s : −(s − a) = 1 + :− = CAc : Ac B,
a a
s−b s−b
CBc : Bc A =s : −(s − b) = 1 + :− = ABa : Ba C,
b b
s−c s−c
ACa : Ca B =s : −(s − c) = 1 + :− = BCb : Cb A.
c c
Furthermore, the centroids of the four triangles XY Z, X Y Z , Ab Bc Ca and
Ac Ba Cb form a parallelogram. See Figure 2.
128 M. Dalcı́n
Cb
Bc
Z Y
Z Y
B
Ac X X C Ab
Ba
Ca
Figure 2
2. Triangles of residual centroids
For an inscribed triangle XY Z, we call the triangles AY Z, BZX, CXY its
residuals. From (2, 5), we easily determine the centroids of these triangles.
1
GAY Z = ((2 + y − z)A + zB + (1 − y)C),
3
1
GBZX = ((1 − z)A + (2 + z − x)B + xC),
3
1
GCXY = (yA + (1 − x)B + (2 + x − y)C).
3
We call these the residual centroids of the inscribed triangle XY Z.
The following two propositions are very easily to established, by making the
interchanges (x, y, z) ↔ (1 − x, 1 − y, 1 − z).
Proposition 5. The triangles of residual centroids of isotomic inscribed triangles
have equal areas.
Proof. From the coordinates given above, we obtain the area of the triangle of
residual centroids as
2 + y − z z 1 − y
1
1−z 2+z−x x
27 y 1−x
2 + x − y
1
= (3 − x − y − z + xy + yz + zx)
9
1
= (2 + xyz + (1 − x)(1 − y)(1 − z))
9
Isotomic inscribed triangles and their residuals 129
By effecting the interchanges (x, y, z) ↔ (1 − x, 1 − y, 1 − z), we obtain the area
of the triangle of residual centroids of the isotomic inscribed triangle X Y Z . This
clearly remains unchanged.
Proposition 6. Let XY Z and X Y Z be isotomic inscribed triangles of ABC.
The centroids of the following five triangles are collinear:
• G of triangle ABC,
• GXY Z and GX Y Z of the inscribed triangles,
• G and G of the triangles of their residual centroids.
Furthermore,
GXY Z G : GG
: GG GX Y Z = 1 : 2 : 2 : 1.
: G
Z Y
Z G Y
GXY Z GX Y Z
G
G
B X X C
Figure 3
Proof. The centroid G is the point
= 1 ((3 + 2y − 2z)A + (3 + 2z − 2x)B + (3 + 2x − 2y)C).
G
9
We obtain the centroid G by interchanging (x, y, z) ↔ (1 − x, 1 − y, 1 − z). From
these coordinates and those given in (6,7), the collinearity is clear, and it is easy to
figure out the ratios of division.
3. Triangles of residual orthocenters
Proposition 7. The triangles of residual orthocenters of isotomic inscribed trian-
gles have equal areas.
See Figure 4. This is an immediate corollary of the following proposition (see
Figure 5), which in turn is a special case of a more general situation considered in
Proposition 8 below.
130 M. Dalcı́n
Proposition 8. An inscribed triangle and its triangle of residual orthocenters have
equal areas.
A A
Z
Ha Ha
Ha Y Y
Y
Z Hc Z
Hc
Hb Hc Hb
Hb
B X X C B X C
Figure 4 Figure 5
Proposition 9. Given a triangle ABC, if pairs of parallel lines L1B , L1C through
B, C, L2C , L2A through C, A, and L3A , L3B through A, B are constructed, and
if
Pa = L2C ∩ L3B , Pb = L3A ∩ L1C , Pc = L1B ∩ L2A ,
then the triangle Pa Pb Pc has the same area as triangle ABC.
Proof. We write Y = L2C ∩ L3A and Z = L2A ∩ L3B . Consider the parallelogram
AZPa Y in Figure 6. If the points B and C divide the segments ZPa and Y Pa in
the ratios
ZB : BPa = v : 1 − v, Y C : CPa = w : 1 − w,
then it is easy to see that
1 + vw
Area(ABC) = · Area(AZPa Y ). (8)
2
A
Pb
Pc
C
B
Pa
Figure 6
Now, Pb and Pc are points on AY and AZ such that BPc and CPb are parallel.
If
Y Pb : Pb A = v : 1 − v , ZPc : Pc A = w : 1 − w ,
Isotomic inscribed triangles and their residuals 131
then from the similarlity of triangles BZPc and Pb Y C, we have
ZB : ZPc = Y Pb : Y C.
This means that v : w = v : w and v w = vw. Now, in the same parallelogram
AZPa Y , we have
1 + v w
· Area(AZPa Y ).
Area(Pa Pb Pc ) =
2
From this we conclude that Pa Pb Pc and ABC have equal areas.
4. Triangles of residual circumcenters
Consider the circumcircles of the residuals of an inscribed triangle XY Z. By
Miquel’s theorem, the circles AY Z, BZX, and CXY have a common point. Fur-
thermore, the centers Oa , Ob , Oc of these circles form a triangle similar to ABC.
See, for example, [1, p.134]. We prove the following interesting theorem.
Theorem 10. The triangles of residual circumcenters of the isotomic inscribed
triangles are congruent.
A A
Oa
Oa
Z Z
Y Y
Z Y Z Y
Oc
Ob Ob
Oc
B C B X C
X X X
Figure 7A Figure 7B
We prove this theorem by calculations.
Lemma 11. Let X, Y , Z be points on BC, CA, AB such that
BX : XC = w : v, CY : Y A = uc : w, AZ : ZB = v : ub .
The distance between the circumcenters Ob and Oc is the hypotenuse of a right
triangle with one side a2 and another side
(v − w)(ub + v)(uc + w)a2 + (v + w)(w − uc )(ub + v)b2 + (v + w)(w + uc )(ub − v)c2
· a.
8(ub + v)(v + w)(w + uc )
(9)
132 M. Dalcı́n
Proof. The distance between Ob and Oc along the side BC is clearly a2 . We calcu-
late their distance along the altitude on BC. The circumradius of BZX is clearly
Rb = 2 ZXsin B . The distance of Ob above BC is
ZX cos BZX 2BZ · ZX cos BZX BZ 2 + ZX 2 − BX 2
Rb cos BZX = = =
2 sin B 4BZ sin B 4BZ sin B
BZ 2 + BZ 2 + BX 2 − 2BZ · BX cos B − BX 2
=
4BZ sin B
BZ − BX cos B c(BZ − BX cos B)
= = ·a
2 sin B 4
c ubu+v
b w
c − v+w a cos B
= ·a
4
ub (v + w)2c2 − w(ub + v)(c2 + a2 − b2 )
= ·a
8(ub + v)(v + w)
−(ub + v)w(a2 − b2 ) + (2ub v + ub w − vw)c2
= ·a
8(ub + v)(v + w)
By making the interchanges b ↔ c, v ↔ w, and ub ↔ uc , we obtain the distance
of Oc above the same line as
−(uc + w)v(a2 − c2 ) + (2uc w + uc v − vw)b2
· a.
8(uc + w)(v + w)
The difference between these two is the expression given in (9) above.
Consider now the isotomic inscribed triangle X Y Z . We have
BX : X C =v : w,
vw
CY : Y A =w : uc = : v,
uc
vw
AZ : Z B =ub : v = w : .
ub
Let Ob and Oc be the circumcenters of BZ X and CX Y . By making the
following interchanges
vw vw
v ↔ w, ub ↔ , uc ↔
ub uc
in (9), we obtain the distance between Ob and Oc along the altitude on BC as
(w − v)( vw
u
+ w)( vw
uc
+ v)a2 + (v + w)(v − vw
uc
)( vw
ub
+ w)b2 + (v + w)(v + vw
uc
)( vw
ub
− w)c2
b
·a
8( vw
ub
+ w)(v + w)(v + vw
uc
)
(w − v)(v + ub )(w + uc )a2 + (v + w)(uc − w)(v + ub )b2 + (v + w)(w + uc )(v − ub )c2
= · a.
8(v + ub )(v + w)(uc + w)
Except for a reversal in sign, this is the same as (9).
Isotomic inscribed triangles and their residuals 133
From this we easily conclude that the segments Ob Oc and Ob Oc are congruent.
The same reasoning also yields the congruences of Oc Oa , Oc Oa , and of Oa Ob ,
Oa Ob . It follows that the triangles Oa Ob Oc and Oa Ob Oc are congruent. This
completes the proof of Theorem 9.
5. Isotomic conjugates
Let XY Z be the cevian triangle of a point P , i.e., X, Y , Z are respectively
the intersections of the line pairs AP , BC; BP , CA; CP , AB. By the residual
centroids ( (respectively orthocenters, circumcenters) of P , we mean those of its
cevian triangle. If we construct points X , Y , Z satisfying (1), then the lines AX ,
BY , CZ intersect at a point P called the isotomic conjugate of P . If the point
P has homogeneous barycentric
coordinates
(x : y : z), then P has homogeneous
barycentric coordinates x1 : y1 : 1z . All results in the preceding sections apply to
the case when XY Z and X Y Z are the cevian triangles of two isotomic conju-
gates. In particular, in the case of residual circumcenters in §4 above, if XY Z is
the cevian triangle of P with homogeneous barycentric coordinates (u : v : w),
then
BX : XC = w : v, CY : Y A = u : w, AZ : ZB = v : u.
By putting ub = uc = u in (9) we obtain a necessary and sufficient condition for
the line Ob Oc to be parallel to BC, namely,
(v−w)(u+v)(u+w)a2 +(v+w)(w−u)(u+v)b2 +(v+w)(w+u)(u−v)c2 = 0.
This can be reorganized into the form
(b2 +c2 −a2 )u(v 2 −w2 )+(c2 +a2 −b2 )v(w2 −u2 )+(a2 +b2 −c2 )w(u2 −v 2 ) = 0.
This is the equation of the Lucas cubic, consisting of points P for which the line
joining P to its isotomic conjugate P passes through the orthocenter H. The
symmetry of this equation leads to the following interesting theorem.
Theorem 12. The triangle of residual circumcenters of P is homothetic to ABC
if and only if P lies on the Lucas cubic.
It is well known that the Lucas cubic is the locus of point P whose cevian
triangle is also the pedal triangle of a point Q. In this case, the circumcircles of
AY Z, BZX and CXY intersect at Q, and the circumcenters Oa , Ob , Oc are the
midpoints of the segments AQ, BQ, CQ. The triangle Oa Ob Oc is homothetic to
ABC at Q.
For example, if P is the Gergonne point, then Oa Ob Oc is homothetic to ABC
at the incenter I. The isotomic conjugate of P is the Nagel point, and Oa Ob Oc is
homothetic to ABC at the reflection of I in the circumcenter O.
References
[1] R. A. Johnson, Advanced Euclidean Geometry, 1925, Dover reprint.
[2] C. Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998) 1 –
285.
134 M. Dalcı́n
[3] C. Kimberling, Encyclopedia of Triangle Centers, May 23 edition, available at
http://www2.evansville.edu/ck6/encyclopedia/.
[4] P. Yiu, The uses of homogeneous barycentric coordinates in plane euclidean geometry, Int. J.
Math. Educ. Sci. Technol., 31 (2000) 569 – 578.
Mario Dalcı́n: Caribes 2364, C.P.11.600, Montevideo, Uruguay
E-mail address: filomate@adinet.com.uy
