Forum Geometricorum

Volume 19 (2019) 17–28.

FORUM GEOM
ISSN 1534-1178

Adjugate Points and Adjugate Triangle

Sándor Nagydobai Kiss

Abstract. In this paper we introduce two new notions in triangle geometry: the
adjugate points and the adjugate triangle of a point with respect to a given trian-
gle. These notions are used to better characterize the anticomlementary triangles.

1. Preliminaries
Consider a triangle ABC with lengths of sides BC = a, CA = b, and AB = c.
Denote by s its semiperimeter and Δ its area. It is well known that the circumra-
dius, inradius, and exradii are given by

abc Δ Δ Δ Δ
R= , r= , ra = , rb = , rc = .
4Δ s s−a s−b s−c

We work with barycentric coordinates, absolute and homogeneous, of points

with reference to the triangle. Every finite point P is given by its absolute barycen-
tric coordinates (xP , yP , zP ) with xP + yP + zP = 1. It is more convenient to
work with homogeneous barycentric coordinates. Thus, the same point P is also
given by P = (x : y : z), where x : y : z = xP : yP : zP .
To express the coordinates more succinctly, we also make use of the following
notations:

b2 + c2 − a2 c2 + a2 − b2 a2 + b2 − c2
SA = , SB = , SC = .
2 2 2

These satisfy the following relations, where S = 2Δ, and for convenience, we
write SBC for SB SC etc.

Publication Date: January 16, 2019. Communicating Editor: Paul Yiu.

The author thanks Editor Paul Yiu for his help in the preparation of this paper.
18 S. N. Kiss

SA = S cot A, SB = S cot B, SC = S cot C;

SBC + SCA + SAB = S 2 ,
a2 SA + b2 SB + c2 SC = 2S 2 ,
a2 SA + SBC = b2 SB + SCA = c2 SC + SAB = S 2 ;
bc + ca + ab = s2 + r2 + 4Rr,
a2 + b2 + c2 = 2(s2 − r2 − 4Rr),
a3 + b3 + c3 = 2s(s2 − 3r2 − 6Rr).

2. Adjugate points and adjugate triangles

Suppose M is a center of triangle ABC, and let f (a, b, c) be a center function
for M (given for example by the barycentric coordinates of M ):
M = (f (a, b, c) : f (b, c, a) : f (c, a, b)).
Definition. The points
Ma = (f (−a, b, c) : f (b, c, −a) : f (c, −a, b)),
Mb = (f (a, −b, c) : f (−b, c, a) : f (c, a, −b)),
Mc = (f (a, b, −c) : f (b, −c, a) : f (−c, a, b))
are called the adjugate points, and ΔMa Mb Mc the adjugate triangle of M with
respect to the reference triangle ABC.
Example 1. The adjugate points of the incenter I = (a : b : c) are the centers of
excircles, i.e. the points
Ia = (−a : b : c), Ib = (a : −b : c), Ic = (a : b : −c).
The adjugate triangle of the incenter is the excentral triangle Ia Ib Ic of ABC.

Let X, Y , Z be the tangency points of incircle with the sides of triangle ABC:
X = (0 : s − c : s − b), Y = (s − c : 0 : s − a), Z = (s − b : s − a : 0).
Similarly we denote the points of tangency of the A-excircle, B-excircle, C-excircle
respectively, with the sides of triangle ABC:
Xa = (0 : s − b : s − c), Ya = (−(s − b) : 0 : s), Za = (−(s − c) : s : 0);
Xb = (0 : −(s − a) : s), Yb = (s − a : 0 : s − c), Zb = (s : −(s − c) : 0);
Xc = (0 : s : −(s − a)), Yc = (s : 0 : −(s − b)), Zc = (s − a : s − b : 0).
Theorem 1. Each of the triplets of lines (AX, BY, CZ), (AXa , BYa , CZa ),
(AXb , BYb , CZb ), (AXc , BYc , CZc ) are concurrent. Equivalently, ABC is
perspective with each of the triangles XY Z, Xa Ya Za , Xb Yb Zb , and Xc Yc Zc . See
Figure 1.
Adjugate points and adjugate triangle 19

Zb

Ib

Yc

Ic Gec
Zc
Y
Geb
Z Na Yb
I
Ge

X Xa

Xc B C Xb

Gea Ya

Za

Ia

Figure 1. The Gergonne point and its adjugate triangle

Proof. We begin with the equations of the lines:

AX : −(s − b)y + (s − c)z = 0, AXa : (s − c)y − (s − b)z = 0,
BY : (s − a)x − (s − c)z = 0, BYa : (s − b)x + sz = 0,
CZ : −(s − a)x + (s − b)y = 0; CZa : sx + (s − c)y = 0;
AXb : sy + (s − a)z = 0, AXc : (s − a)y + sz = 0,
BYb : (s − c)x − (s − a)z = 0, BYc : (s − c)z + sz = 0,
CZb : (s − c)x + sy = 0; CZc : −(s − b)x + (s − a)y = 0.

It is well known that AX, BY , CZ are concurrent at the Gergonne point

 
1 1 1
Ge = : : .
s−a s−b s−c
20 S. N. Kiss

With the above equations, it is easy to verify that the concurrency of each of the
triples:

Triple of lines Point of

 concurrency 
AXa ∩ BYa ∩ CZa Gea =  − 1s : s−c
1 1
: s−b 
1
AXb ∩ BYb ∩ CZb Geb =  s−c : − 1s : s−a
1

1 1
AXc ∩ BYc ∩ CZc Gec = s−b : s−a : − 1s
This completes the proof of the theorem. The points Gea , Geb , Gec are the
adjugate points of the Gergonne point Ge. 
Theorem 2. Each of the triplets of lines (AXa , BYb , CZc ), (AX, BYc , CZb ),
(AXc , BY, CZa ), (AXb , BYa , CZ) are concurrent. Equivalently, ABC is per-
spective with each of the triangles Xa Yb Zc , XYc Zb , Xc Y Za , and Xb Ya Z. See
Figure 2.

Proof. Again, we begin with the equations of the lines:

AXa : (s − c)y − (s − b)z = 0, AX : −(s − b)y + (s − c)z = 0,
BYb : (s − c)x − (s − a)z = 0, BYc : (s − b)x + sz = 0,
CZc : −(s − b)x + (s − a)y = 0; CZb : (s − c)x + sy = 0;
AXc : (s − a)y + sz = 0, AXb : sy + (s − a)z = 0,
BY : (s − a)x − (s − c)z = 0, BYa : sx + (s − b)z = 0,
CZa : sx + (s − c)y = 0; CZ : −(s − a)x + (s − b)y = 0.

It is well known that AXa , BYb , CZc are concurrent at the Nagel point
Na = (s − a : s − b : s − c).
The other three points of concurrency are the adjugate points of the Nagel point
Na :

Triple of lines Point of concurrency

AX ∩ BYc ∩ CZb Naa = (−s : s − c : s − b)
AXc ∩ BY ∩ CZa Nab = (s − c : −s : s − a)
AXb ∩ BYa ∩ CZ Nac = (s − b : s − a : −s)
This completes the proof of the theorem. The points Naa , Nab , Nac are the
adjugate points of the Nagel point N a. 
Theorem 3. The triangle ABC and the adjugate triangle Gea Geb Gec of the Ger-
gonne point are perspective, and the perspector is the Nagel point. See Figure
1.
y z z x
Proof. The line AGea , BGeb , CGec have equations s−b − s−c = 0, s−c − s−a = 0,
x y
and s−a − s−b = 0. They are concurrent at (s − a : s − b : s − c), the Nagel
point. 
Adjugate points and adjugate triangle 21

Naa

Zb

Ib

Yc
A

Ic Gec
Zc
YN Geb
Z a Yb
I
Ge
X C

Xc B Xa Xb
Gea
Ya
Nac

Za

Ia

Nab

Figure 2. The Nagel point and its adjugate triangle

Theorem 4. The triangle ABC and the adjugate triangle Naa Nab Nac of the Nagel
point are perspective, and the perspector is the Gergonne point.

Proof. The line ANaa , BNab , CNac have equations (s − b)y − (s − c)z = 0,
(s
 − c)z − (s − a)x
 = 0, and (s − a)x − (s − b)y = 0. They are concurrent at
1 1 1
s−a : s−b : s−c , the Gergonne point. See Figure 2. 

Example 2. The incircle is internally tangent to the nine-point circle. Its point of
tangency is the Feuerbach point F of the triangle ABC:

F = ((b − c)2 (s − a) : (c − a)2 (s − b) : (a − b)2 (s − c)).

22 S. N. Kiss

The nine-point circle is also tangent externally to the excircles at the following
points:
Fa = (−(b − c)2 s : (c + a)2 (s − c) : (a + b)2 (s − b)),
Fb = ((b + c)2 (s − c) : −(c − a)2 s : (a + b)2 (s − a)),
Fc = ((b + c)2 (s − b) : (c + a)2 (s − a) : −(a − b)2 s).
Since the coordinates of F can be rewritten as
((b − c)2 (b + c − a) : (c − a)2 (c + a − b) : (a − b)2 (a + b − c)),
its A-adjugate point is
((b − c)2 (b + c − (−a)) : (c − (−a))2 (c + (−a) − b)
: ((−a) − b)2 ((−a) + b − c))
= ((b − c)2 (a + b + c) : −(c + a)2 (a + b − c) : −(a + b)2 (c + a − b))
= (−(b − c)2 s : (c + a)2 (s − c) : (a + b)2 (s − b)).
This is the point Fa above; similarly for the B- and C-adjugate points. Therefore,
the Feuerbach triangle Fa Fb Fc is the adjugate triangle of the Feuerbach point. See
Figure 3.

Zb

Ib

Yc

Ic F
Fb
Zc Y

I Yb
Fc
Z
N
X

Xc B Xa C Xb
Fa
Ya

Za

Ia

Figure 3. The Feuerbach point and its adjugate triangle

Adjugate points and adjugate triangle 23

3. Anticomplementary triangle
The anticomplementary triangle is the triangle A B  C  which has the given tri-
angle ABC as its medial triangle. Consider the homothety h with center in trian-
gle centroid G and ratio −2, i.e. M  G = 2GM , G ∈ (M M  ). This homothety
h = h(G, −2) transforms the reference triangle ABC into the anticomplementary
triangle A B  C  . Indeed, if (xM , yM , zM ) are the absolute barycentric coordinates
of M then
h(M ) = M  = 3G − 2M = (1 − 2xM , 1 − 2yM , 1 − 2zM ).
Therefore,
h(A) = A = (−1, 1, 1), h(B) = B  = (1, −1, 1), h(C) = C  = (1, 1, −1).
It is easy to verify that the points A, B, C are the midpoints of segments B  C  ,
C  A , A B  respectively.
For h = h(G, −2), we shall call M  = h(M ) the anticomplement of M . In
homogeneous barycentric coordinates, if M = (x : y : z), then
M  = (−x + y + z : x − y + z : x + y − z).
Here are the coordinates of some common triangle centers and their comple-
ments. (The notations follow [2] and [3]).

M M
I = (a : b : c) Na = (s − a : s − b : s − c)
O = (a2 SA : b2 SB : c2 SC ) H = (SBC : SCA : SAB )
H = (SBC : SCA : SAB ) X(20) = (S 2 − 2SBC : S 2 − 2SCA
: S 2 − 2SAB )
N = (S 2 + SBC : S 2 + SCA O = (a2 SA : b2 SB : c2 SC )
: S 2 + SAB )
K = (a : b : c2 )
2 2 X(69)= (SA : SB : SC)
1 1 1
X(9) = (a(s − a) : b(s − b) : c(s − c)) Ge = s−a : s−b : s−c
Sp = (b + c : c + a : a + b) I = (a : b : c)
Theorem 5. The Euler line of the anticomplementary triangle A B  C  coincide
with the Euler line of the reference triangle ABC .
Proof. The Euler line is determined by the circumcenter, the orthocenter and the
centroid of a triangle. Since O = H, the Euler lines of A B  C  and ABC coincide.

Lemma 6 (see [1]). The anticomplement of the Feuerbach point F is the triangle
center  
a b c
X(100) = : : .
b−c c−a a−b
Proof. In homogeneous barycentric coordinates,
F = ((b − c)2 (s − a) : (c − a)2 (s − b) : (a − b)2 (s − c)).
24 S. N. Kiss

The A-coordinate of its anticomplement is

− (b − c)2 (s − a) + (c − a)2 (s − b) + (a − b)2 (s − c)

= (−(b − c)2 + (c − a)2 + (a − b)2 )s − (−a(b − c)2 + b(c − a)2 + c(a − b)2 )
= (2a2 − 2ab − 2ac + 2bc)s − (a2 (b + c) + a(−(b − c)2 − 2bc − 2bc) + bc2 + b2 c)
= 2(a − b)(a − c)s − a2 (b + c) + a(b + c)2 − bc(b + c)
= 2(a − b)(a − c)s − (b + c)(a2 − a(b + c) + bc)
= 2(a − b)(a − c)s − (b + c)(a − b)(a − c)
= (2s − (b + c))(a − b)(a − c)
= a(a − b)(a − c).

Similarly, the B- and C-coordinates are b(b − c)(b − a) and c(c − a)(c − b).
Therefore,
 
 a b c
F = (a(a−b)(a−c) : b(b−c)(b−a) : c(c−a)(c−b) = : : .
b−c c−a a−b

Lemma 7. The anticomplement of Fa is the point

Fa = (−a(a + b)(a + c) : b(b − c)(b + a) : c(c − b)(c + a)).

Proof. From

Fa = (−(b − c)2 s : (c + a)2 (s − c) : (a + b)2 (s − b)),

the A-coordinate of Fa is

(b − c)2 s + (c + a)2 (s − c) + (a + b)2 (s − b)

= ((b − c)2 + (c + a)2 + (a + b)2 )s − (c(c + a)2 + b(a + b)2 )
= 2(a2 + b2 + c2 − bc + ca + ab)s − (a2 (b + c) + 2a(b2 + c2 ) + b3 + c3 )
= (a + b + c)(a2 + a(b + c) + (b2 − bc + c2 )) − (a2 (b + c) + 2a(b2 + c2 ) + (b3 + c3 ))
= a3 + a2 (b + c) + a(b2 − bc + c2 ) + a2 (b + c) + a(b + c)2 + (b3 + c3 )
− a2 (b + c) − 2a(b2 + c2 ) − (b3 + c3 )
= a3 + a2 (b + c) + abc
= a(a + b)(a + c).
Adjugate points and adjugate triangle 25

The B-coordinate of the anticomplement of Fa is

− (b − c)2 s − (c + a)2 (s − c) + (a + b)2 (s − b)
= −(b − c)2 s − (c + a)2 (s − c) − (a + b)2 (s − b) + 2(a + b)2 (s − b)
= −a(a + b)(a + c) + 2(a + b)2 (s − b)
= (a + b)[−a(a + c) + 2(s − b)(a + b)]
= (a + b)[−a(a + c) + (c + a − b)(a + b)]
= (a + b)[−a2 − ac + a2 − b2 + ac + bc]
= −b(a + b)(b − c).
Similarly, the C-coordinate is −c(a + c)(c − b). Therefore,
Fa = (−a(a + b)(a + c) : b(b − c)(b + a) : c(c − b)(c + a)).

Similarly the anticomplements of Fb and Fc are
Fb = (a(a − c)(a + b) : −b(b + c)(b + a) : c(c − a)(c + b)),
Fc = (a(a − b)(a + c) : b(b − a)(b + c) : −c(c + a)(c + b)).
Lemma 8. The points F  , Fa , Fb , Fc are on the circumcircle O(R).
Proof. These points are the anticomplements of F , Fa , Fb , Fc , which are on the
nine-point circle. Since the anticomplement of the nine-point circle is the circum-
cle, the result follows. 
Theorem 9. (a) The anticomplements of the excenters Ia , Ib , Ic are the adjugate
points Naa , Nab , Nac of the Nagel point.
(b) The triplets of points (F, Na , O), (Fa , Naa , O), (Fb , Nab , O), (Fc , Nac , O)
are collinear, and F N  OF  , Fa N  OFa , Fb N  OFb , Fc N  OFc .
(c) The circle Na (2r) is the incircle; the circles Naa (2ra ), Nab (2rb ), Nac (2rc )
are the excircles of the anticomplementary triangle A B  C  .
(d) The incircle Na (2r), and respectively the excircles Naa (2ra ), Nab (2rb ),
Nac (2rc ) and the nine-point circle O(R) of the anticomplementary triangle A B  C 
are tangent at the anticomplements of Feuerbach point F  ≡ X(100) and Fa , Fb ,
Fc .
Proof. (a) Since Ia = (−a : b : c), we have:
Ia = (−(−a) + b + c : −a − b + c : −a + b − c)
= (a + b + c : −(a + b − c) : −(c + a − b))
= (−s : s − c : s − b)
= Naa .
Similarly, Ib = Nab and Ic = Nac .
(b) The Feuerbach point F , the incenter I, and the nine-point center N , are
collinear. Since the homothety preserve the collinearity, the points F  , Na , and
26 S. N. Kiss

O as anticomplements of F , I, N are collinear, too. The points Fa , Naa , O are

collinear since they are the images under h of the collinear points Fa , Ia , N . Since
OG F G 
GN = 2 = GF , the lines F N and OF are parallel.
(c) Denote by D(P, L) the distance from the point P to the straight line L. The
equations of the sidelines of A B  C  are:
B  C  : y + z = 0, C  A : z + x = 0, A B  : x + y = 0.
We prove that
D(Na , B  C  ) = D(Na , C  A ) = D(Na , A B  ) = 2r
and
D(Naa , B  C  ) = D(Naa , C  A ) = D(Naa , A B  ) = 2ra .
Indeed,
 s−b s−c 
 + s  S 2sr
D(Na , B  C  ) = S s = = = 2r,
2 2
b + c − 2SA s s
 
 s−c s−b 
− s−a − s−a  S 2(s − a)ra
D(Naa , B  C  ) = S = = = 2ra .
b2 + c2 − 2SA s−a s−a
(d)
(ONa )2
1
2 
= R2 − 2
a (s − b)(s − c) + b2 (s − c)(s − a) + c2 (s − a)(s − b)
s
1

= R − 2 −s2 (a2 + b2 + c2 ) + s(a3 + b3 + c3 ) + abc · 2s
2
s
1

= R2 − 2 −2s2 (s2 − r2 − 4Rr) + 2s2 (s2 − 3r2 − 6Rr) + 4Rsr · 2s
s
= R2 − 2(−2r2 + 2Rr)
= (R − 2r)2 ;
i.e. ONa = R − 2r.
Similarly, ONaa = R + 2ra , ONab = R + 2rb , and ONac = R + 2rc . 
Remarks. (1) The excentral triangle of the anticomplementary triangle A B  C  is
the adjugate triangle Naa Nab Nac of the Nagel point.
(2) Since homotheties preserve parallelism, the sides of the excentral triangles
Ia Ib Ic and Naa Nab Nac are parallel, i.e.
Ib Ic  Nab Nac , Ic Ia  Nac Naa , Ia Ib  Naa Nab ,
and
Nab Nac = 2Ib Ic , Nac Naa = 2Ic Ia , Naa Nab = 2Ia Ib .
Furthermore, we have
AIa  A Naa , BIb  B  Nab , CIc  C  Nac .
Adjugate points and adjugate triangle 27

Corollary 10. The anticomplementary triangle is the orthic triangle of the adju-
gate triangle Naa Nab Nac of the Nagel point.
Theorem 11. The Feuerbach triangle of the anticomplementary triangle A B  C 
is the adjugate triangle of X(100), the anticomplement of the Feuerbach point F .
Proof. A center function of F  is f (a, b, c) = a(a − b)(a − c). The points Fa , Fb ,
Fc are the adjugate points of X(100). Indeed,
Fa = (−a(a + b)(a + c) : b(b − c)(b + a) : c(c + a)(c − b))
= (f (−a, b, c) : f (b, c, −a) : f (c, −a, b)).

Theorem 12. The adjugate triangles Gea Geb Gec and Naa Nab Nac of the Gergonne
and Nagel points are perspective, and the perspector is the symmedian point of the
anticomplementary triangle A B  C  , i.e. the point K  = X(69).
Proof. It is easy to verify that the lines through the corresponding pairs of points
have equations
a(b − c)sx+b(c + a)(s − c)y −c(a + b)(s − b)z =0,
−a(b + c)(s − c)x +b(c − a)sy+c(a + b)(s − a)z =0,
a(b + c)(s − b)x−b(c + a)(s − a)y +c(a − b)sz =0,
and that each of these lines contains the point
X(69) = (b2 + c2 − a2 : c2 + a2 − b2 : a2 + b2 − c2 ).

Theorem 13. The Gergonne point Ge , the symmedian point of the anticomplemen-
tary triangle K  , and the Nagel point Na are collinear.
Proof. The line containing these points has equation
a(b − c)(s − a)x + b(c − a)(s − b)y + c(a − b)(s − c)z = 0.

Theorem 14. The pairs of perspective triangles
(ABC, Gea Geb Gec ) (ABC, Naa Nab Nac ) (Gea Geb Gec , Naa Nab Nac )
have a common perspectrix, which is the trilinear polar of the isotomic conjugate
of incenter with equation ax + by + cz = 0.
Proof. (a) The sidelines of triangle Gea Geb Gec have equations
Geb Gec : −(b + c)(s − b)(s − c)x + bs(s − a)y + cs(s − a)z = 0,
Gec Gea : as(s − b)x − (c + a)(s − c)(s − a)y + cs(s − b)z = 0,
Gea Geb : as(s − c)x + bs(s − c)y − (a + b)(s − a)(s − b)z = 0.
These lines intersect BC, CA, AB respectively at the three points
(0 : c : −b), (−c : 0 : a), (b : −a : 0)
28 S. N. Kiss

collinear on the line ax + by + cz = 0. This shows  that ABC  and Gea Geb Gec are
perspective with perspectrix the trilinear polar of a1 : 1b : 1c .
(b) The sidelines of triangle Naa Nab Nac have equations
Nab Nac : (b + c)x + by + cz = 0,
Nac Naa : ax + (c + a)y + cz = 0,
Naa Nab : ax + by + (a + b)z = 0.
These lines intersect BC, CA, AB respectively at the same three points
(0 : c : −b), (−c : 0 : a), (b : −a : 0)
on the line ax + by + cz = 0, showing that ABC and Naa Nab Nac are perspective
with the same perspectrix.
(c) For the triangles Gea Geb Gec and Naa Nab Nac , we have
Geb Gec ∩ Nab Nac = (0 : c : −b),
Gec Gea ∩ Nac Naa = (−c : 0 : a),
Gea Geb ∩ Naa Nab = (b : −a : 0).
Again, the corresponding lines intersect at the same three collinear points on ax +
by + cz = 0.
This shows that the three pairs of perspective triangles have the same perspec-
trix. 

References
[1] N.Dergiades and Q. H. Tran, Simple proofs of Feuerbach’s theorem and Emelyanov’s theorem,
Forum Geom., 18 (2018) 353–359.
[2] C. Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998)
1–285.
[3] C. Kimberling, Encyclopedia of Triangle Centers,
http://faculty.evansville.edu/ck6/encyclopedia/ETC.html.
[4] E. W. Weisstein, Anticomplementary Triangle, from MathWorld–A Wolfram Web Resource,
http://mathworld.wolfram.com/AnticomplementaryTriangle.html

Sándor Nagydobai Kiss: Satu Mare, Romania

E-mail address: d.sandor.kiss@gmail.com