Barycentric Coordinates

Zachary Abel
August 17, 2007
1 Barycentric Coordinates: Denition
1.1 Denition
Consider placing masses of 2, 3, and 7 at vertices A, B, and C of a non-degenerate triangle. Letting D be the point
of BC so that BD/DC = 7/3, we see that lever BC balances at fulcrum D, meaning the triangle ABC balances
along cevian AD. Likewise, it balances along cevians BE and CF where CE/EA = 2/7 and AF/FB = 3/2. With
a bit of physical intuition, these balancing lines should all pass through the center of mass of the system, and we
dene this point of concurrency P (the center of mass) to have barycentric coordinates (2 : 3 : 7) with respect to
triangle ABC. More generally,
A
B
C
D
E
F
P = (2 : 3 : 7)
7u 3u
2v
7v
3w
2w
Figure 1: Barycentric coordinates denition
Barycentric Coordinates Denition. The point with coordinates (x : y : z) is the center of mass of the system
when masses of x, y, and z (which may be zero or even negative!) are placed at the vertices A, B and C of the
reference triangle.
Equivalently, as explained above, we may say that
Denition

. The point P = (x : y : z) is the point whose traces D, E, F satisfy

BD
DC
=
z
y
,
CE
EA
=
x
z
, and
AF
FB
=
y
x
,
using signed ratios.
Using this formulation, we can give a quick proof of Cevas theorem, which states that cevians AD, BE, and
CF are concurrent if and only if the three ratios
BD
DC
= r
a
,
CE
EA
= r
b
, and
AF
FB
= r
c
have product 1. The point
P = (r
a
r
b
: 1 : r
a
) can be seen to have traces D, E, and F

, where
AF

B
=
1
rar
b
. Cevians AD, BE, CF are concurrent
if and only if F = F

, i.e. if and only if r

c
=
1
rar
b
, QED. (The barycentric coordinates of P can be written more
symmetrically as P = (
3
_
r
a
r
2
b
:
3
_
r
b
r
2
c
:
3
_
r
c
r
2
a
).)
1
There is one more description of barycentric coordinates that is often much more useful that the previous two:
Denition

. The point P = (x : y : z) is the point in the plane of triangle ABC so that the three (signed) areas
[PBC], [PCA], and [PAB] are in the ratio x : y : z (which explains the chosen notation).
Indeed, we have
[PAB]
[PCA]
=
[PAD]
[PDC]
=
PD
DC
=
z
y
,
and likewise for [PBC].
A
B
C
D
E
F
P
x
y
z
Figure 2: Barycentric coordinates through areas
Note that these coordinates are homogeneous. That is, the points (x : y : z) and (kx : ky : kz) for any nonzero
constant k are the same. For this reason, it is often desirable to normalize so that the coordinates have sum 1.
When P is in normalized form, we will use the following notation:
P = (x : y : z) normalized =
1
x +y +z
(x, y, z) =
_
x
x +y +z
,
y
x +y +z
,
z
x +y +z
_
.
(Note the commas instead of colons in the last two representations.) It can be veried that such a scaling is not
possible, i.e. x + y + z = 0, if and only if AD, BE, CF are parallel, which means the point of concurrency P
corresponds to a point at innity in the projective plane.
1.2 Examples
Lets calculate the normalized barycentric coordinates for a few common triangle centers.
1.2.1 Centroid
The centroid G is the point of concurrency of the medians of ABC, i.e. the point of concurrency when D, E, F
are taken as midpoints. As
BD
DC
=
CE
EA
=
AF
FB
= 1, the second denition above shows that
G = (1 : 1 : 1) normalized =
1
3
(1, 1, 1).
2
A
B
C
Ma
M
b
Mc
G
Figure 3: Centroid
r
r
r
A
B
C
I
Figure 4: Incenter
1.2.2 Incenter
With I as the incenter of ABC and r the inradius, we see that [IBC] =
1
2
ar, so we can write I = (
1
2
ar :
1
2
br :
1
2
cr),
or after normalizing,
I =
1
2s
(a, b, c)
(where s =
a+b+c
2
is the semiperimeter).
1.2.3 Spieker Center
If the sides of triangle ABC are traced with a uniform wire, the Spieker center S
p
is the center of mass of the wire.
As the wire BC has mass proportional to a and has center of mass M
a
, the Spieker center it is the center of mass
when weights of a, b, and c are placed at the midpoints M
a
, M
b
, M
c
of BC, CA, AB respectively. This implies
two important facts about this center. First, basing our barycentric coordinate system around the medial triangle
M
a
M
b
M
c
gives S
p
the coordinates (a : b : c), and since M
a
M
b
M
c
ABC, we nd that S
p
is the incenter of the
medial triangle. Secondly, the center of mass of the system with weights a, b, c at M
a
, M
b
, M
c
is the same as
the center of mass of the system which has weights of
b+c
2
,
c+a
2
,
a+b
2
at A, B, C respectively, so the barycentric
coordinates of S
p
with respect to ABC are
S
p
=
1
4s
(b +c, c +a, a +b).
A
B
C
Ma
M
b
Mc
Sp
Figure 5: Speiker Center

A
B
C
D
E
F
H
Figure 6: Orthocenter
3
1.2.4 Orthocenter
In Figure 6, notice that BHD = 90 DBH = ECB = , so
[HBC] =
1
2
aHD =
1
2
aADtan =
1
2
ac cos tan =
c
2 sin
a cos cos = Ra cos cos .
Thus,
H = (a cos cos : b cos cos : c cos cos ) normalized.
By the Law of Cosines, a cos cos =
1
4abc
(c
2
+a
2
b
2
)(a
2
+b
2
c
2
), so we also have
H =
_
(c
2
+a
2
b
2
)(a
2
+b
2
c
2
) : (a
2
+b
2
c
2
)(b
2
+c
2
a
2
) : (b
2
+c
2
a
2
)(c
2
+a
2
b
2
)
_
normalized.
But what are the normalizing factors? The sum of the coordinates in the second expression can be expanded to

cyc
(c
2
+a
2
b
2
)(a
2
+b
2
c
2
) = 2a
2
b
2
+ 2b
2
c
2
+ 2c
2
a
2
a
4
b
4
c
4
= (a +b +c)(a +b c)(b +c a)(c +a b) = 16(s)(s a)(s b)(s c) = 16[ABC]
2
= 16r
2
s
2
,
so we nd the two equivalent normalized forms
H =
1
16r
2
s
2
_
(c
2
+a
2
b
2
)(a
2
+b
2
c
2
), ,
_
=
R
rs
(a cos cos , , )
(we used the well-known formula abc = 4srR to obtain the second expression from the rst).
2 Collinearity
Barycentric coordinates can be used to detect when three points are on a line. Suppose we have two normalized
points P = (x
1
, y
1
, z
1
) and Q = (x
2
, y
2
, z
2
), which means x
1
= [PBC]/[ABC], etc. Let P
x
, P
y
, P
z
be the
projections from P to lines BC, CA, AB respectively, and likewise for Q. Since [PBC] =
1
2
PP
x
a, we have
PP
x
= 2[PBC]/a =
2[ABC]
a
x
1
, and likewise QQ
x
=
2[ABC]
a
x
2
.
A
B
C
P
Q
R
Px
Py Pz
Qx
Qy
Qz
Rx
Ry
Rz
Figure 7: Collinearity condition
Consider the point R = (x
3
, y
3
, z
3
) on line BC so that PR/PQ = k for some real number k, and let R
x
be its
projection onto BC. By right trapezoid PP
x
Q
x
Q, it can be seen that RR
x
= (1 k) PP
x
+ (k) QQ
x
, i.e. that
4
x
3
= (1 k) x
1
+ (k) x
2
. As the same holds for y
3
and z
3
, we nd that
R = (1 k)P + (k)Q,
so R is simply a weighted average of P and Q. When homogeneity is taken into account (i.e. without assuming P
and Q have been normalized), the criterion for collinearity becomes the following:
Collinearity Condition. Points P = (x
1
: y
1
: z
1
), Q = (x
2
: y
2
: z
2
), and R = (x
3
: y
3
: z
3
) are collinear if and
only if the vectors (x
1
, y
1
, z
1
), (x
2
, y
2
, z
2
), and (x
3
, y
3
, z
3
) are linearly dependent, i.e. the determinant

x
1
y
1
z
1
x
2
y
2
z
2
x
3
y
3
z
3

is zero. Furthermore, if P, Q, and R are normalized, the value of k for which R = (1 k)P + (k)Q corresponds to
the ratio k = PR/PQ.
2.1 Examples
2.1.1 Euler Line
The points H, G, and O lie on a line in that order with OH = 3OG. To see why, we need the normalized
barycentric coordinates for O (the other two were calculated above in sections 1.2.1 and 1.2.4). The area of OBC
is
1
2
R
2
sin2 = R
2
sincos =
1
2
Ra cos , so (with a bit of work) we obtain the following equivalent normalized
expressions:
O =
R
2rs
(a cos , , ) =
1
4s
2
r
2
_
a
2
(b
2
+c
2
a
2
), ,
_
.
We simply need to illustrate that 3G = H + 2O, i.e. that
1 =
Ra
rs
cos cos +
Ra
rs
cos .
Notice that cos + cos cos = cos + cos( + ) + sin sin =
bc
4R
2
, so the right side of the previous equation
equals
abc
4srR
= 1, as needed.
A
B
C
Ma
M
b
Mc
H
G
O
Figure 8: Euler line
A
B
C
Na
Sp
G
I
Figure 9: Nagel line
5
2.1.2 Nagel Line
The Nagel point N
a
, Spieker center S
p
, centroid G, and incenter I are collinear along the so-called Nagel Line in
that order, with S
p
and G respectively bisecting and trisecting segment N
a
I. The Nagel point is dened as the
point of concurrency of AD, BE, and CF where DEF is the extouch triangle, meaning the excircle opposite vertex
A is tangent to BC at D, and similarly for E and F. Since BD/DC = (s c)/(s b), and likewise for E and F,
we obtain N
a
=
1
s
(s a, s b, s c). Now the statement above is easy to verify:
1
2
_
N
a
+I
_
=
_
sa
s
+
a
2s
2
, ,
_
=
_
b +c
4s
, ,
_
= S
p
and
1
3
N
a
+
2
3
I =
_
s a
3s
+
a
3s
, ,
_
=
_
1
3
, ,
_
= G.
3 Area
Barycentric coordinates can also be used to calculate triangle areas, as follows:
Area Formula. For three points P = (x
1
, y
1
, z
1
), Q = (x
2
, y
2
, z
2
), R = (x
3
, y
3
, z
3
) written in normalized barycen-
tric coordinates with respect to triangle ABC, we have
[PQR]
[ABC]
=

x
1
y
1
z
1
x
2
y
2
z
2
x
3
y
3
z
3

.
Proof. Choose a point O not in the plane of ABC, and set up a three dimensional coordinate system with
O = (0, 0, 0), A = (1, 0, 0), B = (0, 1, 0), and C = (0, 0, 1) (note that this need not be an orthonormal frame!). It is
not dicult to show that for a point X = (x, y, z) with x +y +z = 1, i.e. in the plane of ABC, the coordinates
(x, y, z) in the coordinate system correspond to its normalized barycentric coordinates X. Letting P
ABC
be the
parallelepiped spanned by vectors

OA,

OB,

OC, and likewise for P
PQR
, the denition of determinant via volume
gives
vol(P
PQR
)
vol(P
ABC
)
=

x
1
y
1
z
1
x
2
y
2
z
2
x
3
y
3
z
3

.
We have vol(P
ABC
) = 6 vol(OABC) = 2[ABC] h, where h is the length of the height from O to plane ABC, and
likewise, vol(P
PQR
) = 2[PQR] h. Thus, the ratio above equals
[PQR]
[ABC]
,
as claimed.
6
3.1 Example: Triangle OIH
Problem 1. The area of triangle OIH is
1
8r
(a b)(b c)(c a).
A
B
C
O I
H
Figure 10: Triangle OIH
Solution. According to the previous formula, the area of triangle OIH is
[ABC]
R
2rs

1
2s

R
rs

a cos b cos c cos

a b c
a cos cos b cos cos c cos cos

=
R
2
abc
4rs
2

cos cos c cos

1 1 1
cos cos cos cos cos cos

=
R
3
s
(cos cos )(cos cos )(cos cos ).
Finally, using the formula
cos cos =
c
2
+a
2
b
2
2ca

b
2
+c
2
a
2
2bc
=
(a b)(a +b +c)(a +b c)
2abc
=
(a b)(s c)
2rR
,
the above simplies to
1
8sr
3
(s a)(s b)(s c)(a b)(b c)(c a) =
1
8r
(a b)(b c)(c a),
as claimed. (The last simplication is due to Herons formula: rs =
_
s(s a)(s b)(s c).)
4 Problems
Problem 2 (MOP 2006). Triangle ABC is inscribed in circle . Point P lies on line BC such that line PA is
tangent to . The bisector of APB meets segments AB and AC at D and E respectively. Segments BE and CD
meet at Q. Given that line PQ passes through the center of , compute BAC.
Solution. By similar triangles PBA and PAC,
PB
PA
=
PA
PC
=
c
b
, so
BD
DA
=
c
b
and
AE
EC
=
c
b
. This is enough to identify
D = (c : b : 0), E = (b : 0 : c), and Q = (bc : b
2
: c
2
). Point P, lying on line BC, has the form P = (0 : x : y)
for some x and y, and since P, D, and E are collinear, we have

0 x y
c b 0
b 0 c

= 0, i.e.
x
y
=
b
2
c
2
. So P = (0 : b
2
: c
2
).
7
A
B
C
P
D
E
Q
O
Figure 11: Problem 2
Finally, since P, Q, and O = (a cos : : ) are collinear, we nd that

0 b
2
c
2
bc b
2
c
2
a cos b cos c cos

= bc

0 b c
bc b c
a cos cos cos

= 0
which simplies to 2abc cos = bc
2
cos +b
2
c cos = bc(c cos +b cos ) = abc, i.e. cos =
1
2
and = 60

.
Problem 3 (USAMO 2001 #2). Let ABC be a triangle and let be its incircle. Denote by D
1
and E
1
the points
where is tangent to sides BC and AC, respectively. Denote by D
2
and E
2
the points on sides BC and AC,
respectively, such that CD
2
= BD
1
and CE
2
= AE
1
, and denote by P the point of intersection of segments AD
2
and BE
2
. Circle intersects segment AD
2
at two points, the closer of which to the vertex A is denoted by Q.
Prove that AQ = D
2
P.
A
B
C
Ma
M
b
Mc
D
1
E
1
F
1
D
2
E
2
F
2
Na
Q
I
Figure 12: Problem 3
A
B C
D
1
E
1
F
1
D
2
Q
Ia
I
Figure 13: A, Q, D
1
collinearity
Solution. We can locate most of the points in the diagram: I =
1
2s
(a, b, c), D
1
=
1
a
(0, sc, sb), D
2
=
1
a
(0, sb, sc)
(the point of tangency of BC with As excircle), E
2
=
1
b
(s a, 0, s c), and P = N
a
=
1
s
(s a, s b, s c) (this
is the Nagel point). To nd Q, we note that the homothecy at A taking As excircle to the incircle must take D
2
to Q. This means the radius IQ is parallel to I
a
D
2
ID
1
, i.e. Q is diametrically opposite to D
1
along the incircle:
Q = 2I D
1
=
_
a
s
,
b
s

sc
a
,
c
s

sb
a
_
. Now, all we must show is AQ = PD
2
.
While distances in general are not pretty in barycentric coordinates, we are saved by the fact that all four points
are on a line. We oer two ways to nish. The rst is to appeal to the 3D coordinate frame view developed in
8
A
B
C
P
Py
Pz Q
Qy
Qz

Figure 14: Isogonal conjugates

A
B
C
Na
N

a
I
O G

e
Figure 15: Problem 4
section 3 and claim it suces to show that

AQ =

PD
2
(since these points are already in normalized form), i.e. that
_
a
s
1,
b
s

s c
a
,
c
s

s b
a
_
=
_

s a
s
,
s b
a

s b
s
,
s c
a

s c
s
_
.
This can be checked directly. The rst coordinates are clearly equal; equality of the second coordinates boils down
to 2s = a +b +c; and likewise for the third coordinates. Another method is to show that [AQB] = [PD
2
B]. This
becomes

1 0 0
a
s
b
s

sc
a
c
s

sb
a
0 1 0

sa
s
sb
s
sc
s
0
sb
a
sc
a
0 1 0

,
i.e.
sb
a

c
s
=
sa
s

sc
a
. But again, this is just a rewriting of a +b +c = s.
Problem 4. Show that the isogonal conjugate of the Nagel point is the center of positive homothecy between the
incircle and circumcircle. Likewise, the isogonal conjugate of the Gergonne point is the center of negative homothecy
between the two circles.
Solution. To do this, rst we need to calculate the isogonal conjugate P

of a general point P = (x : y : z). Let P

y
and P
z
be the projections from P to AC and AB respectively, and likewise for P

y
and P

z
. Setting BAP = and
PAC = , we have
[P

CA]
[P

AB]
=
b
c

P

y
P

z
=
b
c

sin
sin
=
b
c

PP
z
PP
y
=
b
2
c
2

[PAB]
[PCA]
=
b
2
/y
c
2
/z
,
so we nd P

=
_
a
2
x
:
b
2
y
:
c
2
z
_
.
Now, since N
a
= (s a : s b : s c) normalized, the above gives
N

a
=
_
a
2
s a
: :
_
normalized =
_
a
2
(s b)(s c) : :
_
normalized.
9
Using the identity sin
2
2
=
(sb)(sc)
bc
, the above can be manipulated as follows:
N

a
=
_
a
2
bc sin
2
2
: :
_
normalized
= (a(1 cos ) : : ) normalized
=
_
(a : b : c) (a cos : b cos : c cos )
_
normalized
=
_
2s I
2rs
R
O
_
normalized
=
_
R I r O
_
normalized
=
R
R r
I
r
R r
O.
This means that N

a
is on line OI such that N

a
I/N

a
O = r/R, i.e. N

a
is exactly the center of homothecy taking I
to O with positive ratio R/r. Likewise, it may be calculated that G

e
=
R
R+r
I +
r
R+r
O, i.e. G

e
is the center of
homothecy taking I to O with negative ratio R/r.
10