April 11, 2008
In this article I aim to brieﬂy develop the theory of areal (or ‘barycentric’) co-ordinate methods with a view
to making them accessible to a reader as a means for solving problems in plane geometry. Areal co-ordinate
methods are particularly useful and important for solving problems based upon a triangle, because, unlike
cartesian co-ordinates, they exploit the natural symmetries of the triangle and many of its key points in a
very beautiful and useful way.
Setting up the co-ordinate system
If we are going to solve a problem using areal co-ordinates, the ﬁrst thing we must do is choose a triangle
ABC, which we call the triangle of reference, and which plays a similar role to the axes in a cartesian
co-ordinate system. Once this triangle is chosen, we can assign to each point P in the plane a unique triple
(x, y, z) ﬁxed such that x + y + z = 1, which we call the areal co-ordinates of P . The way these numbers are
assigned can be thought of in three diﬀerent ways, all of which are useful in diﬀerent circumstances. I shall
reserve the proofs that these three conditions are equivalent, along with a proof of the uniqueness of areal
co-ordinate representation, for the appendix. The ﬁrst deﬁnition we shall see is probably the most intuitive
and most useful for working with. It also explains why they are known as ‘areal’ co-ordinates.
1st Deﬁnition: A point P internal to the triangle ABC has areal co-ordinates
If a sign convention is adopted, such that a triangle whose vertices are labelled clockwise has negative area,
Figure 1: The ‘areal’ deﬁnition of areal co-ordinates
resulting system has centre of mass P , then (x, y, z) are the areal co-ordinates of P (hence the alternative
3rd Deﬁnition: If we take a system of vectors with arbitrary origin (not on the sides of triangle ABC) and
let a,b,c,p be the position vectors of A, B, C, P respectively, then p = xa + yb + zc for some triple (x, y, z)
such that x + y + z = 1. We deﬁne this triple as the areal co-ordinates of P .
There are some remarks immediately worth making:
• The vertices A, B, C of the triangle of reference have co-ordinates (1, 0, 0), (0, 1, 0), (0, 0, 1) respectively.
• All the co-ordinates of a point are positive if and only if the point lies within the triangle of reference,
and if any of the co-ordinates are zero, the point lies on one of the sides (or extensions of the sides) of
The Equation of a Line
would therefore expect the equation of a line to be linear, such that any pair of simultaneous line equations,
together with the condition x + y + z = 1, can be solved for a unique triple (x, y, z) corresponding to the areal
co-ordinates of the point of intersection of the two lines. Indeed, it follows (using the equation x + y + z = 1
to eliminate any constant terms) that the general equation of a line is of the form
lx + my + nz = 0
where l, m, n are constants and not all zero. Clearly there exists a unique line (up to multiplication by a
constant) containing any two given points P (x
). This line can be written explicitly as
This equation is perhaps more neatly expressed in the determinant form (see Appendix 1):
While the above form is useful, it is often quicker to just spot the line automatically. For example try to spot
the equation of the line BC, containing the points B(0,1,0) and C(0,0,1), without using the above equation.
Of particular interest (and simplicity) are Cevian lines, which pass through the vertices of the trian-
gle of reference. We deﬁne a Cevian through A as a line whose equation is of the form my = nz.
Clearly any line containing A must have this form, because setting y = z = 0, x = 1 any equation with
a nonzero x coeﬃcient would not vanish. It is easy to see that any point on this line therefore has form
(x, y, z) = (1 − mt − nt, nt, mt) where t is a parameter. In particular, it will intersect the side BC with
equation x = 0 at the point U (0,
implies that the ratio BU/U C = [ABU ]/[AU C] = m/n.
Example 1: Ceva’s Theorem
We are now in a position to start using areal co-ordinates to prove useful theorems. In this section we shall
state and prove (one direction of) an important result of Euclidean geometry known as Ceva’s Theorem. The
for themselves using the ideas introduced above, before reading the proof given.
Ceva’s Theorem: Let ABC be a triangle and let L, M, N be points on the sides BC, CA, AB respec-
tively. Then the cevians AL,BM ,CN are concurrent at a point P if and only if
Partial proof: Suppose ﬁrst that the cevians are concurrent at a point P , and let P have areal co-ordinates
(p, q, r). Then AL has equation qz = ry (following the discussion of Cevian lines above), so L(0,
which implies BL/LC = r/q. Similarly, CM/M A = p/r, AN/N B = q/p. Taking their product we get
The above proof was very typical of many areal co-ordinate proofs. We only had to go through the de-
tails for one of the three cevians, and then could say ‘similarly’ and obtain ratios for the other two by
symmetry. This is one of the great advantages of the areal co-ordinate system in solving problems where
such symmetries do exist (particularly problems symmetric in a triangle ABC: such that relabelling the
triangle vertices would result in the same problem).
Areas and Parallel Lines
One might expect there to be an elegant formula for the area of a triangle in areal co-ordinates, given they are
a system constructed on areas. Indeed, there is. If P QR is an arbitrary triangle with P (x
An astute reader might notice that this seems like a plausible formula, because if P, Q, R are collinear, it
tells us that the triangle P QR has area zero, by the line formula already mentioned. It should be noted that
the area comes out as negative if the vertices P QR are labelled in the opposite direction to ABC.
It is now fairly obvious what the general equation for a line parallel to a given line passing through two
) should be, because the area of the triangle formed by any point on such a line
and these two points must be constant, having a constant base and constant height. Therefore this line has
Where k is a real constant.
Exercise: (BMO1 2007/8 Q5) Given a triangle ABC and an arbitrary point P internal to it, let the
line through P parallel to BC meet AC at M , and similarly let the lines through P parallel to CA,AB meet
AB,BC at N ,L respectively. Show that
Before we start looking at some more deﬁnite speciﬁc useful tools (like the positions of various interesting
points in the triangle), we round oﬀ the general theory with a device that, with practice, greatly simpliﬁes
if we are just intersecting lines with lines or lines with conics, and not trying to calculate any ratios, it is
legitimate to ignore this constraint and to just consider the points (x, y, z) and (kx, ky, kz) as being the same
point for all k = 0. This is because areal co-ordinates are a special case of a more general class of co-ordinates
called projective homogeneous co-ordinates
, where here the projective line at inﬁnity is taken to be
in x, y, z), so, for example, x + y = 1 and x
+ y = z are not homogeneous, whereas x + y − z = 0 and
yz + b
zx + c
xy = 0 are homogeneous. We can therefore, once all our line and conic equations are happily
in this form, no longer insist on x + y + z = 1, meaning points like the incentre (
just be written (a, b, c), a signiﬁcant advantage for the practical purposes of doing manipulations. However,
if any ratios or areas are to be calculated, it is imperative that the co-ordinates are normalised again to
make x + y + z = 1. This process is easy: just apply the map
(x, y, z) → (
x + y + z
Signiﬁcant areal points and formulae in the triangle
We have seen that the vertices are given by A(1, 0, 0), B(0, 1, 0), C(0, 0, 1), and the sides by x = 0, y = 0, z = 0.
In the section on the equation of a line we examined the equation of a cevian, and this theory can, together
with other knowledge of the triangle, be used to give areal expressions for familiar points in Euclidean
triangle geometry. We invite the reader to prove some of the facts below as exercises.
• Triangle centroid: G(1, 1, 1). The midpoints of the sides BC, CA, AB are given by (0, 1, 1), (1, 0, 1), (1, 1, 0)
• Centre of the inscribed circle: I(a, b, c) (hint: use the angle bisector theorem)
• Centres of escribed circles: I
(−a, b, c), I
(a, −b, c), I
(a, b, −c)
• Centre of the circumcircle: O(sin 2A, sin 2B, sin 2C)
It should be noted that the rather nasty trigonometric forms of O and H mean that they should be ap-
proached using areals with caution, preferably only if the calculations will be relatively simple.
If the reader is familiar with isogonal and isotomic conjugation, it is interesting to ﬁnd that the isogonal
conjugate of a point (x, y, z) is (a
/z) (verify with G, K, I, I and O, H above), and the isotomic
conjugate (1/x, 1/y, 1/z).
Exercise: Let D, E be the feet of the altitudes from A and B respectively, and P, Q the meets of the
angle bisectors AI,BI with BC,CA respectively. Show that D, I, E are collinear if and only if P, O, Q are.
The author regrets that, in the interests of concision, he is unable to deal with these co-ordinates in this document, but
also with a more detailed account of areals and a plethora of applications of the methods touched on in this document. Even
better, though only for projectives and lacking in the wealth of fascinating modern examples, is E.A.Maxwell’s The methods
of plane projective geometry based on the use of general homogeneous coordinates, recommended to the present author by the
author of the ﬁrst book.
We ﬁnally quickly outline some slightly more advanced theory, which is occasionally quite useful in some
problems, We show how to manipulate conics (with an emphasis on circles) in areal co-ordinates, and how to
ﬁnd the distance between two points in areal co-ordinates. These are placed in the same section because the
formulae look quite similar and the underlying theory is quite closely related. Derivations can be found in .
Firstly, the general equation of a conic in areal co-ordinates is, since a conic is a general equation of the
second degree, and areals are a homogeneous system, given by
+ 2dyz + 2ezx + 2f xy = 0
Since multiplication by a nonzero constant gives the same equation, we have ﬁve independent degrees of
freedom, and so may choose the coeﬃcients uniquely (up to multiplication by a constant) in such a way as
to ensure ﬁve given points lie on such a conic.
In Euclidean geometry, the conic we most often have to work with is the circle.
The most important
circle in areal co-ordinates is the circumcircle of the reference triangle, which has the equation (with a, b, c
equal to BC, CA, AB respectively)
zx + c
xy = 0
with the above, a general circle is just a variation on this theme, being
xy + (x + y + z)(ux + vy + wz) = 0
We can, given three points, solve the above equation for u, v, w substituting in the three desired points to
obtain the equation for the unique circle passing through them.
Now, the areal distance formula looks very similar to the circumcircle equation.
If we have a pair of
points P (x
) and Q(x
), which must be normalised, we may deﬁne the displacement P Q :
) = (u, v, w), and it is this we shall measure the distance of. So the distance of a
displacement P Q(u, v, w), u + v + w = 0 is given by
vw − b
wu − c
Since u + v + w = 0 this is, despite the negative signs, always positive unless u = v = w = 0.
Matrix determinants play an important role in areal co-ordinate methods. We deﬁne the determinant of a
3 by 3 square matrix A as
) + a
z) + a
This can be thought of as (as the above equation suggests) multiplying each element of the ﬁrst column by the
determinants of 2x2 matrices formed in the 2nd and 3rd columns and the rows not containing the element of
the ﬁrst column. Alternatively, if you think of the matrix as wrapping around (so b
is in some sense directly
in the above matrix) you can simply take the sum of the products of diagonals running from
to top-right (so think of the above RHS as (a
) − (a
)). In any
with areal co-ordinates.
All circles have two (imaginary) points in common on the line at inﬁnity. It follows that if a conic is a circle, its behaviour
Here we attach a selection of problems compiled by Tim Hennock, largely from UK IMO activities in 2007
and 2008. None of them are trivial, and some are quite diﬃcult, with diﬃculty roughly proportional to
number of asterisks.
1. (Pre-IMO training 2007) *
Let ABC be a triangle. Let D, E, F be the reﬂections of A, B, C in BC, AC, AB respectively. Show that
D, E, F are collinear if and only if OH = 2R.
2. (Balkan MO 2005) **
Let ABC be an acute-angled triangle whose inscribed circle touches AB and AC at D and E respectively.
Let X and Y be the points of intersection of the bisectors of the angles ∠ACB and ∠ABC with the line
DE and let Z be the midpoint of BC. Prove that the triangle XY Z is equilateral if and only if ∠A = 60
3. (NST 2007) **
meet the circumcircle of ABC again at A . Lines CA and AB intersect at D and BA and AC intersect at
E. Prove that the circumcentre of triangle ADE lies on the circumcircle of ABC.
4. (IMO 2007) **
In triangle ABC the bisector of ∠BCA intersects the circumcircle again at R, the perpendicular bisector of
BC at P , and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is
L. Prove that the triangles RP K and RQL have the same area.
5. (RMM 2008) ***
Let ABC be an equilateral triangle. P is a variable point internal to the triangle, and its perpendicular
distances to the sides are denoted by a
for positive real numbers a, b and c. Find the locus of
points P such that a, b and c can be the side lengths of a non-degenerate triangle.
6. (ISL 2006) ***
Let ABC be a triangle such that ∠C < ∠A <
ABC touches AB at K and AC at L. Let J be the incentre of triangle BCD. Prove that KL bisects AJ .
7. (NST 2007) ***
The excircle of a triangle ABC touches the side AB and the extensions of the sides BC and CA at points
M, N and P , respectively, and the other excircle touches the side AC and the extensions of the sides AB
and BC at points S, Q and R, respectively. If X is the intersection point of the lines P N and RQ, and Y
the intersection point of RS and M N , prove that the points X, A and Y are collinear.
8. (Sharygin GMO 2008) ***
Let ABC be a triangle and let the excircle opposite A be tangent to the side BC at A
. N is the Nagel
such that AP = N A
. Prove that P lies on the incircle of ABC.
9. (NST 2007) ****
Let ABC be a triangle with ∠B = ∠C. The incircle I of ABC touches the sides BC, CA, AB at the points
D, E, F , respectively. Let AD intersect I at D and P .
Let Q be the intersection of the lines EF and the line passing through P and perpendicular to AD, and let
X, Y be intersections of the line AQ and DE, DF , respectively. Show that the point A is the midpoint of
Given a triangle ABC. Point A
is chosen on the ray BA so that the segments BA
and BC are equal.
is chosen on the ray CA so that the segments CA
and BC are equal. Points B
are chosen similarly. Prove that the lines A
The author owes thanks to Christopher Bradley, Geoﬀ Smith and David Loeﬄer, all of whom provided
invaluable advice on the presentation of this article, and to Tim Hennock for typesetting the exercises.
The author suggests the following resources (particularly 1 and 2) for a more detailed and rigorous exposi-
tion of these modern methods, and would also invite an interested reader to visit reference 4 to get a feeling
for how useful areal co-ordinates are in the classiﬁcation of triangle centres, one of the great geometrical
endeavours of recent times. The author is a frequent user of the AskNRICH forum, as are other competent
users of areal co-ordinates, so any questions on the subject posted there should receive a reply.
 C.J. Bradley - Challenges in Geometry (OUP)
 C.J. Bradley - The Algebra of Geometry (Highperception)
 E.A. Maxwell - The methods of plane projective geometry based on the use of general homogeneous
 Clark Kimberling - Encyclopedia of Triangle Centers (http://faculty.evansville.edu/ck6/encyclopedia/ETC.html)
c Thomas Lovering, 2008 Permission is granted for this material to be downloaded and used for any non-proﬁt