Graphics Reference
In-Depth Information
therefore
area
∆
ABP
area
∆
ABC
t
=
similarly
x
A
y
A
1
area
∆
BCP
=
2
r
x
B
y
B
1
=
r
×
area
∆
ABC
x
C
y
C
1
r
=
area
∆
BCP
area
∆
ABC
and
x
A
y
A
1
area
∆
CAP
=
2
s
x
B
y
B
1
=
s
×
area
∆
ABC
x
C
y
C
1
s
=
area
∆
CAP
area
∆
ABC
thus, we see that the areas of the internal triangles are directly proportional
to the barycentric coordinates of
P
.
This is quite a useful relationship and can be used to resolve various geo-
metric problems. For example, let's use it to find the radius and centre of the
inscribed circle for a triangle. We could approach this problem using classical
Euclidean geometry, but barycentric coordinates provide a powerful analytical
tool for resolving the problem very quickly.
Consider triangle ∆
ABC
with sides
a
,
b
,and
c
as shown in Figure 11.17.
The point
P
is the centre of the inscribed circle with radius
R
.Fromour
knowledge of barycentric coordinates we know that
P
=
r
A
+
s
B
+
t
C
C
b
a
P
R
c
A
B
Fig. 11.17.
The inscribed circle in triangle ∆
ABC
.
