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where
r
+
s
+
t
= 1 (11.12)
We also know that the area properties of barycentric coordinates permit us
to state that
area
∆
ABC
=
1
area
∆
BCP
=
r
×
2
aR
area
∆
ABC
=
1
area
∆
CAP
=
s
×
2
bR
area
∆
ABC
=
1
area
∆
ABP
=
t
×
2
cR
therefore
aR
bR
cR
r
=
s
=
t
=
2
×
area
∆
ABC
2
×
area
∆
ABC
2
×
area
∆
ABC
substituting
r
,
s
and
t
in (11.11) we get
R
area
∆
ABC
(
a
+
b
+
c
)=1
2
×
and
area
∆
ABC
a
+
b
+
c
Substituting
R
in the definitions of
r
,
s
and
t
we obtain
R
=
2
×
a
a
+
b
+
c
b
a
+
b
+
c
c
a
+
b
+
c
r
=
s
=
t
=
and
x
P
=
rx
A
+
sx
B
+
tx
C
y
P
=
ry
A
+
sy
B
+
ty
C
To tes
t th
is solution, consider the right-angled triangle in Figure 11.18, where
a
=
√
200
,b
=10
,c
=10and
area
∆
ABC
= 50. Therefore
2
×
50
10 + 10 +
√
200
=2
.
929
R
=
and
√
200
34
.
1421
10
34
.
1421
10
34
.
1421
r
=
=0
.
4142
s
=
=0
.
2929
t
=
=0
.
2929
therefore
x
P
=0
.
4142
×
0+0
.
2929
×
10 + 0
.
2929
×
0=2
.
929
y
P
=0
.
4142
×
0+0
.
2929
×
0+0
.
2929
×
10 = 2
.
929
