25.2 Routh theorem 703
25.2 Routh theorem .................
λ^1
μ
1
ν
1
A
B X C
Z
Y
P Q
R
We make use ofhomogeneous barycentric coordinateswith respect
toABC.
X=(0:1:λ),Y=(μ:0:1),Z=(1:ν:0).
Those ofP,Q,Rcan be worked out easily:
P=BY∩CZ Q=CZ∩AX R=AX∩BY
Y =(μ:0:1) Z=(1:ν:0) X=(0:1:λ)
Z=(1:ν:0) X=(0:1:λ) Y =(μ:0:1)
P=(μ:μν:1) Q=(1:ν:νλ) R=(λμ:1:λ)
This means that theabsolute barycentric coordinatesofX,Y,Zare
P=
1
μν+μ+1
(μA+μνB+C),
Q=
1
νλ+ν+1
(A+νB+νλC),
R=
1
λμ+λ+1
(λμA+B+λC).
From these,
Area(PQR)=
∣ ∣ ∣ ∣ ∣ ∣
μμν 1
1 ννλ
λμ 1 λ
∣ ∣ ∣ ∣ ∣ ∣
(μν+μ+1)(νλ+ν+1)(λμ+λ+1)
·
=
(λμν−1)^2
(μν+μ+1)(νλ+ν+1)(λμ+λ+1)
·
.