000RM.dvi

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)

·

.