30 solution analytique de la question 344 (Mannheim).
mais les points
B
{\displaystyle \mathrm {B} }
C
{\displaystyle \mathrm {C} }
O
{\displaystyle \mathrm {O} }
|
1
x
2
y
2
1
x
3
y
3
1
x
4
y
4
|
=
0
,
{\displaystyle {\begin{vmatrix}1&x_{2}&y_{2}\\1&x_{3}&y_{3}\\1&x_{4}&y_{4}\end{vmatrix}}=0,}
c’est-a-dire
|
x
2
−
x
1
y
2
−
y
1
λ
h
−
μ
m
λ
k
−
μ
n
|
−
λ
μ
|
k
h
n
m
|
=
0
,
{\displaystyle {\begin{vmatrix}x_{2}-x_{1}&y_{2}-y_{1}\\\lambda h-\mu m&\lambda k-\mu n\end{vmatrix}}-\lambda \mu {\begin{vmatrix}k&h\\n&m\end{vmatrix}}=0,}
par conséquent,
1
2
(
1
A
B
O
+
1
A
O
C
)
=
|
k
h
n
m
|
|
x
2
−
x
1
y
2
−
y
1
h
k
|
|
x
1
−
x
2
y
1
−
y
2
m
n
|
{\displaystyle {\frac {1}{2}}\left({\frac {1}{\mathrm {ABO} }}+{\frac {1}{\mathrm {AOC} }}\right)={\frac {\begin{vmatrix}k&h\\n&m\end{vmatrix}}{{\begin{vmatrix}x_{2}-x_{1}&y_{2}-y_{1}\\h&k\end{vmatrix}}{\begin{vmatrix}x_{1}-x_{2}&y_{1}-y_{2}\\m&n\end{vmatrix}}}}}
quantité indépendante de
λ
{\displaystyle \lambda }
μ
{\displaystyle \mu }
Théorème analogue dans l’espace.
Par un point
O
{\displaystyle \mathrm {O} }
A
{\displaystyle \mathrm {A} }
B
{\displaystyle \mathrm {B} }
C
{\displaystyle \mathrm {C} }
D
{\displaystyle \mathrm {D} }
ν
1
{\displaystyle \nu _{1}}
ν
2
{\displaystyle \nu _{2}}
ν
3
{\displaystyle \nu _{3}}
A
O
C
D
{\displaystyle \mathrm {AOCD} }
A
O
D
B
{\displaystyle \mathrm {AODB} }
A
O
B
C
{\displaystyle \mathrm {AOBC} }
ν
1
ν
2
ν
3
+
ν
2
ν
3
ν
1
+
ν
3
ν
1
ν
2
{\displaystyle {\sqrt {\frac {\nu _{1}}{\nu _{2}\nu _{3}}}}+{\sqrt {\frac {\nu _{2}}{\nu _{3}\nu _{1}}}}+{\sqrt {\frac {\nu _{3}}{\nu _{1}\nu _{2}}}}}
est constante, de quelque manière qu’on mène le plan sécant.
Soient
x
1
,
y
1
,
z
1
,
x
2
,
y
2
,
z
2
,
…
,
x
5
,
y
5
,
z
5
{\displaystyle x_{1},y_{1},z_{1},x_{2},y_{2},z_{2},\ldots ,x_{5},y_{5},z_{5}}
A
{\displaystyle \mathrm {A} }
O
{\displaystyle \mathrm {O} }
B
{\displaystyle \mathrm {B} }
C
{\displaystyle \mathrm {C} }
D
{\displaystyle \mathrm {D} }
x
1
,
y
1
,
z
1
,
x
2
,
y
2
,
z
2
{\displaystyle x_{1},y_{1},z_{1},x_{2},y_{2},z_{2}}
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
γ
{\displaystyle \gamma }
x
3
−
x
1
α
1
=
y
3
−
y
1
β
1
=
z
3
−
z
1
γ
1
=
λ
,
{\displaystyle {\frac {x_{3}-x_{1}}{\alpha _{1}}}={\frac {y_{3}-y_{1}}{\beta _{1}}}={\frac {z_{3}-z_{1}}{\gamma _{1}}}=\lambda ,}
x
4
−
x
1
α
2
=
y
4
−
y
1
β
2
=
z
4
−
z
1
γ
2
=
μ
,
{\displaystyle {\frac {x_{4}-x_{1}}{\alpha _{2}}}={\frac {y_{4}-y_{1}}{\beta _{2}}}={\frac {z_{4}-z_{1}}{\gamma _{2}}}=\mu ,}
x
5
−
x
1
α
3
=
y
5
−
y
1
β
3
=
z
5
−
z
1
γ
3
=
ν
,
{\displaystyle {\frac {x_{5}-x_{1}}{\alpha _{3}}}={\frac {y_{5}-y_{1}}{\beta _{3}}}={\frac {z_{5}-z_{1}}{\gamma _{3}}}=\nu ,}
