SUR LES QUESTIONS 321 ET 322.7
Nouvelles Annales de Mathématiques , 1.re
Soient
a
r
{\displaystyle a_{r}}
b
r
{\displaystyle b_{r}}
c
r
{\displaystyle c_{r}}
r
{\displaystyle r}
rième
l
r
{\displaystyle l_{r}}
(
r
,
r
+
1
)
{\displaystyle (r,r+1)}
α
r
{\displaystyle \alpha _{r}}
β
r
{\displaystyle \beta _{r}}
γ
r
{\displaystyle \gamma _{r}}
a
2
=
a
1
+
α
1
l
1
{\displaystyle a_{2}=a_{1}+\alpha _{1}l_{1}}
,
b
2
=
b
1
+
β
1
l
1
{\displaystyle b_{2}=b_{1}+\beta _{1}l_{1}}
,
c
2
=
c
1
+
γ
1
l
1
{\displaystyle c_{2}=c_{1}+\gamma _{1}l_{1}}
a
3
=
a
1
+
α
1
l
1
+
α
2
l
2
{\displaystyle a_{3}=a_{1}+\alpha _{1}l_{1}+\alpha _{2}l_{2}}
,
...
,
...
a
4
=
a
1
+
α
1
l
1
+
α
2
l
2
+
α
3
l
3
{\displaystyle a_{4}=a_{1}+\alpha _{1}l_{1}+\alpha _{2}l_{2}+\alpha _{3}l_{3}}
,
...
,
...
a
5
=
a
1
+
α
2
l
2
+
α
3
l
3
{\displaystyle a_{5}=a_{1}+\alpha _{2}l_{2}+\alpha _{3}l_{3}}
,
...
,
...
a
6
=
a
1
+
α
3
l
3
{\displaystyle a_{6}=a_{1}+\alpha _{3}l_{3}}
,
...
,
...
Par conséquent, l’équation du plan passant par les milieux des côtés (1, 2), (2, 3), (3, 4) sera
|
1
1
1
1
2
x
2
a
1
+
α
1
l
1
2
a
1
+
2
α
1
l
1
+
α
2
l
2
2
a
1
+
2
α
1
l
1
+
2
α
2
l
2
+
α
3
l
3
2
y
2
b
1
+
β
1
l
1
2
b
1
+
2
β
1
l
1
+
β
2
l
2
2
b
1
+
2
β
1
l
1
+
2
β
2
l
2
+
β
3
l
3
2
z
2
c
1
+
γ
1
l
1
2
c
1
+
2
γ
1
l
1
+
γ
2
l
2
2
c
1
+
2
γ
1
l
1
+
2
γ
2
l
2
+
γ
3
l
3
|
=
0
,
{\displaystyle {\begin{vmatrix}1&1&1&1\\2x&2a_{1}+\alpha _{1}l_{1}&2a_{1}+2\alpha _{1}l_{1}+\alpha _{2}l_{2}&2a_{1}+2\alpha _{1}l_{1}+2\alpha _{2}l_{2}+\alpha _{3}l_{3}\\2y&2b_{1}+\beta _{1}l_{1}&2b_{1}+2\beta _{1}l_{1}+\beta _{2}l_{2}&2b_{1}+2\beta _{1}l_{1}+2\beta _{2}l_{2}+\beta _{3}l_{3}\\2z&2c_{1}+\gamma _{1}l_{1}&2c_{1}+2\gamma _{1}l_{1}+\gamma _{2}l_{2}&2c_{1}+2\gamma _{1}l_{1}+2\gamma _{2}l_{2}+\gamma _{3}l_{3}\end{vmatrix}}=0,}
ou, en transformant ce déterminant par des théorèmes très-connus,
|
1
0
1
0
4
(
x
−
a
1
)
α
2
l
2
+
α
3
l
3
α
3
l
3
+
α
1
l
1
α
1
l
1
+
α
2
l
2
4
(
y
−
b
1
)
β
2
l
2
+
β
3
l
3
β
3
l
3
+
β
1
l
1
β
1
l
1
+
β
2
l
2
4
(
z
−
c
1
)
γ
2
l
2
+
γ
3
l
3
γ
3
l
3
+
γ
1
l
1
γ
1
l
1
+
γ
2
l
2
|
=
0.
{\displaystyle {\begin{vmatrix}1&0&1&0\\4(x-a_{1})&\alpha _{2}l_{2}+\alpha _{3}l_{3}&\alpha _{3}l_{3}+\alpha _{1}l_{1}&\alpha _{1}l_{1}+\alpha _{2}l_{2}\\4(y-b_{1})&\beta _{2}l_{2}+\beta _{3}l_{3}&\beta _{3}l_{3}+\beta _{1}l_{1}&\beta _{1}l_{1}+\beta _{2}l_{2}\\4(z-c_{1})&\gamma _{2}l_{2}+\gamma _{3}l_{3}&\gamma _{3}l_{3}+\gamma _{1}l_{1}&\gamma _{1}l_{1}+\gamma _{2}l_{2}\end{vmatrix}}=0.}
