En géométrie , le triangle heptagonal est le triangle , unique à similitude près, d'angles de mesures en radians π/7 , 2π/7 et 4π/7 , soit environ 26°, 51° et 103°. C'est l'unique triangle dont les angles sont dans des rapports 4:2:1.
L'heptagone régulier convexe (côtés rouges), avec ses diagonales longues (en vert) et courtes (en bleu). On compte quatorze triangles heptagonaux congruents constitués d'un côté vert, un bleu et un rouge. Triangle heptagonal. On l’obtient dans l'heptagone régulier convexe en partant d'un des sommets et en prenant les deuxième et quatrième sommets. Ses côtés sont donc constitués d'un côté de l'heptagone régulier, et deux de ses diagonales (une longue et une courte).
Comme le triangle d'or , dont les angles sont dans les rapports 2:2:1, le triangle heptagonal a de nombreuses propriétés remarquables.
Le triangle heptagonal et ses centres
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Le centre du cercle d'Euler du triangle heptagonal est aussi son premier point de Brocard [ 1] :Propos. 12 . Le second point de Brocard se trouve sur le cercle d'Euler[ 2] :p. 19 .
Le centre du cercle circonscrit et les points de Fermat du triangle heptagonal forment un triangle équilatéral [ 1] :Thm. 22 .
En notant R le rayon du cercle circonscrit et r le centre du cercle inscrit , on peut exprimer la distance entre le centre du cercle circonscrit O et l'orthocentre H par[ 2] :p. 19
O
H
=
R
2
,
{\displaystyle OH=R{\sqrt {2}},}
et la distance entre le centre du cercle circonscrit I à l'orthocentre par[ 2] :p. 19
I
H
2
=
R
2
+
4
r
2
2
.
{\displaystyle IH^{2}={\frac {R^{2}+4r^{2}}{2}}.}
Les deux tangentes au cercle circonscrit issues de l'orthocentre sont perpendiculaires [ 2] :p. 19 .
Relations entre les distances
modifier
Longueurs des côtés
modifier
Les côtés du triangle heptagonal a < b < c [ 3] :Lemma 1
a
2
=
c
(
c
−
b
)
,
b
2
=
a
(
c
+
a
)
,
c
2
=
b
(
a
+
b
)
,
1
a
=
1
b
+
1
c
{\displaystyle a^{2}=c(c-b),\ b^{2}=a(c+a),\ c^{2}=b(a+b),\ {\frac {1}{a}}={\frac {1}{b}}+{\frac {1}{c}}}
(la dernière est connue sous le nom d'équation optique (en) [ 2] :p. 13 ) et donc
a
b
+
a
c
=
b
c
,
{\displaystyle ab+ac=bc,}
et[ 3] :Coro. 2
b
3
+
2
b
2
c
−
b
c
2
−
c
3
=
0
,
{\displaystyle b^{3}+2b^{2}c-bc^{2}-c^{3}=0,}
c
3
−
2
c
2
a
−
c
a
2
+
a
3
=
0
,
{\displaystyle c^{3}-2c^{2}a-ca^{2}+a^{3}=0,}
a
3
−
2
a
2
b
−
a
b
2
+
b
3
=
0.
{\displaystyle a^{3}-2a^{2}b-ab^{2}+b^{3}=0.}
Ainsi, les rapports –b /c , c /a a /b équation cubique
t
3
−
2
t
2
−
t
+
1
=
0.
{\displaystyle t^{3}-2t^{2}-t+1=0.}
Il n'existe aucune expression algébrique réelle pour les solutions de cette équation, car c'est un exemple de casus irreducibilis
b
≈
1
,
80193
⋅
a
,
c
≈
2
,
24698
⋅
a
.
{\displaystyle b\approx 1,80193\cdot a,\qquad c\approx 2,24698\cdot a.}
On a aussi[ 4] , [ 5]
a
2
b
c
,
−
b
2
c
a
,
−
c
2
a
b
{\displaystyle {\frac {a^{2}}{bc}},\quad -{\frac {b^{2}}{ca}},\quad -{\frac {c^{2}}{ab}}}
qui vérifient l'équation cubique
t
3
+
4
t
2
+
3
t
−
1
=
0.
{\displaystyle t^{3}+4t^{2}+3t-1=0.}
On a [ 4]
a
3
b
c
2
,
−
b
3
c
a
2
,
c
3
a
b
2
{\displaystyle {\frac {a^{3}}{bc^{2}}},\quad -{\frac {b^{3}}{ca^{2}}},\quad {\frac {c^{3}}{ab^{2}}}}
qui vérifient l'équation cubique
t
3
−
t
2
−
9
t
+
1
=
0.
{\displaystyle t^{3}-t^{2}-9t+1=0.}
On a [ 4]
a
3
b
2
c
,
b
3
c
2
a
,
−
c
3
a
2
b
{\displaystyle {\frac {a^{3}}{b^{2}c}},\quad {\frac {b^{3}}{c^{2}a}},\quad -{\frac {c^{3}}{a^{2}b}}}
qui vérifient l'équation cubique
t
3
+
5
t
2
−
8
t
+
1
=
0.
{\displaystyle t^{3}+5t^{2}-8t+1=0.}
On a [ 2] :p. 14
b
2
−
a
2
=
a
c
,
c
2
−
b
2
=
a
b
,
a
2
−
c
2
=
−
b
c
,
{\displaystyle b^{2}-a^{2}=ac,\ c^{2}-b^{2}=ab,\ a^{2}-c^{2}=-bc,}
et[ 2] :p. 15
b
2
a
2
+
c
2
b
2
+
a
2
c
2
=
5.
{\displaystyle {\frac {b^{2}}{a^{2}}}+{\frac {c^{2}}{b^{2}}}+{\frac {a^{2}}{c^{2}}}=5.}
On a aussi[ 4]
a
b
−
b
c
+
c
a
=
0
,
{\displaystyle ab-bc+ca=0,}
a
3
b
−
b
3
c
+
c
3
a
=
0
,
{\displaystyle a^{3}b-b^{3}c+c^{3}a=0,}
a
4
b
+
b
4
c
−
c
4
a
=
0
,
{\displaystyle a^{4}b+b^{4}c-c^{4}a=0,}
a
11
b
3
−
b
11
c
3
+
c
11
a
3
=
0.
{\displaystyle a^{11}b^{3}-b^{11}c^{3}+c^{11}a^{3}=0.}
Il n'existe aucun autre couple d'entiers strictement positifs (m , n ), m , n > 0, m, n < 2000 tel que [réf. nécessaire]
a
m
b
n
±
b
m
c
n
±
c
m
a
n
=
0.
{\displaystyle a^{m}b^{n}\pm b^{m}c^{n}\pm c^{m}a^{n}=0.}
Les hauteurs ha , hb et hc vérifient[ 2] :p. 13-14
h
a
=
h
b
+
h
c
{\displaystyle h_{a}=h_{b}+h_{c}}
et
h
a
2
+
h
b
2
+
h
c
2
=
a
2
+
b
2
+
c
2
2
{\displaystyle h_{a}^{2}+h_{b}^{2}+h_{c}^{2}={\frac {a^{2}+b^{2}+c^{2}}{2}}}
[ 2] :p. 14 .La hauteur pour le côté b (d'angle opposé B ) est la moitié de la bissectrice interne wA de A [ 2] :p. 19 :
2
h
b
=
w
A
.
{\displaystyle 2h_{b}=w_{A}.}
Ici, l'angle A est le plus petit angle, et B le second plus petit.
Bissectrices internes
modifier
Les longueurs des bissectrices internes wA , wB et wC (bissectrices des angles A , B et C respectivement) vérifient[ 2] :p. 16 :
w
A
=
b
+
c
,
w
B
=
c
−
a
,
w
C
=
b
−
a
.
{\displaystyle w_{A}=b+c,\ w_{B}=c-a,w_{C}=b-a.}
Rayons des cercles circonscrit, inscrit et exinscrits
modifier
On note R le rayon du cercle circonscrit au triangle heptagonal. Son aire vaut alors[ 6]
A
=
7
4
R
2
.
{\displaystyle A={\frac {\sqrt {7}}{4}}R^{2}.}
On a aussi[ 2] :p. 12,15 , [ 7]
a
2
+
b
2
+
c
2
=
7
R
2
,
a
4
+
b
4
+
c
4
=
21
R
4
,
a
6
+
b
6
+
c
6
=
70
R
6
,
{\displaystyle a^{2}+b^{2}+c^{2}=7R^{2},\quad a^{4}+b^{4}+c^{4}=21R^{4},\quad a^{6}+b^{6}+c^{6}=70R^{6},}
1
a
2
+
1
b
2
+
1
c
2
=
2
R
2
,
1
a
4
+
1
b
4
+
1
c
4
=
2
R
4
,
1
a
6
+
1
b
6
+
1
c
6
=
17
7
R
6
.
{\displaystyle {\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}+{\frac {1}{c^{2}}}={\frac {2}{R^{2}}},\quad {\frac {1}{a^{4}}}+{\frac {1}{b^{4}}}+{\frac {1}{c^{4}}}={\frac {2}{R^{4}}},\quad {\frac {1}{a^{6}}}+{\frac {1}{b^{6}}}+{\frac {1}{c^{6}}}={\frac {17}{7R^{6}}}.}
De façon générale, pour tout entier n ,
a
2
n
+
b
2
n
+
c
2
n
=
g
(
n
)
(
2
R
)
2
n
{\displaystyle a^{2n}+b^{2n}+c^{2n}=g(n)(2R)^{2n}}
avec
g
(
−
1
)
=
8
,
g
(
0
)
=
3
,
g
(
1
)
=
7
{\displaystyle g(-1)=8,\quad g(0)=3,\quad g(1)=7}
et
g
(
n
)
=
7
g
(
n
−
1
)
−
14
g
(
n
−
2
)
+
7
g
(
n
−
3
)
,
{\displaystyle g(n)=7g(n-1)-14g(n-2)+7g(n-3),}
on a[ 7]
2
b
2
−
a
2
=
7
b
R
,
2
c
2
−
b
2
=
7
c
R
,
2
a
2
−
c
2
=
−
7
a
R
.
{\displaystyle 2b^{2}-a^{2}={\sqrt {7}}bR,\quad 2c^{2}-b^{2}={\sqrt {7}}cR,\quad 2a^{2}-c^{2}=-{\sqrt {7}}aR.}
On a aussi[ 4]
a
3
c
+
b
3
a
−
c
3
b
=
−
7
R
4
,
{\displaystyle a^{3}c+b^{3}a-c^{3}b=-7R^{4},}
a
4
c
−
b
4
a
+
c
4
b
=
7
7
R
5
,
{\displaystyle a^{4}c-b^{4}a+c^{4}b=7{\sqrt {7}}R^{5},}
a
11
c
3
+
b
11
a
3
−
c
11
b
3
=
−
7
3
17
R
14
.
{\displaystyle a^{11}c^{3}+b^{11}a^{3}-c^{11}b^{3}=-7^{3}17R^{14}.}
Le rapport r /R [ 6]
8
x
3
+
28
x
2
+
14
x
−
7
=
0.
{\displaystyle 8x^{3}+28x^{2}+14x-7=0.}
Le rayon du cercle exinscrit au côté a est égal au rayon du cercle d'Euler du triangle heptagonal[ 2] :p. 15 .
Triangle orthique
modifier
Les nombreuses identités trigonométriques associées au triangle heptagonal incluent[ 2] :pp. 13–14 , [ 6]
A
=
π
7
,
B
=
2
π
7
=
2
A
,
C
=
4
π
7
=
4
A
=
2
B
{\displaystyle A={\frac {\pi }{7}},\quad B={\frac {2\pi }{7}}=2A,\quad C={\frac {4\pi }{7}}=4A=2B}
cos
A
=
b
2
a
,
cos
B
=
c
2
b
,
cos
C
=
−
a
2
c
,
{\displaystyle \cos A={\frac {b}{2a}},\quad \cos B={\frac {c}{2b}},\quad \cos C=-{\frac {a}{2c}},}
[ 4] :Proposition 10
cos
A
cos
B
cos
C
=
−
1
8
.
{\displaystyle \cos A\cos B\cos C=-{\frac {1}{8}}.}
Démonstration
On applique la loi des sinus au triangle heptagonal :
a
sin
(
A
)
=
b
sin
(
B
)
=
c
sin
(
C
)
=
2
R
{\displaystyle {\frac {a}{\sin(A)}}={\frac {b}{\sin(B)}}={\frac {c}{\sin(C)}}=2R}
D'autre part, la valeur des angles donne :
sin
(
4
A
)
=
sin
(
3
A
)
⟹
cos
2
(
A
)
−
cos
(
A
)
cos
(
2
A
)
=
1
4
,
sin
(
A
)
=
sin
(
6
A
)
,
cos
(
C
)
=
−
cos
(
3
A
)
.
{\displaystyle \sin(4A)=\sin(3A)\Longrightarrow \cos ^{2}(A)-\cos(A)\cos(2A)={\frac {1}{4}},\sin(A)=\sin(6A),\cos(C)=-\cos(3A).}
On a donc :
b
a
=
sin
(
2
B
)
sin
(
A
)
=
2
cos
(
A
)
,
c
b
=
sin
(
2
B
)
sin
(
B
)
=
2
cos
(
B
)
,
a
c
=
sin
(
A
)
sin
(
C
)
=
sin
(
6
A
)
sin
(
3
A
)
=
2
cos
(
3
A
)
=
−
2
cos
(
C
)
.
{\displaystyle {\frac {b}{a}}={\frac {\sin(2B)}{\sin(A)}}=2\cos(A),\ {\frac {c}{b}}={\frac {\sin(2B)}{\sin(B)}}=2\cos(B),\quad {\frac {a}{c}}={\frac {\sin(A)}{\sin(C)}}={\frac {\sin(6A)}{\sin(3A)}}=2\cos(3A)=-2\cos(C).}
Le produit de ces trois identités donne :
cos
A
cos
B
cos
C
=
b
2
a
×
c
2
b
×
(
−
a
c
)
=
−
1
8
.
{\displaystyle \cos A\cos B\cos C={\frac {b}{2a}}\times {\frac {c}{2b}}\times \left(-{\frac {a}{c}}\right)=-{\frac {1}{8}}.}
Par différentes méthodes (comme l'utilisation judicieuse de la formule de Moivre ), on peut trouver les égalités suivantes :
cos
2
A
+
cos
2
B
+
cos
2
C
=
5
4
,
{\displaystyle \cos ^{2}A+\cos ^{2}B+\cos ^{2}C={\frac {5}{4}},}
cos
4
A
+
cos
4
B
+
cos
4
C
=
13
16
,
{\displaystyle \cos ^{4}A+\cos ^{4}B+\cos ^{4}C={\frac {13}{16}},}
cot
A
+
cot
B
+
cot
C
=
7
,
{\displaystyle \cot A+\cot B+\cot C={\sqrt {7}},}
cot
2
A
+
cot
2
B
+
cot
2
C
=
5
,
{\displaystyle \cot ^{2}A+\cot ^{2}B+\cot ^{2}C=5,}
csc
2
A
+
csc
2
B
+
csc
2
C
=
8
,
{\displaystyle \csc ^{2}A+\csc ^{2}B+\csc ^{2}C=8,}
csc
4
A
+
csc
4
B
+
csc
4
C
=
32
,
{\displaystyle \csc ^{4}A+\csc ^{4}B+\csc ^{4}C=32,}
sec
2
A
+
sec
2
B
+
sec
2
C
=
24
,
{\displaystyle \sec ^{2}A+\sec ^{2}B+\sec ^{2}C=24,}
sec
4
A
+
sec
4
B
+
sec
4
C
=
416
,
{\displaystyle \sec ^{4}A+\sec ^{4}B+\sec ^{4}C=416,}
sin
A
sin
B
sin
C
=
7
8
,
{\displaystyle \sin A\sin B\sin C={\frac {\sqrt {7}}{8}},}
sin
2
A
sin
2
B
sin
2
C
=
7
64
,
{\displaystyle \sin ^{2}A\sin ^{2}B\sin ^{2}C={\frac {7}{64}},}
sin
2
A
+
sin
2
B
+
sin
2
C
=
7
4
,
{\displaystyle \sin ^{2}A+\sin ^{2}B+\sin ^{2}C={\frac {7}{4}},}
sin
4
A
+
sin
4
B
+
sin
4
C
=
21
16
,
{\displaystyle \sin ^{4}A+\sin ^{4}B+\sin ^{4}C={\frac {21}{16}},}
tan
A
tan
B
tan
C
=
tan
A
+
tan
B
+
tan
C
=
−
7
,
{\displaystyle \tan A\tan B\tan C=\tan A+\tan B+\tan C=-{\sqrt {7}},}
tan
2
A
+
tan
2
B
+
tan
2
C
=
21.
{\displaystyle \tan ^{2}A+\tan ^{2}B+\tan ^{2}C=21.}
La racine positive de l'équation cubique[ 8] :p. 186–187
x
3
+
x
2
−
2
x
−
1
=
0
{\displaystyle x^{3}+x^{2}-2x-1=0}
est égale à
2
cos
2
π
7
.
{\displaystyle 2\cos {\frac {2\pi }{7}}.}
Les racines de l'équation cubique[ 4]
x
3
−
7
2
x
2
+
7
8
=
0
{\displaystyle x^{3}-{\frac {\sqrt {7}}{2}}x^{2}+{\frac {\sqrt {7}}{8}}=0}
sont
sin
(
2
π
7
)
,
sin
(
4
π
7
)
,
sin
(
8
π
7
)
.
{\displaystyle \sin \left({\frac {2\pi }{7}}\right),\sin \left({\frac {4\pi }{7}}\right),\sin \left({\frac {8\pi }{7}}\right).}
Les racines de l'équation cubique[ 2] :p. 14
64
y
3
−
112
y
2
+
56
y
−
7
=
0
{\displaystyle 64y^{3}-112y^{2}+56y-7=0}
sont
sin
2
(
π
7
)
,
sin
2
(
2
π
7
)
,
sin
2
(
4
π
7
)
.
{\displaystyle \sin ^{2}\left({\frac {\pi }{7}}\right),\sin ^{2}\left({\frac {2\pi }{7}}\right),\sin ^{2}\left({\frac {4\pi }{7}}\right).}
On a aussi[ 7]
sin
A
−
sin
B
−
sin
C
=
−
7
2
,
{\displaystyle \sin A-\sin B-\sin C=-{\frac {\sqrt {7}}{2}},}
sin
A
sin
B
−
sin
B
sin
C
+
sin
C
sin
A
=
0
,
{\displaystyle \sin A\sin B-\sin B\sin C+\sin C\sin A=0,}
sin
A
sin
B
sin
C
=
7
8
.
{\displaystyle \sin A\sin B\sin C={\frac {\sqrt {7}}{8}}.}
Pour un entier n , on pose S (n ) = (–sin A )n n B + sinn C
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
S (n )
3
7
2
{\displaystyle {\tfrac {\sqrt {7}}{2}}}
7
2
2
{\displaystyle {\tfrac {7}{2^{2}}}}
7
2
{\displaystyle {\tfrac {\sqrt {7}}{2}}}
7
⋅
3
2
4
{\displaystyle {\tfrac {7\cdot 3}{2^{4}}}}
7
7
2
4
{\displaystyle {\tfrac {7{\sqrt {7}}}{2^{4}}}}
7
⋅
5
2
5
{\displaystyle {\tfrac {7\cdot 5}{2^{5}}}}
7
2
7
2
7
{\displaystyle {\tfrac {7^{2}{\sqrt {7}}}{2^{7}}}}
7
2
⋅
5
2
8
{\displaystyle {\tfrac {7^{2}\cdot 5}{2^{8}}}}
7
⋅
25
7
2
9
{\displaystyle {\tfrac {7\cdot 25{\sqrt {7}}}{2^{9}}}}
7
2
⋅
9
2
9
{\displaystyle {\tfrac {7^{2}\cdot 9}{2^{9}}}}
7
2
⋅
13
7
2
11
{\displaystyle {\tfrac {7^{2}\cdot 13{\sqrt {7}}}{2^{11}}}}
7
2
⋅
33
2
11
{\displaystyle {\tfrac {7^{2}\cdot 33}{2^{11}}}}
7
2
⋅
3
7
2
9
{\displaystyle {\tfrac {7^{2}\cdot 3{\sqrt {7}}}{2^{9}}}}
7
4
⋅
5
2
14
{\displaystyle {\tfrac {7^{4}\cdot 5}{2^{14}}}}
7
2
⋅
179
7
2
15
{\displaystyle {\tfrac {7^{2}\cdot 179{\sqrt {7}}}{2^{15}}}}
7
3
⋅
131
2
16
{\displaystyle {\tfrac {7^{3}\cdot 131}{2^{16}}}}
7
3
⋅
3
7
2
12
{\displaystyle {\tfrac {7^{3}\cdot 3{\sqrt {7}}}{2^{12}}}}
7
3
⋅
493
2
18
{\displaystyle {\tfrac {7^{3}\cdot 493}{2^{18}}}}
7
3
⋅
181
7
2
18
{\displaystyle {\tfrac {7^{3}\cdot 181{\sqrt {7}}}{2^{18}}}}
7
5
⋅
19
2
19
{\displaystyle {\tfrac {7^{5}\cdot 19}{2^{19}}}}
S (-n )
3
0
23
−
2
3
⋅
3
7
7
{\displaystyle -{\tfrac {2^{3}\cdot 3{\sqrt {7}}}{7}}}
25
−
2
5
⋅
5
7
7
{\displaystyle -{\tfrac {2^{5}\cdot 5{\sqrt {7}}}{7}}}
2
6
⋅
17
7
{\displaystyle {\tfrac {2^{6}\cdot 17}{7}}}
−
2
7
7
{\displaystyle {\scriptstyle -2^{7}{\sqrt {7}}}}
2
9
⋅
11
7
{\displaystyle {\tfrac {2^{9}\cdot 11}{7}}}
−
2
10
⋅
33
7
7
2
{\displaystyle -{\tfrac {2^{10}\cdot 33{\sqrt {7}}}{7^{2}}}}
2
10
⋅
29
7
{\displaystyle {\tfrac {2^{10}\cdot 29}{7}}}
−
2
14
⋅
11
7
7
2
{\displaystyle -{\tfrac {2^{14}\cdot 11{\sqrt {7}}}{7^{2}}}}
2
12
⋅
269
7
2
{\displaystyle {\tfrac {2^{12}\cdot 269}{7^{2}}}}
−
2
13
⋅
117
7
7
2
{\displaystyle -{\tfrac {2^{13}\cdot 117{\sqrt {7}}}{7^{2}}}}
2
14
⋅
51
7
{\displaystyle {\tfrac {2^{14}\cdot 51}{7}}}
−
2
21
⋅
17
7
7
3
{\displaystyle -{\tfrac {2^{21}\cdot 17{\sqrt {7}}}{7^{3}}}}
2
17
⋅
237
7
2
{\displaystyle {\tfrac {2^{17}\cdot 237}{7^{2}}}}
−
2
17
⋅
1445
7
7
3
{\displaystyle -{\tfrac {2^{17}\cdot 1445{\sqrt {7}}}{7^{3}}}}
2
19
⋅
2203
7
3
{\displaystyle {\tfrac {2^{19}\cdot 2203}{7^{3}}}}
−
2
19
⋅
1919
7
7
3
{\displaystyle -{\tfrac {2^{19}\cdot 1919{\sqrt {7}}}{7^{3}}}}
2
20
⋅
5851
7
3
{\displaystyle {\tfrac {2^{20}\cdot 5851}{7^{3}}}}
Avec les cosinus
modifier
Les cosinus aux angles cos A , cos B , cos C sont les racines de l'équation cubique :
x
3
+
1
2
x
2
−
1
2
x
−
1
8
=
0.
{\displaystyle x^{3}+{\frac {1}{2}}x^{2}-{\frac {1}{2}}x-{\frac {1}{8}}=0.}
Pour un entier n , on pose C (n ) = (-cos A )n n B + cosn C
n
0
1
2
3
4
5
6
7
8
9
10
C (n )
3
−
1
2
{\displaystyle -{\frac {1}{2}}}
5
4
{\displaystyle {\frac {5}{4}}}
−
1
2
{\displaystyle -{\frac {1}{2}}}
13
16
{\displaystyle {\frac {13}{16}}}
−
1
2
{\displaystyle -{\frac {1}{2}}}
19
32
{\displaystyle {\frac {19}{32}}}
−
57
128
{\displaystyle -{\frac {57}{128}}}
117
256
{\displaystyle {\frac {117}{256}}}
−
193
512
{\displaystyle -{\frac {193}{512}}}
185
512
{\displaystyle {\frac {185}{512}}}
C (-n )
3
-4
24
-88
416
-1824
8256
-36992
166400
-747520
3359744
Avec les tangentes
modifier
Les tangentes aux angles tan A , tan B , tan C sont les racines de l'équation cubique :
x
3
+
7
x
2
−
7
x
+
7
=
0.
{\displaystyle x^{3}+{\sqrt {7}}x^{2}-7x+{\sqrt {7}}=0.}
Les carrés des tangentes aux angles tan2 A , tan2 B , tan2 C sont les racines de l'équation cubique :
x
3
−
21
x
2
+
35
x
−
7
=
0.
{\displaystyle x^{3}-21x^{2}+35x-7=0.}
Pour un entier n , on pose T (n ) = tann A + tann B + tann C
n
0
1
2
3
4
5
6
7
8
9
10
T (n )
3
−
7
{\displaystyle -{\sqrt {7}}}
7
⋅
3
{\displaystyle 7\cdot 3}
−
31
7
{\displaystyle -31{\sqrt {7}}}
7
⋅
53
{\displaystyle 7\cdot 53}
−
7
⋅
87
7
{\displaystyle -7\cdot 87{\sqrt {7}}}
7
⋅
1011
{\displaystyle 7\cdot 1011}
−
7
2
⋅
239
7
{\displaystyle -7^{2}\cdot 239{\sqrt {7}}}
7
2
⋅
2771
{\displaystyle 7^{2}\cdot 2771}
−
7
⋅
32119
7
{\displaystyle -7\cdot 32119{\sqrt {7}}}
7
2
⋅
53189
{\displaystyle 7^{2}\cdot 53189}
T (-n )
3
7
{\displaystyle {\sqrt {7}}}
5
25
7
7
{\displaystyle {\frac {25{\sqrt {7}}}{7}}}
19
103
7
7
{\displaystyle {\frac {103{\sqrt {7}}}{7}}}
563
7
{\displaystyle {\frac {563}{7}}}
7
⋅
9
7
{\displaystyle 7\cdot 9{\sqrt {7}}}
2421
7
{\displaystyle {\frac {2421}{7}}}
13297
7
7
2
{\displaystyle {\frac {13297{\sqrt {7}}}{7^{2}}}}
10435
7
{\displaystyle {\frac {10435}{7}}}
On a aussi[ 7] , [ 9]
tan
A
−
4
sin
B
=
−
7
,
{\displaystyle \tan A-4\sin B=-{\sqrt {7}},}
tan
B
−
4
sin
C
=
−
7
,
{\displaystyle \tan B-4\sin C=-{\sqrt {7}},}
tan
C
+
4
sin
A
=
−
7
.
{\displaystyle \tan C+4\sin A=-{\sqrt {7}}.}
On a aussi[ 4]
cot
2
A
=
1
−
2
tan
C
7
,
{\displaystyle \cot ^{2}A=1-{\frac {2\tan C}{\sqrt {7}}},}
cot
2
B
=
1
−
2
tan
A
7
,
{\displaystyle \cot ^{2}B=1-{\frac {2\tan A}{\sqrt {7}}},}
cot
2
C
=
1
−
2
tan
B
7
.
{\displaystyle \cot ^{2}C=1-{\frac {2\tan B}{\sqrt {7}}}.}
cos
A
=
−
1
2
+
4
7
sin
3
C
,
{\displaystyle \cos A=-{\frac {1}{2}}+{\frac {4}{\sqrt {7}}}\sin ^{3}C,}
cos
2
A
=
3
4
+
2
7
sin
3
A
,
{\displaystyle \cos ^{2}A={\frac {3}{4}}+{\frac {2}{\sqrt {7}}}\sin ^{3}A,}
cot
A
=
3
7
+
4
7
cos
B
,
{\displaystyle \cot A={\frac {3}{\sqrt {7}}}+{\frac {4}{\sqrt {7}}}\cos B,}
cot
2
A
=
3
+
8
7
sin
A
,
{\displaystyle \cot ^{2}A=3+{\frac {8}{\sqrt {7}}}\sin A,}
cot
A
=
7
+
8
7
sin
2
B
,
{\displaystyle \cot A={\sqrt {7}}+{\frac {8}{\sqrt {7}}}\sin ^{2}B,}
csc
3
A
=
−
6
7
+
2
7
tan
2
C
,
{\displaystyle \csc ^{3}A=-{\frac {6}{\sqrt {7}}}+{\frac {2}{\sqrt {7}}}\tan ^{2}C,}
sec
A
=
2
+
4
cos
C
,
{\displaystyle \sec A=2+4\cos C,}
sec
A
=
6
−
8
sin
2
B
,
{\displaystyle \sec A=6-8\sin ^{2}B,}
sec
A
=
4
−
16
7
sin
3
B
,
{\displaystyle \sec A=4-{\frac {16}{\sqrt {7}}}\sin ^{3}B,}
sin
2
A
=
1
2
+
1
2
cos
B
,
{\displaystyle \sin ^{2}A={\frac {1}{2}}+{\frac {1}{2}}\cos B,}
sin
3
A
=
−
7
8
+
7
4
cos
B
,
{\displaystyle \sin ^{3}A=-{\frac {\sqrt {7}}{8}}+{\frac {\sqrt {7}}{4}}\cos B,}
On a aussi[ 10]
sin
3
B
sin
C
−
sin
3
C
sin
A
−
sin
3
A
sin
B
=
0
,
{\displaystyle \sin ^{3}B\sin C-\sin ^{3}C\sin A-\sin ^{3}A\sin B=0,}
sin
B
sin
3
C
−
sin
C
sin
3
A
−
sin
A
sin
3
B
=
7
2
4
,
{\displaystyle \sin B\sin ^{3}C-\sin C\sin ^{3}A-\sin A\sin ^{3}B={\frac {7}{2^{4}}},}
sin
4
B
sin
C
−
sin
4
C
sin
A
+
sin
4
A
sin
B
=
0
,
{\displaystyle \sin ^{4}B\sin C-\sin ^{4}C\sin A+\sin ^{4}A\sin B=0,}
sin
B
sin
4
C
+
sin
C
sin
4
A
−
sin
A
sin
4
B
=
7
7
2
5
,
{\displaystyle \sin B\sin ^{4}C+\sin C\sin ^{4}A-\sin A\sin ^{4}B={\frac {7{\sqrt {7}}}{2^{5}}},}
sin
11
B
sin
3
C
−
sin
11
C
sin
3
A
−
sin
11
A
sin
3
B
=
0
,
{\displaystyle \sin ^{11}B\sin ^{3}C-\sin ^{11}C\sin ^{3}A-\sin ^{11}A\sin ^{3}B=0,}
sin
3
B
sin
11
C
−
sin
3
C
sin
11
A
−
sin
3
A
sin
11
B
=
7
3
⋅
17
2
14
.
{\displaystyle \sin ^{3}B\sin ^{11}C-\sin ^{3}C\sin ^{11}A-\sin ^{3}A\sin ^{11}B={\frac {7^{3}\cdot 17}{2^{14}}}.}
Identités de type Ramanujan
modifier
On peut également obtenir des identités similaires à celles découvertes par Srinivasa Ramanujan [ 7] , [ 11]
2
sin
(
2
π
7
)
3
+
2
sin
(
4
π
7
)
3
+
2
sin
(
8
π
7
)
3
=
(
−
7
18
)
−
7
3
+
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
{\displaystyle {\sqrt[{3}]{2\sin \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{2\sin \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{2\sin \left({\frac {8\pi }{7}}\right)}}=\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}
1
2
sin
(
2
π
7
)
3
+
1
2
sin
(
4
π
7
)
3
+
1
2
sin
(
8
π
7
)
3
=
(
−
1
7
18
)
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
{\displaystyle {\frac {1}{\sqrt[{3}]{2\sin \left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{2\sin \left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{2\sin \left({\frac {8\pi }{7}}\right)}}}=\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}
4
sin
2
(
2
π
7
)
3
+
4
sin
2
(
4
π
7
)
3
+
4
sin
2
(
8
π
7
)
3
=
(
49
18
)
49
3
+
6
+
3
(
12
+
3
(
49
3
+
2
7
3
)
3
+
11
+
3
(
49
3
+
2
7
3
)
3
)
3
{\displaystyle {\sqrt[{3}]{4\sin ^{2}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{4\sin ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{4\sin ^{2}\left({\frac {8\pi }{7}}\right)}}=\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}}
1
4
sin
2
(
2
π
7
)
3
+
1
4
sin
2
(
4
π
7
)
3
+
1
4
sin
2
(
8
π
7
)
3
=
(
1
49
18
)
2
7
3
+
6
+
3
(
12
+
3
(
49
3
+
2
7
3
)
3
+
11
+
3
(
49
3
+
2
7
3
)
3
)
3
{\displaystyle {\frac {1}{\sqrt[{3}]{4\sin ^{2}\left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{4\sin ^{2}\left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{4\sin ^{2}\left({\frac {8\pi }{7}}\right)}}}=\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{2{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}}
2
cos
(
2
π
7
)
3
+
2
cos
(
4
π
7
)
3
+
2
cos
(
8
π
7
)
3
=
5
−
3
7
3
3
{\displaystyle {\sqrt[{3}]{2\cos \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{2\cos \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{2\cos \left({\frac {8\pi }{7}}\right)}}={\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}}
1
2
cos
(
2
π
7
)
3
+
1
2
cos
(
4
π
7
)
3
+
1
2
cos
(
8
π
7
)
3
=
4
−
3
7
3
3
{\displaystyle {\frac {1}{\sqrt[{3}]{2\cos \left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{2\cos \left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{2\cos \left({\frac {8\pi }{7}}\right)}}}={\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}}
4
cos
2
(
2
π
7
)
3
+
4
cos
2
(
4
π
7
)
3
+
4
cos
2
(
8
π
7
)
3
=
11
+
3
(
2
7
3
+
49
3
)
3
{\displaystyle {\sqrt[{3}]{4\cos ^{2}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{4\cos ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{4\cos ^{2}\left({\frac {8\pi }{7}}\right)}}={\sqrt[{3}]{11+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}}
1
4
cos
2
(
2
π
7
)
3
+
1
4
cos
2
(
4
π
7
)
3
+
1
4
cos
2
(
8
π
7
)
3
=
12
+
3
(
2
7
3
+
49
3
)
3
{\displaystyle {\frac {1}{\sqrt[{3}]{4\cos ^{2}\left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}\left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}\left({\frac {8\pi }{7}}\right)}}}={\sqrt[{3}]{12+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}}
tan
(
2
π
7
)
3
+
tan
(
4
π
7
)
3
+
tan
(
8
π
7
)
3
=
(
−
7
18
)
7
3
+
6
+
3
(
5
+
3
(
7
3
−
49
3
)
3
+
−
3
+
3
(
7
3
−
49
3
)
3
)
3
{\displaystyle {\sqrt[{3}]{\tan \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\tan \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\tan \left({\frac {8\pi }{7}}\right)}}=\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}}
1
tan
(
2
π
7
)
3
+
1
tan
(
4
π
7
)
3
+
1
tan
(
8
π
7
)
3
=
(
−
1
7
18
)
−
49
3
+
6
+
3
(
5
+
3
(
7
3
−
49
3
)
3
+
−
3
+
3
(
7
3
−
49
3
)
3
)
3
{\displaystyle {\frac {1}{\sqrt[{3}]{\tan \left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{\tan \left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{\tan \left({\frac {8\pi }{7}}\right)}}}=\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{-{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}}
tan
2
(
2
π
7
)
3
+
tan
2
(
4
π
7
)
3
+
tan
2
(
8
π
7
)
3
=
(
49
18
)
3
49
3
+
6
+
3
(
89
+
3
(
3
49
3
+
5
7
3
)
3
+
25
+
3
(
3
49
3
+
5
7
3
)
3
)
3
{\displaystyle {\sqrt[{3}]{\tan ^{2}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\tan ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\tan ^{2}\left({\frac {8\pi }{7}}\right)}}=\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{3{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}}
1
tan
2
(
2
π
7
)
3
+
1
tan
2
(
4
π
7
)
3
+
1
tan
2
(
8
π
7
)
3
=
(
1
49
18
)
5
7
3
+
6
+
3
(
89
+
3
(
3
49
3
+
5
7
3
)
3
+
25
+
3
(
3
49
3
+
5
7
3
)
3
)
3
{\displaystyle {\frac {1}{\sqrt[{3}]{\tan ^{2}\left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}\left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}\left({\frac {8\pi }{7}}\right)}}}=\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{5{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}}
On a aussi[ 10]
cos
(
2
π
7
)
/
cos
(
4
π
7
)
3
+
cos
(
4
π
7
)
/
cos
(
8
π
7
)
3
+
cos
(
8
π
7
)
/
cos
(
2
π
7
)
3
=
−
7
3
.
{\displaystyle {\sqrt[{3}]{\cos \left({\frac {2\pi }{7}}\right)/\cos \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos \left({\frac {4\pi }{7}}\right)/\cos \left({\frac {8\pi }{7}}\right)}}+{\sqrt[{3}]{\cos \left({\frac {8\pi }{7}}\right)/\cos \left({\frac {2\pi }{7}}\right)}}=-{\sqrt[{3}]{7}}.}
cos
(
4
π
7
)
/
cos
(
2
π
7
)
3
+
cos
(
8
π
7
)
/
cos
(
4
π
7
)
3
+
cos
(
2
π
7
)
/
cos
(
8
π
7
)
3
=
0.
{\displaystyle {\sqrt[{3}]{\cos \left({\frac {4\pi }{7}}\right)/\cos \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\cos \left({\frac {8\pi }{7}}\right)/\cos \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos \left({\frac {2\pi }{7}}\right)/\cos \left({\frac {8\pi }{7}}\right)}}=0.}
2
sin
(
2
π
7
)
3
+
2
sin
(
4
π
7
)
3
+
2
sin
(
8
π
7
)
3
=
(
−
7
18
)
−
7
3
+
6
+
3
(
5
−
3
7
3
3
+
4
−
3
7
3
3
)
3
{\displaystyle {\sqrt[{3}]{2\sin \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{2\sin \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{2\sin \left({\frac {8\pi }{7}}\right)}}=\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}
cos
4
(
4
π
7
)
/
cos
(
2
π
7
)
3
+
cos
4
(
8
π
7
)
/
cos
(
4
π
7
)
3
+
cos
4
(
2
π
7
)
/
cos
(
8
π
7
)
3
=
−
49
3
/
2.
{\displaystyle {\sqrt[{3}]{\cos ^{4}\left({\frac {4\pi }{7}}\right)/\cos \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{4}\left({\frac {8\pi }{7}}\right)/\cos \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{4}\left({\frac {2\pi }{7}}\right)/\cos \left({\frac {8\pi }{7}}\right)}}=-{\sqrt[{3}]{49}}/2.}
cos
5
(
2
π
7
)
/
cos
2
(
4
π
7
)
3
+
cos
5
(
4
π
7
)
/
cos
2
(
8
π
7
)
3
+
cos
5
(
8
π
7
)
/
cos
2
(
2
π
7
)
3
=
0.
{\displaystyle {\sqrt[{3}]{\cos ^{5}\left({\frac {2\pi }{7}}\right)/\cos ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{5}\left({\frac {4\pi }{7}}\right)/\cos ^{2}\left({\frac {8\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{5}\left({\frac {8\pi }{7}}\right)/\cos ^{2}\left({\frac {2\pi }{7}}\right)}}=0.}
cos
5
(
4
π
7
)
/
cos
2
(
2
π
7
)
3
+
cos
5
(
8
π
7
)
/
cos
2
(
4
π
7
)
3
+
cos
5
(
2
π
7
)
/
cos
2
(
8
π
7
)
3
=
−
3
7
3
/
2.
{\displaystyle {\sqrt[{3}]{\cos ^{5}\left({\frac {4\pi }{7}}\right)/\cos ^{2}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{5}\left({\frac {8\pi }{7}}\right)/\cos ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{5}\left({\frac {2\pi }{7}}\right)/\cos ^{2}\left({\frac {8\pi }{7}}\right)}}=-3{\sqrt[{3}]{7}}/2.}
cos
14
(
2
π
7
)
/
cos
5
(
4
π
7
)
3
+
cos
14
(
4
π
7
)
/
cos
5
(
8
π
7
)
3
+
cos
14
(
8
π
7
)
/
cos
5
(
2
π
7
)
3
=
0.
{\displaystyle {\sqrt[{3}]{\cos ^{14}\left({\frac {2\pi }{7}}\right)/\cos ^{5}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{14}\left({\frac {4\pi }{7}}\right)/\cos ^{5}\left({\frac {8\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{14}\left({\frac {8\pi }{7}}\right)/\cos ^{5}\left({\frac {2\pi }{7}}\right)}}=0.}
cos
14
(
4
π
7
)
/
cos
5
(
2
π
7
)
3
+
cos
14
(
8
π
7
)
/
cos
5
(
4
π
7
)
3
+
cos
14
(
2
π
7
)
/
cos
5
(
8
π
7
)
3
=
−
61
7
3
/
8.
{\displaystyle {\sqrt[{3}]{\cos ^{14}\left({\frac {4\pi }{7}}\right)/\cos ^{5}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{14}\left({\frac {8\pi }{7}}\right)/\cos ^{5}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{14}\left({\frac {2\pi }{7}}\right)/\cos ^{5}\left({\frac {8\pi }{7}}\right)}}=-61{\sqrt[{3}]{7}}/8.}
Notes et références
modifier
↑ a et b (en) Paul Yiu, « Heptagonal Triangles and Their Companions », Forum Geometricorum vol. 9, 2009 , p. 125–148 (lire en ligne ) ↑ a b c d e f g h i j k l m n o et p (en) Leon Bankoff et Jack Garfunkel, « The heptagonal triangle », Mathematics Magazine no 46 (1), janvier 1973 , p. 7–19 (DOI https://doi.org/10.1080/0025570X.1973.11976267 lire en ligne ) ↑ a et b (en) Abdilkadir Altintas, « Some Collinearities in the Heptagonal Triangle », Forum Geometricorum vol. 16, 2016 , p. 249–256 (lire en ligne ) ↑ a b c d e f g et h (en) Kai Wang, « Heptagonal Triangle and Trigonometric Identities », Forum Geometricorum vol. 19, 2019 , p. 29–38 (lire en ligne ) ↑ (en) Kai Wang, « On cubic equations with zero sums of cubic roots of roots ↑ a b et c (en) Eric W. Weisstein , « Heptagonal Triangle MathWorld ↑ a b c d et e (en) Kai Wang, « Trigonometric Properties For Heptagonal Triangle ↑ (en) Andrew Mattei Gleason , « Angle trisection, the heptagon, and the triskaidecagon », The American Mathematical Monthly vol. 95, no 3, mars 1988 , p. 185–194 (DOI 10.2307/2323624 lire en ligne [archive du 19 décembre 2015 ] ↑ (en) Victor Hugo Moll , An elementary trigonometric equation https://arxiv.org/abs/0709.3755 , 2007.↑ a et b (en) Kai Wang, « Topics of Ramanujan type identities for PI?7 (consulté le 14 février 2022 ) ↑ (en) Roman Wituła et Damian Słota, « New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7 », Journal of Integer Sequences vol. 10, 2007 
![{\displaystyle {\sqrt[{3}]{2\sin \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{2\sin \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{2\sin \left({\frac {8\pi }{7}}\right)}}=\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a4e877048f553e7b9271acf915f1b6d37554eb5)
![{\displaystyle {\frac {1}{\sqrt[{3}]{2\sin \left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{2\sin \left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{2\sin \left({\frac {8\pi }{7}}\right)}}}=\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27e5d1129cdfc831a7fab209dd942e06fde56edc)
![{\displaystyle {\sqrt[{3}]{4\sin ^{2}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{4\sin ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{4\sin ^{2}\left({\frac {8\pi }{7}}\right)}}=\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ff1b4ebfacb0135780d8dc69244c21ee9c0be61)
![{\displaystyle {\frac {1}{\sqrt[{3}]{4\sin ^{2}\left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{4\sin ^{2}\left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{4\sin ^{2}\left({\frac {8\pi }{7}}\right)}}}=\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{2{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1b8ebcccd6e8105768ced47cdda5fab2856954a)
![{\displaystyle {\sqrt[{3}]{2\cos \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{2\cos \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{2\cos \left({\frac {8\pi }{7}}\right)}}={\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0013ea29caeef7a813bc5cd9fb59ff975bec43f2)
![{\displaystyle {\frac {1}{\sqrt[{3}]{2\cos \left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{2\cos \left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{2\cos \left({\frac {8\pi }{7}}\right)}}}={\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/470af8134ce51cebf4971ae14999df13b2ac736e)
![{\displaystyle {\sqrt[{3}]{4\cos ^{2}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{4\cos ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{4\cos ^{2}\left({\frac {8\pi }{7}}\right)}}={\sqrt[{3}]{11+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d94ede6e2bda2099929174e1b2f96b0decd87f2)
![{\displaystyle {\frac {1}{\sqrt[{3}]{4\cos ^{2}\left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}\left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}\left({\frac {8\pi }{7}}\right)}}}={\sqrt[{3}]{12+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/332448690ce6aa0036a6fbff1fbe006b70c44487)
![{\displaystyle {\sqrt[{3}]{\tan \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\tan \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\tan \left({\frac {8\pi }{7}}\right)}}=\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aac0981c1bc3146ef0ee033e769ae853f0690c9e)
![{\displaystyle {\frac {1}{\sqrt[{3}]{\tan \left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{\tan \left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{\tan \left({\frac {8\pi }{7}}\right)}}}=\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{-{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d09893a09e288638ffeb6b39fae206d9e78fbb8)
![{\displaystyle {\sqrt[{3}]{\tan ^{2}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\tan ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\tan ^{2}\left({\frac {8\pi }{7}}\right)}}=\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{3{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50ed6e73952bb90faf9ffb874c6adf6d3bc4d156)
![{\displaystyle {\frac {1}{\sqrt[{3}]{\tan ^{2}\left({\frac {2\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}\left({\frac {4\pi }{7}}\right)}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}\left({\frac {8\pi }{7}}\right)}}}=\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{5{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e979bde464e8ccd9576234c933a4243631ed3919)
![{\displaystyle {\sqrt[{3}]{\cos \left({\frac {2\pi }{7}}\right)/\cos \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos \left({\frac {4\pi }{7}}\right)/\cos \left({\frac {8\pi }{7}}\right)}}+{\sqrt[{3}]{\cos \left({\frac {8\pi }{7}}\right)/\cos \left({\frac {2\pi }{7}}\right)}}=-{\sqrt[{3}]{7}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee377333b16e239879fa79f04ae65dedb3d1022f)
![{\displaystyle {\sqrt[{3}]{\cos \left({\frac {4\pi }{7}}\right)/\cos \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\cos \left({\frac {8\pi }{7}}\right)/\cos \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos \left({\frac {2\pi }{7}}\right)/\cos \left({\frac {8\pi }{7}}\right)}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8efecd5dbd17774539e92d224bd830100fc5fd3)
![{\displaystyle {\sqrt[{3}]{\cos ^{4}\left({\frac {4\pi }{7}}\right)/\cos \left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{4}\left({\frac {8\pi }{7}}\right)/\cos \left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{4}\left({\frac {2\pi }{7}}\right)/\cos \left({\frac {8\pi }{7}}\right)}}=-{\sqrt[{3}]{49}}/2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d3ab817cdb166be3d94b0d3e87d1bc63da01cd0)
![{\displaystyle {\sqrt[{3}]{\cos ^{5}\left({\frac {2\pi }{7}}\right)/\cos ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{5}\left({\frac {4\pi }{7}}\right)/\cos ^{2}\left({\frac {8\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{5}\left({\frac {8\pi }{7}}\right)/\cos ^{2}\left({\frac {2\pi }{7}}\right)}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16cb6cdb9516783bcce7009854f287e8ec0acf04)
![{\displaystyle {\sqrt[{3}]{\cos ^{5}\left({\frac {4\pi }{7}}\right)/\cos ^{2}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{5}\left({\frac {8\pi }{7}}\right)/\cos ^{2}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{5}\left({\frac {2\pi }{7}}\right)/\cos ^{2}\left({\frac {8\pi }{7}}\right)}}=-3{\sqrt[{3}]{7}}/2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edc1fae4dd4058614edd40e0a967ad2a557406fa)
![{\displaystyle {\sqrt[{3}]{\cos ^{14}\left({\frac {2\pi }{7}}\right)/\cos ^{5}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{14}\left({\frac {4\pi }{7}}\right)/\cos ^{5}\left({\frac {8\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{14}\left({\frac {8\pi }{7}}\right)/\cos ^{5}\left({\frac {2\pi }{7}}\right)}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25fb0acdbbc6ac2b205c3193736f8c2a0ed335bf)
![{\displaystyle {\sqrt[{3}]{\cos ^{14}\left({\frac {4\pi }{7}}\right)/\cos ^{5}\left({\frac {2\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{14}\left({\frac {8\pi }{7}}\right)/\cos ^{5}\left({\frac {4\pi }{7}}\right)}}+{\sqrt[{3}]{\cos ^{14}\left({\frac {2\pi }{7}}\right)/\cos ^{5}\left({\frac {8\pi }{7}}\right)}}=-61{\sqrt[{3}]{7}}/8.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05ca1a43e2a96d62376da6d141f565c6b76f2ddf)
