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In geometry , the Conway triangle notation allows trigonometric functions of a triangle to be managed algebraically. Given a reference triangle whose sides are a, b and c and whose internal angles are A, B and C then the Conway triangle notation is simply represented as follows:-
S
=
b
c
sin
A
=
a
c
sin
B
=
a
b
sin
C
{\displaystyle S=bc\sin A=ac\sin B=ab\sin C\,}
S
{\displaystyle S\,}
and
S
ϕ
=
S
cot
ϕ
.
{\displaystyle S_{\phi }=S\cot \phi .\,}
in particular
S
A
=
S
cot
A
=
b
c
cos
A
=
b
2
+
c
2
−
a
2
2
{\displaystyle S_{A}=S\cot A=bc\cos A={\frac {b^{2}+c^{2}-a^{2}}{2}}\,}
S
B
=
S
cot
B
=
a
c
cos
B
=
a
2
+
c
2
−
b
2
2
{\displaystyle S_{B}=S\cot B=ac\cos B={\frac {a^{2}+c^{2}-b^{2}}{2}}\,}
S
C
=
S
cot
C
=
a
b
cos
C
=
a
2
+
b
2
−
c
2
2
{\displaystyle S_{C}=S\cot C=ab\cos C={\frac {a^{2}+b^{2}-c^{2}}{2}}\,}
S
ω
=
S
cot
ω
=
a
2
+
b
2
+
c
2
2
{\displaystyle S_{\omega }=S\cot \omega ={\frac {a^{2}+b^{2}+c^{2}}{2}}\,}
ω
{\displaystyle \omega \,}
Brocard angle .
S
π
3
=
S
cot
π
3
=
S
3
3
{\displaystyle S_{\frac {\pi }{3}}=S\cot {\frac {\pi }{3}}=S{\frac {\sqrt {3}}{3}}\,}
Hence:
sin
A
=
S
b
c
cos
A
=
S
A
b
c
tan
A
=
S
S
A
{\displaystyle \sin A={\frac {S}{bc}}\quad \quad \cos A={\frac {S_{A}}{bc}}\quad \quad \tan A={\frac {S}{S_{A}}}\,}
Some important identities:
∑
c
y
c
l
i
c
S
A
=
S
A
+
S
B
+
S
C
=
S
ω
{\displaystyle \sum _{cyclic}S_{A}=S_{A}+S_{B}+S_{C}=S_{\omega }\,}
S
2
=
b
2
c
2
−
S
A
2
=
a
2
c
2
−
S
B
2
=
a
2
b
2
−
S
C
2
{\displaystyle S^{2}=b^{2}c^{2}-S_{A}^{2}=a^{2}c^{2}-S_{B}^{2}=a^{2}b^{2}-S_{C}^{2}\,}
S
B
S
C
=
S
2
−
a
2
S
A
S
A
S
C
=
S
2
−
b
2
S
B
S
A
S
C
=
S
2
−
c
2
S
C
{\displaystyle S_{B}S_{C}=S^{2}-a^{2}S_{A}\quad \quad S_{A}S_{C}=S^{2}-b^{2}S_{B}\quad \quad S_{A}S_{C}=S^{2}-c^{2}S_{C}\,}
S
A
S
B
S
C
=
S
2
(
S
ω
−
4
R
2
)
{\displaystyle S_{A}S_{B}S_{C}=S^{2}(S_{\omega }-4R^{2})\,}
R
{\displaystyle R\,}
circumcenter and
a
b
c
=
2
S
R
{\displaystyle abc=2SR\,}
Some useful trigonometric conversions:
sin
A
sin
B
sin
C
=
S
4
R
2
cos
A
cos
B
cos
C
=
S
ω
−
4
R
2
4
R
2
{\displaystyle \sin A\sin B\sin C={\frac {S}{4R^{2}}}\quad \quad \cos A\cos B\cos C={\frac {S_{\omega }-4R^{2}}{4R^{2}}}}
∑
c
y
c
l
i
c
sin
A
=
S
2
R
r
∑
c
y
c
l
i
c
cos
A
=
S
2
−
4
S
ω
r
2
−
12
R
r
3
4
R
r
3
{\displaystyle \sum _{cyclic}\sin A={\frac {S}{2Rr}}\quad \quad \sum _{cyclic}\cos A={\frac {S^{2}-4S_{\omega }r^{2}-12Rr^{3}}{4Rr^{3}}}\,}
r
{\displaystyle r\,}
incenter and
a
+
b
+
c
=
S
r
{\displaystyle a+b+c={\frac {S}{r}}\,}
∑
c
y
c
l
i
c
a
2
S
A
=
a
2
S
A
+
b
2
S
B
+
c
2
S
C
=
2
S
2
∑
c
y
c
l
i
c
a
4
=
2
(
S
ω
2
−
S
2
)
{\displaystyle \sum _{cyclic}a^{2}S_{A}=a^{2}S_{A}+b^{2}S_{B}+c^{2}S_{C}=2S^{2}\quad \quad \sum _{cyclic}a^{4}=2(S_{\omega }^{2}-S^{2})\,}
∑
c
y
c
l
i
c
S
A
2
=
S
ω
2
−
2
S
2
∑
c
y
c
l
i
c
S
B
S
C
=
S
2
∑
c
y
c
l
i
c
b
2
c
2
=
S
ω
2
+
S
2
{\displaystyle \sum _{cyclic}S_{A}^{2}=S_{\omega }^{2}-2S^{2}\quad \quad \sum _{cyclic}S_{B}S_{C}=S^{2}\quad \quad \sum _{cyclic}b^{2}c^{2}=S_{\omega }^{2}+S^{2}\,}
Let D be the distance between two points P and Q whose trilinear coordinates are p a p b p c q a q b q c K p ap a bp b cp c K p aq a bq b cq c
D
2
=
∑
c
y
c
l
i
c
a
2
S
A
(
p
a
K
p
−
q
a
K
q
)
2
{\displaystyle D^{2}=\sum _{cyclic}a^{2}S_{A}({\frac {p_{a}}{K_{p}}}-{\frac {q_{a}}{K_{q}}})^{2}\,}
Using this formula it is possible to determine OH, the distance between the circumcenter and the orthocenter as follows:
For the circumcenter
p
a
=
a
S
A
{\displaystyle p_{a}=aS_{A}\,}
q
a
=
S
B
S
C
a
{\displaystyle q_{a}={\frac {S_{B}S_{C}}{a}}\,}
K
p
=
∑
c
y
c
l
i
c
a
2
S
A
=
2
S
2
K
q
=
∑
c
y
c
l
i
c
S
B
S
C
=
S
2
{\displaystyle K_{p}=\sum _{cyclic}a^{2}S_{A}=2S^{2}\quad \quad K_{q}=\sum _{cyclic}S_{B}S_{C}=S^{2}\,}
Hence:
D
2
=
∑
c
y
c
l
i
c
a
2
S
A
(
a
S
A
2
S
2
−
S
B
S
C
a
S
2
)
2
{\displaystyle D^{2}=\sum _{cyclic}a^{2}S_{A}({\frac {aS_{A}}{2S^{2}}}-{\frac {S_{B}S_{C}}{aS^{2}}})^{2}\,}
=
1
4
S
4
∑
c
y
c
l
i
c
a
4
S
A
3
−
S
A
S
B
S
C
S
4
∑
c
y
c
l
i
c
a
2
S
A
+
S
A
S
B
S
C
S
4
∑
c
y
c
l
i
c
S
B
S
C
{\displaystyle ={\frac {1}{4S^{4}}}\sum _{cyclic}a^{4}S_{A}^{3}-{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{cyclic}a^{2}S_{A}+{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{cyclic}S_{B}S_{C}\,}
=
1
4
S
4
∑
c
y
c
l
i
c
a
2
S
A
2
(
S
2
−
S
B
S
C
)
−
2
(
S
ω
−
4
R
2
)
+
(
S
ω
−
4
R
2
)
{\displaystyle ={\frac {1}{4S^{4}}}\sum _{cyclic}a^{2}S_{A}^{2}(S^{2}-S_{B}S_{C})-2(S_{\omega }-4R^{2})+(S_{\omega }-4R^{2})\,}
=
1
4
S
2
∑
c
y
c
l
i
c
a
2
S
A
2
−
S
A
S
B
S
C
S
4
∑
c
y
c
l
i
c
a
2
S
A
−
(
S
ω
−
4
R
2
)
{\displaystyle ={\frac {1}{4S^{2}}}\sum _{cyclic}a^{2}S_{A}^{2}-{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{cyclic}a^{2}S_{A}-(S_{\omega }-4R^{2})\,}
=
1
4
S
2
∑
c
y
c
l
i
c
a
2
(
b
2
c
2
−
S
2
)
−
1
2
(
S
ω
−
4
R
2
)
−
(
S
ω
−
4
R
2
)
{\displaystyle ={\frac {1}{4S^{2}}}\sum _{cyclic}a^{2}(b^{2}c^{2}-S^{2})-{\frac {1}{2}}(S_{\omega }-4R^{2})-(S_{\omega }-4R^{2})\,}
=
3
a
2
b
2
c
2
4
S
2
−
1
4
∑
c
y
c
l
i
c
a
2
−
3
2
(
S
ω
−
4
R
2
)
{\displaystyle ={\frac {3a^{2}b^{2}c^{2}}{4S^{2}}}-{\frac {1}{4}}\sum _{cyclic}a^{2}-{\frac {3}{2}}(S_{\omega }-4R^{2})\,}
=
3
R
2
−
1
2
S
ω
−
3
2
S
ω
+
6
R
2
{\displaystyle =3R^{2}-{\frac {1}{2}}S_{\omega }-{\frac {3}{2}}S_{\omega }+6R^{2}\,}
=
9
R
2
−
2
S
ω
{\displaystyle =9R^{2}-2S_{\omega }\,}
Which gives:
O
H
=
(
9
R
2
−
2
S
ω
)
{\displaystyle OH={\sqrt {(}}9R^{2}-2S_{\omega })\,}
References
