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Conway triangle notation

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In geometry, the Conway triangle notation, named after John Horton Conway, allows trigonometric functions of a triangle to be managed algebraically. Given a reference triangle whose sides are a, b and c and whose corresponding 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\,}

where S = 2 × area of reference triangle and

S φ = S cot ⁡ φ . {\displaystyle S_{\varphi }=S\cot \varphi .\,}

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}}\,}      where ω {\displaystyle \omega \,} is the Brocard angle. The law of cosines is used: a 2 = b 2 + c 2 − 2 b c cos ⁡ A {\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos A} .
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}}\,}
S 2 φ = S φ 2 − S 2 2 S φ S φ 2 = S φ + S φ 2 + S 2 {\displaystyle S_{2\varphi }={\frac {S_{\varphi }^{2}-S^{2}}{2S_{\varphi }}}\quad \quad S_{\frac {\varphi }{2}}=S_{\varphi }+{\sqrt {S_{\varphi }^{2}+S^{2}}}\,}    for values of   φ {\displaystyle \varphi }   where   0 < φ < π {\displaystyle 0<\varphi <\pi \,}
S ϑ + φ = S ϑ S φ − S 2 S ϑ + S φ S ϑ − φ = S ϑ S φ + S 2 S φ − S ϑ . {\displaystyle S_{\vartheta +\varphi }={\frac {S_{\vartheta }S_{\varphi }-S^{2}}{S_{\vartheta }+S_{\varphi }}}\quad \quad S_{\vartheta -\varphi }={\frac {S_{\vartheta }S_{\varphi }+S^{2}}{S_{\varphi }-S_{\vartheta }}}\,.}

Furthermore the convention uses a shorthand notation for S ϑ S φ = S ϑ φ {\displaystyle S_{\vartheta }S_{\varphi }=S_{\vartheta \varphi }\,} and S ϑ S φ S ψ = S ϑ φ ψ . {\displaystyle S_{\vartheta }S_{\varphi }S_{\psi }=S_{\vartheta \varphi \psi }\,.}

Hence:

sin ⁡ A = S b c = S S A 2 + S 2 cos ⁡ A = S A b c = S A S A 2 + S 2 tan ⁡ A = S S A {\displaystyle \sin A={\frac {S}{bc}}={\frac {S}{\sqrt {S_{A}^{2}+S^{2}}}}\quad \quad \cos A={\frac {S_{A}}{bc}}={\frac {S_{A}}{\sqrt {S_{A}^{2}+S^{2}}}}\quad \quad \tan A={\frac {S}{S_{A}}}\,}
a 2 = S B + S C b 2 = S A + S C c 2 = S A + S B . {\displaystyle a^{2}=S_{B}+S_{C}\quad \quad b^{2}=S_{A}+S_{C}\quad \quad c^{2}=S_{A}+S_{B}\,.}

Some important identities:

∑ cyclic S A = S A + S B + S C = S ω {\displaystyle \sum _{\text{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 C = S B S C = S 2 − a 2 S A S A C = S A S C = S 2 − b 2 S B S A B = S A S B = S 2 − c 2 S C {\displaystyle S_{BC}=S_{B}S_{C}=S^{2}-a^{2}S_{A}\quad \quad S_{AC}=S_{A}S_{C}=S^{2}-b^{2}S_{B}\quad \quad S_{AB}=S_{A}S_{B}=S^{2}-c^{2}S_{C}\,}
S A B C = S A S B S C = S 2 ( S ω − 4 R 2 ) S ω = s 2 − r 2 − 4 r R {\displaystyle S_{ABC}=S_{A}S_{B}S_{C}=S^{2}(S_{\omega }-4R^{2})\quad \quad S_{\omega }=s^{2}-r^{2}-4rR\,}

where R is the circumradius and abc = 2SR and where r is the incenter,   s = a + b + c 2 {\displaystyle s={\frac {a+b+c}{2}}\,}    and   a + b + c = S r . {\displaystyle a+b+c={\frac {S}{r}}\,.}

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}}}}
∑ cyclic sin ⁡ A = S 2 R r = s R ∑ cyclic cos ⁡ A = r + R R ∑ cyclic tan ⁡ A = S S ω − 4 R 2 = tan ⁡ A tan ⁡ B tan ⁡ C . {\displaystyle \sum _{\text{cyclic}}\sin A={\frac {S}{2Rr}}={\frac {s}{R}}\quad \quad \sum _{\text{cyclic}}\cos A={\frac {r+R}{R}}\quad \quad \sum _{\text{cyclic}}\tan A={\frac {S}{S_{\omega }-4R^{2}}}=\tan A\tan B\tan C\,.}


Some useful formulas:

∑ cyclic a 2 S A = a 2 S A + b 2 S B + c 2 S C = 2 S 2 ∑ cyclic a 4 = 2 ( S ω 2 − S 2 ) {\displaystyle \sum _{\text{cyclic}}a^{2}S_{A}=a^{2}S_{A}+b^{2}S_{B}+c^{2}S_{C}=2S^{2}\quad \quad \sum _{\text{cyclic}}a^{4}=2(S_{\omega }^{2}-S^{2})\,}
∑ cyclic S A 2 = S ω 2 − 2 S 2 ∑ cyclic S B C = ∑ cyclic S B S C = S 2 ∑ cyclic b 2 c 2 = S ω 2 + S 2 . {\displaystyle \sum _{\text{cyclic}}S_{A}^{2}=S_{\omega }^{2}-2S^{2}\quad \quad \sum _{\text{cyclic}}S_{BC}=\sum _{\text{cyclic}}S_{B}S_{C}=S^{2}\quad \quad \sum _{\text{cyclic}}b^{2}c^{2}=S_{\omega }^{2}+S^{2}\,.}

Some examples using Conway triangle notation:

Let D be the distance between two points P and Q whose trilinear coordinates are pa : pb : pc and qa : qb : qc. Let Kp = apa + bpb + cpc and let Kq = aqa + bqb + cqc. Then D is given by the formula:

D 2 = ∑ cyclic a 2 S A ( p a K p − q a K q ) 2 . {\displaystyle D^{2}=\sum _{\text{cyclic}}a^{2}S_{A}\left({\frac {p_{a}}{K_{p}}}-{\frac {q_{a}}{K_{q}}}\right)^{2}\,.}

Using this formula it is possible to determine OH, the distance between the circumcenter and the orthocenter as follows:

For the circumcenter pa = aSA and for the orthocenter qa = SBSC/a

K p = ∑ cyclic a 2 S A = 2 S 2 K q = ∑ cyclic S B S C = S 2 . {\displaystyle K_{p}=\sum _{\text{cyclic}}a^{2}S_{A}=2S^{2}\quad \quad K_{q}=\sum _{\text{cyclic}}S_{B}S_{C}=S^{2}\,.}

Hence:

D 2 = ∑ cyclic a 2 S A ( a S A 2 S 2 − S B S C a S 2 ) 2 = 1 4 S 4 ∑ cyclic a 4 S A 3 − S A S B S C S 4 ∑ cyclic a 2 S A + S A S B S C S 4 ∑ cyclic S B S C = 1 4 S 4 ∑ cyclic a 2 S A 2 ( S 2 − S B S C ) − 2 ( S ω − 4 R 2 ) + ( S ω − 4 R 2 ) = 1 4 S 2 ∑ cyclic a 2 S A 2 − S A S B S C S 4 ∑ cyclic a 2 S A − ( S ω − 4 R 2 ) = 1 4 S 2 ∑ cyclic a 2 ( b 2 c 2 − S 2 ) − 1 2 ( S ω − 4 R 2 ) − ( S ω − 4 R 2 ) = 3 a 2 b 2 c 2 4 S 2 − 1 4 ∑ cyclic a 2 − 3 2 ( S ω − 4 R 2 ) = 3 R 2 − 1 2 S ω − 3 2 S ω + 6 R 2 = 9 R 2 − 2 S ω . {\displaystyle {\begin{aligned}D^{2}&{}=\sum _{\text{cyclic}}a^{2}S_{A}\left({\frac {aS_{A}}{2S^{2}}}-{\frac {S_{B}S_{C}}{aS^{2}}}\right)^{2}\\&{}={\frac {1}{4S^{4}}}\sum _{\text{cyclic}}a^{4}S_{A}^{3}-{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{\text{cyclic}}a^{2}S_{A}+{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{\text{cyclic}}S_{B}S_{C}\\&{}={\frac {1}{4S^{4}}}\sum _{\text{cyclic}}a^{2}S_{A}^{2}(S^{2}-S_{B}S_{C})-2(S_{\omega }-4R^{2})+(S_{\omega }-4R^{2})\\&{}={\frac {1}{4S^{2}}}\sum _{\text{cyclic}}a^{2}S_{A}^{2}-{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{\text{cyclic}}a^{2}S_{A}-(S_{\omega }-4R^{2})\\&{}={\frac {1}{4S^{2}}}\sum _{\text{cyclic}}a^{2}(b^{2}c^{2}-S^{2})-{\frac {1}{2}}(S_{\omega }-4R^{2})-(S_{\omega }-4R^{2})\\&{}={\frac {3a^{2}b^{2}c^{2}}{4S^{2}}}-{\frac {1}{4}}\sum _{\text{cyclic}}a^{2}-{\frac {3}{2}}(S_{\omega }-4R^{2})\\&{}=3R^{2}-{\frac {1}{2}}S_{\omega }-{\frac {3}{2}}S_{\omega }+6R^{2}\\&{}=9R^{2}-2S_{\omega }.\end{aligned}}}

This gives:

O H = 9 R 2 − 2 S ω . {\displaystyle OH={\sqrt {9R^{2}-2S_{\omega }\,}}.}

References

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