# Conway triangle notation

In geometry, the Conway triangle notation, named after John Horton Conway, allows trigonometric functions of a triangle to be managed algebraically. However, though the notation was promoted by Conway, a much earlier reference to the notation goes back to the Spanish nineteenth century mathematician gl:Juan Jacobo Durán Loriga.
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:
    $S=bc\sin A=ac\sin B=ab\sin C\,$
where S = 2 × area of reference triangle and
    $S_{\varphi }=S\cot \varphi .\,$
in particular
    $S_{A}=S\cot A=bc\cos A={\frac {b^{2}+c^{2}-a^{2}}{2}}\,$
    $S_{B}=S\cot B=ac\cos B={\frac {a^{2}+c^{2}-b^{2}}{2}}\,$
    $S_{C}=S\cot C=ab\cos C={\frac {a^{2}+b^{2}-c^{2}}{2}}\,$
    $S_{\omega }=S\cot \omega ={\frac {a^{2}+b^{2}+c^{2}}{2}}\,$ where $\omega \,$ is the Brocard angle. The law of cosines is used: $a^{2}=b^{2}+c^{2}-2bc\cos A$.
    $S_{\frac {\pi }{3}}=S\cot {\frac {\pi }{3}}=S{\frac {\sqrt {3}}{3}}\,$
    $S_{2\varphi }={\frac {S_{\varphi }^{2}-S^{2}}{2S_{\varphi }}}\quad \quad S_{\frac {\varphi }{2}}=S_{\varphi }+{\sqrt {S_{\varphi }^{2}+S^{2}}}\,$ for values of $\varphi$ where $0<\varphi <\pi \,$
    $S_{\vartheta +\varphi }={\frac {S_{\vartheta }S_{\varphi }-S^{2}}{S_{\vartheta }+S_{\varphi }}}\quad \quad S_{\vartheta -\varphi }={\frac {S_{\vartheta }S_{\varphi }+S^{2}}{S_{\varphi }-S_{\vartheta }}}\,.$
Furthermore the convention uses a shorthand notation for $S_{\vartheta }S_{\varphi }=S_{\vartheta \varphi }\,$ and $S_{\vartheta }S_{\varphi }S_{\psi }=S_{\vartheta \varphi \psi }\,.$
Hence:
    $\sin A={\frac {S}{bc}}={\frac {S}{\sqrt {S_{A}^{2}+S^{2}}}}\quad \quad \cos A={\frac {S_{A}}{bc}}={\frac {S_{A}}{\sqrt {S_{A}^{2}+S^{2}}}}\quad \quad \tan A={\frac {S}{S_{A}}}\,$
    $a^{2}=S_{B}+S_{C}\quad \quad b^{2}=S_{A}+S_{C}\quad \quad c^{2}=S_{A}+S_{B}\,.$
Some important identities:
    $\sum _{\text{cyclic}}S_{A}=S_{A}+S_{B}+S_{C}=S_{\omega }\,$
    $S^{2}=b^{2}c^{2}-S_{A}^{2}=a^{2}c^{2}-S_{B}^{2}=a^{2}b^{2}-S_{C}^{2}\,$
    $S_{BC}=S_{B}S_{C}=S^{2}-a^{2}S_{A}\quad \quad S_{AC}=S_{A}S_{C}=S^{2}-b^{2}S_{B}\quad \quad S_{AB}=S_{A}S_{B}=S^{2}-c^{2}S_{C}\,$
    $S_{ABC}=S_{A}S_{B}S_{C}=S^{2}(S_{\omega }-4R^{2})\quad \quad S_{\omega }=s^{2}-r^{2}-4rR\,$
where R is the circumradius and abc = 2SR and where r is the incenter, $s={\frac {a+b+c}{2}}\,$ and $a+b+c={\frac {S}{r}}\,.$
Some useful trigonometric conversions:
    $\sin A\sin B\sin C={\frac {S}{4R^{2}}}\quad \quad \cos A\cos B\cos C={\frac {S_{\omega }-4R^{2}}{4R^{2}}}$
    $\sum _{\text{cyclic}}\sin A={\frac {S}{2Rr}}={\frac {s}{R}}\quad \quad \sum _{\text{cyclic}}\cos A={\frac {r+R}{R}}\quad \quad \sum _{\text{cyclic}}\tan A={\frac {S}{S_{\omega }-4R^{2}}}=\tan A\tan B\tan C\,.$
Some useful formulas:
    $\sum _{\text{cyclic}}a^{2}S_{A}=a^{2}S_{A}+b^{2}S_{B}+c^{2}S_{C}=2S^{2}\quad \quad \sum _{\text{cyclic}}a^{4}=2(S_{\omega }^{2}-S^{2})\,$
    $\sum _{\text{cyclic}}S_{A}^{2}=S_{\omega }^{2}-2S^{2}\quad \quad \sum _{\text{cyclic}}S_{BC}=\sum _{\text{cyclic}}S_{B}S_{C}=S^{2}\quad \quad \sum _{\text{cyclic}}b^{2}c^{2}=S_{\omega }^{2}+S^{2}\,.$
Some examples using Conway triangle notation:
Let D be the distance between two points P and Q whose trilinear coordinates are p$_a$ : p$_b$ : p$_c$ and q$_a$ : q$_b$ : q$_c$. Let K$_p$ = ap$_a$ + bp$_b$ + cp$_c$ and let K$_q$ = aq$_a$ + bq$_b$ + cq$_c$. Then D is given by the formula:
    $D^{2}=\sum _{\text{cyclic}}a^{2}S_{A}\left({\frac {p_{a}}{K_{p}}}-{\frac {q_{a}}{K_{q}}}\right)^{2}\,.$
Using this formula it is possible to determine OH, the distance between the circumcenter and the orthocenter as follows:
For the circumcenter p$_a$ = aS$_A$ and for the orthocenter q$_a$ = S$_B$S$_C$/a
    $K_{p}=\sum _{\text{cyclic}}a^{2}S_{A}=2S^{2}\quad \quad K_{q}=\sum _{\text{cyclic}}S_{B}S_{C}=S^{2}\,.$
Hence:
    ${\begin{aligned}D^{2}&=\sum _{\text{cyclic}}a^{2}S_{A}\left({\frac {aS_{A}}{2S^{2}}}-{\frac {S_{B}S_{C}}{aS^{2}}}\right)^{2}\\&={\frac {1}{4S^{4}}}\sum _{\text{cyclic}}a^{4}S_{A}^{3}-{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{\text{cyclic}}a^{2}S_{A}+{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{\text{cyclic}}S_{B}S_{C}\\&={\frac {1}{4S^{4}}}\sum _{\text{cyclic}}a^{2}S_{A}^{2}(S^{2}-S_{B}S_{C})-2(S_{\omega }-4R^{2})+(S_{\omega }-4R^{2})\\&={\frac {1}{4S^{2}}}\sum _{\text{cyclic}}a^{2}S_{A}^{2}-{\frac {S_{A}S_{B}S_{C}}{S^{4}}}\sum _{\text{cyclic}}a^{2}S_{A}-(S_{\omega }-4R^{2})\\&={\frac {1}{4S^{2}}}\sum _{\text{cyclic}}a^{2}(b^{2}c^{2}-S^{2})-{\frac {1}{2}}(S_{\omega }-4R^{2})-(S_{\omega }-4R^{2})\\&={\frac {3a^{2}b^{2}c^{2}}{4S^{2}}}-{\frac {1}{4}}\sum _{\text{cyclic}}a^{2}-{\frac {3}{2}}(S_{\omega }-4R^{2})\\&=3R^{2}-{\frac {1}{2}}S_{\omega }-{\frac {3}{2}}S_{\omega }+6R^{2}\\&=9R^{2}-2S_{\omega }.\end{aligned}}$
This gives:
    $OH={\sqrt {9R^{2}-2S_{\omega }\,}}.$