Conway triangle notation explained

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 = 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 = (b ^{2+c} ^{2-a} ^{2})/(2) S _{B} = S \cot B = ac \cos B = (a ^{2+c} ^{2-b} ^{2})/(2) S _{C} = S \cot C = ab \cos C = (a ^{2+b} ^{2-c} ^{2})/(2) S _{\omega} = S \cot \omega = (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 _{(\pi)/(3)} = S \cot { (\pi)/(3) } = S \frac \,S _{2\varphi} = (S ^{2} _{\varphi} - S ^{2})/(2 S _{\varphi}) S _{(\varphi)/(2)} = S _{\varphi} + \sqrt {S ^{2} _{\varphi} + S ^{2}}    for values of   \varphi   where   0 < \varphi < \pi S _{\vartheta} = (S _{\vartheta} S _{\varphi} - S ^{2})/(S _{\vartheta} + S _{\varphi}) S _{\vartheta} = (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 = (S)/(bc) = (S)/(\sqrt {S ^{2} _{A} + S ^{2}) } \quad\quad \cos A = \frac = \frac \quad\quad \tan A = \frac \, a ^{2} = S _{B} + S _{C} b ^{2} = S _{A} + S _{C} c ^{2} = S _{A} + S _{B} .

Some important identities:

\sum _{cyclic} S _{A} = S _{A+S} _{B+S} _{C} = S _{\omega} S ^{2} = b ^{2c} ^{2} - S ^{2} _{A} = a ^{2c} ^{2} - S ^{2} _{B} = a ^{2b} ^{2} - S ^{2} _{C} S _{BC} = S _{BS} _{C} = S ^{2} - a ^{2 S} _{A} S _{AC} = S _{AS} _{C} = S ^{2} - b ^{2 S} _{B} S _{AB} = S _{AS} _{B} = S ^{2} - c ^{2 S} _{C} S _{ABC} = S _{AS} _{BS} _{C} = S ^{2 )} _{\omega -4 R} S ^{2- r} _{\omega = s} ^{2-4rR}

where R is the circumradius and abc = 2SR and where r is the incenter,  

s = (a + b + c)/(2)    and   a + b + c = (S)/(r) .

Some useful trigonometric conversions:

\sin A \sin B \sin C = (S)/(4 R ^{2}) \cos A \cos B \cos C = (S ^{2} _{\omega -4 R})/(4 R ^{2}) \sum _{cyclic} \sin A = (S)/(2 Rr) = (s)/(R) \sum _{cyclic} \cos A = (r + R)/(R) \sum _{cyclic} \tan A = (S)/(S ^{2} _{\omega -4 R}) = \tan A \tan B \tan C .

Some useful formulas:

\sum _{cyclic} a ^{2 S} _{A} = a ^{2 S} _{A} + b ^{2 S} _{B} + c ^{2} S _{C} = 2 S ^{2} \sum _{cyclic} a ^{4} = 2 (S ^{2- S} _{\omega} ^{2)} \sum _{cyclic} S ^{2} _{A} = S ^{2} _{\omega} - 2 S ^{2} \sum _{cyclic} S _{BC} = \sum _{cyclic} S _{BS} _{C} = S ^{2} \sum _{cyclic} b ^{2c} ^{2} = S ^{2} _{\omega} + 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 _{cyclic} a ^{2 S} _{A\left( (p _{a})/(K _{p})} - (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_BS_C/a

K _{p=} \sum _{cyclic} a ^{2 S} _{A} = 2 S ^{2} K _{q=} \sum _{cyclic} S _{BS} _{C} = S ^{2} .

Hence:

\begin{align} D ^{2} & {} = \sum _{cyclic} a ^{2 S} _{A\left( (aS _{A})/(2 S ^{2})} - (S _{BS} _{C})/(aS ^{2}) \right) ^{2} \\ & {} = (1)/(4 S ^{4}) \sum _{cyclic} a ^{3} _{A} - (S _{AS} _{BS} _{C})/(S ^{4}) \sum _{cyclic} a ^{2 S} _{A} + (S _{AS} _{BS} _{C})/(S ^{4}) \sum _{cyclic} S _{BS} _{C} \\ & {} = (1)/(4 S ^{4}) \sum _{cyclic} a ^{2 (S} _{A} ^{2- S} ^{BS} _{C)} - 2 (S ^{2 )} _{\omega -4 R} + (S ^{2 )} _{\omega -4 R} \\ & {} = (1)/(4 S ^{2}) \sum _{cyclic} a ^{2} _{A} - (S _{AS} _{BS} _{C})/(S ^{4}) \sum _{cyclic} a ^{2 S} _{A} - (S ^{2 )} _{\omega -4 R} \\ & {} = (1)/(4 S ^{2}) \sum _{cyclic} a ^{2(b} ^{2c} ^{2-S} ^{2)} - (1)/(2) (S ^{2 )} _{\omega -4 R} - (S ^{2 )} _{\omega -4 R} \\ & {} = (3 a ^{2b} ^{2c} ^{2})/(4 S ^{2}) - (1)/(4) \sum _{cyclic} a ^{2} - (3)/(2) (S ^{2 )} _{\omega -4 R} \\ & {} = 3 R ^{2-} (1)/(2) S _{\omega} - (3)/(2) S _{\omega} + 6 R ^{2} \\ & {} = 9 R ^{2-} 2 S _{\omega. \end{align}}

This gives:

OH = \sqrt{ 9 R ^{2-} 2 S _{\omega} }.