Conway triangle notation

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 &times; 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:


 * $$ \sum_\text{cyclic} S_A = S_A+S_B+S_C = S_\omega \, $$


 * $$ S^2 = b^2c^2 - S_A^2 = a^2c^2 - S_B^2 = a^2b^2 - S_C^2 \, $$


 * $$ S_{BC} = S_BS_C = S^2 - a^2S_A \quad\quad S_{AC} = S_AS_C = S^2 - b^2S_B \quad\quad S_{AB} = S_AS_B = S^2 - c^2S_C \, $$


 * $$ S_{ABC} = S_AS_BS_C = S^2(S_\omega-4R^2)\quad\quad S_\omega=s^2-r^2-4rR \, $$

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:


 * $$ \sum_\text{cyclic} a^2S_A = a^2S_A + b^2S_B + c^2 S_C = 2S^2 \quad\quad \sum_\text{cyclic} a^4 = 2(S_\omega^2-S^2) \, $$


 * $$ \sum_\text{cyclic} S_A^2 = S_\omega^2 - 2S^2 \quad\quad \sum_\text{cyclic} S_{BC} = \sum_\text{cyclic} S_BS_C = S^2 \quad\quad \sum_\text{cyclic} b^2c^2 =  S_\omega^2 + S^2 \, .$$

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:


 * $$ D^2= \sum_\text{cyclic} a^2S_A\left(\frac {p_a}{K_p} - \frac {q_a}{K_q}\right)^2 \, .$$

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
 * $$ K_p= \sum_\text{cyclic} a^2S_A = 2S^2 \quad\quad K_q= \sum_\text{cyclic} S_BS_C = S^2 \, .$$

Hence:



\begin{align} D^2 & {} = \sum_\text{cyclic} a^2S_A\left(\frac {aS_A} {2S^2} - \frac {S_BS_C} {aS^2}\right)^2 \\ & {} = \frac {1} {4S^4} \sum_\text{cyclic} a^4S_A^3 - \frac {S_AS_BS_C} {S^4} \sum_\text{cyclic} a^2S_A + \frac {S_AS_BS_C} {S^4} \sum_\text{cyclic} S_BS_C \\ & {} = \frac {1} {4S^4} \sum_\text{cyclic} a^2S_A^2(S^2-S_BS_C) - 2(S_\omega-4R^2) + (S_\omega-4R^2) \\ & {} = \frac {1} {4S^2} \sum_\text{cyclic} a^2S_A^2 - \frac {S_AS_BS_C} {S^4} \sum_\text{cyclic} a^2S_A - (S_\omega-4R^2) \\ & {} = \frac {1} {4S^2} \sum_\text{cyclic} a^2(b^2c^2-S^2) - \frac {1} {2}(S_\omega-4R^2) -(S_\omega-4R^2) \\ & {} = \frac {3a^2b^2c^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{align} $$

This gives:


 * $$ OH = \sqrt{9R^2- 2S_\omega \,}.$$