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 × 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.

$S_{\frac {\pi} {3}} = S \cot {\frac {\pi} {3}} = S \frac {\sqrt 3}{3} \,$

$S_{2\varphi} = \frac {S_\varphi^2 - S^2} {2S_\varphi} \quad\quad S_{ \frac {\varphi} {2}} = S_\varphi + \sqrt {S_\varphi^2 + S^2} \,$    for values of   $\varphi$  where   $0 < \varphi < \pi \,$

$S_{\vartheta + \varphi} = \frac {S_\vartheta S_\varphi - S^2} {S_\vartheta + S_\varphi} \quad\quad S_{\vartheta - \varphi} = \frac {S_\vartheta S_\varphi + S^2} {S_\varphi - S_\vartheta} \,$

Hence:

$\sin A = \frac {S} {bc} = \frac {S} {\sqrt {S_A^2 + S^2}} \quad\quad \cos A = \frac {S_A} {bc} = \frac {S_A} {\sqrt {S_A^2 + S^2}} \quad\quad \tan A = \frac {S} {S_A} \,$

Some important identities:

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

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

$S_BS_C = S^2 - a^2S_A\quad\quad S_AS_C = S^2 - b^2S_B\quad\quad S_AS_B = S^2 - c^2S_C \,$

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

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:

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

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

Some examples using Conway triangle notation:

Let D be the distance between two points P and Q whose trilinear coordinates are 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:

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

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

For the circumcenter p_a = aS_A and for the orthocenter q_a = S_BS_C/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:

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

References

• Weisstein, Eric W., "Conway Triangle Notation", MathWorld.