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 = b c sin ⁡ A = a c sin ⁡ B = a b sin ⁡ C

where S = 2 × area of reference triangle and

S φ = S cot ⁡ φ .

in particular

S A = S cot ⁡ A = b c cos ⁡ A = b 2 + c 2 − a 2 2 S B = S cot ⁡ B = a c cos ⁡ B = a 2 + c 2 − b 2 2 S C = S cot ⁡ C = a b cos ⁡ C = a 2 + b 2 − c 2 2 S ω = S cot ⁡ ω = a 2 + b 2 + c 2 2      where ω is the Brocard angle. S π 3 = S cot ⁡ π 3 = S 3 3 S 2 φ = S φ 2 − S 2 2 S φ S φ 2 = S φ + S φ 2 + S 2    for values of   φ   where   0 < φ < π 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 .

Furthermore the convention uses a shorthand notation for S ϑ S φ = S ϑ φ and S ϑ S φ S ψ = S ϑ φ ψ .

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 a 2 = S B + S C b 2 = S A + S C c 2 = S A + S B .

Some important identities:

∑ cyclic S A = S A + S B + S C = S ω 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 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

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 ω − 4 R 2 4 R 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 .

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 ) ∑ 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 .

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 = ∑ cyclic a 2 S A ( p a K p − q a K q ) 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 = ∑ cyclic a 2 S A = 2 S 2 K q = ∑ 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 .

This gives:

O H = 9 R 2 − 2 S ω .