Kept on Wikipedia: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

$S_{\phi }=S\cot \phi .$

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: $a^2=b^2+c^2-2bc\cos A$.

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

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

$S_{\vartheta +\phi }=\frac{S_{\vartheta }{S}_{\phi }-S^2}{S_{\vartheta }+S_{\phi }}{S}_{\vartheta -\phi }=\frac{S_{\vartheta }{S}_{\phi }+S^2}{S_{\phi }-S_{\vartheta }}.$

Furthermore the convention uses a shorthand notation for $S_{\vartheta }{S}_{\phi }=S_{\vartheta \phi }$ and $S_{\vartheta }{S}_{\phi }{S}_{\psi }=S_{\vartheta \phi \psi }.$

Hence:

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

$a^2=S_B+S_C{b}^2=S_A+S_C{c}^2=S_A+S_B.$

Some important identities:

$\sum _{\text{cyclic}}{S}_A=S_A+S_B+S_C=S_{\omega }$

$S^2=b^2{c}^2-S_A^2=a^2{c}^2-S_B^2=a^2{b}^2-S_C^2$

$S_{BC}=S_B{S}_C=S^2-a^2{S}_A{S}_{AC}=S_A{S}_C=S^2-b^2{S}_B{S}_{AB}=S_A{S}_B=S^2-c^2{S}_C$

$S_{ABC}=S_A{S}_B{S}_C=S^2(S_{\omega }-4R^2)S_{\omega }=s^2-r^2-4rR$

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:

$\sin A\sin B\sin C=\frac{S}{4R^2}\cos A\cos B\cos C=\frac{S_{\omega }-4R^2}{4R^2}$

$\sum _{\text{cyclic}}\sin A=\frac{S}{2Rr}=\frac{s}{R}\sum _{\text{cyclic}}\cos A=\frac{r+R}{R}\sum _{\text{cyclic}}\tan A=\frac{S}{S_{\omega }-4R^2}=\tan A\tan B\tan C.$

Some useful formulas:

$\sum _{\text{cyclic}}{a}^2{S}_A=a^2{S}_A+b^2{S}_B+c^2{S}_C=2S^2\sum _{\text{cyclic}}{a}^4=2(S_{\omega }^2-S^2)$

$\sum _{\text{cyclic}}{S}_A^2=S_{\omega }^2-2S^2\sum _{\text{cyclic}}{S}_{BC}=\sum _{\text{cyclic}}{S}_B{S}_C=S^2\sum _{\text{cyclic}}{b}^2{c}^2=S_{\omega }^2+S^2.$

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:

$D^2=\sum _{\text{cyclic}}{a}^2{S}_A{(\frac{p_a}{K_p}-\frac{q_a}{K_q})}^2.$

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

$K_p=\sum _{\text{cyclic}}{a}^2{S}_A=2S^2{K}_q=\sum _{\text{cyclic}}{S}_B{S}_C=S^2.$

Hence:

$\begin{array}{cc}{D}^2& =\sum _{\text{cyclic}}{a}^2{S}_A{(\frac{a{S}_A}{2S^2}-\frac{S_B{S}_C}{a{S}^2})}^2\\ & =\frac{1}{4S^4}\sum _{\text{cyclic}}{a}^4{S}_A^3-\frac{S_A{S}_B{S}_C}{S^4}\sum _{\text{cyclic}}{a}^2{S}_A+\frac{S_A{S}_B{S}_C}{S^4}\sum _{\text{cyclic}}{S}_B{S}_C\\ & =\frac{1}{4S^4}\sum _{\text{cyclic}}{a}^2{S}_A^2(S^2-S_B{S}_C)-2(S_{\omega }-4R^2)+(S_{\omega }-4R^2)\\ & =\frac{1}{4S^2}\sum _{\text{cyclic}}{a}^2{S}_A^2-\frac{S_A{S}_B{S}_C}{S^4}\sum _{\text{cyclic}}{a}^2{S}_A-(S_{\omega }-4R^2)\\ & =\frac{1}{4S^2}\sum _{\text{cyclic}}{a}^2(b^2{c}^2-S^2)-\frac{1}{2}(S_{\omega }-4R^2)-(S_{\omega }-4R^2)\\ & =\frac{3a^2{b}^2{c}^2}{4S^2}-\frac{1}{4}\sum _{\text{cyclic}}{a}^2-\frac{3}{2}(S_{\omega }-4R^2)\\ & =3R^2-\frac{1}{2}{S}_{\omega }-\frac{3}{2}{S}_{\omega }+6R^2\\ & =9R^2-2S_{\omega }.\end{array}$

This gives:

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

References

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

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