Conway triangle notation

Short description: Notation for trigonometric relationships

In geometry, the Conway triangle notation simplifies and clarifies the algebraic expression of various trigonometric relationships in a triangle. Using the symbol S for twice the triangle's area, the symbol S_φ is defined to mean S times the cotangent of any arbitrary angle φ.

The notation is named after English mathematician John Horton Conway, who promoted its use, but essentially the same notation (using p instead of S) can be found in an 1894 paper by Spanish mathematician Juan Jacobo Durán Loriga (gl).

Definition

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 $$\begin{array}{rl}S& =2\times \text{reference triangle area},\\ S_{\phi }& =S\cot \phi .\end{array}$$

Basic formulas

In particular: $$\begin{array}{rl}{S}_A& =S\cot A=bc\cos A=\frac{b^2+c^2-a^2}{2},\\ S_B& =S\cot B=ac\cos B=\frac{a^2+c^2-b^2}{2},\\ S_C& =S\cot C=ab\cos C=\frac{a^2+b^2-c^2}{2},\\ S_{\omega }& =S\cot \omega =\frac{a^2+b^2+c^2}{2},\end{array}$$

where ω is the Brocard angle. The law of cosines is used: $a^2=b^2+c^2-2bc\cos A.$

Third-, double-, and half-angle identities: $$\begin{array}{rl}{S}_{\frac{\pi }{3}}& =S\cot \frac{\pi }{3}=S\frac{\sqrt{3}}{3}\\ S_{2\phi }& =\frac{S_{\phi }^2-S^2}{2S_{\phi }}\\ S_{\frac{\phi }{2}}& =S_{\phi }+\sqrt{S_{\phi }^2+S^2}\end{array}$$

for values of φ where 0 < φ < π,

$$\begin{array}{rl}{S}_{\vartheta +\phi }& =\frac{S_{\vartheta }{S}_{\phi }-S^2}{S_{\vartheta }+S_{\phi }},\\ S_{\vartheta -\phi }& =\frac{S_{\vartheta }{S}_{\phi }+S^2}{S_{\phi }-S_{\vartheta }}.\end{array}$$

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

Trigonometric relationships

$$\begin{array}{rl}\sin A& =\frac{S}{bc}=\frac{S}{\sqrt{S_A^2+S^2}}\\ \cos A& =\frac{S_A}{bc}=\frac{S_A}{\sqrt{S_A^2+S^2}}\\ \tan A& =\frac{S}{S_A}\\ a^2& =S_B+S_C\\ b^2& =S_A+S_C\\ c^2& =S_A+S_B\end{array}$$

Important identities

$$\begin{array}{rl}\sum _{\text{cyclic}}{S}_A& =S_A+S_B+S_C=S_{\omega }\\ S^2& =b^2{c}^2-S_A^2=a^2{c}^2-S_B^2=a^2{b}^2-S_C^2\\ S_{BC}& =S_B{S}_C=S^2-a^2{S}_A\\ S_{AC}& =S_A{S}_C=S^2-b^2{S}_B\\ S_{AB}& =S_A{S}_B=S^2-c^2{S}_C\\ S_{ABC}& =S_A{S}_B{S}_C=S^2(S_{\omega }-4R^2)\\ S_{\omega }& =s^2-r^2-4rR\end{array}$$

where

R is the circumradius

r is the incenter

$abc=2SR$

$s=\frac{a+b+c}{2}$

$\frac{S}{r}=a+b+c$

Trigonometric conversions

$$\begin{array}{rl}\sin A\sin B\sin C& =\frac{S}{4R^2}\\ \cos A\cos B\cos C& =\frac{S_{\omega }-4R^2}{4R^2}\end{array}$$

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

Useful formulas

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

Applications

Let D be the distance between two points P and Q whose trilinear coordinates are $$\begin{array}{r}P=p_a:p_b:p_c,\\ Q=q_a:q_b:q_c.\end{array}$$ Let $$\begin{array}{rl}{K}_p& =a{p}_a+b{p}_b+c{p}_c,\\ K_q& =a{q}_a+b{q}_b+c{q}_c.\end{array}$$ Then D is given by the formula: $$D^2=\sum _{\text{cyclic}}{a}^2{S}_A{(\frac{p_a}{K_p}-\frac{q_a}{K_q})}^2$$

Distance between circumcenter and orthocenter

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

For the circumcenter $p_a=a{S}_A,$

and for the orthocenter $q_a=\frac{S_B{S}_C}{a},$

$$\begin{array}{rl}{K}_p& =\sum _{\text{cyclic}}{a}^2{S}_A=2S^2,\\ K_q& =\sum _{\text{cyclic}}{S}_B{S}_C=S^2.\end{array}$$

Hence:

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

Thus,

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

See also

• Brocard angle
• Orthocenter
• Circumcenter
• Trilinear coordinates