Conway triangle notation

In geometry, the Conway triangle notation simplifies and clarifies the algebraic expression of various trigonometric relationships in a triangle. Using the symbol for twice the triangle's area, the symbol is defined to mean times the cotangent of any arbitrary angle . The notation is named after English mathematician John Horton Conway, who promoted its use, but essentially the same notation (using instead of ) can be found in an 1894 paper by Spanish mathematician .

In geometry, the Conway triangle notation simplifies and clarifies the algebraic expression of various trigonometric relationships in a triangle. Using the symbol for twice the triangle's area, the symbol is defined to mean times the cotangent of any arbitrary angle . The notation is named after English mathematician John Horton Conway, who promoted its use, but essentially the same notation (using instead of ) can be found in an 1894 paper by Spanish mathematician . S = bc \sin A = ac \sin B = ab \sin C , where \begin{align} S &= 2 \times \text{reference triangle area}, \\[2pt] S_\varphi &= S \cot \varphi. \end{align} \begin{align} S_A &= S \cot A = bc \cos A= \frac {b^2+c^2-a^2} {2}, \\[2pt] S_B &= S \cot B = ac \cos B= \frac {a^2+c^2-b^2} {2}, \\[2pt] S_C &= S \cot C = ab \cos C= \frac {a^2+b^2-c^2} {2}, \\[2pt] S_\omega &= S \cot \omega = \frac {a^2+b^2+c^2} {2}, \end{align} where 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{align} S_{\frac {\pi} {3}} &= S \cot {\frac {\pi} {3}} = S \frac {\sqrt 3}{3} \\[2pt] S_{2\varphi} &= \frac {S_\varphi^2 - S^2} {2S_\varphi} \\[2pt] S_{ \frac {\varphi} {2}} &= S_\varphi + \sqrt {S_\varphi^2 + S^2} \end{align} for values of where , \begin{align} S_{\vartheta + \varphi} &= \frac {S_\vartheta S_\varphi - S^2} {S_\vartheta + S_\varphi}, \\[2pt] S_{\vartheta - \varphi} &= \frac {S_\vartheta S_\varphi + S^2} {S_\varphi - S_\vartheta}. \end{align} 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} . \begin{align} \sin A &= \frac {S} {bc} = \frac {S} {\sqrt {S_A^2 + S^2}} \\[2pt] \cos A &= \frac {S_A} {bc} = \frac {S_A} {\sqrt {S_A^2 + S^2}} \\[2pt] \tan A &= \frac {S} {S_A} \\[6pt] a^2 &= S_B + S_C \\[2pt] b^2 &= S_A + S_C \\[2pt] c^2 &= S_A + S_B \end{align} \begin{align} \sum_\text{cyclic} S_A &= S_A+S_B+S_C = S_\omega \\[2pt] S^2 &= b^2c^2 - S_A^2 = a^2c^2 - S_B^2 = a^2b^2 - S_C^2 \\[6pt] S_{BC} &= S_BS_C = S^2 - a^2S_A \\[2pt] S_{AC} &= S_AS_C = S^2 - b^2S_B \\[2pt] S_{AB} &= S_AS_B = S^2 - c^2S_C \\[8pt] S_{ABC} &= S_AS_BS_C = S^2(S_\omega-4R^2) \\[8pt] S_\omega &= s^2-r^2-4rR \end{align} where is the circumradius is the incenter abc = 2SR s = \frac{a+b+c}{2} \frac{S}{r} = a+b+c \begin{align} \sin A \sin B \sin C &= \frac {S} {4R^2} \\[2pt] \cos A \cos B \cos C &= \frac {S_\omega-4R^2} {4R^2} \end{align} \begin{align} \sum_\text{cyclic} \sin A &= \frac {S} {2Rr} = \frac {s}{R} \\[2pt] \sum_\text{cyclic} \cos A &= \frac {r+R} {R} \\[2pt] \sum_\text{cyclic} \tan A &= \frac {S}{S_\omega-4R^2}=\tan A \tan B \tan C \end{align} \begin{array}{ll} \displaystyle \sum_\text{cyclic} a^2S_A \!\! &=& a^2S_A + b^2S_B + c^2 S_C = 2S^2 \\[2pt] \displaystyle \sum_\text{cyclic} a^4 &=& 2(S_\omega^2-S^2) \\[2pt] \displaystyle \sum_\text{cyclic} S_A^2 &=& S_\omega^2 - 2S^2 \\[2pt] \displaystyle \sum_\text{cyclic} S_{BC} &=& \displaystyle \sum_\text{cyclic} S_BS_C = S^2 \\[2pt] \displaystyle \sum_\text{cyclic} b^2c^2 &=& S_\omega^2 + S^2 \end{array} \begin{align} P = p_a : p_b : p_c, \\ Q = q_a : q_b : q_c. \end{align} Let \begin{align} K_p &= ap_a + bp_b + cp_c, \\ K_q &= aq_a + bq_b + cq_c. \end{align} Then 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 For the circumcenter p_a = a S_A, and for the orthocenter q_a = \frac{S_B S_C}{a}, \begin{align} K_p &= \sum_\text{cyclic} a^2S_A = 2S^2, \\[2pt] K_q &= \sum_\text{cyclic} S_BS_C = S^2. \end{align} 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 \\[2pt] &= \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 \\[2pt] &= \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) \\[2pt] &= \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) \\[2pt] &= \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) \\[2pt] &= \frac {3a^2b^2c^2} {4S^2} - \frac {1} {4} \sum_\text{cyclic} a^2 - \frac {3} {2}(S_\omega-4R^2) \\[2pt] &= 3R^2- \frac {1} {2} S_\omega - \frac {3} {2} S_\omega + 6R^2 \\[10pt] &= 9R^2- 2S_\omega. \end{align} Thus, |OH| = \sqrt{9R^2- 2S_\omega}.

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