Problem 920

Theorem of Mercedes: As shown in the figure, $O$ is a point inside $\triangle ABC$, then we have
$$S_{\triangle BOC} \overrightarrow{OA} + S_{\triangle COA} \overrightarrow{OB} + S_{\triangle AOB} \overrightarrow{OC} = \mathbf{0}.$$

Proof: Since $\frac{BD}{DC} = \frac{S_{\triangle AOB}}{S_{\triangle COA}}$,
$$\begin{array}{l} \Rightarrow \overrightarrow{BD} = \frac{S_{\triangle AOB}}{S_{\triangle AOB} + S_{\triangle COA}} \overrightarrow{BC} \\ \Rightarrow \overrightarrow{AD} = \frac{S_{\triangle COA}}{S_{\triangle AOB} + S_{\triangle COA}} \overrightarrow{AB} + \frac{S_{\triangle AOB}}{S_{\triangle AOB} + S_{\triangle COA}} \overrightarrow{AC}, \end{array}$$

and $\frac{AO}{AD} = \frac{S_{\triangle ABC} + S_{\triangle COA}}{S_{\triangle ABC}} \Rightarrow \overrightarrow{AO} = \frac{S_{\triangle COA}}{S_{\triangle ABC}} \overrightarrow{AB} + \frac{S_{\triangle AOB}}{S_{\triangle ABC}} \overrightarrow{AC}$.
Similarly, $\overrightarrow{BO} = \frac{S_{\triangle AOB}}{S_{\triangle ABC}} \overrightarrow{BC} + \frac{S_{\triangle BOC}}{S_{\triangle ABC}} \overrightarrow{BA}$, $\overrightarrow{CO} = \frac{S_{\triangle BOC}}{S_{\triangle ABC}} \overrightarrow{CA} + \frac{S_{\triangle COA}}{S_{\triangle ABC}} \overrightarrow{CB}$,
thus $S_{\triangle BOC} \overrightarrow{AO} + S_{\triangle COA} \overrightarrow{BO} + S_{\triangle AOB} \overrightarrow{CO} = \mathbf{0}$, so the original proposition holds.
When $O$ is the orthocenter $H$, $\frac{BD}{DC} = \frac{\tan C}{\tan B}$, then $S_{\triangle BOC} : S_{\triangle COA} : S_{\triangle AOB} = \tan A : \tan B : \tan C \Rightarrow \tan A \overrightarrow{HA} + \tan B \overrightarrow{HB} + \tan C \overrightarrow{HC} = \mathbf{0}$.
Thus, $\tan A : \tan B : \tan C = 1 : 2 : 6 \Rightarrow \tan B = 2 \tan A, \tan C = 6 \tan A$, substituting into $\tan A + \tan B + \tan C = \tan A \tan B \tan C \Rightarrow 9 \tan A = 12 \tan^3 A$
$\Rightarrow \tan A = \frac{\sqrt{3}}{2} \Rightarrow \tan B = \sqrt{3} \Rightarrow B = \frac{\pi}{3}$.