1022

在 ${\displaystyle \mathrm{\Delta }ABC}$ 中，设 ${\displaystyle A,B,C}$ 所对的边分别为 ${\displaystyle a,b,c}$, 且 ${\displaystyle b\ne c,\tan A=\sin B+\sin C}$，则以下结论正确的有_____.

1. ${\displaystyle {\displaystyle a\in (0,\frac{2}{\frac{1}{b}+\frac{1}{c}})}}$;
2. ${\displaystyle {\displaystyle a\in (\frac{2}{\frac{1}{b}+\frac{1}{c}},\sqrt{bc})}}$;
3. ${\displaystyle {\displaystyle a\in (\sqrt{bc},\frac{b+c}{2})}}$;
4. ${\displaystyle a\in {\displaystyle (\frac{b+c}{2},\sqrt{\frac{b^2+c^2}{2}})}}$;
5. ${\displaystyle a\in {\displaystyle (\sqrt{\frac{b^2+c^2}{2}},+\mathrm{\infty })}}$.

依据正弦定理和余弦定理，原式等价于$$\begin{array}{r}\frac{a}{\cos A}=\frac{2abc}{b^2+c^2-a^2}=b+c⟺((b+c)a-(b^2+c^2))(a+(b+c))=0,\end{array}$$
从而 ${\displaystyle a={\displaystyle \frac{b^2+c^2}{b+c}}}$. 而不难验证$$\begin{array}{r}2(b^2+c^2)>(b+c{)}^2\Rightarrow \frac{b^2+c^2}{b+c}>\sqrt{\frac{b^2+c^2}{2}},\end{array}$$
因此只有 5 正确.