本帖最后由 青青子衿 于 2020-3-15 22:07 编辑
第一费马点到三顶点的距离:
\begin{align*}
\begin{split}
\left|AF_{2}\right|&=\dfrac{\sqrt{\,3}\,(b^2+c^2-a^2)+4\Delta}{\sqrt{6\Big(a^2+b^2+c^2+4\sqrt{3}\,\Delta \Big)}}\\
\left|BF_{1}\right|&=\dfrac{\sqrt{\,3}\,(a^2+c^2-b^2)+4\Delta\,}{\sqrt{6\Big(a^2+b^2+c^2+4\sqrt{3}\,\Delta \Big)}}\\
\left|CF_{1}\right|&=\dfrac{\sqrt{\,3}\,(a^2+b^2-c^2)+4\Delta\,}{\sqrt{6\Big(a^2+b^2+c^2+4\sqrt{3}\,\Delta \Big)}}\\
\\
4\Delta&=\sqrt{\big(a+b+c\big)\big(b+c-a\big)\big(a+c-b\big)\big(a+b-c\big)}\>
\end{split}
\end{align*}
第二费马点到三顶点的距离:
\begin{align*}
\begin{split}
\left|AF_{2}\right|&=\dfrac{\Bigg|\sqrt{\,3}\,(b^2+c^2-a^2)-4\Delta \Bigg|}{\sqrt{6\Big(a^2+b^2+c^2-4\sqrt{3}\,\Delta \Big)}}\\
\left|BF_{2}\right|&=\dfrac{\Bigg|\sqrt{\,3}\,(a^2+c^2-b^2)-4\Delta \Bigg|\,}{\sqrt{6\Big(a^2+b^2+c^2-4\sqrt{3}\,\Delta \Big)}}\\
\left|CF_{2}\right|&=\dfrac{\Bigg|\sqrt{\,3}\,(a^2+b^2-c^2)-4\Delta \Bigg|\,}{\sqrt{6\Big(a^2+b^2+c^2-4\sqrt{3}\,\Delta \Big)}}\\
\\
4\Delta&=\sqrt{\big(a+b+c\big)\big(b+c-a\big)\big(a+c-b\big)\big(a+b-c\big)}\>
\end{split}
\end{align*} |