三角形外接椭圆的一点有趣结论

creasson

如果外接椭圆以重心为中心，那么半轴长是
\[\frac{1}{3}\sqrt {{a^2} + {b^2} + {c^2} \pm 2\sqrt {{a^4} + {b^4} + {c^4} - {a^2}{b^2} - {b^2}{c^2} - {c^2}{a^2}} } \]
如果以内心为中心，半轴长是
\[\frac{{abc}}{{4{S_{ABC}}}}\left( {1 \pm \sqrt {\frac{{{a^3} + {b^3} + {c^3} - {a^2}b - {b^2}c - {c^2}a - a{b^2} - b{c^2} - c{a^2} + 3abc}}{{abc}}} } \right)\]

数学星空

1.若以外心，垂心为中心的椭圆，其半轴长各为多少？
2.若将外接椭圆改成内切椭圆，其四心为中心的时，椭圆半轴长为多少？
3.若固定椭圆\(\frac{x^2}{m^2}+\frac{y^2}{n^2}=1\),试分别给出椭圆所有外接和内切三角形重心，外心，内心，垂心构成的轨迹？

creasson

以$$O = uA + vB + wC\left( {u + v + w = 1} \right)$$为中心的$ABC$的外接椭圆，其半轴长为：
\[{L^2} = \frac{{uvw}}{{\left( {2u - 1} \right)\left( {2v - 1} \right)\left( {2w - 1} \right)}}\left\{ {\left( {2u - 1} \right)\left( {2v - 1} \right){c^2} + \left( {2v - 1} \right)\left( {2w - 1} \right){a^2} + \left( {2w - 1} \right)\left( {2u - 1} \right){b^2} \pm 2\sqrt T } \right\}\]
\[T = v\left( {2v - 1} \right)w\left( {2w - 1} \right){a^4} + u\left( {2u - 1} \right)w\left( {2w - 1} \right){b^4} + u\left( {2u - 1} \right)v\left( {2v - 1} \right){c^4} + \left( {2u - 1} \right)\left( {2v - 1} \right)\left( {2w - 1} \right)\left( {u{b^2}{c^2} + v{c^2}{a^2} + w{a^2}{b^2}} \right)\]
又记号
\[{X^2} = \frac{{ - uvw\left( {2u - 1} \right){{\left( {{a^2}\left( {2v - 1} \right) + u\left( {{a^2} + {b^2} - {c^2}} \right)} \right)}^2}}}{{2uT \pm \left( {\left( {u - 1} \right)\left( {2v - 1} \right)\left( {1 - 2w} \right){a^2} + u\left( {2u - 1} \right)\left( {1 - 2v} \right){c^2} + u\left( {2u - 1} \right)\left( {1 - 2w} \right){b^2}} \right)\sqrt T }}\]
\[Y = \frac{{{b^2}u - 2{b^2}{u^2} - {a^2}v + 2{a^2}{v^2} \pm \sqrt T }}{{\left( {2u - 1} \right)\left( {{a^2}\left( {2v - 1} \right) + u\left( {{a^2} + {b^2} - {c^2}} \right)} \right)}}X\]
那么$\alpha  = u + X,\beta  = v + Y,\gamma  = 1 - \alpha  - \beta ,P = \alpha A + \beta B + \gamma C$是四个轴端点。
椭圆面积为$$4\pi {S_{ABC}}\frac{{uvw}}{{\sqrt {\left( {1 - 2u} \right)\left( {1 - 2v} \right)\left( {1 - 2w} \right)} }}$$
并知最小面积的外接椭圆的中心是重心。



，

creasson

对于以$$O = uA + vB + wC\left( {u + v + w = 1} \right)$$为中心的$ABC$的内切椭圆，其三个切点是
\[\left\{ \begin{array}{l}
D = \frac{{1 - 2w}}{{2u}}B + \frac{{1 - 2v}}{{2u}}C \\
E = \frac{{1 - 2u}}{{2v}}C + \frac{{1 - 2w}}{{2v}}A \\
F = \frac{{1 - 2v}}{{2w}}A + \frac{{1 - 2u}}{{2w}}B \\
\end{array} \right.\]
并
\[\begin{array}{l}
D{E^2} = \left\{ {\frac{{1 - 2w}}{{2v}}\frac{{1 - 2w}}{{2u}}{c^2} - \frac{{1 - 2w}}{{2u}}\left( {\frac{{1 - 2v}}{{2u}} - \frac{{1 - 2u}}{{2v}}} \right){a^2} + \left( {\frac{{1 - 2v}}{{2u}} - \frac{{1 - 2u}}{{2v}}} \right)\frac{{1 - 2w}}{{2v}}{b^2}} \right\} \\
E{F^2} = \left\{ {\left( {\frac{{1 - 2w}}{{2v}} - \frac{{1 - 2v}}{{2w}}} \right)\frac{{1 - 2u}}{{2w}}{c^2} + \frac{{1 - 2u}}{{2w}}\frac{{1 - 2u}}{{2v}}{a^2} - \frac{{1 - 2u}}{{2v}}\left( {\frac{{1 - 2w}}{{2v}} - \frac{{1 - 2v}}{{2w}}} \right){b^2}} \right\} \\
F{D^2} = \left\{ { - \frac{{1 - 2v}}{{2w}}\left( {\frac{{1 - 2u}}{{2w}} - \frac{{1 - 2w}}{{2u}}} \right){c^2} + \left( {\frac{{1 - 2u}}{{2w}} - \frac{{1 - 2w}}{{2u}}} \right)\frac{{1 - 2v}}{{2u}}a + \frac{{1 - 2v}}{{2u}}\frac{{1 - 2v}}{{2w}}{b^2}} \right\} \\
\end{array}\]
外接椭圆的相应式子可以移成内切椭圆的。
内切椭圆的面积是\[{S_{ABC}}\pi \sqrt {\left( {1 - 2u} \right)\left( {1 - 2v} \right)\left( {1 - 2w} \right)} \]
知最小面积的内切椭圆也是以重心为中心。

陈九章

以\(\triangle ABC\)的重心为中心的外接椭圆即为外接Steiner椭圆.
Creasson 老师把\(\triangle ABC\)的外接Steiner椭圆推广了.

陈九章

知最小面积的内切椭圆也是以重心为中心?

mathe

仿射变换保持线段比和面积比不变。所以任意三角形仿射为正三角形后问题就清楚了