寻找三角形内的一点

hejoseph

$\triangle ABC$ 已知，求点 $D$ 的位置

lsr314

椭圆的离心率只有一个变量，D应该有无穷多个选择（不一定）

creasson

可令
\[\mathop {DF}\limits^ \to   = \frac{{\left( {s + t} \right)\left( {1 - st} \right)}}{{s{{\left( {1 + it} \right)}^2}}}\mathop {DA}\limits^ \to  \]
\[\mathop {DF}\limits^ \to   = \frac{{\left( {s - u} \right)\left( {1 + su} \right)}}{{s{{\left( {1 - iu} \right)}^2}}}\mathop {DC}\limits^ \to  \]
\[\mathop {DG}\limits^ \to   = \frac{{\left( {p + q} \right)\left( {1 - pq} \right)}}{{p{{\left( {1 - iq} \right)}^2}}}\mathop {DA}\limits^ \to  \]
\[\mathop {DG}\limits^ \to   = \frac{{\left( {p - u} \right)\left( {1 + pu} \right)}}{{p{{\left( {1 + iu} \right)}^2}}}\mathop {DB}\limits^ \to  \]
\[\mathop {DE}\limits^ \to   = \frac{{\left( {v + t} \right)\left( {1 - vt} \right)}}{{v{{\left( {1 - it} \right)}^2}}}\mathop {DB}\limits^ \to  \]
\[\mathop {DE}\limits^ \to   = \frac{{\left( {v - q} \right)\left( {1 + qv} \right)}}{{v{{\left( {1 + iq} \right)}^2}}}\mathop {DC}\limits^ \to  \]
其中：\[q = \frac{{1 - tu}}{{t + u}}\]
\[\frac{{\left( {s + t} \right)\left( {1 - st} \right)}}{{s{{\left( {1 + it} \right)}^2}}}\frac{{\left( {p - u} \right)\left( {1 + pu} \right)}}{{p{{\left( {1 + iu} \right)}^2}}}\frac{{\left( {v - q} \right)\left( {1 + qv} \right)}}{{v{{\left( {1 + iq} \right)}^2}}} = \frac{{\left( {s - u} \right)\left( {1 + su} \right)}}{{s{{\left( {1 - iu} \right)}^2}}}\frac{{\left( {p + q} \right)\left( {1 - pq} \right)}}{{p{{\left( {1 - iq} \right)}^2}}}\frac{{\left( {v + t} \right)\left( {1 - vt} \right)}}{{v{{\left( {1 - it} \right)}^2}}}\]
后一式又可化为
\[T = p s u^3 t^4-p^2 s u^2 t^4+s u^2 t^4-p s u^3 v^2 t^4+p^2 s u^2 v^2 t^4-s u^2 v^2 t^4+p s u v^2 t^4-p s u t^4-p s^2 u^3 v t^4+p u^3 v t^4-p^2 u^2 v t^4+p^2 s^2 u^2 v t^4-s^2 u^2 v t^4+u^2 v t^4+p s^2 u v t^4-p u v t^4+2 p s^2 u^3 t^3-2 p u^3 t^3+2 p^2 u^2 t^3-2 p^2 s^2 u^2 t^3+2 s^2 u^2 t^3-2 u^2 t^3-2 p s^2 u^3 v^2 t^3+2 p u^3 v^2 t^3-2 p^2 u^2 v^2 t^3+2 p^2 s^2 u^2 v^2 t^3-2 s^2 u^2 v^2 t^3+2 u^2 v^2 t^3+2 p s^2 u v^2 t^3-2 p u v^2 t^3-2 p s^2 u t^3+2 p u t^3-p s^2 u^4 v t^3+p u^4 v t^3-p^2 s u^4 v t^3+s u^4 v t^3-2 p^2 u^3 v t^3+2 p^2 s^2 u^3 v t^3-2 s^2 u^3 v t^3+2 u^3 v t^3-p s^2 v t^3+6 p s^2 u^2 v t^3-6 p u^2 v t^3-2 p^2 s u^2 v t^3+2 s u^2 v t^3+p v t^3-p^2 s v t^3+s v t^3+2 p^2 u v t^3-2 p^2 s^2 u v t^3+2 s^2 u v t^3-2 u v t^3+p s^2 u^4 t^2-p u^4 t^2+p^2 s u^4 t^2-s u^4 t^2+2 p^2 u^3 t^2-2 p^2 s^2 u^3 t^2+2 s^2 u^3 t^2+2 p s u^3 t^2-2 u^3 t^2+p s^2 t^2-6 p s^2 u^2 t^2+6 p u^2 t^2-p s^2 u^4 v^2 t^2+p u^4 v^2 t^2-p^2 s u^4 v^2 t^2+s u^4 v^2 t^2-2 p^2 u^3 v^2 t^2+2 p^2 s^2 u^3 v^2 t^2-2 s^2 u^3 v^2 t^2-2 p s u^3 v^2 t^2+2 u^3 v^2 t^2-p s^2 v^2 t^2+6 p s^2 u^2 v^2 t^2-6 p u^2 v^2 t^2+p v^2 t^2-p^2 s v^2 t^2+s v^2 t^2+2 p^2 u v^2 t^2-2 p^2 s^2 u v^2 t^2+2 s^2 u v^2 t^2+2 p s u v^2 t^2-2 u v^2 t^2-p t^2+p^2 s t^2-s t^2-2 p^2 u t^2+2 p^2 s^2 u t^2-2 s^2 u t^2-2 p s u t^2+2 u t^2+6 p s^2 u^3 v t^2-6 p u^3 v t^2+6 p^2 u^2 v t^2-6 p^2 s^2 u^2 v t^2+6 s^2 u^2 v t^2-6 u^2 v t^2-6 p s^2 u v t^2+6 p u v t^2-2 p s^2 u^3 t+2 p u^3 t-2 p^2 u^2 t+2 p^2 s^2 u^2 t-2 s^2 u^2 t+2 u^2 t+2 p s^2 u^3 v^2 t-2 p u^3 v^2 t+2 p^2 u^2 v^2 t-2 p^2 s^2 u^2 v^2 t+2 s^2 u^2 v^2 t-2 u^2 v^2 t-2 p s^2 u v^2 t+2 p u v^2 t+2 p s^2 u t-2 p u t+p s^2 u^4 v t-p u^4 v t+p^2 s u^4 v t-s u^4 v t+2 p^2 u^3 v t-2 p^2 s^2 u^3 v t+2 s^2 u^3 v t-2 u^3 v t+p s^2 v t-6 p s^2 u^2 v t+6 p u^2 v t+2 p^2 s u^2 v t-2 s u^2 v t-p v t+p^2 s v t-s v t-2 p^2 u v t+2 p^2 s^2 u v t-2 s^2 u v t+2 u v t+p s u^3-p^2 s u^2+s u^2-p s u^3 v^2+p^2 s u^2 v^2-s u^2 v^2+p s u v^2-p s u-p s^2 u^3 v+p u^3 v-p^2 u^2 v+p^2 s^2 u^2 v-s^2 u^2 v+u^2 v+p s^2 u v-p u v = 0 \]
各小三角形内心分别为：
\[\mathop {DM}\limits^ \to   = \frac{{1 - st}}{{1 + it}}\mathop {DA}\limits^ \to  \]
\[\mathop {DL}\limits^ \to   = \frac{{s - u}}{{s\left( {1 - iu} \right)}}\mathop {DC}\limits^ \to  \]
\[\mathop {DH}\limits^ \to   = \frac{{1 - pq}}{{1 - iq}}\mathop {DA}\limits^ \to  \]
\[\mathop {DI}\limits^ \to   = \frac{{p - u}}{{p\left( {1 + iu} \right)}}\mathop {DB}\limits^ \to  \]
\[\mathop {DJ}\limits^ \to   = \frac{{1 - vt}}{{1 - it}}\mathop {DB}\limits^ \to  \]
\[\mathop {DK}\limits^ \to   = \frac{{v - q}}{{v\left( {1 + iq} \right)}}\mathop {DC}\limits^ \to  \]
由前述关系式可得：
\[\mathop {DH}\limits^ \to   = \frac{{p - t - u - ptu}}{{\left( {1 - st} \right)\left( {i - u} \right)}}\mathop {DM}\limits^ \to  \]
\[\mathop {DI}\limits^ \to   = \frac{{\left( {1 + pt + pu - tu} \right)\left( { - p + t + u + ptu} \right)}}{{p\left( {i - t} \right)\left( {1 - st} \right)\left( {i - u} \right)\left( {1 + pu} \right)}}\mathop {DM}\limits^ \to  \]
\[\mathop {DJ}\limits^ \to   = \frac{{\left( {1 + pt + pu - tu} \right)\left( { - p + t + u + ptu} \right)\left( {1 - tv} \right)}}{{\left( {1 + {t^2}} \right)\left( {1 - st} \right)\left( {p - u} \right)\left( {1 + pu} \right)}}\mathop {DM}\limits^ \to  \]
\[\mathop {DL}\limits^ \to   = \frac{{\left( {s + t} \right)\left( {i + u} \right)}}{{s\left( {i - t} \right)\left( {1 + su} \right)}}\mathop {DM}\limits^ \to  \]
\[\mathop {DK}\limits^ \to   = \frac{{\left( {s + t} \right)\left( {i + u} \right)\left( {1 - tu - tv - uv} \right)}}{{\left( {1 + {t^2}} \right)\left( {s - u} \right)\left( {1 + su} \right)v}}\mathop {DM}\limits^ \to  \]
这样我们只需根据这几个式子，在条件$T=0$下，证明六内心构成的六边形确实外切于某椭圆曲线即可。
这可归为一般性问题：给定六边形，已知其外切于椭圆，求椭圆的表示。
目前暂时没想到好办法，暂时先解决圆的情形：
若椭圆退化为圆，设圆的中心为$O$，
\[\mathop {DO}\limits^ \to   = \left( {\lambda  + i\mu } \right)\mathop {DM}\limits^ \to  \]
(以下计算开始繁琐了，待有时间再仔细计算)
我们可根据marden定理导出三角形内心所满足的方程，见 https://bbs.emath.ac.cn/thread-17622-1-1.html
由$IH,LM,JK$的延长线交点构成的三角形，与由$IJ,KL,HM$的延长线交点构成的三角形，其内心是相同的，这样可得到两个等式：
$ eq1(s,t,p,u,v) = 0 $
$ eq2(s,t,p,u,v) = 0 $
这两个方程与前面的$T=0$联立。

另一方面，我们也可以根据
\[\frac{{\left( {s + t} \right)\left( {1 - st} \right)}}{{s{{\left( {1 + it} \right)}^2}}}\mathop {DA}\limits^ \to   = \frac{{\left( {s - u} \right)\left( {1 + su} \right)}}{{s{{\left( {1 - iu} \right)}^2}}}\mathop {DC}\limits^ \to  \]
\[\frac{{\left( {p - u} \right)\left( {1 + pu} \right)}}{{p{{\left( {1 + iu} \right)}^2}}}\mathop {DB}\limits^ \to   = \frac{{\left( {p + q} \right)\left( {1 - pq} \right)}}{{p{{\left( {1 - iq} \right)}^2}}}\mathop {DA}\limits^ \to  \]
反解$A，D$为关于$B,C$的表示式，再改写成重心坐标形式即可。

hujunhua

lsr314 发表于 2021-3-2 18:59
椭圆的离心率只有一个变量，D应该有无穷多个选择（不一定）

明显HK、IL、JM三线相交于D，由布利安桑逆定理知六边形HIJKLM共切于一条二次曲线。

假定D的坐标为 (u,v), 由它确定的椭圆方程为\[a(u,v)x^2+b(u,v)xy+c(u,v)y^2+dx+ey+f=0\]椭圆特化为圆的条件为 \[a(u,v)=c(u,v), b(u,v)=0\]两个坐标两个约束方程，故解存在并有限。

hejoseph

D 应该是确定的，至少应该是有限个，可能得到的会是一个很高次的方程

yigo

内切换成外接，命题也成立。

creasson

前面已得到了六边形的表示，现在来求六边形内切椭圆的表示。
已知凸六边形$HIJKLM$，三条对角线共点。

设
\[\left( {H,I,J,K,L,M} \right) = {x_k} + i{y_k},\left( {k = 1,...6} \right)\]
可求得延长线交点X,Y,Z
\[X = \frac{{{x_2}{x_5}{y_1} - {x_2}{x_6}{y_1} - {x_1}{x_5}{y_2} + {x_1}{x_6}{y_2} - {x_1}{x_6}{y_5} + {x_2}{x_6}{y_5} + {x_1}{x_5}{y_6} - {x_2}{x_5}{y_6} + i\left( {{x_2}{y_1}{y_5} - {x_6}{y_1}{y_5} - {x_1}{y_2}{y_5} + {x_6}{y_2}{y_5} - {x_2}{y_1}{y_6} + {x_5}{y_1}{y_6} + {x_1}{y_2}{y_6} - {x_5}{y_2}{y_6}} \right)}}{{{x_5}{y_1} - {x_6}{y_1} - {x_5}{y_2} + {x_6}{y_2} - {x_1}{y_5} + {x_2}{y_5} + {x_1}{y_6} - {x_2}{y_6}}}\]
\[Y = \frac{{{x_2}{x_3}{y_1} - {x_2}{x_4}{y_1} - {x_1}{x_3}{y_2} + {x_1}{x_4}{y_2} - {x_1}{x_4}{y_3} + {x_2}{x_4}{y_3} + {x_1}{x_3}{y_4} - {x_2}{x_3}{y_4} + i\left( {{x_2}{y_1}{y_3} - {x_4}{y_1}{y_3} - {x_1}{y_2}{y_3} + {x_4}{y_2}{y_3} - {x_2}{y_1}{y_4} + {x_3}{y_1}{y_4} + {x_1}{y_2}{y_4} - {x_3}{y_2}{y_4}} \right)}}{{{x_3}{y_1} - {x_4}{y_1} - {x_3}{y_2} + {x_4}{y_2} - {x_1}{y_3} + {x_2}{y_3} + {x_1}{y_4} - {x_2}{y_4}}}\]
\[Z = \frac{{{x_4}{x_5}{y_3} - {x_4}{x_6}{y_3} - {x_3}{x_5}{y_4} + {x_3}{x_6}{y_4} - {x_3}{x_6}{y_5} + {x_4}{x_6}{y_5} + {x_3}{x_5}{y_6} - {x_4}{x_5}{y_6} + i\left( {{x_4}{y_3}{y_5} - {x_6}{y_3}{y_5} - {x_3}{y_4}{y_5} + {x_6}{y_4}{y_5} - {x_4}{y_3}{y_6} + {x_5}{y_3}{y_6} + {x_3}{y_4}{y_6} - {x_5}{y_4}{y_6}} \right)}}{{{x_5}{y_3} - {x_6}{y_3} - {x_5}{y_4} + {x_6}{y_4} - {x_3}{y_5} + {x_4}{y_5} + {x_3}{y_6} - {x_4}{y_6}}}\]
三角形$XYZ$的内切椭圆上的点$T$可表示为:
\[T = \frac{{q{u^2}X + {{\left( {1 - u} \right)}^2}Y + pZ}}{{q{u^2} + {{\left( {1 - u} \right)}^2} + p}}\]
其中$u$为自变量。
分别由$IJ,KL,MH$与椭圆相切，即交点唯一可得三个等式，由此可得
\[p = \frac{{\left( {{x_2}{y_1} - {x_4}{y_1} - {x_1}{y_2} + {x_4}{y_2} + {x_1}{y_4} - {x_2}{y_4}} \right)\left( {{x_3}{y_2} - {x_5}{y_2} - {x_2}{y_3} + {x_5}{y_3} + {x_2}{y_5} - {x_3}{y_5}} \right)\left( {{x_5}{y_3} - {x_6}{y_3} - {x_5}{y_4} + {x_6}{y_4} - {x_3}{y_5} + {x_4}{y_5} + {x_3}{y_6} - {x_4}{y_6}} \right)}}{{\left( {{x_3}{y_1} - {x_4}{y_1} - {x_3}{y_2} + {x_4}{y_2} - {x_1}{y_3} + {x_2}{y_3} + {x_1}{y_4} - {x_2}{y_4}} \right)\left( {{x_4}{y_2} - {x_5}{y_2} - {x_2}{y_4} + {x_5}{y_4} + {x_2}{y_5} - {x_4}{y_5}} \right)\left( {{x_5}{y_3} - {x_6}{y_3} - {x_3}{y_5} + {x_6}{y_5} + {x_3}{y_6} - {x_5}{y_6}} \right)}}\]
\[q = \frac{{\left( {{x_2}{y_1} - {x_4}{y_1} - {x_1}{y_2} + {x_4}{y_2} + {x_1}{y_4} - {x_2}{y_4}} \right)\left( { - {x_3}{y_2} + {x_5}{y_2} + {x_2}{y_3} - {x_5}{y_3} - {x_2}{y_5} + {x_3}{y_5}} \right)\left( {{x_5}{y_1} - {x_6}{y_1} - {x_5}{y_2} + {x_6}{y_2} - {x_1}{y_5} + {x_2}{y_5} + {x_1}{y_6} - {x_2}{y_6}} \right)}}{{\left( {{x_3}{y_1} - {x_4}{y_1} - {x_3}{y_2} + {x_4}{y_2} - {x_1}{y_3} + {x_2}{y_3} + {x_1}{y_4} - {x_2}{y_4}} \right)\left( {{x_2}{y_1} - {x_5}{y_1} - {x_1}{y_2} + {x_5}{y_2} + {x_1}{y_5} - {x_2}{y_5}} \right)\left( {{x_5}{y_2} - {x_6}{y_2} - {x_2}{y_5} + {x_6}{y_5} + {x_2}{y_6} - {x_5}{y_6}} \right)}}\]
并有
\[ \left(x_2 x_5 y_1 y_3-x_4 x_5 y_1 y_3-x_2 x_6 y_1 y_3+x_4 x_6 y_1 y_3-x_1 x_5 y_2 y_3+x_4 x_5 y_2 y_3+x_1 x_6 y_2 y_3-x_4 x_6 y_2 y_3+x_1 x_4 y_5 y_3-x_2 x_4 y_5 y_3-x_1 x_6 y_5 y_3+x_2 x_6 y_5 y_3-x_1 x_4 y_6 y_3+x_2 x_4 y_6 y_3+x_1 x_5 y_6 y_3-x_2 x_5 y_6 y_3-x_2 x_5 y_1 y_4+x_3 x_5 y_1 y_4+x_2 x_6 y_1 y_4-x_3 x_6 y_1 y_4+x_1 x_5 y_2 y_4-x_3 x_5 y_2 y_4-x_1 x_6 y_2 y_4+x_3 x_6 y_2 y_4-x_2 x_3 y_1 y_5+x_2 x_4 y_1 y_5+x_3 x_6 y_1 y_5-x_4 x_6 y_1 y_5+x_1 x_3 y_2 y_5-x_1 x_4 y_2 y_5-x_3 x_6 y_2 y_5+x_4 x_6 y_2 y_5-x_1 x_3 y_4 y_5+x_2 x_3 y_4 y_5+x_1 x_6 y_4 y_5-x_2 x_6 y_4 y_5+x_2 x_3 y_1 y_6-x_2 x_4 y_1 y_6-x_3 x_5 y_1 y_6+x_4 x_5 y_1 y_6-x_1 x_3 y_2 y_6+x_1 x_4 y_2 y_6+x_3 x_5 y_2 y_6-x_4 x_5 y_2 y_6+x_1 x_3 y_4 y_6-x_2 x_3 y_4 y_6-x_1 x_5 y_4 y_6+x_2 x_5 y_4 y_6\right) \left(x_3 x_4 y_1 y_2-x_3 x_5 y_1 y_2-x_4 x_6 y_1 y_2+x_5 x_6 y_1 y_2+x_1 x_5 y_3 y_2-x_4 x_5 y_3 y_2-x_1 x_6 y_3 y_2+x_4 x_6 y_3 y_2-x_1 x_3 y_4 y_2+x_3 x_5 y_4 y_2+x_1 x_6 y_4 y_2-x_5 x_6 y_4 y_2+x_1 x_3 y_6 y_2-x_3 x_4 y_6 y_2-x_1 x_5 y_6 y_2+x_4 x_5 y_6 y_2-x_2 x_4 y_1 y_3+x_4 x_5 y_1 y_3+x_2 x_6 y_1 y_3-x_5 x_6 y_1 y_3+x_1 x_2 y_3 y_4-x_1 x_5 y_3 y_4-x_2 x_6 y_3 y_4+x_5 x_6 y_3 y_4+x_2 x_3 y_1 y_5-x_3 x_4 y_1 y_5-x_2 x_6 y_1 y_5+x_4 x_6 y_1 y_5-x_1 x_2 y_3 y_5+x_2 x_4 y_3 y_5+x_1 x_6 y_3 y_5-x_4 x_6 y_3 y_5+x_1 x_3 y_4 y_5-x_2 x_3 y_4 y_5-x_1 x_6 y_4 y_5+x_2 x_6 y_4 y_5-x_2 x_3 y_1 y_6+x_2 x_4 y_1 y_6+x_3 x_5 y_1 y_6-x_4 x_5 y_1 y_6-x_1 x_2 y_4 y_6+x_2 x_3 y_4 y_6+x_1 x_5 y_4 y_6-x_3 x_5 y_4 y_6+x_1 x_2 y_5 y_6-x_1 x_3 y_5 y_6-x_2 x_4 y_5 y_6+x_3 x_4 y_5 y_6\right) = 0 \]
这两个因式的其中一个即为三对角线共点的条件：
\[ x_3 x_4 y_1 y_2-x_3 x_5 y_1 y_2-x_4 x_6 y_1 y_2+x_5 x_6 y_1 y_2+x_1 x_5 y_3 y_2-x_4 x_5 y_3 y_2-x_1 x_6 y_3 y_2+x_4 x_6 y_3 y_2-x_1 x_3 y_4 y_2+x_3 x_5 y_4 y_2+x_1 x_6 y_4 y_2-x_5 x_6 y_4 y_2+x_1 x_3 y_6 y_2-x_3 x_4 y_6 y_2-x_1 x_5 y_6 y_2+x_4 x_5 y_6 y_2-x_2 x_4 y_1 y_3+x_4 x_5 y_1 y_3+x_2 x_6 y_1 y_3-x_5 x_6 y_1 y_3+x_1 x_2 y_3 y_4-x_1 x_5 y_3 y_4-x_2 x_6 y_3 y_4+x_5 x_6 y_3 y_4+x_2 x_3 y_1 y_5-x_3 x_4 y_1 y_5-x_2 x_6 y_1 y_5+x_4 x_6 y_1 y_5-x_1 x_2 y_3 y_5+x_2 x_4 y_3 y_5+x_1 x_6 y_3 y_5-x_4 x_6 y_3 y_5+x_1 x_3 y_4 y_5-x_2 x_3 y_4 y_5-x_1 x_6 y_4 y_5+x_2 x_6 y_4 y_5-x_2 x_3 y_1 y_6+x_2 x_4 y_1 y_6+x_3 x_5 y_1 y_6-x_4 x_5 y_1 y_6-x_1 x_2 y_4 y_6+x_2 x_3 y_4 y_6+x_1 x_5 y_4 y_6-x_3 x_5 y_4 y_6+x_1 x_2 y_5 y_6-x_1 x_3 y_5 y_6-x_2 x_4 y_5 y_6+x_3 x_4 y_5 y_6 = 0 \]
另一个因式也许属于双曲线的情形，暂不做讨论。

可以将$T$的参数表示转为直角坐标方程而求得椭圆成为圆的条件，
但这里有更简便的求法，根据marden定理，椭圆的焦点满足方程：
\[pq\left( {z - Z} \right)\left( {z - X} \right) + q\left( {z - X} \right)\left( {z - Y} \right) + p\left( {z - Y} \right)\left( {z - Z} \right) = 0\]
若两焦点重合，则椭圆成为圆，即有方程判别式为$0$, 分离出虚部和实部得到两个条件。

lsr314

直接令六条边的中垂线交于一点，可以考虑用复数表示