过四点的所有二次曲线公共的一对共轭方向的夹角

hbghlyj

为了从斯坦纳提出的几何性质证明3#需要解决以下平面几何题, 来求助大家
四边形$ABCD$的内角是$α,β,γ,δ$
直线$AD,BC$交于$E$.
点$F$在直线$AB$上, 使得$EF\px CD$
点$G,H$在直线$AB$上, 使得$FG=FH=\sqrt{FA·FB}$
求证\begin{equation}\label1
4\cot^2\left(\angle GEH\right)={\sin^2(α + γ) \over \sin(α) \sin(β) \sin(γ) \sin(δ)}\end{equation}
过四点的所有二次曲线公共的一对共轭方向的夹角.ggb (30.29 KB, 下载次数: 1)

2023-2-18 17:07 上传

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自动导出Asymptote代码, (手动修改了线条粗细) 如下
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = blue; /* point style */
real xmin = -2.314085986018613, xmax = 1.6173022634046543, ymin = -1.7496048483670998, ymax = 2.248276225570726;  /* image dimensions */

/* draw figures */
draw((xmin, 1.7113964211768826*xmin + 0.922435531051931)--(xmax, 1.7113964211768826*xmax + 0.922435531051931), linewidth(.5)); /* line */
draw((xmin, -0.019554150693591826*xmin + 0.9757950686529901)--(xmax, -0.019554150693591826*xmax + 0.9757950686529901), linewidth(.5)); /* line */
draw((xmin, -12.572036731667295*xmin + 16.359884185334643)--(xmax, -12.572036731667295*xmax + 16.359884185334643), linewidth(.5)); /* line */
draw((xmin, 0.46590909090909094*xmin + 1.3224612435349916)--(xmax, 0.46590909090909094*xmax + 1.3224612435349916), linewidth(.5)); /* line */
draw((xmin, -12.572036731667296*xmin-7.987851915726761)--(xmax, -12.572036731667296*xmax-7.987851915726761), linewidth(.5)); /* line */
draw(circle((-0.6238197323732662,-0.14516732639119737), 1.5590266044278045), linewidth(.5));
draw((xmin, 0.2408170769739092*xmin + 1.1617244857739186)--(xmax, 0.2408170769739092*xmax + 1.1617244857739186), linewidth(.5) + linetype("6 6")); /* line */
draw((xmin, 3.5633108519847383*xmin + 3.534295896638695)--(xmax, 3.5633108519847383*xmax + 3.534295896638695), linewidth(.5) + linetype("6 6")); /* line */
/* dots and labels */
dot((0.3211800736640583,1.472101959673928),dotstyle);
label("$A$", (0.22926455166324664,1.558413171232182), NE * labelscalefactor);
dot((0.03082672519261977,0.9751922782231838),dotstyle);
label("$B$", (0.03,0.7771224831861204), NE * labelscalefactor);
dot((1.2255813953488315,0.9518298653610766),dotstyle);
label("$C$", (1.2599033316389023,1.0347821781800344), NE * labelscalefactor);
dot((1.1533582932796813,1.8598213574493887),dotstyle);
label("$D$", (1.1850989040600242,1.940746912190893), NE * labelscalefactor);
dot((-0.7140935608997624,0.9897585617521477),dotstyle);
label("$E$", (-1,1.1012750026945928), NE * labelscalefactor);
dot((-0.6238197323732662,-0.14516732639119737),dotstyle);
label("$F$", (-0.5769609455757744,-0.3283207243684135), NE * labelscalefactor);
dot((0.1627174729913523,1.2009096319922772),dotstyle);
label("$G$", (0.1,1.0181589720513948), NE * labelscalefactor);
dot((-1.4103569377378844,-1.4912442847746716),dotstyle);
label("$H$", (-1.6740925500659884,-1.5999959932093435), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);

hbghlyj

import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = blue; /* point style */
real xmin = -2.314085986018613, xmax = 1.6173022634046543, ymin = -1.7496048483670998, ymax = 2.248276225570726;  /* image dimensions */

/* draw figures */
draw((xmin, 1.7113964211768826*xmin + 0.922435531051931)--(xmax, 1.7113964211768826*xmax + 0.922435531051931), linewidth(.5)); /* line */
draw((xmin, -0.019554150693591826*xmin + 0.9757950686529901)--(xmax, -0.019554150693591826*xmax + 0.9757950686529901), linewidth(.5)); /* line */
draw((xmin, 0.46590909090909094*xmin + 1.3224612435349916)--(xmax, 0.46590909090909094*xmax + 1.3224612435349916), linewidth(.5)); /* line */
draw((xmin, -12.572036731667296*xmin-7.987851915726761)--(xmax, -12.572036731667296*xmax-7.987851915726761), linewidth(.5)); /* line */
draw(circle((-0.6238197323732662,-0.14516732639119737), 1.5590266044278045), linewidth(.5));
draw((xmin, 0.2408170769739092*xmin + 1.1617244857739186)--(xmax, 0.2408170769739092*xmax + 1.1617244857739186), linewidth(.5) + linetype("6 6")); /* line */
draw((xmin, 3.5633108519847383*xmin + 3.534295896638695)--(xmax, 3.5633108519847383*xmax + 3.534295896638695), linewidth(.5) + linetype("6 6")); /* line */
draw((-0.7140935608997624,0.9897585617521477)--(-0.1524295769015068,0.6615680986611859), linewidth(.5) + linetype("6 6"));
/* dots and labels */
dot((0.3211800736640583,1.472101959673928),dotstyle);
label("$A$", (0.22926455166324664,1.558413171232182), NE * labelscalefactor);
dot((0.03082672519261977,0.9751922782231838),dotstyle);
label("$B$", (0.03,0.7771224831861204), NE * labelscalefactor);
dot((-0.7140935608997624,0.9897585617521477),dotstyle);
label("$E$", (-1,1.1012750026945928), NE * labelscalefactor);
dot((-0.6238197323732662,-0.14516732639119737),dotstyle);
label("$F$", (-0.5769609455757744,-0.3283207243684135), NE * labelscalefactor);
dot((0.1627174729913523,1.2009096319922772),dotstyle);
label("$G$", (0.1,1.0181589720513948), NE * labelscalefactor);
dot((-1.4103569377378844,-1.4912442847746716),dotstyle);
label("$H$", (-1.6740925500659884,-1.5999959932093435), NE * labelscalefactor);
dot((-0.1524295769015068,0.6615680986611859),dotstyle);
label("$I$", (-0.08,0.511151185127887), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
首先$CD$是多余的, 擦掉$CD$只保留$EF$. 直线$AB,AE,BE,EF$之间的夹角都能由$\alpha,\beta,\gamma,\delta$表示.
作$EI\perp AB$, 设$EI=1$, 则$AI=\cot(\pi-\alpha)=-\cot(\alpha),BI=\cot(\beta)$.
$\angle EFI=\alpha+\delta-\pi\implies FI=\cot(\alpha+\delta)\implies FG^2=FA\cdot FB=[\cot(\alpha+\delta)-\cot(\alpha)][\cot(\alpha+\delta)+\cot(\beta)]$
$GI=\tan\angle GEI=FG-FI$
$HI=\tan\angle HEI=FG+FI$
$$\tan\angle GEH=\tan(\angle GEI+\angle HEI)=\frac{2FG}{1-FG^2+FI^2}$$
$$\implies\frac14\tan^2\angle GEH=\frac{FG^2}{(1-FG^2+FI^2)^2}=\frac{[\cot(\alpha+\delta)-\cot(\alpha)][\cot(\alpha+\delta)+\cot(\beta)]}{(1-[\cot(\alpha+\delta)-\cot(\alpha)][\cot(\alpha+\delta)+\cot(\beta)]+\cot^2(\alpha+\delta))^2}$$
将$γ=-(α+β+δ)$代入\eqref{1}, 要证明的是\begin{equation}\label2
-{\sin(α) \sin(β) \sin(α+β+δ) \sin(δ)\over\sin^2(β + δ)}=\frac{[\cot(\alpha+\delta)-\cot(\alpha)][\cot(\alpha+\delta)+\cot(\beta)]}{(1-[\cot(\alpha+\delta)-\cot(\alpha)][\cot(\alpha+\delta)+\cot(\beta)]+\cot^2(\alpha+\delta))^2}\end{equation}

1. In[]:= ((Cot[\[Alpha]+\[Delta]]-Cot[\[Alpha]])(Cot[\[Alpha]+\[Delta]]+Cot[\[Beta]]))/(1-(Cot[\[Alpha]+\[Delta]]-Cot[\[Alpha]])(Cot[\[Alpha]+\[Delta]]+Cot[\[Beta]])+Cot[\[Alpha]+\[Delta]]^2)^2+(Sin[\[Alpha]]Sin[\[Beta]]Sin[\[Alpha]+\[Beta]+\[Delta]]Sin[\[Delta]])/Sin[\[Beta]+\[Delta]]^2//FullSimplify
2. Out[]= 0

复制代码

FullSimplify发现恒等式成立这样就证明了3#.
问题:如何手工化简\eqref{2}?

hbghlyj

代入$\begin{aligned}\cot(\alpha+\delta)-\cot\alpha&=-\frac{\sin(\delta)}{\sin(\alpha+\delta)\sin(\alpha)}\\\cot(\alpha+\delta)+\cot\beta&=\frac{\sin(α+β+δ)}{\sin(α+δ)\sin(β)}\end{aligned}$ 从\eqref{2}两边消去$-\sin(α+β+δ) \sin(δ)$
\[{\sin(α) \sin(β) \over\sin^2(β + δ)}=\frac{\frac1{\sin(\alpha+\delta)\sin(\alpha)}\cdot\frac1{\sin(α+δ)\sin(β)}}{(1-[\cot(\alpha+\delta)-\cot\alpha][\cot(\alpha+\delta)+\cot\beta]+\cot^2(\alpha+\delta))^2}\]
即\[{\sin^2(α)\sin^2(β)\sin(α+δ)^2\over\sin^2(β+δ)}=\frac1{[1+\cot(α+δ)(\cot(α)-\cot(β))+\cot(α)\cot(β)]^2}\]
只需证明\begin{equation}\sin(α) \sin(β) \sin(α + δ) [1 + \cot(α + δ) (\cot(α) - \cot(β))+\cot(α) \cot(β)]=\sin(β + δ)\end{equation}把$\sin(α) \sin(β)$乘进去\[\sin(α + δ) [\cos(α-β)-\cot(α + δ) \sin(α-β)]=\sin(β + δ)\]把$\sin(α + δ)$乘进去\[\sin(α + δ) \cos(α-β)-\cos(α + δ) \sin(α-β)=\sin((α + δ)-(α-β))=\sin(β + δ)\]证毕.

hbghlyj

我转发到 math.stackexchange.com/questions/4832646/express-the-minimum-ecc ... quadrilateral-in-te/
希望看到更好的证明