完全四边形

hbghlyj

目的是重画page19

首先尝试标记点
unitsize(1cm);
pair E=(-2.3529374913291847,3.535264190734489),
A=(1.900326185442464,-1.9907402932984364),
D=(-2.0,-2.0),
B=(-1.2127830569016493,2.053931502252112),
F=(-4.4517975913995596,-2.0058207763937017),
C=extension(B,F,D,E),
N=(-0.1403864824950639,-0.5752438254319003),
T=(-3.402367541364372,0.7647217071703936),
W=(-1.6063915284508248,0.026965751126055926),
J=(1.5403148320771507,-1.2656469244942739),
K=(-4.40425862856662,1.1762812778865392),
G=(-2.870718204081655,0.5463293331703352);
draw(A--E--D--B--F--A--C);
label("R",A,Fill(white));
label("Q",B,Fill(white));
label("P",C,Fill(white));
label("S",D,Fill(white));
label("B",E,Fill(white));
label("C",F,Fill(white));
label("A",extension(A,C,B,D),Fill(white));
需要标注点和线的坐标

hbghlyj

标记了点坐标后，它已经变得有点混乱了
unitsize(1cm);
pair E=(-2.3529374913291847,3.535264190734489),
A=(1.900326185442464,-1.9907402932984364),
D=(-2.0,-2.0),
B=(-1.2127830569016493,2.053931502252112),
F=(-4.4517975913995596,-2.0058207763937017),
C=extension(B,F,D,E),
N=(-0.1403864824950639,-0.5752438254319003),
T=(-3.402367541364372,0.7647217071703936),
W=(-1.6063915284508248,0.026965751126055926),
J=(1.5403148320771507,-1.2656469244942739),
K=(-4.40425862856662,1.1762812778865392),
G=(-2.870718204081655,0.5463293331703352),
Z=extension(A,C,B,D);
draw(F--E--Z--cycle,red);
draw(A--E--D--B--F--A--C);
label("R",A,Fill(white));
label("(f:$-$g:$-$h)",A,(0,-1));

label("Q",B,Fill(white));
label("(f:g:$-$h)",B,(1,0));

label("P",C,Fill(white));
label("(f:g:h)",C,(-1,0));

label("S",D,Fill(white));
label("(f:$-$g:h)",D,(0,-1));

label("B",E,Fill(white));
label("(0:1:0)",E,(1,0));

label("C",F,Fill(white));
label("(0:0:1)",F,(0,-1));

label("A",Z,Fill(white));
label("(1:0:0)",Z,(1,0));

hbghlyj

I find the locus of the pole of a given line $\lambda \alpha+\mu \beta+\nu \gamma=0$, with respect to a conic which passes through four fixed points, $(f: \pm g: \pm h)$.

By proper choice of the reference triangle the conics of this pencil may be expressed by the equation $a \alpha^2+b\beta^2+c \gamma^2=0$, where $a, b$, and $c$, are constants having different values for each conic of the pencil.
The polar of any point $\left(\alpha_1: \beta_1: \gamma_1\right)$ will have the form
\begin{equation}
a \alpha \alpha_1+b\beta \beta_1+c \gamma \gamma_1=0
\end{equation}
Since the conics pass through $(f: \pm g: \pm h)$ we have the condition
\begin{equation}
a f^2+b g^2+c h^2=0
\end{equation}
The polar (1) is to be indentified with the line $\lambda \alpha+\mu \beta+\nu \gamma=0$, since $\left(\alpha_1: \beta_1: \gamma_1\right)$ is to be the pole of this line. Therefore, identifying coefficients,
\[
\begin{array}{lll}
\lambda=a \alpha_1 & \text { or } & a=\frac{\lambda}{\alpha_1} \\
\mu=b \beta_1 & \text { or } & b=\frac{\mu}{\beta_1} \\
\nu=c \gamma_1 & \text { or } & c=\frac{\nu}{\gamma_1}
\end{array}
\]Putting these values for $a, b$, and $c$, in (2) gives
\[\frac{\lambda f^2}{\alpha_1}+\frac{\mu g^2}{\beta_1}+\frac{\nu h^2}{\gamma_1}=0 .
\]
since $\left(\alpha_1: \beta_1: \gamma_1\right)$ was any point, by removing the subscripts and letting the point vary we have the locus:
\[
\frac{\lambda f^2}{\alpha}+\frac{\mu g^2}{\beta}+\frac{\nu h^2}{\gamma}=0, \quad \text { which is evidently a conic. }
\]