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[几何] $OI^2=R(R-2r)$推广到“三角形外接椭圆”

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hbghlyj 发表于 2023-4-18 04:41 |阅读模式
本帖最后由 hbghlyj 于 2024-12-29 01:30 编辑 guide circumcircle(pair A, pair B, pair C)
{
        pair M,N,P,Q,O;
        M=midpoint(A--B);
        N=midpoint(B--C);
        P=rotate(90,M)*A;
        Q=rotate(90,N)*B;
        O=extension(M,P,N,Q);return circle(O,abs(O-A));
}
guide incircle(pair A, pair B, pair C)
{
        pair P,Q,I;
        P=rotate((angle(C-A)-angle(B-A))*90/pi,A)*B;
        Q=rotate((angle(A-B)-angle(C-B))*90/pi,B)*C;
        I=extension(A,P,B,Q);
        real a,b,c,s;
        a=abs(B-C);
        b=abs(A-C);
        c=abs(B-A);
        s=(a+b+c)/2;
        return circle(I,sqrt((s-a)*(s-b)*(s-c)/s));
}
size(200);
pair A=(1,4),B=(-5,-2),C=(5,-1);
draw(scale(1.5,1)*circumcircle(scale(2/3,1)*A,scale(2/3,1)*B,scale(2/3,1)*C));
draw(incircle(A,B,C));
draw(A--B--C--cycle);
dot(A,Fill(white));
dot(B,Fill(white));
dot(C,Fill(white));

纯几何吧 7514
椭圆 $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$ 内接三角形 $\triangle A B C$. 设 $\left(x-x_0\right)^2+\left(y-y_0\right)^2=r^2$ 是其内切圆, 则
$$\tag1\label1
\left(a^2-b^2\right)^2 r^4+2\left[\left(b^2 x_0^2-a^2 y_0^2\right)\left(a^2-b^2\right)-\left(a^2+b^2\right) a^2 b^2\right] r^2+\left(b^2 x_0^2+a^2 y_0^2-a^2 b^2\right)^2=0
$$

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 楼主| hbghlyj 发表于 2024-12-29 07:59

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 楼主| hbghlyj 发表于 2024-12-29 08:09
Poncelet闭合定理(夹有一个 $n$ 边形)的充要条件:
$n=3$时,Cayley closure condition为$a_2=0$.
将本题中的曲线方程代入其中计算
  1. SeriesCoefficient[Sqrt[Det[t{{1/a^2,0,0},{0,1/b^2,0},{0,0,-1}}+{{1,0,-x0},{0,1,-y0},{-x0,-y0,-r^2+x0^2+y0^2}}]],{t,0,2}]
复制代码

0=a^4 b^4-2 a^4 b^2 r^2-2 a^2 b^4 r^2+a^4 r^4-2 a^2 b^2 r^4+b^4 r^4-2 a^2 b^4 x0^2+2 a^2 b^2 r^2 x0^2-2 b^4 r^2 x0^2+b^4 x0^4-2 a^4 b^2 y0^2-2 a^4 r^2 y0^2+2 a^2 b^2 r^2 y0^2+2 a^2 b^2 x0^2 y0^2+a^4 y0^4
整理得\eqref{1}式

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 楼主| hbghlyj 发表于 2024-12-29 08:09
Theorem 4.4 (Cayley's theorem). Let $C$ and $D$ two non-degenerate conics of $\mathbb{P}_{\mathbb{C}}^2$ meeting at four different points, and let
\[
\sqrt{\operatorname{det}(t C+D)}=A_0+A_1 t+A_2 t^2+\ldots
\]
be the Taylor expansion, at the point $t=0$, of the function $\sqrt{\operatorname{det}(t C+D)}$. Then, there exists a $n$-sided polygon inscribed in $C$ and circumscribed about $D$ if and only if,
\[
\begin{aligned}
& \left|\begin{array}{ccc}
A_2 & \ldots & A_{m+1} \\
\vdots & & \vdots \\
A_{m+1} & \ldots & A_{2 m}
\end{array}\right|=0, \quad \text { when } n \text { is odd and } n=2 m+1 \text {, for some } m \geq 1 \\
& \left|\begin{array}{ccc}
A_3 & \ldots & A_{m+1} \\
\vdots & & \vdots \\
A_{m+1} & \ldots & A_{2 m-1}
\end{array}\right|=0, \quad \text { when } n \text { is even and } n=2 m, \text { for } \text { some } m \geq 2
\end{aligned}
\]
Proof. Recall that we have the curves $E=\left\{(r, u): u_0^2 r_1^3=u_1^2 \cdot \operatorname{det}\left(r_0 C+r_1 D\right)\right\}$ and $\gamma=\{(r, s): H(r, s)=0\}$ in $\mathbb{P}_{\mathbb{C}}^1 \times \mathbb{P}_{\mathbb{C}}^1$, as well as the isomorphisms
\[
G: E \longrightarrow \gamma, \quad G\left(\left(r_0: r_1\right),\left(u_0: u_1\right)\right)=\left(\left(r_0: r_1\right),\left(2 T_2(r) \cdot u_1:-T_1(r) \cdot u_1+u_0 \cdot r_1^2\right)\right)
\]
\[
F: \gamma \longrightarrow \mathfrak{M}, \quad F(r, s)=(p(r), l(s))
\]
Let's consider the isomorphism $\psi=G^{-1} \circ F^{-1}: \mathfrak{M} \longrightarrow E$.
If we take $\theta=\left(p_0, l_0\right)=\left(p((1: 0)), l_0\right)$ as the neutral element on $\mathfrak{M}$ (where $l_0$ is the tangent line to $D$ through $p_0$ ),
\[
\psi(\theta)=G^{-1}\left(F^1\left(p((1: 0)), l_0\right)\right)=G^{-1}\left((1: 0), l^{-1}\left(l_0\right)\right)=\left((1: 0),\left(u_0: u_1\right)\right)
\]
for some $\left(u_0: u_1\right)$ such that $\left((1: 0),\left(u_0: u_1\right)\right) \in E$. That is, $\psi(\theta)=((1: 0),(1: 0))$.
Thus choosing $((1: 0),(1: 0))$ as the neutral element on $E$, by corollary $2.18 \psi$ is also a group isomorphism.
Note that $\eta(\theta)=(\widetilde{p}, \widetilde{l})$, where $\widetilde{p}$ satisfies $C \cap l_0=\left\{p_0, \widetilde{p}\right\}$ and $\widetilde{l}$ is the tangent line to $D$ through $\widetilde{p}$. But $\widetilde{p}=p((0: 1))$ (since $l_0$ is the tangent line to $D=C_{(0: 1)}$ through $p_0$ ), so
\[
\psi(\eta(\theta))=G^{-1}\left(F^1(p((0: 1)), \widetilde{l})\right)=G^{-1}\left((0: 1), l^{-1}(\widetilde{l})\right)=\left((0: 1),\left(u_0: u_1\right)\right)
\]
for some $\left(u_0: u_1\right)$ such that $\left((0: 1),\left(u_0: u_1\right)\right) \in E$. From the equation for $E$, it follows that $\left(u_0: u_1\right)=( \pm \sqrt{\operatorname{det} D}: 1)$.
Using that $\psi$ is a group isomorphism and Lemma 4.3, we deduce that
\[
\begin{aligned}
& \eta^n=I d_{\mathfrak{M}} \Longleftrightarrow \eta(\theta) \text { is a torsion point of } \mathfrak{M} \text { of order } n \Longleftrightarrow \\
& \Longleftrightarrow \psi(\eta(\theta))=((0: 1),( \pm \sqrt{\operatorname{det} D}: 1)) \text { is a } n \text {-torsion point of } E
\end{aligned}
\]
Now, let's study the restriction of the elliptic curve $E$ to the affine chart $A_1=\left\{((x: 1),(y: 1)) \in \mathbb{P}_{\mathbb{C}}^1 \times \mathbb{P}_{\mathbb{C}}^1: x, y \in \mathbb{C}\right\}$ of $\mathbb{P}_{\mathbb{C}}^1 \times \mathbb{P}_{\mathbb{C}}^1$. Namely, consider the plane affine curve
\[
E^{\prime}=\left\{(x, y) \in \mathbb{C}^2:((x: 1),(y: 1)) \in E\right\}=\left\{(x, y) \in \mathbb{C}^2: y^2=\operatorname{det}(x C+D)\right\}
\]
and its projective closure
\[
E^{\prime \prime}=\left\{(x: y: z) \in \mathbb{P}_{\mathbb{C}}^2: y^2 z=\operatorname{det}(x C+D z)\right\}
\]
Since the pair of points at infinity $((1: 0),(1: 0))$ is the neutral element on $E$, the neutral element on $E^{\prime \prime}$ must be on the line at infinity $z=0$ : it's the point $(0: 1: 0)$. Then,
$((0: 1),( \pm \sqrt{\operatorname{det} D}: 1))$ is a $n$-torsion point of $E \Longleftrightarrow(0, \pm \sqrt{\operatorname{det} D})$ is a $n$-torsion point of $E^{\prime} \Longleftrightarrow$
$\Longleftrightarrow(0: \pm \sqrt{\operatorname{det} D}: 1)$ is a $n$-torsion point of $E^{\prime \prime}$
and Cayley's theorem becomes a consequence of theorem 2.19 .

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