两个椭圆的闵可夫斯基和是椭圆的条件

hbghlyj

$a_1,b_1,a_2,b_2>0,$
两个椭圆$X=\left\{x \inR^2: \frac{x_1^2}{a_1^2}+\frac{x_2^2}{b_1^2} \leq 1\right\} \text { and } Y=\left\{x \inR^2: \frac{x_1^2}{a_2^2}+\frac{x_2^2}{b_2^2} \leq 1\right\}$
若$Z=\{x+y:x \in X, y \in Y\}$是一个椭圆，那么它们是位似的

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hbghlyj

如果$X,Y$位似，就会有$Z=\left\{x \inR^2: \frac{x_1^2}{(a_1+a_2)^2}+\frac{x_2^2}{(a_2+b_2)^2} \leq 1\right\}$吧？

hbghlyj

support function 定义为：在$X$中与$l$的内积的最大值。

math.stackexchange.com/questions/31333/
Minkowski sum of two compact convex sets is easily computed if they are represented in terms of support functions, one just adds the two support vectors for each direction.
$$X \oplus Y = \{x+y : x \in X \quad \mathrm{and}\quad y \in Y\}$$
$\rho(l,X\oplus Y) = \rho(l,X) + \rho(l,Y)$ where  $l \in \mathbb{R}^n$

$Z$等于$X,Y$的Minkowski和，则$Z$的support function等于$X,Y$的support function之和。

hbghlyj

设椭圆$Z=\left\{x \inR^2: \frac{x_1^2}{A^2}+\frac{x_2^2}{B^2} \leq 1\right\}$是$X,Y$的Minkowski和，
椭圆$X$与$(\cos\theta,\sin\theta)$的内积的最大值$=\sqrt{a_1^2 \cos ^2 \theta+b_1^2 \sin ^2 \theta}$,
椭圆$Y$与$(\cos\theta,\sin\theta)$的内积的最大值$=\sqrt{a_2^2 \cos ^2 \theta+b_2^2 \sin ^2 \theta}$,
椭圆$Z$与$(\cos\theta,\sin\theta)$的内积的最大值$=\sqrt{A^2 \cos ^2 \theta+B^2 \sin ^2 \theta}$,
那么
$$\forall\theta:\quad\sqrt{a_1^2 \cos ^2 \theta+b_1^2 \sin ^2 \theta}+\sqrt{a_2^2 \cos ^2 \theta+b_2^2 \sin ^2 \theta}=\sqrt{A^2 \cos ^2 \theta+B^2 \sin ^2 \theta}$$
取$\theta=0$得$A=a_1+a_2$,
取$\theta=\frac\pi2$得$B=b_1+b_2$,
式子变成
\[\forall\theta:\sqrt{a_1^2 \cos ^2 \theta+b_1^2 \sin ^2 \theta}+\sqrt{a_2^2 \cos ^2 \theta+b_2^2 \sin ^2 \theta}=\sqrt{\left(a_1+a_2\right)^2 \cos ^2 \theta+\left(b_1+b_2\right)^2 \sin ^2 \theta}\]
除以$\sin^2\theta$,
\[\forall\theta:\sqrt{a_1^2 \operatorname{ctg}^2 \theta+b_1^2}+\sqrt{a_2^2 \operatorname{ctg}^2 \theta+b_2^2}=\sqrt{\left(a_1+a_2\right)^2 \operatorname{ctg}^2 \theta+\left(b_1+b_2\right)^2} .\]

hbghlyj

两边平方，
\begin{multline*}\forall\theta:
\left(a_1 \operatorname{ctg}(\theta)\right)^2+b_1^2+\left(a_2 \operatorname{ctg}(\theta)\right)^2+b_2^2+2 \sqrt{\left(a_1 \operatorname{ctg}(\theta)\right)^2+b_1^2} \sqrt{\left(a_2 \operatorname{ctg}(\theta)\right)^2+b_2^2}= \\
=\left(a_1 \operatorname{ctg}(\theta)\right)^2+\left(a_2 \operatorname{ctg}(\theta)\right)^2+2 a_1 \operatorname{ctg}(\theta) a_2 \operatorname{ctg}(\theta)+b_1^2+b_2^2+2 b_1 b_2
\end{multline*}
消去$\left(a_1 \operatorname{ctg}(\theta)\right)^2+b_1^2+\left(a_2 \operatorname{ctg}(\theta)\right)^2+b_2^2$,
\[\forall\theta:\sqrt{\left(\left(a_1 \operatorname{ctg}(\theta)\right)^2+b_1{ }^2\right)\left(\left(a_2 \operatorname{ctg}(\theta)\right)^2+b_2^2\right)}=a_1 \operatorname{ctg}(\theta) a_2 \operatorname{ctg}(\theta)+b_1 b_2 .\]
再平方，
\[\forall\theta:\left(\left(a_1 \operatorname{ctg}(\theta)\right)^2+b_1^2\right)\left(\left(a_2 \operatorname{ctg}(\theta)\right)^2+b_2^2\right)=\left(a_1 \operatorname{ctg}(\theta) a_2 \operatorname{ctg}(\theta)+b_1 b_2\right)^2 .\]
消去$\left(a_1 \operatorname{ctg}(\theta) a_2\right)^2+\left(\operatorname{ctg}(\theta)+b_1 b_2\right)^2$,
\[\forall\theta:\left(a_1 \operatorname{ctg}(\theta)\right)^2 b_2{ }^2+b_1{ }^2\left(a_2 \operatorname{ctg}(\theta)\right)^2=2 a_1 \operatorname{ctg}(\theta) a_2 \operatorname{ctg}(\theta) b_1 b_2\]
消去$\operatorname{ctg}(\theta)^2$,
\[
a_1^2 b_2^2+b_1^2 a_2^2=2 a_1 a_2 b_1 b_2 \Rightarrow\left(a_1 b_2-b_1 a_2\right)^2=0 \Rightarrow a_1/a_2=b_1/b_2
\]
所以$X,Y$位似

hbghlyj

hbghlyj 发表于 2024-3-21 17:42
椭圆$X$与$(\cos\theta,\sin\theta)$的内积的最大值$=\sqrt{a_1^2 \cos ^2 \theta+b_1^2 \sin ^2 \theta}$,

补充这个结论的证明：
椭圆$X$与$(\cos\theta,\sin\theta)$的内积的最大值显然是在$X$的边界上取到。设$X$边界上一点为$(a_1\cos t,b_1\sin t)$，
$$\max_t(\cos\theta,\sin\theta)\cdot(a_1\cos t,b_1\sin t)=\max_t(a_1\cos\theta,b_1\sin\theta)\cdot(\cos t,\sin t)=|(a_1\cos\theta,b_1\sin\theta)|$$