为什么只看大图模式没有21^#的图片呢? 是因为它在第2页吗

complex.pdf page 4

Lemma 1.4. Let $a, b \in \mathbb{C}$ be distinct complex numbers, and let $\lambda \in(0, \infty)$, $\lambda ≠ 1$. Then the locus of complex numbers satisfying $\lvert z-a\rvert=\lambda\lvert z-b\rvert$ is a circle. Conversely, every circle can be written in this form.

Proof. Without loss of generality, $b=0$ (a translate of a circle is a circle). Now observe the identity$$\lvert t z+a\rvert^2=t(t+1)\lvert z\rvert^2-t\lvert z-a\rvert^2+(t+1)\lvert a\rvert^2,$$valid for all $a, z \in \mathbb{C}$ and all $t \in \mathbb{R}$. This can be checked by a slightly tedious calculation. Applying it with $t=\lambda^2-1$ gives$$\left\lvert\left(\lambda^{2}-1\right) z+a\right\rvert=\lambda\lvert a\rvert,$$which is clearly the equation of a circle. Taking $a=-c\left(\lambda^{2}-1\right)$ and $\lambda=r /\lvert c\rvert$, this gives $\lvert z-c\rvert=r$, and so every circle can be written in this form. Remark. Lemma 1.4 is an interesting and non-obvious fact in classical Euclidean geometry. Phrased in that language, if $A, B$ are points in the plane, and if $\lambda \in(0, \infty), \lambda≠1$, then the set of all points $P$ such that $\lvert P A\rvert /\lvert P B\rvert=\lambda$ is a circle. We have just proven that this is true using complex numbers.

Apollonius' definition of a circle

Main Article: Circle#Circle of Apollonius

Figure 1. Apollonius' definition of a circle.

A circle is usually defined as the set of points P at a given distance \(r\) (the circle's radius) from a given point (the circle's center). However, there are other, equivalent definitions of a circle. Apollonius discovered that a circle could be defined as the set of points P that have a given ratio of distances \(k=\frac{d_1}{d_2}\) to two given points (labeled A and B in Figure 1). These two points are sometimes called the foci.

Proof using vectors in Euclidean spaces

Let $d_1,d_2$ be non-equal positive real numbers. Let C be the internal division point of AB in the ratio $d_1:d_2$ and D the external division point of AB in the same ratio, $d_1:d_2$. $$\overrightarrow{\mathrm{PC}} = \frac{d_{2}\overrightarrow{\mathrm{PA}}+d_{1}\overrightarrow{\mathrm{PB}}}{d_{2}+d_{1}},\ \overrightarrow{\mathrm{PD}} = \frac{d_{2}\overrightarrow{\mathrm{PA}}-d_{1}\overrightarrow{\mathrm{PB}}}{d_{2}-d_{1}}.$$ Then, \begin{aligned} &\mathrm{PA} : \mathrm{PB} = d_{1} : d_{2}. \\ \Leftrightarrow{}& d_{2}|\overrightarrow{\mathrm{PA}}| = d_{1}|\overrightarrow{\mathrm{PB}}|. \\ \Leftrightarrow{}& d_{2}^2|\overrightarrow{\mathrm{PA}}|^2 = d_{1}^2|\overrightarrow{\mathrm{PB}}|^2. \\ \Leftrightarrow{}& (d_{2}\overrightarrow{\mathrm{PA}}+d_{1}\overrightarrow{\mathrm{PB}})\cdot (d_{2}\overrightarrow{\mathrm{PA}}-d_{1}\overrightarrow{\mathrm{PB}})=0. \\ \Leftrightarrow{}& \frac{d_{2}\overrightarrow{\mathrm{PA}}+d_{1}\overrightarrow{\mathrm{PB}}}{d_{2}+d_{1}}\cdot \frac{d_{2}\overrightarrow{\mathrm{PA}}-d_{1}\overrightarrow{\mathrm{PB}}}{d_{2}-d_{1}} = 0. \\ \Leftrightarrow{}& \overrightarrow{\mathrm{PC}} \cdot \overrightarrow{\mathrm{PD}} = 0. \\ \Leftrightarrow{}& \overrightarrow{\mathrm{PC}} = \vec{0} \vee \overrightarrow{\mathrm{PD}} =\vec{0} \vee \overrightarrow{\mathrm{PC}} \perp \overrightarrow{\mathrm{PD}}. \\ \Leftrightarrow{}& \mathrm{P}=\mathrm{C} \vee \mathrm{P}=\mathrm{D} \vee \angle{\mathrm{CPD}}=90^\circ. \end{aligned} Therefore, the point P is on the circle which has the diameter CD.

Proof using the angle bisector theorem

Proof of Apollonius' definition of a circle

First consider the point \(C\) on the line segment between \(A\) and \(B\), satisfying the ratio. By the definition \(\frac{|AP|}{|BP|}=\frac{|AC|}{|BC|}\) and from the angle bisector theorem the angles \(\alpha=\angle APC\) and \(\beta=\angle CPB\) are equal.

Next take the other point \(D\) on the extended line \(AB\) that satisfies the ratio. So \(\frac{|AP|}{|BP|}=\frac{|AD|}{|BD|}.\) Also take some other point \(Q\) anywhere on the extended line \(AP\). Also by the Angle bisector theorem the line \(PD\) bisects the exterior angle \(\angle QPB\). Hence, \(\gamma=\angle BPD\) and \(\delta=\angle QPD\) are equal and \(\beta+\gamma=90^{\circ}\). Hence by Thales's theorem \(P\) lies on the circle which has \(CD\) as a diameter.

Apollonius pursuit problem

The Apollonius pursuit problem is one of finding whether a ship leaving from one point A at speed \(v_\text{A}\) will intercept another ship leaving a different point B at speed \(v_\text{B}\). The minimum time in interception of the two ships is calculated by means of straight-line paths. If the ships' speeds are held constant, their speed ratio is defined by $μ$. If both ships collide or meet at a future point, \(I\), then the distances of each are related by the equation: \[a = \mu b\] Squaring both sides, we obtain: \[a^{2} = b^{2} \mu^{2}\] \[a^{2} = x^{2} + y^{2}\] \[b^{2} = (d-x)^{2} + y^{2}\] \[x^{2} + y^{2} = [(d-x)^{2} + y^{2}]\mu^{2}\] Expanding: \[x^{2}+y^{2} = [d^{2} + x^{2} - 2dx + y^{2}]\mu^{2}\] Further expansion: \[x^{2} + y^{2} = x^{2} \mu^{2} + y^{2}\mu^{2} + d^{2}\mu^{2} - 2dx \mu^{2}\] Bringing to the left-hand side: \[x^{2} - x^{2}\mu^{2} + y^{2} - y^{2}\mu^{2} - d^{2}\mu^{2} + 2dx\mu^{2} = 0\] Factoring: \[x^{2}(1-\mu^{2}) + y^{2}(1-\mu^{2}) - d^{2}\mu^{2} + 2dx\mu^{2} = 0\] Dividing by \(1-\mu^{2}\) : \[x^{2} + y^{2} - \frac{d^{2}\mu^{2}}{1-\mu^{2}} + \frac{2dx\mu^{2}}{1-\mu^{2}} = 0\] Completing the square: \[\left(x+ \frac{d\mu^{2}}{1-\mu^{2}}\right) ^{2}- \frac{d^{2}\mu^{4}}{(1-\mu^{2})^{2}} - \frac{d^{2} \mu^{2}}{1-\mu^{2}} + y^{2} = 0\] Bring non-squared terms to the right-hand side: \begin{align*} \left( x + \frac{d\mu^{2}}{1-\mu^{2}} \right)^{2} + y^{2} &= \frac{d^{2}\mu^{4}}{(1-\mu^{2})^{2}} + \frac{d^{2} \mu^{2}}{1-\mu^{2}}\\ &= \frac{d^{2} \mu^{4}}{(1-\mu^{2})^{2}} + \frac{d^{2} \mu^{2}}{1-\mu^{2}} \frac{(1-\mu^{2})}{(1-\mu^{2})}\\ &= \frac{d^{2}\mu^{4}+d^{2}\mu^{2}-d^{2}\mu^{4}}{(1-\mu^{2})^{2}}\\ &= \frac{d^{2} \mu^{2}}{(1-\mu^{2})^{2}} \end{align*} Then: \[\left( x + \frac{d\mu^{2}}{1-\mu^{2}}\right)^{2} + y^{2} = \left( \frac{d \mu}{1-\mu^{2}} \right)^{2}\] Therefore, the point must lie on a circle as defined by Apollonius, with their starting points as the foci.

Geometric interpretation of the bipolar coordinates.
The angle $σ$ is formed by the two foci and the point $P$, whereas $τ$ is the logarithm of the ratio of distances to the foci. The corresponding circles of constant $σ$ and $τ$ are shown in red and blue, respectively, and meet at right angles (magenta box); they are orthogonal.

Bipolar coordinates

Definition

The system is based on two foci $F_1$ and $F_2$. Referring to the figure at right, the $σ$-coordinate of a point $P$ equals the angle $F_1PF_2$, and the $τ$-coordinate equals the natural logarithm of the ratio of the distances $d_1$ and $d_2$: $$ \tau = \ln \frac{d_1}{d_2}.$$ If, in the Cartesian system, the foci are taken to lie at $(−a,0)$ and $(a,0)$, the coordinates of the point $P$ are $$ x = a \ \frac{\sinh \tau}{\cosh \tau - \cos \sigma}, \qquad y = a \ \frac{\sin \sigma}{\cosh \tau - \cos \sigma}. $$ The coordinate $τ$ ranges from $-\infty$ (for points close to $F_1$) to $\infty$ (for points close to $F_2$). The coordinate $σ$ is only defined modulo $2π$, and is best taken to range from $-π$ to $π$, by taking it as the negative of the acute angle $F_1PF_2$ if $P$ is in the lower half plane.

Proof that coordinate system is orthogonal

The equations for $x$ and $y$ can be combined to give $$ x + i y = a i \cot\left( \frac{\sigma + i \tau}{2}\right). $$ This equation shows that $σ$ and $τ$ are the real and imaginary parts of an analytic function of $x+iy$ (with logarithmic branch points at the foci), which in turn proves (by appeal to the general theory of conformal mapping) (the Cauchy-Riemann equations) that these particular curves of $σ$ and $τ$ intersect at right angles, i.e., that the coordinate system is orthogonal.

Reciprocal relations

The passage from the Cartesian coordinates towards the bipolar coordinates can be done via the following formulas: $$ \tau = \frac{1}{2} \ln \frac{(x + a)^2 + y^2}{(x - a)^2 + y^2} $$ and $$ \pi - \sigma = 2 \arctan \frac{2ay}{a^2 - x^2 - y^2 + \sqrt{(a^2 - x^2 - y^2)^2 + 4 a^2 y^2} }. $$ The coordinates also have the identities: $$ \tanh \tau = \frac{2 a x}{x^2 + y^2 + a^2} $$ and $$ \tan \sigma = \frac{2 a y}{x^2 + y^2 - a^2}. $$ which is the limit one would get a $x = 0$ from the definition in the section above. And all the limits look quite ordinary at $x =0$.

Scale factors

To obtain the scale factors for bipolar coordinates, we take the differential of the equation for $x + iy$, which gives $$ dx + i\, dy = \frac{-ia}{\sin^2\bigl(\tfrac{1}{2}(\sigma + i \tau)\bigr)}(d\sigma +i\,d\tau). $$ Multiplying this equation with its complex conjugate yields $$ (dx)^2 + (dy)^2 = \frac{a^2}{\bigl[2\sin\tfrac{1}{2}\bigl(\sigma + i\tau\bigr) \sin\tfrac{1}{2}\bigl(\sigma - i\tau\bigr)\bigr]^2} \bigl((d\sigma)^2 + (d\tau)^2\bigr). $$ Employing the trigonometric identities for products of sines and cosines, we obtain $$ 2\sin\tfrac{1}{2}\bigl(\sigma + i\tau\bigr) \sin\tfrac{1}{2}\bigl(\sigma - i\tau\bigr) = \cos\sigma - \cosh\tau, $$ from which it follows that $$ (dx)^2 + (dy)^2 = \frac{a^2}{(\cosh \tau - \cos\sigma)^2} \bigl((d\sigma)^2 + (d\tau)^2\bigr). $$ Hence the scale factors for $σ$ and $τ$ are equal, and given by $$ h_\sigma = h_\tau = \frac{a}{\cosh \tau - \cos\sigma}. $$ Many results now follow in quick succession from the general formulae for orthogonal coordinates. Thus, the infinitesimal area element equals $$ dA = \frac{a^2}{\left( \cosh \tau - \cos\sigma \right)^2} \, d\sigma\, d\tau, $$ and the Laplacian is given by $$ \nabla^2 \Phi = \frac{1}{a^2} \left( \cosh \tau - \cos\sigma \right)^2 \left( \frac{\partial^2 \Phi}{\partial \sigma^2} + \frac{\partial^2 \Phi}{\partial \tau^2} \right). $$ Expressions for $\nabla f$, $\nabla \cdot \mathbf{F}$, and $\nabla \times \mathbf{F}$ can be expressed obtained by substituting the scale factors into the general formulae found in orthogonal coordinates.

Circles of Apollonius

A set of Apollonian circles.
Every blue circle intersects every red circle at a right angle, and vice versa. Every red circle passes through the two foci. Every blue circle passes through the two (imaginary) foci.

Indra's Pearls

回复 3# 其妙

与阿氏圆有关，内角平分线定理和外角是平分线定理构造调和点列的一种手段（还有完全四点形（什么塞瓦、美丽 ...
其妙 发表于 2015-6-11 19:05

每日一题[161] 阿波罗尼斯圆（2015年高考数学湖北卷理科数学第14题（填空压轴题））：lanqi.org/everyday/3733/

回复 15# isee
我也看到了

这样的题让考生在高考场上作，出题者扯淡！
乌贼 发表于 2015-6-12 01:02

    初等方法来了：yddy11@人教论坛 4楼：

bbs.pep.com.cn/forum.php?mod=viewthread&t … 986&extra=page=1

    不过，第1个，无须数字计算的，\[ON^2=OT^2=OA\cdot OB=OM^2\Rightarrow \angle OBN=\angle ONM=\angle OMN=\angle OBM.\]

   从而BA是角分线，得到第1个成立。（横向看第1个，分子比分比，分母比分母。）

   不过，此法，第2与3又要麻烦些了，因为比值是多少，个人觉得，要借助解几或者阿氏圆来解决。

回复  活着＆存在
我没看出两者关联
乌贼 发表于 2015-6-13 00:12

    往深处说，就是极线与极点……其实就是这个了。

回复 12# 乌贼


    考试题目中，过A作Y轴的垂线与单位圆相交于点H，然后计算证明BH与OH垂直就可以了。

回复 8# 活着＆存在
我没看出两者关联

回复 10# isee


    那考题考的就是这个，而且不用几何，只要知道结论就可以，用代数更容易证，
前提就是课本题目（圆）。

回复 6# 乌贼

别这么偏激呀，这是几何法，非解析几何法啊，问题的不同方面……

回复 6# 乌贼


    把竞赛题或者研究题放入普通高考卷（不是1B卷），出卷子的在秀。

然而人家却说是回归课本，因为课本有对应的例题和练习。

回复 2# isee
如果$NE$是$\angle ANB$的平分线，那$ME$也是$\angle AMB$的平分线，$E$就是内心了，这样$BE$就平分$\angle MBN$，省去了计算了。

这样的题让考生在高考场上作，出题者扯淡！

题目有没有问题？点A是圆C与Y轴的一个交点，过A的直线还要与圆C相交于M、N两点？