# Little whimsies

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——yet another wired mumbling after failing to fall asleep again.

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# 美国国歌《星条旗》

《星条旗》完整歌唱版

1814年一位律师登上了英国皇家战舰“雷鸣号”，来和英军谈判释放俘虏之事。事情谈妥之后英军却扣押了他们一行人，理由是他们了解了英军的编制和要进攻巴尔的摩的动向。于是在舰队进攻巴尔的摩的那个傍晚，他们只能眼睁睁的看着美军被狂轰滥炸。

I
Oh, say can you see by the dawn’s early light
What so proudly we hailed at the twilight’s last gleaming?
Whose broad stripes and bright stars through the perilous fight,
O’er the ramparts we watched were so gallantly streaming?
And the rocket’s red glare, the bombs bursting in air,
Gave proof through the night that our flag was still there.
Oh, say does that star-spangled banner yet wave
O’er the land of the free and the home of the brave?

II
On the shore, dimly seen through the mists of the deep,
Where the foe’s haughty host in dread silence reposes,
What is that which the breeze, o’er the towering steep,
As it fitfully blows, half conceals, half discloses?
Now it catches the gleam of the morning’s first beam,
In full glory reflected now shines in the stream
‘Tis the star-spangled banner! Oh long may it wave
O’er the land of the free and the home of the brave.

III
And where is that band who so vauntingly swore
That the havoc of war and the battle’s confusion,
A home and a country should leave us no more!
Their blood has washed out of their foul footsteps’ pollution.
No refuge could save the hireling and slave
From the terror of flight or the gloom of the grave
And the star-spangled banner in triumph doth wave
O’er the land of the free and the home of the brave.

IV
Oh! thus be it ever, when freemen shall stand
Between their loved home and the war’s desolation!
Bles’t with victory and peace, may the heav’n rescued land
Praise the Power that hath made and preserved us a nation.
Then conquer we must, when our cause it is just,
And this be our motto: “In God is our trust.”
And the star-spangled banner in triumph shall wave
O’er the land of the free and the home of the brave.

“我们信仰上帝”，此语永矢不忘。

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# 浅议膜蛤文化

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• 传统文化——西方文化：在西化浪潮的一次又一次冲击下，中国人对自身的定位有时很困惑，对传统极其不自信（比如上世纪八九十年代在文化上极其自卑的社会风气），江在其言行中表现出的极其自信的精神，会英语，俄语、熟悉外国的文化，同时又会赋诗作书画，吹拉弹唱无所不包，包容多元文化的同时又很讲究中国原则。这样一个形象对于许多人来说是一个楷模，对于如何对待这一矛盾给了不少启发。这样一个履历光鲜知识分子的形象能让人想起中国人熟悉的“格物致知诚意正心修身齐家治国平天下”的士大夫理想。
• 中国特色社会主义——欧美式的自由主义 ：我们常常会对中国的各种制度感到不满，很羡慕富庶的“外国”所施行的政治制度，渴望能自由浏览各种网页，坦然的发表自己的见解而不用担心被请喝茶。渴望一个清廉高效的政府，而不是经常爆出丑闻的官僚体系，更不喜欢看到学术腐败，公款贪污，强力管控言论，有实权的人不遵循法律等等。面对外国人的批评和抨击，面对国内乱象迭生的社会，很多人认为大力政治改革，施行西方式的政治体制，似乎这些问题就能容易解决。而江的言行从侧面能够展现出他对制度的自信，面对西方、国内的抨击言论，他的回答通常是有力的，有理的，中共的官方言论经常给人呆板僵硬的感觉，但是江通过他的言行很好的解释了中共的态度和原则。这样一个领导人也许的确让许多人对这个政治体制多了份期许。

A. 个人收录的一些江的活动视频

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# 小游郑东新区

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# 使用Stackedit进行写作

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### 什么是Stackedit

Full-featured, open-source Markdown editor based on PageDown, the Markdown library used by Stack Overflow and the other Stack Exchange sites.

• Markdown语法：用Markdown语法写文章会十分的方便，插入数学公式也变得异常容易（配合MathJax就可以直接写$\LaTeX$格式的公式，Stackedit 说明里面介绍了怎么使用MathJax
• 离线与同步：这款编辑器可以直接以插件的方式直接在Chrome浏览器里面离线使用，所有文章都与blog、网盘的版本同步，根本不用担心文章丢失的问题。
• WYSIWYG 所见即所得$\LaTeX$Markdown写文章不能边写边看文章的排版效果，必须等到写完之后才能看，而这个编辑器就实现了所见即所得！左侧是文章的脚本，右侧就是动态显示效果！如图
• 开源，免费

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# How to determine whether 4 points are concyclic

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## Problem

Given 4 points $a,b,c,d$ in plane, how to determine they fall on a common circle?
There are lots of methods can do this, and here I introduce the determinant method, which is very straightforward, elegant and can easily generalized to quadratic curve.
The following content is inspired by this.
First of all, we can write the circle equation defined by $a,b,c$ by a determinant:
$D:=\begin{vmatrix} a_x & a_y & a_x^2+a_y^2 & 1 \\ b_x & b_y & b_x^2+b_y^2 & 1 \\ c_x & c_y & c_x^2+c_y^2 & 1 \\ p_x & p_y & p_x^2+p_y^2 & 1 \end{vmatrix}=0$

#### Proposition 1:

When $D>0$, $p$ is inside the circle defined by $a,b,c$，and outside when $D<0$ , right on the circle when $D=0$,which means a circle go through $a,b,c,p$ simultaneously.

Proof:
Notice that:
$D = -p_x M_{41} + p_y M_{42} - (p_x^2 +p_y^2) M_{43} + M_{44}$
($M_{ij}$ is the minor of the matrix)
and if we regard $p$ as a moving point, then $D=0$ is actually an equation of a circle:

$(p_x+\frac{1}{2}\frac{M_{41}}{M_{43}})^2+(p_y-\frac{1}{2}\frac{M_{42}}{M_{43}})^2=(\frac{1}{2}\frac{M_{41}}{M_{43}})^2+(\frac{1}{2}\frac{M_{42}}{M_{43}})^2+\frac{M_{44}}{M_{43}}$

i.e. the circle locate at $(x_0,y_0)=(\frac{1}{2}\frac{M_{41}}{M_{43}},-\frac{1}{2}\frac{M_{42}}{M_{43}})$
with radius $r_0^2=x_0^2+y_0^2+\frac{M_{44}}{M_{43}}$.

When $D>0$ the circle equation become $(x-x_0)^2+(y-y_0)^2

So it’s clear that if $(p_x,p_y)$ fall inside that circle, then $D>0$ and vice versa.
$\square$

Note: According to Cramer’s rule, $(-\frac{M_{41}}{M_{43}},\frac{M_{42}}{M_{43}},\frac{M_{44}}{M_{43}})$ is exactly the solution of the equation
$\begin{pmatrix} a_x & a_y & 1 \\ b_x & b_y & 1 \\ c_x & c_y & 1 \end{pmatrix} \boldsymbol{x}=\begin{pmatrix}a_x^2+a_y^2 \\ b_x^2+b_y^2 \\c_x^2+c_y^2\end{pmatrix}$

And this 3 parameters perfectly define a circle pass through $a,b,c$.One can easily verify when $p$ is $a,b$ or $c$, $D=0$ by the equation above.

## Generalization

We can generalize this method to the question of determine whether 6 points $a,b,c,d,e,f$ lies on some quadratic curve $\mathcal{C}$
Any equation of quadratic curve can write as (and again, we regard $f$ as a moving point $p$)
$D:=\begin{vmatrix} a_x^2 & a_x a_y & a_y^2 &a_x & a_y & 1 \\ b_x^2 & b_x b_y & b_y^2 &b_x & b_y & 1 \\ c_x^2 & c_x c_y & c_y^2 &c_x & c_y & 1 \\ d_x^2 & d_x d_y & d_y^2 &d_x & d_y & 1 \\ e_x^2 & e_x e_y & e_y^2 &e_x & e_y & 1 \\ p_x^2 & p_x p_y & p_y^2 &p_x & p_y & 1 \\ \end{vmatrix}=0$
and expand the last row of $D$:
$D=−p_x^2 M_{61}+p_x p_y M_{62}−p_y^2 M_{63}+p_x M_{64}−p_y M_{65}+M_{66}$

Definition.
A point is inside of a quadratic curve $\mathcal{C}$ if there are no tangent of that curve pass through $p$. And outside if there is at least 1 tangent pass through it.

And still we can tell whether $p$ lies in or out of $\mathcal{C}$ by check the sign of $D$.
Let $A,B,C,D,E,F$ be the coefficient of the quadratic equation, we have:

Proposition
The point $p$ lies inside of a quadratic curve $\mathcal{C}$ If $D$ has the same sign with the determinant
$\begin{vmatrix}A & \frac{1}{2}B & \frac{1}{2}D\\ \frac{1}{2}B & C & \frac{1}{2}E \\ \frac{1}{2}D & \frac{1}{2}E & F \end{vmatrix}$

But unfortunately I can’t give the proof temporarily. If you have any idea, please let me know, thanks in ahead!

### Check the shape of the curve

If we want to check whether the six point lies on eclipse or hyperbola etc. we can simply check the sign of geometry invariants

Symbols.
$I_1=A+C$
$I_2=\begin{vmatrix} A & \frac{1}{2}B \\\ \frac{1}{2}B & C\end{vmatrix}$
$I_3=\begin{vmatrix}A & \frac{1}{2}B & \frac{1}{2}D\\\ \frac{1}{2}B & C & \frac{1}{2}E \\\ \frac{1}{2}D & \frac{1}{2}E & F \end{vmatrix}$
$K_1=\begin{vmatrix} A & \frac{1}{2}D \\\ \frac{1}{2}D & F\end{vmatrix} + \begin{vmatrix} C & \frac{1}{2}E \\\ \frac{1}{2}E & F\end{vmatrix}$

Theorem.
The curve is classified by the following criterion:
Eclipse : $I_2>0,I_1 I_3<0$
Parabola: $I_2=0,I_3\neq 0$
Hyperbola: $I_2<0,I_3\neq 0$
Two parallel lines : $I_2=I_3,K_1<0$
Two intersecting lines: $I_2<0,I_3=0$
A line( two line coincide): $I_2=I_3=K_1=0$
Imaginary ellipse :$I_2>0, I_1 I_3 >0$
Point(two im. line intersect at real plane) : $I_2>0,I_3=0$
Two parallel imaginary lines: $I_2=I_3=0,K_1>0$

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