Quiz

In the diagram, the circumference of the external large circle is
1) longer, or
2) shorter, or
3) equal to,
the sum of the circumferences of all inner circles centered on the common diameter, tangent to each other.

Facebook & Ranking Elo Formula

Eduardo Saverin (now a Singaporean billionaire investor) gave the wrong Elo formula to his Facebook co-founder Mark Zuckerburg, both of them became 'accidental' billionaire. Watch the video clip in the movie "Social Network":

http://m.youtube.com/#/watch?v=BzZRr4KV59I

The Elo formula is based on the theory of Normal Distribution with Logarithm function, from base of exponential e to base of 10.
The correct Elo Formula should be :
$Latex \boxed
{
E_a =\frac{1}
{1+ \frac{1}{400}.\Huge 10^{(R_b - R_a)}
}
}$

$Latex \boxed
{
E_b =\frac{1}
{1+ \frac{1}{400}.\Huge 10^{(R_a - R_b)}
}
}$

Eduardo had missed the power ^ below:

Sabbath Number 7

Sabbath Number '7':

1. Sunday: God rested on 7th day, a Holy day.

2. Professors rest on 7th year Sabbatical leave to re-charge: overseas exchange-cum-research paid holidays.

3. All farm lands must rest on 7th year before cultivation again.

4. After 6 years (Grade 1 to 6) and the stressful Primary School Leaving Exams (PSLE), relax on Secondary 1 (Grade 7); then after 4+2 years Cambridge 'O' and 'A' levels high-school pressure, 'honey-moon' in Freshman year (1st year University), or 'switch off' from study in the Army National Service.

5.  哈佛一调查报告说，人生平均只有7次决定人生走向的机会，两次机会间相隔约7年，大概25岁后开始出现，75岁以后就不会有什么机会了。这50年里的7次机会，第一次不易抓到，因为太年轻；最后一次也不用抓，因为太老。这样只剩5次，这里面又有两次会不小心错过(*)，所以实际上只有3次机会了。

人生七年之"痒" (机会)
25: 毕业寻职: 第一痒
32: 结婚成家:第二痒
39: 创业: 第三痒 (*)
46: 名利: 第四痒
53: 中年危机: 第五痒(*)
60: 金盆洗手: 第六痒
67: 夕阳无限: 第七痒

Love Math

For those who love Math ...
1x8+1=9
12x8+2=98
123x8+3=987
1234x8+4=9876
12345x8+5=98765
123456x8+6=987654
1234567x8+7=9876543
12345678x8+8=98765432
123456789x8+9=987654321

1x9+2=11
12x9+3=111
123x9+4=1111
1234x9+5=11111
12345x9+6=111111
123456x9+7=1111111
1234567x9+8=11111111
12345678x9+9=111111111
123456789x9+10=1111111111

9x9+7=88
98x9+6=888
987x9+5=8888
9876x9+4=88888
98765x9+3=888888
987654x9+2=8888888
9876543x9+1=88888888
98765432x9+0=888888888

1x1=1
11x11=121
111x111=12321
1111x1111=1234321
11111x11111=123454321
111111x111111=12345654321
1111111x1111111=1234567654321
11111111x11111111=
123456787654321
111111111x111111111=
12345678987654321

100-digit Pi

One fine day when we reach above 80 years old, if the doctor accuses us of having dementia, then prove the doctor wrong by shocking him with 100-digit Pi memory :)

With Chinese single-syllable sound for numbers, better still if can sing it as a song, memorizing 100-digit pi is easy!

Math Chants

Math Chants make learning Math formulas or Math properties fun and easy for memory . Some of them we learned in secondary school stay in the brain for whole life, even after leaving schools for decades.

Math chant is particularly easy in Chinese language because of its single syllable sound with 4 musical tones (like do-rei-mi-fa) - which may explain why Chinese students are good in Math, as shown in the International Math Olympiad championships frequently won by China and Singapore school students.

1. A crude example is the quadratic formula which people may remember as a little chant:
"ex equals minus bee plus or minus the square root of bee squared minus four ay see all over two ay."

$latex \boxed{
x = \frac{-b \pm \sqrt{b^{2}-4ac}}
{2a}
}$

2. $latex \mathbb{NZQRC}$
Nine Zulu Queens Rule China

3. $latex \boxed {\cos 3A = 4\cos^{3} A - 3\cos A }$

cos = $ 1 (kö, Singapore Chinese Fujian福建話 / Taiwan 台湾闽南语dialect) [一元]

cos 3= $1.3
4cos^3 = $4.3
3cos = $3

$1.3 = $4.3 - $3

4. $latex \boxed{\pi = 3.14159\dots}$
Chant in Chinese Mandarin (Beijing):
山顶一寺一壶酒
Chinese sound means: On the mountain (3) top (.) there is one (1) temple (4) with a (1) bottle (5) of wine (9)

Note : Below is the 22 decimal memory chant:
山顶一寺一壶酒，尔乐, 我三壶把酒吃，酒杀尔，杀不死，乐而乐。

$latex \boxed{
\pi = 3.14159 \:26 \:535897 \:932 \:384 \:626 \dots
}$


5. Group Properties:
$latex \boxed{\text{CAN I ?} }$
C: Closure
A: Associative
N: Neutral (e)
I: Inverse

If only 50%, it is Semi-Group(C, A) 半群
If No Inverse, it is MoNoId (C,A,N) 幺群
(MoNoId : No I)

6. The eccentric choice of letters (a,h,g,b,f,c, skipping iand e) for the coefficients in Conic equation: $latex \boxed
{ ax^{2} + 2hxy + 2gx + by^{2}+ 2fy + c = 0
}$
Math Chant for the coefficients:
{a h g b f}
"all hairy guys big feet"

c: constant term (understood)

成语数学

中国的小学离校考试 (PSLE) 「神题」: 猜成语
1) 20 除 3
2）1 除100
3）9寸+1寸=1尺
4）12345609
5）1,3,5,7,9

答案：:
1) 20/3= 6.666 六六大顺
2）百中挑一
3）得寸進尺
4）七零八落
5）举世无双

More on 666

$latex 666 = 1^{6} - 2^{6} + 3^{6}$

$latex 666 = 6 +6 +6 +6^{3}+ 6^{3}+6^{3}$

$latex 666 = 2^{2}+ 3^{2}+ 5^{2}+ 7^{2}+ 11^{2}+ 13^{2}+ 17^{2}$

$latex \phi(666) = 666 $
where
$latex \phi(n) = \text{number of integers less than n and co-prime with n }$

Bible Code

Appeal to my Blog's international readers (currently at 41 countries):
As of today I have verified Bible Code works for 6 languages : English, Chinese, French, Irish, Japanese and Spanish.

Please help to comment below if the "Bible Code" works in your native language (Russian, Spanish, Italian, Korean, Indonesian...) It works best with Bible in King-James version.

Bible Code:

Step 1: From Genesis 1:1 chose any 1 word from the 10 words. (Eg. 'the')

Step 2: Count the length (n1) of the selected word ('the' = n1 = 3)

Step 3: Jump n1 (3) words to the next word (eg. 'created')

Step 4: count the length (n2) of the selected word ('created' = n2 = 7)

Step 5: Jump n2 (7) words to the next word.

...
Repeat till the selected word first appears in Genesis 1:3 verse.

You will always end with the word "God".

Note1: It also works for Chinese Bible with strokes笔划:
( )神 A blank to respect God's name ：can be counted as 1 or 2 words(上帝).

起10-＞ 空8->神9->1:3 神
初7-＞神...是9 -> 神9 ->1:3 ( )神
神9-＞混12-＞水4->1:3( )神
创6-＞空8-＞神9-＞1:3( )神
造10 ->面9->水4 ->1:3( )神
天4-＞空8-＞神9-＞1:3( )神
地6-＞沌7-＞灵7-＞1:3( )神

Note 2: it works for French Bible too!
1.1 Au 2 ->Dieu 4 -> la2 ->Or 2 ->terre5 -> les 3 ->l'abîme 6 ->sur 3
->1.3 Dieu

Note 3: It works for Japanese Bible too!
1:1 さ3->地6 ->な4->み3->の2-> も3
->あ4->の2->が5->て2->お4->い2
->1:3神

Note 4: It also works for Irish Bible !
1:1 báire 5 -> talamh 6 ->an 2-> agus 4 -> aghaidh 7-> ag 2 -> os 2 ->na 2 ->
1:3 (Dúirt : said) Dia (noun God)

Note 5: It works for Spanish too !
(à 1 mot près : +/- 1 word,
God said = Said God
Dios dijo = dijo Dios)

Mozart Pieces

Mozart Piece P

$latex \boxed{P= 0.027465 + 0.157692K + 0.000159446K^{2}}$
where
K = Köchel number (sequenced by time)

Note: This formula is 85% accurate, error not exceeding 2.

Example:
Mozart No. 40 (G Minor) Symphony
is K.550

Background: Wolfgang Amadeus Mozart (1756-1791) was one of the most prolific composers of all time. In 1862, the German musicologist Ludwig von Köchel made a chronological list of Mozart's musical work. This list is the source of the Köchel numbers, or "K numbers", that now accompany the titles of Mozart's pieces (e.g., Sinfonia Concertante in E-flat major, K. 364). The table below gives the Köchel numbers and composition dates of 17 of Mozart's works.

Origamics

Origami + Mathematics = Origamics

The Origamics was invented and coined by its inventor a biologist Prof Kazuo Haga (Japan) in 1994.

Haga's First Theorem

Fold a square paper of size 1 unit x 1 unit: join the right-bottom vertex to the mid-point of the top edge.

This one fold creates respective edge points which mark out various ratios:

$latex \frac{1}{2}, \frac{1}{6}, \frac{1}{8}, \frac{3}{8}, \frac{\sqrt{5}}{2}, \frac{5}{24}, \dots $


Without ruler, this is the most accurate way to obtain length of $latex \frac{1}{3}, \sqrt{5} \dots $
As summarized in the diagram:

Cut a cake 1/5

Visually cut a cake 1/5 portions of equal size:

1) divide into half:




2) divide 1/5 of the right half:




3) divide half, obtain 1/5 = right of (3)

$latex \frac{1}{5}= \frac{1}{2} (\frac{1}{2}(1- \frac{1}{5}))= \frac{1}{2} (\frac{1}{2} (\frac{4}{5}))=\frac{1}{2}(\frac{2}{5})$





4) By symmetry another 1/5 at (2)=(4)




5) divide left into 3 portions, each 1/5

$latex \frac{1}{5}= \frac{1}{3}(\frac{1}{2}+ \frac{1}{2}.\frac{1}{5}) = \frac{1}{3}.\frac{6}{10}$

Mathew Effect of e^x

"For unto every one that hath shall be given, and he shall have abundance: but from him that hath not shall be taken even that which he hath."

—Matthew 25:29, King James Version

Or, "the rich gets richer, the poor gets poorer."

Mathematically, this is $latex e^{x} \text { increases much faster than x increases} $

$latex \frac{d}{dx} e^{x} = e^{x}$

Pigeonhole Principle

$latex \pi = 3.14159265358979323846264 $
$latex \text{Let } a_1, a_2,\dots a_{24} \text{ represent the first 24 digits of } \pi$
Prove:
$latex (a_1 - a_2)(a_3 - a_4) \dots (a_{23} - a_{24}) \text{ is even}$

Proof:

13 Odd digits = {3.14159265358979323846264 }

11 Even digits

$latex \text {12 brackets :}(a_1 - a_2)(a_3 - a_4) \dots (a_{23} - a_{24})$

Put 13 odds into 12 brackets, by Pigeonhole Principle, there is certainly one bracket where
$latex (a_j - a_k) \text{ is a difference of 2 odds, which is an even = 2n}$

2n multiplies with any number will always give even.
The product of 2n with the other 11 brackets will always be even.

Therefore
$latex (a_1 - a_2)(a_3 - a_4) \dots (a_{23} - a_{24}) \text { is even}$

Love + Hate Math

Book on Loving and Hating Mathematics:

ISBN: 978-0-691-142470
Reuben Hershey and Vera John-Steiner
Princeton Press

Φ and 666

We still don't understand these mysterious numbers appearing everywhere in the Nature.
Mathematicians call them 'Transcendental number' (超函数) - basically they are irrational numbers (like $latex \sqrt {2}$ ), in additional, they are NOT solution of any polynomial equations (otherwise, they are called Algebraic number).

Top 3 mysterious numbers are:
$latex \pi, e, \phi$
connected by this formula:
$latex \boxed{e^{i.\pi} + 2\phi = \sqrt{5}}$
$latex \pi \text { involves anything in circle}$

$latex e \text{ anything with exponential growth:}$ $latex \text{ e.g. epidemic, bank interest...}$

$latex \phi \text{ the golden ratio involves symmetry and beauty. }$


Golden Ratio Φ

$latex \boxed{ \phi = 1.61803 \dots = -2 \sin 666 ^\circ}$

Recognize the Satanic number 666?

No wonder we praise in Chinese a gorgeous pretty lady having 魔鬼身材 ('satanic' body shape). It has mathematical truth with the golden ratio of beauty Φ and 666. :)

See also my next blog "More on 666 "

Note:
$latex \boxed{\frac{6}{5} \phi^{2}=\pi }$

Beautiful Trigonometry

Beautiful Trigonometry
tan α +tan β +tan γ = tan α.tan β.tan γ

庖丁解牛数学方法

"庖丁解牛"数学方法
庄子讲庖丁(butcher)解牛有三个功夫階段：
1st Level: 看见一只全牛 (Whole Cow)

2nd Level: 三年后，不见全牛，只见牛的生理结構(Anatomy) ：骨骼，肌肉，筋腱。

3rd Level: 不以目视而是神视，＂与桑林之舞合拍，与经首之会同律。＂达到了"物我"两忘的境界。Intuition.

数学的方法也如此。

1st Level: Whole Math (Primary school to High School)

2nd Level: Component Structure - (Undergraduate Math)：
Macro- structure (Algebra : Group, Ring, Field, Vector Space... );
Micro-structure (Analysis : Calculus, Topology, etc)

3rd Level: 无处不数 Ubiquitous Math - (Graduate Math)
eg. Fermat's Last Theorem used all Math theories available today to prove.

IMO Technique

(a+b)³ = a³ + 3a²b+ 3ab² + b³

Different equivalent forms:
(1)：(a+b)³ = a³ + b³+3ab(a+b)
(2)：a³ + b³ = (a+b)³ - 3ab(a+b)
(3): a³ + b³ = (a+b)(a² -ab + b²)
(4)：(a+b)³ - ( a³ + b³ ) = 3ab(a+b)

1997 USAMO Q5:
Prove:
$latex \frac{1}{a^{3}+b^{3}+abc} +
\frac{1}{b^{3}+c^{3}+abc} +
\frac{1}{c^{3}+a^{3}+abc} \leq
\frac{1}{abc}$

Proof:
Apply (3):
a³ + b³ = (a+b)(a² -ab + b²) ≥ (a+b)ab

Note:
a² -ab + b²= (a-b)² + ab ≥ ab
since (a-b)² ≥ 0

$latex \frac{abc}{a^{3}+b^{3}+abc}
\leq \frac{abc}{(a+b)ab + abc}
= \frac{c}{a+b+c}$


Symmetrically,
$latex \frac{abc}{b^{3}+c^{3}+abc}
\leq \frac{a}{a+b+c}$

$latex \frac{abc}{c^{3}+a^{3}+abc}
\leq \frac{b}{a+b+c}$

Add 3 RHS:
$latex \frac{a+b+c}{a+b+c} = 1$

$latex \frac{abc}{a^{3}+b^{3}+abc} +
\frac{abc}{b^{3}+c^{3}+abc} +
\frac{abc}{c^{3}+a^{3}+abc} \leq 1$

$latex \frac{1}{a^{3}+b^{3}+abc} +
\frac{1}{b^{3}+c^{3}+abc} +
\frac{1}{c^{3}+a^{3}+abc} \leq
\frac{1}{abc}$

[QED]

Generalized Analytic Geometry

Generalized Analytic Geometry

Find the equation of the circle which cuts the tangent 2x-y=0 at M(1,4), passing thru point A(4,-1).

Solution:

1st generalization:
Let the point circle be:
(x-1)² + (y-4)² =0

2nd generalization:
It cuts the tangent 2x-y=0
(x-1)² + (y-4)² +k(2x-y) =0 ...(C)

Pass thru A(4,-1)
x=4, y= -1
=> k= -2
(C): (x-3)² + (y-1)² = 0
[QED]