Math Blog

Having a blog gives me a chance to defend myself against a number of people who took issue with a passage in Mathematics, A Very Short Introduction, where I made the tentative suggestion that an abstract approach to mathematics could sometimes be better, pedagogically speaking, than a concrete one — even at school level. This was part of a general discussion about why many people come to hate mathematics.

The example I chose was logarithms and exponentials. The traditional method of teaching them, I would suggest, is to explain what they mean and then derive their properties from this basic meaning. So, for example, to justify the rule that x^a+b=x^ax^b one would say something like that if you have a xs followed by b xs and you multiply them all together then you are multiplying a+b xs all together.

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Use of mathematics II

Today I had an experience that I have had many times before, and so, I imagine, has almost everybody (at least if they are old enough to be the kind of person who might conceivably read this blog post). I was in a queue in a chemist (=pharmacy=drugstore), and I knew that my particular item would be quick and easy to deal with. But I had to wait a while because in front of me was someone who had an item that was much more complicated and time-consuming. In this instance the complexity of the items was not due to their sizes, but a more common occurrence of the phenomenon is something that often happens to me in a local grocery: I want to buy just a pint of milk, say, and I find myself behind somebody who has a big basket of things, several of which have to be…

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On multiple choice questions in mathematics

Now that the project to upgrade my old multiple choice applet to a more modern and collaborative format is underway (see this server-side demo and this javascript/wiki demo, as well as the discussion here), I thought it would be a good time to collect my own personal opinions and thoughts regarding how multiple choice quizzes are currently used in teaching mathematics, and on the potential ways they could be used in the future. The short version of my opinions is that multiple choice quizzes have significant limitations when used in the traditional classroom setting, but have a lot of interesting and underexplored potential when used as a self-assessment tool.

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2012 H2 Maths Prelim Solution: SRJC/II/8(iv)

8(iv)

Using the model $\displaystyle y=a+\frac{b}{x-2}$ , estimate the total fertility rate for a particular country Z when its GDP per capita is USD$1000, giving your answer to 1 decimal place and comment on the reliability of the estimate.

First, we need to remove the outlier (40,6.6) as mentioned in part iii.

Then, performing linear regression with GC, (with variables $\frac{1}{x-2}$ , $y$ ), we get:

$a=0.974686$

$b=6.86442$

Substituting $x=1$ , we get $\displaystyle y=a+\frac{b}{1-2}=-5.9$ (1 d.p.)

Since we cannot have a negative fertility rate (the average number of children that would be born to a woman ), the estimate obtained for $y$ is not reliable.

Xinmin Secondary 2010 Prelim Paper I Q24 Solution (Challenging/Difficult Probability O Level Question)

A bag A contains 9 black balls, 6 white balls and 3 red balls. A bag B contains 6 black balls, 2 white balls and 4 green balls. Ali takes out 1 ball from each bag randomly. When Ali takes out 1 ball from one bag, he will put it into the other bag and then takes out one ball from that bag. Find the probability that

(a) the ball is black from bag A, followed by white from bag B,
(b) both the balls are white in colour,
(c) the ball is black or white from bag B, followed by red from bag A,
(d) both the balls are of different colours,
(e) both the balls are not black or white in colours.

Solution:

(a) $\displaystyle\frac{9}{18}\times\frac{2}{13}=\frac{1}{13}$

(b) Probability of white ball from bag A, followed by white ball from bag B= $\displaystyle=\frac{1}{2}\times\frac{6}{18}\times\frac{3}{13}=\frac{1}{26}$

Probability of white from B, followed by white from A= $\displaystyle=\frac{1}{2}\times\frac{2}{12}\times\frac{7}{19}=\frac{7}{228}$

Total prob= $\displaystyle\frac{205}{2964}$

$\displaystyle\frac{8}{12}\times\frac{3}{19}=\frac{2}{19}$

(d) Prob of both red = P(red from A, followed by red from B)= $\displaystyle\frac{1}{2}\times\frac{3}{18}\times\frac{1}{13}=\frac{1}{156}$

P(both green)=P(green from B, followed by green from A)= $\displaystyle\frac{1}{2}\times\frac{4}{12}\times\frac{1}{19}=\frac{1}{114}$

P(both black)=P(black from A, followed by black from B)+P(black from B, followed by black from A)= $\displaystyle\frac{1}{2}\times\frac{9}{18}\times\frac{7}{13}+\frac{1}{2}\times\frac{6}{12}\times\frac{10}{19}=\frac{263}{988}$

P(both white)= $\displaystyle\frac{205}{2964}$ (from part b)

$\displaystyle 1-\frac{1}{156}-\frac{1}{114}-\frac{263}{988}-\frac{205}{2964}=\frac{1925}{2964}$

(e)

P(neither black nor white from A, followed by neither black nor white from B)= $\displaystyle\frac{1}{2}\times\frac{3}{18}\times\frac{5}{13}=\frac{5}{156}$

P(neither black nor white from B, followed by neither black nor white from A)= $\displaystyle\frac{1}{2}\times\frac{4}{12}\times\frac{4}{19}=\frac{2}{57}$

$\displaystyle\frac{5}{156}+\frac{2}{57}=\frac{199}{2964}$

Why aren’t all functions well-defined?

I’m in the happy state of just having finished marking exams for this year. There is very little of interest to say about the week that was removed from my life: it would be fun to talk about particularly bizarre mistakes, but I can’t really do that, especially as the results are not yet known (or even fully decided). However, one general theme emerged that made no difference to anybody’s marks. There seems to be a common misconception amongst many Cambridge undergraduates that I’d like to discuss here in the hope that I can clear things up for a few people. (It is an issue that I have discussed already on my web page, but rather than turning that into a blog post I’m starting again.)

The question where the misconception made itself felt was one about functions, injections, surjections, etc. I noticed that a lot of people wrote things…

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Article in the New York Times, and maths education

I’ve received quite a lot of inquiries regarding a recent article in the New York Times, so I am borrowing some space on this blog to respond to some of the more common of these, and also to initiate a discussion on maths education, which was briefly touched upon in the article.

Firstly, some links:

The video for the talk “Structure and randomness in the prime numbers” mentioned in the article can be found here (requires RealPlayer). The slides can be found here. My other expository and research material on number theory can be found here.
I don’t have any specific advice regarding gifted education, though some articles on my own experiences can be found here. I do however have some thoughts on career advice at the undergraduate level and beyond.
I have some responses to several other common queries (e.g. regarding books…

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Displaying maths online, II

As the previous discussion on displaying mathematics on the web has become quite lengthy, I am opening a fresh post to continue the topic. I’m leaving the previous thread open for those who wish to respond directly to some specific comments in that thread, but otherwise it would be preferable to start afresh on this thread to make it easier to follow the discussion.

It’s not easy to summarise the discussion so far, but the comments have identified several existing formats for displaying (and marking up) mathematics on the web (mathML, jsMath, MathJax, OpenMath), as well as a surprisingly large number of tools for converting mathematics into web friendly formats (e.g. LaTeX2HTML, LaTeXMathML, LaTeX2WP, Windows 7 Math Input, itex2MML, Ritex, Gellmu, mathTeX, WP-LaTeX, TeX4ht, blahtex, plastex, TtH, WebEQ, techexplorer…

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Gamifying algebra?

High school algebra marks a key transition point in one’s early mathematical education, and is a common point at which students feel that mathematics becomes really difficult. One of the reasons for this is that the problem solving process for a high school algebra question is significantly more free-form than the mechanical algorithms one is taught for elementary arithmetic, and a certain amount of planning and strategy now comes into play. For instance, if one wants to, say, write $latex {\frac{1,572,342}{4,124}}&fg=000000$ as a mixed fraction, there is a clear (albeit lengthy) algorithm to do this: one simply sets up the long division problem, extracts the quotient and remainder, and organises these numbers into the desired mixed fraction. After a suitable amount of drill, this is a task that can be accomplished by a large fraction of students at the middle school level. But if, for instance, one has to solve…

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How should mathematics be taught to non-mathematicians?

Michael Gove, the UK’s Secretary of State for Education, has expressed a wish to see almost all school pupils studying mathematics in one form or another up to the age of 18. An obvious question follows. At the moment, there are large numbers of people who give up mathematics after GCSE (the exam that is usually taken at the age of 16) with great relief and go through the rest of their lives saying, without any obvious regret, how bad they were at it. What should such people study if mathematics becomes virtually compulsory for two more years?

A couple of years ago there was an attempt to create a new mathematics A-level called Use of Mathematics. I criticized it heavily in a blog post, and stand by those criticisms, though interestingly it isn’t so much the syllabus that bothers me as the awful exam questions. One might…

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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.

Answer: 3) equal

circumference = π. diameter

Let d be the diameter of the external large circle C
Let dj be the diameter of the inner circle Cj

$latex \displaystyle d = \sum_{j} d_j$
$latex \displaystyle \pi. d = \pi. \sum_{j} d_j= \sum_{j}\pi.d_j$

Circumference of the external circle
= sum of circumferences of all inner circles

MI H2 Maths Prelim Solutions 2010 Paper 2 Q7 (P&C)

MI H2 Maths Prelim 2010 Paper 2 Q7 (P&C)

Question:
7) Two families are invited to a party. The first family consists of a man and both his parents while the second family consists of a woman and both her parents. The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if
(i) there is no restriction, [1]
(ii) the men and women are seated alternately, [2]
(iii) members of the same family are seated together and the two other women must be seated separately, [3]
(iv) members of the same family are seated together and the seats are numbered. [2]

Solution:

(i) (10-1)!=9!=362880

(ii) First fix the men’s sitting arrangement: (5-1)!

Then the remaining five women’s total number of arrangements are: 5!

Total=4! x 5!=2880

(iii) Fix the 2 families (as a group) and the 2 men: (4-1)! x 3! x 3!

(3! to permute each family)

By drawing a diagram, the two women have 4 slots to choose from, where order matters: $^4 P_2$

Total = $(4-1)! \times 3! \times 3! \times ^4 P_2 = 2592$

(iv)

We first find the required number of ways by treating the seats as unnumbered: $(6-1)!\times 3!\times 3! =4320$

Since the seats are numbered, there are 10 choices for the point of reference, thus no. of ways = $4320 \times 10 =43200$

What maths A-level doesn’t necessarily give you

I had a mathematical conversation yesterday with a 17-year-old boy who is in his second year of doing maths A-level. Although a sample of size 1 should be treated with caution, I’m pretty sure that the boy in question, who is very intelligent and is expected to get at least an A grade, has been taught as well as the vast majority of A-level mathematicians. If this is right, then what I discovered from talking to him was quite worrying.

The purpose of the conversation was to help him catch up with some work that he had missed through illness. The particular topics he wanted me to cover were integrating $latex \log x$, or $latex \ln x$ as he called it, and integration by parts. (Actually, after I had explained integration by parts to him, he told me that that hadn’t been what he had meant, but I don’t think…

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Stratified sampling vs. quota sampling

SOC 382: the blog

Shawn asked a good question in class yesterday about the differences between stratified sampling and quota sampling. In terms of sampling mechanism (i.e. the actual process by which cases are chosen from the population), it is clear that these two samples are different. Unclear, however, is why they would lead to different results.

Recall that stratified sampling is conducted by dividing a population into two or more strata by virtue of some characteristic, and taking random samples from each strata. This is done when a simple random sample of an entire population will likely not generate enough analyzable cases for a given group of particular interest.

Let’s say we want to study the income differences between blacks and whites in the United States. Unfortunately, we only have enough funding to distribute 500 questionnaires. Given that 10% of the population is black (made up, but reasonably approximate), a simple random…

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H2 JC Maths Tuition Foot of Perpendicular 2007 Paper 1 Q8

One of my students asked me how to solve 2007 Paper 1 Q8 (iii) using Foot of Perpendicular method.

The answer given in the TYS uses a sine method, which is actually shorter in this case, since we have found the angle in part (ii).

Nevertheless, here is how we solve the question using Foot of Perpendicular method.

(Due to copyright issues, I cannot post the whole question here, so please refer to your Ten Year Series.)

Firstly, let F be the foot of the perpendicular.

Then, $\vec{AF}=k\begin{pmatrix}3\\-1\\2\end{pmatrix}$ ——– Eqn (1)

$\vec{OF}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17$ ——– Eqn (2)

From Eqn (1), $\vec{OF}-\vec{OA}=\begin{pmatrix}3k\\-k\\2k\end{pmatrix}$

$\begin{array}{rcl}\vec{OF}&=&\vec{OA}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\ &=&\begin{pmatrix}1\\2\\4\end{pmatrix}+\begin{pmatrix}3k\\-k\\2k\end{pmatrix}\\ &=&\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\end{array}$

Substituting into Eqn (2),

$\begin{pmatrix}1+3k\\2-k\\4+2k\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=17$

$14k+9=17$

$k=4/7$

Substituting back into Eqn (1),

$\displaystyle\vec{AF}=\frac{4}{7}\begin{pmatrix}3\\-1\\2\end{pmatrix}$

$\displaystyle|\vec{AF}|=\frac{4}{7}\sqrt{14}$

Multiplication Worksheet Generator for Grade 1 – Grade 3

Multiplication Worksheet Generator for Grade 1 – Grade 3

Generate hundreds of Multiplication Questions automatically, with answers!
Suitable for Grade 1 – Grade 3 (ages 7 to 9)

JC Junior College H2 Maths Tuition

If you or a friend are looking for Maths tuition: O level, A level H2 JC (Junior College) Maths Tuition, IB, IP, Olympiad, GEP and any other form of mathematics you can think of.

Experienced, qualified (Raffles GEP, NUS Maths 1st Class Honours, NUS Deans List) and most importantly patient even with the most mathematically challenged.

So if you are in need of the solution to your mathematical woes, drop me a message!

Tutor: Mr Wu

Email: mathtuition88@gmail.com

Website: Singapore Maths Tuition | Patient and Dedicated Maths Tutor in Singapore

Landau’s Beautiful Proofs

Landau’s beautiful proofs:
1= cos 0 = cos (x-x)

Opening cos (x-x):
1 = cos x.cos (-x) – sin x.sin (-x)
=> 1= cos² x + sin² x
[QED]

Let cos x= b/c, sin x = a/c
1= (b/c)² + (a/c)²
c² = b² + a²
=> Pythagoras Theorem
[QED]

Landau (1877-1938) was the successor of Minkowski at the Gottingen University (Math) before WW II.

Open Cubic Root

Mental Trick

It was discovered by the Martial Art writer Liang Yusheng 武侠小说家梁羽生 (《白发魔女传》作者), who met Hua Luogeng (华罗庚) @1979 in England:
2³= [8]
8³= 51[2]
3³= 2[7]
7³= 34[3]

The last digit pairs :
[2 <->8] , [3 <-> 7]
Others unchanged.

Example:

$latex \sqrt[3]{658503} = N$
Last three digits 503 <-> …[7]
First three digits 658:
(8³ =512)< 658 < (729 = 9³)
=> 8…
Answer : $latex \sqrt[3]{658503} = N$= 87
Note: Similar trick for opening $latex \sqrt[23] {200 digits}$ by an indian lady Ms Shakuntala (83) dubbed “Human computer”.

Quick verify multiplication

Verify:2,578,314 x 6,321,737 = 16,299,423,011,418 ?

2,578,314 => (2+5+7+8+3+1+4)) =30 => (3+0) = 3

6,321,737 => (6+3+2+1+7+3+7)=29 => 2+9 = 11 => (1+1) = 2

3 x 2 = 6

16,299,423,011,418 => (1+6+2+9+9+4+2+3+0+1+1+4+1+8) = 51 => (5+1) = 6

Correct !

3D Trigonometry Maths Tuition

Solution:

(a) Draw a line to form a small right-angled triangle next to the angle $18^\circ$

Then, you will see that

$\angle ACD=90^\circ-18^\circ=72^\circ$ (vert opp. angles)

$\angle BAC=180^\circ-72^\circ=108^\circ$ (supplementary angles in trapezium)

By sine rule,

$\displaystyle \frac{\sin\angle ABC}{30}=\frac{\sin 108^\circ}{40.9}$

$\sin\angle ABC=0.697596$

$\angle ABC=44.23^\circ$

$\angle ACB=180^\circ-44.23^\circ-108^\circ=27.77^\circ=27.8^\circ$ (1 d.p.)

(b) By Sine Rule,

$\displaystyle\frac{AB}{\sin\angle ACB}=\frac{30}{\sin 44.23^\circ}$

$AB=\frac{30}{\sin 44.23^\circ}\times\sin 27.77^\circ=20.04=20.0 m$ (shown)

(c)

$\angle BCD=\angle ACD-\angle ACB=72^\circ-27.77^\circ=44.23^\circ$

By Cosine Rule,

$BD^2=40.9^2+50^2-2(40.9)(50)\cos 44.23^\circ=1242.139$

$BD=35.24=35.2 m$

(d)

$\displaystyle\frac{\sin\angle BDC}{40.9}=\frac{\sin 44.23^\circ}{35.24}$

$\sin\angle BDC=0.80957$

$\angle BDC=54.05^\circ$

angle of depression = $90^\circ-54.05^\circ=35.95^\circ=36.0^\circ$ (1 d.p.)

(e)

Let X be the point where the man is at the shortest distance from D. Draw a right-angle triangle XDC.

$\displaystyle\cos 72^\circ=\frac{XC}{50}$

$XC=50\cos 72^\circ=15.5 m$

Rectangle Maths Tuition

Solution:

(a) $\angle SPC=\angle SRQ=90^\circ$

$\angle PCS=\angle SQR$ (given)

Thus $\triangle PCS$ is similar to $\triangle RQS$ (AAA)

(b)(I)

$\angle KRQ=90^\circ /2=45^\circ$ (KR bisects $\angle QRS$ )

Thus $\angle QKR=180^\circ-90^\circ-45^\circ=45^\circ$

Hence $\triangle KQR$ is an isosceles triangle.

So $KQ=QR=PS=\frac{1}{2}PQ$

Hence,

$\begin{array}{rcl}PK&=&PQ-KQ\\ &=&PQ-\frac{1}{2}PQ\\ &=&\frac{1}{2}PQ\\ &=&\frac{1}{2}(2PS)\\ &=&PS \end{array}$

(shown)

(ii) By similar triangles,

$\displaystyle\frac{PC}{PS}=\frac{RQ}{RS}=\frac{PS}{PQ}=\frac{PS}{2PS}=\frac{1}{2}$

Thus $PC=\frac{1}{2}PS=\frac{1}{2} PK$

Hence

$\begin{array}{rcl}CK&=&PK-PC\\ &=&PK-\frac{1}{2}PK\\ &=&\frac{1}{2}PK\\ &=&PC \end{array}$

(shown)

Tips on attempting Geometrical Proof questions (E Maths Tuition)

Tips on attempting Geometrical Proof questions (O Levels E Maths/A Maths)

1) Draw extended lines and additional lines. (using pencil)

Drawing extended lines, especially parallel lines, will enable you to see alternate angles much easier (look for the “Z” shape). Also, some of the more challenging questions can only be solved if you draw an extra line.

2) Use pencil to draw lines, not pen

Many students draw lines with pen on the diagram. If there is any error, it will be hard to remove it.

3) Rotate the page.

Sometimes, rotating the page around will give you a fresh impression of the question. This may help you “see” the way to answer the question.

4) Do not assume angles are right angles, or lines are straight, or lines are parallel unless the question says so, or you have proved it.

For a rigorous proof, we are not allowed to assume anything unless the question explicitly says so. Often, exam setters may set a trap regarding this, making the angle look like a right angle when it is not.

5) Look at the marks of the question

If it is a 1 mark question, look for a short way to solve the problem. If the method is too long, you may be on the wrong track.

6) Be familiar with the basic theorems

The basic theorems are your tools to solve the question! Being familiar with them will help you a lot in solving the problems.

Hope it helps! And all the best for your journey in learning Geometry! Hope you have fun.

“There is no royal road to Geometry.” – Euclid

Animation of a geometrical proof of Phytagoras... — Animation of a geometrical proof of Pythagoras theorem (Photo credit: Wikipedia)

Solution 2 (Eigenvalue): Monkeys & Coconuts

Solution 2: Use Linear Algebra Eigenvalue equation: A.X = λ.X

A =S(x)= $Latex \frac{4}{5}(x-1)$ where x = coconuts

S(x)=λx

Since each iteration of the transformation caused the coconut status ‘unchanged’, which means λ = 1 (see remark below)

$Latex \frac{4}{5}(x-1)=x$
We get
x = – 4

S¹(x) = ⅘ (x-1)

S²(x) = ⅘( [⅘ (x-1)]-1)

= ⅘ (⅘ x -⅘ -1)

= (⅘)² x – (⅘)² – ⅘

Also by recursive, after the fifth monkey:

$Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x-1)- (\frac{4}{5})^4-(\frac{4}{5})^3- (\frac{4}{5})^2- \frac{4}{5}$

$Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x) – (\frac{4}{5})^5 – (\frac{4}{5})^4 – (\frac{4}{5})^3+(\frac{4}{5})^2 – \frac{4}{5}$

$Latex 5^5$ divides (x)

Minimum positive x= – 4 mod ($Latex 5^{5}$ )= $Latex 5^{5} – 4$= 3,121 [QED]

Note: The meaning of eigenvalue λ in linear transformation is the change by a scalar of λ factor (lengthening or shortening by λ) after the transformation. Here
λ = 1 because…

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Solution 1 (Sequence): Monkeys & Coconuts

Monkeys & Coconuts Problem

Solution 1 : iteration problem => Use sequence
$Latex U_{j} =\frac {4}{5} U_{j- 1} -1 $

(initial coconuts)
$Latex U_0 =k$
Let
$Latex f(x)=\frac{4}{5}(x-1)=\frac{4}{5}(x+4)-4$
$Latex U_1 =f(U_0)=f(k)= \frac{4}{5}(k+4)-4$

$Latex U_2 =f(U_1)=f(\frac{4}{5}(k+4)-4)= \frac{4}{5}((\frac{4}{5}(k+4)-4+4)-4$

$Latex U_2=(\frac{4}{5})^2 (k+4)-4$

$Latex U_3=(\frac{4}{5})^3 (k+4)-4$

$Latex U_4=(\frac{4}{5})^4 (k+4)-4$

$Latex U_5=(\frac{4}{5})^5 (k+4)-4$

Since
$Latex U_5$ is integer ,
$Latex 5^5 divides (k+4)$
k+4 ≡ 0 mod($Latex 5^5$)
k≡-4 mod($Latex 5^5$)
Minimum {k} = $Latex 5^5 -4$= 3121 [QED]

Note: The solution was given by Paul Richard Halmos (March 3, 1916 – October 2, 2006)

Monkeys & Coconuts Problem

5 monkeys found some coconuts at the beach.

1st monkey came, divided the coconuts into 5 groups, left 1 coconut which it threw to the sea, and took away 1 group of coconuts.
2nd monkey came, divided the remaining coconuts into 5 groups, left 1 coconut again thrown to the sea, and took away 1 group.
Same for 3rd , 4th and 5th monkeys.

Find: how many coconuts are there initially?

Note: This problem was created by Nobel Physicist Prof Paul Dirac (8 August 1902 – 20 October 1984). Prof Tsung-Dao Lee (李政道) (1926 ~) , Nobel Physicist, set it as a test for the young gifted students in the Chinese university of Science and Technology （中国科技大学－天才儿童班）.

Reason for Maths Tuition

My take is that Maths tuition should not be forced. The child must be willing to go for Maths tuition in the first place, in order for Maths tuition to have any benefit. Also, the tuition must not add any additional stress to the student, as school is stressful enough. Rather Maths tuition should reduce the student’s stress by clearing his/her doubts and improving his/her confidence and interest in the subject. There is a quote “One important key to success is self-confidence. An important key to self-confidence is preparation.“. Tuition is one way to help the child with preparation.

Parallelogram Maths Tuition: Solution

Solution:

(a) We have $\angle APQ=\angle ARQ$ (opp. angles of parallelogram)

$AP=RQ$ (opp. sides of parallelogram)

$AR=PQ$ (opp. sides of parallelogram)

Thus, $\triangle APQ\equiv\triangle QRA$ (SAS)

Similarly, $\triangle ABC\equiv\triangle CDA$ (SAS)

$\triangle CHQ\equiv\triangle QKC$ (SAS)

Thus, $\begin{array}{rcl}\text{area of BPHC}&=&\triangle APQ-\triangle ABC-\triangle CHQ\\ &=&\triangle QRA-\triangle CDA-\triangle QKC\\ &=& \text{area of DCKR} \end{array}$

(proved)

(b)

$\angle ACD=\angle HCQ$ (vert. opp. angles)

$\angle ADC=\angle CHQ$ (alt. angles)

$\angle DAC=\angle CQH$ (alt. angles)

Thus, $\triangle ADC$ is similar to $\triangle QHC$ (AAA)

Hence, $\displaystyle\frac{AC}{DC}=\frac{QC}{HC}$

Thus, $AC\cdot HC=DC\cdot QC$

(proved)

Congruent Triangles Geometry Maths Tuition: Solution

Solution:

(a) $\angle EBD=\angle BEC$ (given)

BE is a common side for both triangles $\triangle BCE$ and $\triangle EDB$

BD=EC (given)

Therefore, $\triangle BCE \equiv \triangle EDB$ (SAS)

(proved)

(b)

Since $\triangle BCE\equiv\triangle EDB$ we have $\angle CBE=\triangle DEB$

Thus, $\begin{array}{rcl}\angle ABE&=&180^\circ - \angle CBE\\ &=&180^\circ -\angle DEB\\ &=& \angle AEB \end{array}$

Therefore, $\triangle ABE$ is an isosceles triangle.

Thus, $AB=AE$

(proved)

Congruent Triangles Maths Tuition: Solution

Solution:

BS=BE+ES=ST+ES=ET

$\angle DES=\angle ESA=90^\circ$

BA=DT (given)

Thus, $\triangle ASB$ is congruent to $\triangle DET$ (RHS)

Hence $\angle DTE=\angle SBA$

Thus DT//BA (alt. angles)

(Proved)

By Pythagoras’ Theorem, we have

$\begin{array}{rcl}DB&=&\sqrt{DE^2+BE^2}\\ &=&\sqrt{SA^2+ST^2}\\ &=&TA \end{array}$

Hence $\triangle DEB$ and $\triangle AST$ are congruent (SSS).

Hence $\angle DBE=\angle STA$

Thus DB//TA (alt. angles)

Therefore, ABDT is a parallelogram since it has two pairs of parallel sides.

(shown)

French Curve

The French method of drawing curves is very systematic:

“Pratique de l’etude d’une fonction”

Let f be the function represented by the curve C

Steps:

1. Simplify f(x). Determine the Domain of definition (D) of f;
2. Determine the sub-domain E of D, taking into account of the periodicity (eg. cos, sin, etc) and symmetry of f;
3. Study the Continuity of f;
4. Study the derivative of fand determine f'(x);
5. Find the limits of fwithin the boundary of the intervals in E;
6. Construct the Table of Variation;
7. Study the infinite branches;
8. Study the remarkable points: point of inflection, intersection points with the X and Y axes;
9. Draw the representative curve C.

Example:

$latex \displaystyle\text{f: } x \mapsto \frac{2x^{3}+27}{2x^2}$
Step 1: Determine the Domain of Definition D
D = R* = R –…

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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

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

guanyinmiao's musings (Archived: July 2009 to July 2019)

Tuition That We May Have To Believe In

This insightful article makes a really good read.

Quotes from the article:

To be honest, the amount to be learnt at each level of education is constantly increasing, and tuition could just help you get that edge over others. After all, it was meant to be supplementary in nature.

The toughest part at the end of the day however, is probably this: getting the right tutor.

This commentary, “Tuition That We May Have To Believe In”, is a reply to a previous article on tuition by Howard Chiu (Mr.), “Tuition We Don’t Have To Believe In” (Read).

I must say Howard’s article had me on his side for a moment. He appealed to me emotively. Nothing like a mental picture of some kid attending hours and hours of tuition immediately after school when he could well be enjoying himself thoroughly with… an iPhone or iPad (I highly doubt kids these days still indulge their time at playgrounds). But the second time I read his article, I silenced the part of my brain which still prays the best for children, so do pardon me if I sound a tad too pragmatic at times.

The overarching assertion that Howard projects his points from is that there is “huge over consumption of this good”. Firstly, private tutoring…

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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 =…
[QED]

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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}…

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What is “sin A”

What is “sin A” concretely ?

1. Draw a circle (diameter 1)
2. Connect any 3 points on the circle to form a triangle of angles A, B, C.
3. The length of sides opposite A, B, C are sin A, sin B, sin C, respectively.

Proof:
By Sine Rule:

$latex \frac{a}{sin A} = \frac{b}{sin B} =\frac{c}{sin C} = 2R = 1$
where sides a,b,c opposite angles A, B, C respectively.
a = sin A
b = sin B
c = sin C

成语数学

An alternative answer to Q1) 20 除 3 is “陆续不断”.

20除以3，因为它的答案接近于6.6666，所以这道题的答案是陆续不断，或者是六六大顺都行，百分之一就是百里挑一，9寸加1寸等于一尺即是得寸进尺，12345609，七零八落，1、3、5、7、9无双数所以叫做举世无双，或者你把它答出天下无双都行，如此小升初的难题您答对了吗？

Source: http://politics.people.com.cn/n/2013/0528/c70731-21638931.html

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