Article in the New York Times, and maths education

What's new

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

What's new

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?

What's new

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?

Gowers's Weblog

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

Math Online Tom Circle

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

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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

Gowers's Weblog

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

Math Online Tom Circle

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.

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Open Cubic Root

Math Online Tom Circle

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”.

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Quick verify multiplication

Math Online Tom Circle

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 !

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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

Math Online Tom Circle

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

Math Online Tom Circle

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)

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Monkeys & Coconuts Problem

Math Online Tom Circle

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）.

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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

Math Online Tom Circle

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

Math Online Tom Circle

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

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

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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.

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

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

Math Online Tom Circle

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 Online Tom Circle

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”

Math Online Tom Circle

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

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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

Math Online Tom Circle

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

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O Level E Maths Tuition: Statistics Question

Solution:

From the graph,

Median = 50th percentile = $22,000 (approximately)

The mean is lower than $22000 because from the graph, there is a large number of people with income less than $22000, and fewer with income more than $22000. (From the wording of the question, calculation does not seem necessary)

Hence, the median is higher.

The mean is a better measure of central tendency, as it is a better representative of the gross annual income of the people. This is because more people have an income closer to the mean, rather than the median.

Secondary Maths Tuition: Kinematics Question

Solution:

acceleration of car $=\frac{12}{6}=2m/s^2$

$\frac{v}{15-5}=2$

$v=2\times 10=20$

Let $T$ be the time (in seconds) when the car overtakes the truck.

Total distance travelled by car at T seconds = area under graph = $\frac{1}{2}(T-5)(2(T-5))$

( $2(T-5)$ is the velocity of car at T seconds, it is obtained in the same way as we calculated v.)

Total distance travelled by truck at T seconds = $\frac{1}{2}(6)(12)+\frac{1}{2}(15-6)(12+16)+16(T-15)$

Equating the two distances will lead to a quadratic equation $T^2-26T+103=0$

Solving that gives $T=21.12s$ or $T=4.876s$ (rejected as car only starts at t=5)

$21.12-5=16.1s$ (3 s.f.)

Maths Resources available for sale!

Do check out our Maths Resources available for sale! All Maths notes and worksheets are priced affordably, starting from just $0.99!
All the resources are personally written by our principal tutor Mr Wu.

https://mathtuition88.com/maths-notes-worksheets-sale/

Tip: Using Google “filetype” filter to search for Free Exam Papers

This is a handy tip to search for Free Exam Papers on Google.

(Note: Google is very powerful, but it can only search for exam papers that are already online in the first place)

Since most exam papers are in PDF format, we can restrict our search to PDF files by adding the phrase “filetype:pdf” to the Google search.

For example, searching “hwa chong maths sec 2” in Google does not yield many exam papers.

Searching “hwa chong maths sec 2 filetype:pdf” returns a much better result, including some worksheets and test papers.

Screenshot:

Make (y) the subject of the formulae (O Level Maths Tuition)

Question:

Make (y) the subject of the formulae

(1) $\displaystyle x =\frac{y}{y+1}$

Solution:

$\displaystyle \frac{x}{1}=\frac{y}{y+1}$

Cross multiply,

$xy+x=y$

$xy-y=-x$

$y(x-1)=-x$

$\displaystyle y=\frac{-x}{x-1}=\frac{x}{1-x}$

O Level 2007 E Maths Paper 2 Q3 Solution (Geometry Question)

(a)
$\angle DAB=\angle DCB$
(opposite angles of parallelogram)

$\displaystyle\angle PAD=\frac{1}{2}\angle DAB=\frac{1}{2}\angle DCB=\angle RCB$
(shown)

(b)
$AD=BC$

$\angle PAD=\angle RCB$
(from part a)

$\angle ADC=\angle ABC$
(opposite angles of parallelogram)

$\displaystyle\angle ADP=\frac{1}{2}\angle ADC=\frac{1}{2} ABC=\angle RBC$

Thus, triangles ADP and CBR are congruent (ASA).

(c)(i)
$\angle ADC+\angle DAB=180^\circ$
$\angle DAP+\angle ADP=90^\circ$

Considering the triangle ADP,
$\angle DPA=180^\circ-90^\circ=90^\circ$
(shown)

(ii)
$\angle DAB+\angle ABC=180^\circ$

$\angle BAQ+\angle ABQ=90^\circ$

Considering the triangle ABQ,
$\angle PQR=180^\circ-90^\circ=90^\circ$
(shown)

Logarithm and Exponential Question: A Maths Question

Question:

Solve $(4x)^{\lg 5} = (5x)^{\lg 7}$

Solution:

$4^{\lg 5}\cdot x^{\lg 5}=5^{\lg 7}\cdot x^{\lg 7}$

$\displaystyle\frac{4^{\lg 5}}{5^{\lg 7}}=\frac{x^{\lg 7}}{x^{\lg 5}}=x^{\lg 7-\lg 5}$

Using calculator, and leaving answers to at least 4 s.f.,

$0.6763=x^{0.1461}$

Lg both sides,

$\lg 0.6763=0.1461\lg x$

$\lg x=\frac{\lg 0.6763}{0.1461}=-1.1626$

$x=10^{-1.1626}=0.0688$ (3 s.f.)

Check answer (to prevent careless mistakes):

$LHS=(4\times 0.0688)^{\lg 5}=0.406$

$RHS=(5\times 0.0688)^{\lg 7}=0.406$

Since LHS=RHS, we have checked that our answer is valid.

Maths Tutor Singapore, H2 Maths, A Maths, E Maths

If you or a friend are looking for maths tuition: o level, a level, IB, IP, olympiad, GEP and any other form of mathematics you can think of

Experienced, qualified (Raffles GEP, Deans List, NUS Deans List, Olympiads etc) 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: https://mathtuition88.com/

Sec 3 Hwa Chong Institution Maths Test Papers and Resources

Site: http://anglc.wiki.hci.edu.sg/space/content

Description: Includes Sec 3 Hwa Chong Institution Maths Test Papers and Resources

Please visit our site https://mathtuition88.com/free-exam-papers for updates on more Free Exam Papers and Maths Tips.

Top Maths Tuition

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Please continue to visit our site for Free Exam Papers, and tips regarding Maths.

Free Exam Paper: CHIJ Katong Convent Sec 3 Express Additional Mathematics

Katong Convent Exam Paper

(Sec 3 A Maths, with answers)

H2 Maths Tuition: Complex Numbers Notes

H2 Maths: Complex Numbers 1 Page Notes

	Modulus	Argument
Cartesian Form		Draw diagram first, then find the appropriate quadrant and use (can use GC to double check)
Polar Form
Exponential Form

☺ When question involvespowers, multiplication or division, it may be helpful toconvert to exponential form.

☺ Please write Ƶ and 2 differently.

☺ De Moivre’s Theorem

Equivalent to

☺

Memory tip: Notice that arg behaves similarly to log.

☺ Locusof z is aset of pointssatisfying certain given conditions.

☺ in English means:The distance between (the point representing)and (the point representing)

☺ Means the distance offromis a constant,.

So this is acircular loci.

Centre:, radius =

☺ means that the distance offromis equal to its distance from

In other words, the locus is theperpendicular bisectorof the line segment joiningand.

☺ represents ahalf-linestarting frommaking an anglewith the positive Re-axis.

(Exclude the point (a,b) )

Common Errors

– Some candidates thought thatis the same asand thatis the same as.

– The “formula”for argumentsdoes not workfor points in the 2^ndand 3^rdquadrant.

– Very many candidates seem unaware that their calculators will work in radians mode and there were many unnecessary “manual” conversions from degrees to radians.

Maths Tuition Free Exam Papers (Primary, Secondary, O Levels, A Levels) Links

The Free Exam Papers Database has moved to:

https://mathtuition88.com/free-exam-papers

(Click here to go to Free Exam Papers Database)

Ten Year Series: How many questions or papers to practice for Maths O Levels / A Levels?

This is a question to ponder about, how many questions or papers to practice for Maths O Levels / A Levels for the Ten Year Series?

If you searched Google, you will find that there is no definitive answer of how many questions to practice for Maths O Levels/ A Levels anywhere on the web.

For O Level / A Level, practicing the Ten Year Series is really helpful, as it helps students to gain confidence in solving exam-type questions.

Here are some tips about how to practice the Ten Year Series (TYS):

1) Do a variety of questions from each topic. This will help you to gain familiarity with all the topics tested, and also revise the older topics.

2) Fully understand each question. If necessary, practice the same question again until you get it right. There is a sense of satisfaction when you finally master a tough question.

3) Quality is more important than quantity. It is better to do and understand 1 question completely than do many questions but not understanding them.

Back to the original query of how many questions or papers to practice for Maths O Levels / A Levels for the Ten Year Series, I will attempt to give a rough estimate here, based on personal experience.

5 Questions done (full questions worth more than 5 marks) will result in an improvement of roughly 1 mark in the final exam.

(The 5 Questions must be fully understood. )

So, if a student wants to improve from 40 marks to 70 marks, he/she should try to do 30×5=150 questions (around 7 years worth of past year papers). Repeated questions are counted too, so doing 75 questions (around 3 years worth of past year papers) twice will also count as doing 150 questions. In fact, that is better for students with weak foundation, as the repetition reinforces their understanding of the techniques used to solve the question.

If the student starts revision early, this may work out to just 1 question per day for 5 months. Of course, the 150 questions must be varied, and from different subject topics.

Marks improved by	Long Questions to be done	Approx. Number of years of TYS	OR (even better)
10	50	2	1 year TYS practice twice
20	100	4	2 year TYS practice twice
30	150	6	3 year TYS practice twice
40	200	8	4 year TYS practice twice
50	250	10	5 year TYS practice twice

This estimate only works up to a certain limit (obviously we can’t exceed 100 marks). To get the highest grade (A1 or A), mastery of the subject is needed, and the ability to solve creative questions and think out of the box.

When a student practices TYS questions, it is essential that he/she fully understands the question. This is where a tutor is helpful, to go through the doubts that the student has. Doing a question without understanding it is essentially of little use, as it does not help the student to solve similar questions should they come out in the exam.

Hope this information will help your revision.

积少成多: How can doing at least one Maths question per day help you improve! (Maths Tuition Revision Strategy)

We all know the saying “an apple a day keeps the doctor away“. Many essential activities, like eating, exercising, sleeping, needs to be done on a daily basis.

Mathematics is no different!

Here is a surprising fact of how much students can achieve if they do at least one Maths question per day. (the question must be substantial and worth at least 5 marks)

This study plan is based on the concept of 积少成多, or “Many little things add up“. Also, this method prevents students from getting rusty in older topics, or totally forgetting the earlier topics. Also, this method makes use of the fact that the human brain learns during sleep, so if you do mathematics everyday, you are letting your brain learn during sleep everyday.

Let’s take the example of Additional Mathematics.

Exam is on 24/25 October 2013.

Let’s say the student starts the “One Question per day” Strategy on 20 May 2013

Days till exam: 157 days (22 weeks or 5 months, 4 days)

So, 157 days = 157 questions (or more!)

Each paper in Ten Year Series has around 25 questions (Paper 1 & Paper 2), so 157 questions translates to more than 6 years worth of practice papers! And all that is achieved by just doing at least one Maths question per day!

A sample daily revision plan can look like this. (I create a customized revision plan for each of my students, based on their weaknesses).

Suggested daily revision (Additional Mathematics):

	Topic
Monday	Algebra
Tuesday	Geometry and Trigonometry
Wednesday	Calculus
Thursday	Algebra
Friday	Geometry and Trigonometry
Saturday	Calculus
Sunday	Geometry and Trigonometry

(Calculus means anything that involves differentiation, integration)

(Geometry and Trigonometry means anything that involves diagrams, sin, cos, tan, etc. )

(Algebra is everything else, eg. Polynomials, Indices, Partial Fractions)

By following this method, using a TYS, the student can cover all topics, up to 6 years worth of papers!

Usually, students may accumulate a lot of questions if they are stuck. This is where a tutor comes in. The tutor can go through all the questions during the tuition time. This method makes full use of the tuition time, and is highly efficient.

Personally, I used this method of studying and found it very effective. This method is suitable for disciplined students who are aiming to improve, whether from fail to pass or from B/C to A. The earlier you start the better, for this strategy. For students really aiming for A, you can modify this strategy to do at least 2 to 3 Maths questions per day. From experience, my best students practice Maths everyday. Practicing Ten Year Series (TYS) is the best, as everyone knows that school prelims/exams often copy TYS questions exactly, or just modify them a bit.

The role of the parent is to remind the child to practice maths everyday. From experience, my best students usually have proactive parents who pay close attention to their child’s revision, and play an active role in their child’s education.

This study strategy is very flexible, you can modify it based on your own situation. But the most important thing is, practice Maths everyday! (For Maths, practicing is twice as important as studying notes.) And fully understand each question you practice, not just memorizing the answer. Also, doing a TYS question twice (or more) is perfectly acceptable, it helps to reinforce your technique for answering that question.

If you truly follow this strategy, and practice Maths everyday, you will definitely improve!

Hardwork $\times$ 100% = Success! (^_^)

“There is no substitute for hard work.” – Thomas Edison

A Maths Tuition: Trigonometry Formulas

Many students find Trigonometry in A Maths challenging.

This is a list of Trigonometry Formulas that I compiled for A Maths. Students in my A Maths tuition class will get a copy of this, neatly formatted into one A4 size page for easy viewing.

A Maths: Trigonometry Formulas

$\mathit{cosec}x=\frac{1}{\sin x}$ ; $\mathit{sec}x=\frac{1}{\cos x}$

$\cot x=\frac{1}{\tan x}$ ; $\tan x=\frac{\sin x}{\cos x}$

(All Science Teachers Crazy)

$y=\sin x$

$y=\cos x$

$y=\tan x$

$\frac{d}{\mathit{dx}}(\sin x)=\cos x$

$\frac{d}{\mathit{dx}}(\cos x)=-\sin x$

$\frac{d}{\mathit{dx}}(\tan x)=\mathit{sec}^{2}x$

$\int {\sin x\mathit{dx}}=-\cos x+c$

$\int \cos x\mathit{dx}=\sin x+c$

$\int \mathit{sec}^{2}x\mathit{dx}=\tan x+c$

Special Angles:

$\cos 45^\circ=\frac{1}{\sqrt{2}}$

$\cos 60^\circ=\frac{1}{2}$

$\cos 30^\circ=\frac{\sqrt{3}}{2}$

$\sin 45^\circ=\frac{1}{\sqrt{2}}$

$\sin 60^\circ=\frac{\sqrt{3}}{2}$

$\sin 30^\circ=\frac{1}{2}$

$\tan 45^\circ=1$

$\tan 60^\circ=\sqrt{3}$

$\tan 30^\circ=\frac{1}{\sqrt{3}}$

$y=a\sin (\mathit{bx})+c$ Amplitude: $a$ ; Period: $\frac{2\pi }{b}$

$y=a\cos (\mathit{bx})+c$ Amplitude: $a$ ; Period: $\frac{2\pi }{b}$

$y=a\tan (\mathit{bx})+c$ Period: $\frac{\pi }{b}$

$\pi \mathit{rad}=180^\circ$

Area of $\triangle \mathit{ABC}=\frac{1}{2}\mathit{ab}\sin C$

Sine Rule: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Cosine Rule: $c^{2}=a^{2}+b^{2}-2\mathit{ab}\cos C$

Exam Time Management and Speed in Maths (Primary, O Level, A Level)

Time management is a common problem for Maths, along with careless mistakes.

For Exam Time management, here are some useful tips:
1) If stuck at a question for some time, it is better to skip it and go back to it later, rather than spend too much time on it. I recall for PSLE one year, there was a question about adding 1+2+…+100 early in the paper, and some children unfortunately spent a lot of time adding it manually.
2) Use a exam half-time strategy. At the half-time mark of the exam, one should finish at least half of the paper. If no, then need to speed up and skip hard questions if necessary.

To improve speed:
1) Practice. It is really important to practice as practice increases speed and accuracy.
2) Learn the faster methods for each type of question. For example, guess and check is considered a slower method, as most questions are designed to make guess and check difficult.

Sincerely hope it helps.
For dealing with careless mistakes (more for O and A levels), you may read my post on How to avoid Careless Mistakes for O-Level / A-Level Maths?

English: an animated clock (Photo credit: Wikipedia)

Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question

Question:

ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

(a) Prove that $\Delta CXN$ and $\Delta MXB$ are similar.

(b) Given that area of $\triangle CXN$ : area of $\triangle MXB$ =9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of $\triangle XBC$ . (Challenging)

[Answer Key] (b) (i) 1:3

(ii) 20:3

Suggested Solutions:

(a)
$\angle MXB=\angle NXC$ (vert. opp. angles)

$\angle MBX = \angle XNC$ (alt. angles)

$\angle BMX = \angle XCN$ (alt. angles)

Therefore, $\Delta CXN$ and $\Delta MXB$ are similar (AAA).

(b) (i) $\displaystyle\frac{NC}{BM}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

Let $BM=2u$ and $NC=3u$

Then $DC=2\times 2u=4u$

So $DN=4u-3u=u$

Thus, $DN:NC=1u:3u=1:3$

(ii)

We now have a shorter solution, thanks to a visitor to our site! (see comments below)

From part (a), since $\Delta CXN$ and $\Delta MXB$ are similar, we have $MX:XC=2:3$

This means that $MC:XC=5:3$

Thus $\triangle MBC:\triangle XBC=5:3$ (the two triangles share a common height)

Now, note that $\displaystyle\frac{\text{area of }ABCD}{\triangle MBC}=\frac{BC\times AB}{0.5 \times BC \times MB}=\frac{AB}{0.5MB}=\frac{2MB}{0.5MB}=4$

Hence area of $ABCD=4\times\triangle MBC$

We conclude that area of rectangle ABCD: area of $\triangle XBC=4(5):3=20:3$

Here is a longer solution, for those who are interested:

Let area of $\triangle XBC =S$

Let area of $\triangle MXB=4u$

Let area of $\triangle CXN=9u$

We have $\displaystyle\frac{S+9u}{S+4u}=\frac{3}{2}$ since $\triangle NCB$ and $\triangle CMB$ have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get $2S+18u=3S+12u$

So $\boxed{S=6u}$

$\displaystyle\frac{\triangle BCN}{\triangle BDC}=\frac{3}{4}$ since $\triangle BCN$ and $\triangle BDC$ have the same base BC and their heights have ratio 3:4.