mig newsletter - nus high school students' website...3-1 8 = 6- 2 4 . 4 the next step we have...

MIG Newsletter DECEMBER 2017

Math Interest Group | NUS High School | 2017

MIG NEWSLETTER DEC 2017

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Table of Content

Year in Review by President of MIG 2

Geometry and Trigonometry by Matthew Fan 4

A Refinement of the Fejér-Jackson Inequality by Soh Jing Ren

11

Finding Area of Polygon through Shoelace Formula by Cao Wenhui and Zhang Jingwen

15

Did you know? by Raghavendra Narayan Rao 17

Sudoku by Gnoh Cheng Yi 18

Crossword by Gnoh Cheng Yi 19

Answers to Sudoku and Crossword 20

MIG Newsletter Team

21

2

Year in Review Written by: Chan Wayde, M17607, President of Math Interest Group

Year 2017 saw huge changes for Math Interest Group, with Mr Chan Yu Ming as MIG’s teacher in-charge. Ms Lee Sheau Huey stayed on with MIG to oversee outreach opportunities. On the leadership front, Matthew Fan and I were promoted to Overall Vice President and President respectively, with Joel Tan and Theo Rajan Terence being welcomed into the Executive Committee.

Under this new leadership team, the popularity and influence of Math has grow in the school in my opinion. The NUS High Open House, co-organized with the other Interest Groups, Subject Departments, and Administrative Department, saw an overwhelming number of participants, well over 8000 visitors. We have had many applications applying to join MIG, of which we welcomed 8 of them to our MIG family. In our annual SMO Preparatory Course, we have seen well over 160 participants across all 3 divisions, more than double from last year.

We have also made changes to the events MIG organizes. On top of the Einstein Workshops and Math Masterclass, we continued organizing the Inter Class Math Challenge and AMC 8 Preparatory session, which will be held by the end of this term. We have launched our MIG Magazine (at the time you are reading this), and debuted a new MIG T-Shirt, as you will see some of us wearing at future MIG events. We also launched the Math Problem of the Month (MPOM), an initiative to encourage problem solving in NUS High Students and Staff.

Our members have also done well in their various pursuits of mathematics. Joel Tan won Gold in the International Math Olympiad in Brazil, while Matthew Fan won Gold in the TUYMAADA Olympiad held in Republic of Sakha, Russia. Li Yuelin won Bronze in TUYMAADA as well. Our members have also taken part in the Singapore Math Olympiad, American Math Competition 10A/12A/12B, American Invitational Mathematics Examination, Australian Mathematics Challenge, and many other competitions and have done very well in them, but it would take too much space to list all the Olympiad achievements.


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In the Singapore Mathematics Project Festival, Timothy Ho (with his project “Trigonometrical Inequalities”) won Gold with Excellent Presentation and Joshua Chin (with his project “Gambler’s Ruin”) won Silver for the Junior Category. Soh Jing Ren (with his project “A refinement of Fejer-Jackson inequality”) and Chong Jing Quan (with his project “An elementary approach to Euler’s sums via trigonometric series”) won double Golds for the Senior category. They also credit their research partners and school mentors for their outstanding results for this competition.

This is not to say that MIG is only about Olympiads and Research. We have members who specialize in other areas, such as Sudoku, Rubik’s Cube, or just enjoy spreading the love for mathematics with their peers.

I extend my humble invitation to all who have the passion in mathematics to join our family, to continue our efforts in developing the interest of mathematics to NUS High. Next year, under the new leadership, MIG will not only continue to develop its outreach, but also focus on internal development too.

Matthew Fan and I are stepping down after 4 years in MIG as we are graduating. It has been a huge honour serving.

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Geometry and Trigonometry Written by: Matthew Fan, M17604

All diagrams are constructed with the help of Geogebra and/or

Microsoft Paint.

1 Introduction and Definition

Hi everyone! Geometry and Trigonometry are important topics in

Mathematics. We spend a lot of time in school learning about certain

properties. They have many applications in real life that you may

search up on your own. We will go back and look at the basics and

hopefully come up with something interesting.

Let us look at sin 𝜃, a pretty basic trigonometrical function. How do we

define it? Well, we create a right-angled triangle ABC with angle

𝐴𝐵𝐶 = 𝜃(Fig. 1)

Now sin 𝜃 is 𝐴𝐶

𝐴𝐵, or you can say that it is the opposite side over the

hypotenuse, whichever version you prefer. Are you able to define the

trigonometrical ratios for cos, tan, cot, csc, sec, cot of 𝜃?


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What if 𝜃 > 90°? Well then we are unable to create a right-angled

triangle with q as one of the two other angles. So instead we go with the

unit-circle definition:

The unit circle (Fig. 2) is the circle with centre (0,0) and radius 1. If the

ray from the origin to point P on the unit circle forms an angle of 𝜃 with

the positive x-axis, then we let point P have coordinates (cos 𝜃, sin 𝜃).

This will be able to cover all values of 𝜃.

2 Values for Certain Angles

Those who have learnt would know that for these angles: 𝜃 = 0°, 30°,

45°, 60°, 90°, sin 𝜃 would give a certain value, which you probably had

to memorise by heart. Let us try to derive them. We can assume that

BC=1, as we can scale the triangle bigger or smaller while keeping the

angles and the ratio the same.

For 𝜃 = 45°, we would have an isosceles

right-angled triangle. So AC=BC=1 and

hence AB= AC2+BC2= 2. Then sin 45° =1

√2=

√2

2. Pretty simple right?

For 𝜃 = 30°, 60°, we would have a 30-60-90 triangle. We can reflect

point A over BC to get A' (Fig. 3). Then ∆𝐴𝐵𝐴′ has all angles 60o and is

equilateral. So sin30o= ACAB=

12AA'

AB = 12.

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If we let AC=s (Fig. 4), Then we have (2s)2=11+s2 or 3s2=1 giving

3s=1. Then sin60o= 12s=

3s2s =

32 .

3 Two Methods

A natural question to ask next is what about the angles 15o and 75o?

Well sin215o+sin275o=sin215o+cos215o=1. So if we know the value

of one we know the value of both. There are two methods that can help

us.

Method 1

Start from a 30-60-90 triangle as above. We extend BC such that B lies

between C and D and AB=BD. Then ∠𝐷𝐴𝐶 = 15°. If we let AC=s and


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AD=1 then 𝐵𝐶 = √3𝑠, DB=AB=2s. So (2+ 3)2s2+s2=1 or

(8+4 3)s2=1 and s= 1

8+4 3.

If you do a quick search online most resources would give

sin15o= 6- 2

4 . These two answers have the same value, which can be

seen from

8+4 3= 2 4+2 3= 2( (1+ 3)2)= 2(1+ 3)= 6+ 2.

Now

1

6+ 2=

6- 2

( 6- 2)( 6+ 2)=

6- 24 .

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Method 2

Start from a 30-60-90 triangle as above (Fig. 3). Let ▵ADE be congruent

to ▵BAC, with C,A,E on a line in that order. Then ∠𝐵𝐴𝐷 = 90° and

BA=AD means ∠𝐷𝐵𝐴 = 45° and so ∠𝐷𝐵𝐶 = 75° . Now let the foot of the

perpendicular from D be BC, we have a 15-75-90 triangle ((Fig. 4).

If AC=1, DF=EA+AC= 3+1, BF=BC-DE= 3-1. Then

BD2=( 3+1)2+( 3-1)2=8 gives BD= 8. So

sin15o= BFBD=

3-1

8=

6- 24 .

4 The Next Step

We have now obtained the values of sine for 𝜃 = 0°, 15°, 30°, 45°, 60°,

75°, 90°. For these values, we go in increments of 15o. Can we go one

step further and find the values of sine for

7.5o,22.5o,37.5o,52.5o,67.5o,82.5o? You can try deriving it by using the

same ideas as the two methods above (at least one method will work).


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5 Generalisation

Let us take a look at method 2 again. We combined 30o and 45o to make

75o. What will happen if we replace them with a,b and a+b?

Now if BD=1, we have AD=sinb and AB=cosb. Then DE=sinb sina,

AE=sin b cos a, AC=cos b sin a and BC=cos b cos a.

Readers are encouraged to explore further with sin(a+b) and cos(a+b).

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6 An Interesting Question

Given that 0° < 𝐴, 𝐵, 𝐶 < 90°, tanA=1, tanB=2, tanC=3, show that

A+B+C=180o.

Solution:

7 Additional Questions

Readers are encouraged to try the following questions, have fun!

1) Show that (sin 𝜃)2 + (cos 𝜃)2 = 1.

2) Show that tan𝜃

2=

sin 𝜃

1+cos 𝜃.

3) Show that the area of a triangle, with any two sides a and b is given

by 1

2𝑎𝑏 sin 𝜃 where 𝜃, is the angle between the two sides a and b of the

triangle.

5) Show the law of sines in a triangle ABC: 𝐴𝐵

sin ∠𝐴𝐶𝐵=

𝐵𝐶

sin ∠𝐵𝐴𝐶=

𝐶𝐴

sin ∠𝐶𝐵𝐴.

6) Show the law of cosines in a triangle ABC:

𝐴𝐵2 = 𝐴𝐶2 + 𝐶𝐵2 + 2 ∙ 𝐴𝐶 ∙ 𝐶𝐵 cos ∠𝐴𝐶𝐵.

7) Given that 0° < 𝐴, 𝐵 < 90°, tanA= 17 and sinB=

1

10, find the length

of A+2B.


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A Refinement of the Fejér-

Jackson Inequality Written by: Soh Jing Ren M17506

In 1910, a mathematician named Lipót Fejér had a conjecture. He conjectured that

∑𝑠𝑖𝑛(𝑘𝑥)

𝑘

𝑛𝑘=1 > 0 for all 𝑛 ∈ ℕ and 𝑥 ∈ (0, 𝜋). (1)

Looking at the graphs of the sum for 𝑛 = 1, … ,6, inequality (1) seems to hold:

Figure 1: Graphs of ∑sin(𝑘𝑥)

𝑘

𝑛𝑘=1 for 𝑛 = 1, … ,6.

However, Fejér was unable to prove the inequality at the time. The task of proving the conjecture fell into the hands of Dunham Jackson, who proved it one year later. Inequality (1) is now known as the Fejér-

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Jackson inequality. Since then, many mathematicians have sought to improve the inequality, by replacing the lower bound in (1) with a positive function. The first non-trivial generalisation of the inequality was given by Pál Turán in 1952. Turán used Complex Analysis to obtain the following inequality:

∑sin(𝑘𝑥)

𝑘

𝑛𝑘=1 > 4 sin2 (

𝑥

2) (cot (

𝑥

2) −

𝜋−𝑥

2).

The above inequality remained the best refinement of the Fejér-Jackson inequality until 2003, where two mathematicians, Alzer and Koumandos, gave an improved version of the refinement using Real Analysis and a substantial amount of computer calculations. Their refinement gives

∑sin(𝑘𝑥)

𝑘

𝑛𝑘=1 > 𝑥2 (cot (

𝑥

2) −

𝜋−𝑥

2) for 𝑛 = 2,3,4, … and 𝑥 ∈ (0, 𝜋).

However, unlike Turán’s inequality their result was not generalizable.

The aim of this project is to provide a better refinement to the Fejér-

Jackson inequality that is easily generalizable. We have obtained the

following inequality: Let 𝑛 be an integer such that 𝑛 ≥ 2. Then for all

𝑥 ∈ (0, 𝜋), we have

∑sin(𝑘𝑥)

𝑘

𝑛

𝑘=1

>2

3sin(𝑥) +

2

5sin(2𝑥) +

1

9sin(3𝑥) +

1

20sin(4𝑥)

The function on the right-hand side is called a degree 4 sine polynomial.

A verification of our result is given below.


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Figure 2: Graphs of ∑sin(𝑘𝑥)

𝑘

𝑛𝑘=1 for 𝑛 = 1, … ,6,

graph of 𝑥2 (cot (𝑥

2) −

𝜋−𝑥

2),

graph of 2

3sin(𝑥) +

2

5sin(2𝑥) +

1

9sin(3𝑥) +

1

20sin(4𝑥)

Our method of proof was vastly different from Turán’s, Alzer’s, and

Koumandos’ approach. We separated the proof of our result into

different cases, starting with the lower values of 𝑛. For 𝑛 = 2, … , 6, we

converted the sine polynomial

∑sin(𝑘𝑥)

𝑘

𝑛

𝑘=1

− (2

3sin(𝑥) +

2

5sin(2𝑥) +

1

9sin(3𝑥) +

1

20sin(4𝑥))

into a regular polynomial. For example, the sine polynomial for 𝑛 =

2 was converted to

2

9+

8

5𝑡 +

4

9𝑡2 +

2

5𝑡3

Subsequently we showed that the corresponding polynomial in 𝑡 is

indeed positive. For 𝑛 ≥ 7 however, the degree of the polynomial

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became too large to work with, so instead we used a piecewise function

𝜙 to separate the sum ∑sin(𝑘𝑥)

𝑘

𝑛𝑘=1 and the degree 4 sine polynomial. In

other words, the function 𝜙 satisfies

∑sin(𝑘𝑥)

𝑘

𝑛

𝑘=1

> 𝜙(𝑥) >2

3sin(𝑥) +

2

5sin(2𝑥) +

1

9sin(3𝑥) +

1

20sin(4𝑥)

for all 𝑛 = 7,8,9, …

Our result has various applications in the field of Thermodynamics,

Harmonic Oscillations, and more. For example, calculations for heat

distribution on a piece of metal and position of a vibrating string involve

functions derived from Fourier series, and our result is able to provide

an approximation for such functions.


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Finding Area of Polygon

through Shoelace Formula Written by: Cao Wenhui M17601 and Zhang Jingwen M17607

The shoelace formula is a well-known formula such that it can be used to determine the area of a polygon. Motivated by the shoelace multiplication, this project is an investigation on the use of shoelace formula where the vertices of a polygon are arranged in random order and classify various polygons according the number of intersections that they have. (The random orders that we arrange must form closed shapes.)

Shoelace Formula & Notations:

1 2

1 2 1 2 2 3 1 2 1 3 2 1

1 2

...| ... | ( ... ) ( ... )

... y

n

n n n n n

n

x x xPP P x y x y x y x y x y x y

y y

Here, (𝑥𝑖 , 𝑦𝑖) represents the coordinates of the vertices 𝑃𝑖 , where 𝑖 =1, … , 𝑛. |𝑃1𝑃2 … 𝑃𝑛| is referred to as Generalized Shoelace Multiplication.

Properties:

Property 1

1 2 1 2| | | |,

1

r n r r nPP P P PP P P P

where r n

Property 2

1 2 1 2| | | ... |,

1

r r n r s s r n

m

PP P P P PP P P P P P

where r n

Property 3

1 2 1 2| | | | | |,

1

r r n r r nPP P P P PP P P P

where r n

Property 4

1 2 1 1

1 2 1

| | | |,

, 3

n n n

n n

PP P P PP

where P P P P are on the same

straight line n

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Since this formula does not require the length of each side of a polygon to solve area problems, it has promising application values in areas such as surveying and forestry.

For example, in forestry, it is difficult to measure the exact areas of forests using conventional area measurement methods, since the length of forests will be too large to measure precisely. After modeling the forest into a graph and obtaining the GPS coordinates of the vertices of the forest area, we can easily calculate the area using the shoelace formula.

To illustrate, consider the forest area with the coordinates (3,4), (5,11), (12,8), (9,5), and (5,6). The calculation may be tedious using conventional methods. However, with shoelace formula, we can easily get:

Area = ½ |3*11+5*8+12*5+9*6+5*4-4*5-11*12-8*9-5*5-6*3|

=60/2=30

Image of a forest that can use

shoelace formula to calculate its

area easily


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Did You Know?

Written by: Raghavendra Narayan Rao M17407

1) The only number that is spelt in alphabetical order is Forty

2) The only numbers to have 3 factors are squares of prime numbers

3) eiτ – 1 = 0. This a variation of Euler’s famous “ei π + 1 = 0”. This identity is beautiful as it links five fundamental Mathematical constants, 1, 0, i, e and τ, where τ = 2 π.

4) 148

296 +

35

70 = 1. This equation uses all the digits from 0 to 9 only once.

5) 18 is the only number twice the sum of its digits

6) 𝜋

4= 1 −

1

3+

1

5−

1

7+ ⋯

7) 1.3712885742 = 100.13712885742

8) A sequence of six ‘9’s occurs in the decimal representation of π, starting at the 763rd digit. This sequence is called the “Feynman point”.

9) Both −1

1+𝑥 and

𝑥

1+𝑥 have the same derivative with respect to x.

10) Given a fixed perimeter, a cyclic quadrilateral has the largest area

11) The mathematical name for division sign is Obelus

12) In a class of 23 students, there is about a 50% chance that at least 2 of them will share the same birthday.

13) Mathematically speaking, there are 177147 ways to tie a tie.

14) Pascal’s triangle – powers of 11

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Sudoku Compiled by:

Gnoh Cheng Yi M17404

Fill all empty squares so that the numbers 1 to 9 appear exactly once in each row, column and 3x3 box.

If you think that was easy, try the following Sudoku.

This was known as the “World’s Hardest” Sudoku.


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Crossword Created by: Gnoh Cheng Yi M17404

Fill in the correct answers, one letter per square, both across and down, from the given clues.

ACROSS

3 The mathematics of working with variables 5 The method for proving a proposition that is valid for infinitely many different values of a variable 7 The study of geometric figures 9 A relation for which each element of the domain corresponds to exactly one element of the range 10 The process of finding a derivative DOWN 1 The mathematics of counting

2 The branch of mathematics dealing with limits, derivatives, integrals and power series 4 A polynomial with two terms which are not like terms 6 The study of triangles 8 A “number” which indicates a quantity, size, or magnitude that is larger than any real number

1

2

3

4

5

6

7

8

9

10

20

Answers


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MIG Newsletter Team

From left to right : Gnoh Cheng Yi M17404, Raghavendra Narayan Rao M17407 , Soh Jing Ren M17506, William Chia M17207, Chan Wayde M17607, Joel Tan Junyao M17401, Matthew Fan M17604

Teachers-in-charge: Mr Chan Yu Ming, Ms Lee Sheau Huey Editorial team: Chan Wayde, Matthew Fan Xin, Yu, Joel Tan Junyao, Soh Jing Ren, Zhang Jingwen, Cao Wenhui, Raghavendra Narayan Rao, Gnoh Cheng Yi, William Chia and Tasha Annakin S Idnani Newsletter designed by: William Chia and Tasha Annakin S Idnani For suggestions and feedback, please contact us at [email protected] and [email protected]

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Not in photo : Tasha Annakin S Idnani M17305, Cao Wenhui M17601 and Zhang Jingwen M17607 Sources for Images Used in Newsletter Front cover (The Dancing Mathematician): https://gaurish4math.files.wordpress.com/2014/11/wpid-wp-1416560443609.png Tree image (pg 3) : https://previews.123rf.com/images/kudryashka/kudryashka1111/kudryashka111100074/11264060-Art-tree-with-math-symbols-for-your-design-Stock-Vector-mathematics.jpg Tree image (pg 16): https://www.google.com.sg/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwiWxubF573XAhUT448KHenGDqQQjRwIBw&url=https%3A%2F%2Fwww.istockphoto.com%2Fphotos%2Fjungle-top-view&psig=AOvVaw1WxwaaJ3svcY2tr3bOvf47&ust=1510740083303358 Sources for Did You Know?

https://www.popsci.com/article/science/there-are-177147-ways-tie-tie

https://www.slideshare.net/sureshmurthy2/e-book-final1v3

https://www.youtube.com/user/numberphile

https://gaurish4math.files.wordpress.com/2014/11/wpid-wp-1416560443609.png

https://gaurish4math.files.wordpress.com/2014/11/wpid-wp-1416560443609.png

https://previews.123rf.com/images/kudryashka/kudryashka1111/kudryashka111100074/11264060-Art-tree-with-math-symbols-for-your-design-Stock-Vector-mathematics.jpg



https://www.google.com.sg/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwiWxubF573XAhUT448KHenGDqQQjRwIBw&url=https%3A%2F%2Fwww.istockphoto.com%2Fphotos%2Fjungle-top-view&psig=AOvVaw1WxwaaJ3svcY2tr3bOvf47&ust=1510740083303358






https://www.popsci.com/article/science/there-are-177147-ways-tie-tie

https://www.slideshare.net/sureshmurthy2/e-book-final1v3

https://www.youtube.com/user/numberphile

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