The Mathematics Society Board is located in front of the snack bar in the cover playground.

𝑙

2l

𝑙 3𝑙

𝐴 =3 3

2𝑙2

Hexagon (六邊形) In geometry, a hexagon is a polygon with six edges and six vertices. This type of

polygon is commonly found in both nature and human-made structures.

Following are some examples:

Honeycomb(蜂巢)

Honeycomb has a hexagonal

structure in order to use space

more fully

Turtle's carapace(龜殼) Snowflake(雪花)

Carbon Fort Jefferson in Dry

Tortugas National Park

Flag of Israel(以色列)

Mathematics behind regular hexagon

An internal angle of a regular hexagon (one where all sides and all angles are

equal) is 120° thus the sum of the interior angles is 720° degrees. It has

rotational symmetry(旋轉對稱) of order 6 and 6 axes of reflectional symmetry

(反射對稱). The longest diagonals of a regular hexagon, connecting diametrically

opposite vertices (對角), are twice its sides in length. Like squares and

equilateral triangles, regular hexagons fit together without any gaps to tile the

plane (密鋪平面) (three hexagons meeting at every vertex), and so are useful for

constructing tessellations (嵌石裝飾). The cells of a beehive honeycomb are

hexagonal for this reason and because the shape makes

effective use of space and building materials.

The area of a regular hexagon of side length l is

given by

𝐴 =3 3

2𝑙2

The perimeter of a regular hexagon of side length l is,

of course, 6l, its maximal diameter (對角長) 2l, and its minimal diameter (對邊長) 3𝑙.

120∘

The Mathematics Society Board is located in front of the snack bar in the cover playground.

Proofs of Pythagoras’ Theorem

Most of you have learned Pythagoras’ theorem, but how many proofs do you know?

Mathematicians found over 300 ways to prove Pythagoras theorem. We are going

to show 2 interesting proofs.

Proof 1

We can make an algebraic proof by similarity. It is a property of right-angled

triangles, as the one shown in the above figure. The right-angled triangle with

sides x, a, and d (small triangle on the right) is similar to the right-angled triangle

with sides d, b, and y (large triangle in the middle), giving 𝑥

𝑎=

𝑎

𝑐 and

𝑦

𝑏=

𝑏

𝑐

𝑥 =𝑎2

𝑐 and 𝑦 =

𝑏2

𝑐

𝑐 = 𝑥 + 𝑦 =𝑎2

𝑐+

𝑏2

𝑐=

𝑎2 + 𝑏2

𝑐

∴ 𝑐2 = 𝑎2 + 𝑏2

Proof 2

Construct a circle with radius c and a right triangle with sides a and b as

shown in above figure. In this situation, a few well known facts can be applied.

For example, in the diagram three points F, G, H located on the circle form

another right triangle. (Angle in semi-circle)

Thus, we have triangle GFK similar to triangle FHK so FK

GK=

HK

FK

𝑎

𝑐 + 𝑏=

𝑐 − 𝑏

𝑎

𝑎2 = 𝑐 + 𝑏 𝑐 − 𝑏 = 𝑐2 − 𝑏2 ∴ 𝑐2 = 𝑎2 + 𝑏2

If you are interested, please join the Origami Workshop which is jointly organized by Maths Society and Art Club in November. Anyone who wants to join this workshop, please feel free to contact Maths Society. You can find

Lo Chak Hei 6C(18) or e-mail to lochakhei1@hotmail.com for details.

Snow-Capped Sonobe

The Sonobe unit is one of the foundations of modular origami. There are

many variations with Sonobe units. In this issue, we will teach you a simple

one called Snow-Capped Sonobe.

The above three are some ways to assemble these Sonobe units. You may

even find other ways to assemble them. You can assemble with 3, 6, 12, 30,

60, 90 (or larger) units to make different modular origami.

If you have any enquires of this newsletter, please feel free to find our chairman, Lo Chak Hei 6C (18).

Permutations A chairman, a vice chairman and a treasurer are chosen at random from

a committee of 8 people. How many ways are there to choose them? Need

some time to think? Read the definition below and you will know more.

Permutation: Permutation means arrangement of certain objects in a

definite order. Number of permutations of ‘n’ different things taken ‘r’ at a

time is given by:

𝑃 𝑟𝑛 =

𝑛!

𝑛 − 𝑟 !

where n! stand for the Factorial of n: The product of first ‘n’ natural

numbers is denoted by n!

𝑛! = 𝑛 𝑛 − 1 𝑛 − 2 … 3 × 2 × 1

Proof: Say we have ‘n’ different things a1, a2……, an.

Clearly there are ‘n’ ways to make the first choice (eg. a2). The number of

possibilities left after the first selection = n – 1

So, to choose the second item, we have n – 1 ways. Number of selections

left after selecting the second item = n – 2. Similarly, the third item that

can be selected has n – 2 ways.

Thus number of ways of filling-up first-place = n.

Number of ways of filling-up second-place = n – 1.

Number of ways of filling-up third-place = n – 2.

Number of ways of filling-up r-th place = n – (r – 1) = n – r + 1

So the total number of ways of filling up, first, second,…,rth-place

together: 𝑛 𝑛 − 1 𝑛 − 2 … 𝑛 − 𝑟 + 1

Hence, 𝑃𝑟𝑛 = 𝑛 𝑛 − 1 𝑛 − 2 … 𝑛 − 𝑟 + 1

= 𝑛 𝑛 − 1 𝑛 − 2 … 𝑛 − 𝑟 + 1 𝑛 − 𝑟 𝑛 − 𝑟 − 1 … 3 × 2 × 1

𝑛 − 𝑟 𝑛 − 𝑟 − 1 … 3 × 2 × 1

∴ 𝑃 𝑟𝑛 =

𝑛!

𝑛 − 𝑟 !

Now do you know the answer of the question above? It is simple, there

are 𝑷𝟑𝟖 =

𝟖!

(𝟖−𝟑)!=

𝟖× 𝟕× 𝟔× 𝟓× 𝟒× 𝟑× 𝟐× 𝟏

𝟓× 𝟒× 𝟑× 𝟐× 𝟏= 𝟑𝟑𝟔 ways to select a chairman, vice

chairman and the treasurer in a committee of 8 people!

SPC Mathematics Society will hold many activities this year including the Orgami Workshop and the inter-class competitions.

SUDOKURO

A Kakuro consists of a playing area of filled and empty

cells similar to a crossword puzzle. Some black cells

contain a diagonal slash from top left to bottom right

with the numbers in them called ‘the clues’. A number

in the top right corner relates to an ‘across’ clue and the

one in the bottom left a ‘down’ clue. The objective of a

Kakuro is to insert digits from 1-9 into the white cells to

total the clue associated with it. However no digit can

be duplicated in an entry.

Sudoku

Every column, row and box

(3x3 marked by heavier lines)

must contain all digits from

1-9.

Hypotrochoid

If any student answers the questions correctly, his/her name will be posted on the next issue of this newsletter and he/she will get a prize.

Warming up Level 1. If n leaves a remainder of 13 when divided by 2008, what is the

remainder when n3 is divided by 2008?

2. If 𝑥 +1

𝑥= 5, find

𝑥2

𝑥4+𝑥2+1.

Elementary Level 1. There are two positive integers, one of which is a square number. If the

sum of twice the square number and the other integer is 2008 less than

twice their product, find the difference between the two numbers.

2. How often does the minute hand pass the hour hand on an ordinary

clock on average for a day?

Intermediate Level 1. In the following trapezium, 𝐷𝐶 is parallel to 𝐴𝐵; 𝐴𝐶 ⊥ 𝐶𝐵; 𝐴𝐶 = 𝐶𝐵

and 𝐵𝐴 = 𝐵𝐷. Find ∠𝐴𝐵𝐷.

2. What is the least number of weights that can be used on a scale pan to

weigh against any integral number of pounds from 1 to 40 inclusive, if

the weights can be placed in either sides of the scale pan?

Olympic Level 1. How many nine-digit positive integers consist of nine pairwise distinct

(兩兩互不相同) digits and are divisible by 4950?

2. Find the value of

1

1+12+14 +

2

1+22+24 +

3

1+32+34 + ⋯ +

100

1+1002+1004

Polynomial Division (多項式除法)

(for fx-3650p,fx-3950p,fx-50FH)

The programme can calculate the quotient (商式) and remainder (餘數式) with a polynomial (多項式)

divided by a linear or quadratic polynomial (一次或二次多項式).

In order to enter this programme, we must first enter the progamme mode, then select a programme

number, followed by following codes:

Programme set 66 bytes

1 Mem clear : ? → A : ? → B : ? → C : ? → M :

2 Lbl 1 : ? → D : ( D – BX – CY )」A → D :

3 X → Y : D → X : 1 M– : M => Goto 1 :

4 AD ◢ ? → D : D – CY

MODE MODE MODE 2

Mem clear : shift mode 1 (fx-3650p,fx-3950p),shift 9 (fx-50FH)

: :exe

?:shift 3 1

→:shift RCL

A,B,C,D,X,Y,M: alpha (-), °’”, hyp, sin, ), ,,M+

」: a b/c

◢:shift 3 4

Lbl: shift 3 right right right 2 (fx-50FH)

Example:

Find the quotient and remainder of ( x4 + 4x3 + 6x2 + 5x + 2 ) ÷ ( x2 + 2x + 1 ) Step Button to be pressed (meaning) Display (meaning)

1 Prog → corresponding programme number A?

2 1 EXE 2EXE 1EXE (coefficient of the divisor)

3 4 EXE (highest degree of dividend)

4 1 EXE (1st coefficient if dividend, which is coefficient of x4 in this example

1 (coefficient of x2 in quotient)

5 4 EXE (2nd coefficient if dividend, which is coefficient of x3 in this example

2 (coefficient of x in quotient)

6 6 EXE (3rd coefficient if dividend, which is

coefficient of x2 in this example

1 (constant term in quotient)

7 5 EXE (4th coefficient if dividend, which is

coefficient of x in this example)

1 (coefficient of x in remainder)

8 EXE D?

9 2 EXE (the fifth coefficient in dividend,

constant term in this example)

1 (constant term of remainder)

So the quotient is x2 + 2x + 1, remainder is x+1

If the divisor is a linear polynomial (0x2+ax+b), then enter 0 EXE a EXE b EXE in step 2

Reference:

Book: Marvelous Modular Origami by Meenakshi Mukerji

Website: http://en.wikipedia.org/wiki/Hexagon

http://www.fiendishsudoku.com/

http://www.kakuro.ws/

http://lpl.hkcampus.net/

Teacher advisors:

Mr. WONG Kam Wing

Mr. POON Wai Hoi, Bobby

Mr. NGAN Full

Editors:

Chief Editor

Lo Chak Hei Hugo(6C)

Assistant Chief Editors

Tam Kin Boon Alex(6C)

Ko Ka Long Jacky(6C)

Kwan Ming Tak Milton(6C)

Chief Art Designer

Chan Ming Hong Benjamin(6C)

Editors

Mak Hang Kin(4F)

Sit Ka Nap Caleb(4F)

Chan Wai Cheung Adrian(3A)

__________________

K. W. Wong C. H. Mak Lo Chak Hei

Chief adviser Vice principal Chairman