Content
INTRODUCTORY WORDS
3
From the President of the
MATH ARTICLES
10
DSE Corner: Solving circle problems
HKJSMS
using constructions of cyclic
4
From Dr. Koopa Koo
quadrilaterals
5
From Mr. Andy Loo
5
From committees of the
14
IMO Zone: Looking the question from another perspective
HKJSMS
9
YEAR PLAN
2 HKJSMS, Nov 2015
16
PROBLEM ARENA
From the President of JSMS First of all, it is my great honor to become the 28th President of Hong Kong joint School Mathematics Society. During the past 27 years, our society has been trying its very best arousing students’ interest towards mathematics and cultivating friendships among schools. We would like to further attain this goal in the coming year. Last year, our society has held several activities successfully. The Inter-School Mathematics Contest (ISMC) has broadened participants’ horizon, and the HKJSMS Day Camp “Choice” has aroused interests among students. Here, I would like to thank all the executive committee members and the sub-committee members of 27th HKJSMS achievements possible.
for
making
such
In the coming year, not only will we continue to hold the major events last year, but we will also organize some new and unique activities, including a series of bimonthly competitions for members and students of our member schools. Different interesting topics will also be introduced, and participants will be guided to discover the interesting parts. This newsletter includes not only a summary of this year’s plan, but also articles on DSE-level and IMO-level mathematical concepts and problems tailored for you to learn more. Please enjoy it and we hope our coming activities interest you. Mr. Leung Arvin Yui Hin, the current president of the HKJSMS
3 HKJSMS, Nov 2015
From Dr. Koopa Koo Dear HKJSMS, It is with great pleasure to see that the new Executive Committee of the Hong Kong Joint School Mathematics Society (HKJSMS) carries forward the enduring spirit of its predecessor in promoting public access to the appreciation of mathematical beauty. Mathematics, as a study of pattern, is the highest form of supreme beauty. Elegant proofs transform complicated concepts into simple, yet powerful ideas accessible to a wide audience. Those who possess magic lenses that see through hidden patterns hold the key to unveiling a vivid vision beyond the naked eyes that penetrate the haven of Mother Nature. It is my utmost aspiration that the HKJSMS committee continues to organize innovative and thought-provoking activities to equip students with the special lenses conducive to appreciating the aesthetic and expressional beauty Mathematics offers. Mathematics
is
also
the
embodiment
of
insurmountable logical beauty. Its essence is about identifying and tackling the kernel of a problem. The transition from high school to university mathematics learning is marked by a tremendous inflow of knowledge and choices. The more knowledge you acquire, the more you find
Dr. Koopa Koo, a teacher advisor and
yourselves overwhelmed with infinite choices. Which concept(s) should I apply in a particular
tutorial teacher of Beacon Group
also a well-renowned mathematics
Holdings Limited.
scenario? How do the theorems I have learnt integrate with each other? Is this information relevant to my solution or merely misleading? Through challenging events like the Inter School Mathematics Competition and the Math Camps, it is hoped that students can develop the analytical problem-solving skills necessary to identify the crux of any new problem, thereby facilitating the smooth transition from high school to university. All in all, I look forward to seeing a major breakthrough in the development of the HKJSMS. Best Wishes, Dr. Koopa Koo 4 HKJSMS, Nov 2015
From Mr. Andy Loo It is my honor to be appointed as a Teacher Advisor to the JSMS at a time when Mathematics education is so crucial to the success of our hometown. In the first place, the breadth and depth of mathematical applications in Hong Kong’s increasingly knowledge-based economy will be hard to overstate. Moreover, with the citizenry’s growing enthusiasm in public controversies, what is vitally needed is our ability to think independently and logically, which is precisely what mathematical learning cultivates. Last but not least, Mathematics instills us with a spirit of pursuing and upholding the truth, which will be a source of courage if we are to preserve Hong Kong’s core principles in the face of changes and challenges.
Mr. Andy Loo, a teacher advisor and also an undergraduate student in Princeton University He was a medalist in the International Mathematics Olympiad and
My congratulations go to the Executive Committee and Sub-Committee members of the 28th JSMS, who, I am sure, will go from strength to strength in promoting Mathematics among the future pillars of
International Physics Olympiad, received Sir Edward Youde Memorial Medal, and even published as an individual in the International
Journal
of
Contemporary
Mathematical Sciences when he was 17.
our society.
From committees of the HKJSMS
Hello everybody! I am Chan Hiu Chun, Sunny from STFA Leung Kau Kui College. This year is my third year to be working in the Society. During my work from the past two years, the mission of the JSMS to promote mathematics to all students was gradually imprinted in my mind. This time, as the Internal Vice President, I will lead the internal departments to continue working for the mission.
5 HKJSMS, Nov 2015
Chan Hiu Chun, Sunny Internal Vice President
I am Joey Yip from St. Paul’s Convent School and I am honored to be the External Vice President this year. JSMS will provide a good platform for cultivating the interest in mathematics and I hope we can enjoy this year’s events together. Yip Cheuk Yi, Joey External Vice President
Hello everyone! I am Poey Wong from St. Mark’s School. It is my pleasure to serve as the Internal Secretary in JSMS this year. I believe that JSMS is a platform for you to unveil your interest in mathematics. Hope you enjoy our activities with us this year!
Wong Ching Kei, Poey Internal Secretary
Hello dear reader! I am Allie Poon from Diocesan Girls’ School, and I am honored to be JSMS's External Secretary this year. JSMS is an
Poon Ho Kiu, Allie External Secretary
excellent platform for students to express their passion for mathematics, and I am grateful for this opportunity to be able to share my interest with you. With our engaging activities, I am sure that we will be having a fantastic time exploring the fun in mathematics together!
Hello everyone! I am Justin Leung from Diocesan Boys’ School, the Financial Secretary of JSMS this year. I believe that JSMS would be a great platform for us to learn more about mathematics, as well as finding our interest in it. I hope that I could grasp this opportunity to promote and share the love of mathematics in JSMS. Together, we make JSMS a better place.
6 HKJSMS, Nov 2015
Leung Chak Yin, Justin Financial Secretary
I am Ip Chak Ming, Ocean, F.5 from Wah Yan College, Kowloon, and a green hand in IT. Merely judging by appearance, it is only the tip of the iceberg; but after closer examinations, I discovered that it is of vital importance for some place. I am desperate to learn more. I want to dedicate my skills and try to cultivate a better JSMS. From my point of view, JSMS is IT Officer definitely expedient to my wish. As the old saying goes, action speaks louder than words, so I hope 28th JSMS can evolve Ip Chak Ming, Ocean
continuously. Hello everyone I am Li Shing Yan, Kobe, studying F.4 in STFA Leung Kau Kui College. This is the first year for me to enter the Ex-Co of JSMS. Actually, I do not have such rich backgrounds in website maintenance and other IT stuffs, but I hope I can still bring you a good platform. I also wish to contribute much to JSMS and enhance my skills in different aspects. Enjoy our activities this year!
Li Shing Yan, Kobe IT Officer
Hi everyone! I am Natalie Fung from Belilios Public School. It is my pleasure to serve you as the Promotion Officer of JSMS this year. With passion and commitment, we hope to inspire more friends to discover their interest towards the mathematics world, and we believe that it will be an unforgettable experience to engage in JSMS.
Fung Wing Suet, Natalie Promotion Officer
Hi everyone! I am Audrey Li from Good Hope School. It is with immense pleasure and excitement that I invite you to participate in our exhilarating activities. Think out of the box and never limit your imagination! Li Cheuk Yee, Audrey Public Relations Officer
7 HKJSMS, Nov 2015
A warm greeting to you all! I am Tai Wai Ting from Queen Elizabeth School, and I am one of the two Publication Officers this year. It is both my honor and a pleasure to be able to share my passion in mathematics alongside with other fellow enthusiasts here. We shall demonstrate to you now how mathematics can be both useful and amusing!
Tai Wai Ting, Terrence Publications Officer
Nice to meet you! I am Tiffany Wong from St. Paul’s Co-eduational College, one of the Publication Officers of JSMS this year. I am very glad to be able to share my interest and passion in Mathematics with you all! I wish you can learn Wong Ka Wai, Tiffany Publications Officer
the fun side of this world of mathematics and be inspired with distinctive thoughts. Please enjoy the reading!
Greetings. I’m Harry, currently F.5 in La Salle College. As a Mathematics Coordinator in JSMS this year and the previous one, I hope that we Math Co.s will be able to lead JSMS to greater heights than before. This year’s activities are guaranteed to be plentiful and fun - stay tuned!
Yu Hoi Wai, Harry Math Coordinator
Hi, everyone. I am Anson Au, one of the Math Coordinators this year. This year, we will launch new events apart from the ISMC and day camp so that your passion in mathematics can flourish. As you engage yourself in them, feel the heartfelt Au Chak Him, Anson Math Coordinator
8 HKJSMS, Nov 2015
joy of mathematics!
Warm greetings! I'm Dorothy, a F.5 student in Diocesan Girls' School. This is my second year in JSMS - and my second year as one of the Math Coordinators. I absolutely adore this post and it's a pleasure to share my passion in mathematics with my fellow JSMS committees and all of you! I sincerely hope you will enjoy the activities in the coming year!
Cheng Wai Chung, Dorothy Math Coordinator
Activity highlights Year Plan A new school year laden with even more activities to kindle your math spirit Inter-school (ISMC)
Mathematics
Contest Publication of HKDSE Mock Exam Paper
The ISMC has been the annual highlight In order to provide opportunities for of JSMS and it will be held in mid-March. The competition comprised of Individual Event, divided to the Junior Group (S1-S2) and the Senior Group (S3-S4),
students to practice more for their upcoming public exams, a set of mock paper will be published. Mock papers will be approved by experienced teachers.
and the Group Event for all S1-S4 students. This is a great opportunity for Bimonthly Competition students to experience a formal mathematics competition, to train their In order to provide opportunities for problem solving skills, as well as to students to understand different topics of develop their interests in mathematics. Joint-School Mathematics Day Camp
mathematics, a small-scaled competition will be held bimonthly. All members and students of member schools are invited.
Participants will be guided to explore The day camp will be held in August. different ideas in the subject. Various interesting math-related activities, such as treasure hunts and detective Mathematics Quiz games, will be held during the camp. In the camp, we introduce mathematics to The mathematics quiz will be a fast-paced students in a fun and interesting way. competitive event. Participants will Participants will have the opportunity to compete with other teams in answering learn mathematics and share their questions about mathematical facts or experiences in studying mathematics with arithmetic. Students are introduced to 9 HKJSMS, Nov 2015
students from other schools in this these different facts of mathematics via relaxing and friendly environment. this fun event.
DSE Corner Solving circle problems using constructions of cyclic quadrilaterals A comprehensive guide on tackling cyclic quadrilateral-related problems in DSE with three real problems as examples By Tai Wai Ting quadrilateral is cyclic: Basic concept A cyclic quadrilateral is, just like what its name suggests, a quadrilateral whose vertices are circumscribed, like the one below: Two of the most important properties are:
(
+
= ∠ ℎ
= 180° ( .∠ = ( . ∠ =
.
! "! .) . ∠)
)
You may wish to warm your brain up and try the following questions. +
= 180° ( . ∠ , = ( . ∠,
.)
Question 1 (HKCEE 2008 Paper II 50.) In the figure, .) O is the centre of the circle ABCD. If These two properties are very frequently ∠ADC = 84° and ∠CBO = 38°, then ∠ used in problem solving, so you may AOB = wish to remember it. Quite often, you need to construct such cyclic A. 64° quadrilaterals and use these two B. 88° C. 104° properties. D. 168° On the other hand, it is also not uncommon in the DSE math questions to ask for proof of a quadrilateral being cyclic. In that case, you have three At first glance, the angle we need to find common methods, all of which are is nowhere convenient to be calculated. converses of theorems you have learnt. Hardly enough angles are given, and the Consider the quadrilateral below, if one to be helpful. Maybe OA=OB, given O is the center… But then what? of the three conditions are true, then the condition that O is the center doesn’t seem 10 HKJSMS, Nov 2015
How about joining AB, then? Then we The conditions given here are even more have a=b easily. And here is the trick: awkward However, given AD is the remember that we are discussing the diameter, it immediately results that construction of cyclic quadrilaterals. f=90°. Therefore, we have ∠ CBA+ ∠ CDA=180°, and we can find out a easily. But then … to divide the large and Finally, with a and b, c can be found out inconvenient angles x and z, we can join with ease. AD together. Then, we have b+d=90°. Meanwhile, since ABCD is a cyclic quadrilateral, y+c=180°. That leaves a alone. It should be time to make use of the conditions AB = BC = CD. It doesn’t seem easy to construct isosceles triangles,
(38° + ) + 84°
. ∠s, cyclic quad. ) = 58° ∵ 01 = 02( ) = = 58°( ∠s, isos. ∆)
= 180° ( ∴
6 ∆120, +
+ = 180°
so let’s try other conditions. Given that we know f, and AB, BC, and CD are opposite to the angle, maybe we can use the fact that angles with equal chords are equal to each other? This leaves g=h=30°, and it immediately results that a=g+h=60°. Adding a, b, y, d, e together… problem solved.
∴ ∠102 = c = 180 7 a 7 b = 64° Therefore, the correct answer is A. Question 2 (HKCEE 2003 Paper II 50.) The figure shows a circle with diameter AD. If AB = BC = CD find x + y + z. A. 315° B. 324° C. 330° D. 360° 11 HKJSMS, Nov 2015
) ∵ 1: ℎ ! (" ∴ = 90° 6 ∆1:<, + + = 180° ∴ + = 90° + = 180° ( .∠ , .)
∵ 12 = 2= = =: ("
∴"=ℎ=>= = 30° (
3 ℎ
∴ ="+ℎ = 60° (∠ ℎ ∴
+
+? =
,
)
quadrilateral, then ∠ABQ=∠PQD=90°, which… is true, given AD is a diameter ) ∵ 1: ! ("
∠ )
! "!
)
∴ ∠12@ = 90° (∠ ! 7 ∴ ∠12@ = ∠A@: ∴ 1, @, A, 2 ( = . . ∠)
.∠
)
+ + + + = 330° Consequently, the correct answer is B. Part (b) QED. The position of θ is quite awkward, so why don’t we move it out to a by angles in the same segment using the cyclic Question 3 (HKCEE 1993 Paper I 11.) The figure quadrilateral just proved? That shows a semicircle with diameter AD and immediately solves b(ii), since ∠BOC is center O. The chords AC and BD meet at just the twice of a. But how b(i)? Since P. Q is the foot of the perpendicular from we know ∠ BQP, and ∠ BQC is P to AD. composed of ∠BQP and ∠PQC, all we (a) Show that A, Q, P, B are concyclic. (b) Let ∠BQP = θ. Find the following in terms of θ
need to find is ∠PQC. Again, since its position
is
pretty
awkward, maybe we should move it out, but how? In fact, what part (a) shows is a similar logic that we can apply to PQDC, and so PQDC is a cyclic quadrilateral,
(i) ∠BQC, (ii) ∠BOC. © Let ∠ CAD = φ. Find ∠ CBQ in and thus ∠PQC=b using angles in the terms of φ. same segment. Since a=b by angles in the same segment of the original semicircle, the problem is solved.
The diagram may look like that you. should use the converse of angles in the ( )∠:=@ = 90° (∠ ! 7 ∴ ∠:=@ = ∠A@1 same segment, but it doesn’t look easy… Well, if AQPB is a concyclic ∴
= (∠ ℎ
12 HKJSMS, Nov 2015
! "!
)
)
∵ 1, @, A, 2
= ∠2@A = B (∠ ℎ
(
! "!
6 ℎ " ! , = (∠ ℎ ! "! ∴
=
=
∴ ∠2@= = B + ( )∠20= = 2 = 2B(∠
D
=B
= 2B
)
)
)
∠ ⊙FG )
Part (c) Observe how φ can be easily translated to c and d using two circles? It is then straightforward to prove that c = d = φ, and so ∠CBQ = 2φ. QED.
6 ℎ
" !
,
= H (∠ ℎ ! "! ) 6 ℎ " ℎ "ℎ 1, 2, A, @, = H (∠ ℎ ! "! ) ∴ ∠=2@ = +
= 2H
Hopefully you can now use cyclic quadrilaterals to tackle some of the most difficult geometric problems in real DSE. See you next time!
13 HKJSMS, Nov 2015
Puzzle: The queen (♕) is the most powerful piece in the game of chess, able to move any number of squares vertically, horizontally or diagonally. Can you... a) Place eight chess queens on an 8×8 chessboard so that no two queens threaten each other. b) Same as a), but no three queens are in a straight line. The one below is an example solution of a 5x5 chessboard with 5 queens. (Answer is given on P.16)
IMO Zone Looking the question from another perspective How to think in geometry when facing an apparently algebraic problem By Au Chak Him, Anson
Find √
J
the
minimum
7 6 + 25 + √
J
value
+ 6 + 13
triangular inequality states, 12 + 1= ≥
of 2= = L(3 + 3)J +(0 7 6)J = 6√2 . The minimum occurs when A,B,C are collinear, which means
= 71.
It seems like an algebraic problem. And indeed it can be solved with certain This method has shown its power in 3 non-simple algebraic methods. Yet, if we are willing to alter our viewpoint, a simple geometric formula can be our savior.
years of IMOPSC (International Mathematical Olympiad Preliminary Selection Contest). Here come the contest problems and the solution.
Before looking into the solution, we first Question 1 need to have a basic understanding of the (IMOPSC2013, Q8) Find the smallest distance formula. On the Cartesian value of coordinate plane, the distance between L J + 4 + 5 + L J 7 8 + 25 two points ( K , K ) and ( J , J ) can be found by L( K 7 J )J +( K 7 J )J . For This seems extremely similar to the above example, the distance between (3,1) and problem. (5,4) is L(3 7 5)J +(1 7 4)J = √13 . We can make a slight manipulation first. And also please bear in mind that the sum of any two sides in a triangle is larger than the third side. Back to the above problem, we would like to apply the distance formula to the problem-solving process. To achieve this, we can manipulate the stuff inside the root, i.e. L
7 6 + 25 + L J + 6 + 13 = L( 7 3)J +4J + L( + 3)J +2J = L( 7 3)J +(4 7 0)J + L( + 3)J +(4 7 6)J J
Let 1( , 4), 2(3,0), =(73,6) 14 HKJSMS, Nov 2015
L
+ 4 + 5 + L J 7 8 + 25 = L( + 2)J +1J + L( 7 4)J +3J = L( + 2)J +(0 7 1)J J
+ L( 7 4)J +(0 + 3)J
Let 1( , 0), 2(72,1), =(4, 73) This becomes 12 + 1=.
12 + 1= ≥ 2= =
L(4 + 2)J +(73 7 1)J = 2√13. Question 2 (IMOPSC2014,Q3) Find the minimum value of L4 + √
J
J
+L
J
7 8 + 17
+
J
74 74 +8+
Then this becomes AB+AC. As the The expression becomes 1= + 2=. Our aim is to have a pair of square inside 1= + 2= ≥ 12 = L(1 7 0)J +(2 + 1)J each root. J + J 7 4 7 4 + 8 can be = √10 J J written as ( 7 2) +( 7 2) while The equality is attained when C is taken √ J 7 8 + 17 as ( 7 4)J +1J. as the intersection between AB and the Let 1(0,0), 2(2, ), =( , 2), :(4,3)
curve
The expression in the question becomes
= 1 in the first quadrant,
which gives
=
KQ√KW Y
. Therefore, the
minimal when 1, 2, =, : are collinear. minimum value of the expression = The collinearity can be achieved by √10. taking B as the intersection between AD and the line = 2 and C as the From the above examples, I hope you can intersection between AD and the line y=2. master this special technique and become Therefore, the minimum value of the capable of solving relevant problems in expression = 1: = √4J +3J = 5
the future. If you have any enquiry, feel free to contact us at any time.
Question 3 (IMOPSC2007, Q23) Find the minimum Fun fact: In number theory, Fermat's Last Theorem √O P QOR QJOQKQ√O P SJOT QUO R SVOQK value of O (sometimes called Fermat's conjecture, especially in older texts) states that no three First step is to put the x in the positive integers a, b, and c can satisfy the denominator into the root. equation Z + Z = Z for any integer V
+
=X
J
√
=X
J
+2 +1+√
+1+ J
2
1
+
1
J
V
+X
+ ( + 1)J
72 J
W
+5
J
72 +57
74 +1 4
+
1
J
value of n greater than two. However, did you know that Fermat wrote about it in 1637 inside his copy of a book called Arithmetica and wrote "I have a proof of this theorem, but
there is not enough space in this margin." However, no correct proof was found for 357 years. An English mathematician named Andrew Wiles found a solution in 1995. The proof took eight years of research. He received the Wolfskehl Prize from Göttingen Academy in June 1997: it amounted to about $50,000 U.S. dollars.
15 HKJSMS, Nov 2015
Problem Arena 1. How many ways are there to fill the following 7 boxes with 1 of 1,2,3,4,5,6,7 such that all greater than ‘>’ relations are satisfied?
One of the possible configurations
2. Let a,b be 2 positive real numbers. Consider
Y J
+
Y J
and 2
them always larger than the other one regardless of the values of prove it? 3. AB is a diameter of a circle.
V V
, Is one of
and ? Can you
is a line not touching the circle and perpendicular to
AB. Let D be a point on . AD intersect the circle at C (and A).Prove that regardless of the choice of point D, the value AC[AD will always be the same. 4. How many 6-digit integers with each digit being either 1 or 2 are divisible by 7? (For example, 121121 is one of them) Hints: 1. What must go into the middle cell? 2. Is the inequality J + J ≥ 2 always true for all real numbers c and d? 3. Can you spot 4 points that are concyclic from the diagram?
4. What does it mean to the remainder when divided by 7 if one of the ‘1’ is changed to ‘2’?
Answer of P.11 16 HKJSMS, Nov 2015