HKJSMS newsletter may 2013

Hong Kong Joint School Mathematics Society HKJSMS Newsletter

Issue 3 12-13

May 2013

From the Editor Dear readers, it is a busy month for all of us after the Easter Holiday. However, please don't forget the big day for HKJSMS, that is the annual 25th ISMC (Inter -School Mathematics Contest) which will be hold on 4/5. HKJSMS and us strongly encourage you to support this big activity!

Eileen Tam & Jeff Siu

Interview with hang Lung Mathematics Awards winner P.2 – 4 // 5 nonspherical shapes that have π in their area P.5 – 8 // Introduction to Simson Line P.9-11 // Beauty of Geometry in Architecture P.11—15 // Rene Descartes—I think, therefore I am P.15—17 // Proof without words P.18-19 // Fun with Math P.19 // Solutions for Challenging Times of last issue P.20

Geometry

I have created a new universe from nothing— Janos Bolyai

1

Interview with Hang Lung Mathematics Awards winner Ewina Pun Bronze winner of Hang Lung Mathematics Awards 2012 School: St. Mary's Canossian College (Form 4 student when entering competition) Topic: Complexity Reduction of Graphs P: T:

Ewina Pun Eileen Tam

Background of Hang Lung Mathematics Awards The Hang Lung Mathematics Awards is a mathematics research competition for secondary school students in Hong Kong which is held every 2 years. The latest one was in 2012 and the following one will be held in 2014. It is coorganized by Hang Lung Properties, The Institute of Mathematical Sciences, and Department of Mathematics of The Chinese University of Hong Kong. The competition provides a great platform for competitors spending around half a

Interview

year time to study topics they are interested.

T:

sult! First of all, I am interested in

A two-stage assessment process is adopted by

knowing why you joined the Hang Lung

Hang Lung Mathematics Awards, the "research report review" and the "oral defense". For details, please refer to the page of Assessment Pro-

Mathematics Awards (HLMA).

P:

provide me a platform to motivate myself

will be sent to the Scientific Committee for re-

to do something out of my boundary.

view. Shortlisted teams will then be invited to T:

cation panel.

I see, so did HLMA fulfill your expectation?

vited team need to give a brief presentation and then respond to questions raised by the adjudi-

It is because I like to give challenges to myself. Also, this competition can exactly

cess. Reports that reach the basic requirements

participate in the oral defense session. Each in-

Congratulations on your splendid re-

P:

Yes, it did. At first, I was kind of worry that I would be unable to finish the re search as the deadline was approaching.

For

more

details,

please

hlma.math.cuhk.edu.hk/.

visit

https://

But then I managed to finish it. I was really happy and proud that I can achieve this task. I also learnt a lot during the

2

process. The award was just like a bonus

P:

mathematics or science or even other

to me, I didn't really expect so.

T:

subjects, but it should definitely be

I know your topic is Complexity Reduc-

mathematics related.

tion of Graphs. Could you first briefly talk about your topic which you study

T:

To be simple, it is to find out a new

P:

a year. And during summer, I spent more

to lower degree. It will be useful to apply

time to deal with the proposal and re

to many areas, as many network in our

search writing.

road distribution and etc. Reduction of

T:

know, you did not have any group-

der to investigate their properties easier.

mates. Would

people working on

you did for HLMA?

just on your own?

When I was a form 3 student, I partici-

P:

research

or

A group of people would be nice. I did try to find group mates at the beginning.

course held by The Chinese University of

However, they turned out rejected so I

Hong Kong. It focused on graph theory,

had to do it on my own. If I had a partner

thus I have some basic knowledge about

or two, a more complete and extended

that. I noticed that many of the graphs

result could definitely be obtained.

T:

Working alone was not enough for a tight schedule, just half a year. But you

the graphs be found if they are very com-

still did a great job in the end. How

plex, which is the topic that I chose for

nice! As you mentioned before, you

HLMA.

think your research is not perfect yet.

I see. So after the study, are you sure

So, I would like to know whether you

that you would like to study graph the-

came across any difficulties during the

ory further?

project studying. If yes, how did you overcome them?

Maybe later, but not now. As you know, I still have to deal with DSE in the coming

T:

the

pated in a mathematics enrichment

I then wondered how the properties of

P:

you prefer a group of

What inspired you to study the topic

demonstrated are very simple and direct.

T:

Wow. That's a lot of time. Was it hard for you to finish the whole project? You

such complex graphs is necessary in or-

P:

I had roughly spent half an hour a day for

method to reduce the complexity of graph

life, like the brain neuron transmission,

T:

Back to the project, how much time had you spent on the project?

for HLMA? P:

I am not too sure. I mean I may choose

P:

Yes, I did. I had to input a lot of data to

year. *(Miss Pun is now a Form 5 stu-

check my hypothesis just on my own.

dent.)

Moreover, I had been very desperate for

I bet you would like to study mathematics or something related to it in university, right?

several times because of failing to obtain something meaningful or useful for my research. But then I reminded myself to think positively and tried to think in dif-

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ferent angles, which helped me to solve

P:

pecially when I realized that every one of

the problem at last.

T:

them is mathematics expert.

I am glad to hear you overcame the diffi -culties in the end. Effort does have re

T:

nounced, I guess you started to prepare for the presentation. How was the

P:

about their research topic and the process

I started making the slides using latex

of doing research, which helps them in the

and write my speech for each slide. I then

future too.

asked my school mathematics teachers if they wanted to have a preview. They said yes and gathered to watch my perform ance. They also gave concrete suggestions

T:

what have you learnt?

P:

Many people think the most difficult

all very crucial and challenging.

is

the

Q&A

session

in

the

sis, proving, writing the paper and etc. are T:

Yes, this is true. I did prepare by predicting what questions might be asked.

enroll Hang Lung Mathematics Awards?

P:

For those who like to explore new things are welcome to take part in HLMA. Of

Though I prepared, the questions they

course, if you are interested in mathemat-

raised were quite strange and hard to

ics, you should join the competition too.

comprehend (maybe because of my own problem, ha-ha). Nevertheless, I under-

It must be a precious experience. What kind of people would you encourage to

you prepare for that?

T:

As I know, the competition is held every

stood that more questions they raised,

2 years. That means there will be an

more interested in the topic they were. So

other HLMA in 2014. Would you like to

I didn't really care much, I just answered

join the competition again in 2014?

as long as I knew how to respond them.

P:

The process of doing research, from title and direction setting to making hypothe-

presentation. Is this true? And how did

T:

Besides the knowledge of your project,

about my delivery and presenting skills. part

P:

Their effort is not in vain even if they didn't get prizes. I believe they did learn a lot

preparation?

T:

What comment would you like to give to your past opponents?

turn. Well, after the result had been an

P:

I was pretty overwhelmed! You know, es

P:

I am afraid that I may not participate in

Did you guess some of the questions

the coming HLMA as I have to prepare for

right?

DSE and university stuff.

Err, I don’t think so. The questions they raised were pretty simple and not into details.

T:

How you felt when you were facing a group of top mathematicians to present?

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5 nonspherical shapes that have π in their area/volume

By Tsang Chi Cheuk The pi day was about a month ago, math lovers all around the world celebrated it in different ways.

Some ate pi that day, or pizza, or even cake! No matter how they celebrate, it seems like the element of circles cannot be escaped. Of course, I heard you say, π is the ratio of circumference to diameter of a circle, hence one cannot associate π to anything else than circles or spheres. Well, unfortunately you’re wrong, there are a lot of shapes in which π appears in the formula for the shapes’ area and volume. We shall introduce you to 6 be-

Surprisingly enough, the area of the shaped bounded by the tractrix and the y-axis is actually . Readers who know calculus are encouraged to try to find this out themselves!

low…

The tractrix is first studied by Leibniz, and later 1. Tractrix

found to be significant in mathematics and physics. One of its properties is that by drawing out

Imagine you put a weight at (1,0) on a co-

the normals at all of it points, a catenary, that is, a

ordinate plane, the weight is connected to a

hyperbolic cosine function, will emerge. One

chain of length 1. The other end of the chain is at

more fact: when two end of a string are fixed, the

(0,0). Now pull the chain down, along the y-axis,

string will hang under its own weight and form

and picture how the weight moves. You should

the catenary.

obtain the lower half of the graph. After solving a few equations (readers who know differentiation and integration can attempt to derive so), we can get the equation:

where cosh t is the hyperbolic cosine function, tanh t is the hyperbolic tangent function. In function terms, it’s :

(only for top part of graph)

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2. Piriform

3. Sine surface

Let’s play a game: the graph below is a piriform, can you write out its equation?

Moving on from 2D to 3D now, we introduce to you the sine surface. Its equation is :

Or

Or

The shape may be difficult to visualize; you may want to use a computer plotting program (eg.

Ok, I’m just kidding you, you have to be an abso-

java) to help you fully understand it. But for

lute genius if you managed to do so, because its

those who don’t have a computer handy, just im-

formula looks like this:

agine a cube with its sides squashed in. While you’re still busy digesting the shape of the sine surface, let me give you another mind-boggling

or

if you

prefer the parametric form.

fact: the volume of the sine surface is simply π. If you still cannot imagine the shape of the graph, here it is:

The piriform means ‘pear-shaped’ in Latin (quite fitting, don’t you agree), and is first studied by G. de Longchamps in 1886 as an algebraic curve.

What is an algebraic curve? It is just a curve in which its formula can be expressed as a polynomial of x and y. Don’t think of it as a difficult term; circles and parabolas are common examples of algebraic curves. You may have heard of elliptic curves, in which Andrew Wiles used to prove Fermat’s Last Theorem. Those are algebraic curves too!

Oh, yes, I forgot, the area of a piriform, if you use the same equation given above, is

6

4. Cross cap

strip, while 2 cross caps with their boundaries

No, a cross cap doesn’t look like a cap, as in a hat,

glued is topologically equivalent to a Klein bottle.

but looks more like a fortune cookie. This is its parametric equation:

It can also be expressed as:

This is the appearance of a typical cross cap: Mobius strip (above), Klein bottle (below)

But here’s something topologists may not know: the volume enclosed by the cross cap is Topologist are quite interested in this shape because it exhibits interesting properties. First, let

Next time you see a topologist, try beating him by this fact!

us familiarize ourselves with the notion of topologically equivalence: two ‘shapes’ are topologically equivalent when one can be transformed into another by purely squeezing or twisting them, and without cutting and gluing; just imagine the shape made out of dough. Therefore a pencil is topologically equivalent to a rubber, and a donut is topologically equivalent to a tea cup (both have a hole or handle, depending on how you look at it). Here’s an interesting fact: the cross cap is topologically equivalent to a Mobius

5. Reuleaux tetrahedron Let us return to the 2D world for now, and think about this question: can we find a shape to replace the circle as wheels? You may think this question is absurd, but there is actually an answer if we remove only one small requirement: we won’t secure the shape to the car with an axis.

Still think it’s impossible? Then let me introduce the Reuleaux triangle. You can construct prisms

7

with cross sections of this shape, then push a block on top of the prisms smoothly, just like how you push blocks on top of cylinders. That’s because the unique way we construct the Reuleaux triangle in the first place: draw an equilateral triangle, draw circles with centre at each of the vertexes, so that those circles have radius same as the length of sides of the equilateral triangle, abandon the original sides of the equilateral triangle and use the arcs of the circles as the new sides. The fol-

inherit the property of its 2D counterpart, that is,

lowing diagram should illustrate clearly:

it cannot substitute spheres to cushion a block for smooth movement. However, Meissener and Schilling showed that it is possible to modify it slightly to form the Meissener tetrahedron, which can serve the function we mentioned above, also called a ‘surface of constant width’ given this property.

Along the many shapes we introduced here, the Reuleaux tetrahedron is properly the most obvious in containing π in its volume. That’s why we decided to not give you the formula, and you have to derive it yourself. Don’t worry, you don’t Now let us think in 3D again. We take the Reuleaux triangle and make it the Reuleaux tetrahedron. Construct a tetrahedron (a pyramid with equilateral triangles as each and every face),

need any advanced mathematical knowledge such as calculus, and we will provide you the answer at the end of this newsletter. Have fun calculating!

then draw spheres with centre at each of the vertexes, so that those spheres have radius same as the length of sides of the tetrahedron, abandon the original faces of the tetrahedron and use the curves of the spheres as the new faces. Here is a diagram of the Reuleaux tetrahedron: Surprisingly, the Reuleaux tetrahedron does not

8

Introduction to Simson line

By Brian Lui Ching Yin

.

As PLCM is a cyclic quadrilateral, we have

1. Simson line Given a triangle ABC and a point P on its circumcircle, the three closest points to P on lines AB,

As ANPL is a cyclic quadrilateral, we have

AC, and BC are collinear. The line joining these three points (L, M and N) is called the Simson line of P. Hence, we have continued to modify our aim in this proof. Now, it remains to prove that

Clearly, the above condition is satisfied if and only if LMN is a straight line. We conclude that the claim is proved.

3. Problems related to Simson’s Theorem The simson-related topics are highlighted in the The following proves the existence of such a Sim-

following 2 IMO problems.

son line.

1) 32nd IMO 1991 shortlist problem 3

2. Simson’s Theorem The points L, M and N are collinear if and only if P lies on the circumcircle of ABC.

If S is a point on the circumcircle of the triangle PQR, show that the feet of the perpendiculars from S to the lines PQ, QR, RP are collinear. De-

Proof: The point P lies on the circumcircle of ABC if and only if

note the line by [S, PQR]. If ABCDEF is a hexagon whose vertices lie on a circle, show that the lines [A, BDF], [B, ACE], [D, ABF], [E, ABC] are concurrent if CDEF is a rectangle.

Since the opposite angles at X and Z in the quadrilateral BPMN are right angles, we have

Therefore, it remains to prove that , which is equivalent to:

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Both HU and SV are perpendicular to PQ, so F is the midpoint of SV. Let SV meet the circle again at R'. Then ∠SR'R = ∠SUR (circumcircle) = ∠WUH (same angle) = ∠WHU (U reflection of H) = ∠HVS (HU parallel to SV). So HV is parallel to RR'.

∠ SR'R = ∠SQR (circumcircle) = ∠SQD (same angle) = ∠SFD (cyclic). So RR' is parallel to the Simson line (that is also a well-known result). Hence

Solution The fact that the feet of the perpendiculars are collinear is well-known (the line is known as the Simson line after Robert Simson 1687-1768, although the result seems to have been first

HV is parallel to XF. But F is the midpoint of SV, so X is the midpoint of HS. 2) 48th IMO 2007 Problem 2

proved by William Wallace in 1797). Label the feet of the perpendiculars D, E, F as shown (D on

Consider five points A, B, C, D and E such that

QR etc).

ABCD is a parallelogram and BCED is a cyclic quadrilateral. Let i be a line passing through A.

The proof is straightforward. ∠QSR =

180o

- P (S

lies on circumcircle) = angle ESF (SEPF cyclic).

Subtracting angle FSR, we get ∠QSF = ∠RSE. But SDRE is cyclic, so ∠RSE = ∠RDE, and QFDS is cyclic, so ∠QSF = ∠QDF. Hence ∠RDE = ∠QDF. Hence EDF is a straight line.

Suppose that l intersects the interior of the segment DC at F and intersects line BC at G. Suppose also that EF = EG = EC. Prove that l is the bisector of angle DAB. Solution Based on Simson’s Theorem, the feet of projections of E down to the three sides of triangle BCD (denoted M, L and K) are colinear as seen on the

Next we show that [S, PQR] passes through X, the

diagram.

midpoint of HS, where H is the orthocenter of PQR. This is also a well-known result.

Let the altitude from R meet the circumcircle again at U. Let SU meet PQ at W and let the lines HW and SF meet at V. Then PQ bisects HU (that is a well-known result, easily proved, ∠QPU = ∠QRU = 90o - Q = ∠QPH, so U is the reflection of H in PQ).

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Since L and K are the midpoints of FC and CG, respectively, LK||FG, and in triangle ACG, LK intersects AC at its midpoint M. Since ABCD is a parallelogram, M is also the midpoint of DB.

Therefore, there is only one unique point M to satisfy conditions that M is on DB and also collinear with K and L, and M is also the foot of E to DB.

Beauty of Geometry in Architecture

By Irene Kwok When talking about Olympics, you may immediately think of various sport games. However, have you ever noticed the Olympic Stadiums? Besides sports, the design of the Olympic Stadium is always an eye-catching point in Olympics. I believe you have also been stunned by their amaz-

Extend EM to cut the circle at I. Since EI is perpendicular to DB and DM = BM, it is understood that I is midpoint of arc DB. Therefore ∠DCI = ∠BCI = ∠4, as denoted on the graph. EI is also the

ing designs, such as the one in Beijing and London. However, do you wonder what elements in these buildings make them gorgeous? The answer is very simple – it is all about geometry.

diameter of the circle.

You may question how geometry is related to all It follows that ∠ECI = 90°. Further, ∠ICB = ∠CEK

these astonishing buildings. Indeed, architecture

(they both have sides perpendicular to one an-

is all about geometry -geometry of space, struc-

other), or (*)∠4 = ∠5.

ture and circulation. It is this understanding of geometry and shapes to act in a way that dictates us to design and build. For example, looking at

In triangle EFL:

∠1 + ∠2 + ∠3 = 90°

the Beijing and London Olympic Stadiums, we

In triangle EFG:

2(∠1 + ∠2 + ∠5) = 180°

can find out many different polygons. In addition,

Or ∠1 + ∠2 + ∠5 = 90°.

Therefore

∠3 = ∠5.

the London Olympic Stadium is symmetrical in shape. All of these are related to geometry. Of course, they are only some very basic ones. In a more advanced level, we can talk about golden ratio, fractal geometry and discrete differential geometry.

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1) Golden Ratio Definition of golden ratio The golden ratio, which is denoted as τ, is the number 1/2+1/2√5. It is approximately equal to 1.618. A numeric definition of golden ratio can be explained by 'Fibonacci series’ as following: Here is a 'Fibonacci series' : 0, 1, 1, 2, 3, 5, 8, 13,

In the pictures below, it shows how golden spiral is applied to the design of buildings:

21, 34, 55, 89, 144, …… If we take the ratio of two successive numbers in this series and divide each by the number before it, we will find the following series of numbers, i.e.

1/1 = 1 2/1 = 2 3/2 = 1.5 5/3 = 1.6666...

8/5 = 1.6 13/8 = 1.625 21/13 = 1.61538...

34/21 = 1.61904..

The ratio seems to be settling down to a particular value, which we call the golden ratio (Phi=1.618..).

Application of golden ratio 1. The golden spiral The Golden Spiral is created by making adjacent squares of Fibonacci dimensions and is based on the pattern of squares that can be constructed

2. Golden Rectangle

with the golden rectangle. If you take one point,

A Golden Rectangle is a rectangle with propor-

and then a second point on-quarter of a turn

tions that are two consecutive numbers from the

away from it, the second point is Phi times far-

Fibonacci sequence.

ther from the center than the first point. The spiral increases by a factor of Phi.

The Golden Rectangle has been said to be one of the most visually satisfying of all geometric

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forms. We can find many examples in art masterpieces such as in edifices of ancient Greece, such as the Parthenon showed in the picture below:

The Federation Square buildings in Melbourne applies the concept of Penrose Tilings. Here is the outlook of the building:

3. Penrose Tilings

2) Fractal geometry

The British physicist and mathematician, Roger Penrose, has developed an aperiodic tiling which Definition of fractal geometry incorporates the golden section. The tiling is comprised of two rhombi, one with angles of 36 Fractal geometry is the study of shapes made up and 144 degrees (figure A), which is two Golden of smaller repeating patterns. These patterns Triangles, base to base) and one with angles of called fractals repeat themselves through the use 72 and 108 degrees (figure B).

of self-similarity. For instance as seen in Chapter 1, a tree exhibits a similar structure at different levels of magnification. A close examination of the veins of a leaf reveals a branching pattern similar to the whole tree. Thus, we call the leaf a "self-similar" component of the tree. Just as a

tree’s overall pattern is carried by its genetic When a plane is tiled according to Penrose's di- code, so too is a fractal's shape carried by its rections, the ratio of tile A to tile B is the Golden mathematical code, the equation. In both cases Ratio. The diagram on the upper-right corner we refer this generating code as its “seed”. indicates this property:

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Application of fractal geometry

3) Discrete Differential Geometry Discrete differential geometry is the study of dis-

A prime example of this adaptation is illustrated in the Eiffel Tower built in 1889. Gustave Eiffel and his engineers purposely incorporated an interwoven design of steel girders supported by a continuing lattice of smaller trusses and beams into his structure, a structure that incorporates self-similarity to make effective use of material

crete counterparts of notions in differential geometry. Instead of smooth curves and surfaces, there are polygons, meshes, and simplicial complexes. It is used in the study of computer graphics and topological combinatorics. The pictures on show cured surfaces with discrete differential geometry.

and minimize weight.

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Application of discrete differential geometry An application of discrete differential geometry

Rene Descartes — I think, therefore I am

is the multi-layer freeform structure in the Opus By Ian Kwan building: 1. Introduction: Descartes, the father of analytical geometry (or Coordinate geometry) We often use Pythagoras Theorem, but do we really know who Pythagoras is? How about Gaussian Elimination? Who actually is Gauss? And Heron’s Formula? Who exactly is Heron? Its funny how we use the work of someone else without even realizing who the hell he or she is, no offence. And the one most ironic example of all is the Coordinate System. Oh, is there even an inventor to the coordinate system, I see you are wondering. Of course, the Coordinate System did not just popped out from midair as you may sugThe Opus, a mixed-use commercial and retail development in Dubai designed by Zaha Hadid, is being launched in London today. The £235 million project by developer Omniyat Properties will be located at the Business Bay development on the Dubai waterfront. The picture below shows the structural design of the building:

gest, to be more specific, the Cartesian Coordinate System. Who the heck is Cartesian, or who is Cartesy, or Cart? I am sorry to tell you that you are terribly wrong if the three names stated above are your guesses. The inventor of Cartesian Coordinate System is called Rene Descartes (you see –cartes, Cartesian). I am quite sure that you are trying to pronounce the word in your mind now, but I am again sorry if your guess is (Des-cart-tes), in fact, the name is pronounced as (Day-cart), a far more beautiful pronunciation, at least in my point of view.

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2. Descartes, the talented mathematician.

Descartes is the father of Analytical Mathematics (Coordinate Geometry, if you prefer that way.) it includes a, lot of the same things we use in Algebra, using equations, terms and polynomials that leads us back to Geometry. For example, for calculating circles, we just need to apply the equation of circle into the Coordinate system and al-

surdity. Here is the story: So there is a friend of Descartes, heading towards the home of his mathematician friend, who had gotten a bit sick, in fact, he was barely alive, with only a slight sense of consciousness. What a loyal friend, anyways, back to our story, Descartes, who was lying on the bed like a dead dog, while chit chatting with his loyal friend, suddenly let out a scream.

most anything can be found by this mean. The invention of Cartesian Coordinate System helped Don’t worry, he wasn’t going to die of a heart atfound Scientific Revolution and is still being used tack yet, he still had a long way to go, at least not in construction nowadays, where people uses it before he discovered the Cartesian Coordinate in blueprints, because the distance between any System. Guess what horrific sight did he saw that

points can be easily yet accurately measured, as- made him screamed like crazy? Yes, a spider, a sisting in measurement and architecture. spider as “large” as the tip of a pencil. (How could he even see that spider when he was being that sick? And it is also hard to believe that the mathematics giant was such a coward.) He was suddenly as alive as a jumpy kid, and kept pointing at the spider and asked his friend to kill it. His friend was really loyal, I must admit, but not really observant and couldn’t see the spider. Poor him, he couldn’t find the spider and had to bear the continuous screaming of Descartes. Seeing that his friend is not really that observant, and the ceiling Blueprint of an iPhone 5, showing the coordinate plane in behind

was made by pieces of white square tiles, Descartes’ mind lightened up and, first time ever in history, he gave the specific location of the spider using a set of coordinates, with the white square

3. Myth of discovery: Descartes, the innovative spider-phobic.

tiles as the coordinate system. Spider got squashed, happy, funny ending, but not really convincing, never mind.

Despite me being a bit of a lunatic, the unrealistic myth of how Descartes discovered the Cartesian Coordinate System simply is too absurd to con- 4. Descartes, the great philosopher vince me, and even if it was true, it is the second If you consider Descartes to be a great mathemafunniest story in mathematics history, only be- tician, then you will have to acknowledge that he hind the Fermat’s “this margin is too small to is actually a greater philosopher, a more widely contain” thingy, in terms of amusement and ab- known identity for him (Go and search the page

16

“philosophy” on Facebook and you will find his you know Blaise Pascal, the inventor of Pascal’s picture as the main picture of the page) . Rene triangle (you would or maybe you had already Descartes was concerned with the uncertainty in learnt about it when learning Binomial Theothe sciences and the skepticism caused from the rem.), the name would be familiar to you too if publications of "Sextus Empiricus", he rejected you study physics, found in the unit of pressure, anything which seemed uncertain and decided to Pa, or kPa that stand for Pascal and kiloPascal only accept "apodictic" knowledge as the truth. respectively. In 1647, Descartes met Pascal and Descartes discovered at least one thing could be argued with him that a vacuum could not exist, known apodictically. If he was doubting, then he and then again in 1648. For your information, had to exist. "Cogito Ergo Sum" - I think, there- Blaise Pascal and Pierre de Fermat (mentioned fore I am - became the first principle of before) were his mortal enemies in the mathe"Cartesianism", his home-made theory in philos- matics and science field. But clearly, we know that Descartes was wrong because vacuum do

ophy.

exists, in many parts of space. But his theory of (For more information of Cartesianism: Mind, for

vortices was really a contribution to physics.

Descartes, was a thing apart from the physical universe, a separate substance, linking human beings to the mind of God. The non-human, on the other hand, is nothing but complex automata, with no souls, minds, or reason. They can see, hear, and touch, but they are not, in any sense, conscious, and are unable to suffer or even to feel

Blaise Pascal and Rene Descartes.

pain. That was why he said, “I think therefore I am.” Also a spirit worth applying in mathematics, by the way.)

6. Conclusion: Descartes, the all time legend.

No matter how Descartes was successful in his philosophy career, and how he sucked at physics at some point of view, let’s not forget that this essay should be about mathematics, so lets again draw our attention back to mathematics. Rene Descartes introduced the Cartesian coordinate

I think, therefore I am.

system to mathematics, and made several notable contributions to the studies of imaginary numbers and trigonometric functions. But after

5. Descartes, the “not-so-great” physicist.

Despite having great discoveries and laying numerous milestones in Mathematics and Philosophy history, his physics ability is not bad, but comparatively, not so great. I am sure that all of

all, his biggest contribution of all: the Cartesian Coordinate System, left his name unburied by time in the great field of Mathematics, and I am quite confident his name will stay in eternity, just as the Cartesian Coordinate System, because they both are all-time mathematical legends.

17

Proof without words

Let

. By multiplying both

By Andrew Ying . Next, As with the previous issue of the newsletter on sides by 2, we get March, it has already been covered with the defi- subtract the second equation by the first equanition of a “proof without words” (which should tion, we get . Also simple and easy if you be very obvious by referring to the phrase itself) know the method, but as always, getting the and what’s so special about that. method is quite difficult.

1) Infinite Geometric Series Sum Formula

2) Viviani's Theorem

Consider the geometric series sum formula as According to the Viviani’s Theorem, the sum of listed below (which is regularly seen by believa- the distances from any interior point to the sides bly a lot of people):

of an equilateral triangle equals the length of the triangle's altitude. Not quite easy to proof with-

out a lot of words, but check out the following Of course everybody knows the above formula or figure. otherwise equality is always true from our everyday life. But how to prove that is a question.

That’s where a “proof without words” works:

By some simple triangle drawing and rearranging the triangles, the proof is already done. Easy, simple, no words involved. Evidently another great example of “proof without words”.

It is possible that you may want to know how to proof the theorem with words, but unfortunately you would still have to construct an equilateral triangle and then three more triangles inside.

Clean, clear and no words involved. Proof done.

And by considering their area, you could then prove the theorem. Certainly not that difficult, but all those words just made things a lot more

Of course, for every geometric sequence, there’s complicated. always an algebraic proof. We’ll be using the method similar to that of used in the last issue.

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3) Intersecting Chords Theorem The Intersecting Chords Theorem says that when

the experiment is going to happen there is a table full of food and a chair.

two chords intersect each other in a circle, the products of their segments are equal. Seems For the experiment the researcher told the Matheven difficult to prove. Again, a “proof without ematician that he was going to put him in the word” is better choice now.

chair and move him half way closer to the table every time the chair moved. Outraged the Mathematician stormed out of the room.

Next, the researcher explained the experiment to the physicist, and the physicist happily agreed to participate. Perplexed the researcher asked, "How come you aren't upset? Don't you realize that you will never actually get to the food?" The physicist replied, "But I will get close enough for

Yet another simple proof. Of course, with great

all practical purposes.

deal of geometric knowledge, you can probably still prove this. Warning: it is difficult.

2. Setting up a fenced-in area for sheep An engineer, a physicist, and a mathematician are

Really, majority of the “proof without words” just trying to set up a fenced-in area for some sheep, requires a little more drawing and also some but they have a limited amount of building matethinking. Next time, take a bit of time before ac- rial. The engineer gets up first and makes a tually trying to prove something using algebraic square fence with the material, reasoning that it's

proof or even word proof, and see if a “proof a pretty good working solution. "No no," says the physicist, "there's a better way." He takes the without words” would work. fence and makes a circular pen, showing how it

encompasses the maximum possible space with

Fun with Math

By Kwun Chun Hin

the given material.

Then the mathematician speaks up: "No, no, there's an even better way." To the others'

After reading serious articles about Geometry, amusement he proceeds to construct a little tiny fence around himself, then declares: let’s have 2 mathematical jokes to relax : 1. A mathematician and a physicist

"I define myself to be on the outside."

A mathematician and a physicist are both invited to partake in an experiment. In the room where

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Solutions for Challenging Time of last issue 1.

APMO 1991 Q3

mine those integers.

Solution Let a, b, c denote BC, CA, AB respectively. Since the line from a triangle's incenter to one of its vertices

Let a1, a2, … , an, b1, b2, … , bn be positive real num-

bisects the angle at the triangle's vertex, the condi-

bers such that a1+a2+…+an = b1+b2+…+bn. Show

tion of the problem is equivalent to the

Solution

By the Cauchy-Schwarz Inequality, we know

By Cauchy-Schwarz inequality,

But since

Then

we have: with equality only when

2.USAMO 2002 Q2 Let ABC be a triangle such that

are directly proportional to 36, 9, 4, respectively.

Therefore (clearing denominators and taking square roots) our problem requires that (s-a)(s-b)

, where s and r denote its semi perimeter, inradius respectively. Prove that triangle ABC is similar to a triangle T whose side lengths are all posi-

(s-c) be directly proportional to 36, 9, 4, and since a=(s-b)+(s-c) etc., this is equivalent to the condition that a, b, c be in proportion with 13, 40, 45.

tive integers with no common divisor and deter-

Upcoming competition 4/5/2013 (Sat) Inter-school Mathematics Contest 2013 18/5/2013 (Sat) International Mathematical Olympiad Preliminary Selection Contest - Hong Kong 2013 18/5/2013 (Sat) 希望杯國際數學競賽

Acknowledgement Ms Ewina Pun, St. Mary's Canossian College

Publication Team Andrew Ying // Brian Lui // Eileen Tam // Ian Kwan // Irene Kwok // Jeff Siu // Kwan Chun Hin // Tsang Chi Cheuk

Need your contributions The HKJSMS welcomes contributions to the upcoming issues of our newsletters from teachers and fellow students of our member schools. Please feel free to send your submission, such as event write-ups, articles or jokes related to Mathematics, to jsms.hk@gmail.com. We need your work to make the newsletter more rich!

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