Monday, May 25, 2015

[OlymPri_20150524MSCR] Round and Round, Twice or Thrice?

Question


Introduction
     If you think the answer is  2 cm (since the wheel rotates two rounds), you are wrong!  This tricky question is taken from a previous year Asia Pacific Mathematical Olympiad for Primary Schools competition.  In the original question, there was no colouring.  I added colours just to make the distinction between the wheel and the track a bit clearer.

Visualisation
     There is some subtlety in this question.  It is something we often do not notice unless we really think hard about it.  To get a handle on what is really happening, let us imagine we made a mark on the wheel in its original position, indicated by a blue dot at a twelve o’clock position.  Let us imagine rolling the wheel clockwise.  Note that the centre of the wheel will move in a bigger circle in an anti-clockwise (American: counter-clockwise) direction within the orange circular track.  Actually it does not matter which way you roll, the centre of the wheel goes round the centre of the track in a direction opposite to that of the turning wheel, which always remains in contact with the track.

Since the blue dot on the wheel turns two rounds, by the time the wheel reaches the bottom half, the blue dot must again be on top or at a twelve o’clock position.  Note however, the point of contact between the wheel and the circular track (indicated by a green dot whenever possible) is not the same as the blue dot!  The green dot is actually a dynamic dot (it is not always the same point on the wheel) whereas the blue dot is always the same dot on the wheel but the wheel is being rotated.

Note that at the halfway point, the green dot is at the bottom of the wheel while the blue dot is on top of the wheel.  Notice also that, relative to the wheel, the green dot goes in the opposite direction as the blue dot and they actually crossed over somewhere along the way!  Actually the green dot has made one-and-a-half turns with respect to the wheel.  Remember that the green dot is not a fixed point on the wheel, but it measures how much the wheel and the track have been in contact.


As the wheel continues to roll back up, the green dot makes another  1½  rounds.  So altogether the green dot moves through  3  rounds while the blue dot rotates only  2  rounds around the centre of the wheel!  The green dot is the one that matters.

Solution
     circumference of track (measured by green dot) : circumference of wheel = 3 : 1
     Since radius is proportional to circumference,  the radius of the track is  3  cm.

Remarks
     Yes, the solution is that short.  But the thinking behind it is profound.  But do we need to draw all the diagrams as in the visualisation above?  I did that to explain to you.  Actually I imagined it in my mind.  If imagination is difficult, you can act it out by drawing a big circle and then using a small coin to simulate the rotation around the track.  I actually drew a rough sketch by hand  to convince myself that my thinking was accurate.

     Notice that I used a proportionality argument.  If you know how to use the concept of proportion, you can make the working short and sweet.  There is nothing wrong in using the formula                     circumference = 2p ´ radius.
This formula just says that the circumference of a circle is proportional to its radius, and the constant of proportionality is  2p.  Your working would look like this
                                  circumference of track = 3 ´ circumference of wheel
                                      2p  ´ radius of track = 3 ´ 2p  ´ radius of wheel
After cancelling out the  2p,  you would get
                                       radius of track = 3 ´ radius of wheel = 3 cm
You get the same conclusion, but using the proportionality method, you do not need to bother about the  2p.

     By the way, the centre of wheel traces out the path of a circle of radius  2 cm.  See the red arrow in the first diagram on top.  The path traced out by the blue dot looks like an  epicycleIn the old days,people thought that the sun, moon and planets rotated around the earth in epicycles.  This also reminds me of Spirograph,which is a toy that allows you to use your coloured pencils to create very beautiful patterns.  [Click to search for images of Spirograph and patterns produced.]


H01. Act it out
H02. Use a diagram / model
H04. Look for pattern(s)
H05. Work backwards
H06. Use before-after concept
H09. Restate the problem in another way

Suitable Levels
Primary School Mathematics Olympiad
* anybody game for a challenge relating to imagination, circumference and lengths





Sunday, May 24, 2015

[MathsEd] The Secret to Mathematical Problem Solving

     Many students treat mathematics problem solving either as a mystery, or they like to shoot at random from a set of formulas or recipes that they have memorised and just see whether it works or not.  When this strategy does not work and after working for a few minutes, they just give up.  They wonder how all those maths geniuses get it.  Many do not know that even the professional mathematicians take many years or evencenturies to solve mathematics problems.  The mathematician George Polya has written a book “How to Solve It” in 1945, describing how mathematics problems are solved.  Since then many people have adapted and modified the steps slightly but they basically boil down to the following steps.

Step 1: Understanding
     Try to make sure you really understand the problem first.  If it is an examination question, read and take note of the given information.  Ask what is known, what can be known and what is to be found.

Step 2: Planning
     This is where you plan your strategy of attack.  Can you organise the information into a table or a diagram?  Heuristics(rules-of-thumb or guidelines) are usually useful to help you formulate yourstrategy, especially if you have not seen this type of question before.

Step 3: Execution
     Carry out your plan.  Make sure you are conscious of what you are doing.  Are you able to explain it to yourself, or younger brother/sister?

Step 4: Evaluation
     Check your calculations and logic.  Are there any careless mistakes?  Is your answer plausible or believable?  Are you on the right track?  Are you getting somewhere?  You also need good number sense.  For example, if you are calculating with a triangle with sides  4 cm,  5 cm  and  6 cm,  and you get an area like 1 000 cm ²,  does your answer even taste and smell right?  Do not continue doing the same wrong thing.  If you catch yourself making a mistake, go back to step 2 and change your strategy.  Try another approach.

Step 5: Reflection
     After solving the problem, think back at what lessons you have learnt by attempting / solving this problem.  How could you have done better?  Did you discover anything that can be applied in other problems?  What if the numbers are changed?  What if the conditions are changed?  You can test your own understanding by setting yourself a similar or modified question.  Can you generalise your results?  Can you link what you have learnt to daily life or to other subjects?



Suitable Levels
all levels, all topics !!!

[Pri20150523WNCA] Trees arranged in a Hexagon Outline

Question

Introduction
     This is real eeeasy peasy lemon squeezy, isn’t it?  54 ¸ 6 = 9  Ta da!  The answer, right? Wrong!  You got tricked!  Ha!  Ha!
     Always tryto understand the question and do the planning first.  Never be in a hurry and jump to thecalculation stage.  So what went wrong?  Well, the tree at each vertex is counted twice.
     Huh?
     Sometimes to understand the situation, it may be easier to consider a simpler problem.  Let us say there are four trees per side.  This is how it looks like from above.

     You can see that the corner trees (coloured in orange instead of brown) are counted twice, because they each serve as an extreme marker of two of the sides of the hexagon.  There are 18 trees and if you divide by  6,  you get  3  and not  4.  One way to count properly is to start from one corner tree and count groups of three trees, either in a clockwise or anti-clockwise (American: counter-clockwise) direction.

     Notice that the number of trees on one edge of the hexagon is equal to the number of trees in one group plus one (the corner tree for the next group).  So for  18  trees, the correct calculation is  18 ¸ 6 + 1 = 3 + 1 = 4  for the number of trees along one edge.  We use the same procedure for  54  trees.

Solution
     number of trees on each side = 54 ¸ 6 + 1 = 9 + 1 = 10

Final Remarks
     You may want to generalise it into a formula
                    # trees on each side = total # trees ¸ #sides + 1
However, I do not recommend that you purposely memorise this formula.  Mathematics is not about memorisation.  It is about understanding.  Once you understand it, the formula comes out automatically.  You may test yourself or get a friend to test your understanding by setting a similar question but changing the number of trees and number of sides.

H02. Use a diagram / model
H04. Look for pattern(s)
H10. Simplify the problem

Suitable Levels
Primary School Mathematics
* other syllabuses that involve whole numbers







Saturday, May 23, 2015

[S2_20150523XFDS] Numbers that can be Difference of Squares

Question

Introduction
     This is likely an primary mathematics olympiad-type of question, but lower secondary pupils can also try this.  It involves deeper thinking.  But where do we begin?  Sometimes it is good to begin from the beginning, and then follow your nose. 

Reminders

Solution
     Suppose  N  is a whole number such that  1 < N < 1000  and  N  can be expressed as
                                        N = a2b2  = (ab)(a + b)
a difference of squares.  So  N  can be split as a product of two factors  (a + b)  and  (ab).  Observe that     (a + b) – (ab) = 2b,       which is an even number.
     The difference between the two factors is an even number.  This can only mean that the two factors are  both odd  or  both even.  You cannot have one of them odd and the other even, because when you subtract them, you would get an odd number.  We now have three cases:-
     Case 1a:  N  is even but not divisible by 4.
     Case 1b:  N  is divisible by 4 (and, of course, is even)
     Case 2:    N  is odd  i.e. both  (a + b)  and  (ab)  are odd



Ans:  750

Remarks
     In the foregoing, it is possible for  b  to be zero.  0 happens to be a perfect square, because  02 = 0.  However, we need not worry about this, because the above algebra is general enough to cover the case where  b  is  0.
   We have solved the problem using logic, even-vs-odd analysis and the three important algebraic identities under reminders (highlighted in orange).  We also used the special cases (highlighted in light blue) and made observations based on them.

H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem
H12* Think of a related problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
Primary School Mathematics Olympiad
Secondary 2 Mathematics » grade 8 (expansion and factorisation)
* anyone who is game for a challenge in algebra and number theory





Friday, May 22, 2015

[Pri20150522RPRW] More Hands make Light Work?

Question

Introduction
     The key to solving this is to find the total rate of work for Ahmad, Kumar and Calvin.  There is a relationship Work = Rate ´ Time, similar to the relationship Distance = Speed ´ Time, and you can use a similar triangle mnemonic for it.  For example if you cover ‘R’ with your finger, you get the relation Rate = Work / Time.  To organise your information, you may use a table to tabulate the given data.

     Do you believe “more hands make light work”?  Or is it “too many cooks spoil the broth”?  In real life, people may not work harmoniously together, or work at the same rate (never getting tired).  They may get distracted by Facebook, mobile phones or office gossip.  In school mathematics problems, we assume that when people work together, we can just add up their rates of work.  Yes, it is quite funny, but let us just assume.

Solution

Ans: 10 days

Commentary
     When we add up the given rates, we get  1/5,  which is the rate that 2 Ahmads, 2 Kumars and 2 Calvins would work.  Unfortunately we do not have the clones.  We just have one Ahmad, one Kuman and one Calvin.  So we divide that by 2 to get  1/10  and this is their combined rate.  This means that they can complete  1  house in  10 days.
     It is possble to do this question without the table, for example by taking the LCM of the denominators and considering how many houses can the guys build in  60  days.  However, the table is a good way of organising information and you can solve the problem by considering the total rate.

H02. Use a diagram / model   (including table)
H03. Make a systematic list
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem

Suitable Levels
Primary School Mathematics
* other syllabuses that involve fractions and ratios




Thursday, May 21, 2015

[S2_20150520SERC] Simultaneous Equations with Reciprocals

Question


Introduction
     This is a secondary 2 (» grade 8) problem that involves linear simultaneous equations in two unknown.  It looks non-linear and indeed if you try to solve for  x  and  y  directly,  you might get into a big mess with extraneous solutions to boot.  What should our approach be?

Observations
     It is always a good idea to stare at the question for a little while longer before jumping in to try to solve it.  Let us (mentally) reformat the equations a little bit.
     Do you notice any repeated chunks?  Chunking is very useful.  Let use substitutions for those chunks to simplify the equations.

Solution
Final Remarks
     To recap: It is a good learn to make observations first before attempt to solve the problem.  If you see any repeated chunks, it is a good idea to use substitution to simplify the problem.  Once the reciprocals have been substituted, we try to use elimination (which is usually the more effective method).  Also, try as far as possible to avoid fractions.  By multiplying equation [2] by 5,  we get – 30Y, which can be cancelled if we multiply  15Y by 2.  Thus  Y  is eliminated.  The problem becomes easy from this point onwards.


H04. Look for pattern(s)
H09. Restate the problem in another way
H10. Simplify the problem
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence


Suitable Levels
Lower Secondary Mathematics (Secondary 1)
GCE ‘O’ Level “Elementary” Mathematics (revision)
* other syllabuses that involve algebra and simultaneous linear equations





[IB-HL H&H_8G Q16] Sum of Squares of some Binomial Coefficients

Question

Introduction
     This problem is taken from the Haese textbook for International Baccalaureate, 3rd Edition, page 262.  It looks pretty daunting doesn’t it?  Where do we even begin?  The key to solving this problem is to realise that the binomial coefficients are coefficients of (numbers attached to) certain powers of  x  in the expansion.  The question is:  which power or powers?
     Before we go into that, let us review some important relevant facts.

Reminders
Solution


Final Remarks
     This problem was solved by using the symmetry property and treating binomial coefficients as coefficients of certain powers of  x.  We also worked backwards by noting that the RHS of the equation to be proven is the coefficient of  xn.  This suggests that we compare this with the coefficients of  xn  on the LHS.


H03. Make a systematic list
H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
International Baccalaureate Mathematics (HL)
GCE ‘A’ Levels H2 Mathematics
* other syllabuses that involve complex numbers and polynomials