A blog about Mathematics and Mathematics Education in Singapore, as well as Mathematics Education in general. Written for students, parents, educators and other stakeholders in Singapore, and around the world.
More information here. Check out my Education and Technology blog. Follow me on Twitter.
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.
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
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.
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.
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.
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
of track = 3 ´ circumference of wheel
2p´ radius of track = 3 ´ 2p´ radius of wheel
After cancelling out the 2p, you would get
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.
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
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.
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 cm²,
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?
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.
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
number of trees on each side = 54 ¸ 6 + 1 = 9 + 1 = 10
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.
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.
N is a whole number such that 1 <N< 1000 and N can be expressed as
N = a2 – b2 = (a – b)(a + b)
a difference of squares. So N
can be split as a product of two factors (a +
b) and (a – b). Observe that (a
+ b) – (a – b) = 2b, which is an even number.
difference between the two factors is an even number. This can only mean that the two factors are both odd
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:-
1a: N is even but not divisible by 4.
is divisible by 4 (and, of course, is
2: N is odd i.e. both (a +
b) and (a – b)
foregoing, it is possible forbto 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 wherebis 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.
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.
Ans: 10 days
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.
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
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
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.
notice any repeated chunks? Chunking is
very useful.Let use substitutions for
those chunks to simplify the equations.
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 
by 5,we get – 30Y, which can be cancelled if we multiply15Y
by 2.ThusYis eliminated.The problem becomes easy from this point
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.
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.