Solutions for Mathematics Enrichment

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IMOK Key Fob Problems

Hints and outline solutions for the mathematical problems on the IMOK key fobs.

What are the key fob problems?

Each year since 2005 the United Kingdom Mathematics Trust has produced a key fob, given to participants in the IMOK competition. The key fob contains a mathematical problem at an appropriate level.

2009 problem

Hints

Not a lot of information is given in the question. In particular no lengths are given, so you need to introduce some symbols. There are several ways to do this, but be careful—don't introduce too many letters.

It is also possible to simplify the working by choosing the units of measurement so that one convenient length is equal to 1.

You may also like to consider the region which is not the central square.

2008 problem

Hints

You may be lucky and find one time at which this happens by trial and error. However, in order to find all possible times, you need to be more systematic. In this case the only sensible approach is to introduce some letters and use an algebraic method. What are the unknowns here?

2007 problem

Hints

You may prefer to think in terms of cuts rather than pieces: how many ways are there to form a cut from the centre to the edge?

First it is necessary to decide what exactly is meant by "how many ways?". How are we counting? In other words, when are two ways considered to be the same?

2006 problem

Hints

As far as solving the problem goes, there is nothing special about the number 2006, so you may like to use a letter in place of the number.

You may also find it easier to consider the area which is not shaded.

2005 problem

This type of problem is sometimes called an alphametic:

• each letter stands for a one of the digits 0–9;
• different letters stand for different digits;
• the first digit of any number is never 0.

Hints

The problem is presented as a subtraction. It may be easier to think of it as an addition, in other words, in the form RING + ROO = KANGA.

What can you deduce from the sizes of the numbers?

Outline solutions

2009

There are several approaches to this problem, including similar triangles, coordinate geometry and trigonometry. Some of these methods use the fact that the region outside the central square is composed of four triangles whose area is known. We shall, instead, outline a proof using areas and Pythagoras’ theorem.

Firstly, let the outer square have sides of length 1 and use the notation shown. We may find the side length s of the central square from the information about areas given in the question. Also, using Pythagoras’ theorem, we may find b in terms of m.

Consider the parallelogram shown shaded in the diagram. We may express the area of the parallelogram in two different ways: base 1 – m times height 1; and base b times height s. Equating these two expressions for the area, squaring, and substituting for s and b, we obtain a quadratic equation for m. We may determine m by solving the quadratic equation. (Why can one of the two solutions be rejected?)

All that remains is to find the required area, that is, the area of the original shaded triangles. One way to do this is as follows. Note that the central square is the region common to two copies of the parallelogram. Hence the area of the outer square is equal to the required area plus twice the area of the parallelogram minus the area of the central square. We may use this relationship to calculate the area of the original shaded triangles, since we know all the other areas.

2008

Let the time required be h hours and m minutes.

What is the angle between the hands at this time? We can find this angle by comparing each hand with its position at 12:00. Since the hour hand turns through 360º in 12 hours, and the minute hand turns through 360º in 60 minutes, we can find how many degrees, H and M, they each turn through in h hours and m minutes, remembering that the number of minutes need to be included when dealing with the hour hand. The difference between these two angles is one angle between the hands; the other angle is obtained by subtracting from 360º.

But according to the problem, the angle between the hands is also h × m degrees. This gives an equation connecting h and m. In fact there is more than one version of the equation to solve: the difference referred to in the discussion above may be H – M or M – H, and the second angle also has to be considered, by subtracting from 360º.

To complete the problem, it is necessary to find all solutions to these equations, given that both h and m are integers and that h lies between 0 and 23, and m lies between 0 and 59. One way to do this is to find an expression for m and find which values of h give an integral value in the range 0 to 59. More sophisticated methods are also possible.

2007

We consider two ways of dividing up the figure to be the same if one can be rotated on reflected to give the other.

Can you see why this means we only need to consider a cut which starts along the line AB shown in the diagram on the right?

Once the cut reaches B, it can then either turn left, go straight on, or turn right. Proceeding in this way determines all the possible cuts.

Notice that not all routes are possible, since we also need to consider the other three cuts.

For example, in the diagram on the left, the black route is not a possible cut, since it collides with the next cut, shown in red.

2006

First observe that the small unshaded triangles in the right-hand corners are isosceles and right-angled. It follows that all the angles in the figure are either right angles or 45º, and so all the unshaded triangles are isosceles and right-angled.

The two left-hand triangles may be placed together to form a square of side 1003.

The two corner triangles may be placed together to form a square of side x.

The remaining unshaded triangle may be cut into two pieces, which may be placed together to form a square of side (1003 – x).

We can then write down an expression for the total unshaded area in terms of x. But this area is equal to three-eighths of the whole area, and the whole area is 2006 × (1003 + 2x). We can therefore find an equation, which may be solved for x. The equation is quadratic and so has two solutions, but only one of these makes sense in the context.

2005

In the form RING + ROO = KANGA we have a four-digit number added to a three-digit number. The largest possible value of the left-hand side is 9999 + 999 = 10998. Since the right-hand side is a five-digit number, what does this tell us about the values of K and A?

Once K and A are known, the value of R is also determined.

One way to complete the problem is to consider the last digits: since A is known we know what digit G + O ends in, and there are then only a few possibilities for the pair G, O. Considering each in turn shows that only one of them works, so that there is only one solution to the given problem.

Answers

2009

Together, the shaded triangles make up one fifteenth of the whole square.

2008

There are six times when this occurs: 00:00, 02:08, 02:56, 03:36, 11:20 and 12:00 (and also 08:48, 13:44, 14:40 and 22:24 if angles greater than 360º are allowed).

2007

There are seven ways, up to symmetry.

2006

The value of x is 501¹/₂.

2005

There is one solution, 10580 – 922 = 9658.

© A K Jobbings 2004-2009