$$\def\degrees{^\circ}$$

# IMOK Key Fob Problems

## 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.

## 2015 problem

### Hints

When a polygon is regular and has 36 sides, what relevant angles can you find?

Consider some particular polygon, and suppose it has more than one neighbour. How many edges can there be between the two adjacent polygons?

## 2014 problem

### Hints

Find a way to express the number of cubes on the outside of the cuboid in terms of $$n$$.

## 2013 problem

### Hints

You may find it helpful to find what sets of four consecutive digits are possible.

Then you need to realise that the counting is different if 0 is one of the digits.

## 2012 problem

### Hint

You may find it helpful to think about what could happen in one corner.

## 2011 problem

### Hints

Draw a careful diagram, showing the circle, the trapezium and its diagonals; it may also be helpful to join the centre of the circle to some of the vertices of the trapezium.

There are several methods, but you will need to use at least one property of circles.

## 2010 problem

### Hints

The triangle formed by the centres is isosceles and it is possible to find its height. Can you see how this height relates to the height of the rectangle?

Now all you need to find is the width of the rectangle.

## 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 $$\text{RING} + \text{ROO} = \text{KANGA}$$.

What can you deduce from the sizes of the numbers?

## Outline solutions

### 2015

We shall consider two arrangements of polygons to be the same if one can be rotated and/or enlarged to look like the other.

Because the question refers to polygons in the plural, we assume that $$m$$ is at least two. The case $$m=2$$ clearly always works, so let us suppose that $$m$$ is greater than two.

Consider one polygon $$\mathcal{P}$$. Because we are supposing that $$m$$ is at least three, there are two distinct polygons adjacent to $$\mathcal{P}$$.

Let $$O$$ be the centre of the circle, let $$M$$ be the midpoint of the edge common to $$\mathcal{P}$$ and one adjacent polygon, and let $$M'$$ be the midpoint of the edge common to $$\mathcal{P}$$ and the other adjacent polygon. Then $$\angle M'OM = 360^\circ \div m$$.

Joining the middle of each edge to the centre $$C$$ of $$\mathcal{P}$$, we can divide $$\mathcal{P}$$ into 36 identical pieces, each of which has an angle of $$10^\circ$$ at $$C$$, as shown.

Suppose there are $$v$$ vertices of $$\mathcal{P}$$ between $$M'$$ and $$M$$. Then, from the sum of the angles of the quadrilateral $$OMCM'$$, we get $\frac{360}{m}+90+10v+90 = 360.$

Rearranging, we obtain the equation $$m(18-v) = 36$$. (Can you see how?)

Now $$m$$ is a positive integer and $$v$$ is a non-negative integer (it may be zero). Therefore, from the above equation, $$m$$ is an integer divisor of 36. Moreover, $$m$$ is greater than one. So the possible values of $$m$$ are 2, 3, 4, 6, 9, 12, 18 and 36.

Hence there are eight possible values of $$m$$.

Can you see how to check that each of the eight values of $$m$$ that we found does indeed lead to a configuration of the required sort?

### 2014

The number of cubes that have at least one red face is the same as the number of cubes on the outside of the cuboid. One way to find the number of cubes on the outside is to subtract the number of cubes on the inside from the total number of cubes.

Now the whole cuboid measures $$n \times (n+1) \times (n+3)$$, so the inside is a cuboid measuring $$(n-2) \times (n-1) \times (n+1)$$. Therefore the total number of cubes is $$n(n+1)(n+3)$$ and the number of cubes on the inside is $$(n-2)(n-1)(n+1)$$. Hence the number of cubes on the outside is $$n(n+1)(n+3) - (n-2)(n-1)(n+1)$$.

It is possible to multiply out all the brackets in order to rewrite this expression. But there is a better approach, which makes use of the fact that $$(n+1)$$ appears in both terms, so that $$(n+1)$$ is a common factor. We therefore factorise the expression, and so obtain the number of cubes on the outside in the form $$(n+1)[ n(n+3) - (n-2)(n-1) ]$$, which we may simplify to $$2(n+1)(3n-1)$$—can you see how?

But we are told that there are 2014 cubes on the outside. Thus $$2(n+1)(3n-1) = 2014$$, so that $$(n+1)(3n-1) = 1007$$. We could solve this equation for $$n$$ using one of the standard techniques for solving a quadratic equation, but in this case there is another method, because we know that $$n$$ is a positive integer. It is also helpful to know that $$1007 = 19 \times 53$$.

Can you see how to find $$n$$ and hence complete the solution?

### 2013

For a four-digit integer, without any restrictions, the first digit may be 1 to 9, giving nine possibilities, and each of the other digits may be 0 to 9, so there are ten possibilities for each of these. Thus there are $$9\times10^3=9000$$ in total, which, of course, is equal to the number of integers in the range from 1000 to 9999 (inclusive).

If the digits are consecutive then there are far fewer possibilities. Ignoring for a moment the order in which they appear, what could the four digits be? The lowest possible sequence of four consecutive digits is 0, 1, 2 and 3; the highest is 6, 7, 8 and 9. So there are seven possible sets of four digits. Each of these sets may be ordered in $$4\times3\times2\times1=24$$ ways. Thus the total number of arrangements altogether is $$7\times24=168$$.

However, we have included the cases in which the first digit is zero. In each of these cases, we know that the other digits are 1, 2 and 3 in some order. So the number of such cases is $$3\times2\times1=6$$.

Hence the number of integers of the required type is $$168-6=162$$.

### 2012

We claim that the L-shapes fit together in pairs to form $$2\times3$$ blocks. These blocks fit into the region of white cells in only two possible ways, the arrangement shown on the right, and its mirror image.

Now there are four blocks, and a pair of L-shapes can be placed in each block in exactly two different ways—can you see why? So altogether we may fit L-shapes into the blocks in $$2\times2\times2\times2=2^4$$ ways. Since there are 2 ways to fit the blocks in the region, you should now be able to find the total number of ways of placing the L-shapes in the region of white cells.

To see why the L-shapes fit together in pairs in $$2\times3$$ blocks, consider how two L-shapes can be placed in the bottom left corner of the region. There is only one possibility which does not lead to a $$2\times3$$ block, the arrangement shown on the left. Can you verify this?

You should now be able to see that, after starting in this way, it is not possible to continue to fill the region of white cells. So our assertion that the L-shapes form $$2\times3$$ blocks is justified.

### 2011

There are several possible methods. We present one which makes use of various geometrical properties of the configuration.

Label the trapezium $$ABCD$$ and draw in the diagonals, as shown on the right.

Can you explain why the three marked angles at $$A$$, $$B$$ and $$C$$ are equal? Since the diagonals are perpendicular it follows that angle $$ABD$$ is $$45\degrees$$.

Now join $$A$$ and $$D$$ to the centre $$O$$ of the circle, as shown on the left.

Then angle $$AOD$$ is $$90\degrees$$ since it is an angle at the centre of the circle and therefore twice angle $$ABD$$.

Finally, draw the perpendiculars from the centre $$O$$ to the chords $$AB$$ and $$CD$$, as shown on the right. These perpendiculars meet the chords at their midpoints $$M$$ and $$N. Can you see how to prove that the two shaded triangles are congruent? You can now use Pythagoras' theorem in either shaded triangle to find the radius of the circle, and hence its area. ### 2010 The width of the triangle formed by the centres is equal to the width of the rectangle plus two radii, hence the width of the rectangle is \(40 - 29 = 11$$.

Slide the shaded rectangle up through a distance of one radius to see that its height is equal to the height $$h$$ of the triangle.

Since the triangle is isosceles this height divides the triangle into two congruent right-angled triangles, as shown. Now use Pythagoras' theorem to show that $$h^2=29^2-20^2=(29-20)(29+20)=9\times 49$$, from which you can find $$h$$.

Finally, the area of the rectangle is $$11 \times h$$.

### 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 $$360\degrees$$ in 12 hours, and the minute hand turns through $$360\degrees$$ 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 $$360\degrees$$.

But according to the problem, the angle between the hands is also $$h \times m$$ 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 $$360\degrees$$.

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 $$45\degrees$$, 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 $$2006 \times (1003 + 2x)$$. 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 $$\text{RING} + \text{ROO} = \text{KANGA}$$ 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 $$\text{K}$$ and $$\text{A}$$?

Once $$\text{K}$$ and $$\text{A}$$ are known, the value of $$\text{R}$$ is also determined.

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

### 2015

There are eight possible values of $$m$$.

### 2014

There are 5168 cubes with no red faces.

### 2013

There are 162 such 4-digit integers.

### 2012

There are 32 ways of covering the region by L-shapes.

### 2011

The area of the circle is $$\frac{521}{4}\pi$$ (square units).

### 2010

The area of the shaded rectangle is 231 (square units).

### 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 $$360\degrees$$ are allowed).

### 2007

There are seven ways, up to symmetry.

### 2006

The value of $$x$$ is $$501\negthinspace\frac{1}{2}$$.

### 2005

There is one solution, $$10580 - 922 = 9658$$.

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##### UKMT

The United Kingdom Mathematics Trust organises the Maths Challenges and enrichment activities for schools and colleges.

United Kingdom Mathematics Trust
##### IMOK

The IMOK competition (the Intermediate Mathematical Olympiad and Kangaroo) is the follow-on round for the IMC (the Intermediate Maths Challenge).

Pupils who do well in the IMC are invited to take either an Olympiad paper or a multiple-choice Kangaroo paper

The Kangaroo is a European competition with over five million entrants worldwide.

European Kangaroo