Parallelogram 32 Level 3 11 Apr 2024Olympic mascots

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteau word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.
  • Tackle each Parallelogram in one go. Don’t get distracted.
  • Finish by midnight on Sunday if your whole class is doing parallelograms.
  • Your score & answer sheet will appear immediately after you hit SUBMIT.
  • Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.


5 marks

1.1 One of the mascots for the 2012 Olympic Games is called "Wenlock" because the town of Wenlock in Shropshire first held the Wenlock Olympian Games in 1850. How many years before the 2012 Olympics was that?

  • 62
  • 152
  • 158
  • 162
  • 172
  • (Not answered)

All you have to do is a subtraction: 2012 - 1850 = 162.


5 marks

2.1 The diagram shows three squares of the same size. What is the value of x?

  • 105
  • 120
  • 135
  • 150
  • 165
  • (Not answered)

The triangle in the centre of the diagram is equilateral since each of its sides is equal in length to the side of one of the squares. The sum of the angles at a point is 360°, so x=36090+90+60=120.


6 marks

3.1 Mathias is given a grid of twelve small squares.

He is asked to shade grey exactly four of the small squares so that his grid has two lines of reflection symmetry.

How many different grids could he produce?

  • 2
  • 3
  • 4
  • 5
  • 6
  • (Not answered)

The two lines of reflection symmetry of the grid are shown in the figure on the right as broken lines.

We see from this that if the grid with some of the small squares shaded grey has both these lines of reflection symmetry, then all the four squares labelled P must be the same colour. Similarly, all the four squares labelled Q must be the same colour, the two squares labelled R must be the same colour, and the two squares labelled S must be the same colour.

It follows that there are only three ways in which Mathias can shade exactly four of the small squares of the grid grey so that the result has two lines of reflection symmetry. These are:

(i) shade grey all the squares labelled P, and no others,
(ii) shade grey all the squares labelled Q, and no others,
and (iii) shade grey all the squares labelled R and all those labelled S, and no others.

The three different grids that Mathias could produce are shown in the figure below.


6 marks

4.1 Dominic wants to place the six dominoes above in a hexagonal ring so that, for every pair of adjacent dominoes, the numbers of pips match. The ring on the right indicates how one adjacent pair match.

In a completed ring, how many of the other five dominoes can he definitely not place adjacent to ?

  • 1
  • 2
  • 3
  • 4
  • 5
  • (Not answered)

There are only two 1-pip dominoes among the six Dominic has. These must therefore be adjacent. Likewise for the two 2-pips, the two 6-pips and the two 4-pips. So the ring must include the three adjacent dominoes:

and also the three adjacent dominoes:

The ring may now be completed two ways, i.e., by connecting top-left 5-pip to bottom-left 5-pip and the top-right 5-pip to bottom-right 5-pip … or by connecting top-left 5-pip to bottom-right 5-pip and the top-right 5-pip to bottom-left 5-pip.

We therefore see that Dominic can create two different rings of six dominoes, as shown below.

We now see that there are just 2 dominoes that the [1:5] domino cannot be adjacent to, namely the domino [2:5] and, of course, the domino [4:6].


6 marks

5.1 A die has the shape of a regular tetrahedron, with the four faces having 1, 2, 3 and 4 pips. The die is placed with 4 pips ‘face down’ in one corner of the triangular grid shown, so that the face with 4 pips precisely covers the triangle marked with 4 pips.

The die is now ‘rolled’, by rotating about an edge without slipping, so that 1 pip is face down. It is rolled again, so that 2 pips are face down, as indicated. The rolling continues until the die rests on the shaded triangle in the opposite corner of the grid.

How many pips are now face down?

  • 1
  • 2
  • 3
  • 4
  • it depends on the route taken
  • (Not answered)

We first consider what happens if one vertex of the tetrahedral die remains fixed on the grid and the die rolls about this fixed point. In doing this the die covers in succession six faces making up a hexagon. The faces adjacent to the fixed vertex are each face down twice in this hexagon. If the die is rolled about any of the edges on the perimeter of the hexagon, the remaining face is then face down. This is shown in the diagram on the right for the case where the faces with 1, 2 and 3 pips are adjacent to the fixed vertex, and the remaining face has 4 pips on it.

The pattern shown above can be extended to the whole of the given triangular grid, as shown in the diagram on the right. (Indeed, this pattern could be extended to a triangular grid filling the entire plane.)

For any triangle in the grid, the face then face down together with the two preceding faces form part of a hexagon, and determine the orientation of the die. (Here, by the orientation of the die, we mean the positions of all four of its faces.) The die can roll to this triangle across any of its edges. However, the pattern implies that the same face is face down in all these cases. In particular, the face with 1 pip is face down when the die reaches the shaded triangle.

Before you hit the SUBMIT button, here are some quick reminders:

  • You will receive your score immediately, and collect your reward points.
  • You might earn a new badge... if not, then maybe next week.
  • Make sure you go through the solution sheet – it is massively important.
  • A score of less than 50% is ok – it means you can learn lots from your mistakes.
  • The next Parallelogram is next week, at 3pm on Thursday.
  • Finally, if you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.

Cheerio, Simon.