Parallelogram 30 Level 3 28 Mar 2024Wombat Geometry

This is a preview of Parallel. You have to login or create an account, to be able to answer questions and submit answers.

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 Which of the following has exactly one factor other than 1 and itself?

  • 6
  • 8
  • 13
  • 19
  • 25
  • (Not answered)

The factors of 6 are 1, 2, 3 and 6; the factors of 8 are 1, 2, 4 and 8; the factors of 13 are 1 and 13; the factors of 19 are 1 and 19; and the factors of 25 are 1, 5 and 25. We see from this that, of the numbers we are given as options, only 25 has exactly one factor other than 1 and itself. It is worth noticing that 25 is the only square number.

2. Wombat Geometry

Take a look at this short video by YouTuber Vsauce.

(If you have problems watching the video, right click to open it in a new window)

2 marks

2.1 Yes, you did just watch a wombat farting. If the wombat went on to do a poo, what shape would its poo be?

  • Sphere
  • Cube
  • Cylinder
  • Klein bottle
  • Stellated rhombic dodecahedron
  • (Not answered)

Wombat poo is roughly cubic, and the BBC explains why this is the case.


5 marks

3.1 The figure on the right shows an arrangement of ten square tiles.

Which labelled tile could be removed, but still leave the length of the perimeter unchanged?

  • A
  • B
  • C
  • D
  • E
  • (Not answered)

Removing tile A or tile B or tile D has the effect of reducing the perimeter by a distance equal to twice the side of one tile, whilst removing tile C increases the perimeter by that same distance.

Removing tile E, however, leaves the length of the perimeter unchanged.

4. Wombats in art

Dante Gabriel Rossetti attended King’s College School in London, where he developed a passion for art.

After he became a world famous painter, he developed an interest in exotic pets, and in 1869 he acquired his first pet wombat.

We might have veered away from our main topic of mathematics, but that happens sometimes.

The world is an extraordinary place and it is good learn about everything.

2 marks

4.1 Like all marsupials, a wombat female has a pouch. Where is the opening to the pouch?

  • At the top of the pouch, nearer the wombat’s front legs.
  • At the bottom of the pouch, nearer the wombat’s back legs.
  • (Not answered)

Unlike kangaroos, wombats have their pouches upside-down, with the opening at the bottom, nearer to their back legs.

This is because wombats dig with their front legs, and having the pouch opening at the bottom avoids soil been thrown into the pouch.

Koalas are close relatives of wombats, and they also the opening to the pouch at the bottom.

This sounds risky, because koalas spend their time upright, high in trees, but the opening to the pouch is probably quite tight, so there is no risk of the baby koala falling out.


5 marks

5.1 In the multiplication grid on the right, the input factors (in the first row and the first column) are all missing and only some of the products within the table have been given.

What is the value of A + B + C + D + E ?

  • 132
  • 145
  • 161
  • 178
  • 193
  • (Not answered)

Let the input factors be p,q,r,s and t along the top row, and v,w,x,y and z in the first column, as shown in the grid on the right.

We see that wp=15 and xp=18, so that p is a divisor of both 15 and 18, and hence is either 1 or 3. However, if p=1, then it would follow that w=15. But this is not possible as wr=40. We deduce that p=3. It follows that w=5 and x=6. Since w=5 and wr=40, we have that r=8.

Therefore, as zr=56, z=7. Also, as x=6 and xs=60, s=10. Since vs=20, it follows that v=2. Then, as vq=10, q=5. Hence, as yq=20, we have that y=4 . Finally, as yt=24, we deduce that t=6.

This enables us to complete the table, as shown on the left. (Though, really, we need only to calculate the diagonal entries that we have shown in bold.)

Therefore, we have A+B+C+D+E=6+25+48+40+42=161.


6 marks

6.1 In rectangle PQRS, the ratio of PSQ to PQS is 1:5 . What is the size of QSR?

  • 15°
  • 18°
  • 45°
  • 72°
  • 75°
  • (Not answered)

Let QSR =x°. Since PS is parallel to QR, the alternate angles PQS and QSR are equal. So PQS =x°. Therefore, as PSQ:PQS =1:5, PSQ =15x°. Therefore, from the right angled triangle PQS we deduce that x+15x+90=180 and so 65x=90. Therefore x=56×90=75.

So QSR is 75°.


6 marks

7.1 Sam wants to complete the diagram so that each of the nine circles contains one of the digits from 1 to 9 inclusive and each contains a different digit.

Also, the digits in each of the three lines of four circles must have the same total.

What is this total?

  • 17
  • 18
  • 19
  • 20
  • 21
  • (Not answered)

We suppose that x, a, b, c, d, e, and f are the numbers in the circles, as shown in the diagram. These numbers are the digits 1, 3, 4, 6, 7, 8, and 9, in some order.

Let the common total of the digits in the three lines be k.

Then: x+a+b+5+x+e+f+2+5+c+d+2=3k,

that is, x+a+b+c+d+e+f+x+14=3k.

Now x+a+b+c+d+e+f=1+3+4+6+7+8+9=38

and therefore


It follows that 52+x is divisible by 3. Now x is one of 1, 3, 4, 6, 7, 8, 9, and the only one of these possible values for which 52 + x is divisible by 3 is x = 8. This gives 52 + x = 60.

Hence 60=3k, and so k=20.

In the context of the JMC, we can stop here. But to give a full mathematical solution we should satisfy ourselves that it is possible to place the digits in the circles so that each line adds up to 20.

One way this can be done is shown in the diagram opposite.

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.