Parallelogram 18 Year 7 6 Feb 2020It’s all downhill from here

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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.

Last week there was a scoring issue, which meant users were not awarded all of the points for their correct answers. To make up for it, this week's Parallelogram will be worth double points!.

1. Winter Olympics Puzzle (A)

Every four years, the world’s leading winter athletes – the skiers, skaters, sledders and snowboarders – gather to compete for medals in the Winter Olympic Games.

Of course, among the drama, bravery and triumph, there is plenty of room for maths! So, while you’re cheering on your favourites, don’t miss the chance to tackle these Winter Olympics maths puzzles.

The Jamaican bobsled team was one of the most famous feel-good stories of the 1988 Games in Calgary. Jamaica’s Caribbean climate is not the most helpful for bobsledding, but the team borrowed equipment and asked for help. While the four-man bob team was unplaced after crashing in qualifying, the two-man team finished a creditable 30th out of 41. The movie Cool Runnings is – very loosely – based on their experience.

In the bobsled (or bobsleigh), teams of two or four pack into a streamlined sled and slide down a twisting, icy course with steep, banked corners as fast as they possibly can – often as fast as 75mph. It gets its name from early competitors who realised they could go faster by bobbing back and forth inside the sled!

The 2010 Winter Olympics at Whistler, Canada, took place on a track 1,450 metres long. Each team had four runs at the course, and the winners were the team with the smallest total time.

Feel free to use a calculator for these puzzles!

1 mark

1.1 Bronze medal puzzle: The run times for the gold medal-winning team (USA) were:

  • 50.89 seconds
  • 50.86 seconds
  • 51.19 seconds
  • 51.52 seconds

What is the total time for the four runs in seconds? (please answer to two decimal places)

Correct Solution: 204.46

1 mark

1.2 Silver medal puzzle: After their final heat, Canada were in second place with total time of 3 minutes, 24.85 seconds – but Germany, in third place, still had one run to go.

In the first three heats, Germany had a total time of 2 minutes, 33.48 seconds.

How quickly, in seconds, did the German team have to complete their final run to win at least a share of the silver medal?


Correct Solution: 51.37

3:24.85 is 204.85 seconds; 2:33:48 is 153.48 seconds. The difference, and target time, is 51.37 seconds. In fact, Germany’s final heat time was 51:36!

1 mark

1.3 Gold medal puzzle: The USA’s second heat time of 50.86 seconds was a track record.

What was their average speed in metres per second? Give your answer to the nearest whole number, and remember that the track was 1,450 metres long.

Correct Solution: 29 m/s

Solutions 1,450/50.86 = 28.5096, which is closest to 29 m/s.

2. Winter Olympics Puzzle (B)

The most famous ski jumper in history is not exactly famous for being good. Eddie “The Eagle” Edwards, self-funded and short-sighted, famously finished in last place in both the 70m and 90m events at Calgary, Canada in 1988, scoring fewer than half as many points as his nearest competitor. However, these scores are still the British records for ski-jumping! In 2016, a film of his exploits, Eddie The Eagle, was released.

In the ski jump, competitors ski down a huge ramp, reaching speeds of up to 60mph before jumping as far down the hill as they possibly can.

Each jump is awarded points based on two factors: the distance jumped, and the jumper’s style.

Here is the recipe (we've simplified it a bit for this puzzle!):

  • Take 60 metres away from the distance jumped (this could be a negative number!)
  • Double the result
  • Add on the style score.

A group of judges awards up to 60 points for style – how well the jumper flies and lands.

For example, to work out the score for a ski jumper who jumped 95 metres with enough grace to pick up 40 style points, you would calculate:

95 – 60 = 35
35 × 2 = 70
70 + 40 = 110 the jump would be worth 110 points.

Similarly, a jumper who manages 80 metres with 25 style points would score 65 points altogether.

1 mark

2.1 Bronze medal puzzle: Albert the Albatross (of Albania) jumps exactly 90 metres and scores a perfect 60 for style. How many points does he earn?

Correct Solution: 120

60 for distance, 60 for style = 120 altogether

1 mark

2.2 Silver medal puzzle: Kelly the Kestrel (of Kazakhstan) jumps 95 metres and scores a total of 120 points. How many style points was she awarded?

Correct Solution: 50

She has 70 for distance, so she must have another 50 for style.

1 mark

2.3 Gold medal puzzle: Lemmy the Lemming (of Latvia) scores a total of zero, including no style points at all. How far did he jump in metres?

Correct Solution: 60

To get 0 points for distance, Lemmy must have jumped exactly 60 metres.

This puzzle and the previous one come from the Komodo website, which offers a programme of maths, largely aimed at younger children. I’ve used it with my son, and I think it offers a great way to practice core arithmetic skills.

The puzzles were created by Colin Beveridge - a maths tutor, former NASA researcher, and writer of maths books including Basic Maths for Dummies.

3. The physics of ski jumping

To understand the underlying physics in the previous question, take a look at this video that explains how to be a ski jump champion... and then answer this question.

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

1 mark

3.1 In the 1960s, what formation did ski jumpers make with their skies?

  • V
  • X
  • Parallel
  • Perpendicular
  • Sierpinski Triangle

4. Junior Maths Challenge Problem (UKMT)

The diagram shows a cuboid in which the area of the shaded face is one-quarter of the area of each of the two visible unshaded faces. The total surface area of the cuboid is 72 cm².

3 marks

4.1 What, in cm², is the area of one of the visible unshaded faces of the cuboid?

  • 16
  • 28.8
  • 32
  • 36
  • 48

Let the area of the shaded face be ⲭcm². Then each of the visible unshaded faces has area 4ⲭcm². Hence the total surface area of the cuboid in cm² is ⲭ + ⲭ + 4ⲭ + 4ⲭ + 4ⲭ + 4ⲭ = 18ⲭ (because there are two ends, each of area ⲭ, and four sides, each of area 4ⲭ).

So 18ⲭ = 72. Therefore ⲭ = 4, and the area of one of the visible unshaded faces, in cm² is 4ⲭ = 16.

5. The affine cipher

Over the last couple of weeks, we discussed how you can encode a message by turning each letter into a number according to the alphabet-number pairings below, and either (1) adding a fixed value to each letter-number or (2) multiplying each letter-number by a fixed value. The result is then turned back into a letter.

00 01 02 03 04 05 06 07 08 09 10 11 12
13 14 15 16 17 18 19 20 21 22 23 24 25

The affine cipher uses both multiplication and addition to encipher a letter, word or whole message. For example, let’s multiply each letter in the message (LM) by 5 and then add 2 to get each letter in the code (LC). We could say:

LC = (5 × LM) + 2

Let’s apply this affine cipher to the word CAB. If we turn this into numbers, we obtain 2 ¦ 0 ¦ 1.

If we multiply by 5 and add 2, we obtain 12 ¦ 2 ¦ 7.

If we translate back into letters, the code is MCH.

Now, let’s encipher the same word, but this time we will change the affine cipher so that we multiply by 9 and add 11:

LC = (9 × LM) + 11

Again we will start with the word CAB. If we turn this into numbers, we obtain 2 ¦ 0 ¦ 1.

If we multiply by 9 and add 11, we obtain 29 ¦ 11 ¦ 20.

Remember, we cannot have numbers bigger than 25, so we subtract 26 from any problematic numbers, so that 29 becomes 3. So, the new letter sequence is 3 ¦ 11 ¦ 20.

If we translate back into letters, the code is DLU.

If you are confused (perhaps you missed some of the earlier Parallelograms), then take a look at this video from Chris at The explanation is fairly slow and lengthy, but it should clear up any confusion.

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

Now that you have mastered the affine cipher, answer these questions.

00 01 02 03 04 05 06 07 08 09 10 11 12
13 14 15 16 17 18 19 20 21 22 23 24 25
1 mark

5.1. If we are encoding by multiplying by [3] and adding [4], what letter do we get if we encipher the letter A?

Correct Solution: E

1 mark

5.2. If we are encoding by multiplying by [3] and adding [4], what letter do we get if we encipher the letter B?

Correct Solution: H

1 mark

5.3. If we are encoding by multiplying by [3] and adding [4], what letter do we get if we encipher the letter Z?

Correct Solution: B

3 marks

5.4. If we are encoding by multiplying by [11] and adding [6], what do we get if we encipher the word CAT?

Correct Solution: CGH

3 marks

5.5. You have intercepted the enciphered message FCLUO. You know that the word has been enciphered with affine cipher, and in particular each letter was multiplied by [3] and then [2] was added. Can you work out what the coded word FCLUO means?

Correct Solution: BADGE

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.