PG 19 14 Feb 2019Collision Mystery

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteaux 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.
  • 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.
  • As it is half-term, you have a bit of extra time to complete this Parallelogram. You should finish it by Sunday 24. And the next Parallelogram will appear on Thursday 28 February.

1. Collision mystery

This is a video from the brilliant “3 Blue 1 Brown” YouTube channel posing an interesting puzzle with a very unexpected answer. Take a look and answer the questions below:

1 mark

1.1. The video shows how you can calculate the value of a number by bouncing two blocks against each other. What is that number?

  • Pi
  • Square root of pi
  • Square root of 2
  • Square root of -1
  • The golden ratio
2 marks

1.2. If you want to calculate 20 digits of pi, and one block weights 1 Kg, how much must the other block weigh?

  • About 10 times the weight of a bowling ball
  • About 10 times the weight of a human
  • About 10 times the weight of an elephant
  • About 10 times the weight of Earth
  • About 10 times the weight of the Sun
  • About 10 times the weight of a supermassive black hole

2. Intermediate Maths Challenge Problem (UKMT)

2 marks

2.1. How many quadrilaterals are there is this diagram, which is constructed using 6 straight lines?

  • 4
  • 5
  • 7
  • 8
  • 9

In a question of this type we need to try and find a systematic approach so that we count all the quadrilaterals but do not count any of them twice. The method we use (there are others) is to count all the quadrilaterals according to how many of the four smaller quadrilaterals shown in the diagram make them up.

We see that there are 4 quadrilaterals each made up of just one of the small quadrilaterals, 4 made up of two of the smaller quadrilaterals, and 1 made up of all four of the smaller quadrilaterals.

This makes a total of 4 + 4 + 1 = 9 quadrilaterals altogether. These are shown in the diagram below.

3. The Mathematical Art Of M.C. Escher

If you have never seen any of M.C. Escher’s art, then I think you will find this video fascinating. Just take a look and then answer the two questions below.

2 marks

3.1. Which object, invented by Roger Penrose, inspired Escher?

  • Impossible line
  • Impossible triangle
  • Impossible square
  • Impossible quadrilateral
  • Impossible cube
2 marks

3.2. Escher was intrigued by the Möbius strip. What property does it have?

  • It has only one dimension
  • It has only one side
  • It is infinitely long
  • It has three ends
  • It cannot be constructed

4. Intermediate Maths Challenge Problem (UKMT)

3 marks

4.1. I have a bag of coins. In it, one third of the coins are gold, one fifth of them are silver, two sevenths are bronze and the rest are copper.

My bag can hold a maximum of 200 coins. How many coins are in my bag?

  • 101
  • 105
  • 153
  • 195
  • more information is needed

Since one third of the coins are gold, and we cannot have fractions of a coin, the number of coins in the bag must be divisible by 3.

Similarly, this number must also be divisible by 5 and by 7. A number is a multiple of 3, 5 and 7 if, and only if, it is a multiple of 3 × 5 × 7 = 105.

Since my bag holds at most 200 coins, the only possibility is that there are 105 coins in my bag.

3 marks

4.2. I have another bag of coins. In it, one quarter of the coins are gold, one third of them are silver, one fifth of them are bronze and the rest are copper.

What is the smallest number of coins I could have in my bag?

Correct Solution: 60

The bag must have a number of coins that is a multiple of 3, 4 and 5, and the smallest such number is 60.

3 marks

4.3. I have yet another bag of coins. In it, one quarter of the coins are gold, one twelfth of them are silver, three tenths are bronze and the rest are copper.

What is the smallest number of coins I could have in my bag?

Correct Solution: 60

The bag must have a number of coins that is a multiple of 4, 10 and 12, but any number that is a multiple of 12 is also a multiple of 4, so we only need to worry about 10 and 12.

You might think the answer is 120 (ie 10 x 12), but 10 and 12 have a common factor of 2, so we only need to worry about 5 and 12. So the answer is 60 (ie 5 x 12).

5. More Escher

No questions and no marks – just take a look at this video which takes you on a flight through weird world of M.C. Escher.

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

  • You will receive your score immediately, and 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.
  • Finally, if you missed any earlier Parallelograms, don't forget to go back and complete them.
  • You earn reward points and badges from completing missed Parallelogams.
  • As it is half-term, you have a bit of extra time to complete this Parallelogram. You should finish it by Sunday 24. And the next Parallelogram will appear on Thursday 28 February.

Cheerio, Simon.