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Parallelogram 39 • Level 4 • 30 May 2024Richard Feynman - Superhero

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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.

1. An end of year message from Simon Singh

As this is the penultimate Parallelogram of the year, Simon has a special end of year message for you... and a quiz question for you to answer:

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

2 marks

1.1 Which mathematical constant was Simon asking about?

  • e
  • π
  • Τ
  • i
  • (Not answered)

June 28th is Tau (Τ) day, because Tau is equal to 2π, which is approximately 6.28.

2. Richard Feynman

Richard Feynman is one of my heroes. He was a brilliant physicist (and won a Nobel Prize) and he was a terrific writer and speaker. Here he is talking about science and how it makes the world even more amazing and beautiful.

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

One of the first science books I ever read was “Surely You're Joking, Mr. Feynman!” by Richard Feynman. Maybe borrow it from your local library and read it over the summer.

3 marks

2.1 An English poet wrote about science “unweaving the rainbow” in a poem titled “Lamia”. He meant that science analyses natural phenomena and thereby reduces the beauty of nature. Who was this misguided (but brilliant) poet, who took the opposite view from Feynman?

  • William Wordsworth
  • John Keats
  • Elizabeth Barrett Browning
  • Lord Byron
  • Percy Bysshe Shelley
  • (Not answered)
3 marks

2.2 Take a look at Richard Feynman’s Wikipedia page or any page that tells you about his life. Which project did Feynman work on when he was young man?

  • Sending humans to the moon (the Apollo Project)
  • Designing the first atomic bomb (the Manhattan Project).
  • Understanding the structure of DNA (the Double-Helix Project).
  • Creating genetically modified crops (the Super-Plant Project).
  • (Not answered)

3. Intermediate Maths Challenge Problem (UKMT)

3 marks

3.1 A sector of a disc is removed by making two straight cuts from the circumference to the centre. The perimeter of the sector has the same length as the circumference of the original disc.

What fraction of the area of the disc is removed?

  • π−1π
  • π1
  • π360
  • 13
  • 12
  • (Not answered)
Show Hint (–1 mark)
–1 mark

Suppose that the disc has radius r, so that its circumference is 2πr. The ratio of the area of the circle to that of the sector is the same as the ratio of the circumference of the circle to the length of the arc which is part of the perimeter of the sector. Let this common ratio be 1:k. Then the arc has length 2πrk.

The perimeter of the sector is made up of this arc and two radii of the circle and so its length is 2πrk+2r. Since this is equal to the circumference of the circle, we have

2πrk+2r=2πr.

Suppose that the disc has radius r, so that its circumference is 2πr. The ratio of the area of the circle to that of the sector is the same as the ratio of the circumference of the circle to the length of the arc which is part of the perimeter of the sector. Let this common ratio be 1:k. Then the arc has length 2πrk.

The perimeter of the sector is made up of this arc and two radii of the circle and so its length is 2πrk+2r. Since this is equal to the circumference of the circle, we have

2πrk+2r=2πr.

Therefore

2πrk=2πr−2r

and hence

k=2πr−2r2πr=π−1π.

4. Intermediate Maths Challenge Problem (UKMT)

4 marks

4.1 How many 4-digit integers (from 1000 to 9999) have at least one digit repeated?

  • 62 × 72
  • 52 × 72
  • 52 × 82
  • 42 × 82
  • 42 × 92
  • (Not answered)
Show Hint (–1 mark)
–1 mark

There are 9000 integers from 1000 to 9999. Instead of counting those with at least one digit repeated, it is easier to count the number with all different digits, and then subtract this total from 9000.

Show Hint (–1 mark)
–1 mark

Consider an integer in this range all of whose digits are different. Its first (thousands) digit may be chosen from any of the 9 different non-zero digits. Thereafter we have all 10 digits to choose from other than those that have already been used. So the second (hundreds) digit may be chosen in 9 ways, the third (tens) in 8 ways and the last (units) in 7 ways. So there are 9 × 9 × 8 × 7 four-digit numbers with no digit repeated.

Therefore the number of 4-digit numbers which have at least one digit repeated is 9000 − 9 × 9 × 8 × 7.

Show Hint (–1 mark)
–1 mark

It remains only to show that this can be written as one of the five products given as options.

Since 1000 = 8 × 125, we have 9000 = 9 × 8 × 125 and therefore

9000 − 9 × 9 × 8 × 7 = 9 × 8 × 125 − 9 × 9 × 8 × 7 = (9 × 8) × (125 − 9 × 7) = 72 × (125 − 63)

There are 9000 integers from 1000 to 9999. Instead of counting those with at least one digit repeated, it is easier to count the number with all different digits, and then subtract this total from 9000.

Consider an integer in this range all of whose digits are different. Its first (thousands) digit may be chosen from any of the 9 different non-zero digits. Thereafter we have all 10 digits to choose from other than those that have already been used. So the second (hundreds) digit may be chosen in 9 ways, the third (tens) in 8 ways and the last (units) in 7 ways. So there are 9 × 9 × 8 × 7 four-digit numbers with no digit repeated.

Therefore the number of 4-digit numbers which have at least one digit repeated is 9000 − 9 × 9 × 8 × 7.

It remains only to show that this can be written as one of the five products given as options.

Since 1000 = 8 × 125, we have 9000 = 9 × 8 × 125 and therefore

9000 − 9 × 9 × 8 × 7
= 9 × 8 × 125 − 9 × 9 × 8 × 7
= (9 × 8) × (125 − 9 × 7)
= 72 × (125 − 63)
= 72 × 62
= 62 × 72.

5. Intermediate Maths Challenge Problem (UKMT)

4 marks

5.1 The diagram shows two concentric circles with radii of 1 and 2 units, together with a shaded octagon, all of whose sides are equal.

What is the length of the perimeter of the octagon?

  • 82
  • 83
  • 83π
  • 25+22
  • 85−22
  • (Not answered)
Show Hint (–1 mark)
–1 mark

We need to find the length of one of the sides of the octagon. We consider the side PQ as shown in the diagram on the left. This is a side of the triangle PQO, where O is the centre of both circles. In this triangle OP has length 2, OQ has length 1 and ∠POQ=45°.

Show Hint (–1 mark)
–1 mark

For clarity we have drawn this triangle in a larger scale below.

Show Hint (–1 mark)
–1 mark

We let T be the point where the perpendicular from Q to OP meets OP.

In the right-angled triangle QTO, ∠TOQ=45°. Therefore ∠TQO=45° and so the triangle is isosceles. So OT=TQ. Since OQ=1, we have, using Pythagoras’ Theorem, that OT2+OT2=1. Hence OT2=12 and therefore OT=TQ=12.

Show Hint (–1 mark)
–1 mark

It follows that PT=2−12. Therefore, by Pythagoras’ Theorem applied to triangle PTQ,

PQ2=PT2+TQ2
=2−122+122
=4−22+12+12
=5−22.

It follows that PQ=5−22 and hence the length of the perimeter of the octagon is eight times the length of PQ.

We need to find the length of one of the sides of the octagon. We consider the side PQ as shown in the diagram on the left. This is a side of the triangle PQO, where O is the centre of both circles. In this triangle OP has length 2, OQ has length 1 and ∠POQ=45°. For clarity we have drawn this triangle in a larger scale below.

We let T be the point where the perpendicular from Q to OP meets OP.

In the right-angled triangle QTO, ∠TOQ=45°. Therefore ∠TQO=45° and so the triangle is isosceles. So OT=TQ. Since OQ=1, we have, using Pythagoras’ Theorem, that OT2+OT2=1. Hence OT2=12 and therefore OT=TQ=12.

It follows that PT=2−12. Therefore, by Pythagoras’ Theorem applied to triangle PTQ,

PQ2=PT2+TQ2
=2−122+122
=4−22+12+12
=5−22.

It follows that PQ=5−22 and hence the length of the perimeter of the octagon is 85−22.

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.