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Parallelogram 12 • Level 5 • 21 Nov 2024Waterfalls

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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.
  • When you finish, remember to hit the SUBMIT button.
  • Finish by Sunday night if your whole class is doing parallelograms.

IMPORTANT – it does not really matter what score you get, because the main thing is that you think hard about the problems... and then examine the solution sheet to learn from your mistakes.

1. Waterfalls

2 marks

1.1. What is the opposite of a waterfall?

  • Ice fountain
  • Steam spout
  • Rock slide
  • Firefly
  • Spider
  • (Not answered)

The opposite of water is fire, and the opposite of to fall is to fly.

2. Peaceable Queens

Watch this video, which reveals a mathematics problem based on chess. Don’t worry – you don’t need to play chess to understand this problem. All you need to know is that the Queen can move forwards, backwards, left, right and diagonally one, two or several squares. Watch carefully and answer the questions below.

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

2 marks

2.1 What is the maximum number of black and white peaceable queens that a 3x3 board can support?

  • 0
  • 1 of each
  • 2 of each
  • 3 of each
  • 4 of each
  • (Not answered)
2 marks

2.2 What is the maximum number of black and white peaceable queens that an 8x8 board can support?

  • 5 of each
  • 6 of each
  • 7 of each
  • 8 of each
  • 9 of each
  • (Not answered)
4 marks

2.3 For a 20x20 board, nobody knows for certain the maximum number of black and white peaceable queens that can be supported. Which of these numbers is a possible answer?

  • 20
  • 21
  • 40
  • 41
  • 58
  • 59
  • (Not answered)

In the video, Neil Sloane explained that the number of peaceable queens was at least 7n248, where n is the number of squares along one side of the board.

7×20248=58.3.

We know this is a minimum, so the answer must be 59.

3. Intermediate Maths Challenge Problem (UKMT)

3 marks

3.1 How many squares have 7 as their units digit?

  • 0
  • 1
  • 2
  • 3
  • 4
  • (Not answered)

The units digit of the square of an integer n is the same as that of the square of the units digit of n.

For example, the units digit of 2372 is the same as that of 72, namely 9.

Therefore, to find the possible units digits of squares, we need only consider the squares of the one-digit numbers. We have 02=0, 12=1, 22=4, 32=9, 42=16, 52=25, 62=36, 72=49, 82=64 and 92=81. This shows that the units digit of a square can only be 0, 1, 4, 5, 6 or 9.

In particular, there are no squares which have 7 as their units digit.

4. Intermediate Maths Challenge Problem (UKMT)

4 marks

4.1 Two semicircles are drawn in a rectangle as shown.

What is the width of the overlap of the two semicircles?

  • 3 cm
  • 4 cm
  • 5 cm
  • 6 cm
  • 7 cm
  • (Not answered)
Show Hint (–2 mark)
2 mark

Let P, Q be the midpoints of the longer sides of the rectangle. Let R, S be the points where the semicircles meet. Let T be the point where PQ meets RS.

Each of the semicircles has radius 5 cm. Therefore PR, PS, QR and QS all have length 5 cm. Therefore PSQR is a rhombus. Hence the diagonals PQ and RS bisect each other at right angles. It follows that PT and QT each have length 4 cm. Let the common length of RT and ST be x cm.

We now apply Pythagoras’ Theorem to the right-angled triangle PTR. This gives 42+x2=52, and hence x2=52−42=25−16=9. Therefore x=3. It follows that both RT and ST have length 3 cm. Hence the length of RS is 6 cm. Therefore the width of the overlap of the two semicircles is 6 cm.

5. Intermediate Maths Challenge Problem (UKMT)

5 marks

5.1 A Saxon silver penny, from the reign of Ethelbert II in the eighth century, was sold in 2014 for £78 000.

A design on the coin depicts a circle surrounded by four equal arcs, each a quarter of a circle, as shown.

The width of the design is 2 cm.

What is the radius of the small circle, in centimetres?

  • 12
  • 2−2
  • 122
  • 5−32
  • 22−2
  • (Not answered)
Show Hint (–1 mark)
1 mark
Show Hint (–2 mark)
2 mark

We see from the diagram in the previous hint that the quarter circles touch each other tangentially. Let the points where they touch be K, L, M and N. Let the centres of the quarter circles be P, Q, R and S, arranged as shown in the diagram.

Because the circles touch, PKQ, QLR, RMS and SNP are straight lines.

Since S is the centre of the quarter circle that goes through M and N, we have SM=SN and ∠MSN=90°. Therefore MSN is a right-angled isosceles triangle. The length of the hypotenuse NM of this triangle is the width of the design, that is, 2 cm.

Therefore, by Pythagoras’ Theorem applied to the triangle SMN, we see that SM and SN each have length 2 cm.

We see from the diagram that the quarter circles touch each other tangentially. Let the points where they touch be K, L, M and N. Let the centres of the quarter circles be P, Q, R and S, arranged as shown in the diagram.

Because the circles touch, PKQ, QLR, RMS and SNP are straight lines.

Since S is the centre of the quarter circle that goes through M and N, we have SM=SN and ∠MSN=90°. Therefore MSN is a right-angled isosceles triangle. The length of the hypotenuse NM of this triangle is the width of the design, that is, 2 cm.

Therefore, by Pythagoras’ Theorem applied to the triangle SMN, we see that SM and SN each have length 2 cm.

Similarly MR and NP both have length 2 cm. Therefore SPR is a right-angled isosceles triangle in which both SR and SP have length 22 cm. Therefore, by Pythagoras’ Theorem applied to the triangle SPR, the hypotenuse PR of this triangle has length 4 cm.

[Alternatively, we could argue that SNM and SPR are similar triangles in which SP is twice the length of SN. Therefore PR is twice the length of NM.]

The line segment PR is made up of two radii of the quarter circles with centres P and R, which have total length 22 cm, and the diameter of the small circle. It follows that the diameter of the small circle has length 4 cm −22 cm. The radius of the small circle is half of this, that is, 2−2 cm.

There will be more next week, and the week after, and the week after that. So check your email or return to the website on Thursday at 3pm.

In the meantime, you can find out your score, the answers and go through the answer sheet as soon as you hit the SUBMIT button below.

When you see your % score, this will also be your reward score. As you collect more and more points, you will collect more and more badges. Find out more by visiting the Rewards Page after you hit the SUBMIT button.

It is really important that you go through the solution sheet. Seriously important. What you got right is much less important than what you got wrong, because where you went wrong provides you with an opportunity to learn something new.

Cheerio, Simon.