PG 24 28 Mar 2019Teacher’s Pets

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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.
  • 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. Some riddles

2 marks

1.1 Which word in the dictionary has 11 letters and is spelled incorrectly?

Correct Solution: incorrectly

The word is "incorrectly".

2 marks

1.2 Which of the following statements is correct?

  • The yolk of an egg is white
  • The yolk of an egg are white
  • Neither

Neither, because the yolk of an egg is yellow.

2. Intermediate Maths Challenge Problem (UKMT)

3 marks

2.1 Jim rolled some dice and was surprised that the sum of the scores on the dice was equal to the product of the scores on the dice. One of the dice showed a score of 2, one showed 3 and one showed 5. The rest showed a score of 1. How many dice did Jim roll?

  • 10
  • 13
  • 17
  • 23
  • 30

Suppose there were n dice which showed a score of 1. Then the sum of the scores on the dice was n×1+2+3+5=n+10. The product of the scores was 1n×2×3×5=1×2×3×5=30.

Therefore, n+10=30 and hence n=20.

So, including the three dice which showed the scores of 2, 3 and 5, there were 23 dice altogether.

3. Teacher’s pets

A teacher tells her class that she has some pets at home. She explains that all of the pets are cats, dogs or rabbits, and that all except two of the pets are cats, all except two are dogs, and all except two are rabbits.

1 mark

3.1 How many cats does she have at home?

Correct Solution: 1

1 mark

3.2 How many dogs does she have at home?

Correct Solution: 1

1 mark

3.3 How many rabbits does she have at home?

Correct Solution: 1

If C, D and R represent the number of cats, dogs and rabbits, then we can summarise the information in the questions as:

(a) D + R = 2 -> “all except two of the pets are cats”
(b) C + R = 2 -> “all except two of the pets are dogs”
(c) C + D = 2 -> “all except two of the pets are rabbits”

Substituting (b) into (a) tells us that [D + (2 – C) = 2] or D = C.

And (c) tells us C + D = 2, so C = 1, D = 1... and therefore R = 1.

4. Space science

Here are a couple of videos to stretch your brain in the direction of science. Just watch these short videos and answer the questions that follow.

Is there life on Mars?

1 mark

4.1. Which musician is mentioned in the video?

  • Bruno Mars
  • David Bowie

Dark side of the Moon

2 marks

4.2. Which band recorded the album “The Dark side of the Moon”? The answer is not in the video.

  • Pink Floyd
  • Red Hot Chili Peppers
  • The White Stripes
  • Deep Purple
  • The Moody Blues

5. Intermediate Maths Challenge Problem (UKMT)

6 marks

5.1. Jack’s teacher asked him to draw a triangle of area 7cm2. Two sides are to be of length 6cm and 8cm.

How many possibilities are there for the length of the third side of the triangle?

  • 1
  • 2
  • 3
  • 4
  • more than 4

This problem could be tackled in several different ways.

Method 1: This is based on the fact that if a triangle has sides of lengths a and b and the angle between the sides is θ, then the area of the triangle is given by 12absinθ. If you are not familiar with this, ask your teacher.

Let the angle between the sides of length 8 cm and 6 cm be θ. Then the area of the triangle is 128×6sinθ=24sinθcm2. So we need to have 24sinθ=7, and hence sinθ=724. This equation has two solutions in the range 0<θ<180, say φ and ψ as we see from the graph. Note that φ+ψ=180, and so we can represent the two resulting triangles as shown in the diagram on the right below (for clarity, this is not drawn to scale).

In the diagram above on the right PQ = QR = 8 cm and QS = 6 cm. So PQS and QRS are the two triangles with sides of lengths 6 cm and 8 cm, and area 7 cm2. It is evident that the third sides of these triangles, PS and RS have different lengths. So there are two possibilities for the length of the third side.

Method 2: Here we use the formula 12 (base × height) for the area of a triangle. We consider triangles which have a base, say PQ, of length 8 cm, and where the third vertex, R, is such that PR has length 6 cm. By symmetry we need only consider triangles where R lies above PQ.

The third vertex R lies on a semicircle with centre P and radius 6 cm. The area of the triangle PQR is determined by the vertical height of R above PQ. So the largest area is when R is at the point S vertically above P, and the area is then 128×6=24cm2. The minimum area is when the height is zero. This occurs when R coincides with either the point T or the point U which are the endpoints of the diameter of the semicircle. In these cases the area of the triangle is 0 cm2.

Since the area drops as the height drops, as R moves clockwise around the semicircle from S to U there will be exactly one point, say R1, where the area becomes 7 cm2, and when R moves anticlockwise from S to T there will be exactly one point, say R2, where the area becomes 7 cm2. So there are precisely two values for the length of QR.

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.