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Parallelogram 20 • Level 5 • 18 Jan 2024Hypatia

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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. Assumptions

Take a look at this mind-blowing 43 second video from the psychologist Professor Richard Wiseman. If you like it, then take a look at the Prof’s YouTube channel afterwards.

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

2. The letter A in numbers

2 marks

2.1 What is the first number that is less than 200 and which contains the letter A when it is spelt out? Write the answer as a number.

Correct Solution: 101

2 marks

2.2 What is the first number that contains the letter A when it is spelt out? In this case the word AND does NOT count. Write the answer as a number.

Correct Solution: 1,000

3. Intermediate Maths Challenge Problem (UKMT)

3 marks

3.1 Four different positive integers are to be chosen so that they have a mean of 2017.

What is the smallest possible range of the chosen integers?

  • 2
  • 3
  • 4
  • 5
  • 6
  • (Not answered)

We recall that the range of a set of numbers is the difference between the largest and smallest numbers in the set.

The smallest range of a set of four different positive integers is 3. This occurs when the integers are consecutive, and only in this case. We first show that we cannot have four consecutive integers with a mean of 2017.

Assume, to the contrary, that n is a positive integer such that the consecutive integers n, n+1, n+2 and n+3 have 2017 as their mean.

Then 14n+n+1+n+2+n+3=2017,

and hence n+n+1+n+2+n+3=4×2017.

It follows that 4n+6=4×2017,

and hence 4n=4×2017−6.

Therefore n=2017−32, contradicting our assumption that n is an integer. This contradiction shows that there do not exist four consecutive integers whose mean is 2017. So the smallest possible range is not 3.

On the other hand, we see that the four integers 2015, 2016, 2018 and 2019 have mean 2017, because

2015+2016+2018+2019=2015+2019+2016+2018
=2×2017+2×2017
=4×2017.

The range of these numbers is 2019 − 2015, that is, 4. We deduce that the smallest possible range is 4.

4. Intermediate Maths Challenge Problem (UKMT)

4 marks

4.1 For what value of x is 64x equal to 5125?

  • 6
  • 7.5
  • 8
  • 16
  • 40
  • (Not answered)
Show Hint (–1 mark)
–1 mark

Convert everything into powers of 2.

Show Hint (–1 mark)
–1 mark

64x=26x=26x

We note first that 64=26 and 512=29. Therefore 64x=26x=26x and 5125=295=29×5=245.

It follows that the equation 64x=5125 is equivalent to the equation 26x=245. We deduce from this that 6x=45. Hence

x=456=152=7.5.

5. Intermediate Maths Challenge Problem (UKMT)

5 marks

5.1 The diagram shows an arc PQ of a circle with centre O and radius 8.

Angle QOP is a right angle, the point M is the midpoint of OP and N lies on the arc PQ so that MN is perpendicular to OP.

Which of the following is closest to the length of the perimeter of triangle PNM?

  • 17
  • 18
  • 19
  • 20
  • 21
  • (Not answered)
Show Hint (–2 mark)
–2 mark

Because NM is perpendicular to OP, the triangle OMN has a right angle at M. Therefore, by Pythagoras’ Theorem,

OM2+NM2=ON2.

Since M is the midpoint of OP and the circle has radius 8, it follows that OM=4. Also ON=8.

Therefore, by the above equation NM2=82−42=64−16=48.

It follows that NM=48.

Because NM is perpendicular to OP, the triangle OMN has a right angle at M. Therefore, by Pythagoras’ Theorem,

OM2+NM2=ON2.

Since M is the midpoint of OP and the circle has radius 8, it follows that OM=4. Also ON=8. Therefore, by the above equation NM2=82−42=64−16=48. It follows that NM=48.

In the right-angled triangle NMP, we have NM2=48 and MP2=42=16. Therefore, by Pythagoras’ Theorem applied to this triangle, NP2=MN2+MP2=48+16=64, and therefore NP=8.

It follows that the perimeter of the triangle PNM is MP+PN+NM=4+8+48=12+48.

Because 48 is close to 7, the perimeter is close to 12+7=19.

6. Hypatia

If you travelled back in time 1,600 years or so, then one of the greatest mathematicians in the world was Hypatia. She lived a courageous life and who ultimately faced a terrible death, as revealed in this short video.

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

2 marks

6.1 One of Hypatia’s interests was “conic sections”, which is all about the different curves that can be created by slicing through a cone. Imagine taking an axe chopping through a cone.

Which of these is NOT a conic section? You might want to google for clues.

  • Circle
  • Spiral
  • Ellipse
  • Parabola
  • Hyperbola
  • (Not answered)

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.