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Parallelogram 29 • Level 5 • 21 Mar 2024Asteroid Impact

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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. Asteroid sizes

This short video shows the sizes of various asteroids that have been spotted in outer space. In fact, some of them have actually struck Earth, such as 2008 TC3.

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

1 mark

1.1 Asteroid 99942 Apophis is the same shape and density as 2004 FH, but 10 times taller. How much bigger is its mass?

  • 10
  • 100
  • 1,000
  • 10,000
  • 100,000
  • 1,000,000
  • (Not answered)

If we multiply the length by x, then we multiply the volume (and therefore the mass by x3, which is 103, which is 1,000.

2 marks

1.2 The formula for kinetic energy (the energy of a moving object) is 12mv2, where
m = mass and v = velocity. If 99942 Apophis is not only 10 times taller, but also moving 10 times faster than 2004 FH, then how much bigger is its kinetic energy?

  • 10
  • 100
  • 1,000
  • 10,000
  • 100,000
  • 1,000,000
  • (Not answered)

The mass is 1,000 (103) times more massive and then we have to take into account v2 (102), so the energy is 105 times greater, ie 100,000 times greater.

2. Intermediate Maths Challenge Problem (UKMT)

3 marks

2.1 If the following numbers are arranged in increasing order of size, which one is in the middle?

  • 4.04
  • 4.004
  • 4.4
  • 4.44
  • 4.044
  • (Not answered)

In increasing order, the numbers are 4.004, 4.04, 4.044, 4.4, 4.44.

3. Intermediate Maths Challenge Problem (UKMT)

4 marks

3.1 If two of the sides of a right-angled triangle are 5 cm and 6 cm long, how many possibilities are there for the length of the third side?

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

The side of length 5cm cannot be the hypotenuse of the right-angled triangle as it is shorter than the side of length 6cm.

If the 6cm side is the hypotenuse, then the third side of the triangle has length 11 cm.

If the 6cm side is not the hypotenuse, then the hypotenuse has length 61 cm.

These are the only two possibilities.

4. The shortest known maths paper

A “paper” is a published piece of research, and mathematics papers usually run to a few pages, and sometimes to over 100 pages (e.g. the proof of Fermat’s Last Theorem).

This is the shortest-known paper published in a serious mathematics journal.

The mathematician Euler had conjectured (suggested) that you need at least four 4th powers to sum to a 4th power, or at least five 5th powers to sum to a 5th power, and so on. However, this paper gives an example where only four 5th powers sum to a 5th power (but I have replaced one of the numbers with a question mark). So the conjecture is wrong.

3 marks

4.1 What is the missing number?

Clue: the answer is a two-digit number.

x5+845+1105+1335=1445

Show Hint (–1 mark)
–1 mark

The numbers involved are very big, so your calculator might struggle. The best approach is to use your calculator’s memory function.

Correct Solution: 27

5. Intermediate Maths Challenge Problem (UKMT)

5 marks

5.1 Two circles with radii 1 cm and 4 cm touch at a single point.

The point P is on the smaller circle, Q is on the larger circle, and PQ is a tangent to both circles.

What is the length of PQ?

  • 17 cm
  • 3 cm
  • 23 cm
  • 32 cm
  • 4 cm
  • (Not answered)
Show Hint (–1 mark)
–1 mark

This diagram takes you a long way towards the answer. I must admit that, at first, I did not notice that PA is parallel to QB, because both radii must be perpendicular to the tangent PQ. That means PA=QC.

The diagram shows points A and B, which are the centres of the two circles, and C, the point on BQ such that AC is parallel to PQ.

As radii PA and QB are both perpendicular to tangent PQ, APQC is a rectangle. So ACB is a right angled triangle.

The length of AB=1+4=5; the length of BC=4−1=3.

So, by Pythagoras' Theorem, AC=52−32=4, which, therefore, is also the length of PQ.

There will be more next week, 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.