Parallelogram 28 Year 10 14 May 2020McNuggets

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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.
  • 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. McNuggets

Take a look at this video from Numberphile, presented by Brady Haran and James Grime.

(If you have any problems watching the video then just right click and open it in a new window)

1 mark

1.1 McNugget numbers are properly know as F-O-E-I-S numbers. Type the full 9-letter word here.

Correct Solution: Frobenius

2 marks

1.2 True or false? If you cannot make a number of nuggets from a combination of 6, 9 & 20 nuggets, then that number must be prime.

  • True
  • False

For example, you cannot make 4 nuggets from 6, 9 & 20, but 4 is not prime.

2 marks

1.3 True or false? If you can make a number of nuggets from a combination of 6, 9 & 20 nuggets, then that number cannot be prime.

  • True
  • False

For example, you can make 29 nuggets from 6, 9 & 20, because 20 + 9 = 29, and 29 is prime.

2. Intermediate Maths Challenge Problem (UKMT)

3 marks

2.1 Which of the following expressions is equal to 2005?

  • (12 + 1)(102 + 1)
  • (22 + 1)(202 + 1)
  • (32 + 1)(302 + 1)
  • (42 + 1)(402 + 1)
  • (52 + 1)(502 + 1)

The values of the expressions are, respectively, 202, 2005, 9010, 27217 and 65026.

3. Intermediate Maths Challenge Problem (UKMT)

4 marks

3.1 Ten stones, of identical shape and size, are used to make an arch, as shown in the diagram.

Each stone has a cross section in the shape of a trapezium with three equal sides.

What is the size of the smallest angles of the trapezium?

  • 72°
  • 75°
  • 81°
  • 83°
  • 85°
Show Hint (–2 mark)
–2 mark

Let ABCD be the cross-section of one of the stones, as shown.

As AD=BC, ABCD is an isosceles trapezium with ADC=BCD.

If ADC=BCD=θ, then 2θ is the interior angle of a regular 20-sided polygon, namely 18036020°, which equals 162°.

So θ is 81°.

4. Intermediate Maths Challenge Problem (UKMT)

This is a very tough question, but give it your best shot and then try the hint if you are stuck. Even with the hint, it is a tough question.

5 marks

4.1 This regular hexagon has been divided into four trapezia and one hexagon.

If each of the five sections has the same perimeter, what is the ratio of the lengths p, q and r ?

  • 8 : 2 : 1
  • 12 : 4 : 1
  • 9 : 3 : 1
  • 6 : 3 : 1
  • 9 : 4 : 1
Show Hint (–2 mark)
–2 mark

Each exterior angle of a regular hexagon = 360° ÷ 6 = 60°, so when sides HB and IC are extended to meet at A, an equilateral triangle, ABC, is created.

Let the sides of this triangle be of length x. As BC, DE and FG are all parallel, triangles ABC, ADE and AFG are all equilateral. So DE=DA=p+x; FG=FA=q+p+x.

Each exterior angle of a regular hexagon = 360° ÷ 6 = 60°, so when sides HB and IC are extended to meet at A, an equilateral triangle, ABC, is created.

Let the sides of this triangle be of length x. As BC, DE and FG are all parallel, triangles ABC, ADE and AFG are all equilateral. So DE=DA=p+x; FG=FA=q+p+x.

The perimeter of trapezium BCED=x+p+x+2p=2x+3p; the perimeter of trapezium DEGF=p+x+q+p+x+2q=2x+2p+3q; the perimeter of hexagon FGIKJH=2q+p+x+2r=2x+2p+2q+4r.

So 2x+3p=2x+2p+3q; hence p=3q. Also 2x+2p+3q=2x+2p+2q+4r; hence q=4r.

So p:q:r=12r:4r:r=12:4:1.

5. Professor Wiseman’s puzzles

Another puzzle from Professor Wiseman, whose website is worth a visit when you’ve completed this Parallelogram.

3 marks

5.1 There are four suspects and one of them has committed a crime. They make the following statements, but only one of them is telling the truth. Who committed the crime?

  • Jon: James did it.
  • James: Bob did it.
  • Sid: I didn’t do it.
  • Bob: James is lying.
  • Jon
  • James
  • Sid
  • Bob

If anyone other than Bob is telling the truth, the other statements produce a contradiction.

If Sid committed the crime, then he is clearly lying, and so is James (who says Bob did it), and so is Jon (who says James did it). But Bob is telling the truth, because James is indeed lying.

Everything is compatible with the question, which says there is only one criminal and one liar.

6. Intermediate Maths Challenge Problem (UKMT)

3 marks

6.1 In the diagram, what is the sum of the marked angles?

  • 180°
  • 360°
  • 450°
  • 540°
  • 720°
Show Hint (–2 mark)
–2 mark

Let the sizes of the interior angles of the quadrilateral in the centre of the figure be a°, b°, c° and d°. Then a+b+c+d=360°.

Each of these angles is vertically opposite to one of the unmarked angles in the four triangles, as shown. So the sum of the marked angles plus the four angles of the quadrilateral is equal to the sum of the interior angles of four triangles, that is 720°.

Hence the sum of the marked angles is 720° − 360° = 360°.

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.