Parallelogram 15 Year 10 16 Jan 2020Optical Illusions

This is a preview of Parallel. You have to login or create an account, to be able to answer questions and submit answers.

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. Mind Blowing Optical Illusions

Take a look at these 10 amazing optical illusions.

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

2 marks

1.1 The circle illusion is the result of 8 dots moving...

  • in circles
  • in spirals
  • in straight lines
  • in ellipses
  • randomly

My favourite among these illusions is the dragon whose eyes follow you around the room.

You can download a printout via this link and the following video shows how to make your own dragon.

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

2. Probability puzzle

3 marks

2.1 It’s dark. Your sock drawer contains 10 white socks and 10 black socks. You take out one sock, but you don’t know what colour it is. You take out a second sock – what is the probability that the second sock is white?

Write down your probability as a decimal, accurate to one decimal place.

Correct Solution: 0.5

Show Hint (–1 mark)
–1 mark

The first sock is equally likely to be black or white.

Because the first sock could have been black or white, with equal probability, then it has no real impact on working out the probability of the second.

If the first sock was black, then the chance of picking a white sock would be 10/19. If the first sock was white, then the chance of picking a white sock would be 9/19. But we don’t know whether the first sock is black or white, so we take the average, which is 9.5/19 = ½.

3. Intermediate Maths Challenge Problem (UKMT)

3 marks

3.1 The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6.

What is the difference between the largest angle and the smallest angle?

  • 30°
  • 40°
  • 50°
  • 60°
  • 70°

Because the angles are in the ratio 3 : 4 : 5 : 6, and 3 + 4 + 5 + 6 = 18, the largest angle makes up 618, that is, 13 of the sum of the angles of the quadrilateral. Similarly, the smallest angle makes up 318, that is, 16 of the sum of the angles of the quadrilateral.

It follows that the difference between the largest and the smallest angle is 1316, that is, 16 of the sum of the angles of the quadrilateral.

The sum of the angles of a quadrilateral is 360°. It follows that the difference between the largest angle and the smallest angle of the quadrilateral is

16×360°=60°.

4. Intermediate Maths Challenge Problem (UKMT)

4 marks

4.1 The diagram shows a quadrilateral PQRS made from two similar right-angled triangles, PQR and PRS.

The length PQ is 3, the length QR is 4 and PRQ=PSR.

What is the perimeter of PQRS?

  • 22
  • 22 5/6
  • 27
  • 32
  • 45 1/3
Show Hint (–2 mark)
–2 mark

It’s easy to work out PR. Then you can try two approaches.

(1) Work out the perimeter of the small triangle. Then you know the perimeter of the large triangle is 5/3 bigger, as QP corresponds to PR in the two similar triangles. So you can work out the perimeter of the small triangle plus the big triangle, but you have to subtract the lengths of the two common sides, which are not part of the perimeter of the quadrilateral.

(2) The longer way is to work out each of the lengths one by one, using the ratio 5/3 and Pythagoras' theorem.

Applying Pythagoras’ Theorem to the right-angled triangle PQR gives RP2=PQ2+QR2=32+42=9+16=25. Therefore RP has length 5.

The corresponding sides PQ and PR of the similar triangles PQR and PRS have lengths 3 and 5.

It follows that the ratio of the perimeter of PQR to that of PRS is 3 : 5.

The perimeter of the triangle PQR is PQ+QR+RP=3+4+5=12. Let the perimeter of PRS be p. Then 12 : p = 3 : 5. Hence

12p=35.

Hence

p=53×12=603=20.

The perimeter of PQRS is PQ+QR+RS+SP. This is equal to PQ+QR+RPRP+RS+SP+RPRP. Thus the perimeter equals the sum of the perimeters of PQR and PRS less twice the length of RP. We deduce that the perimeter of PQRS is 12+202×5=3210=22.

5. Intermediate Maths Challenge Problem (UKMT)

5 marks

5.1 Two brothers and three sisters form a single line for a photograph. The two boys refuse to stand next to each other.

How many different line-ups are possible?

  • 24
  • 36
  • 60
  • 72
  • 120
Show Hint (–2 mark)
–2 mark

There are only 6 ways to arrange the brothers and sisters that keep the brothers apart. Now the challenge is to work out how many ways you can position the brothers and sisters within each arrangement.

B, S, B, S, S
B, S, S, B, S
B, S, S, S, B
S, B, S, B, S
S, B, S, S, B
S, S, B, S, B

We first count the number of ways in which two brothers and three sisters can stand in line without two brothers being next to each other, without regard to which particular brother and which particular sister is in any given position.

Counting from the left, the first brother can be in the first, second or third position. (He cannot be in fourth position as then the second brother from the left would be next to him in fifth position.)

If the first brother is in first position, the second brother can be in third, fourth or fifth position. If the first brother is in second position, the second brother can be in fourth or fifth position. If the first brother is in third position, the second brother can only be in fifth position.

It follows that there are six ways to arrange where the brothers and the sisters are positioned.

These are listed below using B to represent a brother and S to represent a sister.

B, S, B, S, S
B, S, S, B, S
B, S, S, S, B
S, B, S, B, S
S, B, S, S, B
S, S, B, S, B

For any two given positions which the two brothers must occupy, they can stand in one of two ways. If the brothers are B1 and B2, these are B1, B2 and B2, B1.

For any three given positions which the three sisters must occupy, they can stand in one of six ways. If the sisters are S1, S2 and S3, these are S1, S2, S3; S1, S3, S2; S2, S1, S3; S2, S3, S1; S3, S1, S2 and S3, S2, S1.

Since the 6 choices of which positions the brothers occupy, the 2 choices of how the brothers are put in these positions, and the 6 choices of how the sisters are put in the remaining positions may be made independently, the total number of different line-ups is given by 6 × 2 × 6 = 72.

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.