Parallelogram 8 Year 11 5 Nov 2020Borromean Rings

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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.

These challenges are a random walk through the mysteries of mathematics, most of which will be nothing to do with what you are doing at the moment in your classroom. Be prepared to encounter all sorts of weird ideas, including a few questions that appear to have nothing to do with mathematics at all.

  • 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. Borromean rings

Borromean rings are a construction from Topology (a branch of mathematics concerned with deforming shapes, for example by stretching, twisting or crumpling).

In a set of Borromean rings, if you remove any one of the rings, all of the others come apart.

Watch this video from the YouTube channel Numberphile.

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

3 marks

1.1 Tadashi's borromean rings consist of 3 rings - one made of pink ribbon, one made of blue ribbon, and what is the third ring made of?

  • Black ribbon
  • Green ribbon
  • Tadashi's fingers
  • The space within the room
  • A steel ring
  • (Not answered)

2. Intermediate Maths Challenge Problem (UKMT)

3 marks

2.1 Which of the following is exactly halfway between 14 and 18?

  • 316
  • 16
  • 524
  • 15
  • 732
  • (Not answered)

Since 14=416 and 18=216, the number exactly halfway between them is 316.

3. Impossible dart scores

This puzzle is all about the game of darts. If you’ve never played darts before, here are the rules:

The board has 20 sectors numbered 1-20. Within each sector, there are smaller sections which double or triple the value of that throw. For example, throwing a dart in the double section of the sector labelled 3 gives you 6 points. In addition, there is the bullseye in the middle worth 50 points, and a section outside the bullseye worth 25 points.

On each turn, a player throws 3 darts, and their score is the sum of the scores for each throw.

180 is the largest possible score, obtained by throwing three triple 20s.
179 is an impossible score.

There are eight other impossible scores.

3 marks

3.1 Which of these is an impossible score?

  • 158
  • 168
  • 178
  • (Not answered)
Show Hint (–1 mark)
–1 mark

158 = 3 × 20 + 50 + 3 × 16, so a triple 20, followed by a bullseye, followed by triple 16 gets you 158. Can you improve this score by 10 or 20?

158 = 3 × 20 + 50 + 3 × 16. So, a triple 20, followed by a bullseye, followed by triple 16 gets you 158.

Swapping the bullseye for another triple 20 give you 168.

168 = 3 × 20 + 3 × 20 + 3 × 16.

Given that the question implies that one of them is impossible, we can assume that 178 is impossible.

3 marks

3.2 Which of these is an impossible score?

  • 163
  • 164
  • 165
  • (Not answered)
Show Hint (–1 mark)
–1 mark

165 = 3 × 20 + 3 × 20 + 3 × 15. So, two triple 20s followed by a triple 15 gets you 165.

Can you reduce this score by 1 or 2?

164 = 3 × 20 + 50 + 3 × 18.

165 = 3 × 20 + 3 × 20 + 3 × 15.

Given that the question implies that one of them is impossible, we can assume that 163 is impossible.

5 marks

3.3 Of the 9 impossible scores, which one is smallest?

Correct Solution: 163

Show Hint (–1 mark)
–1 mark

It’s best to start by considering the largest score, 180 = 3 × 20 + 3 × 20 + 3 × 20. We can’t reduce this by 1 or 2, but we can reduce it by 3 by changing any of the 20s to 19s.

180 = 3 × 20 + 3 × 20 + 3 × 20.
179 impossible
178 impossible
177 = 3 × 20 + 3 × 20 + 3 × 19.

Continuing in this way will give us ‘gaps’ of two. When will we be able to ‘fill’ these gaps, reducing by 1 or 2?

Show Hint (–1 mark)
–1 mark

171 = 3 × 20 + 3 × 20 + 3 × 17. The last throw, 3 × 17, is equal to 51. You can reduce this by 1 by replacing this with a bull’s eye.
168 = 3 × 20 + 3 × 20 + 3 × 16. You can reduce this by 1 by reducing the second throw by 10 using the bull’s eye, and increasing the third throw by 9.

Show Hint (–1 mark)
–1 mark

167 = 3 × 20 + 50 + 3 × 19.
You can’t use the same trick as with 168 to reduce this by 1, as you can’t have 3 × 22, so 166 is impossible.
You can use the same trick to reduce 165 by 1.
165 = 3 × 20 + 3 × 20 + 3 × 15, and
164 = 3 × 20 + 50+ 3 × 18.

Again, you can’t increase the last throw by 9 to get 163, as you can’t have 3 × 21.

Show Hint (–1 mark)
–1 mark

163 is impossible.
162 = 3 × 20 + 3 × 20 + 3 × 14.
161 = 3 × 20 + 50 + 3 × 17.

Can you reduce this by 1, starting by replacing the first throw with a bull’s eye?

It's all about the spacing between triples. There's always a gap of two. But the bullseye becomes useful at 170, by making one of the gaps between the triples disappear. Then another bull comes into play at 160 so below 160 there is always a combination of triple 20s and 50s that fills in the space between adjacent triples.

180 = 3 × 20 + 3 × 20 + 3 × 20
179 impossible
178 impossible
177 = 3 × 20 + 3 × 20 + 3 × 19
176 impossible
175 impossible
174 = 3 × 20 + 3 × 20 + 3 × 18
173 impossible
172 impossible
171 = 3 × 20 + 3 × 20 + 3 × 17
170 = 3 × 20 + 3 × 20 + 50 in creeps the bull
169 impossible
168 = 3 × 20 + 3 × 20 + 3 × 16
167 = 3 × 20 + 50 + 3 × 19
166 impossible
165 = 3 × 20 + 3 × 20 + 3 × 15
164 = 3 × 20 + 50 + 3 × 18
163 impossible
162 = 3 × 20 + 3 × 20 + 3 × 14
161 = 3 × 20 + 50 + 3 × 17
160 = 50 + 50 + 3 × 20 another bull can be used
159 = 3 × 20 + 3 × 20 + 3 × 13
158 = 3 × 20 + 50 + 3 × 16
157 = 50 + 50 + 3 × 19
There are 9 impossible scores:
163 166 169 172 173 175 176 178 179

Some scores can be obtained in more than one way.

For example, 100 = 3 × 20 + 20 + 20, and 100 = 2 × 20 + 2 × 20 + 20.

3 marks

3.4 Which is the largest score which can be obtained in two different ways?

Correct Solution: 174

Start by considering the largest score, 180 = 3 × 20 + 3 × 20 + 3 × 20. We can reduce it by 3 by changing any of the 20s to 19s. This gives us only one way to score 177.

Now we have 177 = 3 × 20 + 3 × 20 + 3 × 19, which we can reduce by 3 in two different ways, by changing a 20 to 19, or a 19 to 18. This gives us two ways to score 174.

174 = 3 × 20 + 3 × 20 + 3 × 18 = 3 × 20 + 3 × 19 + 3 × 19.

4. Intermediate Maths Challenge Problem (UKMT)

4 marks

4.1 Equal regular pentagons are placed together to form a ring in the manner shown. The diagram shows the first three pentagons. How many more are needed to complete the ring?

  • 6
  • 7
  • 8
  • 9
  • 10
  • (Not answered)
Show Hint (–1 mark)
–1 mark

The angle sum of a pentagon is 540°.

Show Hint (–1 mark)
–1 mark

This means the interior angle of a regular pentagon is 540 ÷ 5 = 108°. Therefore, each interior angle of the regular polygon formed by the inner sides of the pentagon is 360 – 2 × 108 = 144°.

Show Hint (–1 mark)
–1 mark

The exterior angle of the polygon is 180° - 144° = 36°, and the sum of exterior angles in any polygon is 360°.

The interior angle of a regular pentagon is 108°.

Therefore each interior angle of the regular polygon formed by the inner sides of the pentagons is (360 - 2 × 108)° = 144°.

The exterior angle of this regular polygon is 36° and hence it has 360 ÷ 36, ie 10 sides.

Therefore 7 more pentagons are required.

5. Intermediate Maths Challenge Problem (UKMT)

5 marks

5.1 AB is a diameter of a circle of radius 1 cm.

Two circular arcs of equal radius are drawn with centres A and B.

These arcs meet on the circle, as shown.

What is the shaded area?

  • π2 cm2
  • 1 cm2
  • π1 cm2
  • 2 cm2
  • 2π3 cm2
  • (Not answered)
Show Hint (–1 mark)
–1 mark

Let O be the centre of the circle and let the points where the arcs meet be C and D respectively.

ADBC is a square since its sides are all equal to the radius of the arc CD and ACB=90° (angle in a semicircle).

In triangle OCB, CB2=OC2+OB2, hence CB=2 cm.

Show Hint (–1 mark)
–1 mark

The area of the segment bounded by arc CD and diameter CD is equal to the area of sector BCD - the area of triangle BCD, ie:

14π2212×2×2 cm2

Show Hint (–2 mark)
–2 mark

The area of the segment bounded by arc CD and diameter CD is therefore 12π1 cm2.

The unshaded area in the original figure is, therefore, π2 cm2.

Let O be the centre of the circle and let the points where the arcs meet be C and D respectively.

ADBC is a square since its sides are all equal to the radius of the arc CD and ACB=90° (angle in a semicircle).

In triangle OCB, CB2=OC2+OB2, hence CB=2 cm.

The area of the segment bounded by arc CD and diameter CD is equal to the area of sector BCD - the area of triangle BCD, ie:

14π2212×2×2 cm2, ie 12π1 cm2.

The unshaded area in the original figure is, therefore, π2 cm2.

Now the area of the circle is π cm2.

Hence the shaded area is 2 cm2.

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.