Week 17Sundogs and Random Dots

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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.
  • Finish by midnight on Sunday if your whole class is doing parallelograms.
  • Your score & answer sheet will appear immediately after you hit SUBMIT.
  • Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

1. Four random dots create something wonderful

Okay. Grab a pencil, paper and ruler.

  • Make 4 random dots (or ask someone else to make them for you).
  • Join up the dots to make a quadrilateral.
  • Put a mark at the midpoint of each line in this first quadrilateral.
  • Join up the four new points to make a second quadrilateral.
2 marks

1.1 What shape is this second quadrilateral? We are looking for a single word answer, and make sure you spell it correctly. (Clue: your answer should not have an H in it. If it does, then you might need to look for a more generalised answer.)

Correct Solution: Parallelogram

The reason for the pretty mathematical image above is that I wanted to create a buffer between the question and the answer... which I am now about to tell you.

If you followed the instructions above, you should have ended up with a parallelogram. I find it shocking that you can start with four random dots and then end up with a perfect parallelogram, as shown here.

You can test this by playing around with the interactive graphic on this page on the Math Open Reference website. You can drag the four orange dots wherever you want, and it automatically generates the first quadrilateral, the midpoints and the second internal quadrilateral, which is always a parallelogram.

2. Sundogs over Sweden

I spotted this on a NASA website called Astronomy Picture of the Day, even though it is a video, not a picture. It shows the Sun as it was observed in Sweden in 2017.

According to the NASA website, the Sun looks weird because the air is full of ice crystals, which behave like millions of tiny lenses:

“Water may freeze in the atmosphere into small, flat, six-sided, ice crystals. As these crystals flutter to the ground, much time is spent with their faces flat and parallel to the ground. An observer may find themselves in the same plane as many of the falling ice crystals near sunrise or sunset. During this alignment, each crystal can act like a miniature lens, refracting sunlight into our view and creating phenomena like “parhelia”, the technical term for sundogs. Visible in the centre is the most direct image of the Sun, while two bright sundogs glow prominently from both the left and the right. Also visible is the bright 22-degree halo -- as well as the rarer and much fainter 46-degree halo – also created by sunlight reflecting off of atmospheric ice crystals.”

There is a different Astronomy Picture of the Day each day, so visit regularly if you are interested in the universe.

3. Junior Maths Challenge Problem (UKMT)

2 marks

3.1 If ▲ + ▲ = ◻ and ◻ + ▲ = ◯ and ◆ = ◯ + ◻ + ▲, how many ▲s are equal to ◆?

  • 2
  • 3
  • 4
  • 5
  • 6

We have that ◻ = ▲ + ▲ (1), and ◯ = ◻ + ▲ (2). Substituting from (1) into (2) gives:

◯ = ▲ + ▲ + ▲.

Hence from ◆ = ◯ + ◻ + ▲, we deduce that:

◆ = (▲ + ▲ + ▲) + (▲ + ▲) + ▲.

So ◆ is equal to six ▲s.

4. The three siblings riddle

Can you solve the following riddle?

1 mark

4.1 John’s mother had three children. The first was named April and the second was named May. What was the name of the third child?

Correct Solution: John

The natural inclination is to continue the sequence – April, May and... June. But read the question carefully, because it starts with “John’s mother had three children...”, so John is one of the children, so John must be the answer.

5. Multiplicative ciphers

Over the last couple of weeks, we discussed how you can encode a message by turning each letter into a number according to the alphabet-number pairings below, adding a fixed number to each letter-number and then turning the new numbers back into letters.

If you want to decode a message, then you just reverse the process, so you subtract the fixed number, rather than adding it. If any of this sounds unfamiliar or confusing, then revisit Parallelogram 2 by way of revision.

A B C D E F G H I J K L M
00 01 02 03 04 05 06 07 08 09 10 11 12
N O P Q R S T U V W X Y Z
13 14 15 16 17 18 19 20 21 22 23 24 25

Perhaps you are already asking the following question: instead of adding a number, can we multiply by a number?

Let’s give it a go and find out what happens.

Let’s encode the word BAT by multiplying by 2. First, turn BAT into numbers: 1 ¦ 0 ¦ 19.

If we multiply each number by 2, then the result is 2 ¦ 0 ¦ 38, but we don’t allow numbers bigger than 25, because our alphabet only stretches from 0 to 25. Therefore, we subtract 26 from 38, and the result is 2 ¦ 0 ¦ 12, which translates to the letters CAM.

BAT is encoded as CAM.

So far, so good. Now let’s encode the letters of the word BAG by multiplying by 2. BAG is first turned into the numbers 1 ¦ 0 ¦ 6.

If we multiply each number by 2, then the result is 2 ¦ 0 ¦ 12, which translates to the letters CAM.

BAG is encoded as CAM.

Hang on! BAT is encoded as CAM, but BAG is also encoded as CAM. How can two different words be encoded into the same word? This is a serious problem. If someone sent you the coded word CAM, then you would not know if they meant BAT or BAG.

The problem arises because there are two ways of encoding something into M:

G = 6 2 × 6 = 12 12 = M
T = 19 2 × 19 = 38 38 – 26 = 12 12 = M

Actually, there are two ways of encoding into every letter. For example,

A = 0 2 × 0 = 0 0 = A
N = 13 2 × 13 = 26 26 – 26 = 0 0 = A

And:

B = 1 2 × 1 = 2 2 = C
O = 14 2 × 14 = 28 28 – 26 = 2 2 = C

And:

C = 2 2 × 2 = 4 4 = E
P = 15 2 × 15 = 30 30 – 26 = 4 4 = E

There is a pattern here. The two letters that are encoded in the same way (A & N), (B & O), (C & P) are all 13 letters apart. Why 13? Well, we are multiplying by 2 and there are 26 letters in the alphabet, and 26/2 = 13.

The problem of different letters being encoded in the same way arises because we are multiplying by 2, and 2 is factor of 26, which is the size of our alphabet.

A similar problem would arise if we were multiplying by 13, because 13 is a factor of 26. This time the problem is much worse, because A, C, E, G,…. are all encoded in the same way and become A.

A = 0 13 × 0 = 0 0 = A
C = 2 13 × 2 = 26 26 – 26 = 0 0 = A
E = 4 13 × 4 = 52 52 – 26 – 26 = 0 0 = A
G = 6 13 × 6 = 78 78 – 26 – 26 – 26 = 0 0 = A

Also, B, D, F, H,... are all encoded in the same way and become N.

Everything in this section is probably new maths to you, or at least applied in a new way, so don’t worry if you are a bit baffled. If you are a bit confused, then just read this section again – slowly – and try to make sense of it. Then have a go at the questions below.

By the way, the notion that encoded numbers must be between 0 and 25 is called modulo 26 arithmetic or MOD (26), and this topic was covered in Parallelogram 15. It might help to review this if you are unsure why, for example, 30 is the same as 4 modulo (26).

A B C D E F G H I J K L M
00 01 02 03 04 05 06 07 08 09 10 11 12
N O P Q R S T U V W X Y Z
13 14 15 16 17 18 19 20 21 22 23 24 25
5 marks

5.1. If we are encoding by multiplying by 5, which letter is each of the following letters transformed into?

The letter A becomes Correct Solution: A
The letter B becomes Correct Solution: F
The letter D becomes Correct Solution: P
The letter G becomes Correct Solution: E
The letter M becomes Correct Solution: I

3 marks

5.2. If we multiply by 5, then the word "OK" is encoded as:

  • KO
  • SY
  • RY
  • SB
  • US

O = 14, and 5 × 14 = 70, which is bigger than 26 so we have to start subtracting multiples of 26 to bring it within the range of 0 to 25. Thus, 70 – 26 – 26 = 18, and 18 is the letter S.

K = 10, and 5 × 10 = 50. To bring it within the range of 0 – 25, we work out 50 – 26 = 24, and 24 is the letter Y.

We found that encoding by multiplying by 2 or 13 was a problem, because these numbers are both factors of 26, the size of the English alphabet. In fact, there is also a problem when you multiply by any even number, because all the even numbers have a common factor with 26, namely 2.

Here is the rule if you want to encode a message by multiplying: “You can choose to multiply by any number, as long as the number has no factors in common with the number of letters in the alphabet that you are using.”

With this in mind, answer the questions below.

2 marks

5.3. The Thai alphabet has 72 letters. If I am encoding a message in Thai by multiplying, which of the following would be the best number to multiply by?

  • 2
  • 3
  • 4
  • 5
  • 6
Show Hint (–1 mark)
–1 mark

Remember the golden rule: you can multiply by any number “as long as the number has no factors in common with the size of the alphabet that you are using.”

The answer is 5, because it is the only number that has no factors in common with 72 (the size of the alphabet).

2 marks

5.4. The Albanian alphabet has 36 letters. If I am encoding a message in Albanian by multiplying, which of the following would be the best number to multiply by?

  • 2
  • 3
  • 4
  • 5
  • 6

The answer is 5, because the Albanian alphabet has 36 letters, and all the other options not only share a common factor with 36, they are factors of 36.

2 marks

5.5. Consider the following 8 alphabets, listed with the number of letters that each contain. If I wanted to encode by multiplying by 13, how many of these alphabets would turn out to be problematic?

(The number of letters in an alphabet can vary, but please assume these numbers below)

Thai - 72 Early Abkhaz - 37
Sinhala - 60 Albanian - 36
Bangla - 52 Russian - 33
Hungarian - 44 English - 26
  • 0
  • 1
  • 2
  • 3
  • 4

The answer is 2, because English has 26 letters, and 13 is a factor of 26 AND Bangla has 52 letters, and 13 is a factor of 52 (4 × 13 = 52).

2 marks

5.6. Which of the eight alphabets above would allow me to multiply by almost any number I want?

  • Thai
  • Abkhaz
  • Sinhala
  • Albanian
  • Bangla
  • Russian
  • Hungarian
  • English
Show Hint (–1 mark)
–1 mark

Alphabets with a prime number of letters are generally good, because prime numbers have no factors apart from 1 and the number itself.

Abkhaz has 37 letters, which is a prime number of letters, so you could multiply by anything and be sure of choosing a number that has no common factors with 37. Of course, if you chose to multiply by 1, then that would not be very clever, as every letter would remain the same. And if you multiplied by 37 there would also be a problem. But otherwise, there are no problems.

6. University Challenge – modular arithmetic

Take a look at this.

How did “Magdalen - Binnie” work it out so quickly?

As he mentions at the end, it’s all down to modular arithmetic, which is something we’ve been using to work out some of the answers to do with codes over the last week or two.

In this case, the question is what day is 100 days after Monday? The days of the week operate according to modulo 7, so we just need to work out 100 in MOD(7).

100 ÷ 7 = 14 remainder 2, so 100 in MOD(7) is 2.

In other words, 100 days is 14 weeks + 2 days.

14 weeks after Monday bring us back to Monday, and 2 more days take us to Wednesday.

Before you hit the SUBMIT button, here are some quick reminders:

  • You will receive your score immediately, and collect your reward points.
  • You might earn a new badge... if not, then maybe next week.
  • Make sure you go through the solution sheet – it is massively important.
  • A score of less than 50% is ok – it means you can learn lots from your mistakes.
  • The next Parallelogram is next week, at 3pm on Thursday.
  • Finally, if you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.

Cheerio, Simon.

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