Spot the PatternOn the next slide is a grid and on each subsequent slide there are 4 pieces of information.

Can you work out how the grid should be coloured in?

Spot the PatternThere are 4 red squares (arranged in a square) in the middle of the design.

There are 7 red squares in the bottom right hand quarter of the design.

There is one square of each colour in the top row of the design (the rest are blank).

No blue square is directly next to a yellow square.

Spot the PatternThere are 9 blank squares (arranged in a square) in the bottom right hand corner

of the design.

There is one square of each colour in the first column of the design (the rest are blank).

The top left corner to the bottom right corner is a line of reflection symmetry.

The blue square in the top row has two blank squares between it and the yellow square in the top right hand corner.

Spot the PatternThe yellow squares are only on the top right to bottom left diagonal.

There are 5 more red squares than blue squares.

The design has one line of reflection symmetry.

There are 6 blue squares in the top left hand quarter of the design

Spot the PatternThere are 6 blue squares in the top left hand quarter of the design

There are 13 red squares in the design.

Some of the squares are not coloured in.

The design does not have rotation symmetry.

Spot the PatternThe top left corner to the bottom right corner diagonal has 5 red squares on it (the

rest are blank)

The design uses 3 different colours.

The top right corner to bottom left corner diagonal has red and yellow squares only

in the ratio 1:3.

The ratio of yellow squares to blue squares is 3:4.

Algebra with cards and paper

Three rectangular business cards are shown.What is the perimeter of each?

7cm

5 cm

7cm

a

b

a

Algebra with cards and paper

5 cm+7cm+5 cm+7cm = 24 cm7cm

5 cm

7cm

a

b

a

a cm+7cm+a cm+7cm = 14+2a cm

a cm+ b cm+ a cm+ b cm=2a + 2b cm

Algebra with cards and paperUsing this business card, three arrangements of 2 cards are shown below. What is the perimeter of each?

Algebra with cards and paper

Which is smaller: 2a +4b or 4a+2b?

2a+4b

4a+2b2a+4b

Algebra with cards and paperPutting 2 cards together ‘edge to edge’, what other perimeters can you find?

How might you write an expression for the perimeter of these arrangements?

Algebra with cards and paperCan you describe how to arrange the two cards to obtain:• the maximum perimeter?• the minimum perimeter?Can you explain how you know these are the maximum and minimum values?

How many different arrangements can you find for 3 cards?

Algebra with cards and paper

Explore the maximum and minimum perimeters for:• 3 cards• 4 cards• 5 cards• …• n cardsCan you come up with general algebraic expressions for the maximum and minimum perimeters for n cards?

Algebra with cards and paperDoes using a different sized rectangular card affect the arrangements that give the minimum perimeter?

Does it affect the algebraic value of the minimum perimeter?

Perimeter of Rectangular Rings

Putting 4 cards together, it is possible to make a ring as shown.

Write down an expression for:• the perimeter of the outer rectangle of the ring;• the perimeter of the inner rectangle of the ring;• the total perimeter.

Can you simplify your expressions?

Perimeter of Rectangular Rings

outer rectangle of the ring: 4a + 4b

inner rectangle of the ring: 4b – 4a

the total perimeter: 8b

Perimeter of Rectangular Rings

Putting 6 cards together there are 2 possible rings. Find them and write expressions for:• the perimeter of the outer rectangle of the

ring;• the perimeter of the inner rectangle of the

ring;• the total perimeter.

Simplify the expressions where possible

Perimeter of Rectangular Rings

outer rectangle of the ring: 4a + 6b

inner rectangle of the ring: 6b – 4a

the total perimeter: 12b

Perimeter of Rectangular Rings

outer rectangle of the ring: 4a + 6b

inner rectangle of the ring: 6b – 4a

the total perimeter: 12b

Perimeter of Rectangular Rings

Explore for different numbers of cards.

What do you notice each time?

Can you explain why?

Monty Hall Problem Monty Hall was a U.S. game show host in the 1970s.

His show provides us with a probability

problem.Contestants on the show would either win a car…

…or a goat.

Monty Hall ProblemMonty presents the contestant with a choice of 3 doors. Behind one of them is a car, behind the other two are goats.

Green!

Monty Hall ProblemHaving chosen a door, Monty shows her what is behind one of the other doors – he knows where the car is and always shows her a goat.

Monty Hall ProblemHe now asks her whether she wants to stick with the green door or switch to the pink one.

Monty Hall ProblemShould she stick with the door she chose first or switch? What’s your initial instinct?

Try it out several times with a partner to see what happens. Do you win more times if you stick or switch?

Monty Hall ProblemLet’s look at the problem mathematically.Supposing the car is behind the green door.

Fill in the table on the following slide and decide whether it’s generally better to stick or switch.

Monty Hall Problem

Door chosen by you:

Behind it is a..

Door Monty would then show you:

Stick, and you win a…

Switch and you win a…

Monty Hall Problem

Door chosen by you:

Behind it is a..

Door Monty would then show you:

Stick, and you win a…

Switch and you win a…

or

Monty Hall ProblemSo if you switch, you can expect to win a car 2 out of 3 times, whereas if you stick you would only win the car 1 out of 3 times.

This problem is famous for puzzling mathematicians during the last century and illustrates that although probability questions can seem confusing andeven counter-intuitive sometimes, using a logical approach helps to unravel them.

Teacher notes

In this edition, 4 short activities from the MEI conference are used.

All four activities could be used with a wide range of students, although they are in approximate order of ‘age appropriateness’.

The full session PowerPoint and materials can be downloaded from the conference page

‘Spot the Pattern’ problem by Phil Chaffé was in Session B

‘Algebra with cards and paper’ by Kevin Lord was in Session C‘Perimeter of Rectangular Rings’ is from the same session.

‘Introduction to Probability (S1)’ by Clare Parsons was in session I

Teacher notes: Problem Solving, Phil Chaffé

During his session, Phil led teachers to consider what is meant by ‘problem solving’, why it is important, how we might enable students to improve their problem solving skills and he also looked at a range of resources and their sources.

One of his problems is presented here which requires students to discern a pattern detailed by a series of snippets of information.

There are 20 information cards and a blank grid for students to work on, which also appear on the PPT slides.

This activity can be tackled in small groups with rules for collaborative work imposed to prevent some students from dominating and others from not participating, or it could be tackled in pairs or individually.

Teacher notes: Problem Solving, Phil Chaffé

Either present the whole class with the slides one at a time so that they have 4 bits of information to work with at a time (some to-ing and fro-ing may be necessary to check wording) or print the slides for groups to use.

Extension ideas: • Ask if there are any pieces of information that are unnecessary• Ask students to come up with their own designs (on a 4x4 grid,

perhaps) and describe them with only 8 pieces of information.

Teacher notes: Algebra with cards and paper, Kevin

LordKevin’s session began very simply and built up to more complex use of algebra. The activities can be used to support students with reasoning, justification and proof as well as use of algebra. Just two of his activities are used here; several more are available on the conference page.

His activities all used business cards as a starting point, but identical rectangles of paper or card would work just as effectively.

The first activity looks at the maximum and minimum areas for two or more cards put together, the second at perimeters of rings of cards.

Teacher notes: Algebra with cards and paper, Kevin

LordSlide 13 Since b is the long edge, 4a + 2b will be smaller in value than 2a + 4b

Slide 14 The maximum perimeter will be when the two cards are almost separate (as shown) so the limit is 4a+4b

The minimum occurs when the cards have the long edges together:

Teacher notes: Algebra with cards and paper, Kevin

LordSlide 16When working with n cards, the maximum perimeter will be n(2a + 2b)

‘Taking out’ the longest edges by putting them together will always minimise the perimeter, reducing n(2a + 2b) by 2b each time, thus the minimum perimeter is n(2a + 2b) – (n-1)2b = 2na + 2b

Explanation:Placing one card initially, the perimeter is 2a + 2b; another card makes the perimeter 2(2a+2b), but by placing a long edge of the second card against the first one this is reduced by a maximum of 2b, there being a ‘b’ ‘taken out’ on each side of the join.Adding a third card, the maximum perimeter is 3(2a +2b),this is reduced by a maximum of 2(2b) by ensuring that joins are made at 2 long edges.

Teacher notes: Algebra with cards and paper, Kevin

LordSlide 17The ratio of the sides of the rectangle affects the arrangements that are possible to obtain the minimum perimeter, but the algebraic value of the minimum perimeter is unchanged; being 6a+2b for 3 cards .Similar arrangements for 3 different rectangles are shown.

Arrangement for minimum perimeter

Arrangement for minimum perimeter

Not an arrangement for minimum perimeter

Teacher notes: Perimeter of Rectangular Rings, Kevin

LordSlides 19-24

When creating a ring of cards, if they are always placed so that the long edge is on the outside then the following occurs:

Length b from each rectangle on the outside.

Length a from 4 of the rectangles on the outside at the ‘corners’.

Total outside perimeter is 4a + nb

Teacher notes: Perimeter of Rectangular Rings, Kevin

LordSlides 19-24

When creating a ring of cards, if they are always placed so that the long edge is on the outside then the following occurs:

Length b-a from 4 rectangles on the inside.

Length b from n-4 of the rectangles on the inside’.

Total inside perimeter is (n-4)b + 4(b-a)= nb - 4a

Teacher notes: Perimeter of Rectangular Rings, Kevin

LordSlides 19-24

When creating a ring of cards, if they are always placed so that the long edge is on the outside then the following occurs:

Total perimeter is inside + outside: nb - 4a + 4a + nb =2nb

Teacher notes: Perimeter of Rectangular Rings, Kevin

LordSlides 19-24

When creating a ring of cards, if they are always placed so that the long edge is on the outside then the following occurs:

An alternative way of getting to this result is to consider the perimeter of all the cards used: 2n(a+b) = 2na + 2nb

At each ‘join’, 2a is lost from the perimeter.There are n such joins.

Perimeter = 2na + 2nb – 2naPerimeter = 2nb

Teacher notes: Introducing Probability (S1), Clare

ParsonsTo begin her session, Clare used a short activity which highlighted the challenge that understanding probability presents. Not wishing to spoil the session for her with future TAM teachers, I have instead used the ‘Monty Hall’ problem, of which it was reminiscent.

During the session Clare used several approaches to teaching probability which made solving problems much more straight-forward, retaining understanding and insight whilst giving a very helpful structure.