Spot the Pattern
On 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 Pattern
There 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 Pattern
There 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 Pattern
The 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 Pattern
There 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 Pattern
The 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
7cm
b
7cm
5 cm
a
Three rectangular business cards are shown.
What is the perimeter of each?
a
Algebra with cards and paper
7cm
5 cm
5 cm+7cm+5 cm+7cm = 24 cm
a
a cm+7cm+a cm+7cm = 14+2a cm
7cm
b
a
a cm+ b cm+ a cm+ b cm=2a + 2b cm
Algebra with cards and paper
Using this business card, three
arrangements of 2 cards are shown
below. What is the perimeter of
each?
Algebra with cards and paper
2a+4b
4a+2b
2a+4b
Which is smaller: 2a +4b or 4a+2b?
Algebra with cards and paper
Putting 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 paper
Can 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 cards
Can you come up with general algebraic
expressions for the maximum and minimum
perimeters for n cards?
Algebra with cards and paper
Does 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 Problem
Monty 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 Problem
Having 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 Problem
He now asks her whether she wants to stick with the
green door or switch to the pink one.
Monty Hall Problem
Should 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 Problem
Let’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 Switch and
you win
you win
a…
a…
Monty Hall Problem
Door
chosen by
you:
Behind it is
a..
Door Monty
would then
show you:
or
Stick, and
you win
a…
Switch and
you win
a…
Monty Hall Problem
So 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 and
even 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 Lord
Kevin’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 Lord
Slide 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 Lord
Slide 16
When 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 Lord
Slide 17
The 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 Lord
Slides 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 a from 4 of the rectangles
on the outside at the ‘corners’.
Length b from each rectangle
on the outside.
Total outside perimeter is
4a + nb
Teacher notes:
Perimeter of Rectangular Rings, Kevin Lord
Slides 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 n-4 of the rectangles
on the inside’.
Length b-a from 4 rectangles
on the inside.
Total inside perimeter is
(n-4)b + 4(b-a)
= nb - 4a
Teacher notes:
Perimeter of Rectangular Rings, Kevin Lord
Slides 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 Lord
Slides 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 – 2na
Perimeter = 2nb
Teacher notes:
Introducing Probability (S1), Clare Parsons
To 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.