United Kingdom Mathematics Trust

York Teacher Meeting7 July 2015

Introduction

Steve Mulligan– Chairman of the Team Maths Challenges Subtrust

and member of UKMT Council– (day job – Deputy Headteacher in Bradford and ex-

head of maths)– email s.mulligan@talktalk.net

Aims of the Session

• Brief overview of the activities of the UKMT• Opportunities to try out lots of questions from

TMC• Hopefully have something to take back and

use or develop for use in the classroom

Warm up question...

Warm up question...

Take 1000 and add 40 to it.Add another 1000. Add 30.Add another 1000.Add 20.Add another 1000.Add 10

Now write

your answer

down!

(don’t show

anyone)

Warm up question...

Take 1000 and add 40 to it. 1040Now add another 1000. 2040Now add 30. 2070Now add another 1000. 3070Now add 20. 3090Now add another 1000. 4090Now add 10 4100

UKMT Activities

Challenges

New

A problem from the Junior Kangaroo

A problem from the Junior Kangaroo

• The triangle has to equal 1• Square + Square + Circle < 27• Either 1 or 2 is carried from the units column• Circle must be 8 or 9• Square + Square = 111 – 88 = 23 or 111 – 99 = 12• Square is a single digit so must equal 6

Challenges

Challenges

Team Competitions

Outreach

Other UKMT Activities

Team Maths Challenge (TMC)

Group Circus

1

1 2

1 2

1

1 2

1

2

1

1 2

1

2

16

1 2

1

2

161/3

1 2

1

2

161/3

2 8

1 2

1

2

161/3

2 8

1/3

1 2

1

2

16

2 8

3

1/3

1 2

1

2

16

2 8

3

1

1 2

1

2

16

2 8

3

1

Back in the classroom...

• Primes• Factors• Multiples• Square numbers• Logical deduction• Problem solving• Proof• Teamwork/communication

Shuttle

• Replaced the “Head-to-head” round

Head to Head

2 7 12 9 5 42 150 999

5 15 25 19 85 30111 1999

f(n) = 2n + 1

Head to Head

2 5 10 4 7 30 0.5 200

7 28 103 19 903 3.2552 40 003

f(n) = n2 + 3

Head to Head

4 6 14 8 15 11 98 128

2 3 7 2 11 7 5 2

f(n) = largest prime factor of n

Head to Head

8 12 16 24 40 6 99 5000

2 3 4 4 2 9 6

70

f(n) = integer part of square root of n

Head to Head

6 12 16 1 3 8 11 20

3 6

3 5 65 6

Six Twelve Sixteen One Three Eight Eleven Twenty

f(n) = number of letters in the word n

7

Algebra Cards

• The aim is to work out the value represented by the question mark

Question 1

36 24

?

?=6

Question 2

42 15

?

?=1

Question 3

27 19

? ?

?=5

Question 4

30 15

?

? ?

?=3

Question 5

39 59

?

?

?=23

Question 6

40 29

? ? ? ?

?=9

Question 7

81 43

???

?

??

??

???

?

?=4

Question 8

5

?

?

?

96

?=5

Question 9

20

?

?? ?

2011

?=3

Question 10

10

??

1417

?=5

TMC Shuttle Round

TMC Shuttle Round

• Similar to loop cards• Two pairs of students; one teacher• One pair gets Q1 and Q3; other pair gets Q2

and Q4• 8 minutes per round• 6 minute bonus whistle

8 576

24 360

6 12

50 3750

48 54

156

9 2

12067

Changes to GCSE

• Greater emphasis on problem solving• Shift of “higher” material to the foundation

tier• More complex proofs (including proofs of

circle theorems)

Proof

Proof

Can you prove that the blue region and green region have the same area?

Proof

Introduce some dimensions. Let the radius of the larger circle be 2r2r

2r

Proof

Let the areas of the four regions be A, B, C and D

2r

2r

A

BC

D

Proof

“A + B + C + D” is a quarter circle

So A + B + C + D = ¼ π(2r)2

= ¼ π . 4r2

= πr2

2r

2r

A

BC

D

Proof

“A + B” and “B + C” are identical semicircles

SoA + B = ½πr2

andC + B = ½πr2

2r

2r

A

BC

D

Proof

Adding these gives:

A + 2B + C = πr2 2r

2r

A

BC

D

Proof

So we now have two equations:

A + B + C + D = πr2

andA + 2B + C = πr2

2r

2r

A

BC

D

Proof

So we now have two equations:

A + B + C + D = πr2

andA + 2B + C = πr2

We conclude that B and D must be equal

2r

2r

A

BC

D

JMC 2013 – Question 24

JMC 2013 – Question 24

Each of the overlapping regions contributes to the area of two squares

Total area of the three squares = (117 + 2(2 + 5 + 8) ) cm2

= 147 cm2

Area of one square = 147 ÷ 3 = 49 cm2

So length of each square is 7cm

IMC 2013 – Question 21

IMC 2013 – Question 21

Large square area 196 cm2 so length is 14cm

Ratio of areas of squares is 4:1, so ratio of sides of squares is 2:1

Let side of larger inner square be 2x, then side of smaller inner square is x.

Height is 2x – 1 + x = 3x – 1 = 14 and so x = 5cm

Shaded regions each have dimensions 4cm by 9cm, so total shaded area = 2 × 4 × 9 = 72 cm2

JMO 2011 – Question B1

Every digit of a given positive integer is either a 3 or a 4 with each occurring at least once. The integer is divisible by both 3 and 4.

What is the smallest such integer?

UKMT website: www.ukmt.org.uk .

Extended challenge solutions, including some extension material. All primary team maths resources available here.

Resources website: www.ukmt-resources.org.uk

All Challenge papers since 2004; TMC materials from 2010; many other resources (including this presentation!)

UKMT Website - Resources

Thanks for listening!