1. United Kingdom Mathematics Trust
York Teacher Meeting
7 July 2015

2. 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)

3. 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

4. Warm up question...

5. Now write your answer down! (don’t show anyone)
Warm up question...
Now write your answer down! (don’t show anyone)
Take 1000 and add 40 to it.
Add another 1000.
Add 30.
Add 20.
Add 10

6. Warm up question... Take 1000 and add 40 to it. 1040
Now add another
Now add
Now add another
Now add
Now add another
Now add

7. UKMT Activities

8. Challenges
Y7/8
JMC
JMO
Junior Kangaroo
New

9. A problem from the Junior Kangaroo

10. 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

11. Challenges IMC Y9/10/11 IMOK Cayley Y9 Hamilton Y10 Maclaurin Y11
Summer Schools
Pink Kangaroo Y10/11
Grey Kangaroo Y9
Kangaroo

12. Challenges
Y11/12/13
SMC
round 1
BMO
round 2
Senior Kangaroo

13. Team Competitions
TMC (Y8/9)
STMC (Y11/12/13)
PTMR

14. Outreach
Teacher meetings
Summer schools
Mathematical Circles

15. Other UKMT Activities
Mentor schemes
IMO selection
Publications

16. Team Maths Challenge (TMC)
Regional Finals
National Final
Group Round
Poster Round
Crossnumber
Group Circus
Shuttle
Relay

17. Group Circus

20. 1

21. 1 2

22. 1 2 1

23. 1 1 2 2 1

24. 6 1 1 2 2 1

25. 6 1
1/3 1 2 2 1

26. 6 1
1/3 1 2 2 1 2 8

27. 6 1
1/3 1 2 2 1
1/3 2 8

28. 6 1
1/3 1 2 2 3 1 2 8

29. 1 6 1 1 2 2 3 1 2 8

30. 1 6 1 1 2 2 3 1 2 8

31. Back in the classroom... Primes Factors Multiples Square numbers
Logical deduction
Problem solving
Proof
Teamwork/communication

32. Shuttle
Replaced the “Head-to-head” round

33. Head to Head 19 11 85
301
1999
f(n) = 2n + 1

34. Head to Head 19 52
903
3.25
40 003
f(n) = n2 + 3

35. f(n) = largest prime factor of n
Head to Head 2 5 11 7 2
f(n) = largest prime factor of n

36. f(n) = integer part of square root of n
Head to Head 4 6 2 9 70
f(n) = integer part of square root of n

37. f(n) = number of letters in the word n
Head to Head 7 3 5 5 6 6
Six Twelve Sixteen One Three Eight Eleven Twenty
f(n) = number of letters in the word n

38. Algebra Cards
The aim is to work out the value represented by the question mark

39. Question 1 ? 36 24
?=6

40. Question 2 ? 42 15
?=1

41. Question 3
? ? 27 19
?=5

42. Question 4
? ? ? 30 15
?=3

43. Question 5 ? ? 39 59
?=23

44. Question 6
? ?
? ? 40 29
?=9

45. Question 7 ? ??
??? ? ??
??? 81 43
?=4

46. Question 8 ? ? ? 6 9 5
?=5

47. Question 9 ? ?
? ? 11 20 20
?=3

48. Question 10 ? ? 17 14 10
?=5

49. TMC Shuttle Round

50. 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

51. 8
576 24
360

52. 6 12
3750 50

53. 48 54 15 6

54. 9 2
120 67

55. Changes to GCSE Greater emphasis on problem solving
Shift of “higher” material to the foundation tier
More complex proofs (including proofs of circle theorems)

56. Proof

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

58. Proof
Introduce some dimensions. Let the radius of the larger circle be 2r 2r 2r

59. Proof
Let the areas of the four regions be A, B, C and D A D 2r B C 2r

60. Proof “A + B + C + D” is a quarter circle So A + B + C + D = ¼ π(2r)2

61. Proof “A + B” and “B + C” are identical semicircles So A + B = ½πr2 A
C + B = ½πr2 A D 2r B C 2r

62. Proof
Adding these gives:
A + 2B + C = πr2 A D 2r B C 2r

63. Proof So we now have two equations: A + B + C + D = πr2 and
A + 2B + C = πr2 A D 2r B C 2r

64. Proof So we now have two equations: A + B + C + D = πr2 and
A + 2B + C = πr2
We conclude that B and D must be equal A D 2r B C 2r

65. JMC 2013 – Question 24

66. JMC 2013 – Question 24
Each of the overlapping regions contributes to the area of two squares Total area of the three squares = ( ( ) ) cm2 = 147 cm2 Area of one square = 147 ÷ 3 = 49 cm2 So length of each square is 7cm

67. IMC 2013 – Question 21

68. 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

69. 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?

70. UKMT Website - Resources
UKMT website: .
Extended challenge solutions, including some extension material. All primary team maths resources available here.
Resources website:
All Challenge papers since 2004; TMC materials from 2010; many other resources (including this presentation!)

71. Thanks for listening!