Download huddersfield nov 2015

curiosity &
conviction
curiosity &
conviction
hmmm
mmmm
aha
ha
thanks to Rob Eastaway
play
around;
explore
a pattern
begins to
emerge
and a
relationship
seems to
hold
reasons are
sought and
may be
found
×2
+3
+3
×2
7
×2
+3
+3
10
14
17
×2
20
?
×2
+3
+3
×2
37
×2
+3
+3
40
74
77
×2
80
you are testing out
the generalisation
(rule) that the
difference is always
try out two easy
starting numbers,
two harder numbers
and
two mega - hard
starting numbers
3½
×2
+3
+3
×2
.
2 1
×2
+3
+3
47.52
×2
10.2
× 2 – 10
–5
+3
+3
–7
–2
×2
–4
×2
+3
+3
65
× 2 68
m
×2
+3
+3
×2
d
×2
+3
+3
×2
Beth’s number
b
×2
+3
+3
2b + 3
×2
2b + 6
2(b + 3)
that’s a
proof
found a rule (generalisation)
checked it out with a range of numbers
tested it widely; thrashed it in fact
done reversing (inverses) to get back to the
starting number
built algebraic expressions with ‘m’ for a million
proved that the rule always works using
someone’s (unknown) number
here’s another proof
any old number
2m + 3
×2
+3
+3
×2
5p – 1
×2
+3
+3
×2
×2
+ 10
+ 10
×2
×2
+b
+b
×2
×5
+3
+3
×5
×t
+a
+a
tn + a
×t
tn + ta
ta – a
a(t – 1)
8n + 1
numbers
can’t ever be the square of an even number
for what integer values of n is 8n + 1 a square number?
0
1
3
8n + 1 = 25
8n + 1 = 36
and a
relationship
seems to
hold
6
10
getting all
the squares
of the odds
for what integer values of n is 8n + 1 a square number?
3
for what integer values of n is 8n + 1 a square number?
3
for what integer values of n is 8n + 1 a square number?
3
n = 10
n=3
n=6
n = 10
numbers
2
7
–
2
5
24
2
3
–
2
1
8
2
10
–
2
8
36
2
11
–
40
2
9
2
6
–
2
4
20
notice?
algebra
(n +
2
2)
–
2
n
(n + 2)(n + 2) –
2
n
2
–
n
(n + 2)(n + 2)
2
n
2
–
n
+ 2n + 2n + 4
2
n
4n
+4 –
4(
4n + 1)
4
+
2
n
proof
diagram
diagram
diagram
diagram
proof
utilising this diagram:
use this multi-purpose diagram
to try to prove that:
1
x ≥ 0, x + x ≥ 2
1.
for
2.
where 8n + 1 is a square number, n is
a triangular number
3.
the arithmetic mean of two numbers, ½ (a + b)
greater than the geometric mean, √ab
2
a–b
2
4.
a+b
5.
a2 – b2 = (a – b)(a + b)
2
–
2
= a.b
Babylonian
multiplication
is
for x ≥ 0,
1
x+ x ≥ 2
1
x
x
x+
x+
1
x
1
x
2
≥ 4
≥ 2
√b
√a
a.m. > g.m.
(√a + √b )2
>
4√ ab
Babylonians
a
b
b
a–b
they used this
to multiply
and to solve a
quadratic
equation
(a + b)2 – (a – b)2 = 4ab
8 × ½ (1 × 2) + 1
9
8 × ½ (2 × 3) + 1
25
8 × ½ (3 × 4) + 1
49
8 × ½ (4 × 5) + 1
81
8 × ½ (5 × 6) + 1
121
8 × ½ (6 × 7) + 1
169
…..
n+1
n
a – b = 2w
a
b
area and
perimeter
of
rectangles
the area is 18
the perimeter is 22
the area is 24
the perimeter is 28
the area has the same value
as the perimeter
the area has double the value
of the perimeter
perimeter
area = 24 cm²
= 20 cm
= 22 cm
= 28 cm
= 50 cm
= 97 cm
= 35 cm
= 21.4 cm
area and
perimeter
this shape can be
described as
(20 , 18)
can you see why?
(1) try to find some rectangles (or squares) that fit the rule:
P–A=4
(2) try to find some rectangles (or squares) that fit the rule:
P – 2A = 2
(3) try to find some rectangles (or squares) that fit the rule:
3P – 2A = 18
(A, P)
area and
perimeter
P – 2A = 2
2PP––2A
A = 48
3P – 2A = 18
next?
P–A= 4
and
in general?
2P – A = 16
what size (dimensions) rectangle fits the rules
3P – 2A = 18
and
P– A= 4
?
2 wide rectangles
‘A’ squares
A – 4 single units
plus 4 lots of 2
P=A–4+8
P–A= 4
2 wide rectangles, by induction
start with 4 squares
P=A+4
every time you add in 2 squares
you add 2 onto the perimeter
so P still = A + 4
P–A= 4
3 wide rectangles
P=⅔A+ 6
4 × 2 + A – 4 – (⅓A – 2)
⅔A + 8 – 4 + 2
P = ⅔A + 6
4 wide rectangles
P=½A+ 8
4 × 2 + A – 4 – 2(¼A – 2)
½A+8–4+4
P=½A+8
P
A
area and perimeter (i)
this shape can be described as
(20 , 18)
can you see why?
(a) try to sketch or write down the dimensions (length and
width) of the following rectangles (or squares):
(1)
(1 , 4)
(9)
(9 , 12)
(2)
(2 , 6)
(10)
(6 , 14)
(3)
(3 , 8)
(11)
(10 , 14)
(b)
which rectangles fit the rule:
P–A=4 ?
(c)
which rectangles fit the rule:
P – 2A = 2 ?
(4)
(4 , 8)
(12)
(12 , 14)
(d)
(5)
(4 , 10)
(13)
(7 , 16)
(6)
(6 , 10)
(14)
(12 , 16)
(7)
(5 , 12)
(15)
(15 , 16)
(8)
(8 , 12)
(16)
(16 , 16)
which rectangles fit the rule:
2P – A = 16 ?
(e)
which rectangles fit the rule:
3P – 2A = 18 ?
in general?
(A, P)
area and perimeter (ii)
(1)
this shape can be described as (20 , 18)
can you see that P = A – 2 ?
or that A – P = 2 ?
there is only one more rectangle (with whole number
sides) that fits this rule
can you find it?
(2)
what size (dimensions) would the rectangle, (24 , 20) be?
this fits the rule A – P = 4
try to find the other rectangle (with integer sides) that fits this rule
(3)
what size (dimensions) rectangles fit the rule 16A = P2 ?
(4)
what size (dimensions) rectangles fit the rule P = 4A + 1 ?
(5)
what size (dimensions) rectangle fits the rules 3P – A = 36 and 2P – A = 16 ?
(6)
what size (dimensions) rectangle fits the rules 3P – 2A = 18 and P – A = 4 ?
angle bisector
angle bisector
angle bisectors
and the incircle
d in terms of
a , b and c?
b
d in terms of b?
d
a
c
90 + ½a
90 + ½b
90 + ½c
b
+
270 + ½(a + b + c)
90 + ½c
90 + ½a
90 + ½b
a
c
angle bisectors
what is the relationship
between the (angles) dots?
two interior bisectors and
an exterior bisector
intersect the lines (one
extended) of the triangle
what is the relationship between the red dots?
any quadrilateral
does this
always
happen?
when you bisect
all of the angles
show that the resulting
quadrilateral must be cyclic
place all of the digits
1 to 9 in the cells
so that,
the rows
the columns and
the two diagonals
all sum (add) to the
same number
1
2
3
4
5
6
7
8
9
the total of the digits 1 to 9 is
45
the total of the
three lines is 45
so the total of each line
must be 45  3 =
15
each of the three
lines has the same
total
4 lines passing through the central cell
the total of these four lines is 4 × 15 = 60
but there is only a total of 45 available
the central cell is counted 4 times (3 too many)
so 3 × central cell = 60 – 45
central cell = 5
9
5
1
the 9 must go somewhere – try it in a corner cell
need the 1, to make 15
6
9
5
1
consider the 6
it can’t go in the same line as 9
6
9
8
7
5
1
consider the 6, 7 and 8
neither of them can go in the same line as 9
6
9
8
7
5
1
two places for three numbers
eh?
9
5
1
the 9 cannot go in a corner
the 9 must go in one of the middle cells
9
5
1
need to sum to 6
only 4 and 2
4
9
2
3
5
7
8
1
6
there is essentially only one magic square using the 1 to 9
the Lo Shu
magic square
a magical turtle emerged from
the water with the curious and
decidedly unnatural (for a turtle
shell) Lo Shu pattern on its shell
early records are ambiguous,
referring to a "river map", and
date to 650 BCE
but clearly refer to a magic
square by 80 CE, and explicitly
give one since 570 CE