CCSP03 Mathematical Ideas

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```					   Singapore Mathematical Society
Association of Mathematics Educators

Peter Pang
15/2/03
The Constant 

All circles are similar.
C=circumference
The relative size of
the circumference C
to diameter D will be
exactly the same,            D=diameter
that is,
C/D = constant
=
 C = D = 2R
The Constant  (2)

Thus  provides the
connection between      C=circumference
two lengths, which
are the circumference           R=radius
and the diameter.
D=diameter
This same constant
connects the circle’s        area
Approximating the Circle

The critical idea is to
approximate the circle    C=circumference
with an inscribed

D=diameter

area
Approximating the Circle
(2)

We inscribe a regular
pentagon in a circle
Break up the
pentagon into five
triangular pieces.          r
h
Each triangle has           b
base b and height h.
The dotted line is
called the apothem.
Connecting Length and
Area

Area of each triangle
=(base)(height)/2
=bh/2
Area of inscribed
pentagon
=5(Area of triangle)        r
h
=5(bh/2)                    b
=(h/2)(5b)
=(h/2)(Perimeter of
inscribed pentagon)
Connecting Length and
Area (2)

Area of inscribed
pentagon
=(h/2)(Perimeter)
What if we choose
any other regular
polygon?                r
h
b
Connecting Length and
Area - Hexagon

If we inscribe a
regular hexagon, then
the hexagon will be
divided into 6 little
triangles.
Area of inscribed hexagon           r
h
=6(Area of triangle)            b
=6(bh/2)
=(h/2)(6b)
=(h/2)(Perimeter)
Connecting Length and
Area - Octagon

If we inscribe a
regular octagon, then
the octagon will be
divided into 8 little
triangles.
Area of inscribed octagon
=8(Area of triangle)        h       r
b
=8(bh/2)
=(h/2)(8b)
=(h/2)(Perimeter)
Inscribing a regular n-gon

If we inscribe a regular n-gon, we obtain
Area of polygon = (h/2)(Perimeter)
If we increase the number of sides from 10 to
10,000 to 10,000,000, what happens?
The polygons will gradually “fill up” the circle.
Area of circle
= lim (Area of regular inscribed polygon)
= lim [(h/2)(Perimeter)]
Two Questions

What happens to the apothem and to the
perimeter as the number of polygonal
sides increases indefinitely?
Clearly h will have as its limit the radius of
the circle.
The limiting value of the perimeters of the
inscribed regular n-gons will be the circle’s
circumference.
The Connection Between
Length and Area

Hence
Area of circle = lim [(h/2)(Perimeter)]
= [(r/2)(C)]
= rC/2
= r(2r)/2
= 2r2/2
= r2.
Approximating 

The simplest way to approximate the ratio C/D
is to measure the circumference and diameter
of a particular circle and divide the former by
the latter.
However physical measurements introduce
inaccuracies and in any case tangible objects
like bicycle wheels and coffee cans are not
perfect, mathematical circles.
Archimedes’
Approximation

Requires only algebra and the
Pythagorean theorem.
Pythagorean theorem:
a2+b2=c2.

c
a

b
Inscribing a Square

circle r =1.
A   s       B
Pythagorean theorem
tells us that s2+s2=22
s   2
1
2s2=4
s =2
D           C   Perimeter of the
square P =4s=42
Inscribing a Square (2)

We get
A   s       B        circumfere nce
 
diameter
perimeter
s   2              
1            diameter
4 2

D           C          2
2 2
 2.828427125
A Better Approximation

Improve upon the first estimate by doubling the
number of sides of the inscribed polygon to get
an octagon and let its perimeter be our next
estimate of the circle’s circumference.
Double again to get a 16-gon, then a 32-gon,
and so on.
The major obstacle is how to determine the
relationship between the perimeter of one these
polygons to the next.
Overcoming the Obstacle

A                      Line segment AB of
b             length a is a side of a
C       regular inscribed n-
a/2                 gon.
1                 1-x
D           Segment AC, a side of
x             a/2
a regular inscribed
B
2n-gon is generated.
1
O
Overcoming the Obstacle
(2)

A                       Pythagorean Theorem
b              tells us that
C            12 =(a/2)2+x2
a/2                         =a2/4 + x2
1                 1-x
D                            a2
x             a/2            x2 1
B                4
a2
1                  x  1
O                                             4
Overcoming the Obstacle
(2)

A                        Pythagorean Theorem
b                tells us that
2
C            a 
a/2                 b     (1  x )
2                 2

1-x            2
1             D                   a2
x             a/2                  1  2x  x 2
4
B
a 2
a 2
a 2

     1  2 1      1 
1                  4              4        4
O
a 2

 22 1
4
Overcoming the Obstacle
(3)

A                                         a2
b
b2  2  2 1 
4
C                      a2 
a/2                      2  4 1  
1                 1-x                      4
D
x             a/2            2  4 a2
B
1
b  2  4  a 2 .
O
Back to Approximating  -
Octagon

Apply the above
C             formula to calculate
A       s       B
1
the side of a regular
2         inscribed octagon as
s
O              b  2  4  a2
 2  4   2
2

D               C’
 2 42
 2 2
Back to Approximating  -
Octagon (2)

We get
C
A       s       B         circumfere nce
 
1                        diameter
2
perimeter
s                       
O                 diameter
8 2 2

D               C’            2
4 2 2
 3.061467459
Back to Approximating  -
16-gon

Apply the formula
C             again to calculate the
A       s       B
1
side of a regular
2         inscribed 16-gon as
s
O             b  2  4  a2
 2 4        2 2   
2

D               C’
 2  4  2  2 
 2 2 2
Back to Approximating  -
16-gon (2)

We get
C
A       s       B         circumfere nce
 
1                        diameter
2
perimeter
s                       
O                 diameter
16 2  2  2

D               C’               2
8 2 2 2
 3.121445153
Back to Approximating  -
32-gon

Another doubling and
C             application of the formula
A       s       B
1
yields the perimeter of a
2         regular inscribed 32-gon as
s                     32 2  2  2  2
O
Now we get
D               C’          perimeter
 
diameter
 16 2  2  2  2
 3.136548491
Estimate of 

We double seven more times, to a 64-gon, a
128-gon, a 256-gon, a 512-gon, a 1024-gon, a
2048-gon and a 4096-gon.
The 4096-gon yields an estimate of
perimeter
 
diameter

 3.141594618
Estimate of  (2)

Thus  can be approximated as closely as we
wish.
This basic approach dates back 22 centuries to
Archimedes.
It however has one liability: the need to
calculate square roots of square roots.
Archimedes pushed through the calculations
with bare hands, without the benefit of the
decimal notation, up to the 96-gon.
Methodologies highlighted

Iterative process

Limiting process
Computation of √2

It is well-known that 2 is irrational, i.e., it has
an infinite non-repeating decimal expansion.
The following algorithm appeared in the
Jiuzhang Suanshu for obtaining this decimal
expansion.
An Iterative Approach

First, it is clear that the integer part of the
decimal expansion is 1, i.e., the decimal
expansion is of the form 1.abc…. We shall refer
to a as the first decimal digit, b as the second
decimal digit, etc. The strategy is to obtain one
decimal digit at a time.
Write 2 as 1 + x. Then,
2  (1  x)  1  2 x  x
2                2
 2 x  x  1.
2

Find an x that has only 1 decimal digit, i.e., x =
0.a, such that 2x + x2 is as close to 1 as possible
but without exceeding 1 (note that this x is not a
solution to the equation above). When you carry
this out, you will find the answer to be a = 4, or x
= 0.4.
Indeed (1.4)2 is less than 2, but (1.5)2 exceeds 2.
Thus, the decimal expansion is of the form 1.4bc…,
and our next step is to find the decimal digit b.
Now, given that the integer part is 1 and the
first decimal digit is 4, we write 2 as 1.4 + x.
Then,
2  (1.4  x)  1.4  21.4x  x
2        2               2

 2(1.4) x  x 2  2  (1.4) 2 .

We now seek a number x = 0.0b, i.e., only the
second decimal digit is possibly non-zero, such
that 2(1.4)x + x2 is as close to 2  (1.4)2 as
possible but without exceeding it.
We will get b = 1.
Squaring the Circle:
The Transcendence of              
Let us start with a straightedge of unit length.
Obviously, by putting line segments drawn using
such a straightedge together, one can make line
segments of any (positive) integer length.

Now, suppose we have two such line segments
of lengths x and y (x and y being positive
whole numbers), what are all the possible
lengths of new segments we can construct
using the straightedge and compass only?
Here are five possibilities:

It is possible to make a segment length x+y by
lying the two next to each other.
Likewise, drawing segment x starting at the
right end of y and heading left, the distance
from the left endpoint of y to the left endpoint
of x is x y.
Through a more sophisticated construction, one
can make segments of length xy, x/y, and the
square root of x or y .

y          x

x+ y
Subtraction:

x-y       y

x
Multiplication:

xy

y

1           x
Division:

y

y/x

1
x
Square Root:

a

1       a
Thus we can add, subtract, multiply, divide, and/or
take square roots to obtain any line segment whose
length is the sum, difference, product and/or
quotient of any (positive) rational number and/or
square root of rational number.
At the same time, using a straightedge and compass,
we can construct two perpendicular lines on which a
scale of length is imposed. This of course is nothing
else but a (2-dimensional) Cartesian coordinate
system, sometimes also known as the x-y plane.
Thus, using addition, subtraction, multiplication,
division and the extraction of square roots, we can
construct geometrically points on the plane whose
coordinates are sum/difference/product/quotient of
any (positive) rational number and/or their square
roots.
Geometric Constructions
(Using Compass and Straightedge Only)

The five constructions above cover five powerful
operations on the coordinates of points in the
plane. But are any additional operations possible
by geometric construction? The answer is no. To
see this, let us scrutinize what we can do using
a straightedge or a compass.
 Since the only shapes one can draw with a
straightedge or a compass are line segments
and circles (or portions thereof), the following
is an exhaustive list of techniques for
constructing new points out of old ones:

1. Draw a line segment through two points.
2. Find the intersection point of two (non-parallel)
lines.
3. Draw a circle of a given radius with a given
centre, or equivalently, draw a circle with a
given centre through another given point.
4. Find the intersection points of a circle with
another circle or line.
We have already seen that geometric
construction allows us to reach all points with
rational coordinates. Starting with points with
rational coordinates, and using 1 – 4 above,
what points can we reach?
If two points have rational coordinates (a, b)
and (c, d), i.e., a, b, c, d are rational numbers
(below, as an abbreviation, we shall simply call
such points rational points), then points on the
line segment between them satisfy the equation
 x(bd)  y(ac) = (bcad),
which is of the form
jx+ky = m
where j, k, m are rational numbers.
Given two non-parallel lines of this form:
jx+ky = m and nx+py = q
where j, k, m, n, p, q rational,
the lines intersect at the unique point
 x = (mp-kq) / (jp-kn), y = (jq-mn) / (jp-kn),
which again has rational coordinates.
Thus 1 and 2 do not give any additional points.
Next, points on a circle with rational centre (a, b)
through a rational point (c, d) satisfy the equation
 (xa)2+(yb)2 = (ca)2+(db)2
which is of the form
 x2+y2+jx+ky = m
where j, k, m are rational numbers.
Given a circle of this form and the line nx+py=q
where n, p, q are rational numbers, their
intersection points will be solutions of the

 n2    2       2nq nk     q 2 kq  
 2  1 x   j  2   x   2   m   0

p


          p   p
p
      p  

or
 p2  2         2 pq pj    q 2 jq  
 2  1 y   k  2   y   2   m   0,
n                          n        
                n   n         n   
which are of the form

()       a  b   c , d e f         
where a, b, c, d, e, f are rational.

Similarly, we see that coordinates of points of
intersection of two circles are again solutions of
quadratic equations and hence also of the form
().
Finally, it remains to examine points constructed
from 1 – 4 starting with points whose
coordinates are of the form ().
It is not hard to see that coordinates of such
points are again combinations of
sum/difference/product/ quotient of rational
numbers and their square roots.
After knowing exactly which numbers are
"geometrically constructible", the important
point really is this:
 Since a "geometrically constructible" number is
obtainable from the rationals using only
addition, subtraction, multiplication, division,
and the taking of square roots, such a number
must be a solution to some polynomial equation
with rational coefficients.
For example, take x = 6+2. Then x  6=2, or
(x6)2  2=0. Thus, x is a solution of the
equation x2  12x + 34 = 0.
In fact, more is true: If one factors the
polynomial until one gets down to an "irreducible"
polynomial equation (one that cannot be factored
any further and still has rational coefficients), the
degree of this polynomial will always be a power
of 2.
In the previous example, we have seen that since
x contains a square root, we need to square x in
order to arrive at a polynomial equation satisfied
by x. Now, intuitively speaking, every additional
square root that appears in the number requires
you to double the degree of the polynomial
equation satisfied by that number. Thus, a
number obtained by taking a square root twice is
typically the root of an irreducible 4th degree
polynomial; a number obtained by taking a
square root thrice is typically the root of an
irreducible 8th degree polynomial, etc.

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