CCSP03 Mathematical Ideas

					   Singapore Mathematical Society
Association of Mathematics Educators

                               Peter Pang
The Constant 

All circles are similar.
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.
This same constant
 connects the circle’s        area
 area and radius.
Approximating the Circle

The critical idea is to
 approximate the circle    C=circumference
 with an inscribed
 regular polygon.                  radius


Approximating the Circle

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

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

Area of inscribed
What if we choose
 any other regular
 polygon?                r
Connecting Length and
Area - Hexagon

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

If we inscribe a
 regular octagon, then
 the octagon will be
 divided into 8 little
Area of inscribed octagon
 =8(Area of triangle)        h       r
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
The Connection Between
Length and Area

   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.

Requires only algebra and the
 Pythagorean theorem.
Pythagorean theorem:


Inscribing a Square

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

                  We get
  A   s       B        circumfere nce
                    
  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
                                2n-gon is generated.
Overcoming the Obstacle

         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
                   1                  x  1
 O                                             4
Overcoming the Obstacle

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

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

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

                                    22 1
Overcoming the Obstacle

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

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

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

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

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

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

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

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

   3.141594618
Estimate of  (2)

Thus  can be approximated as closely as we
This basic approach dates back 22 centuries to
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
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…. 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.

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.
   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

          x-y       y




            1           x



Square Root:


                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
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)
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
 quadratic equation

   n2    2       2nq nk     q 2 kq  
   2  1 x   j  2   x   2   m   0
                   p   p
                                     p  
   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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