Singapore Mathematical Society Association of Mathematics Educators Peter Pang 15/2/03

The Constant

  All circles are similar .

 The to relative size the circumference diameter that is, D of will be exactly the same, C  C C / D =  D = constant =  = 2  R C =circumference R =radius D =diameter

The Constant



(2)

 Thus two  provides the connection between lengths , which are the circumference and the diameter .

 This same constant connects the circle’s area and radius .

C =circumference R =radius D =diameter area

Approximating the Circle

 The critical idea is to approximate the circle with an inscribed regular polygon.

C =circumference radius D =diameter area

Approximating the Circle (2)

 We inscribe a regular pentagon in a circle of radius r .

 Break up the pentagon into five triangular pieces.

 Each triangle base b and called the has height The dotted line is apothem h . .

h b r

Connecting Length and Area

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

Connecting Length and Area (2)

 Area of inscribed pentagon =( h /2)(Perimeter)  What if we choose any other regular polygon?

h b r

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 =6(Area of triangle) =6( =( =( h h bh /2) /2)(6 b ) /2)(Perimeter) h b r

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) =8( =( =( h h bh /2) /2)(8 b ) /2)(Perimeter) h b r

Inscribing a regular

n

-gon

 If we inscribe a regular n Area of polygon = ( -gon, we obtain 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 lim (Area of regular inscribed polygon) [( h /2)(Perimeter)]

Two Questions

 What happens to the apothem and to the perimeter as the number of polygonal sides increases indefinitely?

 Clearly h the circle.

will have as its limit the radius of  The limiting value of the perimeters of the inscribed regular circumference.

n -gons will be the circle’s

The Connection Between Length and Area

 Hence Area of circle = lim = [( = = = 2  =  [( r 2 /2 r 2 .

h /2)(Perimeter)] r /2)(C)] rC /2 r (2  r )/2

Approximating

  The simplest way to approximate the ratio is to measure the circumference and diameter of a particular circle and divide the former by the latter.

C / D  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: a 2 + b 2 = c 2 .

c a b

Inscribing a Square

A s D s 2 1 B C  Choose radius of circle r =1.

 Pythagorean theorem tells us that  2  s s 2 = s =4  2 2 + s 2 =2 2  Perimeter of the square P =4 s =4  2

Inscribing a Square (2)

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

O 1 A x a /2 b C D 1 x a /2 1 B  Line segment length a AB is a side of a regular inscribed gon.

of n  Segment a regular inscribed 2 n AC , a side of -gon is generated.

Overcoming the Obstacle (2)

O 1 A x a /2 b C D 1 x a /2 1 B  Pythagorean Theorem tells us that 1 2 =( a /2) 2 + x 2 = a 2 /4 + x 2   x x 2   1 1   a 2 4 a 2 4

Overcoming the Obstacle (2)

O 1 A x a /2 b C D 1 x a /2 1 B b 2  Pythagorean Theorem tells us that   a a 2 2  2 1   ( 1 2  x x  ) x 2 2  a 4 2 4  1  2 1  a 4 2  1   2  2 1  a 2 4 a 2 4

Overcoming the Obstacle (3)

O 1 A x a /2 b C D 1 x a /2 1 B b 2    2  2  2  2 4 1      a 4 a 2 4 2 4  a 2     b  2  4  a 2 .

Back to Approximating



Octagon -

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

Back to Approximating



Octagon (2) -

A s 1 C s 2 O D B C’  We get    circumfere diameter perimeter diameter nce   8 4 2  2 2  2 2  3 .

061467459

Back to Approximating



16-gon -

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

Back to Approximating



16-gon (2) -

A s 1 C s 2 O D B C’  We get    circumfere diameter perimeter diameter nce  16  8 2 2   2  2 2  2 2  3 .

121445153

Back to Approximating



32-gon -

A s 1 C s 2 O D B C’  Another doubling and application of the formula yields the perimeter of a regular inscribed 32-gon as 32 2  2  2  Now we get   perimeter diameter  16 2  2    3 .

136548491 2 2  2

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  wish.

can be approximated as closely as we  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 expansion. 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 1.

to a decimal digit at a time.

abc as the first decimal digit, b …. We shall refer as the second decimal digit, etc. The strategy is to obtain one

 Write  2 as 1 + x . Then, 2  ( 1 

x

) 2  1  2

x



x

2  2

x



x

2  1 .

 Find an 0.

a x that has only 1 decimal digit, i.e., , such that 2 = 0.4.

x + x 2 x = is as close to 1 as possible but without exceeding 1 (note that this this out, you will find the answer to be x is not a solution to the equation above). When you carry a = 4, or x  Indeed (1.4) 2 is less than 2, but (1.5) 2 exceeds 2.

 Thus, the decimal expansion is of the form 1.4

and our next step is to find the decimal digit b .

bc …,

 Now, given that the integer part is 1 and the first decimal digit is 4, we write Then,  2 as 1.4 + x . 2  ( 1 .

4 

x

) 2    2  2  

x



x

2  2 ( 1 .

4 )

x



x

2  2  ( 1 .

4 ) 2 .

 We now seek a number second decimal digit is possibly non-zero, such that 2(1.4) x + x 2 x = 0.0

b is as close to 2 , i.e., only the  (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 lying the two next to each other.

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

xy , x/y , and the

 Addition:

y x x+ y

 Subtraction:

x-y x y

 Multiplication: xy

y

1

x

 Division:

y/x y

1

x

 Square Root: 1

a 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.

2.

3.

4.

Draw a line segment through two points. Find the intersection point of two (non-parallel) lines.

Draw a circle of a given radius with a given centre, or equivalently, draw a circle with a given centre through another given point.

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 ( and ( c , d ), i.e., such points a , b , c , d rational points a , b ) are rational numbers (below, as an abbreviation, we shall simply call ), 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 ( through a rational point ( c, d a, b ) ) satisfy the equation  (x  a) 2 +(y  b) 2 = (c  a) 2 +(d  b) 2  which is of the form  x 2 +y 2 +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  

n

2

p

2   1 

x

2   

j

 2

nq p

2 

nk p

 

x

  

q

2

p

2 

kq p



m

   0  or  

p n

2 2   1 

y

2 

k

 2

pq n

2 

pj n y

  

q n

2 2 

jq n



m

   0 ,

 which are of the form (  ) 

a



b c

,

d



e

 where a , b , c , d , e , f are rational.

f

  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  equation x 2  12 x 2=0. Thus, = 6+ x  2. Then x + 34 = 0.

x  6=  2, or is a solution of the

 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 by x in order to arrive at a polynomial equation satisfied 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.