an algorithm to construct the circle and find the value of pi

8/7/2019 an algorithm to construct the circle and find the value of pi

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AN ALGORITHM TO CONSTRUCT A CIRCLE AND FIND THE VALUE OF PI

Sujeet Kumar Mehta

Laxmi Neewas, Pugmil Road, Hazaribag, Jharkhand

Annamalai University (B.E ECE 4th semester)

[email protected]

ABSTRACT

There is a new algorithm to construct a circle and to evaluate the value of pi. This value of pi is in term of 2 and

square roots. A theorem says about the conversion of a circle into rectangle. The rectangle formed by the circle is

transformed into a square to proof how pi can have two values. There is a quadratic equation whose coefficient is

irrational which give the value of pi. A pi function has been defined using the above theorem. The symbol of pi

contains geometrical meaning.

(1) ALGORITHM TO CONSTRUCT A CIRCLE

Definition: - Circle, in geometry, plane curve such that each point on the curve is the same distance from a fixed

pointcalled the centre of the circle.

THEOREM: - In an isosceles triangle the unequal side is bisected by the line of length of equal sides of isosceles

triangle and joins the vertex of unequal side to the end of bisector line. These processes construct two isosceles

triangles. Proceeding with the same process continuously will result in construction of an arc of circle.

PROOF:-

An isosceles right angle triangle (IRT) ABC in that equal sides are AB=BC=L and longest side is AC=R.

The length of R according to Pythagoras theorem

R= L i.e. AC= L (1.1)

Now bisect the longest side of IRT i.e. (AC) R by a line of the length L say BE so that the bisecting length will

exceed from bisecting point D of AC and could form two new isosceles triangles (IT) after joining the line to AE

and EC. The exceeded length from bisected point D is



DE=BE-BD

DE=L- =

L= L (1.2) {BE=L, BD=

L (AD=DC=BD=AC/2= L) from (1.1)}

AE=EC

In an isosceles triangle the unequal side is bisected at right angle [1].

AE= from (1.2)

AE= AE= L (1.3)

Now again bisect the newly formed IT (ABE & CBE) i.e. side AE & EC by line of the length L say BG1 & BG2 so

that the bisecting length will exceed from bisecting point F1 & F2 of AE & EC respectively and could form four new

IRT after joining the line to AG1,G1E,EG2 & G2C and. The exceeded length from bisected point F1 & F2 are

F1G1=F2G2

F1G1=BG1-BF1 (Now calculate BF1 & BG1)

BF1=

Where AF1=AE/2= L/2 (1.4) .... from (1.3) put this value in upper equation

BF1=

BF1= (1.5)

Now F1G1=BG1-BF1=L- = (1- ) L= L (1.6)

Now calculate the length of the unequal side of newly formed IT i.e.AG1, G1E, EG2 & G2C



AG1= AG1= {from ((1.3) AF1=AE/2 = & (1.6))}

AG1=L ) (1.7)

Now again bisect the newly formed IT (ABG1, BG1E, BEG2, & CBG2) i.e. side AG1,G1E,EG2 & G2C by line of

the length L say AI1, I1G1, G1I2,I2E, EI3, I3G2, G2I4 & I4C so that the bisecting length will exceed from bisecting

point H1, H2, H3 & H4 of AE & EC respectively and could form eight new IT after joining the line to AI1, I1G1,

G1I2,I2E, EI3, I3G2, G2I4 & I4C and. The exceeded lengths from bisected point H1, H2, H3 & H4 are

H1I1=H2I2=H3I3=H4I4

H1I1=BI1-BH1 (BI1=L)

BH1= (AB=L)

BH1= {AH1=AG1/2= (from (1.7)}

BH1= (1.8)

H1I1=BI1-BH1

H1I1=L- = L (1.9) From (1.8)

Now calculate the length of the unequal sides of newly formed IT i.e. AI1, I1G1, G1I2, I2E, EI3, I3G2, G2I4 & I4C.

AI1=



AI1= From (1.7) & (1.9)

AI1= ) L (1.10)

OBSERVATION

EXCEEDED LENGTH

Now observe the exceeded length from the bisecting point after each operation of formation of new IT.

1)

DE=(

)L (1.2)

2) F1G1= ( )L (1.6)

3) H1I1=( )L (1.9)

The process involved in calculating the exceeded length is the same. Observing the sorted list of the exceeded

length then we find after each calculation that root of 2 is get added in the last 2 under the root and constituting

the preceding values.

One root of 2 is subtracted in 2 in first exceeded length and one root of 2 is get added in 2 under whole root is

subtracted in 2nd

and 3 in 3rd

and so on. So nth exceeded length is

Nth exceeded length = ) L

UNEQUAL SIDE OF IT

Now observe the unequal side of newly formed IT.



1) AE= L (1.3)

2) AG1= L (1.7)

3) AI1=

L (1.10)

The process involved in calculating the unequal side of IT is the same. Observing the sorted list of the unequal side

of IT then we find after each calculation that root of 2 is get added in the last 2 under the root and constituting the

preceding values.

One root of 2 is subtracted in 2 under whole root in first unequal side of IT and two root of 2 in 2nd

and 3 in 3rd

and

so on. So nth unequal side of IT is

Nth unequal side of IT= ) L

The most basic concept of circle is satisfied only by the statement of the theorem that in an isosceles triangle the

unequal side is bisected by the line of length of equal sides of isosceles triangle and joins the vertex of unequal side

to the end of bisector line. These processes construct two isosceles triangles. Proceeding with the same process

continuously will result in a plane curve such that each point on the curve is the same distance from a fixed point

called the centre of the circle.s

CALCULATING THE VALUE OF PI

Now calculating the number of newly formed unequal sides of IT after each process and together adding they will

give the value of the quarter circumference.

Now the first process gives two, 2nd gives four and 3rd

eight and so on. It is the geometric progression so the nth

numbers of sides are. The total length of the unequal sides of IT that construct the quarter circumference is

given by the product of with the length of unequal side of IT.



Q uarter circumference=

) L

) L

=

=

The number of 2 inside the root will be equal to the power of 2 that means if there are n number of 2 under the

same condition then the power of 2 outside the root will also be n.

=

=3.1415926535897932384626433832795 when n=62 pi gives the most approx value.

Every two n value gives one correct value [observation].

For the most accurate value n=

=



PICTURIAL DEMONSTRATION OF THE ALGORITHUM TO CONSTRUCT THE CIRCLE



Continueit will become



(2) CONVERSION OF CIRCLE INTO RECTANGLE AND TRANSFORMATION OF THAT RECTANGLE INTO SQUARE

THEOREM:-The product of the quarter circumference of a circle & its diameter gives the area of circle.

Let radius of the assumed circle is r. According to the statement the diameter and the sides of rectangle

AB=CD=2rand the quarter circumference of circle CE=ED= .

Area of the circle = (2.1)

Area of rectangle= *2r= (2.2)

Equations (2.1) & (2.2) are equal. Hence proof the upper statement.

TRANSFORMATION OF RECTANGLE INTO SQ UARE

Now transform this rectangle into square with having equal perimeter.

The transformed square is SCFGH with the sides.

4s=r(+4)

Squaring both sides



16= (2.3)

Now multiply both sides by and consider this k as the ratio of area of rectangle to square

k=

k= k (2.4)

Putting this equation in the L.H.S previous equation

16= 16= cancelling 16= k (2.5)

16=k(16+8+)

16=k16+8k+

16k+8k-16+

=0

16k+(8k-16)+

=0 (2.6)

This is a quadratic equation of . So it has two solutions i.e. 1 & 2.

1=

1= (3.7)

2=

2= (2.8)

Equation (2.6) result in two conclusions that is:-

1) If we consider equation (2.5) and calculate the value of k using the value of pi calculated in (1) we can

have the solution of 1 & 2.

16=k

16*3.1415926535897932384626433832795= k

k=0.9855523669887770429282572508757

Using the k value in (2.8)

1= 1=3.1415926535897932384626433832795

Using the k value in (2.7)

2=

2=5.0929581789406507446042804279506



Now observe the equation (2.6). It is a quadratic equation whose solution completely depends on its

coefficient i.e. k. According to (2.5) k value depends on . Both values are interdependent. We had calculated

the value of pi so we got the k value. After putting the k value we got the 1= and

2=5.0929581789406507446042804279506. Here the question arises that if we put 2 in equation (2.5) then

what will be the value of k and its root.

16*5.0929581789406507446042804279506= k

K=0.9855523669887770429282572508757

The most interesting and amazing fact is that using 2 in (2.5) we got the same value of k. So the roots of the

equation (2.6) will be same.

HOW 2 CAN BE TRUE?

When we consider the area of circle with the value of 2 with the radius r then we obviously gets a larger area, so

consider a larger radius say R with using 1. We get:-

2

=

1

R= r

R=1.2732395447351626861510701069838r

The transformed square is CMKL with the side S.

4S=R(+4)

Squaring both sides

16= (2.9)

Now multiply both sides by and consider this k as the ratio of area of rectangle to square

k=

k=

k (2.10)



Putting this equation in the L.H.S previous equation

16=

16= cancelling 16= k (2.11)

Thats how we got to know the value of pi2 is not as 5.096. but the increased radius R can give the same value of

k used in the algorithms of the ttransformation of the circle.

(3) PI FUNCTION

FUNCTION:-A function from S to T, where S and T are non empty sets, is a rule that associates with each element of

S (the domain) a unique element of T (the co-domain)[2].

PI FUNCTION: - The domain of pi function (o<=n<=) and the range (2.8684. <= (n) <), discontinuity at Natural

number.

g (n)=

��



g (1)= �

g (2)= �

g

(3)=

�g (4)= � g (5)= �

The Greek letter was adopted for the number from the Greek word for perimeter íç. The symbol of

pi was firstly given by William Jones in 1707 and popularized by Leonhard Euler in 1737.

To me the symbol of pi is because the upper line shows the diameter and the curved line shows two semi-

Circumference of circle. This will be more relevant according to the meaning and geometrical construction of .

REFERENCE

[1] Challenge and thrill of pre-college Mathematics: V.Krishanmurty, C.R. Praneschar Theorem 12 page no. 59.

[2]Oxford Concise Dictionary of Mathematics: Clapham and Nicholson page no. 189.

[3]Calculation is done on computer scientific calculator.

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an algorithm to construct the circle and find the value of pi