# Altaf Masoodi

1Private school teacher, Sopore, 26/A Model Town ,J &K, India

ABSTRACT  : The research paper brings forth magic of number 1/√2 and value of pie. It also highlights problem with current formulae of all spherical geometric shapes at lower values of radius .Provides alternative solution for all two dimensional & three dimensional spherical shapes. The research paper also highlights the inaccuracy of radian measure & gives accurate value of pie.

### Keywords –

Pie, radian measure, Squaring a circle,.

## 1.Problem

The value of pie fixed at 3 decimals is not accurate. Instead pie assumes different values in different ranges of radii which should rather have a pie value chart. The problem is tackled through magic of 1/√2 and squaring a circle. The basic formulae which we have been using so long and are even now being used, at academic level as well as in field, in all branches of science give bizarre results at lower values of radius. Here we examine two such formulae also:

(1) Circumference of a circle = 2πr &

(2) Area = πr 2

Example1 At r = 0.1 mm we have,

Circumference = 2 (3.414…..) (0.1) = 0.62831mm & area = (3.414…..) (0.1)2 = 0.031415 sq. mm = 0.00098 mm .Thus area is less than the circumference. This is bizarre. The immediacy of problem is obvious to all fields of science. I present in this paper alternative to the above formulae. Proof to the fact that pie assumes value of a pie chart and also a proof to the inaccuracy of radian measure tables.

## 2.Magic of 1/√2

The problem just highlighted points to two facts;

(a) Either the value of π is inaccurate

(b) Or the formulation of formulae is wrong.

Thus in developing the formulae for basic geometric shapes, we avoid both in our solution i.e π as well as use of current formulae. Again as geometric reality is most accurate, we proceed from geometric to algebraic reality.

### 2.1 Generation of circles

For a square we have diagonal greater than side. A circle of radius 1/√2 r has an inscribed square of side r along vertices of its perpendicular diameters “Fig 1.1”.

Fig 1.1 a circle with inscribed square of side r

(6) (1/√ 2r) 2 + (1/√ 2r) 2 = s2

(7) 1/ 2 r2 + 1/ 2 r2 = s2

(8) (1/ 2 + 1/ 2) r2 = s2

(9) r2 = s2

(10) S = r

(Note; Figures not drawn to scale for better visibility.)

Suppose the circle changes into a square [1] along same diagonals or diameters. Its diagonals become larger & so do the sides of the square. Suppose diagonal changes by 1/√2 units “Fig 1.2”

 1/√2r

Fig 1.2 As circle changes to square of side 2, radius increases by 1/√2r.

In such a square we have:

(9) 2 (2/√2 r)2 = (2r)2

Proof Diagonals of a square bisect each other at right angles, thus by Pythagoras theorem we have,

Perpendicular (P)2 + base (B)2 = hypotenuse (H)2

(10) (1/√2r + 1/√2r)2 + (1/√2r + 1/√2r )2 = Side2

(11) 2 (1/√2r + 1/√2r)2 = Side2

(12) √ 2 (1/√2r + 1/√2r) = Side

(13) 2r = side “hence proved”

This means as the circle changes into a square, radius changes by 1/√2 units & side by 1 unit of the inscribed square. The vice versa is also true.

(14) 2 (r + r)2 = (√2r)2

i.e. as the radius changes by I unit, side changes by √2.This means Area of the circle becomes, (√2r)2 if radius changes by unit and Qr2 if radius changes by 1/√2 where Q is a real no..

Thus any circle is a square with 1/√2 unit increase in radius and the area of such circle is area of the inscribed circle with side increased by a unit.

Thus we can consider generation of circles by 1/√2 increase in radius and having area equal to inscribed square with increase in side by a unit. And we also have every square of the previous circle as an inscribed square of next circle “fig 2.3”.

1/√2 Fig 2.3 generation of circles with 1/√2 increase in radius each circle is a square with 1r increase in side

### 2.2 Assumptions

a) Suppose the smallest possible circle has circumference equal to area i.e. there is no internal region and circumference encompasses the whole area.

b) The circle has an inscribed square of unit radius i.e. side 1/√2 “fig 1.2”

This means the inscribed square is generated from a circle of radius almost nil i.e. a point, as 0 + 1/√2 = 1 is the situation “fig 2.3”

We have area of the inscribed square = 1 sq. units

Now there is a unit increase in radius i.e. side changes by 1/√ 2 (9). Thus,

Area of the generated square from the circle = area of circle= (2)2 = 4

So that the areas of subsequent circles become 9, 16, 25…………………………….

.This we can write as,

Area of the circle = r2 + 2 r + 1 = (r + 1)2

Example 02 + 2 × 0 + 1 = (0 + 1)2 = 1

12 + 2 × 1 + 1= (1 + 1)2 = 4

22 + 2 × 2 + 1= (2 + 1)2 = 9 and so on.

Now since radius of our circle increases by 1/√2 times each time, we have r = 1/√2

I.e. (r + 1)2 = (1/√2 + 1)2 = (½ + 1 + 2/√2) = 3/2 + √2 =2.914213562……….

This gives value of π since πr2 = π (1)2 = π when radius changes 1 times each time.

Now we can visualize smaller circles as the first generated circle plus area increase subsequently with each unit increase in radius i.e.

Area of first generated circle + increase in area with increase in radius

1 r2 + (3/2 + √2)r2

We have solved 1r2 = 3/2 + √2

We have formula for the area of the circle as follows

### 2.3 Area of the circle = t + tr2

t = 3/2 + √2 ± 0.20 {2.80 ≤ t ≤ 3 for 0.1< r ≤ 0.75, 3 ≤ t ≤ 3.14 for 0.8 ≥ r and stabilizes at this value for certain range and then increases again at very large radii}.This I am going to prove next.

## 3 MAGIC OF 1/ √2 SHOWS OUR RADIAN MESURE TABLES ARE INACCURATE

In a unit circle hypotenuse rotates as a unit [2] while the base (cos θ ) and the perpendicular (sin θ) change .

For base = perpendicular = x, by Pythagoras we have ,

(P)2 + (B)2 = (H)2

(x)2 + (x)2 = (1)2

2(x)2 = 1

(x)2 = ½

(16) X = 1/√2

OR

(sinθ)2 + (cosθ)2 = 1

For sinθ = cosθ , We have,

(sinθ)2 + (sinθ)2 = 1

2(sinθ)2 = 1

(sinθ)2 = ½

(17) Sinθ = 1/√2

Comparing (16) & (17), we get

sinθ = cosθ = = arc cosθ = sinx = cosx = arc cosx = x = 1/√2 = 0.707

But the radian measure tables give different values [2].While tables show sin 45° = cos 45° = 1/√2 = 0.707

But they show sin x =0.765, Cos x = 0.644 for x =1/√2 = 0.707

This is magic of number 1/√2.This can also be tested from the graph of the above functions.

Again when the case is as it is, we also get value of 1 radian in above situation

X = 45o = 1/√2 = 0.707

1o = 0.707/45 = 0.015711111 Which shows our radian measure tables are inaccurate

## 4 ACCURATE VALUES OF PIE (π)

### 4.1 The possible range of π

We start again with geometric reality. First the possible range of π “fig 4.1” on next page.

The circle can not have area greater than 4r2 i.e. the area of circumscribed square and not less than 2r2 i.e. area of inscribed square.

This gives range of π as 2 < π < 4.

Now as the circle becomes smaller and smaller the four arcs in the middle of two squares come closer and closer to the inscribed circle .When circle becomes very small they approach very close to the sides of inscribed square.

Fig 4.1

As the circle radius increases, the arcs get closer and closer to the circumscribed square i.e. area approaches closer to 4r2. That this is the case is evident from “fig 4.2” the graph below.

Fig 4.2 circles of radii 0.15cm , 0.225, 0.30, 0.375, 0.525, 0.60, 0.675, 0.7, 0.975, 1.50 and 2.25cm drawn to scale to show exact values of ‘t’, “ Scale: one small square = 0.15cm × 0.15 cm)

I have included the first circle in 2 × 2 grid, second in 3 × 3, third in 4 × 4, fourth in 5 × 5, fifth in 7 × 7grid and so on. The area of 2 × 2 grid circle= 4 × (0.15)2 = 0.09 sq.cm. Each circle has area less than 4r2 (area of circumscribed square). When we watch first circle very closely, we can locate four corners outside the area of the circle. Giving each corner a value of only 0.3 of one small square, the area outside the circle is 4 × 0.3 = 1.2 small squares.

Now, 1 small square = 1 r2

0.2 small squares = 0.2 r2

1.2 small squares = 1.2 r2

The area of circle = area of circumscribed square – area outside the area of circle.

This gives area of the circle = 4r2 – 1.2r2 = 2.8 r2 i.e. 2.8 × (0.15)2 = 0.063 sq. cm. Thus value of “t” is 2.80 for radius of ‘0.15’ (Hence proved)

The third circle has area < 4 × 4 grid square= 16 small squares = 4r2 = 0.36sq. cm. i.e.. if we have a closer look at the graph there are one almost complete square and two very small corners from two squares at each corner of the circle outside its area. Taking the value to be only 1.1 the area outside the area of circle = 4 × 1.1 = 4.4 sq. units.

we have, 4 sq. units = 1r2

0.4 square units =0.4/4 r2 = 0.1 r2

and total area outside circle = 1.1 r2

This gives area of circle = 4r2 – 1.1 r2 = 2.90 r2 = 2.90 × (0.225)2 = 0,14681…. Sq. cm.

Hence when radius is 0.225 cm “t” value is 2.90. (Hence proved)

Working on similar lines with other circles we get a “t value chart” for different radii. See “TABLE 1.1”.

### 4.2 Redefined values of pie (π)

Table 1 ‘t’ values for different radii

 1.Radius 2.No. of small squares outside area of circle at each corner (sq. units) 3.Total area outside area of circle (r2) 4.Area of circle (4r2-column 3) 5.‘t value’ 0.15 0.3 1.2 2.80 r2 2.80 0.225 0.66 1.1734 2.8226 r2 2.8226 0.30 1.1 1.10 2.90 r2 2.90 0.375 1.68 1.0752 2.9248 r2 2.9248 0.525 3.125 1.05 2.95 r2 2.95 0.60 4.12 1.032 2.968 r2 2.968 o.675 5.11 1.02 2.98 r2 2.98 0.70 6.3 1.002 2.992 r2 2.992 0.975 10 0.94675 3.05325 r2 3.05325 1.50 23.5 0.94 3.06 r2 3.06 2.25 49.5 0.88 3.12 r2 3.12

Our data clearly shows that with increase in radius from 0.15 to 2.25 there is an increase of 0.34……… in value of pie. Or even more if very small radii are taken into consideration .This is a clear proof that value of pie changes with change in radius and markedly so.

The circumference of our generated circles = Perimeter of squared circle i.e. 4 times side . If the radius of the circle changes by 1,

The side of the squared circle changes by √2 (14) i.e. √2, 2√2, 3√2, 4√2 ……………………………………………………………

And consequently circumference becomes 4√2, 8√2, 12√2, 16√2 ……………………………………………………………………

Thus circumference is 4 √2 r = 5.656 approximately, this approximates to

### 4.3 Circumference = 2tr

With t = 3/2 + √2 defined already, (16) (t in honor of my teachers)   t = 3/2 + √2 ± 0.20 {2.80 ≤ t ≤ 3 for 0.1< r ≤ 0.75, 3 ≤ t ≤ 3.14 for 0.8 ≥ r and stabilizes at this value for certain range and then increases again at very large radii}.

### 4.4 Check for problem rectification and interpretation

Let us see if the problem raised in introduction is rectified or not At r = 0.1 mm we have, Circumference = 2tr = 2(3/2 + √2) 0.1 mm = 0.5828 mm Area = t + tr2 = 3/2 + √2 + (3/2 + √2) (0.1)2 = 2.96914 mm2 = 8.81579 mm (approximately) I.e. problem is rectified and to interpret circumference is 0.5826 times radius and area 2.9691 sq. unit times.

## 5. Conclusion

The research paper brings forth the need to revise the value of pie in the form of pie chart with proof .This means all the established measurements, both at particle level as well as at space level, based on pie need a revision .The need for revision of radian measure tables is also pointed to with proof. The problem raised and analyzed has got a wide application in all scientific fields of research where ever research is being done in this regard. Moreover in mathematics the basic formulae for two basic geometric shapes have been put forth which obviates the need for unit conversion and also corrects the defects with formulae. The same revision is needed for formulae of sphere, cylinder, cone and all other shapes where circle is involved or pie is involved. The research is of great help to Physics, Engineering and Academics. There is also a centuries old question of squaring a circle for which a solution has been suggested.

## 6. Acknowledgements

I would like to thank all teachers who taught me and boosted my morale to go ahead with the work.

## References

Journal Papers: [1] J.J O’Connor and E F Robertson JOC/EFR April 1999(http://www-history.mcs.st-andrews.ac.uk/HistTopics/Squaring_the_circle.html) Books: [2] Saturno L. Salas, Calculus: One and Several Variables (John Wiley & Sons, Part 1).