Archive for May, 2014

Soul tied to physical body

May 28, 2014

Soul tied to physical body

L V Nagarajan

1.0 Soul, in a spiritual context

Last spring, I attended a spiritual workshop conducted in our neighborhood. In one of the sessions the Guru involved me in a demo conversation to bring out the concept of Soul or Atma.

He asked me : “Who are you?”

I replied, “I am Nagarajan”

“No that is your name. I know it. But who are You?”

After some thought I replied: “I am an Electrical Engineer.”

“No that is your profession. But who are YOU?”

After some more thought I pointed to my body somewhere near the heart and said – “This is Me.”

“No that is your body. But who are Y-O-U? Can you tell me who are you without referring to your extensions like name, profession, your body and such things. They are all temporary and subject to change.”

“How? Body can change ….?”

Guru did not reply. He went on to describe Soul or Atma in a spiritual context and how it will finally merge with the superior SOUL or PARAMATMA. But I was intrigued by the unanswered question – like name and profession, can you change your body also?

2.0 Soul, in Yogic context

I was surprised to find the answer to this question recently through the columns of Times of India. Writing in ‘The Speaking Trees’ of 30th April 2014, Sadhguru Jaggi Vasudev says: “Breath is not just the exchange of oxygen and carbon dioxide. For different levels of thought and emotion that you go through, your breath takes on different types of patterns. When you are angry, peaceful, happy or sad, your breath goes through subtle changes. Whichever way you breathe, that is the way you think. Whichever way you think, that is the way you breathe. Breath can be used as a tool to do many things with body and mind. Pranayama is the science whereby consciously breathing in a particular way, the very way you think, feel, you understand and experience life can be changed”. He further adds: “Breath is like the hand of the Divine. You don’t feel it. It is not the sensations caused by the air. This breath that you do not experience is referred to as Koorma Nadi. It is a string which ties you with this body, an unbroken string. If i take away your breath, you and your body will fall apart because the being and the body are bound by the Koorma Nadi. This is a big deception. There are two, but they are pretending to be one. There are two people here, the body and being, two diametrically opposite ones, but they pretend that they are one. If you travel through breath, deep into yourself, to the deepest core of breath, it will take you to that point where you are actually tied to the body. Once you know where and how you are tied, you can untie it at will. Consciously, you can shed the body as effortlessly as you would shed your clothes. When you know where your clothes are tied, it is easy to drop them. When you don’t know where it is tied, whichever way you pull, it does not come off. You have to tear them apart. Similarly, if you do not know where your body is tied to you, if you want to drop it, you have to damage or break it in some way. But if you know where it is tied, you can very clearly hold it at a distance. When you want to drop it, you can just drop it consciously. Life becomes very different. When somebody willfully sheds the body completely, we say this is mahasamadhi. This is mukti or ultimate liberation. It is a great sense of equanimity where there is no difference between what is inside the body and what is outside the body. The game is up. This is something every yogi longs for. Consciously or not, every human being is working towards this.”

Yes, here we have the answer. You are different from your body. I find this as a Yogic or elemental way looking at your soul. A yogic practice to realize oneself separated from one’s own body. This is perhaps the way Sri Ramana Maharshi found the answer for his monumental question WHO AM I? Subsequently he even achieves out-of-body experience and preaches these concepts to thousands of his followers and devotees. All this said and done, this is still not a complete answer to my question – ‘Like name and profession, can you change your body also?’

3.0 Soul, in a scientific context

Idly I turned my eyes away from ‘The Speaking Trees’ to the next page of the same issue of Times of India. To my surprise I found the missing part of the answer in another news item on science pages. It talks of ‘A device to let you ‘virtually’ swap your body with another’. Here goes the report: 

 “A group of artists based in Barcelona has created an unusual virtual reality device that can allow you to experience what it might be like to step into the skin of another person. The device, called ‘The Machine to be Another’ lets people experience life in another person’s body. Participants in a body swapping experiment at the ‘Be Another’ lab, don an ‘Oculus Rift’ virtual reality headset with a camera rigged to the top of it. The video from each camera is piped to the other person, so what you see is the exact view of your partner. If she moves her arm, you see it. If you move your arm, she sees it. To get used to seeing another person’s body without actually having control of it, participants start by moving their arms and legs very slowly, so that the other can follow along. Eventually, this slow movement becomes comfortable, and participants start to feel as though they are living in another person’s body, BBC News reported.”

Is the above an attempt to look at your soul from scientific aspect? Now we can look up to Jagat Guru Adi Sankara. He had achieved this feat of entering another body but without the aid of such devices as above. Even some lesser mortals have achieved this feat and this is known as one of ashta-ma-siddis, the eight great feats. In Tamil it is known as Koodu-vittu-koodu-paaydal, or ‘from one shell to another’.

We may meditate on this.


  1. THE SPEAKING TREE, Life Breath & The Ultimate Expansion, by Sadhguru Jaggi Vasudev, Time of India 30th April 2014, page 20.
  2. A device to let you virtually swap bodies with another, BBC News report,  Time of India 30th April 2014, page 21

LVN/28 May 2014


Bi-centric Polygons

May 13, 2014

Bi-centric Polygons

L V Nagarajan

Triangle is the basic polygon of 3 sides and 3 angles. Circle is considered as a polygon of infinite sides. In fact any closed figure can be considered as a polygon. Circle has a centre and all points on the circle are equidistant from the centre. Every regular polygon has a unique centre from where all sides are equidistant and all corners are also equidistant. In the case of triangles, the equilateral triangle has a unique centre as above. With this centre we can draw two circles one inscribed within the sides of the regular polygon and the other circumscribing the corners of the polygon. Such polygons which have two centres, a circum-centre and an in-centre are called bi-centric polygons. In case of regular polygons these two centres coincide. Irregular polygons can also be by-centric, but with displaced circum-centre and in-centre.

Bi-Centric Triangles:

For instance all triangles are inherently bi-centric. They have an in-circle tangential to all the three sides with an in-centre, I. This is always interior to the triangle. Triangles also have a circum-circle circumscribing the three corners of the triangle with a circum-centre, O. This can even be exterior to the triangle. In the case of equilateral triangle, as below, these two centres, I and O, coincide, with radius r of the in-circle, being half of radius R of circum-circle.


An interesting feature is: between these two circles you may inscribe infinite number of equilateral triangles, as shown in dotted lines, starting from any arbitrary point on the outer circle.

Now let us take another triangle, an isosceles triangle with its circum-circle and in-circle as shown below. This bi-centric triangle has its circum-centre and in-centre separated by a distance d.


Let OA = R, IE = r and OI = d.

∆ABD ||| ∆AEI. Hence IE/AI = BD/AD

i.e   r/(R+d) = BD/2R  ——–   (1)

As, ∟CBD = ∟CAD = A/2

∟IBD = A/2 + B/2 = ∟BID

(i.e.) BD = ID = R-d

Hence from Eqn (1), r/(R+d)  = (R-d)/2R

i.e.  2Rr = R2 – d2

or, 1/(R+d) + 1/(R-d) = 1/r

This is a handy relationship between R, r and d. Now,  like in the case of equilateral triangles, let us see whether these two circles also give rise to a family of triangles inscribed between them.

Let us take the same circles as above. Let us draw a tangent PQ to inner circle, from any point P on the outer circle. From P let us also draw a tangent PR to inner circle as shown, to a point R on the outer circle. Will QR be also a tangent to inner circle? Let us check. Draw PI and extend it to S on the outer circle.

With PQ and PR as tangents, PI is a bisector of angle P and hence

QS = RS and ∟SRQ = ∟SQR = P/2

Draw SO and extend the same to T on the outer circle. Join TQ.

∟SPQ = ∟STQ = P/2 = ∟SRQ = ∟SQR



PI.IS = AI.ID = (R+d).(R-d) = 2rR, (as derived earlier for these two circles)

i.e., PI/r = 2R/IS  –  – – – – – – (1)

But ∆PIE ||| ∆TSQ and hence,

PI/r = 2R/QS  – – – – – – – – – – (2)

Comparing (1) and (2) above,

IS = QS = RS.

∟IRS = ∟RIS and ∟IQS = ∟QIS.

∟IRQ = ∟IRS – ∟SRQ = ∟RIS – P/2 = ∟IRP

Similarly ∟IQR = ∟IQP

i.e., IR and IQ are respectively the bisectors of angles PRQ and PQR.

Hence QR is also a tangent to the inner circle.

So, if you draw a tangent PQ to inner circle from any point P on the outer circle and continue drawing such tangents it will close on P again forming a triangle. Hence,  like in the case of equilateral triangles, these two circles also give rise to a family of triangles inscribed between them. Amazingly this is true of all bi-centric polygons.

Bi-Centric Polygons:

Jean-Victor Poncelet (1788 – 1867) was a French mathematician who formulated this amazing feature in his Poncelet theorem. He found that this property is true even with polygons inscribed between two ellipses instead of circles as above. In one of its simpler forms, the Poncelet’s Closure Theorem (or Poncelet’s Porism) can be stated as below:

If an n-sided polygon, n-gon, inscribed between two given conic sections, is closed for one point of origin of the polygon, then it is closed for any position of the point of origin.

Given two ellipses, one inside the other, if there exists an n-gon, simultaneously stays inscribed in the outer ellipse and circumscribes the inner ellipse, then any point on the boundary of the outer ellipse is a vertex of another such circum-inscribed n-gon of same number of sides. In the case of circles, their centres will become the circum-centre and in-centre of those n-gons and hence such n-gons are called Bi-centric Polygons.


The following diagram shows the closure feature for ellipses:



Bi-Centric Quadrilaterals:

Coming back to circles, we know of quadrilaterals which are circum-cyclic; i.e., its four vertices lie on a circle. We generally call them as cyclic quadrilateral. But not all quadrilaterals are circum-cyclic or circum-centric. For quadrilateral ABCD  to be circum-centric, necessary and sufficient condition is: the opposite angles should add up to 180 degrees. i.e. Angles A+B = Angle C+D =180 Degrees. Same way, a quadrilateral can also be in-centric or in-cyclic. Necessary and sufficient condition for the same is: the opposite sides of the quadrilateral ‘abcd’ should add up to same value, i.e. a+c = b+d = s. Obviously those quadrilaterals which satisfy both the above conditions will be by-centric having both circum-centre and in-centre. The above conditions are easy to prove and understand. Looking at some of the standard quadrilaterals we can say:

Squares are all bi-centric. Rectangles are only circum-centric. Parallelograms can be only in-centric, as in case of Rhombus. Isosceles Trapeziums are circum-centric and can become bi-centric if opposite sides add up to same value. There are many other quadrilaterals which can qualify to be bi-centric. Another standard quadrilateral ABCD, known as a right-kite is also bi-centric as we will discuss below.


Let us consider the simplest example of a Bi-Centric quadrilateral ABCD, known  as a right-kite, i.e., Angle ABC = Angle ADC = 90 degrees, AD = AB and CD = CB.

I is the centre of the in-circle of radius r. O is the centre of the circum-circle of radius R

‘d’ is the distance between them.

Angles ADC and ABC are 90 deg each and are lying in semi-circles.

Hence, Angles IAE + ICF = θ+ φ = 90 Deg.

Sin2θ + Sin2φ = Sin2θ + Cos2θ = 1

From ∆AIE, Sin2θ = r2/(R-d)2    and   from ∆ICF, sin2φ = r2/(R+d)2

Hence,  r2/(R-d)+   r2/(R+d)2   = 1

(i.e.) 1/r2 = 1/(R+d)2   + 1/(R- d)2

This is again a very handy relationship between R, r and d.

We may write the above in a more interesting way as below

1/ID2 + 1/IB2 = 1/(2r2) +1/(2r2)

= 1/(R+d)2+1/(R- d)2 = 1/IA2 + 1/IC2

For these two circles let us check the Poncelet’s theorem of closure

Now let us construct an arbitrary quadrilateral PQRS inscribing the inner circle. Let PQ and QR be tangents to the inner circle, starting from any arbitrary point P on the outer circle. And let the tangents PS and RS meet at S. For Poncelet’s theorem to be true, S should lie on the outer circle. Draw PI and extend the same to E on the outer circle. Draw RI and extend the same to F on the outer circle. Let angle IPQ = θ and angle IRQ = φ.

∟ EOQ = 2θ and ∟ FOQ = 2φ

∟ EOQ +  ∟ FOQ = 2 (θ +φ)  – – – – – – — – – – – – (1)

Sin θ = r/PI = (r . IE) / (PI . IE) = (r . IE) / (AI . IC)

= (r . IE) / [(R+d).(R-d)]

i.e. Sin2 θ = (r2.IE2) / [(R+d)2.(R-d)2]

Now we know,

1/r2 =  1/(R+d)2 +1/(R-d)2 = 2(R2+d2) / [(R+d)2.(R-d)2]


Hence Sin2 θ = IE2/[2(R2+d2)]

|||ly, Sin2 φ = IF2/[2(R2+d2)]

So, Sin2θ + Sin2φ = (IE2 + IF2)/[2(R2+d2)]

Now, IE2 = EO2 + IO2 – 2EO.IO.Cos(EOI)

= R2 + d2 – 2Rd Cos(EOI)

and   IF2 = EO2 + IO2 – 2FO.IO.Cos(FOI)

= R2 + d2 – 2Rd Cos(FOI)

Hence, Sin2θ + Sin2φ = 1- K[Cos(EOI) + Cos(FOI)] – –  – – (2)

Where K = [Rd/(R2 + d2 )]

∟ EOI + ∟ FOI = 360 – (∟ EOQ +∟ FOQ) = 360 – 2 (θ +φ)

Now, we know θ < 90 Deg and φ < 90 Deg, and (θ +φ) < 180 Deg.

Let us consider three cases in equation 2:

Case-1 :  θ + φ < 90 Deg, and hence ∟EOI + ∟ FOI > 180 Deg

Sin2θ + Sin2φ < 1 and Cos(EOI) + Cos(FOI) < 0, RHS > 1

Case-2 : θ + φ > 90 Deg, and hence ∟ EOI + ∟ FOI < 180 Deg

Sin2θ + Sin2φ > 1 and Cos(EOI) + Cos(FOI) > 0, RHS < 1

Case-3 : θ + φ = 90 Deg, and hence ∟ EOI + ∟ FOI = 180 Deg

Sin2θ + Sin2φ = 1 and Cos(EOI) + Cos(FOI) = 0, RHS = 1

Hence only θ + φ = 90 Deg is admissible.

(i.e.) ∟ SPQ + ∟ SRQ = 180 Deg.

It follows from above, that quadrilateral PQRS is cyclic and hence S should fall on the same circle as PQR.

Thus for any arbitrary initial point P, PQRS will be a closed bi-centric quadrilateral, thus proving the Poncelet’s theorem for bi-centric quadrilaterals also.


In short, let: a = 1/(R+d), b = 1/(R-d) and c = 1/r,

(where R= Radius of Circum-Circle, r = Radius of In-Circle of a bi-centric polygon and d= distance between them)

Then for bi-centric triangle, a+b = c,

for bi-centric quadrilateral, a2 + b2 = c2, and

for bi-centric pentagon,   a3 + b3 + c3  = (a+b)(b+c)(c+a)

or, (a + b + c)3 = 4(a3 + b3 + c3)


For a given R and d, the radius r of the inner circle will keep on increasing from (R2-d2)/2R for a triangle, to (R2-d2)/[√2√( R2+d2)] for a quadrilateral asymptotically approaching R as number sides increases.

With these observations we will end this presentation of bi-centric polygons.