Archive for the ‘Mathematics’ Category


February 27, 2016


Fibonacci – Hemachandra Sequence

Some of my readers will m remember, one Krishnagiri Kittappa, the official percussionist of Oho Productions in the great Tamil romantic comedy of 1960s, Kathalikaa Neramillai (No time for romance). He was initially a self-taught mridangam (Drum) player. He wanted to learn to play Tabla also. He went to a Tabla player to learn the same. He was started on his first lesson, of course, in Teen Tal (or Triputa Tal in Carnatic music) of 8 beats.

Na Din Dinnah – Na Din Dinnah

Na Din Dinnah – Na Din Dinnah

Na Din Dinnah – Na Din Dinnah  . . . . . . .

This went on for quite some time. Our man got bored of playing the same rhythm. It is the same 1,1,2 – 1,1,2 all the time for the 8-beat cycle. Why not 1,2,1, – 1,1,2, he thought.

Din Dinnah Din – Na Dhin Dhinna

Din Dinnah Din – Na Dhin Dhinna

Then, why not 2,1,1-1,2,1

Dinnah Din Din – Din Dinnah Din

Good. Now he further thought about how many such combinations of 1 and 2 (Din and Dinnah), he can make in a cycle of 8 beats. Ancient Indians have already thought about this and so, I gave him the answer as 34 different combinations. He was surprised. So many? How did they calculate?

Ancient Indians always depended on recursive technique in solving such problems. They started from 1-beat, then to 2-beats, 3-beats etc.

No. of Beats (n) Syllables – 1 & 2 Combinations Total Combinations Kn
1 Din 1 1
2 Din, Din






3 Din Din Din

Dinnah Din

Din Dinnah







4 Dinnah (+ 2-beats)

Din  (+ 3-beats)





Now they generalized:
(n+1) Beats Dinnah +  (n-1) beats

Din + (n) beats



K(n+1) = K(n-1) + K(n)


K4 = K2 + K3 = 2 + 3 = 5

K5 = K3 + K4 = 3+ 5 = 8

K6 = K4 + K5 = 5 + 8 = 13

K7 = K5 + K6 = 8 + 13 = 21

K8 = K6 + K7 = 13 + 21 = 34
Hence we have 34 combinations of 1-2 in an 8-beat cycle.

Now look at the series K1, K2, ….. Kn:

1, 2, 3, 5, 8, 13, 21, 34, 55 ……

This is the famous Fibonacci series ‘invented’ by Fibonacci (alias Leonardo Pisano Bogollo) in 13th Century AD. Ancient Indians knew about this, at least, a thousand years before him. Fibonacci himself acknowledges this fact. Fibonacci also helped spread Hindu- Arabic Numerals (like our present numbers 0,1,2,3,4,5,6,7,8,9) through Europe in place of Roman Numerals (I, II, III, IV, V, etc). The olden day knowledge route was from India to Alexandria (Egypt) to Europe.

Susantha Goonatilake (Ref-2) writes that the development of the Fibonacci sequence ” is attributed in part to Pingala (200 BC), later being associated with Virahanka (c. 700 AD), Gopāla (c. 1135), and Hemachandra (c. 1150). Parmanand Singh cites Pingala’s cryptic formula misrau cha (“the two are made together”) and cites scholars who interpret it in the context as saying that the cases for ‘n’ beats (Kn+1) is obtained by adding [Short or 1] to Kn cases and [Long or 2] to the Kn−1 cases. He dates Pingala before 450 BC ”.

“However, the clearest exposition of the sequence arises in the work of Virahanka (c. 700 AD), whose own work is lost, but is available in a quotation by Gopala (c. 1135). The sequence is also discussed by Gopala (before 1135 AD) and by the Jain scholar Hemachandra (c. 1150) “. Fibonacci was born only in 1170 AD.

Prof. Manjul Bhargava (R Brandon Fradd Professor of Mathematics, Princeton University, USA) gave a Lec-Dem on Music & Mathematics at the Music Academy, Madras during their annual conference 2015, on 31st December 2015. Being a Tabla player himself, he dealt with the above aspect of rhythm variations in detail. His talk was the inspiration for me to write this blog.


  3. Manjul Bhargava’s Lec-Dem at the Music Acedemy, Madras (2015)

L V Nagarajan



Meru Prastarah (or Pascal’s Triangle ?!?)

October 21, 2014

Meru Prastarah (or Pascal’s Triangle ?!?)

Let me start with an ancient (1000 CE) Sanskrit text as below:

Anena ekadvyaadilaghukriyasiddhyartham, yaavadabhimatam prathama prastaravat meruprastaram darsayati, uparistadekam chaturasrakoshtam likhitva, tasya adhastat ubhayatordhaniskrantam koshtadwayam likhet, tasyapiadhastatrayam tasyapiadhastatccaturtyamevam yaavadabhimatam sthanamiti meruprastarah tasya prathame koshte ekasamkhyam vyasthapyalakshanamidam pravarttayet, tatra dvikoshtaayaampanktau ubhayo koshtayorekaikamankam dadyaat, tatastritiyaayaam panktau, paryantakoshtayorekaikamankam dadyaat, madhyamakoshtethuparikoshtadvayaankamekikrtya purnam nivesayediti purnasabdarthah, chaturtyampanktavapi, paryantakoshtayorekaikamankam sthapayet, madhyamakoshtayothuparakoshtadvayaankamekikrtya purnam trisamkhya rupam sthapayeth,  uttaraataraapyevameva nyaasah, tatra dwikoshtaayaam pankatauekaakshrasya prastaarah,……. tritiyayaam pankatau, dviakshrasya prastaarah, chaturtyaam pankatau, triakshrasya prastaarah, ….

The above is not in praise of any god of Jain, Budha or Hindu religion. It is not a religious text at all. It is a text describing a method for constructing a mathematical table. Ancient Indian Mathematician Pingala (200 BC) in his Chandahsutra had given the rules for formation of different chandahs (≈ musical meters) for Sanskrit prosody. Another ancient Indian mathematician Halayudha (1000 CE) has given the explanation and commentary on this work by Pingala. Given above is a selected portion of his commentary. For some reasons unknown to me, ancient Sanskrit texts always use composite words very frequently. These words need to be broken into individual words properly to obtain the intended meaning of these words. Here is an attempt to translate the above text into English with proper separation of words.

Anena ekadvyaadi laghu kriya siddhyartham, yaavadabhi matam

(To get every combination of one, two, etc. syllables as required)

Prathama prastaravat meru prastaram darsayati.

(from first row onwards , the meru tabulation will be shown below)

Uparihi tad ekam Chaturasrakoshtam likhitva,

(At the top itself one square cell is drawn)

Tasya adhah tat ubhayato ardhani skrantam

(Below this row let us have a pair, half over lapping)

Koshtadwayam likhet.

(Two cells are drawn)

Tasyapi adhah tat trayam

(Again the row below will have three)

Tasyapi adhah tat chaturtyam,

(Again its next line will have four)

Evam yaavadabhi matam sthanam

(same way, up to the  required stage, cells are constructed)

iti meru prastarah.

(This is called Meru Prastara or Meru-Tabulation)

Tasya prathame koshte eka samkhyam

(Its first stage-cell will hold the number 1)

Vyvasthapya lakshanamidam pravarttayet

(From here on, the following is the way it grows)

Tatra dvikoshtaayaam panktau

(in its twin-cell row)

ubhayo Koshtayoh eka ekam ankam dadyaat

(the pair of cells holds numbers 1,1)

Tatah tritiyaayaam panktau, paryanta Koshtayoh Eka ekam ankam dadyaat

(then in the 3rd row, the extreme cells will hold numbers 1,1)

Madyama koshteth, upari koshtadvayah ankam eki krtya purnam nivesayeth

(middle cell takes the added value of the two cells above)

Iti purnasabdarthah

 (Thus completes the table for 2nd power)

Chaturtyam panktau api, paryanta Koshtayoh eka ekam ankam sthapayet

(then in the 4th row also, the extreme cells will hold numbers 1,1)

Madyama koshtayoth, upara koshtadvayah ankam eki krtya purnam

(middle cells take the added values of the two cells above each)

Trisamkhya rupam sthapayeth

(this completes the 3rd power)

Uttara utaaro api evameva nyaasah,

(next and next stages also follow the same rule)

tatra dwikoshtaayaam pankatau, eka akshrasya prastaarah

(Here the twin-cell row gives one syllable table)

tritiyaam pankatau, dvi akshrasya prastaarah

(the 3rd row gives two syllables table)

chaturtyaam pankatau, tri akshrasya prastaarah

(Thus 4th row gives three syllables table)

And so on.


If we follow the above step by step construction given so clearly by Halayudha (1000 CE), we get the above pyramid or Meru in Sanskrit, (stands for a mountain with a peak). What do we have here? It is the same as Pascal’s Triangle, “discovered” by Blaise Pascal (1623-1662 CE).

This table gives in every nth line the coefficients (a+b)**(n-1). i.e. the second line gives coefficients of (a+b) as 1,1; the second line gives 1,2,1, as coefficients of (a+b)2.; the third line gives 1,3,3,1 as coefficients of (a+b)3 and so on.

However Halayudha gives credit for this table to Pingala (200 BC). He claims to have derived this table from Pingala’s cryptic clue which he translates to a set of rules, as below (with a and b as the two syllables to be combined, in any n-syllable chandah):

  1. First write down all (‘n’ number of)  b’s as the first combination
  2. In the next line, replace the first ‘b’ with an ‘a
  3. At the same line, replace all letters to the left of this new ‘a’ with ‘b
  4. For the next and the subsequent lines repeat the steps 2 & 3.
  5. Continue as above till we arrive at a line with all a’s,

This can be clearly seen as a binomial expansion (a+b)n staring with bn and ending in an. Halayudha later puts these results on a table known as Meru Prasthara. He later gives a step-by-step method as above, for constructing this table without specifically going through the above rules. This Meru Prastarah traveled to China and the Chinese mathematician Yang Hui reported it in the thirteenth century, although his work was unknown in Europe until relatively recent times. The Meru Prastarah traveled to Europe a little later through Arabia, Egypt and Greece and gets “discovered” by Pascal in 17th century CE, 600 years after Halayudha. We are blaming all the time ‘the lack of scientific temper’ among Indians.


  1. Binomial Theorem in Ancient India – By Amulya Kumar Bag, History of Science, Ancient Period Unit II, No.1, Park Street, Calcutta-16 (1966)
  2. Journey Through Genius – The Great Theorems Of Mathematics – by William Dunham – Wiley Science Editions, John Wiley & Sons Inc.(1990)
  3. Probability in Ancient India, by C K Raju.,, 2011.

Connected Topics:

Meru Prastarah

Baudhayana’s Circles

Square Root of Two

Sine of an Angle

LVN/ Oct, 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.


Baudhayana’s Circles

September 21, 2013

Western biographers credit Archimedes of Syracuse (287-212 B.C) with the analytical evaluation of the factor Pi associated with circle, within a close range of 31/7 to 310/71. However in the process of establishing this, they also recognize him as the first person to realize that the same factor is associated with both perimeter and area of the circle. He is also said to be the first person to propose and prove that “Area of the circle = ½ x Perimeter x Radius”. Is it really so? Let us look at another person, one Baudhayana, from ancient India (800BC) who also worked on the circles earlier to Archimedes, by a few centuries.

It was known to Baudhayana, or even to people earlier to him, that the perimeter of the circle depends only on its radius or diameter and that it is actually proportional to the radius or diameter. Though it has not been stated explicitly, it is clear from various sutras that they were well aware that for similar figures, the ratio of the areas equals the square of the ratio of the lengths of the corresponding sides. It was also known that the area of the circle depends only on its radius or diameter and that it is actually proportional to the square of the radius. That is, for a circle, it was known that: – Area = Ka x r2, and Perimeter P = Kp x (2r). But do they know that, Ka = Kp = Pi ?.

During 800 BC, this Indian high priest, Baudhayana, has formulated, in his Sulvasutra (I-48), the so-called Pythagoras theorem, centuries before Pythagoras (572 BC). In another sutra (I-51) he has given a general rule for finding the square root of any number, both geometrically and arithmetically. In his Sutra (I-61) he found the value of √2 to a great accuracy and has given the procedure for the same. This Indian mathematician could construct a circle almost equal in area to a square and vice versa. He has described such procedures in his sutras (I-58 and I-59). All these were achieved in 800BC!

As Baudhayana was designing a religious altar for performing the Hindu rites, he constructed a square within a square as below:


He observed the inner square is exactly half of the bigger square in area.  This led him to formulate, in his Sutra I-48, the so called Pythagoras theorem, which was reinvented by Pythagoras a few centuries later. Please refer to my earlier blog on “Baudhayana’s (Pythagoras) Theorem”.

Subsequently Baudhayana wanted to evolve procedures for constructing circular altars. He constructed two circles circumscribing the two squares shown above. Now, just as the areas of the squares, he realized that the inner circle should be exactly half of the bigger circle in area. Yes, he knows that the area of the circle is proportional to the square of its radius and the above construction proves the same. By the same logic, just as the perimeters of the two squares, the perimeter of outer circle should also be √2 times the perimeter of the inner circle. This proves the known fact, that the perimeter of the circle is proportional to its radius. Now it is known beyond any doubt that for a circle,

Area = Ka x r2, and Perimeter P = Kp x (2r).

But that is not enough. Baudhayana wants to construct circular altars of specific areas. He needs to know the values of Ka and Kp. Baudhayana and his ilk were more interested in the area of the circle than its perimeter.

At this point, Baudhayana would make an important observation. Areas and perimeters of many regular polygons, including the squares above, can be related to each other just as the case of circles. The perimeters and areas of some simple regular polygons are listed below (‘r’ is the distance from the centre of the polygon to its sides):

Equi. Triangle–  Perimeter = K3(2r) & Area = K3(r2); with K3 =  3√3

Square-  Perimeter = K4(2r) & Area = K4(r2); with K4 = 4

Hexagon-  Perimeter = K6(2r) & Area = K6(r2); with K6 =  2√3

Octagon-  Perimeter = K8(2r) & Area = K8 (r2); with K8 =  8(√2-1)

It may also be noticed that the values of the constants, Ki’s, are gradually reducing from about 5 to about 3.3, by the time we reach the octagon. Another fascinating feature with these polygons is: all their areas are ‘½ r’ times their perimeters. Anybody would have been tempted to conclude from above that for circles also, Kp= Ka= K0. This will automatically make Area of the circle = ½ r x (Perimeter). However Baudhayana wanted to prove this.

Let us now consider an N-gon, a regular polygon of N-sides. Let ‘r’ be the distance of the sides from the centre. Let each side be equal to ‘s’. The area of the triangle one side makes along with the centre is (½)sr. Hence the total area of this N-gon is ½Nsr.

So, for N-gon -> Perimeter = Ns & Area = ½Nsr; with Kn = Ns/2r

Here again, Area = (r/2) x Perimeter. In all the above regular polygons, a circle of radius r can be inscribed. Now let us assume that, for this circle, constants Kp and Ka are different. The areas and perimeters of the above polygons are steadily reducing but still remaining more than those of this circle. (i.e.) Kn(2r) > Kp(2r) and Kn(r2) > Ka(r2). Hence any Kn will have to be greater than both K1 and K2. In the N-gon last considered, Ns being the perimeter, it reduces constantly as number of sides N increases. Kn = Ns/2r will also reduce gradually and finally will converge to a finite value, now known as Pi. Baudhayana realized that there is an N-gon with such a perimeter, whose Kn is just more than Kp by any arbitrarily small amount ‘∂’. Similarly Baudhayana concluded that there is an N-gon of such an area, whose Kn is just more than Ka by any small amount ‘ε’.  However both these Kn’s are greater than Kp and Ka. (i.e.) Kp + ∂ > Ka, and,  Ka + ε  > Kp, for any small amounts of ∂’s and ε’s. The above is possible only when Kp=Ka.

Thus Baudhayana concluded, Kp = Ka = Ko, which is now known as Pi. This automatically makes, Area of the circle = ½ r x (Perimeter).

So we may conclude that, the above facts about the circles were already known to Baudhayana and other ancient Indian mathematicians even before Archimedes. Credit is surely due to Archimedes for narrowing down the value of Pi between 31/7  and 310/71. However, Baudhayana on his own has narrowed down the value of Pi to be between 40/12 and 40/13 (i.e. 31/3 and 31/13). It is the value 40/13 he has used in his Sulva Sutra I-58, (for Circling the Square or to find a circle equal in area to a square), as will be demonstrated later. Before going to I-58, let us see how he derived the above values for Pi.

Ancient mathematicians could have already found the value of Pi with limited accuracy, by actual measurements of diameter and perimeter of the circle by using ropes as per the practices existing in those days. It could have given them, at best, a value between 3.11 and 3.17 (i.e. Pi +/- 1%). So, the attempts continued to analytically find the value of Pi.

Even before Baudhayana, the upper limit for the value of Pi was fixed as 4 by considering a circle inscribed in a square. The lower limit was fixed as 3 by considering a regular hexagon along with its circum-circle. But after Baudhayana’s (Pythagoras) theorem, these limits could be narrowed down to be between 3 and 2√3 by considering both circum-circle and in-circle of a hexagon as below.


The circum-radius is ‘a’ and the in-radius is √3/2(a). The perimeter of the hexagon is larger than that of in-circle and less than that of circum-circle. i.e. π(2xa) > 6a and π(2ax√3/2) < 6a. Hence, 2√3 > π > 3.  There is an indirect reference to the value of 3 in an earlier sutra of Baudhayana for constructing altars: “The pits for the sacrificial posts are 1 pada in diameter, 3 padas in circumference.” This gives an approximate value of 3 for Pi. But they knew it is more than 3, as 3 pada is already known as the perimeter for a hexagon inscribed in the above circle. Hence 3 is just the lower limit for value of Pi.

Baudhayana went one step further by considering a regular octagon along with its circum-circle and in-circle. Just as Archimedes found the bounds for value of Pi, by considering 96-gon circum-scribing and in-scribing a given circle, Baudhayana found the bounds, by considering circum-circle and in-circle of an octagon. (yes, a complimentary procedure to what was done by Archimedes, five centuries later). Baudhayana used for √2, the value of 17/12 (normally used in those days) to arrive at these simple fractions, as below. He first considered a square with 12 units as half side. Hence half diagonal will be 17 units considering 17/12 as √2. Referring to the diagram below, the octagon inscribed within this square will have its side as 10.


The in-radius of this octagon will be 12 units and circum-radius will be 13 units. (‘5,12,13’ , the Pythagoras triple, as known even in Baudhayana’s times defines this octagon!). The perimeter of the octagon is 80 units. As this octagon is sandwiched between the circum-circles and in-circle, (ref Figure above), we may compare their perimeters as below:

Pi(2×13) > 80 > Pi(2×12)


Baudhayana’s values of Pi are given by: 40/12 > Pi > 40/13.

Accuracy of Pi was always sought to be improved throughout the history, even after much closer estimates by Archimedes. A later mathematician Manava (650 ~ 300 BC) has stated in his sutra:

Viskambhah pancabhaagasca  viskambhastrigunasca yah.

sa mandalapariksepo na vaalamatiricyate ||      

(Manava  Sulvasutra

(a fifth of the diameter plus three times the diameter, is

the circumference of the circle, not a hair-breadth remains.)

It gives a value of 31/5 as the upper limit for value of Pi, a better value than Baudhayana’s 31/3, and approaching 31/7 of a later day Archimedes: (as per Manava’s statement “not a bit remains” after 3.2 times the diameter).

Baudhayana cleverly used his values of Pi for finding the area of a circle and for drawing a circle of a given area. He found the value of 40/13 to be closer to the eventual value of Pi and has used the same to derive his Sulvasutra I-58 for “Circling the Square”. Baudhayana would have derived this formula as below:

Let ‘a’ be the distance of the sides of the square from its centre. i.e. each side of the square is ‘2a’. To find a circle of radius ‘r’ with an area equivalent to this square, one may write

πr2 = 4a2;        (i.e.), r = (2/√π)a

Square Circle

Baudhayana used 40/13 as value of Pi, and 17/12 as value of √2.

From the picture above we see, a < r < √2a.

Hence he assumed, r = a + (1/x)(√2a-a) = a [1 + (1/x)(√2 – 1)]

For the above square, πr2 = 4a2

r2 = 4a2/π = 4a2 *(13/40) = a2 * 130/100

Hence, r = a * (√130)/10

By Baudhayana’s method (I-51) for finding the square root of any number,

√130 = √(122 – 14)  = 12 – (14/24) = 10 + 17/12

So, r = a(1 + 17/120) = a [1 + (1/3)(51/120)]

≈ a[1 + (1/3)(5/12)] = a[1 + (1/3)(√2 – 1)]

Thus, r = [a + (1/3)(√2a – a)]

So Baudhayana formulates his sulvasūtra I-58 as below:

caturaśraṃ maṇḍalaṃ cikīrṣann

akṣṇayārdhaṃ madhyātprācīmabhyāpātayet |

yadatiśiṣyate tasya saha

tṛtīyena maṇḍalaṃ parilikhet


caturaśraṃ = Square, Mandalam = Circle,

akṣṇayārdhaṃ = Half Diagonal,  

madhyātprācīm = From centre towards east,

abhyāpātayet = laiddown, yadatiśiṣyate = portion in excess,

tasya saha tṛtīyena = using only a third of this, 

parilikhet – Draw around

To make a square into a circle, draw half its diagonal from the centre towards the East; then describe a circle using only a third of the portion which is in excess.

i.e. Using the above formula your are able to draw a circle of given area (=4a2), where a is the measure of half the side of the above square. The radius of this circle is given as:

r = [a+1/3(√2a – a)] = [1+1/3(√2 – 1)] a

Let us see with value of Pi as 40/13, how the areas of square and circle compare.

Area of Square = 4a2

Area of circle = 40/13 x [1 + (1/3)(√2-1)]a2

= (40/13) x (41/36)2 a2  = (67240/16848) a2 = 3.9909 a2

Remarkably close.

As per this construction the value of Pi we obtain as per today’s value of √2 is, Pi = 3.088312

However in the reverse process of squaring the circle, he has gone for corrective fractions as will be demonstrated by Baudhayana’s sutra I-59, as given below:

maṇḍalaṃ caturaśraṃ cikīrṣanviṣkambhamaṣṭau

bhāgānkṛtvā bhāgamekonatriṃśadhā

vibhajyāṣṭāviṃśatibhāgānuddharet |

bhāgasya ca ṣaṣṭhamaṣṭamabhāgonam

viṣkambham = Diameter; aṣṭau bhāgānkṛtvā = making eight parts; 

bhāgaekona = take out one part,

triṃśadhā vibhajya = 29 parts of this part;

āṣṭāviṃśatibhāgānuddharet = of these remove 28 parts; 

bhāgasya ca = from this part also,

ṣaṣṭhamaṣṭamabhāgonam = remove (1/6 minus 1/8 of 1/6).

If you wish to turn a circle into a square, divide the diameter into eight parts and one of these parts into twenty-nine parts: of these twenty-nine parts remove twenty-eight and moreover the sixth part (of the one part left) less the eighth part (of the sixth part). 

The above formula is to make a square of area, equal to a given circle. Baudhayana could have really inverted the earlier formula I-58 to obtain this. But this sutra appears quite complicated. It is so only because Baudhayana in this case used a better value for √2. As per his sutra I-61,

√2 = 1 + (1/3) + (1/3*4) – (1/3*4*34) = 577/408

We may be wondering how handicapped the ancient mathematicians were without the present day decimal point system. However ancient Indians were so facile with fractions they never needed the decimal point system. Even as late as 19th century, Indians were using fractional multiplications tables of ½, ¼, 1/8 , 1/16 , 1/32 and even 3/16  in their every day arithmetic calculations. Baudhayana uses this amazing fractional arithmetic to arrive at the above formula.

We know from I.58, r = a + (1/3)(√2a-a) = [1+1/3(√2 – 1)]a

So we may write, a = r/[1+(1/3)(√2-1)]

With √2 = 577/408,

we get, a = r/[1+(1/3)(169/408)]= (1224/1393)r

Now for fractional magic:

1224/1393 = (1224/1392)*(1392/1393) = (51/58)/(1393/1392)

= [1-(7/56)(56/58)]/[1 + (1/1392)]

= [1-(1/8)(28/29)]*[1- (1/1392)]

= [1-(28/8*29)]*[1 – (1/8*29*6)]

(assuming 28/29 ≈ 1, in the last term),

= 1 – [28/(8*29)] – [1/(8*29*6)] + [1/(8*29*8*6)], .

Thus Baudhayana obtains the final formula of I-59 as:

(With Side of square as ‘s’ and the diameter of the circle as ‘d’)

 s = [1- 28/(8*29) – 1/(8*29*6) + 1/(8*29*6*8)] x d

i.e., s = 0.878682 * d

This increases the value of Pi marginally from 3.088312 to 3.088326.


It was after studying the book “Journey Through Genius” by William Dunham, I got interested in the History of Mathematics. I read several books to know more on this subject. In order to create interest among our genext, I started to write a few blogs on this subject. The present one is on ancient Indian Mathematician Baudhayana (800BC) and his works on Circles. We seem to know him only through his Sulva-sutras. However by reading extensive material on him, I could make out a few narrations of his works. Of course this narration includes a few of my imaginations and intuitions. Ancient Indians in 800BC were well aware of the basic properties of the circle. Baudhayana’s sulvasutras I-58 and I-59 give ample proof of this. Baudhayana was also able to fix the value of Pi to be between 40/12 and 40/13 (i.e. between 3.33 and 3.08).  Baudhayana’s name is still uttered during many Hindu rituals. Even my own family is linked to Baudhayana through his line of disciples as mentioned often by us in our prayers as, Apasthamba, Aangirasa, Baragaspathya and Bharatwaja.


1. A history of Ancient Indian Mathematics – C N Srinivasaiengar, The World Press Private Ltd. Calcutta. (1967)

2. Journey Through Genius – William Dunham, Penguin Books 1990.

3. S.G. Dani, Geometry in Sulvas_sutras, in ‘Studies in the history of Indian mathematics’, Cult. Hist. Math. 5, Hindustan Book Agency, New Delhi, 2010.

Square Root of 2, by Baudayana

June 16, 2013

Square Root of 2, by Baudayana

L V Nagarajan

Baudhayana is a great mathematician of ancient India estimated to have lived during 800BC. He was an expert mathematician, architect, astronomer and a Hindu high priest. He has proposed several mathematical formulas (Sulva Sutras), some of them with proofs. His statement and proof of the so-called Pythagoras theorem is so simple and elegant.

In ancient times, a Square was held as an important geometrical figure. Every area was expressed in so many squares. There was considerable interest in finding an equivalent square for every area, including circle, rectangle, triangle etc.

Baudhāyana, gives the length of the diagonal of a square in terms of its sides, which is equivalent to a formula for the square root of 2:

samasya dvikaraṇī. pramāṇaṃ tṛtīyena vardhayet
tac caturthenātmacatustriṃśonena saviśeṣaḥ

Sama – Square; Dvikarani – Diagonal (dividing the square into two), or Root of Two

Pramanam – Unit measure; tṛtīyena vardhayet – increased by a third

Tat caturtena (vardhayet) – that itself increased by a fourth, atma – itself;

Caturtrimsah savisesah – is in excess by 34th part

In English syntax, it will read as below:

The diagonal of a square of unit measure (is given by) increasing the unit measure by a third and that again by a fourth (of the previous amount). This by itself is in excess by a 34th part (of the previous amount).

That is,

√2 = 1 + 1/3 + ¼ (1/3) = 17/12

But as per the sulba-sutra above, this in excess by a 34th part of the previous amount.


√2 = 1 + 1/3 + ¼ (1/3) – 1/34[(¼ (1/3)]

= 1 + 1/3 + 1/(3*4) + 1/(3*4*34) = 577/408

The above value is correct to five decimals.

There have been several explanations as to how this formula was evolved. Apparently, initially (even on works of late 6th century AD) an approximate value of 17/12 was used for √2, which is nothing but [1+ 1/3 + 1/(3*4)].

1. One theory is they just used actual measurement by ropes to arrive at these fractions – they first tried unit rope length and then 1/2 of the same length. As it was too long they next tried 1/3. It was just short and hence ¼(1/3) was added to it. It was quite close and hence 17/12 was used initially. However it was found slightly longer (savisesah) and when measured by rope again it was found longer by 1/34[1/4(1/3)]. This was also found minutely longer (savisesah) but accepted as a sufficiently accurate value. – That was a simple explanation.

2. Another explanation for the evolution of this formula was based on geometrical construction. – Two equal squares, each with side of one unit were taken. One of the squares was vertically divided into three rectangles. Two pieces of the above was placed along the two adjacent sides of the square to form an approximate square of side 1+1/3, but for missing a small square 1/3×1/3. This was also made up by using a piece from remaining rectangle. Hence we get the first approximate value of (1+1/3) for √2.

Square Root 2

But a small piece is still remaining of size (1/3 x 2/3). This was made into 4 equal strips of size [1/4(1/3) x 2/3]. Two pieces, end to end, were kept along one side of the above augmented square and the other two pieces on the other side. Now we get a total area which is the sum of the two squares. The above augmented square is of side (1 + 1/3 + 1/(3×4)) = 17/12, which was found good enough initially. But, we still miss a small portion of (1/12 x 1/12), to complete the square. Hence the size has to be reduced by this extra area (savisesah). – To find this extra measure, (1/12 x 1/12) should be divided by (17/12 +17/12) and that gives 1/(3x4x34). Of course still we have some minute extra area (savisesah). The above construction is explained in the above figures.

Yes, this is really an interesting explanation. But this amazing formula (sutra) evolved in 800 BC deserves a better explanation which I will offer now.

3. Way back in 1967, I was in a class room of IIT/Kanpur. The teacher was Professor Dr. V. Rajaraman, the pioneer of computer education in India. He was teaching us the basic algorithms for programming in Fortran, a (then) popular programming language. One of the very early recursive algorithms he taught us was, to find the square root of a number. It goes thus:

Let N be the number for which square root is required

Make first guess of the square root as r0 – Later, we will know, the guess may be as bad as 10 times N; still the method works as smoothly as ever.

The next guess can be made as r1 = ½ (r0 + N/r0)

Keep improving this value using the recursion: r(n+1) = ½ [r(n) + N/r(n)]

Surprisingly the value converges very fast to √N, to the required level accuracy.

As you have seen in my earlier blog of Evolution of Sine Table by Hindu Maths, our ancient mathematicians have always preferred recursive steps to solve any problems. Hence, in this case also, Baudayana preferred to use recursive steps, exactly as above. However like others, he preferred to calculate individual step sizes as below:

r(n+1) – r(n) = ½[N/r(n) – r(n)], which is same as the above recursive statement.

To find square root of 2, Boudayana used (1+1/3) = 4/3, as the first guess, r(0).

Hence next step, r(1) – r(0) = ½(3/2 – 4/3) = (3/4 – 2/3) = 1/(3×4)

And hence, r(1) = 1 + 1/3 + 1/(3×4) = 17/12 = 1.4166

Next step, r(2) – r(1) = ½(24/17 – 17/12) = [12/17 – 17/(3x4x2)] =  -1/(3x4x34)

And hence,  r(2) = 1 + 1/3 + 1/(3×4) – 1/(3x4x34) = 577/408 = 1.414216 (correct up to 5 decimals)

Baudayana would have gone to the next minute step also (as his savisesah, indicates), as below

Next step, r(3) – r(2) = ½(816/577 – 577/408) = 408/577 – 577/(3x4x34x2) = -1/(3x4x34x1154)

Hence, r(4) =  1 + 1/3 + 1/(3×4) – 1/(3x4x34) -1/(3x4x34x1157) = 1.41421356237469

The above value is correct up to 13 places.

The recursive algorithm is always the first approach of ancient Indian mathematicians.

Just to satisfy myself that it is not just an isolated case, I tried this logic for finding square root of three also. The step sizes came out to be as below:

Start = 1

First Step = ½             , Root = 1.5

2nd Step = 1/(2×2), Root = 1.75

3rd Step = – 1/(2x2x14), Root = 1.7321429

4th Step = – 1/(2x2x14x194), Root = 1.7320508

The above value is correct up to 7 decimal places.

Again hats-off to Baudayana!

Baudhayana’s (Pythagoras) Theorem

June 12, 2013

                              Baudhayana’s    Pythagoras    Theorem

“Long long ago, so long ago, nobody knows how long ago” – that is how we used to start our stories in our younger days. But this story starts exactly like this.   Long long ago, so long ago, nobody knows how long ago, there lived one Baudhayana, who was an ancient Hindu master. He is dated to have lived during 800BC. He was an expert mathematician, architect, astronomer and a Hindu high priest. Once he was designing a sacrificial alter in the shape of a square. He inscribed another smaller square inside this square as below:


Baudhayana contemplated on this shape and realized the area of the inner square is exactly half the area of the outer square. With the cross-wires drawn as above, it is easy for us also to see this fact.

But the genius of Baudhayana went further. He thought of inscribing an off-set square with in the bigger square as below:


Now he calculated the area of the inner square as:

Area of the inner Square

= Area of the outer square – area of the 4 bordering triangles

= (a + b) – 4 x (ab/2)

(i.e) Area of the inner Square = a2  +  b2

Aaha..! This sounds very familiar. Is this not called Pythagoras Theorem?  But how come, it exists in 800 BC, almost 300 years before Pythagoras (570 -495 BC)? That too found by an ancient Indian? Should we call this then, as Baudhayana’s Theorem. But Baudhayana  proposed  many more such theorems in his Sulva Sutras. His statement of the so called Pythagoras theorem is as below:

 “dīrghasyākṣṇayā rajjuH pārśvamānī, tiryaDaM mānī, 

cha  yatpthagbhUte kurutastadubhayākaroti.”

The above verse can be written again, by separating the combined words and syllables, as below:

“dīrghasya  akṣṇayā  rajjuH – pārśvamānī, tiryaDaM mānī, 

Cha  yat  pthah  bhUte  kurutah – tat ubhayākaroti.”

Below are the meanings of all the words:

Dirgha – Oblong tank or pond

Akshnaya – Diagonally or transversely

Rajjuh – rope

Pārśvamānī = The longer side of the oblong or the side of a square

Tiryak –across, oblique, sideways

Yat (… tat) – Which ( … the same)

Prthah – ( particular) measure

bhūta – become, produce

kurutaha – they (two) do, both do (typical Sanskrit dual verb)

(Yat …) tat –  (Which …) the same

ubhayā – In two ways, two together

Ubhayangkarothi – Produces or effects the two together

Putting the verse in the English language syntax, it reads as below:

In an oblong tank – (what) longer side and (the other) oblique side, the measures (or areas) they produce – (the same) (sum of) both, is effected or produced – by a diagonally held rope.

The natural evolution of this Baudhayana Sutra (Or this Baudhayana Theorem) speaks volumes of its originality. Our salutations to Baudhayana.

In trying to translate this verse into English I was handicapped by two deficiencies – (i) my highly limited knowledge of Sanskrit and, (ii) Non availability of a English-Sanskrit-English technical dictionary.  Such a dictionary is very much a need of the hour, as lot more technical people are now trying to understand and interpret the immense contribution of ancient Indians to Science and Technology. For example in the case of this verse, Deergha, Parsva and Triya may mathematically mean the three sides of a right angled triangle. Experts in this field should take initiative in developing such a technical dictionary for Sanskrit.

Ref : S.G. Dani, On the Pythagorean triples in the ´ Sulvas¯utras, Current Sci. 85(2003), 219-224;

(available at:


L V Nagarajan

12 June 2013

Sine of an angle – by Hindu Maths

December 3, 2012

Sine of an angle – by Hindu Maths

L V Nagarajan

Aryabhata (AD 476-550) was the first in the line of great mathematician-astronomers known to us from the classical age of Indian mathematics and Indian astronomy. His most famous works are the Āryabhaṭīya and the Arya-siddhanta. The following stanza in Āryabhaṭiya gives a series of 24 numbers and calls them as Ardha-Jya differences.

मखि भखि फखि धखि णखि ञखि ङखि हस्झ स्ककि किष्ग श्घकि किघ्व |
घ्लकि किग्र हक्य धकि किच स्ग झश ङ्व क्ल प्त फ छ कला-अर्ध-ज्यास् ||

Makhi Bhakhi Phakhi Dhakhi Nnakhi Nyakhi

Ngakhi Hasjha Skaki Kishga Sghaki Kighva

Ghlaki Kigra Hakya Dhaki Kicha Sga

Jhasa Ngava Kla Ptha Pha Cha kala-ardha-jyas

When decoded into numbers it reads thus:

225, 224, 222, 219, 215, 210; 

                               205, 199, 191, 183, 174, 164

154, 143, 131, 119, 106, 93;  

                               79, 65, 51, 37, 22, 7 : Ardha Jya Differences.

You may imagine a bow. The string tying the two ends of the bow is called Jya (or rope) in Sanskrit. (This sanskrit word Jya, for rope, is the root for Geometry, Geology, Geography, as we know them now). Jya-ardha is half of this length.

Please see the figures below. In Figure-1, ABC is an arc of a circle. AC is its Jya. AM is ardha-Jya or half-Jya. It is seen clearly that half-Jya, AM  is nothing but Sine of angle AOB, multiplied by radius OA.

Considering the Figure-2 above, all the vertical steps from bottom to top add up to respective Jya-ardhas or half-jyas of the increasing angles. In the above diagram, the angles are in steps of 15 degrees. Hence, Step 1 = R Sin 15; step1+ Step2 = R Sin 30; Step 1 +2 +3 = R Sin 45 and so on up to 90 degrees. These steps are called half-Jya differences. The above sanskrit verse gives the step-sizes or half-Jya differences for 24 steps of 3.75 degrees each to add up to 90 degrees. Thus it gives a table RSines for 0 to 90 degrees in steps of 3.75 degrees, with R= R sin90 = Sum of all steps 1 to 24 = 3438. These values are found to be highly accurate with the present day values of Sines, as shown in the table and chart given at the end.

The genius of Aryabhata defined length-MB (refer to Figure-1) as Utkrama-Jya, reverse-Sine or Versine. Aryabhata proposed accumulation of the above Jya-differences in the reverse order to get the successive Utkrama-Jyas, as it is obvious from the above step-diagram where steps are symmetrical about 45 degrees. Hence Koti-Jya or (Cos x) was defined by Aryabhata as (1 – Utkrama-Jya).

Aryabhata has actually devised an algorithm to develop this Sine table. The second section of Āryabhaṭiya, titled Ganitapāda, contains the following stanza indicating a method for the computation of the sine table.

rasi lipthashtamo bhaga: prathamam jya-arda muchyathe

thath dwibhakta labdhon mishritham thath dwitheeyakam

aadyenaivam kramaath pindaan bhaktwa labdhon samyutha:

khandaka: syu: chaturvimsa jya-ardha pinda: kramadami.

There are several ambiguities in correctly interpreting the meaning of this verse. For example, the following is a translation of the verse (by Katz) wherein the words in square brackets are insertions of the translator and not translations of texts in the verse.

“When the second half-chord partitioned is less than the first half-chord, which is [approximately equal to] the [corresponding] arc, by a certain amount, the remaining [sine-differences] are less [than the previous ones] each by that amount of that divided by the first half-chord.”

With my highly limited knowledge of Sanskrit, I can guess a few parts of the above Sutra.

Rasi – 12th part of a full circle (30 degrees).

Ashtomo Bhaga – 8th part (of Rasi), 3.75 degrees

prathamam – First; Jya-ardha – Sine; Muchyate – obtained

Dwibhakta – Double or  Add to itself;

Labdhon – Profit, Dividend, Quotient;

Mishritam – together;  Dwiteeyakam -The second;

aadyenaivam – In a similar way

kramath – Successive; Pindan – difference

chaturvimsa – twenty four; Jya-ardha Pinda – Sine differences;

Kramadami; recursively

Now let me try to give a context-based English translation of the above Sutra:

Arc of an eighth of a rasi gives you the first Jya;

That doubled and divided – together gives the second;

Same way successive Jya-differences, together with quotient,

recurrently give all the twenty four Jya-differences.

There are 12 Rasis in Earth’s trajectory around the Sun. Hence each Rasi is 30 degrees. A quadrant arc of a circle subtends an angle of 90 degrees at the centre, or 3 rasis of 30 degrees each. 8th part of rasi will be 3.75 degrees. There will be twenty four such 3.75 degree arc-sectors in a quadrant, totaling to 90 degrees. An arc (or a bow) of 2 x 3.75 degrees will have a Jya (or a rope). Half of this, is Jya-ardha and corresponds to Sine of 3.75 degrees; which will be same as this arc of 3.75 degrees, as per the Sutra. Yes, in modern mathematics, it only means Sin(x) = x, when the angle is as small as 3.75 degrees. Evidently, a basic circle of certain radius must have been considered for this purpose and hence Jya really means Rsin(x). For some reason (to be explained later) the first Jya is taken as 225 :– 4 x 24 x 225 gives, the total circumference of the circle considered as 21,600, with corresponding radius of 3438.  The value of Jya of the first angle having been initiated as 225, the Jyas of all the twenty four angles can be found following the above Sutra. Interestingly the Sutra above gives different rules for the first two Jyas and prescribes the recursive rule only from Jya3. (Jya is used synonymously with Jya-ardha, which actually represents Sine).

Jya1 = Jya-Diff 1 = 225; Jya2 = (225+225) – (225/225) = 449; Jya-Diff 2 = 224

Jya-Diff 3 (as per the Sutra) =  Jya-Diff 2 – Jya2/225 = 224 – (449/225) = 222.0044; Jya3 = 671.0044

Jya-Diff 4 =  Jya-Diff 3 – Jya3/225 = 222.0044 – (671.0044/225) = 219.0222; Jya4 = 819.0266

The whole table of 24 Jya-diff’s, as developed by using the above sutra is given in a table below. Even this table gives good values for RSines, though not as accurate as the earlier table as could be seen from the plot.

Now let us check this recursive calculation

Difference, D(n) = Sin (n+1)x – Sin nx

= Sin nx Cos x + Cos nx Sin x – Sin nx

Similarly, D(n -1) = Sin nx – Sin (n -1)x

= Sin nx – Sin nx Cos x + Cos nx Sin x

D(n) – D(n-1) = 2 Sin nx Cos x – 2 Sin nx = -2 Sin nx (1 – cos x)

We know,  Jya(n) = J(n) = R Sin nx

Hence Jya Difference, DJ(n) = R * D(n)

Similarly, DJ(n -1) = R * D(n-1)

Hence, DJ(n) – DJ(n-1) = R * [D(n) – D(n-1)] = – 2 J(n) (1- cos x)

i.e., DJ(n) – DJ(n-1) = -J(n)/K,

K being a constant and equals [1/2(1-cosx)]

With x = 3.75 degrees, It works out that K = 233.5374.

If we start the recursive process with J(1) = K = 233.5374, we will get a very accurate table of Jyas, but with R = K x 48/Pi = 3568. However with values of Pi and (Cos x) as available in ancient Hindu period, values of 225 and 3438 for K and R were good enough, as can be seen from the plot below. Or was there a reason to choose these values? The whole circle is 360 degrees. Each degree can be divided into 60 minutes. Now the whole circle is 360×60 = 21600 minutes. Hence perhaps, the radius of the circle was taken as 21600/2Pi = 3438. An arc of 3.75 degrees will be 225 units long.

The ancient Hindus however knew the exact values of Jya for angles of 30, 45, 60 and 90 degrees and Aryabhata could have very well used them to apply corrections to the above table as required. Hence the value of R may not be very critical. Aryabhatta’s table of Sines, given earlier, is the corrected and improved version of the table as developed by his own formula, and hence is much closer to actual values, especially at 30, 45, 60, 90 degrees.

The whole table looks as below:

As per the recursive fromula

Aryabhata’s  Final Table


Angle   x

Quotient   Jx/225

 Ardha Jya Diff RSine(n)  Jya-n  Ardha Jya Diff RSine(n)  Jya-n

R=3438   RSine(n)


0.00     0   0 0


3.75 1.0000 225 225 225 225 224.86
2 7.50 1.9956 224 449 224 449



11.25 2.9822 222.004 671.004 222 671 670.72


15.00 3.9557 219.022 890.027 219 890 889.82
5 18.75 4.9115 215.067 1105.093 215 1105



22.50 5.8455 210.155 1315.248 210 1315 1315.67
7 26.25 6.7536 204.309 1519.558 205 1520



30.00 7.6316 197.556 1717.114 199 1719 1719.00
9 33.75 8.4757 189.924 1907.038 191 1910


10 37.50 9.2822 181.449 2088.486 183 2093



41.25 10.0473 172.166 2260.653 174 2267 2266.83


45.00 10.7679 162.119 2422.772 164 2431 2431.03
13 48.75 11.4405 151.351 2574.123 154 2585



52.50 12.0624 139.911 2714.033 143 2728 2727.55
15 56.25 12.6306 127.848 2841.882 131 2859


16 60.00 13.1427 115.218 2957.099 119 2978



63.75 13.5963 102.075 3059.174 106 3084 3083.45


67.50 13.9896 88.479 3147.653 93 3177


19 71.25 14.3206 74.489 3222.142 79 3256



75.00 14.5880 60.168 3282.310 65 3321


21 78.75 14.7906 45.580 3327.891 51 3372



82.50 14.9275 30.790 3358.681 37 3409


23 86.25


15.862 3374.543 22 3431


24 90.00


0.864 3375.407 7 3438


The above is a complete table of Sines as per ancient Hindu mathematicians. The last column gives the value of Rsine, (i.e. Jya) as calculated using the current accurate values. The closeness of the values can be observed in the following chart.


When I was researching for this blog, I came across the works of late Sri T S Kuppanna Sastri, an expert in Sanskrit and Ancient astronomy. I was naturally feeling proud, since I have met him about 30 years back. He is an uncle of my wife and he is the father of my friend Dr. T K Balasubramanian, a retired scientist of BARC. Sri Kuppanna Sastri was a professor of Sanskrit and Astronomy in several colleges. His renowned major works are two books namely Pañcasiddhāntikā of Varāhamihira and Vedāṅga jyotiṣa of Lagadha.   I dedicate this blog to the memory of late Sri T S Kuppanna Sastri.



2. History of Ancient Indian Mathematics, C N Srinivasiengar, The World Press Private Ltd. Calcutta (1967)

LVN/26 Nov 2012

Evolution of Indo/Arabic Numerals

October 24, 2012

Evolution of Indo/Arabic Numerals

L V Nagarajan

I am not attempting here to give an account of history about development of Number Systems. This is only to wonder how the present Indo/Arabic numerals came in to being. Arabs, the early residents of Arabian Peninsula, are always known to be the link between Europe and ancient India especially in carrying the ancient Indian thought and culture to the elite European community. This is not to say that Arabians themselves were devoid of higher thought and culture. It is the brighter Arabian minds which appreciated the importance of Indian and Oriental contributions and took them to the world along with their own achievements in similar fields. The Arabic numerals are one such great contribution to the world in general, and scientific community in particular. Later day research on ancient India resulted in naming these numerals as Indo/Arabic (or Hindu/Arabic) numerals.

Roman numerals are the earliest system of numerals known to the world. The Romans were active in trade and commerce, and from the time of learning to write they needed a way to indicate numbers. The system they developed lasted many centuries, and still sees some specialized uses today.

seven symbols/letters were used in Roman numerals to indicate following numbers

Roman Numeral Number
I One
V Five
X Ten
L Fifty
C Hundred
D Five Hundred
M Thousand

The list below illustrates how other numbers were constructed using the above 7 symbols/letters:

1 – 10              I,  II,    III,    IV,  V, VI,  VII,    VIII,    IX,   X

10 – 100        X, XX, XXX, XL, L, LX, LXX,  LXXX, XC, C

100 – 1000   C, CC,  CCC, CD, D, DC, DCC, DCCC, CM, M

A string of letters means that their values should be added together. For example, XXX = 10 + 10 + 10 = 30, and LXI = 50 + 10 + 1 = 61. If a smaller value is placed before a larger one, we subtract instead of adding. For instance, IV = 5 – 1 = 4. This is a major difference compared to the modern system. In addition there is no separate symbol for Zero.

The biggest Roman numeral is M, for 1000, so one easy way to write large numbers is to line up the M’s: MMMMMMM would be 7000, for instance. This system gets cumbersome quickly.

The system of numeration employed throughout the greater part of the world today was probably developed in India, but because it was the Arabs who transmitted this system to the West, these numerals have come to be called Arabic. After extending Islam throughout the Middle East, the Arabs began to assimilate the cultures of the peoples they had subdued. One of the great centers of learning was Baghdad, where Arab, Greek, Persian, Jewish, and other scholars pooled their cultural heritages and where in 771CE  an Indian scholar appeared, bringing with him a treatise on astronomy using the Indian numerical system.

Until that time the Egyptian, Greek, and other cultures used their own numerals in a manner similar to that of the Romans. Thus the number 321 was expressed like this:

Egyptian – I nn 999  (Right to left, the Arabic way)

Greek – HHH ÆÆ I

Roman – CCC XX I

The Egyptians actually wrote them from right to left. (Presently with the Indo/Arabic numerals they write numbers only from left to right, may be because of its non-Arabic origin)

The Indian contribution was to substitute a single sign for each cluster of similar signs. In this manner the Indians would render Roman CCC XX I as: 3 2 1. But however CCC I should mean 301, and not 31. Hence the  scholars perceived that a sign representing “nothing” or “naught” was required and Indians are credited with filling this need by inventing the symbol ‘zero’.

If the origin of this new method was Indian, it is not at all certain that the original shapes of the Arabic numerals also were Indian. In fact, it seems quite possible that the Arab scholars used their own numerals but manipulated them in the Indian way. The Indian way had the advantage of using much smaller clusters of symbols and greatly simplifying written computations. Their adoption in Europe began in the tenth century after an Arabic mathematical treatise was translated by a scholar in Spain and spread throughout the West.

Most of the ancient communities traditionally assigned numerical values to their letters and used them as numerals. This alphabetical system is still used by many, much as Roman numerals are used in the West for outlines and in enumerating kings, emperors, and popes. This part of evolution of numerals from letters is not apparently researched and discussed enough. Recently I read an essay about the way the letters of Tamil language were widely used to represent numbers, till as late as 19th century. This might have been the system that existed, since perhaps (not sure?) early Chola period,  9th Century.  This Tamil system of numerals is remarkable not only for using the letters for numerals but also for serving as a precursor for the evolution of modern numerals.

This system of numerals were in use till 19th century in most of the Tamil documents and  records. Tamils used 12 symbols or letters to denote whole numbers 1-9, 10, 100 and 1000. These symbols or letters are given below:.

Numbers Present Digits Tamil Symbols Unicode
One        1        ௧ &#3047
Two        2        ௨ &#3048
Three        3        ௩ &#3049
Four        4        ௪ &#3050
Five        5        ௫ &#3051
Six        6        ௬ &#3052
Seven        7         ௭ &#3053
Eight        8         ௮ &#3054
Nine        9         ௯ &#3055
Ten       10         ௰ &#3056
Hundred      100         ௱ &#3057
Thousand      1000         ௲ &#3058

As these symbols themselves are not important in the present context, I am dealing with Tamil’s number system with known symbols as below

Numerals 1 to 9, X for 10,  C for 100 and M for 1000.

With the above symbols let us see how Tamils wrote other numbers

27 was written as 2X7  –  (More precisely as ௨௰௭)

327 was written as 3C2X7

5327 was written as 5M3C2X7

5307 was written as 5M3C7 – (Symbol for Zero is not used)

234 021 was written as 2C3X4M2X1

1 Million will be  1MM or MM – (Just 2000 in Roman system)

3, 234, 521 will be  3M2C3X4M5C2X1 or (3M 2C3X4)M 5C2X1

1 Billion was written as 1MMM or MMM

and so on.

First let us look at the numbers without a zero in between. If you remove the symbols X,C and M from the sequence it exactly coincides with the present system. This may exactly be the reason why a symbol for zero was invented by Indians because they needed it the most. This symbol zero (0) helped them to totally remove X, C and M from their numerals (but still using their implied presence) . This is perhaps the way the place system of numerals was evolved and gifted to rest of the world. Tamils were definitely a part of the larger Indian Science and Culture and hence this could have been the number system that existed all over ancient India. The ancient Tamil inscriptions just give the evidence of the same. The reason this system was used till 19th century could be the same why English people still call their Queen as Elizabeth II. They find the older systems as authentic in recording history.

It may also be interesting to know that Tamils had letters even for representing about 15 fractions. There are also combination of symbols to represent fractions as low as 1/1,838,400. There are separate names for each of these fractions. After the advent of decimal systems in coinage, weights and measures since 1960s, most of these fractions have gone out of use.

In a lighter vein, there is a poem attributed to Avvayar(?) which mocks at perhaps another inferior poet thus:

௭ட்டேகால் லக்ஷணமே, எமனேறும் வாகனமே

Ettekaal Lakshaname, Emanerum Vahaname

Ettekaal = 8 and 1/4. In Tamil, the symbols are அ and வ

Avalakshaname – Means ‘the ugly one’

Emanerum Vahaname – Lord Yama’s mount, the buffalo

Such pun on number/symbols are aplenty in Tamil Language literature







LVN / 24 Oct 2012



March 23, 2012

Starting from triangle, quadrilateral and pentagon, there are infinite number of polygons. Polygons with all sides equal and all internal angles equal are called regular polygons and there are infinite numbers of regular polygons with 3 to infinite numbers of sides. In the limit the regular polygon becomes a circle.

Extending to the third dimension, there are infinite numbers of polyhedrons, starting from tetrahedron. Regular polyhedrons are those whose faces are all regular polygons congruent to each other, whose polyhedral angels are all equal and which has the same number of faces meet at each vertex. An interesting fact is that there are only five regular polyhedrons: the Tetrahedron (four faces of equilateral triangles), the Cube (six square faces), the Octahedron (eight equilateral triangular faces—think of two pyramids placed bottom to bottom), the Dodecahedron (12 regular pentagonal faces), and the Icosahedron (20 equilateral triangular faces). In this write-up, I want to concentrate on Tetrahedrons, (not necessarily regular ones), as an extension of triangle to three dimensions.

A tetrahedron is some times described as a triangular pyramid. It consists of four triangular faces and four trihedral angles. (Trihedral angles are discussed in good detail in my earlier blog). Even though three intersecting planes are enough to make a trihedral angle, they still do not make a closed polyhedron. A fourth non-parallel plan is needed to get a polyhedron of least order – 4 faces, 6 edges and 4 vertices; this is known as a Tetrahedron.

Let us consider the tetrahedron ABCD as above, for our further discussions, with BCD as the base triangle and A as the peak point. The areas of triangular faces may be denoted by Sa for triangle BCD opposite to vertex A,  by Sb for triangle CDA opposite to vertex B,  by Sc for triangle DAB opposite to vertex C and by Sd for triangle ABC opposite to vertex D. The dihedral angle between faces Sb and Sc and around the edge AD, may be represented by Φbc and respectively the other dihedral angles, Φac, Φad etc.

Volume of a tetrahedron

If we consider the tetrahedron as a triangular pyramid, with a triangular base of area S  and a height h, the volume of the tetrahedron can easily be derived as V = 1/3 S*h (just as area of the triangle is expressed as A = ½ (Base length)(Height)).

A Tetrahedron OABC is also described in terms of three non-co-planar vectors OA(a), OB(b) and OC(c). In this case the volume may be stated as V = 1/6th (a.b x c). (i.e. One sixth of the volume of the parallelepiped a.bxc).

If Tetrahedron OABC is given in terms of lengths as (a, b, c) and angles between them as (α, β, γ) then, the above expression for volume can be expanded and simplified as:

V = (1/6) abc*√(1 + 2 cos α cos β cos γ  – cos2α – cos2 β – cos2 γ)

Two triangles ADB and ADC, with AD as the common edge, and Φbc as the dihedral angle between them, can fully describe the tetrahedron ABCD. Volume of this tetrahedron can be formulated as:

V = 2/3 (Area ADB * Area ADC) (Sin Φbc)/AD;

(or V= 2/3 Sc Sb Sin Φbc/AD)

Given opposite edges, AB and CD and the shortest (perpendicular) distance, EF(=d), between them, we can calculate the volume of the tetrahedron as below. Let Φ be the dihedral angle between the planes ABEF and CDEF. Then the volume of tetrahedron ABCD, V= (1/6) (AB) (CD) (EF) Sin Φ.

Law of Cosines and Sines

Law of cosines for a triangle ABC states: a2 = b2 + c2 – 2bc cos A, where A is the angle between sides ‘b’ and ‘c’, opposite to side ‘a’.

Law of sines for a triangle ABC states: (Sin A)/a = (Sin B)/b = (Sin C)/c = 2S/abc, (where S is the area of the triangle).

Let us derive similar relationships for a Tetrahedron

In my earlier blog on Trihedral angles we have talked briefly about the Pythagoras relation for a right angled tetrahedron. We started off from a semi-right-angled tetrahedron as below: (AD is perpendicular to BCD). Mark an X such that DX and AX are perpendicular to BC.

Consider triangle BCD. We may write,

BC2 = BD2 + CD2 – 2.BD.CD.cos Φ (law of cosines for triangle DBC)

Multiplying both sides by AD2

AD2 * BC2 = AD2 * BD2 + AD2 * CD2 – 2. (AD.BD)(AD.CD). cos Φ

(i.e.) (AX2 – DX2) * BC2

= AD2 *  BD2 + AD2 *  CD2 – 2.(AD.BD)(AD.CD).cos Φ

(i.e.) (2Sd)2 – (2Sa)2 = (2Sc)2 + (2Sb)2 – 2(2Sc)(2Sb) cos Φ,

where Sj is the area of triangular face opposite to vertex j.

(i.e.) Sd2 = Sa2 + Sb2 + Sc2 – 2.Sb.Sc.cos Φ

This is the law of cosines for tetrahedron and in a more general form it is given as:

Sd2 = Sa2 + Sb2 + Sc2

                  – 2.Sa.Sb.cos Φab – 2.Sb.Sc.cos Φbc – 2.Sc.Sa.cos Φca .

In addition to Φab = Φca = 90 degrees as above, if Φbc is also 90degrees (i.e.) for a fully right-angled tetrahedron (right solid-angled at D), we get the Pythogoras relation as,

Sd2 = Sa2 + Sb2 + Sc2

For any trihedral angle Ω, G-sine Ω can be defined as, (refer to my earlier blog on Trihedral Angle)

G-sin Ω = Vp*Vp/ (Product of adjacent areas),

where, Vp is the volume of parallelepiped formed by the arms of the trihedral angle and areas are of adjacent parallelograms. Using this, for a tetrahedron ABCD we can write,

G-sine A = (Vp)2/(2Sb*2Sc*2Sd)

We know that, Volume of a tetrahedron, V, is 1/6th the volume Vp of the extended parallelepiped. Hence,

G-sine A = (6V)2/(2Sb*2Sc*2Sd) = 9/2 (V)2/(Sb*Sc*Sd)

Similarly, G-sine B = 9/2(V)2/(Sc*Sd*Sa), and so on.

Hence we can write,

(G-Sine A)/Sa = (G-Sine B)/Sb = (G-Sine C)/Sc = 9/2 V2/(Sa.Sb.Sc.Sd)

Law of G-Sines for any tetrahedron can be simply written as

(G-Sine A)/Sa = (G-Sine B)/Sb = (G-Sine C)/Sc

      = (9/2) V2/(Sa.Sb.Sc.Sd), where V is the volume of the tetrahedron.


Centres of Tetrahedron

A triangle can be circumscribed by a unique circle known as circum-circle, which passes through all the three vertices of the triangle. The centre of this circle is known as circum-centre. Same way a triangle can inscribe a unique circle known as in-circle, which is tangential to all the three sides of the triangle. The centre of this circle is known as in-centre. In an equilateral triangle the circum-centre and the in-centre coincide to be the same point.

Just as a circle can be drawn through any three non-co-linear points in a plane, a sphere can be constructed to go through any four non co-planar points in space. Hence any tetrahedron can be circumscribed by a unique sphere known as circum-sphere, which passes through all the four vertices of the tetrahedron. Perpendiculars to each of the four triangular faces from their respective circum-centres are concurrent at the centre of circum-sphere, i.e. circum-centre of the tetrahedron.

However the tetrahedron may inscribe two kinds of spheres – one sphere which is tangential to all the four faces of the tetrahedron, and another sphere which is tangential to all its six edges. They are called in-sphere and mid-sphere respectively with their own in-centre and mid-centre.

In-Centre will be equidistant from all the four triangular faces. Consider a tetrahedron ABCD. Consider the dihedral angle Φcd around the edge AB. Consider a plane ABX bisecting this angle. Every point on this plane will be at equal distance from planes ABC and ABD. Similarly consider a plane ACX bisecting the dihedral angle Φbd around the edge AC. These two planes will intersect each other along a straight line AX through A. Any point on this line AX will be at equal distance from all the three planes forming the trihedral angle Ωa at A. One more plane, constructed to bisect the dihedral angle Φad around the edge BC, will intersect the line AX at a point X. This point X will be equidistant from all the four planes of the tetrahedron ABCD. And hence, X will be the in-centre of the In-Sphere which will be tangential to all the four faces of tetrahedron ABCD. Effectively, the in-centre of a tetrahedron is the point of coincidence of all the six planes, bisecting the six dihedral angles around the six edges of the tetrahedron.

Mid sphere itself can be visualized as a sphere jutting out of all the four faces but tightly caged by all the six edges of the tetrahedron. Mid centre will be equidistant from all the six edges. A bisector of a trihedral angle is defined as the locus of points that are equidistant from its (three) edges. Such bisectors of all the four trihedral angles of the tetrahedron will be concurrent at the mid centre. But do the mid-sphere and mid-centre exist for all tetrahedrons? This aspect is not discussed much in the existing literatures and hence I have discussed this in detail below. (This is a result of my own research)

Mid-sphere and Mid-centre of a tetrahedron

Perpendiculars to each of the four triangular faces from their respective in-centres, if they are concurrent, form the centre of the mid-sphere, i.e. mid-centre. Let us consider a tetrahedron ABCD, having a mid sphere tangential to all the six edges.  There will be three tangents to the mid-sphere from A. Let their lengths up to the points of tangency on the sphere be a1, all equal as they should be. Similarly on other three tangents each from B, C and D, let these equal lengths be b1, c1 and d1. We can easily observe,

AB = a1+b1; Similarly, CD = c1+d1; Hence,       AB + CD = a1+b1+c1+d1;

Similarly, AC = a1+c1; BD = b1+d1; and hence, AC + BD = a1+c1+b1+d1;

Similarly, AD = a1+d1; BC = b1+c1; and hence, AD + BC = a1+d1+b1+c1;

We see that, the Mid-sphere exists for a tetrahedron only if lengths of each of the three pairs of opposite edges add up to the same value.

Let us see whether the converse is true. Consider two triangles ADB and ADC, with AD as the common edge, and Φbc as the dihedral angle between them, which can fully describe the tetrahedron ABCD.

AB and CD form one pair of opposite edges of the tetrahedron. AC and BD will form the second pair and AD and BC (when folded across AD) will form the third pair. Let the lengths of each of these pairs of edges add up to the same value, k. (i.e.) AB+CD = AC+BD = AD+BC = k. Let the circles 1 and 2 shown above, be the in-circles of the respective triangles. Let a1 and d1 be the lengths of tangents from A and D to circle-1. Similarly, let a2 and d2 be the lengths of tangents from A and D to circle-2. Same way, the tangents from B and C are also shown above as b1 and c2.

Now, AB + CD = a1 +b1 +c2 +d2 = k.

Same way, AC + BD = a2 +c2 +b1 +d1 = k.

Hence (a1 +d2) = (a2 +d1). This is possible only when circles 1 and 2 touch each other at a common point on AD. This common point and the in-centres of 1 and 2 will all lie in a plane perpendicular to AD, This means, the perpendiculars from the in-centres of 1 and 2 to the respective faces will meet at a point, say Om. Similarly considering other pairs of opposite edges, we may conclude that each adjacent pair of the four in-circles on the four faces of the tetrahedron will be tangential to the respective common edge at the same point. And hence the perpendiculars to the respective faces from their in-centres will all be concurrent at the mid-centre, Om, which will be equidistant from all the edges. Hence we have a mid-sphere also.

Hence a tetrahedron ABCD will have a mid-sphere if and only if the lengths of each pair of opposite edges add up to the same value. Any tetrahedron with equilateral triangle as the base and with other three edges of equal lengths, easily satisfy this requirement and they do have mid-spheres and mid centres. There can be many others also satisfying this condition and having a mid sphere. For a regular tetrahedron, all the three centres, i.e. , Circum-centre, Mid-centre and In-centre will coincide.

The centre of gravity (Centroid) of a tetrahedron is well defined. A median of a tetrahedron is defined as a line from any vertex to the centroid of the opposite triangle. All the four medians of a tetrahedron are concurrent at the centroid of the tetrahedron. Consider the three straight lines connecting the respective mid points of opposite edges of the tetrahedron. It is interesting to note that the centroid is also the point of concurrence of all these three bi-medians, as they are called. The centroid divides all the medians in the ratio of 3:1 (just as ratio of 2:1 for triangles) and divides the bi-medians equally.

However ortho-centre is not defined for a tetrahedron as the four orthogonal lines need not always be concurrent.


Tetrahedron has really turned out to be an interesting solid. There are many more interesting features which may hopefully come out as comments from my readers. Perhaps my next blog could be on parallelepiped (or hexahedrons, in general).

Additional Reference:



There was a request from one of my readers, Shivangi verma

I want a solution for this question
1.Show that the three lines joining the mid points of edges of a tetrahedron meet in a point which bisects them?

Plz provide me with the solution asap…

Here is my proof.

Centre of Gravity CG of a tetrahedron

In the following figures, tetrahedrons are shown like a open triangular book. The first one is shown open along the edge CD. In the second one same tetrahedron is shown open alone the opposite edge AB.

N and M are the midpoints AB and CD, the opposite edges of the tetrahedron. G1 is the cg of ∆ACD and G2 is the cg of ∆BCD, the two adjacent faces of the tetrahedron. |||’ly, in the other figure, G3 and G4 are cgs of ∆s CAB and DAB, the other two adjacent faces of the tetrahedron. The medians of tetrahedron, AG2 and BG1 intersect at Ga, on plane AMB. This is known as cg of the tetrahedron. |||’ly, Gc the cg of tetrahedron as seen from C and D, will be on the plane CND. We are required to prove:

  1. Ga and Gc coincide
  2. They do so, at the midpoint of ‘Bi-median’ MN.
  3. We are also required to prove all the three ‘Bi-medians’ are concurrent at their midpoints which is also the cg of the tetrahedron.
  4. In the process we will also prove that CG of tetrahedron divides each median in the ratio 1:3.

Please refer to the following  of ∆ABM, extracted from the above diagram of the tetrahedron.

Now, Area ∆AMN’/Area ∆BMN’ = Area ∆AGaN’/Area ∆BGaN’ = AN’/BN’

Then it follows, Area ∆AMGa / Area ∆BMGa = AN’/BN’   ——- 1

|||’ly,   Area ∆AMGa / Area ∆ABGa = MG2/G2B = 1/2   ——- 2

And     Area ∆BMGa / Area ∆BAGa = MG1/G1A  = 1/2   ——- 3

G1 and G2 being the centroids of triangular faces, we know MG2/G2B = MG1/G1A = ½,  as written above, and G1G2 is ||lel to AB.

Dividing Eqn-2 by Eqn-3,  We get, Area ∆AMGa / Area ∆BMGa = 1

i.e. AN’ = BN’. (i.e.) N’ coincides with the midpoint N.

Hence Ga lies on Bi-Median MN.

|||’ly, by extracting ∆NCD  from the second diagram of tetrahedron, We can prove Gc, the CG of tetrahedron as seen from C and D will also be on the Bi-Median MN.

Hence We have proved that (a) all four medians of tetrahedron from its vertices A, B, C and D are concurrent on the Bi-Median MN.

Again from Eqns 2 and 3, we know, Area ∆AMGa = Area ∆BMGa = ½ (Area ∆ABGa)

Therefore, Area ∆ABGa = ½ (Area ∆ABM)

Hence we may conclude, Ga is the midpoint of Bi-Median MN

|||’ly, from ∆NCD, We can conclude, Gc also is the midpoint of Bi-Median MN

Hence Ga and Gc coincides with G0, the midpoint of MN, which is the CG of Tetrahedron.

Hence we have proved the CG of tetrahedron coincides with midpoint of the Bi-Median MN.

Similarly if we consider AC and BD, the CG of tetrahedron (obtained from the pairs of medians from A,C and B,D) will again coincide with the midpoint of the line joining the mid-points of edges AC and BD).

Similarly if we consider AD and BC, the cg of tetrahedron (obtained from the pairs of medians from A,D and B,C) will again coincide with the midpoint of the line joining the mid-points of edges AD and BC).

This clearly proves both the facts:

  1. All the 4 medians from the four vertices of tetrahedron are concurrent at one point known as CG of tetrahedron, which also is the concurrent midpoint of all bi-medians.
  2. All the three bi-medians from the three pairs of opposite edges are concurrent at their midpoints which also is the CG of tetrahedron

Now, draw A’B’ parallel to AB through Ga. As Ga is the midpoint of NM, A’ and B’ also become midpoints of AM and BM respectively.

i.e., MA’/A’A = 1. But we know, MG1/G1A = ½

For simplicity, let us assume, MA= 6k.

Then MG1 = 2k and G1A = 4k as G1 is a centroid. And MA’ = 3k as A’ is the midpoint.

Hence, G1A’ = 1k and A’A =  3k.

i.e.,  A’ divides  G1A in the ration 1:3. |||’ly, B’ divides  G2B in the ration 1:3.

Hence we may conclude, Ga also divides the medians BG1 and AG2 in the ration 1:3. Same will be true of Gc, with other medians DG3 and CG4.

Phew! That was interesting!

– L V Nagarajan 27 June 2017




Trihedral Angle

October 15, 2011

Trihedral Angle

L V Nagarajan


Solid geometry has always fascinated me and has fired up my imaginative power. I was introduced to Solid Geometry by my elder brother (L V Sundaram) when I was barely 10 years old. In his university days, he had a habit of reading aloud and explaining to himself as a part of his study process. When I am around as a young boy, he will do the explaining to me, not caring how much of it I understood. But he was rather surprised that I understood a lot of what he told me especially in geometry. As I learnt more of plane geometry, I developed a habit of extending the results of the theorems to the third dimension. Subsequent to my earlier blog on Polyhedrons, I want to now share my research on trihedral angle, as it is the basis of all solid angles

Normal Plane Angles

I am re-stating some fairly obvious facts about plane angles, since we need the same to compare with the three dimensional angle, known as Trihedral Angle. In a plane if two straight lines are drawn, they intersect each other at some point. But when they are parallel to each other, they maintain constant distance between themselves all along their infinite lengths in both directions. Hence they never intersect each other. Two non-parallel straight lines in a plane, intersecting each other, have a rotational distance between them. When one of the straight lines is rotated about the point of intersection, it will coincide with the other straight line after some rotation. This rotational distance between the straight lines is known as Angle. One full rotation is divided into 360 parts for the purpose of quantifying the amount of rotation, each part known as a degree. Consider a circle with O as the centre and R as the radius. As the radius rotates a complete round it completes a circumferential distance of 2πR. Hence a full rotation of 360 degrees is associated with 2π Radians. Any arc of the circle of length L subtends at the centre O an angle of L/R radians. When two straight lines in a plane intersect each other they actually create four angles, with opposite angles of equal magnitudes. It is easy to see that the adjacent angles add up to 180 degrees or half rotation. These four angles are actually the four internal angles of a parallelogram! We may also easily see that two intersecting straight lines in a plane divide the plane into four quadrants. If the straight lines are perpendicular (90 degrees) to each other, they divide the plane into four equal quadrants. The lines themselves become X-axis and Y-axis of the coordinate system.

Trihedral Angle

Now let us move to the third dimension. Consider two planes parallel to each other. They maintain a constant distance between them all along their infinite surface. They never intersect each other. But two non-parallel planes do intersect each other, at some place, along a straight line. They do have a rotational distance between them. This is similar to plane angle and this rotational distance is known as dihedral angle, as the angle between two planes, varying from 0 to 360 degrees (or 2π radians). Now let us consider a third plane non-parallel to either of the earlier two planes. Let this plane intersect each of the earlier two planes along two different and non-parallel straight lines. All these three planes (and their three straight lines of intersection) will meet at a common point. At this point is formed a solid angle, known as trihedral angle. In general an angle represents a rotational distance. Similarly, the trihedral angle also is a measure of the sum of rotational distances, in two directions, i.e., rotations about two of the edges of trihedral intersections. Hence some time it is called as square-angle or square degrees. Trihedral angles vary from 0 to 720 square degrees. Consider a sphere with O as the centre and R as the radius. As the radius rotates a complete round in a, say, vertical plane, we get a vertical circle. When this circle rotates a complete round about any diameter as axis, we get a complete sphere with a surface area 4πR2, covered by rotation. Hence a full rotation of 720 degrees is associated with 4π Stradians. Now consider a solid angle subtended by a partial spherical segment at the centre. It is given as surface area of the segment divided by radius squared. When three planes intersect each other in space, they actually create eight trihedral angles. These eight trihedral angles are actually the eight solid angles of a parallelepiped! We can also see that three intersecting planes divide the 3-dimensional space into eight octants. If these three planes are perpendicular each other, they divide the 3-dimensional space into eight equal octants. The lines of intersections, some time called as edges, form the X-Y-Z axis of the 3-demensional co-ordinate system.

Other angles of trihedral angle

Though we are talking of three planes making a trihedral angle, we feel comfortable to talk of trihedral angle as formed by three (non-planar) straight lines meeting at a point. Hence we talk about vertex angle as the plane-angle between any two edges at the vertex of the trihedral angle. Dihedral angles are the angles between adjacent planes making up any trihedral or polyhedral angles. The solid angle (or any polyhedral angle), the dihedral angle and the vertex angle are normally denoted as, Ω, Φ and θ. It is interesting to see below, how these angles relate to each other, say, for the case of a trihedral angle.

From the theory of spherical excess, it has been derived and stated that, the solid angle at O as enclosed by vectors A, B and C as above, is given by Ω = Φa + Φb + Φc – π

Let us consider the solid angle Ω1, at O, enclosed by OA, OB and OC. The dihedral angle Φa is as shown. This is the angle between two semi circular faces of the orange-like dice ABA1CA. This piece of spherical surface is sometimes known as di-angle. The spherical surface area of this dice is Φa/2π times the whole surface area of the sphere. Hence the solid angle subtended by this area at the centre O is Φa/2π times 4π, i.e., 2Φa.

Hence supplementary solid angle Ω2, enclosed by OA1,OB and OC is given by 2Φa –  Ω1.

We may infer similar results with OB1 and OC1 and get

 Ω2 = 2Φa – Ω1

 Ω3 = 2Φb – Ω1

 Ω4 = 2Φc – Ω1

It is easy to see that Ω1+ Ω2 + Ω3 + Ω4 = a hemisphere = 2π

 i.e. 2(Φa + Φb + Φc – Ω1) = 2π

i.e. Ω = Φa + Φb + Φc – π

This proves the result we got earlier. Ω1, Ω2, Ω3 and Ω4 are the four solid angles of the parallelepiped obtained from OA,OB, OC. The other four solid angles will be equal to their opposite angles. These eight solid angles form the octants in the three dimensional space as mentioned earlier.

Trihedral Angle Ω, in terms of vertex angles α, β and γ

In a parallelogram described by sides ‘a’ and ‘b’ and angle θ, the area is given by (ab sin θ). Referring to the diagram below:

We can write tan θ/2 = b sin θ/ (b + b cos θ)

i.e., tan θ/2 = ab sin θ/ (ab + ab cos θ)

                    = Area of Parallelogram/ ab (1+cos θ)

i.e., tan θ/2 = (a x b)/(ab + a.b)

We have a similar relationship for a parallelepiped described by vectors and angles:  a, b, c, α, β and γ,  given by Oosterom and Strackee (2) as:

Tan(Ω/2) = (a . b x c) / [ abc + (a.b)c + (b.c)a + (c.a)b]

It is easy to see that (a . b x c) represents the volume of the parallelepiped a, b, c.

Volume V = (a . b x c)

     = Det (a / b / c), with  vectors a, b, c represented as row vectors.

     = Det (a , b , c), with  vectors a, b, c represented as column vectors

 i.e., V = a . b x c = Det [a1,a2.a3/ b1,b2,b3/c1,c2,c3]

                                = Det [a1,b1,c1/a2,b2,c2/a3,b3,c3]

Hence, V*V = Det [a.a, a.b, a.c/ b.a, b.b, b.c/ c.a, c.b, c.c]

It will reduce to

V*V =

    a2 b2 c2 (1 + 2 cos α cos β cos γ  – cos2α – cos2 β – cos2 γ )

The solid angle Ω does not depend on magnitudes of vectors a, b, c and hence assuming them to be unit vectors, we get the expression for solid angle Ω, as

tan (Ω/2)

 = (√ (1 + 2 cos α cos β cos γ  – cos2α – cos2 β – cos2 γ)) / (1+cos α+cos β+cos γ)

Now we have expressed Ω in terms of vertex angles α, β and γ. We have also earlier expressed Ω in terms of dihedral angles Φa, Φb, Φc. Later in this write-up, we will also find how α, β and γ relate to Φa, Φb and Φc.

‘Tetragonometric’ ratios of trihedral angle

When we talk of tan (Ω/2) as above, we are not talking of actual trigonometric ratio of trihedral angle Ω. We are only treating the square-degree Ω, as normal angle of the same magnitude and then expressing its trigonometric ratio in the above formula. But do we have ‘trigonometric ratios’, like Sine, Cosine, Tangent, for trihedral angles also? (May be, we should call them as ‘tetragonometric ratios’)

Let us go back to our comparison with a parallelogram.

Area A of the parallelogram a, b, θ, is given by ab sinθ

So we say, Sin θ = A/ab. i.e., Area/(product of adjacent lengths)

Similarly, in case of parallelepiped, (a, b, c, α, β, γ), G-sin Ω has been defined as,

G-sin Ω = V*V/ (Product of adjacent areas)

i.e., G-sin Ω  = V*V/  (ab sin γ)(bc sin α)(ca sin β);

We know V*V =

     a2 b2 c2 (1 + 2 cos α cos β cos γ  – cos2α – cos2 β – cos2 γ )


G-Sin Ω =

(1+2 cosα cosβ cosγ -cos2α -cos2β -cos2γ) /(sinα  sinβ sinγ)

Volume V1 of a unit parallelepiped with angles α, β, γ and with (a=b=c=1) is given by,

V1 = √(1+2 cos α cos β cos γ  – cos2α -cos2 β -cos2 γ)

Hence, we can say,

G-Sin Ω = (V1*V1)/ (sin α sin β sin γ);

It is  easy to see that all the eight solid angles at the eight corners of the parallelepiped will have the same G-sine value, as the same three parallelograms are enclosing them, but in a different order. This is exactly as in the case of a parallelogram where all the four angles have the same sine value.

This G-sine value appears to signify nothing, especially when you see this value not being consistent. Even as two sets of α, β, γ produces same solid angle Ω, their G-sine values will vary. For a particular value of Ω, G-sine will be maximum when α = β = γ.

Let us approach G-sine from a different perspective. We know trigonometric ratios of any angle are derived from a right angled triangle that includes this angle. Please refer to the following diagrams.

Sin θ = BC/AC; Cos θ = AB/AC; Tan θ = BC/AB

Any trihedral angle at O can be represented as a part of a semi-right-angled tetrahedron OABC as above.

Angles OCA and OCB are right angles. i.e. OC is perpendicular to plane ACB.

Angle AOC (= β) and angle COB (=α) are independent and, along with the dihedral angle Φ, completely defines the trihedral angle at O. Angle AOB (=γ) and angle Φ depend on each other. Let OC = c. Now from the definition of G-sin we get,

G-Sin O = V*V/ Product of adjacent areas

V (of parallelepiped) = (AC*BC Sin Φ) * OC

Product of adjacent parallelograms

                      = (OC*AC sin β)*(OC*BC sin α )*(OA*OBsin γ)

(as α =β = 90 degrees);          = (AC*BC*OC*OC)*(OA*OBsin γ)

G-Sin O = AC*BC Sin2 Φ / (OA*OB sin γ)

            = (2*Area ABC/ 2*Area ABO) Sin Φ

i.e. G-Sin O = (Area ABC / Area ABO) * Sin Φ

i.e G- Sin O = (Opposite area/ Hypotenuse area) * Sin Φ

In the special case when Φ = 90 degrees, we get,

G sin O = (Opposite area/ Hypotenuse area)

This is very similar to trigonometric ratio for plane angle. It is easy to see we require a special set of angles α, β, γ (with Φ = 90 deg) to produce a right angled tetrahedron. In all other cases, the relationship,

G- Sin O = (Opposite area/ Hypotenuse area) * Sin Φ, holds good.

For a right angles tetrahedron, it can be easily proved that, the sum of the squares of adjacent areas will be equal to the square of hypotenuse area. This is very similar to Pythagoras theorem of right-angled triangles. Using this property, we can define other ‘tetragonomertic’ ratios for trihedral angles as below:

G-Cos O = (√(Sum of the squares of the two adjacent areas)) / Hypotenuse area.

G-Tan O =  Opposite area / (√(Sum of the squares of adjacent areas))

We can also write G-Sin2 O + G-Cos2 O = 1

We can also note:

G-Sin2 O + G-Sin2 A + G-Sin2 B = G-Sin2 C = 1.

Please note that this right tetrahedron is right-solid-angled at C (i.e. π/2 stradians or 90 sq. degrees).

Other interesting relationships

We started this section with the fact that any trihedral angle can be represented as a part of semi-right-angled tetrahedron as below.

We derived the relationship

G- Sin O = (Opposite area/ Hypotenuse area) * Sin Φc

Now (AC = OA Sin AOC = a sin β) and (BC = OB Sin BOC = b sin α)

Therefore Area ACB

    = (1/2)AC.BC sin Φc = (1/2)(a sin β) (b sin α) Sin Φc

Similarly Area AOB = (1/2)OA.OB. Sin γ = (1/2)a.b. Sin γ

G-Sin O = (Area ACB / Area AOB)* Sin Φc

                 = (Sin α. Sin β. Sin2 Φc) / Sin γ

By extending this logic, we can write,

G-Sin O    = (Sin α. Sin β. Sin2 Φc)/Sin γ

                   = (Sin β. Sin γ. Sin2 Φa)/Sin α

                   = (Sin γ. Sin α. Sin2 Φb)/Sin β

Thus we have established a comprehensive relationship for any trihedral angle in terms of vertex angles and dihedral angles.

Using the above along with the expression for G-Sin O,

G-Sin O = (V1*V1) / (sin α sin β sin γ); we get,

Sin Φa = V1/( sin β sin γ),  Sin Φb = V1/( sin γ sin α) and

      Sin Φc = V1/( sin α sin β)

Remember V1 is volume of unit parallelepiped α, β, γ, with a = b = c =1 given by:

V1 = √(1 + 2 cos α cos β cos γ  – cos2α – cos2 β – cos2 γ )

Once the values of Φa, Φb, Φc are calculated as above, we can find each of the eight solid angles in the eight octants formed by α, β, γ planes. Of course these are same as the eight solid angles of the parallelepiped (a,b,c, α, β, γ). We already know that

Ω1 = Φa +  Φb + Φc – π  = Ω

Ω2 = 2Φa –  Ω1

Ω3 = 2Φb –  Ω1

Ω4 = 2Φc –  Ω1

These solid angles are repeated again as the diametrically opposite angles, thus forming all the eight trihedral angles of the parallelepiped. It may be observed that the above four trihedral angles add up to 2π. All eight trihedral angles add up to 4π.

As stated earlier, G-Sine of all the eight solid angles of the parallelepiped are equal.

G-Sin Ω1 = G-Sin (2Φa –  Ω1)

      = G-Sin (2Φb –  Ω1) = G-Sin (2Φc –  Ω1).

Now let us tabulate all these angles for different types of trihedral angles.






Sq Deg




































































































Types of trihedral angles are as listed below

Type Type of trihedral angle


@ Centre of  a Hemi-sphere


@ Centre of a Reg. Tetrahedron-4


@ Centre of Reg. Octahedron- 8, Vertex of a Cube, one octant of a sphere, Right-angled trihedral angle


@ Centre of reg. Icosahedron – 20


Vertex of a reg. Dodecahedron-12


Vertex of a reg. Tetrahedron – 4


Semi Right angled trihedral angle


Semi Right angled trihedral angle


Semi Right angled trihedral angle


Equilateral Trihedral angle


Isosceles Trihedral angle


Typical acute trihedral angle


The sites given in the references had triggered my interest in this topic of trihedral angle. However most of the concepts developed in this write-up are the results of my own research. I hope this will be useful for students of geometry and for others who are doing further research on the application of solid angles.  



2. A. VAN OOSTEROM, J. STRACKEE: A solid angle of a plane triangle. – IEEE Trans. Biomed. Eng. 30:2 (1983); 125-126.