History of Mathematics Project | Babylonian Nested Triangle Tablet

History of
Mathematics

Project

around 2000–1900 BCE

Babylonian Nested Triangle Tablet

Triangular band divided into trapezoids

This Babylonian tablet gives a figure of two nested equilateral triangles divided into three congruent trapezoids, with numbers annotated along various edges and the number 35 prominently written in cuneiform in the lower right. It is likely this tablet was an unfinished student exercise to compute the area of the band.

The annotations in the figure indicate various lengths. In particular, the inner triangle side length is indicated with the cuneiform glyph for 1 (which actually stands for the number 10—an interpretation that must be inferred due to the lack of a positional zero in Babylonian number notation), the trapezoid side lengths as the glyphs 1 6 4 (meaning 16 + 40/60 = 50/3), and the length of the long trapezoid base as the glyphs 4 3 2 (meaning 43 + 20/60 = 130/3). The number 35 written at the bottom is the central median length of the trapezoid, which gives the trapezoid's area when multiplied by its height.

Artifact dimensions

7.1 cm diam. × 2.5 cm thick

Original artifact location

Babylonia (historical name), Iraq (current name)

Current artifact location

Oslo, Norway

Catalog number

MS 2192

Timeline

Interactive Content

Computational Explanation

This tablet considers a triangular band (or "annulus") consisting of an equilateral triangular boundary. Call the outer edge length

and the inner edge length

, and visualize this "polygon with a hole":

In[]:=

With[{a=1,b=.3,n=3,off=1.5},Graphics[{(*polygon*)LightGray,EdgeForm[Directive[Thick,Black]],Polygon[Rule@@Table[CirclePoints[sCsc[Pi/n]/2,n],{s,{a,b}}]],(*text*)Black,Text[Style["a",Italic],{0,-aCot[Pi/n]/2},{0,off}],Text[Style["b",Italic],{0,-bCot[Pi/n]/2},{0,-off}]}]]

Out[]=

To draw the triangular annulus with correct dimensions, use has been made of the formula for the circumradius

of an

-sided regular polygon having edge length

In[]:=

R[n_][a_]:=aCsc[π/n]/2

where

csc(x)

sin(x)

is the cosecant function. With knowledge of this formula, the area of the annulus can be found directly in the Wolfram Language as:

In[]:=

With[{n=3},Area[RegularPolygon[R[n][a],n]]-Area[RegularPolygon[R[n][b],n]]]//Simplify

Out[]=

(

)

More generally, the area for the

-sided annulus can be given by the beautiful formula:

In[]:=

(Area[RegularPolygon[R[n][a],n]]-Area[RegularPolygon[R[n][b],n]])//Simplify

Out[]=

(

)nCot



A different way to compute the area is to dissect the triangular annulus and add up the contributions from each piece. The triangular annulus can be split into congruent trapezoids in several ways. The tablet uses a dissection in which the inner edges are extended to the right until they touch the outer triangle:

Call the length of the extension

and examine its relationship with the inner and outer edge lengths:

Out[]=

The yellow auxiliary equilateral triangle in the diagram reveals that the long base of each trapezoid has length

b+2c

, the short base has length

b+c

and the edge length of the outer triangle is

ac+(b+c)+cb+3c

. Solving the latter for

immediately gives

c(a-b)/3

The area of the triangular annulus can therefore be found from these values together with the formula for the area of a trapezoid

A

(

)hmh

, where

m

(

)

is the length of the central median and

is the height.

There are many numeric annotations associated with line segments in the diagrams. Fig. 8.2.2.a in Friberg (2007, p .204) gives a schematic of the cuneiform digits together with their translation:

In particular, the tablet annotates the edge length

of the interior triangle as 1 (which actually means 10 since the Babylonians did not have a positional zero; the "implied zero" is denoted using a degree sign ° by Friberg):

In[]:=

ResourceFunction["AncientNumberRepresentation"][10,"Babylonian"]

Out[]=

… the length

of the extension as 1 6 4 (meaning 16; 40  16 + 40/60  50/3):

In[]:=

ResourceFunction["AncientNumberRepresentation"][50/3,"Babylonian"]

Out[]=

… and length

of the long base as 4 3 2 (meaning 43; 20  43 + 20/60  130/3):

In[]:=

ResourceFunction["AncientNumberRepresentation"][130/3,"Babylonian"]

Out[]=

While not indicated on the tablet, the length

b+c

of the short base is given by 80/3  26 + 40/60 (which would be written in cuneiform numbers as 2 6 4). Finding the central median length

m=(

)/2

of the trapezoids then gives:

In[]:=

(b1+b2)/.b1

130

,b2->



Out[]=

Surely not coincidentally, this number corresponds to the symbols in the lower-right corner of the tablet:

In[]:=

ResourceFunction["AncientNumberRepresentation"][35,"Babylonian"]

Out[]=

Since the area of a trapezoid can be found using the formula

, the presence of the numerical value of

exactly corresponding to the trapezoids in the tablet suggests an unfinished solution to the problem, lacking only the calculation and corresponding multiplication by the trapezoid height

to complete the determination of the area.

Note that a more symmetric, easy-to-generalize and natural dissection of the regular

-gonal annulus can be obtained for arbitrary

by connecting corresponding vertices of the inner and outer triangles:

Out[]=

This dissection also gives the area of the annulus by taking

times the area of a single trapezoid. However, this time, the base lengths and height are much easier to find than in the preceding case. To find the area of one of the trapezoids, first note that the formula for the inradius of a regular

-gon of edge length

is given by:

In[]:=

r[n_][a_]:=aCot[π/n]/2

where

Cot[x]1/Tan[x]

is the cotangent function. As can be seen in the following figure showing the incircles of inner and outer triangles, the trapezoid height (shown in red) is simply the difference between the inradii of the two triangles:

Out[]=

Since the area of a single trapezoid is given by

(

)h

(a+b)h

, the height is given by

, and there are

trapezoids, the total area of the annulus is therefore:

In[]:=

(a+b)(r[n,a]-r[n,b])//FullSimplify

Out[]=

(a-b)(a+b)nCot



Plugging in

n3

gives the same area as originally obtained:

In[]:=

(

)nCot

/.n3

Out[]=

(

)

Other Resources

Wolfram Demonstration
Babylonian Numerals

Wolfram Demonstration
Pythagorean Theorem

Additional Reading

Friberg, J. A Remarkable Collection of Babylonian Mathematical Texts: Manuscripts in the Schøyen Collection: Cuneiform Texts I. New York: Springer, pp. 202–205, 2007.
Friberg, J. "Matematiska kilskriftstexter i den norska Schøyensamlingen." Nordisk Matematisk Tidskrift, Vol. 52, pp. 150–151, 2004.

Babylonian Nested Triangle Tablet

Timeline

Interactive Content

Computational Explanation

Other Resources

Wolfram Demonstration

Wolfram Demonstration

Additional Reading

Additional Links