Integrating history of mathematics in a classroom

Pythagorean Knowledge in Babylonia

Pythagorean theorem was found to be used by babylonian centuries before Pythagoras

The Phlimpton 322 Tablet





It contains a highly sophisticated sequence of integer numbers that satisfy the Pythagorean equation a2+b2=c2, known as Pythagorean triples.


Babylonian Problem on circle segment:.

 The following problem and the solution were found on a Babylonian tablet dating from about 2600BC:

Problem:

60 is the Circumference, 2 is the perpendicular, find the chord.






Thou double 2 and get 4
Take 4 from 20, thou gettest 16
Square 16, thou gettest 256
Take 256 from 400, thou gettest 144

Whence the square root of 144, 12 is the chord.

Let's write this in using algebra, with” C ” being the length of the circumference, ”c  ” being the length of the chord and ”s”   being the perpendicular.


Thou double 2 and get 4

That's 2s.

Take 4 from 20, thou gettest 16

20 is C/3. So this is C/3 - 2s.

Square 16, thou gettest 256

That makes (C/3 - 2s)²

Take 256 from 400, thou gettest 144

400 is (C/3)² So we now have (C/3)² - (C/3 - 2s)²

Whence the square root of 144, 12 is the chord.

So they get c = sqrt( (C/3)² - (C/3 - 2s)² ).



Can we derive a formula of chord AB in terms of the sagitta s and circumference C, using Pythagoras’ theorem?








Draw the perpendicular bisector DC of chord  AB, it will be perpendicular to the cord and bisect it.
Then, we have a right triangle BCD, with BD = c/2, CD = r - s, and BC = r

Using Pythagorean formula on triangle BCD,

      BD² = BC² - CD²
     (c/2)² = r² - (r -s)² = (C/(2π))² - (C/(2π) - s)².        

If we multiply by 4 on both sides, we get:

      c²= (C/π)² - (C/(π) - 2s)² or

Take the square root of both sides and we have:

      c= sqrt(C/π)²- (C/π) - 2s)²).



Now we can take the square root of both sides and compare our answer to the Babylonians':

Modern day: c=sqrt( (C/π)² - (C/π - 2s)²).

Babylonians: c=sqrt( (C/3)² - (C/3 - 2s)² ).

As we can see they use a 3 in place of π. It means they assumed C ≈ 3d . So


c=sqrt(d²-(d-2s)²)          Formula 1

And


s= (d- √(d²-s²))/2        Formula 2



Extension: The Susa Tablet (1936 CE)


Problem on the Susa tablet:
A circle centered at O has an isosceles triangle ABC inside, with AC=AB=50, and CB = 60.

Can you find the radius of the circle (AO) using pythagorean theorem?
Hint: Draw AD which it is perpendicular to CB.



ABD is a right angled triangle so, using Pythagoras's theorem:
AD2 = AB2 - BD2
  AD = 40

Let the radius of the circle by x.  Then AO = OB = x and OD = 40 - x.

Using Pythagoras's theorem again on the triangle OBD we have:



x2 = (40-x)2 + 302
x2 = 402 - 80x + x2 + 302
  x =  31.25