that square root problem

At the BAPL 09 banquet Johnny Ball demonstrated a geometric method to determine a square root. A couple of us couldn't quite believe it at the time. Devon came up with a verbal description next morning.

This is what I make of it.

pythag.jpg

In the diagram:

BC=1                   ⍝ by construction
AD=DC                  ⍝ by construction
AD=DE                  ⍝ by construction
Prove:
BE=AB*0.5              ⍝ BE is the square root of AB
Proof:
(BE*2) = (DE*2)-DB*2   ⍝ Pythagorus
       = (DE+DB)×DE-DB ⍝ difference of 2 squares
       =   AB   ×  1
.'. BE = AB*0.5
QED

Does anyone know if it's in Euclid?-- PhilLast 2009-06-10 12:05:14

CategoryBapla09 CategoryMaths

GeometryProblem (last edited 2009-06-20 06:17:13 by KaiJaeger)