How did ancient Greeks use geometry to solve algebraic equations?

  • Context: Undergrad 
  • Thread starter Thread starter Yann
  • Start date Start date
  • Tags Tags
    Algebra
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
2 replies · 3K views
Yann
Messages
48
Reaction score
0
I've heard ancient greeks were able to solve algebra using geometry.

For example, if you have a line between two points AC (B being right in the middle), can you find a point x so;

AB/AX = AX/AC

if a is length of AB, then AB = a and AC = 2a, so the problem is really about;

a/x = x/a2; 2a^2 = x^2; 2^(1/2)a = x, if a = 1 then x = 2^(1/2).

How could they find the point x ?
 

Discussion

Mathematics news on Phys.org
Are you talking about the golden ratio? Where the large length is to the smaller segment as the whole line is to the large part? Let x = long part, and L = total length: x/(L-x) =L/x
 
Yann said:
AB/AX = AX/AC
Cross ratios are easy! Note that's the same as:

AB * AC = AX * AX

So, you just make a line segment:

C ----- D ------------ E

with CD = AB and DE = AC. Then you construct a circle with CE as its diameter. Draw the perpendicular at D, which intersects the circle at points F and G. Note that DF and DG are congruent.

Now, you have two chords intersecting, and thus you have:

DC * DE = DF * DG

and so

AB * AC = DF * DF

So DF is the length you seek.