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Ancient greeks algebra

  1. May 17, 2007 #1
    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 ?
     
  2. jcsd
  3. May 17, 2007 #2
    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
     
  4. May 17, 2007 #3

    Hurkyl

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    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.
     
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