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Math Help - Geometric "Pi Segement Construction" algorithims?

  1. #1
    Member mfetch22's Avatar
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    Geometric "Pi Segement Construction" algorithims?

    Can anybody explain to me some of the methods used to approxiamtely construct a line segement of length \pi ????

    Thanks in advance
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    scroll down to III Exercises/Solutions , part D

    What They Don't Tell You About Pi in High School
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    construct a line segment equal to pi (approx)

    Dear mfetch22,
    Pi is approx 22/7. Draw a line = to 22 inches mark AB. At A draw a ray of an acute angle. Starting at A scribe with a compass 7 equal segments ending in C. Connect BC and then draw lines parallel to BC thru each scribe mark intersecting AB.Each segment on AB = pi


    bjh
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    Hello, mfetch22!

    Can anybody explain to me some of the methods used
    to approxiamtely construct a line segement of length \pi ?

    We can always construct a 30^o angle.
    Construct an equilateral triangle and bisect one of its angles.

    Code:
                    A
                  * o *
              *     |  o  *
            *       |     o *
           *        |1       o
                    |           o
          *         |         *    o
          *         *O        *       o
          *        /|         *          o
                1 / |                       o
           *     /30|        *                 o
            *   /   |1      *                     o
              */    |     *                          o
      P - - - + - * * * - - + - - - - - - + - - - - - - o - - Q
              C     B       D             E             F
              : - -  1  - - : - -  1  - - : - -  1  - - :

    We have a circle with center O and radius 1.
    AB is a diameter.
    Line PQ is tangent to circle O at B.

    Construct segment OC so that \angle COB = 30^o

    From C, mark off three radii on PQ.
    . . CD = DE = EF = 1

    Draw segment AF.

    We have: . AF \;\approx\; \pi.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Proof

    In right triangle OBC\!:\;\tan30^o \:=\: \frac{CB}{1} \quad\Rightarrow\quad CB = \frac{1}{\sqrt{3}} \:=\:\frac{\sqrt{3}}{3}

    Then: . BF \:=\:CF - CB \:=\:3 - \frac{\sqrt{3}}{3}


    In right triangle ABF\!:\;\;AF^2 \:=\:AB^2 + BF^2 \;=\;2^2 + \left(3 - \frac{\sqrt{3}}{3}\right)^2

    . . AF^2 \;=\;4 + 9 - 2\sqrt{3} + \frac{1}{3} \;=\;\frac{40}{3} - 2\sqrt{3} \;=\;\dfrac{40 - 6\sqrt{3}}{3}


    Therefore: . AF \;=\;\sqrt{\dfrac{40-6\sqrt{3}}{3}} \;\approx\; 3.141533339


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    From: .Mathematical Puzzles and Pastimes, Aaron Bakst
    . . . . . . . . D. Van Nostrand Company, Inc., 1954

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