# Thread: Geometric "Pi Segement Construction" algorithims?

1. ## Geometric "Pi Segement Construction" algorithims?

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

Thanks in advance

2. scroll down to III Exercises/Solutions , part D

What They Don't Tell You About Pi in High School

3. ## 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

4. Hello, mfetch22!

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

We can always construct a $\displaystyle 30^o$ 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 $\displaystyle O$ and radius 1.
$\displaystyle AB$ is a diameter.
Line $\displaystyle PQ$ is tangent to circle $\displaystyle O$ at $\displaystyle B.$

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

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

Draw segment $\displaystyle AF.$

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

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Proof

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

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

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

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

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

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