Verified answer

I guess this is not any close to what you hoped for but

If you take a unit circle C, an interior n-regular polygon P_i and the corresponding exterior polygon P_e,

then L_n = (2|P_i| + |P_c|) /3 gives a very good approximation to pi. To obtain something of the same order of accuracy as 22/7, you just need to take dodecagons.

You can explicitely construct a dodecagon of exactly that length L_12 by taking an intermediate dodecagon with a ratio 1/3,2/3.

You get L_12 = 3.142349 as opposed to 22/7 = 3.142857.

By comparison L_24 = 3.141639. Since L_(2^n) converges quickly to pi, this shows that 22/7 is close to pi.

Now if you want to "see 22/7" on any picture related to the unit circle, I can only wish you the best of luck!

edit To contradict my previous statement, let's start with a regular unit hexagon. You glue on each side an isosceles triangle of sides (1,11/21,11/21). You have now a semi-regular dodecagone of length exactly 44/7 and you wonder how close is the length of this dodecagone to 2*pi.

Conversely you may wonder the following: start with a triangle ABC with angles (pi/6, pi/2, pi/3) with AC = 1. Draw the unit circle centered at A. it goes through C and interesects the line (AB) at a point D. How can one choose E on AB so that the length of CE and the arc length CD will be closest? And more specifically is CE = 11/21 a good candidate?

If you consider CB and CD since the accuracy of the approximation is quadratic, you expect

| CD - L(CD)| = 1/4 | CB - L(CD)| since you go from one approximation to the next by halving the angle. Now L(CD) is what you want to reproduce (namely pi/6), so you should shoot for

EC - CB = 4 (EC - CD) which implies 3 EC - 4 CD + CB = 0 or

EC = (4 CD - CB)/3

You get EC = (8 sin 15 -1/2) / 3 = 0.523517 while 11/21 = 0.523809.

at the pi level you get 3.141104 vs 3.142857

So, do you prefer this?

It is easier to geometrically approximate pi as 3: If we draw a circle of radius 1, and divide it into k wedges with angle theta = 2pi/k each, then the triangle approximation for the area of each wedge is:

area of triangle = (1/2)base*height = (1/2)(2cos(theta/2))(sin(theta/2))

So the total area of the triangles is kcos(theta/2)sin(theta/2) = (k/2)sin(theta) = (k/2)sin(2pi/k). As k goes to infinity, (k/2)sin(2pi/k) goes to pi. But if we choose k=12, then sin(2pi/12) = sin(pi/6) = 1/2, and so the area of 12 triangles inside the circle is (12/2)(1/2) = 12/4 = 3. This is easy to draw and you can shade the regions that represent the "error" when approximating pi by 3.

You can get a better approximation by taking k=20, so sin(2pi/20) = sin(pi/10) = (sqrt(5)-1)/4, which gives total area of 20 triangles of 3.0901... (still not very accurate). You can get higher values of k for which sin(2pi/k) can be exactly computed by doing the half-angle formula, but this will just give you more and more complex square roots.

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And here is a joke answer: You can first draw a circle of radius 1. Then around that, draw a circle of radius sqrt(22/(7pi)). They will look pretty much the same and the latter has area 22/7. =)

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Wolfram says that sin(2pi/96) = (1/2)sqrt(2 - sqrt(2 + sqrt(2+sqrt(3)))):

http://www.wolframalpha.com/input/?i=sin%282pi%2F9...

I'm guessing that is from multiple instances of the half-angle formula. From that we get the approximation of pi as the area of 96 triangles, being (48/2)sin(2pi/96)=3.13935...

I always eat pie on July 22.

http://en.wikipedia.org/wiki/Pi#Polygon_approximat...

http://itech.fgcu.edu/faculty/clindsey/mhf4404/arc...

Sad to see that Google is dead.. oh wait, that's what I used to find both those links.

22/7 is the ratio of the circumference of a circle with a diameter of the circle.

22/7 its simplest form. If you have a rope with a length of 44 cm so when fitted into a certain circle of diameter 14 cm, so the ratio becomes 44/14 or 22/7 or 3.14 .... esc