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I have found a proof of the existent of the constant $\pi$ by means of integration. But i could not really understand the proof of this. Is there any elementary way a 11 grader student like me can understand the proof? The proof i saw uses high level integration and arcsin arccos etc.

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    $\begingroup$ You might want to look here for a discussion of how Archimedes proved it. There are many other references on the web as well. $\endgroup$ – rogerl Apr 28 at 15:51
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    $\begingroup$ Specifics, please! We can't help you understand the proof without knowing which proof you're referring to. $\endgroup$ – Cameron Buie Apr 28 at 15:51
  • $\begingroup$ Archimedes proof $\endgroup$ – user607476 Apr 28 at 15:53
  • $\begingroup$ Is this what you're asking: math.stackexchange.com/questions/3198/…? $\endgroup$ – Hans Lundmark Apr 28 at 15:57
  • $\begingroup$ Can you explain the proof of arrchimefes $\endgroup$ – user607476 Apr 28 at 15:58
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You have asked an interesting and deep question.

The fact that the ratio of the circumference of a circle to its diameter is the same for all circles depends on assumptions you make about how geometry works. It's essentially equivalent to the fact that the plane is "flat".

A proof that the ratio is the same for all circles in the Euclidean plane starts with thinking about similar triangles. It's not hard to show that the ratio of the perimeter of a triangle to one of its sides, or to one of its altitudes or to any other linear construction is the same for a pair of similar triangles. Then to prove the same for a circle you have to approximate the circle by polygons. That's Archimedes's proof. The ideas behind calculus come in there in the discussion of the approximation.

On the sphere there are no similar triangles. Any two triangles which have the same angles are congruent. If you try to calculate the ratio of circumference to diameter for circles on a sphere you will find that for larger circles "$\pi$" is smaller.

http://mathforum.org/library/drmath/view/57828.html

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  • $\begingroup$ If you tried defining "a constant $\pi$" from a sphere, you'd probably consider $S/r^2$ with $S$ the surface area, rather than a perimeter-to-diameter ratio for a shape that, however derived it may be from a sphere, is a curve in a plane. $\endgroup$ – J.G. Apr 28 at 16:03
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Hint: $arccos x + arcsin x = \pi/2$. I am sure this was probably the place which was troubling you. This is because $cos x = sin(\pi/2-x)$. Hope it helps you..

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  • $\begingroup$ Yeah thanks! I had trouble in this only.. $\endgroup$ – user607476 Apr 28 at 16:25

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