From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: pi=3,sort of
Date: 6 Aug 1996 16:10:09 GMT

In article <vertu-0208961745300001@cnc80271.concentric.net>,
Michael O <vertu@concentric.net> wrote:
>A spherical contact lens has the following properties:
>
>the circumference = 3
>the maximum distance across the curve = 1
>
>If the lens were thought of as a slice off a larger sphere, what is the
>radius of that larger sphere?
>What is the radius and altitude of the lens?

Where are all the college students who usually answer things like this?

If you slice a sphere of radius  r  at a distance  h  from the origin,
you'll indeed get a shape like a contact lens. The cut edge is a circle
whose radius is  sqrt(r^2-h^2). The angle subtended by the cut edge is
twice arccos(h/r), and so the length across the curve of the lens will
be  2 r arccos(h/r) = 2 r arcsin( sqrt(r^2-h^2) / r ).

From  the data given, we thus have
	r^2-h^2 = (3 / 2pi)^2
	3/ (2pi r) = sin( 1 / 2r )
We can use the first equation to solve for  h  once we get  r. The
second equation will give us  r  but we have to solve it numerically
(rather than expecting a 'closed form'). Maple tells me that the solution 
to this last is about  r = 0.9549, from which I deduce  h=0.8270.

dave