Re: polygons circumscribing a circle
- From: The World Wide Wade <aderamey.addw@xxxxxxxxxxx>
- Date: Mon, 16 Jul 2007 16:37:48 -0700
In article <1184610934.877197.106720@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Musing <generalizing@xxxxxxxxxxx> wrote:
How does one show that the perimeter of a regular polygon
circumscribing a circle is greater than the circumference (of the
circle)? Thanks.
Here's a calculus proof, working on the unit circle. Suppose 0 < b <
1. The line from (0,0) to (1,b) intersects the circle at a point
(sqrt(1-a^2), a). We want to show the arc-length from (1,0) to
(sqrt(1-a^2), a) along the circle is < b.
Similar triangles show b/1 = a/sqrt(1-a^2). The arc in question is
parameterized by (f(y),y), 0 <= y <= a, where f(y) = sqrt(1-y^2). The
calculus definition of the arc-length is int_[0,a] sqrt(1 + f'(y)^2)
dy. A simple computation shows sqrt(1 + f'(y)^2) = 1/sqrt(1-y^2). So
we are trying to show
a/sqrt(1-a^2) > int_[0,a] 1/sqrt(1-y^2) dy.
But the integrand is a strictly increasing positive function of y, so
its integral is less than the length of the interval times its value
at the right-hand end point, namely, a*1/sqrt(1-a^2).
.
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