Polygon circumscribed about circle has higher perimeter than circle? It is quite clear that any (not necessarily regular) convex polygon inscribed in a circle has a perimeter which is lower than that of the circle: between any two adjacent vertices of the polygon the length of the circle segment between these two points exceeds that of the line segment, hence the same inequality holds for the entire perimeter.
But what about circumscribed polygons? I know that for regular polygons the circle always has a lower perimeter (although I can't think of an elementary argument for that either), but what about general ones?
 A: It depends what you are allowed to assume. If you are allowed that a circle is the rectifiable curve of given total length which encloses the greatest area then the circumscribed polygon encloses a strictly greater area than the circle and therefore must have greater length.
This, of course, assumes a result which isn't entirely trivial. I would be interested to see a more elementary proof (or a reason why such doesn't exist - it is not trivial to define length or area). But I wanted to give a reason which depended on a fact which could be simply stated, even if not simply proved.
A: For a polygon $A_1\ldots A_n$, let's call $B_i$ the point where it touches the circle (let $O$ be its centre). ($B_i$ is on $A_iA_{i+1}$, $B_n$ is on $A_nA_1$). Let $C_i$ be the point where $OA_i$ intersects the circle. Note $OB_i=r$ (the radius) for all $i$. Now we have:
$$B_iA_i=r\tan(\angle B_iOA_i)$$
$$\stackrel{\Large\frown}{B_iC_i}=r(\angle B_iOA_i)$$
Thus, $B_iA_i\gt \stackrel{\Large\frown}{B_iC_i}$ because of the (known) inequality $\tan x\gt x$ for $0\lt x\lt \frac{\pi}{2}$.
Similarly, $B_iA_{i-1}\gt \stackrel{\Large\frown}{B_iC}_{i-1}$, for the same reason.
Add up all those inequalities to obtain the desired inequality.
