# 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?

• More generally, the perimeter of a convex curve is larger than that of an enclosed convex curve. That is easy to prove for polygons, then pass to the limit (depending on the formalism you use to define the arc length). – dxiv Dec 26 '17 at 20:57

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.