A simple proof that a polygon circumscribing a circle overestimates its perimeter

Question

Looking at the picture below, it's easy to see why the perimeter of a polygon inscribed in a circle is an underestimation of the circle's perimeter. This follows from the triangle inequality: Any side (say $AB$) of the polygon is shorter than the circular arc with the same endpoints ($\stackrel{\frown}{AB}$). Summing all these inequalities shows the perimeter of the inscribed polygon is indeed smaller than that of the circle.

I'm wondering if there is proof that the perimeter of a circumscribed polygon always overestimates the perimeter of the circle, which is as simple as that of the inscribed polygon case. Thanks!

Answer 1

Pick a point $F$ on one of the arcs of the circle, let us say it's on arc that faces $C$ in your circumscribed quadrilateral. Construct a line segment $GFH$ with $G$ on $BC$ and $H$ on $CD$, tangent to the circle at $F$. $GFH$,being a straight segment, is shorter than $GC+CH$, so the circumscribed pentagon $ABGHD$ has less perimeter than the quadrilateral $ABCD$. Keep adding sides to the polygon by drawing additional tangents and the polygon perimeters will constitute a strictly monotonic decreasing sequence. So the terms of that sequence must be greater than the limiting value which is the circumference of the circle.

Answer 2

You may use a general fact:

If $A,B\subset\mathbb{R}^2$ are two convex bounded shapes and $A\subset B$, the perimeter of $A$ is less than the perimeter of $B$.

Proof: if $A\neq B$, you may "cut out" a slice of $B$ without touching $A$. By convexity, the perimeter of the "reduced set" $B$ is less than the perimeter of the original set $B$. If $A$ is a polygon, by iterating this argument a finite number of times you get that $A$ is a reduced version of $B$, hence $\mu(\partial A)<\mu(\partial B)$ as wanted.

Answer 3

This expands Yves Daoust's comment.

Call the point that the tangent from $D$ touches the circle $P$, and the point where $DE$ intersects the circle $Q$.

Then $DEQ$ is a right triangle.

Let $t = \angle DEP$.

Then $\tan(t) =\dfrac{DP}{PE} $ so $DP =PE \tan(t) $.

We also have the length of arc $QP= t\,PE$.

Therefore $\dfrac{DP}{arc\ QP} =\dfrac{PE \tan(t)}{t\,PE} =\dfrac{\tan(t)}{t} \gt 1 $ for $t > 0$.

Since the same holds on both sides, the sum of the lengths of the two tangents is greater than the length of arc $DE$ by a factor $\dfrac{\tan(t)}{t} $.