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!
4 Answers
$\begingroup$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.
$\endgroup$ 6 $\begingroup$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.
$\endgroup$ $\begingroup$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} $.
$\endgroup$ $\begingroup$This proof assumes the fact (sometimes used as a definition) that the circumference of a circle is the least upper bound of the perimeters of all polygons inscribed within it.
Def.: If $A$ is a polygon, $P(A)$ is its perimeter. If $C$ is a circle, $P(C)$ is its circumference.
Lemma: If $A$ and $B$ are convex polygons with $A \subset B$, then $P(A) \leq P(B)$. (See for a proof.)
Now suppose that $B$ is a polygon circumscribed about circle $C$. Let $A$ be a polygon inscribed in $C$. By the lemma, $P(A) \leq P(B)$. Taking the sup over all such polygons $A$ and using our 'fact', we have that $P(C) \leq P(B)$. Now form a second circumscribed polygon $D$ within and smaller than $B$ (one additional side is enough). Then $P(C) \leq P(D) < P(B)$, showing that $P(B)$ does indeed overestimate $P(C)$.
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