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Is it true that the perimeter of any convex polygon in the unit disk on the Euclidean plane is less than the circumference $2\pi$ of the circle? Thanks. [The OP has solved it]

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Since the upper bound $2\pi$ cannot be attained by the perimeter of a convex polygon, a better word than "maximum" in your title would be "supremum", i.e. least upper bound. (Which is $2\pi$, as you've asked in the body of the question.) –  hardmath Dec 7 '12 at 14:38

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Actually, no maximum exists. The unit circle can be seen as the limit of regular polygons with incenter $(0,0)$ and inradius $1$ as the number of sides goes to infinity. The perimeter of these shapes becomes infinitely close to $2\pi$, but the perimeter $p$ is always less than $2\pi$, since given 2 points on the unit circle, the straight line connecting these point will always be shorter than the arc connecting these points on the unit circle. However, since we can get infinitely close to $2\pi$, $2\pi$ is the supremum of these perimeters.

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Edit: In my first post I proposed projecting the vertices of a convex polygon onto the unit circle from any interior point of the polygon, claiming (without proof) that this derived polygon (inscribed) has at least the perimeter of the original. The OP mathfan, in a comment since removed, pointed out this needs proof because individual edges may shrink under projection along such rays. So here is a more careful construction of an inscribed polygon in which the edges are guaranteed not to shrink.

We may assume without loss of generality that the minimum bounding circle of the given convex polygon $P$ has radius 1, as otherwise dilation $cP$ with $c > 1$ yields such (while increasing the individual edge lengths and perimeter resp.). In this position polygon $P$ must have vertices on the unit circle, either two which form a diameter or three which form an acute inscribed triangle.

In any case between two "consecutive" vertices of $P$ on the circle there may be other vertices of $P$ not on the circle. Our construction is to lengthen the intervening edges of $P$ as necessary until all vertices are pushed out to the unit circle.

Let $\{v_0,v_1,\ldots,v_k\}$ be such a subset of successive vertices of $P$, where $v_0$ and $v_k$ form a chord on the circle (subtending an arc not more than a semicircle) and the intermediate vertices $v_1,\ldots,v_{k-1}$ lie at corresponding perpendicular heights $h_1,\ldots,h_{k-1}$ strictly above the chord and below the subtended arc. Convexity of $P$ implies these heights are unimodal, i.e. that they rise (from zero at one endpoint) to a maximum (which may be unique or shared by two consecutive vertices) and then fall (back to zero at the other endpoint). Since all the intermediate vertices are strictly below the subtended arc, there exists a minimum factor $t \gt 1$ such that stretching their heights above the chord $th_1,\ldots,th_{k-1}$ pushes (at least) one of them out to the circle. This preserves unimodality and hence convexity. The intervening edges do not shrink, and indeed lengths increase with the possible exception of at most one edge parallel to the chord.

Continue in this fashion until all vertices lie on the unit circle, producing a derived polygon $P'$ inscribed therein. Now the sides of the derived polygon are chords whose lengths are exceeded by the circular segments they subtend. Thus the perimeter of the derived polygon is strictly less than the circumference of the unit circle $2\pi$.

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More generally, the perimeter of a convex region $C$ (in Euclidean space of any dimension) does not exceed the perimeter of any bounded region $B$ that contains $C$. This follows from the fact that the nearest-point projection is a nonexpanding surjection $\partial B\to\partial C$. See here and here.

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