Can a convex polygon inside a square with edge length 1 have a perimeter > 4? While featherbrainedly doodling the other day I noticed that it's probably impossible to draw a convex polygon with a greater perimeter then that of the square around it.
Can someone find a counterexample or maybe even find a proof?
 A: If a convex set is contained in another set, the convex set has smaller perimeter.
This can be understood by considering the projection map on the convex set (i.e. the map which sends a point to the closer point on the convex set). This is a map which does not increase distances and sends the perimeter of the enclosing set onto the perimeter of the convex set.
A general formal proof of this fact could be obtained by means of the following facts:


*

*The projection onto a convex set is a $1$-lipschitz map

*The measure of the image of a set does not increase through a $1$-lipschitz map

*The projection of the boundary of a set containing a convex set contains the whole boundary of the convex set.


For your special case (i.e. polygons) you can manage to get an elementary proof by noticing that the projection of a segment on any line is not longer than the segment itself. By subdividing the edges of the outer polygon so that every edge gets projected on a single edge of the inner convex set, you should manage to prove the result.
A: I'll rephrase the argument by Emanuele Paolini without "$1$-Lipschitz" and projection. Place a bomb inside of the convex polygon. When it goes off, the polygon breaks apart at the vertices, with every side of it flying out in the direction in which it was facing: I drew   arrows to show the directions in which two of the sides go.

Flying out, each side  takes out a chunk of the boundary of the square. Use some geometry to prove that


*

*The chunk taken out by each side is at least as long as the side itself. 

*The chunks taken out by different sides do not overlap.


Conclusion follows from 1 and 2.
