Question on the perimeter of any quadrilateral Is it true that the perimeter of any convex quadrilateral inside a unit circle
is no more than $4\sqrt{2}$?
 A: In general: the perimeter of a convex $n$-gon in the unit disc is maximal for a regular $n$-gon with all vertices on the unit circle.  For $n=4$ this gives a maximal perimeter of $4\sqrt{2}$ for a square. To see this make the following observations:


*

*When some vertex is not on the unit circle the perimeter can be increased by moving this vertex slightly outward and away from both neighboring vertices.

*If three consecutive vertices are on the unit circle then the total length of the two segments connecting them is maximal when the middle vertex is exactly halfway between the other two.


The second property may not be so obvious but it can be checked as follows.  Let the middle vertex be $(1,0)$ and the other two $(\cos \alpha, \sin \alpha)$ and $(\cos\beta, -\sin\beta)$ where $\alpha, \beta \geq 0$ and $\alpha+\beta=d$ is the fixed distance along the unit circle between the extreme vertices.  Then the sum of the distances is
$$ 2 \sin \frac{\alpha}{2} + 2 \sin\frac{\beta}{2} = 4 \sin \frac{\alpha+\beta}{4} \cdot\cos \frac{\alpha-\beta}{4}= 4 \sin \frac{d}{4} \cdot\cos \frac{\alpha-\beta}{4}
$$
and this is maximal when $\alpha=\beta$. In general the maximal perimeter for a convex $n$-gon is therefore $$2n \sin\frac{\pi}{n}.$$
