Smallest possible perimeter of a quadriliteral inscribed in a rectangle A rectangle $ABCD$ is given. Let the points $P$ on $AB$, $Q$ on $BC$,$ R $ on $CD$ and $ S$ on $AD$ be inner points of the sides of the rectangle.
For which positions of the points $P, Q, R \ and \ S$ does the quadrilateral $PQ\ RS$ have the smallest perimeter?
I tried mirroring the points to prove that the perimeter is always the same. As it turns out, the perimeter is always $≥2AC$ (the diagonal of the rectangle) but it doesn‘t stay the same.


We have:
$$SD^2+DR^2=SR^2$$
$$RC^2+CQ^2=RQ^2$$
$$PB^2+BQ^2=PQ^2$$
$$AP^2+SA^2=SP^2$$
We sum up both sides, we get:
$(SD^2+SA^2)+ (DR^2+RC^2) +. . .=SR^2+RQ^2+PQ^2+PS^2$
Now consider $SD^2+SA^2$ from  the sum of LHS of above relations , we may write:
$$(SA+SD)^2=SD^2+SA^2+2SA\times SD$$
$SA\times SD$ is maximum if $SD=SA$, because $SD+SA$ is constant. In this case $SD^2+SA^2$ will be minimum, That is if the vertices of quadrilateral is on midpoints of sides of rectangle it's perimeter will be minimum.
Now we show that if in a parallelogram with sides a, b, c, d (a=c and b=d) $(a^2+b^2+c^2+d^2)$ is minimum then $(a+b+c+d)$ is minimum; we have:
$(a+b+c+d)^2=(2a+2b)^2=4(a^2+b^2)+8ab$
Since $a^2$ and $b^2$ and $ab$ are minimum therefore $(a+b+c+d)$ is minimum.
From geometric point of view the resulting parallelogram can be considered as  a transformed  rectangle when vertices move along rectangle sides. the perimeter is maximum when vertices of parallelogram are coincident on vertices of rectangle and becomes minimum when the vertexes of parallelogram are on midpoints and increases when  vertices continue moving toward adjacent vertices.
