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.

 A: 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.
A: Say we have a rectangle with length $AB = CD = a$ and width $BC = DA = b$ and a quadrilateral $PQRS$ inscribed as shown in the diagram. Say, $ \, AP = x, AS = y$.

We reflect point $P$ through both $DA$ and $CB$. So,
$PA = AP'$ and $PB = BP"$. Now $\triangle APS \cong \triangle AP'S$ and that does not change even if we slide point $S$ on line $DA$ up or down.
Same is the case with $\triangle BPQ \cong \triangle BP''Q$.
$RS + SP = P'S + SR \ge P'R$.
The equality occurs when we slide point $S$ on line $DA$ such that $S$ falls on line $P'R$. We do it similarly for point $Q$ such that
$PQ + QR = P''R$.
So, perimeter of quadrilateral reduces to $P'R + RP''$.
The base of the triangle $P'P''R = P'A + AP + PB + BP" = 2 (AP + PB) = 2AB = 2a$.
The height of the triangle is $b$.
Now we know that for a given area of the triangle (fixed base and height), isosceles triangle has the minimum perimeter (we can in fact show that using reflection too).
So, $P'T = P''T = a$. That gives $P'R = P''R = \sqrt{a^2+b^2}$ and hence the minimum perimeter of the quadrilateral is $2\sqrt{a^2+b^2}$.
We can also show that $AS = CQ, AP = CR$ and that PQRS is a parallelogram.
As $\triangle P'SA \sim \triangle PRT$, $\displaystyle \frac{AS}{AP'} = \frac{RT}{P'T} \implies \frac{y}{x} = \frac {b}{a}$. Points P, Q, R, S must meet this condition ensuring parallelogram and for the quadrilateral perimeter to be minimum.
So, $x = \frac {a}{2}, y = \frac{b}{2}$ is definitely one of the solutions but is NOT the only solution.
EDIT:
Here is a diagram of quadrilateral with min perimeter inscribed in a rectangle but its vertices not at the midpoints of the rectangle. Please note it meets the ratio condition I mentioned and the angles above (parallelogram) and hence works.

A: By Optics Fermat principle that light takes minimum time during reflection i.e., when incidence/reflection angles will all be equal, we should have full symmetry with center points of sides as incidence/ bounce off points.
$$ L= \sqrt{(w-a)^2+(h-q)^2+...+...+...+...+...+...} $$
The bounding rectangle measures $(2w\times 2h )$. Partially differentiate the total length $L$ w.r.t variable deviations $(a,b,p,q)$ , equate to zero, so we can establish that  deviations vanish for a minimum total $L$.

