Given a symmetric convex compact set $K\in\mathbb{R}^2$, show there are no sequences $(r_i)_{i=1}^n$ of positive scalars and $(P_i)_{i=1}^n$ of points in $\partial K$, so that $n>1$ and
$$\sum_{i=1}^n r_iP_i\in \mathrm{interior}\bigcap_{i=1}^n r_i(K+P_i)\ \ \ \ \ \ \ \ (*)$$
In other words, given a collection of scaled (not rotated) copies of $K$, each containing the origin in its boundary, is it not possible for the sum of all the centers of the copies to be contained in the interior of the intersection?
Progress:
I think this works.
First, note that the easiest case should be when we shrink the RHS of (*) to $K=\mathrm{symm.conv}\{P_i\}$
So for $n=2$, start with $K$ a parallelogram $ADBC$ and pick $P_1=A,\ \ \ P_2=C,\ \ \ r_1=r_2=1$. 
Since $C\in\partial K$, we have $A+C\in\partial(K+A)$ and similarly $A+C\in\partial (K+C)$. Then, $A+C\in\partial((K+A)\cap(K+C))$ since the intersection of boundaries is contained in the boundary of the intersection. We didn't actually use anything about parallelograms - this argument generalizes to any symmetric convex $K$ with $r_1=r_2$.
On to $r_1\neq r_2$. Working under $r_1>r_2$ and fixing $r_1+r_2=2$, the point $r_1A+r_2C$ travels along the segment from $A+C$ to $A^\prime$. That means it separates from $K+C$, which contains $r_2K+C$. So, $A+C\not\in r_1(K+A)\cap r_2(K+C)$. This also generalizes to any symmetric convex $K$.
I think next is reducing $n=3$ and so on to $n=2$. Not attacking symmetric hexagons and points roaming in triangles.
We might divide through by $\sum r_i$ to assume WLOG $\sum r_i=1$. Then, $\sum r_iP_i$ ranges over the interior $K$. However, that introduces a dependence among the $r_i$.