Given a circle $(O,R)$ and a point $P$ in its interior. For any two points $A,B$ on its perimeter, we define $f(A,B)$ to be the point $C$ such that $APBC$ is a parallelogram.
Determine the image of $f$ when $A,B$ vary over all possible pairs of points on the perimeter. That is, determine the set $S = \{f(A,B) | AO = BO = R\}$
Solution Let $O'$ be a point on the extension of $PO$ such that $OO' = 2PO$. We shall show that $S$ is a circle (including the interior) centered at $O'$ with radius $2R$.

Outline of proof:

Let $T$ to be the circle we claim. Note that $T$ is a 2x dilation of the original circle with center $P$. We show that the boundary of $S$ is boundary of $T$ as well, by letting $A=B$ in the opration

Next we show that any point in $T$ is an image of a specific application of $f$. Let $X' \in T$, then $X'$ is an image of a 2x dilation. Suppose $X$ is the original point. We claim that there exists $A,B$ such that $PX$ is a median of $PAB$. Indeed, if $OAB$ is a triangle such that $OX$ is its median, then $PX$ is median of $PAB$.

Then it is routine to show that $f(A,B) = X'$.

Note: this problem can be extended to any point $P$, and any convex shape $V$ in $R^n$. The sum of vectors $PA + PB$ where $A,B$ are any two points on its boundary is equivalent to the 2x dilation of $V$ with center $P$. However, $V$ must be bounded.

Challenge: prove this generalization. (Hint: use the connectedness of $S^{n-1}$ in $R^n$).

Challenge: give a counter example of this result of $V$ is unbounded