The proposition is missing the three sides of the triangle, as it is possible to inscribe the rectangle in such a way that two opposite vertices lie on a side, otherwise the center lies on the line connecting the midpoint of a side with the midpoint of the corresponding altitude. These are the 3 Schwatt lines of ABC concurring at its symmedian point. See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=598678.

i think, if we mean by the side the strict meaning which is the segment delimited by two vertices the solution is just the three segments concurrent at the symmedian point ;but if we mean by it the line ,it s ok that the solution include certainly the sides

Consider the set of rectangles $TUVW$ with $VW \in BC$ and $T \in AB, U \in AC$. We must have $TU \parallel BC \equiv VW$, and that if $V$ is the projection of $T$ on $BC$, $V \mapsto T \mapsto U$ preserves ratio. Hence, there is a spiral similarity mapping $V \mapsto U$ from the set $BC \mapsto AC$. Hence, the midpoint of $VU$ moves along a line i.e. the centres of the rectangles move along a line. When $T=A=U$, the centre is the midpoint of $AD$, where $D$ is the foot of altitude from $A$ to $BC$. When $T=B, U=C$ then the centre is the midpoint of $BC$. So the locus is the A-Schwatt line. Similarly, we conclude they move on the Schwatt lines. Let the tangents at $B, C$ to $\odot ABC$ meet at $P$, point at infintiy perpendicular to $BC$ be $\ell_A$ and symmedian point of $ABC$ be $L$. If $Q, R$ are midpoints of $BC$ and $AD$, then \[Q(P, L; D, A) \equiv Q(\ell_A, L; D, A) = -1 = Q(\ell_A, R; D, A)\] $\implies L \in QR$. Hence, the concurrence point of the locus lines is the symmedian point.

schwatt lines xd. They concur at the Lemoine point.