The dimensions of a rectangle $R$ are $a$ and $b$, $a < b$. It is required to obtain a rectangle with dimensions $x$ and $y$, $x < a, y < a$, 
so that its perimeter is one-third that of $R$, and its area is one-third that of $R$. The number of such (different) rectangles is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \infty$
Using the perimeter and area formulas,
\[2(x+y) = \frac{2}{3}(a+b)\]
\[x+y = \frac{a+b}{3}\]
\[xy = \frac{ab}{3}\]
Dividing the second equation by the last equation results in
\[\frac1y + \frac1x = \frac1b + \frac1a\]
Since $x,y  \tfrac1a + \tfrac1a > \tfrac1a + \tfrac1b\]
This is a contradiction, so there are $\boxed{\textbf{(A)}\ 0}$ rectangles that satisfy the conditions.