Selected Problems from the Final Round of National Olympiad

The rectangles $ABCD$ and $EFGH$ have a common center $O$ (see Fig. 15).

As rectangles are cyclic quadrangles, the point $F$ lies on the circle with center $O$ and radius $|OE|$. This circle intersects the side $BC$ at two points symmetric with respect to the midpoint of the side. To have exactly one point common to the circle and the side, the side must be tangent to the circle and $F$ must be the midpoint of $BC$. Analogously, $H$ must be the midpoint of $DA$.

In triangle $EFH$, the side $HF$ has length $a$ and the corresponding altitude is $\frac{b}{2}$, giving $\frac{1}{2}\cdot a\cdot \frac{b}{2}=\frac{ab}{4}$ as the area of the triangle. The triangle $GFH$ has the same area. Hence the area of rectangle $EFGH$ is $2\cdot \frac{ab}{4}=\frac{ab}{2}$ that makes up a half of the area $ab$ of the rectangle $ABCD$.

Fig. 15