Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 6.5 National olympiad Prove it Estonia

In a rectangle $ABCD$ we have $|AB| = a$ and $|BC| = b$ , where $a \ge b$ . Let $E$ be a point in the interior of side $AB$ such that there is exactly one possibility to choose points $F, G, H$ on the sides $BC, CD, DA$ , respectively, in such a way that $EFGH$ is a rectangle, too. Find the ratio of the areas of rectangles $EFGH$ and $ABCD$ .

Solution

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

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