## Task 2 - 330842

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0833.jpg?height=348&width=394&top_left_y=1568&top_left_x=317)

Let $ABCD$ be a square with given side length $a$. A line parallel to $AB$ intersects the sides $AD$ and $BC$ at $E$ and $F$, respectively. A line parallel to $AD$ intersects the segments $AB$ and $EF$ at $G$ and $H$, respectively. A line parallel to $AB$ intersects the segments $GH$ and $BC$ at $I$ and $K$, respectively (see figure).

a) In addition to these conditions, it is required that the four rectangles $EFC D$, $AGHE$, $GBK I$, and $IKFH$ are equal in area.

Determine the perimeter of the rectangle $IKFH$ in terms of $a$!

b) Instead of the condition mentioned in a), it is now required that the four rectangles $EFC D$, $AGHE$, $GBK I$, and $IKFH$ are equal in perimeter.

Determine the area of the rectangle $IKFH$ in terms of $a$!
a) Due to the condition to be met, each of the four rectangles EFCD, AGHE, GBKI, IKFH has an area of $\frac{1}{4} a^{2}$. From this it follows that

$$
\begin{gathered}
E D=\frac{1}{4} a^{2}: a=\frac{1}{4} a \quad ; \quad A E=a-\frac{1}{4} a=\frac{3}{4} a \quad ; \quad A G=\frac{1}{4} a^{2}: \frac{3}{4} a=\frac{1}{3} a \\
G B=I K=a-\frac{1}{3} a=\frac{2}{3} a \quad ; \quad G I=I H=\frac{1}{4} a^{2}: \frac{2}{3} a=\frac{3}{8} a
\end{gathered}
$$

The perimeter of the rectangle $I K F H$ is thus $2 \cdot(I K+I H)=\frac{25}{12} a$.

b) For the lengths $E D=F C=x$ and $A G=y$, it follows from the equality of the perimeters of $E F C D$ and $A G H I$ that $2 \cdot(a+x)=2 \cdot(y+a-x)$, hence $y=2 x$.

Due to the equality of the perimeters of $G B K I$ and $I K F H$, we further have $2 \cdot(I K+B K)=2 \cdot(I K+K F)$, hence $B K=K F=\frac{1}{2}(a-x)$.

The equality of the perimeters of $E F C D$ and $G B K I$ thus yields the equation

$$
2 \cdot(a+x)=2 \cdot\left(a-y+\frac{1}{2}(a-x)\right) \quad \text { so } \quad 7 x=a
$$

Thus, we obtain $x=\frac{1}{7} a, y=\frac{2}{7} a, I K=a-y=\frac{5}{7} a, K F=\frac{1}{2}(a-x)=\frac{3}{7} a$; the rectangle $I K F H$ has the area $I K \cdot K F=\frac{15}{49} a^{2}$.