\section*{Problem 5 - 161045}

For a rectangle \(A B C D\), let \(a\) be the length of the segment \(B C\), and let the diagonal \(A C\) be \(q\) times the length of \(B C\) (where \(q\) is real). From the vertices \(B\) and \(D\), perpendiculars are dropped to \(A C\), and their feet are, in this order, \(E\) and \(F\).

Determine the area of the quadrilateral \(FBED\) from the given values \(a\) and \(q\)!
}

Since the diagonal in a square with side length \(a\) has the length \(d=a \sqrt{2}\) and the diagonals in the square are perpendicular to each other, the quadrilateral \(F B E D\) disappears for \(q=\sqrt{2}\). We now distinguish the cases \(q>\sqrt{2}\) and \(q<\sqrt{2}\).

First, we define the following variables: Let \(x:=\) \(|\overline{A C}|, z:=|\overline{E C}|=|\overline{A F}|\) and \(h:=|\overline{F G}|\), where \(G\) is the foot of the perpendicular from \(F\) to the side \(A B\).

By symmetry, it suffices to calculate the area of triangle \(F B E\) and then double it.

For this, we first note that triangles \(A B F\) and \(C E B\) obviously have the same area, as they each form a triangle with the same height and base together with triangle \(F B E\). We now calculate the area of \(\triangle A B C\) in terms of \(a\) and \(q\). According to the given condition, \(x=q \cdot b\). Now, \(x=\sqrt{a^{2}+b^{2}}\). Substituting and squaring yields:

\[
\begin{aligned}
a^{2}+b^{2} & =q^{2} b^{2} \\
a^{2}=q^{2} b^{2}-b^{2} & =b^{2}\left(q^{2}-1\right) \\
b^{2} & =\frac{a^{2}}{q^{2}-1} \Rightarrow b=\frac{a \sqrt{q^{2}-1}}{q^{2}-1}
\end{aligned}
\]

Thus:

\[
A_{\triangle A B C}=\frac{a b}{2}=\frac{a^{2} \sqrt{q^{2}-1}}{2\left(q^{2}-1\right)}
\]

According to the cathetus theorem in triangle \(A B C\), \(b^{2}=z x \Longleftrightarrow z=\frac{b^{2}}{x}\).

By the intercept theorem, \(\frac{b}{h}=\frac{x}{z} \Longleftrightarrow h=\frac{b z}{x}\). Substituting yields:

\[
h=\frac{b \cdot \frac{b^{2}}{x}}{x}=\frac{b^{3}}{x^{2}}=\frac{b^{3}}{q^{2} b^{2}}=\frac{b}{q^{2}}=\frac{a \sqrt{q^{2}-1}}{q^{2}\left(q^{2}-1\right)}
\]

The area of the quadrilateral is therefore:

\[
A=2 \cdot\left(\frac{a^{2} \sqrt{q^{2}-1}}{2\left(q^{2}-1\right)}-2 \cdot \frac{a \cdot a \sqrt{q^{2}-1}}{2 q^{2}\left(q^{2}-1\right)}\right)=\frac{a^{2} \sqrt{q^{2}-1}\left(q^{2}-2\right)}{q^{2}\left(q^{2}-1\right)}
\]

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1386.jpg?height=508&width=397&top_left_y=1765&top_left_x=287)

2. Case: \(q<\sqrt{2}\)

For \(q<\sqrt{2}\), the order of \(E\) and \(F\) on the diagonal changes. Let now \(x:=|\overline{A C}|, z:=|\overline{F C}|=|\overline{A E}|\) and \(h:=|\overline{E G}|\), where \(G\) is the foot of the perpendicular from \(E\) to the side \(A B\).

Now, according to the cathetus theorem, \(a^{2}=z x \Longleftrightarrow z=\frac{a^{2}}{x}\). Substituting:

\[
h=\frac{b \cdot \frac{a^{2}}{x}}{x}=\frac{a^{2} b}{x^{2}}=\frac{a^{2} b}{q^{2} b^{2}}=\frac{a^{2}}{q^{2} b}=\frac{a^{2}\left(q^{2}-1\right)}{q^{2} a \sqrt{q^{2}-1}}=\frac{a \sqrt{q^{2}-1}}{q^{2}}
\]

The area is thus:

\[
A=2 \cdot\left(\frac{a^{2} \sqrt{q^{2}-1}}{2\left(q^{2}-1\right)}-2 \cdot \frac{a \sqrt{q^{2}-1} \cdot a}{2 q^{2}}\right)=\frac{a^{2} \sqrt{q^{2}-1}\left(q^{2}-2\left(q^{2}-1\right)\right)}{q^{2}\left(q^{2}-1\right)}=\frac{a^{2} \sqrt{q^{2}-1}\left(2-q^{2}\right)}{q^{2}\left(q^{2}-1\right)}
\]

Thus, the area \(A=\frac{a^{2} \sqrt{q^{2}-1}\left|q^{2}-2\right|}{q^{2}\left(q^{2}-1\right)}\).