\section*{Problem 3 - 251023}

Let \(A B C\) be an isosceles triangle with base length \(A B=20 \mathrm{~cm}\) and height \(C D=8 \mathrm{~cm}\).

A rectangle \(E F G H\) is to be inscribed in this triangle such that \(E\) and \(F\) lie on \(A B\), \(G\) on \(B C\), and \(H\) on \(A C\), and the area of the rectangle is maximized.

Prove that there is exactly one rectangle with these properties!

Determine the side lengths and the area of this rectangle!
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1500.jpg?height=325&width=460&top_left_y=2259&top_left_x=250)

Since the height \(C D\) in the equilateral triangle \(A B C\) is also a median, its foot \(D\) is the midpoint of \(A B\); thus, \(A D=B D=10 \mathrm{~cm}\).

Let \(A D=a\) and \(D C=b\) (see figure).

For any rectangle \(E F G H\) inscribed in the triangle \(A B C\) (with \(E\) and \(F\) on \(A B\), \(G\) on \(B C\), and \(H\) on \(A C\)), let \(A E=x\). Since \(E H \perp A B\) and \(D C \perp A B\), thus \(E H \| D C\), it follows from the intercept theorem that

\[
E H: D C=A E: A D \quad \text { so } \quad E H=\frac{b x}{a}
\]

Since \(E F G H\) is a rectangle, it follows that \(F G=E H=\frac{b x}{a}\). Since \(F G \| D C\) also holds, it follows from the intercept theorem that

\[
B F: B D=F G: D C
\]

and thus \(B F=x\). Consequently, \(E F=E D+F D=2(a-x)\). The area \(J\) of the rectangle \(E F G H\) is therefore

\[
J=2(a-x) \frac{b x}{a}=\frac{2 b}{a}\left(a x-x^{2}\right)=\frac{2 b}{a}\left(\frac{a^{2}}{4}-\frac{a^{2}}{4}+a x-x^{2}\right)=\frac{a b}{2}-\frac{2 b}{a}\left(x-\frac{a}{2}\right)^{2}
\]

From this, it follows: For \(x \neq \frac{a}{2}\), \(\left(x-\frac{a}{2}\right)^{2}>0\), so \(J<\frac{a b}{2}\); for \(x=\frac{a}{2}\), \(J=\frac{a b}{2}\).

Thus, the area \(J\) is maximized for exactly the rectangle \(E F G H\) that results from \(x=\frac{a}{2}=5 \mathrm{~cm}\). The side lengths of this rectangle are \(E F=10 \mathrm{~cm}, E H=4 \mathrm{~cm}\), and its area is \(J=40 \mathrm{~cm}^{2}\).