\section*{Problem 1 - 121031}

For an isosceles triangle \(\triangle A B C\), let the height \(|C D|=h\) and the base length \(|A B|=g\) be given.

Furthermore, a square \(E F G H\) is inscribed in the triangle such that \(E F\) lies on \(A B\), \(G\) lies on \(B C\), and \(H\) lies on \(A C\).

Determine all ratios \(h: g\) for which the areas of triangle \(\triangle A B C\) and square \(E F G H\) are in the ratio \(9: 4\).
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1317.jpg?height=406&width=257&top_left_y=668&top_left_x=291)

According to the ray theorem, for the base \(g\), the height \(h\), and the length \(x\) of the square sides,

\[
\frac{h-x}{x}=\frac{h}{g} \Rightarrow g=\frac{h x}{h-x}
\]

On the other hand, the area of triangle \(\triangle A B C\) and the area of the square should satisfy:

\[
\frac{A_{\triangle A B C}}{A_{\mathrm{square}}}=\frac{\frac{1}{2} g h}{x^{2}}=\frac{9}{4}
\]

Substituting (1) into (2) yields

\[
18 x^{2}-18 h x+4 h^{2} \Rightarrow x_{1}=\frac{h}{3} ; \quad x_{2}=\frac{2 h}{3}
\]

For the solutions, (2) then becomes \(x_{1}: g=\frac{h}{2}\) and \(x_{2}: g=2 h\).

For the ratio \(g: h=1: 2\) (acute triangle) or \(g: h=2: 1\) (obtuse triangle), the ratio of the areas of triangle \(\triangle A B C\) and square EFGH is equal to \(9: 4\).