\section*{Problem 4 - 221024}

Let \(A B C D E F\) be a regular hexagon, with its area denoted as \(F_{1}\). Let \(F_{2}\) be the area of the (equilateral) triangle \(A C E\), and let \(F_{3}\) be the area of the (equilateral) triangle \(M_{1} M_{2} M_{3}\), where \(M_{1}, M_{2}, M_{3}\) are the midpoints of the sides \(A B, C D\), and \(E F\) respectively.

Calculate the ratio \(F_{1}: F_{2}: F_{3}\)!

(The ratio should be expressed using three smallest possible natural numbers.)
}

The area of an equilateral triangle with side length \(a\) of the hexagon is \(D=\frac{a^{2}}{4} \sqrt{3}\), hence \(F_{1}=6 D=\frac{3}{2} a^{2} \sqrt{3}\).

Each side length of the triangle \(A C E\) is equal to twice the height of an equilateral triangle with side length \(a\). Therefore, \(s=A C=C E=E A=a \sqrt{3}\) and consequently \(F_{2}=\frac{s^{2}}{4} \sqrt{3}=\frac{3}{4} a^{2} \sqrt{3}\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1459.jpg?height=585&width=552&top_left_y=1078&top_left_x=752)

Since \(\angle B A D=60^{\circ}\) and \(\angle A B C=2 \cdot 60^{\circ}\), \(A D \| B C\) (converse of the theorem about opposite angles formed by intersecting parallel lines).

Thus, \(M_{1} M_{2}\) is the midline in a trapezoid whose parallel sides have lengths \(2 a\) and \(a\). The same applies to \(M_{2} M_{3}\) and \(M_{3} M_{1}\); hence

\[
t=M_{1} M_{2}=M_{2} M_{3}=M_{3} M_{1}=\frac{3}{2} a
\]

and consequently

\[
F_{3}=\frac{t^{2}}{4} \sqrt{3}=\frac{9}{16} a^{2} \sqrt{3}
\]

Therefore, \(F_{1}: F_{2}: F_{3}=\frac{3}{2}: \frac{3}{4}: \frac{9}{16}=8: 4: 3\).

Solutions of the II. Round 1982 adopted from [5]

\subsection*{7.24.3 III. Round 1982, Class 10}