10.424 From each vertex of the base of an equilateral triangle with side $a$, two rays are drawn, forming angles of $15^{\circ}$ and $30^{\circ}$ with this base. Find the area of the quadrilateral whose vertices are the points of intersection of the constructed rays.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-255.jpg?height=364&width=377&top_left_y=555&top_left_x=177)

Let $O$ be the center of the given equilateral $\triangle ABC$. Then $AO$ and $CO$ are two given rays forming angles of $30^{\circ}$ with $AC$. Let the other two rays $AN$ and $CM$, forming angles of $15^{\circ}$ with $AC$, intersect at point $E$ (where $N$ and $M$ are the points of intersection with $CO$ and $AO$). $\triangle AMC = \triangle ANC \Rightarrow$ their heights are equal, i.e., $M$ and $N$ are equidistant from $AC \Rightarrow MN \parallel AC$. Let $D$ be the midpoint of $AC$, then $ED$ and $OD$ are the heights of $\triangle AEC$ and $\triangle AOC$ respectively. Therefore, $O, E, D$ lie on the same line and $OE \perp MN$.

\[
\begin{gathered}
\angle NAC = \angle MNA = \angle MAN = 15^{\circ} \Rightarrow AM = MN. \quad \text{By Theorem 4,} \\
\frac{OE}{ED} = \frac{AO}{AD} = \frac{2}{\sqrt{3}}, \frac{AM}{MO} = \frac{AC}{CO} = \sqrt{3} \Rightarrow \frac{OE}{OD - OE} = \frac{2}{\sqrt{3}}, \frac{MN}{AO - MN} = \sqrt{3} \Rightarrow \\
\Rightarrow OE = \frac{a(2 - \sqrt{3}) \sqrt{3}}{3}, MN = \frac{a(\sqrt{3} - 1)}{2} \left(OD = \frac{a}{2 \sqrt{3}}, AO = \frac{a}{\sqrt{3}}\right) \Rightarrow \\
\Rightarrow S_{ONEM} = \frac{MN \cdot OE}{2} = \frac{a^2 (9 - 5 \sqrt{3})}{12}.
\end{gathered}
\]

Answer: $\frac{a^2 (9 - 5 \sqrt{3})}{12}$.