Problem 3. The diameter of a circle is the height of an equilateral triangle. The circle intersects two sides of the triangle. In what ratio are the sides of the triangle divided by the intersection points?
Solution. Let $\triangle A B C$ be equilateral, $C C_{1}$ be the diameter of the circle, and $M$ and $N$ be the intersection points of the circle with sides $A C$ and $B C$, respectively. According to Thales' theorem, $\measuredangle C M C_{1}=90^{\circ}$, which implies that $\measuredangle A M C_{1}=90^{\circ}$. This means that $\triangle A C_{1} M$ is a right triangle and, therefore,

![](https://cdn.mathpix.com/cropped/2024_06_05_a73f0c01d543f9cc2089g-08.jpg?height=368&width=394&top_left_y=1825&top_left_x=1074)

$$
\measuredangle A C_{1} M=90^{\circ}-60^{\circ}=30^{\circ}
$$

Accordingly,

$$
\overline{A M}=\frac{1}{2} \overline{A C_{1}}=\frac{1}{4} a, \overline{C M}=\overline{A C}-\overline{A M}=\frac{3}{4} a \Rightarrow \overline{A M}: \overline{M C}=1: 3
$$

Point $N$ is symmetric to point $M$ with respect to the height, so

$$
\overline{B N}: \overline{N C}=1: 3
$$