A $60^{\circ}$ angle is bisected, and a line passing through a point that is $1 / 2$ unit away from the vertex of the angle on the bisector cuts off a triangle with an area of $\sqrt{3} / 4$ from the angle sector. How far are the intersection points of the cutting line with the sides of the angle from the vertex of the angle?
Let the straight line intersect the sides of the angle at points $A$ and $B$, with $O$ being the vertex of the angle. Let $OA = a$ and $OB = b$ be the lengths of the segments cut off, and $OF = \frac{1}{2}$. We can write the areas of the resulting triangles using the known formula:

$$
\begin{gathered}
T_{AOB} = \frac{a b \sin 60^{\circ}}{2} ; \quad T_{OAF} = \frac{\frac{1}{2} a \sin 30^{\circ}}{2} \\
T_{OBF} = \frac{\frac{1}{2} b \sin 30^{\circ}}{2}
\end{gathered}
$$

On one hand, we know that

$$
T_{AOB} = T_{AOF} + T_{OBF} = \frac{a}{8} + \frac{b}{8} = \frac{\sqrt{3}}{4}
$$

from which $a + b = 2 \sqrt{3}$. On the other hand, from the equation $\frac{a b \frac{\sqrt{3}}{2}}{2} = \frac{\sqrt{3}}{4}$, we get $a b = 1$.

Based on the relationships between the roots and coefficients, $a$ and $b$ are the roots of the following quadratic equation: $x^{2} - 2 \sqrt{3} x + 1 = 0$, from which $x_{1} = \sqrt{3} + \sqrt{2}$ and $x_{2} = \sqrt{3} - \sqrt{2}$.

The lengths of the segments cut off by the sides are: $\sqrt{3} + \sqrt{2}$ and $\sqrt{3} - \sqrt{2}$ units.

![](https://cdn.mathpix.com/cropped/2024_05_02_0ef97803726886413f3ag-1.jpg?height=312&width=501&top_left_y=901&top_left_x=801)