See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]


Triangle $ABC$ and straight line $l$ are given at the plane. Construct using a compass and a ruler the straightline which is parallel to $l$ and bisects the area of triangle $ABC.$
1. **Identify the problem and given elements:**
   - We are given a triangle \(ABC\) and a straight line \(l\).
   - We need to construct a line parallel to \(l\) that bisects the area of triangle \(ABC\).

2. **Initial Assumption:**
   - Assume that for at least one of the vertices (say \(B\)), the line passing through this vertex and parallel to \(l\) intersects the triangle and divides it into two parts.

3. **Intersection Point:**
   - Let \(T\) be the point of intersection of this line (through \(B\) and parallel to \(l\)) with \(AC\).
   - Assume \(AT > TC\). If \(AT = TC\), the problem is trivial as the line through \(B\) parallel to \(l\) already bisects the area.

4. **Constructing the Required Line \(l_1\):**
   - Let \(l_1\) be the line we need to construct.
   - Denote by \(X\) and \(Y\) the points of intersection of \(l_1\) with \(AB\) and \(AC\) respectively.

5. **Area Bisecting Condition:**
   - The line \(l_1\) should bisect the area of \(\triangle ABC\). This implies that the areas of \(\triangle ABX\) and \(\triangle ACY\) should be equal.

6. **Using Proportions:**
   - From the area bisecting condition, we have:
     \[
     \text{Area}(\triangle ABX) = \text{Area}(\triangle ACY)
     \]
   - This implies:
     \[
     \frac{AX}{AB} = \frac{AY}{AC}
     \]
   - Let \(AX = x\) and \(AY = y\). Then:
     \[
     \frac{x}{AB} = \frac{y}{AC}
     \]

7. **Using the Given Line \(l\):**
   - Since \(l_1\) is parallel to \(l\), the ratio of the segments created by \(l_1\) on \(AB\) and \(AC\) should be equal to the ratio of the segments created by the line through \(B\) parallel to \(l\) on \(AB\) and \(AC\).

8. **Equating the Ratios:**
   - Let \(AT = k \cdot TC\). Then:
     \[
     \frac{AX}{AY} = \frac{AB}{AT}
     \]
   - Multiplying the two equations:
     \[
     2 \cdot AX \cdot AY = AB \cdot AC
     \]
   - Substituting \(AX = x\) and \(AY = y\):
     \[
     2x \cdot y = AB \cdot AC
     \]
   - Using the ratio:
     \[
     \frac{x}{y} = \frac{AB}{AT}
     \]

9. **Solving for \(AX\):**
   - From the ratio:
     \[
     x = \frac{AB}{AT} \cdot y
     \]
   - Substituting into the area equation:
     \[
     2 \left( \frac{AB}{AT} \cdot y \right) \cdot y = AB \cdot AC
     \]
   - Simplifying:
     \[
     2 \cdot \frac{AB}{AT} \cdot y^2 = AB \cdot AC
     \]
   - Solving for \(y\):
     \[
     y^2 = \frac{AB \cdot AC \cdot AT}{2 \cdot AB}
     \]
     \[
     y^2 = \frac{AC \cdot AT}{2}
     \]
     \[
     y = \sqrt{\frac{AC \cdot AT}{2}}
     \]
   - Then:
     \[
     x = \frac{AB}{AT} \cdot \sqrt{\frac{AC \cdot AT}{2}}
     \]

10. **Constructing the Line \(l_1\):**
    - With the lengths \(AX = x\) and \(AY = y\) determined, we can construct the points \(X\) and \(Y\) on \(AB\) and \(AC\) respectively.
    - Draw the line \(l_1\) through \(X\) and \(Y\), which will be parallel to \(l\) and bisect the area of \(\triangle ABC\).

The final answer is the line \( \boxed{ l_1 } \) constructed as described above.