Let's mark the following points in a general triangle \(ABC\) according to the diagram. Points \(A_1\) and \(B_1\) are the midpoints of sides \(BC\) and \(AC\), and points \(C_1\), \(C_2\), and \(C_3\) divide side \(AB\) into four equal parts. We connect points \(A_1\) and \(B_1\) with points \(C_1\) and \(C_3\), thus forming a quadrilateral bounded by these lines. What fraction of the area of the entire triangle does this quadrilateral occupy?

(M. Mach)

![](https://cdn.mathpix.com/cropped/2024_04_17_a28d21919b1f57221c31g-3.jpg?height=543&width=574&top_left_y=485&top_left_x=747)
We will show that the quadrilateral $C_{1} C_{3} A_{1} B_{1}$ has half the area of the triangle $A B C$ and that the patch covers exactly half of this quadrilateral. From this, it follows that the patch occupies a quarter of the area of the entire triangle.

Let $c$ be the length of the side $A B$ and $v$ the height of the triangle $A B C$ to this side. With this notation, the area of the triangle is

$$
S_{A B C}=\frac{1}{2} c v
$$

The segment $B_{1} A_{1}$ is the midline of the triangle $A B C$, so it is parallel to the side $A B$ and has half the length. Points $C_{1}, C_{2}$, and $C_{3}$ divide the side $A B$ into four equal parts, so the segment $C_{1} C_{3}$ is half the length of the side $A B$, and thus has the same length as the segment $B_{1} A_{1}$. Since the segments $B_{1} A_{1}$ and $C_{1} C_{3}$ are parallel and of the same length, the quadrilateral $C_{1} C_{3} A_{1} B_{1}$ is a parallelogram. The height of the parallelogram to the side $C_{1} C_{3}$ corresponds to the distance between the side $A B$ and the midline $B_{1} A_{1}$, which is exactly half the height $v$. The area of the parallelogram is therefore

$$
S_{C_{1} C_{3} A_{1} B_{1}}=\frac{c}{2} \cdot \frac{v}{2}=\frac{1}{2} S_{A B C}
$$

The diagonals of the parallelogram $C_{1} C_{3} A_{1} B_{1}$ divide it into four triangles of equal area. This fact can be justified for any two adjacent triangles: The diagonals of the parallelogram bisect each other, from which it follows that $\left|S C_{1}\right|=\left|S A_{1}\right|$, where $S$ is the intersection of the diagonals. Furthermore, the height of the triangle $S C_{1} B_{1}$ to the side $S C_{1}$ is the same as the height of the triangle $S B_{1} A_{1}$ to the side $S A_{1}$. Therefore, the triangles $S C_{1} B_{1}$ and $S B_{1} A_{1}$ have the same area. Similarly, it can be shown that all triangles $S C_{1} B_{1}, S B_{1} A_{1}, S A_{1} C_{3}$, and $S C_{3} C_{1}$ have the same area.

From this, it follows that

$$
S_{\text {patch }}=\frac{1}{2} S_{C_{1} C_{3} A_{1} B_{1}}
$$

which, together with the previous equality, proves that the patch occupies a quarter of the area of the triangle $A B C$.

Evaluation. 2 points for justifying that $C_{1} C_{3} A_{1} B_{1}$ is a parallelogram and expressing its area; 2 points for justifying that the diagonals divide this parallelogram into four equal-area triangles; 2 points for completing the problem and the correct result.

Note. When solving the problem, various other insights can be used, which we do not elaborate here. For example, non-adjacent pairs of triangles in the parallelogram $C_{1} C_{3} A_{1} B_{1}$ are actually congruent, and it is easy to express the height and area of the triangle $S C_{3} C_{1}$ in terms of the triangle $A B C$, etc. The problem can also be solved without the auxiliary consideration of the parallelogram. In any case, adjust the evaluation so that 4 points correspond to the justification of individual observations and 2 points to the expression of the result.