In the equilateral triangle $ABC$ with side length of one unit, we mark the points $A_{1}, B_{1}, C_{1}$ on the sides $BC, CA, AB$ respectively, such that they are $\frac{1}{4}$ of the distance from the vertices $C, A, B$. What is the area of the triangle formed by the segments $AA_{1}, BB_{1}, CC_{1}$?
We denote the intersection of $B B_{1}$ and $A A_{1}$ by $P$, the intersection of $B B_{1}$ and $C C_{1}$ by $Q$, and the intersection of $A A_{1}$ and $C C_{1}$ by $R$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5220be21dbdbedf08e0ag-1.jpg?height=542&width=585&top_left_y=228&top_left_x=759)

The triangles $A A_{1} C$, $B B_{1} A$, and $C C_{1} B$ are clearly congruent due to the equilateral nature of triangle $A B C$. Similarly, the smaller triangles formed are also congruent, one of which is the triangle $A B_{1} P$. To calculate the area of triangle $P Q R$, we subtract three times the area of triangle $A A_{1} C$ from the area of triangle $A B C$; in doing so, we subtract the area of the triangles of type $A B_{1} P$ twice, so we need to add back three times the area of triangle $A B_{1} P$:

$$
T_{P Q R}=T_{A B C}-3 T_{A A_{1} C}+3 T_{A B_{1} P}
$$

The area of triangle $A B_{1} B$ is one-fourth of the area of triangle $A B C$, since they share the height $D B$ corresponding to the sides $A C$ and $A B_{1}$, and $A B_{1} = \frac{1}{4} A C$.

The triangles $A P B_{1}$ and $A B B_{1}$ are similar, as they share the angle $A B_{1} B$, and $\angle P A B_{1} = \angle C_{1} B Q = \angle A B B_{1}$. Let's determine the ratio of their sides:

From the right triangle $B_{1} B D$,

$$
B B_{1} = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2} + \left(\frac{1}{4}\right)^{2}} = \frac{1}{4} \sqrt{13}
$$

Thus, the ratio of similarity is:

$$
B B_{1} : A B_{1} = \frac{1}{4} \sqrt{13} : \frac{1}{4} = \sqrt{13}
$$

The ratio of their areas is

$$
13 = \frac{1}{4} : \frac{1}{52}
$$

Therefore,

$$
T_{P Q R} = \frac{\sqrt{3}}{4} - 3 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{4} + 3 \cdot \frac{1}{52} \cdot \frac{\sqrt{3}}{4} = \frac{1}{13} \sqrt{3} (\approx 0.13 \text{ square units}),
$$

which is $\frac{4}{13}$ of the area of triangle $A B C$, approximately $31\%$.