11.8*. The area of triangle $ABC$ is 1. Let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $BC, CA, AB$ respectively. Points $K, L, M$ are taken on segments $AB_{1}, CA_{1}, BC_{1}$ respectively. What is the minimum area of the common part of triangles $KLM$ and $A_{1}B_{1}C_{1}$?
11.8. Let the intersection point of lines $K M$ and $B C$ be denoted as $T$, and the intersection points of the sides of triangles $A_{1} B_{1} C_{1}$ and $K L M$ as shown in Fig. 11.2. Then $T L: R Z = K L: K Z = L C: Z B_{1}$. Since $T L \geqslant B A_{1} = A_{1} C \geqslant L C$, it follows that $R Z \geqslant Z B_{1}$, i.e., $S_{R Z Q} \geqslant S_{Z B_{1} Q}$. Similarly, $S_{Q Y P} \geqslant S_{Y A_{1} P}$ and $S_{P X R} \geqslant S_{X C_{1} R}$. Adding all these inequalities and the inequality $S_{P Q R} \geqslant 0$, we get that the area of the hexagon $P X R Z Q Y$ is not less than the area of the remaining part of triangle $A_{1} B_{1} C_{1}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_22b0ba5e8c5a1f49bd81g-263.jpg?height=385&width=564&top_left_y=634&top_left_x=1016)

Fig. 11.2 i.e., its area is not less than $S_{A_{1} B_{1} C_{1}} / 2 = 1 / 8$. Equality is achieved, for example, if point $K$ coincides with $B_{1}$, and point $M$ coincides with $B$.